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AN EXPERIMENTAL SCIENCE

CHEMICAL EDUCATION MATERIAL STUDY

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CHEMISTRY:

AN EXPERIMENTAL SCIENCE

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California to commemorate the University s
contribution to tlu eminently successfub :
development of a hujh school chemistry
cutriculum.rm7his project was made-^
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Science Foundation. CRoy aides were-^
returned to tlieU.S.Treasury as complete
repayment f the aran v. <~jlieyrojecZy
directed by faculty of tlu University of
California and of Harvey Mudd College,
resulted iru> tins textbook, a laboratoiy
manual, a teadier'sguidej ancLo
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INTERNATIONAL ATOMIC WEIGHTS

ATOMIC

NAME

SYMBOL

NUMBER

WEIGHT

NAME

SYMBOL

NUMBER

WEIGHT

Actinium

(227)

Mercury

200.6

Aluminum

27.0

Molybdenum

95.9

Amencium

(243)

Neodymium

144.2

Antimony

121.8

Neon

20.2

Argon

39.9

Neptunium

(237)

Arsenic

74.9

Nickel

58.7

Astatine

(210)

Niobium

92.9

Barium

137.3

Nitrogen

14.01

Berkelium

245

Osmium

190.2

Beryllium

9.01

Oxygen

16.00

Bismuth

209.0

Palladium

106.4

Boron

10.8

Phosphorus

31.0

Bromine

79.9

Platinum

195.1

Cadmium

112.4

Plutonium

(242)

Calcium

40.1

Polonium

210

Californium

(251)

Potassium

39.1

Carbon

12.01

Praseodymium

140.9

Cerium

140.1

Promethium

(147)

Cesium

132.9

Protactinium

(231)

Chlorine

35.5

Radium

(226)

Chromium

52.0

Radon

(222)

Cobalt

58.9

Rhenium

186.2

Copper

63.5

Rhodium

102.9

Curium

(247)

Rubidium

85.5

Dysprosium

162.5

Ruthenium

101.1

Einsteinium

(254)

Samarium

150.4

Erbium

167.3

Scandium

45.0

Europium

152.0

Selenium

79.0

Fermium

100

(253)

Silicon

28.1

Fluorine

19.0

Silver

107.9

Francium

(223)

Sodium

23.0

Gadolinium

157.3

Strontium

87.6

Gallium

69.7

Sulfur

32.1

Germanium

72.6

Tantalum

180.9

Gold

197.0

Technetium

(99)

Hafnium

178.5

Tellurium

127.6

Helium

4.00

Terbium

158.9

Holmium

164.9

Thallium

204.4

Hydrogen

1.008

Thorium

232.0

Indium

114.8

Thulium

168.9

Iodine

126.9

Tin

118.7

Iridium

192.2

Titanium

47.9

Iron

55.8

Tungsten

183.9

Krypton

83.8

Uranium

238.0

Lanthanum

138.9

Vanadium

50.9

Lead

207.2

Xenon

131.3

Lithium

6.94

Ytterbium

173.0

Lutetium

175.0

Yttrium

88.9

Magnesium

24.3

Zinc

65.4

Manganese

54.9

Zirconium

91.2

Mendelevium

101

(256)

Parenthetical names refer to radioactive elements; the mass number (not the atomic weight) of the isotope with largest
half-life is usually given.

* Latest values recommended by the International Union of Pure and Applied Chemistry, 1961.

CHEMISTRY

AN EXPERIMENTAL SCIENCE

CHEMISTRY

Prepared by
CHEMICAL EDUCATION MATERIAL STUDY

Under a grant from
THE NATIONAL SCIENCE FOUNDATION

Editor: GEORGE C. PIMENTEL, University of California, Berkeley, California

Associate Editors

BRUCE H. MAHAN, University of California, Berkeley, California

A. L. McCLELLAN, California Research Corporation, Richmond, California

KEITH MacNAB, Sir Francis Drake High School, San Anselmo, California

MARGARET NICHOLSON, Acalanes High School, Lafayette, California

An Experimental Science

Contributors

ROBERT F. CAMPBELL

Miramonte High School, Orinda, California

JOSEPH E. DAVIS, JR.

Miramonte High School, Orinda, California

SAUL L. GEFFNER

Forest Hills High School, Forest Hills, New York

THEODORE A. GE1SSMAN

University of California, Los Angeles, California

MELVIN GREENSTADT

Fairfax High School, Los Angeles, California

CARL GRUHN

South Pasadena High School, South Pasadena, California

EDWARD L. HAENISCH

Wabash College, Crawfordsville, Indiana

ROLFE H. HERBER

Rutgers University, New Brunswick, New Jersey

C. ROBERT HURLEY

Sacramento State College, Sacramento, California

LAWRENCE D. LYNCH, JR.

Beverly Hills High School, Beverly Hills, California

LLOYD E. MALM

University of Utah, Salt Lake City, Utah

CLYDE E. PARRISH

Cubberley Senior High School, Palo Alto, California
ROBERT W. PARRY

University of Michigan, Ann Arbor, Michigan

EUGENE ROBERTS

Polytechnic High School, San Francisco, California

MICHELL J. SIENKO

Cornell University, Ithaca, New York

ROBERT SILBER

American Chemical Society, Washington, D.C.

HARLEY L. SORENSEN

San Ramon Valley Union High School, Danville, California

LUKE E. STEINER

Oberlin College, Oberlin, Ohio

MODDIE D. TAYLOR

Howard University, Washington, D.C.

ROBERT L. TELLEFSEN

Napa High School, Napa, California

Director: J. ARTHUR CAMPBELL, Harvey Mudd College, Claremont, California
Chairman: GLENN T. SEABORG, University of California, Berkeley, California

W. H. FREEMAN AND COMPANY, Cooperating Publishers

SAN FRANCISCO

The University of California reserves all rights to reproduce this book,
in whole or in part, with the exception of the right to use
short quotations for review of the book.

Printed in the United States of America.

Library of Congress Catalog Card Number: 63-18323.

ISBN: 0-7 167-0001 -8

Preface

Chemistry deals with all of the substances that
make up our environment. It also deals with the
changes that take place in these substances —
changes that make the difference between a cold
and lifeless planet and one that teems with life
and growth. Chemistry helps us understand and
benefit from nature's wondrous ways.

Chemistry is an important part of what is
called science. Since every phase of our daily
life is affected by the fruits of scientific activity,
we all should know what scientific activity is,
what it can do, and how it works. The study of
chemistry will help you learn these things.

CHEMISTR Y — An Experimental Science pre-
sents chemistry as it is today. It does so with
emphasis upon the most enjoyable part of chem-
istry: experimentation. Unifying principles are
developed, as is appropriate in a modern chemis-
try course, with the laboratory work providing
the basis for this development. When we are
familiar with these widely applicable principles
we no longer have need for endless memorization
of innumerable chemical facts. To see these prin-
ciples grow out of observations you have made

January 1963

in the laboratory gives you a valid picture of
how all scientific advances begin. It permits you
to engage in scientific activity and thus, to some
extent, to become a scientist yourself.

At the end of this course you won't know all of
chemistry. We hope that you will know enough
chemistry and enough about science to feel that
the part you don't know is understandable, not
mysterious. Perhaps you will appreciate the great
power of scientific methods and appreciate their
limitations. We hope that you will have become
practiced in making unexpected observations, in
weighing facts, and in framing valid conclusions.
We hope that you will have formed the habit of
questioning and of seeking understanding rather
than being satisfied with blind acceptance of
dogmatic assertions. We expect that you will
share in the excitement of science and that you
will feel the rich pleasure that comes with dis-
covery. If most of these hopes are fulfilled, then
you have had an optimum introduction to sci-
ence through chemistry. Nothing could be a
more important part of your education at a time
when science is molding our age.

GEORGE C. PIMENTEL

Editor for the Chemical

Education Material Study

Foreword

This textbook was prepared over a three year
period by a group of university and high school
chemistry teachers under a grant from the Na-
tional Science Foundation. The project, called
CHEM Study, was organized and directed on
broad policy lines by a Steering Committee of
nationally known teachers and pre-eminent sci-
entists from a variety of chemical fields. The
Steering Committee, headed by Nobel Laureate
Glenn T. Seaborg, attempted to staff the study
with the country's most able university scientists
and high school teachers. The university profes-
sors were drawn from all over the United States
on the basis of demonstrated understanding of
science and recognized leadership in teaching it.
The names of the contributors to this text al-
ready appear on more than a dozen widely
accepted college level textbooks. An equal num-
ber of outstanding high school teachers were
named as contributors, each one individually
selected on the basis of enthusiastic recommen-
dations by his peers. These teachers participated
in every phase of the preparation of this course.
The effort of these highly qualified persons,
totaling over fifteen man-years, is summed in the
CHEM Study course. The National Science
Foundation deserves commendation for making
such activities possible; never before has such
an array of talent been assembled to construct
a high school chemistry course.

The textbook, CHEMISTR Y—An Experimen-
tal Science, is designed for a high school intro-
ductory chemistry course and it is meshed closely
with an accompanying Laboratory Manual and a
set of pertinent films. A comprehensive Teachers
Guide is available to aid teachers in gaining
familiarity with the course. The first editions of
the textbook and laboratory manual, written
during the summer of 1960, were used during
1960-1961 in 23 high schools and one junior
college by about 1300 students. During this first
year, there was weekly staff contact with the
pioneering teachers. On the basis of their experi-
ence, the materials were revised during the
summer of 1961 and the Teachers Guide was
written. This second edition was used in 123
high schools and 3 junior colleges scattered over
the country and involving 13,000 students. Again
the closely monitored field experience founded
the third and final revision. The course, essen-
tially in the form presented here, was used during
1962-1963 in 560 high schools in 46 states by
about 45,000 randomly selected students. Its
teachability is assured.

The title, CHEMISTRY— An Experimental
Science, states the theme of this one year course.
A clear and valid picture of the steps by which
scientists proceed is carefully presented and re-
peatedly used. Observations and measurements
lead to the development of unifying principles

vii

VI11

FOREWORD

and then these principles are used to interrelate
diverse phenomena. Heavy reliance is placed
upon laboratory work so that chemical prin-
ciples can be drawn directly from student ex-
perience. Not only does this give a correct and
nonauthoritarian view of the origin of chemical
principles, but it gives maximum opportunity
for discovery, the most exciting part of scientific
activity. This experimental theme is supported
by a number of films to provide experimental
evidence that is needed but not readily available
in the classroom because of inherent danger,
rarity, or expense.

The initial set of experiments and the first few
textbook chapters lay down a foundation for the
course. The elements of scientific activity are
immediately displayed, including the role of un-
certainty. The atomic theory, the nature of
matter in its various phases, and the mole con-
cept are developed. Then an extended section of
the course is devoted to the extraction of im-
portant chemical principles from relevant labo-
ratory experience. The principles considered
include energy, rate and equilibrium character-
istics of chemical reactions, chemical periodicity,
and chemical bonding in gases, liquids, and
solids. The course concludes with several chap-
ters of descriptive chemistry in which the ap-
plicability and worth of the chemical principles
developed earlier are seen again and again.

There are a number of differences from more
traditional courses. The most obvious are, of
course, the shift of emphasis from descriptive
chemistry toward chemical principles to repre-
sent properly the change of chemistry over the
last two decades. Naturally, this reconstruction
of the entire course gives a unique opportunity
to delete obsolete terminology and out-moded
material. Less obvious but perhaps more im-
portant is the systematic development of the
relationship between experiment and theory.
Chemistry is gradually and logically unfolded,
not presented as a collection of facts, dicta, and
dogma. We hope to convey an awareness of the
significance and capabilities of scientific activities
that will help the future citizen assess calmly and
wisely the growing impact of technological ad-
vances on his social environment. Finally, we

have striven for closer continuity of subject
matter and pedagogy between high school and
modern freshman chemistry courses for those
students who will continue their science training.

We do believe that the CHEM Study course
achieves the goals we have set. Experience has
shown that the course is interesting to and within
the grasp of the average high school chemistry
student and that it challenges and stimulates
the gifted student. The course content provides
a strong foundation for the college-bound stu-
dent. Inevitably the question arises, "Is this
course better than (or, as good as) the traditional
one?" An answer is not readily found in com-
parative tests. A CHEM Study student might be
handicapped in a test that has little emphasis
upon principles, that is heavily laden with de-
scriptive "recall questions," or that uses obsolete
terminology. Conversely, a test designed specif-
ically for the modern CHEM Study course
content would surely prejudice against a student
with a traditional preparation. The issue can-
not be completely resolved "objectively" be-
cause value judgments are ultimately involved.
Whether the CHEM Study goals are valid and
the approach is reasonable must be decided with
due consideration to the reported experience of
teachers and to the credentials of those who
developed the materials.

There are numberless ways in which CHEM
Study is indebted to the University of California
and to Harvey Mudd College for contributions
of facilities, personnel, and encouragement. We
acknowledge with thanks the stimulation and
support we have received from the National
Science Foundation. Finally, the Staff feels a
heavy debt of gratitude to all of those who
participated so energetically and enthusiastically
in the preparation of the CHEM Study mate-
rials. We thank the Steering Committee for their
valued and helpful guidance. We thank the con-
tributors listed on the title page for their dedica-
tion of time, interest, and their ample talents to
this effort. We acknowledge especially the key
roles of Mr. Joseph Davis, Mr. Saul Geffner,
Mr. Keith MacNab, Miss Margaret Nicholson,
and Mr. Harley Sorensen. These individuals not
only used the CHEM Study materials in the

FOREWORD

classroom but also served continuously as staff
members. Their contributions and critiques have
greatly increased the teachability of the CHEM
Study course. We thank the many teachers who
used the trial editions in their classrooms; their
careful scrutiny of the text and laboratory man-
ual and their many valuable suggestions pro-

vided a firm basis for revisions. Finally, we thank
the many students who labored through the trial
versions of CHEM Study; their every reaction —
pain or pleasure, enthusiasm or ennui, spark or
sputter— was noted and lent to the improvement
of the course.

J. ARTHUR CAMPBELL,

Director, Chemical Education Material Study
Harvey Mudd College

GEORGE C. PIMENTEL,

Editor, Textbook
University of California

Berkeley, California
January, 1963

LLOYD E. MALM

Editor, Laboratory Manual
University of Utah

A. L. MCCLELLAN

Editor, Teachers Guide

California Research Corporation

DAVID RIDGWAY

Producer, Films

Acknowledgments

Quotations appearing on the following pages are
used, with permission, from the indicated sources.

Page 1 History of Science, W. Dampier. New
York: Cambridge University Press, 1949.
17 Principia, Isaac Newton. Mott's transla-
tion revised by F. Cajori. Berkeley: Uni-
versity of California Press, 1934, p. 673.
38 New Systems of Chemical Philosophy,
John Dalton. Manchester, England, 1810.
49 Readings in the Literature of Science,
W. C. Dampier and M. Dampier. New
York: Harper and Row, 1959, p. 100.
65 Solutions, W. Ostwald. London: Long-
mans, Green and Co., 1891.
Letter by J. A. R. Newlands, Chemical
News, Vol. 10, 1864, p. 94.
Chemical Thermodynamics, A Course of
Study, Frederick T. Wall. San Francisco:
W. H. Freeman and Company, 1958, p. 2.

124 The Drift Toward Equilibrium, H. Ey-
ring, from Science in Progress, Fourth
Series, edited by G. A. Baitsell, New
Haven: Yale University Press, 1945, p.
169.

142 Thermodynamics, G. N. Lewis and
M. Randall. New York: McGraw-Hill
Book Co., Inc., 1923, p. 18.

163 Solubility of Non-electrolytes, J. H. Hilde-
brand. New York: Reinhold Publishing
Corp., 1936, p. 13.

108

179 Elements of Chemistry, A. Lavoisier. New
York, 1806, p. 14.

199 Predictions and Speculation in Chemis-
try, W. M. Latimer, Chemical and Engi-
neering News, Vol. 31, 1953, p. 3366.

224 Textbook of Quantitative Inorganic Anal-
ysis, I. M. Kolthoff and E. B. Sandell.
New York: Macmillan, 1936, p. 2.

233 The Rise of Scientific Philosophy, Hans
Reichenbach. Berkeley: University of
California Press, 1956, p. 168.

252 Valence, C. A. Coulson. New York: Ox-
ford University Press, 1961, p. 3.

274 Chemical Analysis by Infrared, Bryce
Crawford, Jr., New York: Scientific
American, Oct. 1953.

300 The Nature of the Chemical Bond,
L. Pauling. Ithaca: Cornell University
Press, 1939, p. 422.

321 Les Prix Nobel, 1947, Nobel lecture by
R. Robinson. Stockholm: Norstedt and
Soner, 1947, p. 110.

421 From Quantum Chemistry to Quantum
Biochemistry, Alberte Pullman and Ber-
nard Pullman, in Albert Szent-Gyoergyi
and Modern Biochemistry, edited by Rene
Wurmser. Paris: Institute of Biology,
Physics, Chemistry, 1962.

Xll

ACKNOWLEDGMENTS

436 Genesis of Life, J. B. S. Haldane, in The
Earth and Its Atmosphere, edited by D. R.
Bates. New York: Basic Books, Inc.,
1960.

The following photographs are used with permission
from the indicated source.

Frontispiece The Candle— Illuminating Chemistry,
by Bernard Abramson.
Page 5 Ice melting, by Ross H. McGregor.
5 Aluminum melting, courtesy Alumi-
num Corporation of America.
5 Solder melting, by Charles L. Finance.
48 G. N. Lewis, courtesy the Hagemeyer
Collection, Bancroft Library, Univer-
sity of California.
94 Cutting potassium, by Charles L. Fi-
nance.
107 D. Mendeleev, courtesy the Univer-
sity of Leningrad.
141 H. Eyring, courtesy H. Eyring.
198 S. Arrhenius, courtesy The Bettmann
Archive.

299 L. Pauling, courtesy The California
Institute of Technology.

310 Network silicates, by Charles L. Fi-
nance.

312 Sodium chloride crystals, by Charles
L. Finance.

320 P. Debye, courtesy Cornell Univer-
sity.

351 R. Robinson, courtesy Canadian In-
dustries Limited.

386 A. Stock, courtesy The American
Chemical Society.

420 G. T. Seaborg, courtesy California
Research Corporation, Richmond
Laboratory, Richmond, California.

435 R. B. Woodward, courtesy The Am-
erican Chemical Society.

Color plate I Elements and compounds, by
Charles L. Finance.
II Indicator colors, by Charles L.

Finance.
Ill Spectrograph, by Charles L.
Finance.

Contents

Chapter 1. Chemistry: An Experimental Science 1

2. A Scientific Model: The Atomic Theory 17

3. Chemical Reactions 38

4. The Gas Phase: Kinetic Theory 49

5. Liquids and Solids: Condensed Phases of Matter 65

6. Structure of the Atom and the Periodic Table 85

7. Energy Effects in Chemical Reactions 108

8. The Rates of Chemical Reactions 124

9. Equilibrium in Chemical Reactions 142

10. Solubility Equilibria 163

11. Aqueous Acids and Bases 179

12. Oxidation-Reduction Reactions 199

13. Chemical Calculations 224

14. Why We Believe in Atoms 233

15. Electrons and the Periodic Table 252

16. Molecules in the Gas Phase 274

17. The Bonding in Solids and Liquids 300

18. The Chemistry of Carbon Compounds 321

19. The Halogens 352

20. The Third Row of the Periodic Table 364

21. The Second Column of the Periodic Table 377

22. The Fourth-Row Transition Elements 387

XIV CONTENTS

23. Some Sixth- and Seventh-Row Elements 41 1

24. Some Aspects of Biochemistry: An Application of Chemistry 421

25. The Chemistry of Earth, the Planets, and the Stars 436

Appendix 1. A Description of a Burning Candle 449

2. Relative Strengths of Acids in Aqueous Solution 451

3. Standard Oxidation Potentials for Half-Reactions 452

4. Names, Formulas, and Charges of Some Common Ions 454
Index 455

THE CANDLE — ILLUMINATING CHEMISTRY

CHAPTER

Chemistry:

An Experimental

Science

• • • those sciences are vain and full of errors which are not born from
experiment, the mother of all certainty. • • •

LEONARDO DA VINCI, 1452-1519

Many words have been spoken and written in
answer to the questions:

"What is the nature of scientific study?"
"What is the nature of chemistry?"

We shall try to find the answers in this course,
not through words alone, but through experi-
ence. No one can completely convey through
words the excitement and interest of scientific

discovery. Hence we shall see the nature of sci-
ence by engaging in scientific activity. We shall
see the nature of chemistry by considering prob-
lems which interest chemists.

Our starting point will be based on examples
of the activities of science, rather than on defini-
tions. We will perform these activities, beginning
on familiar ground. On such ground, where you
know the answer, you will best see the steps by
which science advances.

1-1 THE ACTIVITIES OF SCIENCE

Every form of life "feels" its surroundings in one
way or another. In response to the feel of the
surroundings, it behaves according to a pattern
which tends to prolong its existence.

A tree is illuminated by the morning sunshine.
In response, the leaves of the tree turn on their

stems to present full surface to the light. This
movement causes the leaves to intercept more
light, and light is the source of energy which runs
the amazing chemical factory operated by the
tree. The tree grows.
A bear feels that summer is over — perhaps by

chemistry: an experimental science I CHAP. 1

the length of the day or by the color of fall
leaves, perhaps by some ursine almanac humans
cannot read. In response, he seeks a secluded
spot and takes a winter-long nap. During this
hibernation, his blood pressure and body tem-
perature drop, his digestion closes shop. The
bear uses the minimum energy necessary to stay
alive. It is not a coincidence that this occurs dur-
ing the season when food is most difficult to find
and the weather is quite unbearable.

Of all living things, man feels his surroundings
and responds to them in the most complex way.
He is more curious than the most inquisitive
kitten. Through his intellect he uses his senses
more effectively than an antelope avoiding a
stalking lion. He has developed communication
far beyond the warning quack of a sentry duck
or the mating call of a lonely moose. Man's in-
tellect, together with his communicative ability,
permit him to respond to his environment in
uniquely beneficial ways. He accumulates infor-
mation about his surroundings, he organizes this
information and seeks regularities in it, he won-
ders why the regularities exist, and he transmits
his findings to the next generation. These are the
basic activities of science:

to accumulate information through observa-
tion;

to organize this information and to seek regu-
larities in it;

to wonder why the regularities exist;

to communicate the findings to others.

Fig. 1-1. A scientist makes careful observations.

So the activities of science begin with observa-
tion. Observation is most useful when the condi-
tions which affect the observation are controlled
carefully. A condition is controlled when it is
fixed, known, and can be varied deliberately if
desired. This control is best obtained in a special
locale — a laboratory. When the observation is
brought under careful control, it is dignified by
a special name — a controlled sequence of observa-
tions is called an experiment. All science is
built upon the results of experiments.

1-1.1 Observation and Description

Everyone thinks of himself as a good observer.
Yet there is much more to it than meets the eye.
It takes concentration, alertness to detail, in-
genuity, and often just plain patience. It even
takes practice! Consider an example from your
own experience. Think how much can be written
about an object as familiar as a burning candle!
Of course, it takes careful observation — a careful
experiment. This means the candle must be ob-
served in a laboratory, that is, in a place where
conditions can be controlled. But, how do we
know which conditions need be controlled? Be
ready for surprises here! Sometimes the impor-
tant conditions are difficult to discover. Here are
some conditions that are important in some ex-
periments but are not important here.

The experiment is done on the second floor.
The experiment is done in the daytime.
The room lights are on.

Here are some conditions that might be impor-
tant here.

The lab bench is near the door.
The windows are open.
You are standing close enough to the candle
to breathe on it.

Why are these conditions important? Do they
have something in common? Yes, there is the
common factor that a candle does not operate
well in a draft. The conditions are important
because they influence ihe result of the experi-

SEC. 1-1 | THE ACTIVITIES OF SCIENCE

ment. Important conditions are often not as
easily recognized as these. A good experimen-
talist pays much attention to the discovery of
conditions that must be controlled. His success
is often determined by his ability to control them.
Review your own description of a burning
candle and compare your essay with the one in
Appendix 1. How many of your observations
are included there? How many listed in the ap-
pendix are not in your description? We see that
the burning candle is a complicated and fascinat-
ing object when subjected to careful observation
and detailed description.

1-1.2 The Search for Regularities

Observation inevitably leads to questions. One of
the first questions that usually arises is "What
regularities appear?" The discovery of regulari-
ties permits simplification of the observations.
Instead of each observation standing alone, sev-
eral observations can be classed together and,
hence, can be used more effectively.

You must become aware of the pitfalls that
exist in the search for regularities. The search is
a meandering one, frequently taking wrong
turns. It is inherent in the exploration of the
unknown that not every step is an advance. Yet
there is no other way to advance than by taking
steps. How the search proceeds is best seen in a
fable. The development of such a transparent
example may help you see how a scientist
searches for regularities.

Fable: A Lost Child Keeping Warm

Once upon a time a small child became
lost. Because the weather was cold, he de-
cided to gather materials for a fire. As he
brought objects back to his campfire, he
discovered that some of them burned and
some of them didn't burn. To avoid collect-
ing useless substances, the child began to
keep track of those objects that burned and
those that did not. (He organized his infor-
mation.) After a few trips, his classification
contained the information that is shown
in Table 1-1.

Table 1-1. flammabilitv

WILL BURN WON'T BURN

Tree limbs
Broom handles
Pencils
Chair legs
Flagpoles

Rocks
Blackberries
Marbles
Paperweights

This organization of the information was
quite an aid in his quest for warmth. How-
ever, as tree limbs and broom handles
became scarce, the child tried to find a
regularity that would guide him to new
burnable materials. Looking at the pile of
objects that failed to burn and comparing
it with the pile of objects that would burn,
the child noticed that a regularity ap-
peared. He proposed a possible "generali-
zation."

Perhaps: "Cylindrical objects burn."

This procedure is one of the elementary logical
thought processes by which information is sys-
tematized. It is called inductive reasoning, and
it means that a general rule is framed on the basis
of a collection of individual observations (or
"facts"). Of what use is the inductive process?
It is an efficient way of remembering.

The next day the child went looking for
burnable materials, but he forgot to bring
along his list. However, he remembered his
generalization. So, he returned to his
hearthside hauling a tree limb, an old cane,
and three baseball bats (successful predic-
tions!). What's more, he reflected with
pleasure that he hadn't bothered to carry
back some other objects: an automobile
radiator, a piece of chain, and a large door.
Since these objects weren't cylindrical there
was no reason to expect them to burn.

No doubt you are ready to complain that this
generalization isn't really true! Quite the oppo-
site! The generalization states a regularity dis-
covered among all the observations available,
and as long as observations are restricted to
objects in the list, the generalization is applicable.
A generalization is reliable within the bounds
defined by the experiments that led to the rule.

chemistry: an experimental SCIENCE I CHAP. 1

As long as we restrict ourselves to the objects
in Table 1-1 (together with canes and baseball
bats) it is surely true that all of the cylindrical
objects burn!

Fig. 1-2. "Cylindrical objects burn."

Because of his successful predictions, the
child became confident of his generaliza-
tion. The next day he deliberately left the
list at his campsite. This time, with the aid
of his rule, he came back heavily laden with
three pieces of pipe, two ginger ale bottles,
and the axle from an old car, while spurn-
ing a huge cardboard box full of news-
papers.

During the long cold night that followed
he drew these conclusions:

(1) The cylindrical shape of a burnable ob-
ject may not be intimately associated
with its fiammability after all.

(2) Even though the "cylindrical" rule is
no longer useful, tree limbs, broom
handles, pencils, and the other burn-
ables in Table 1-1 still burn.

(3) He'd better bring the list along tomor-
row.

But, thinking over the longer list, he saw
a new regularity that fitted Table 1-1 and
the newly acquired information as well:

Perhaps: "Wooden objects burn."

What good is this rule in the light of the earlier
disappointment? Well, it caused the child to go
back and get that door he had passed up two
days earlier, but it didn't lead him to go after
the chain, the automobile radiator, or the card-
board box full of newspapers.

Don't think this is facetious — it is exactly what
science is all about! We make some observations,
organize them, and seek regularities to aid us in
the effective use of our knowledge. The regulari-
ties are stated as generalizations that are called
theories. A theory is retained as long as it is
consistent with the known facts of nature or as
long as it is an aid in systematizing our knowl-
edge. We can be sure that some day a number of
our present scientific views will seem as absurd
as "Cylindrical objects burn." But on that day
we will be proud of better views that have been
substituted. If you are discouraged by the child's
faltering progress — he hasn't yet decided that the
box of newspapers will burn — be reassured. This
child is a scientist and his faltering steps will lead
him to the newspapers. They are the same steps
that led us to our present understanding of rela-
tivity, to our discovery of polio vaccine, and to
our propulsion of rockets to the moon.

A GENERALIZATION ABOUT THE
MELTING OF SOLIDS

Through experiment, you yourself have discov-
ered an important regularity in the behaviors of
solid substances.

A solid melts to a liquid when the temperature
is raised sufficiently. The temperature at which
a solid melts is characteristic. When the warm
liquid is recooled, it solidifies at this same
temperature.

This generalization is of great value. It is based
upon exactly the type of experiment you have
performed. We have confidence in the rule be-
cause this type of experiment has been conducted
successfully on hundreds of thousands of sub-
stances. The melting behavior is one of the most
commonly used methods of characterizing a sub-
stance. It leads us to wonder if every solid can
be converted to a liquid if the temperature is
raised sufficiently. Further, it leads us to wonder

SEC. 1-1 | THE ACTIVITIES OF SCIENCE

Fig. 1-3. A solid melts to a liquid at a characteristic
temperature.

if every liquid can be converted to a solid if the
temperature is lowered sufficiently.

SOMB TERMINOLOGY

We have discovered that a solid can be converted
to a liquid by warming it at or above its melting
point. Then the solid can be restored merely by
recooling. The solid and the liquid are similar
in many respects and one is easily obtained from
the other. Hence they are called different phases
of the same substance. Ice is the solid phase of
water and, at room temperature, water is in the
liquid phase. The change that occurs when a
solid melts or a liquid freezes is called a phase
change.

1-1.3 Wondering Why

We have already experienced some of the activi-
ties of science. First came careful observation
under controlled conditions, then organization
of the information and the search for regularities
of behavior. There is one more activity that, like
dessert, fittingly comes last. This activity may be

called "wondering why," and it arises from our
irresistible urge to know more than merely
"What happens?" We must also seek the answer
to "Why does it happen?" This activity is prob-
ably the most creative and the most rewarding
part of a science. What is the process? What does
it mean to answer a question beginning "Why"?

EXPLANATIONS

Let us see what it means to search for an expla-
nation. Consider a child blowing up a balloon.
As he blows into the balloon again and again it
expands and it becomes "harder." Evidently the
gas is "pushing" on the inside of the balloon,
stretching its elastic walls. Why does the gas push

chemistry: an experimental SCIENCE I CHAP. 1

outward more and more on the walls of the
balloon as it is inflated? Why does the gas con-
tinue to push outward without "tiring" or "run-
ning down"? These are "wondering why" ques-
tions.

There are two ways to proceed in trying to
answer these questions. We have already ex-
amined one of these ways— to look more closely
at the balloon, to record carefully what we see,
and to seek regularities in what we observe. The
second way is to look away from the balloon and
to seek similar behavior in another situation that
we understand better. Maybe this will enable us
to frame an explanation of the gas pressure in
terms of the better-understood situation. Some-
times a useful explanation turns up in a quite
unexpected direction.

Consider the motion of a billiard ball. After
it is struck by the cue, it moves until it strikes a
cushion, from which it bounces, apparently with
undiminished velocity. It rolls along in a new
direction until it strikes another cushion, chang-
ing its direction again. It may continue to roll
until it has hit the cushions six or seven times.
The billiard ball seems almost tireless as it re-
bounds time and again from the "walls" of the
billiard table. Could there be a connection be-
tween the "untiring" motion of a billiard ball
and the "untiring" pressure of a gas in a balloon?

Billiard balls have long fascinated both idle
and curious men. The latter group has found
that the motion of a billiard ball can be described

Fig. 1-4. A rebounding billiard ball suggests a possi-
ble explanation of gas pressure.

by assuming each collision with a cushion is
perfectly elastic. When the ball strikes the cush-
ion, pushing on it, the cilshion pushes back, and
the ball leaves again without any loss of velocity.
Its movement can be reasonably foretold on the
basis of this elastic collision assumption. Perhaps
the behavior of a gas can be explained in these
same terms. Suppose we picture a gas as a col-
lection of particles bouncing around in a con-
tainer, endlessly, making elastic collisions with
the walls, just like miniature billiard balls. Every
time one of these particles strikes a wall, it
"pushes" on the wall and bounces away again.
If there are many particles, there will be many
such collisions per second, which accounts for
the pressure of the gas. If gas is added to the
balloon, there will be even more particles, hence
more wall collisions per second, hence higher
pressure. Thus the billiard ball model does offer
a possible answer to our question.

With this example, we can now see the mean-
ing of an explanation. It began with a "wonder-
ing why" question:

Question: Why does a balloon expand as it is
inflated?

Possible Answer: Perhaps the gas put in the
balloon consists of a collection of small par-
ticles that rebound from the wall of the bal-
loon just as billiard balls rebound from the
cushions of a billiard table. As the gas particles
rebound from the balloon wall, they push on
it. When more gas particles are added, the
number of such wall collisions per second in-
creases, hence the outward push on the bal-
loon wall increases. The balloon expands.

SEC. 1-1 | THE ACTIVITIES OF SCIENCE

This is the characteristic pattern of an explana-
tion. It begins with a "Why?" question that asks
about a process that is not well understood. An
answer is framed in terms of a process that is
well understood. In our example, the origin. of
gas pressure in the balloon is the process we wish
to clarify. It is difficult even to sense the presence
of a gas. The air around us usually cannot be
seen, tasted, nor smelled (take away smog); it
cannot be heard or felt if there is no wind. So we
attempt to explain the properties of a gas in
terms of the behavior of billiard balls. These
objects are readily seen and felt; their behavior
has been thoroughly studied and is well under-
stood.

The search for explanation is, then, the search
for likenesses that connect the system under
study with a model system which has been
studied earlier. The explanation is considered to
be "good" when:

(1) the model system is well understood (that is,
when the regularities in the behavior of the
model system have been thoroughly ex-
plored); and

(2) the connection is a strong one (that is, when
there are close similarities between the stud-
ied system and the model system).

Our example constitutes a good explanation,
because:

(1) how a billiard ball rebounds is well under-
stood; we can calculate in mathematical de-
tail just how much push the billiard ball
exerts on the cushion at each bounce; and

(2) the connection to gas pressure is close; ex-
actly the same mathematics describes the
pressure behavior if the gas is pictured as a
collection of many small particles, endlessly
in motion, bouncing elastically against the
walls of the container.

Therefore, the particle explanation of gas pres-
sure is a good one.

Now, perhaps, you can see that answering the
question "Why?" is merely a highly sophisti-
cated form of seeking regularities. It is indeed a
regularity of nature that gases and billiard balls
have properties in common. The special creativ-

ity shown in the discovery of this regularity is
that the likeness is not readily apparent. Fit-
tingly, there is a special reward for the discovery
of such hidden likeness. The discoverer can bring
to bear on the studied system all of the experience
and knowledge accumulated from the well un-
derstood system.

HEATING STEEL WOOL

These ideas are best learned by using them your-
self. Let us take an example from your own labo-
ratory work. You have observed the heating of a
variety of solid materials: sulfur, wax, tin, lead,
silver chloride, and copper. Each melts at a char-
acteristic temperature. This fact led us to the
generalization "A solid melts to a liquid at a
characteristic temperature." Your confidence in
the generalization was bolstered by the addi-
tional information (communicated to you, not
experienced) that this applies to "hundreds of
thousands of substances."

Your teacher demonstrated the effect of heat-
ing steel wool, and it proved to be a spectacular
exception to the generalization about melting.
An inexperienced observer might note this spe-
cial behavior in his notebook, cross reference it
under "Sparklers," and proceed to the next sub-
stance. A curious person, however, cannot resist
wondering, "Why does steel wool behave in this
special manner?"

In quite another direction of thought you in-
vestigated the burning of a candle. You dis-
covered that air plays a role and that the prod-
ucts are different from the starting materials.

Here we have two areas that have been in-
vestigated thoroughly and that are well under-
stood (and that you have examined to some ex-
tent yourself). Stated briefly, our starting point
of knowledge is:

(1) Solids melt on heating.

(2) A candle burns, consuming oxygen from the
air.

We are wondering why steel wool makes such
a brilliant display on heating. Perhaps the ex-
planation is to be found in 2. Though there are
striking visual differences, perhaps we can ex-
plain the behavior of steel wool by likening it to

chemistry: an experimental science chap. 1

Steel wool hums
■when heated in air-

Fig. 1-5. The behavior of steel wool on heating.

a candle. Can we substitute the words "steel
wool" for "candle" in 2?

Perhaps: (3) Steel wool burns, consuming
oxygen from the air.

If 3 is a useful connection with the behavior
of a burning candle, then we should enjoy the
special reward for this discovery. The connection
we propose implies that the knowledge accumu-
lated on the candle can be brought to bear on
the new system.

Since: A candle does not burn if it is denied
oxygen.

then: steel wool should not sparkle if it is
denied oxygen.

Steel wool melts to

a liquid -when
heated in absence
of oxygen.

dioutd&

Here is a proposal subject to direct test. We
can place steel wool under an atmosphere that
excludes oxygen gas, and look for a change in
the heating behavior. We go to the laboratory,
heat steel wool under carbon dioxide and, lo, the
steel wool melts!

The special behavior of steel wool on heating
can now be said to be explained:

'Steel wool, like a candle, burns when heated

in air.

"Steel wool, like other solids, melts to a liquid
when heated under conditions that prevent burn-
ing."

This type of understanding makes possible the
metallurgical processing of steel (and other met-
als). This type of reasoning makes possible an
increase in our perception of the regularities of
nature. // begins with wondering why.

1-2 UNCERTAINTY IN SCIENCE

Here are three statements concerning the melting
behavior of /wadichlorobenzene:

(1) The melting point is 53°C.

(2) The melting point is 53.2°C.

(3) The melting point is 53.203°C.

Apparently the third statement says more than
the second, and the second statement says more
than the first. Would it surprise you to learn that

the second statement could be the most informa-
tive of the three? It is so. To understand why. we
must consider uncertainty in measurement.

1-2.1 Uncertainty in a Measurement

A scientific statement conveys knowledge about
the environment. The statement is careless if it
says less than is known. It is misleading if it says

SEC. 1-2 I UNCERTAINTY IN SCIENCE

more than is known. The most accurate statement
clearly conveys just what is known and no more.
Thus, a scientist will decide whether to list the
melting point as 53°C, 53.2°C or 53.203°C by
considering which value tells just what is known
about the melting behavior and no more.

Consider your own laboratory measurement
of the melting point of paradichlorobenzene
(Experiment 3). Do your temperature measure-
ments permit you to say that the melting point
is 53°C, not 54°C? Probably, yes. It isn't too
difficult to read the thermometer with this cer-
tainty. Can you read the thermometer so as to
distinguish 53.0°C and 53.2°C? This is more dif-
ficult. It depends upon the thermometer and
your skill in its use. It also depends upon whether
the temperature of the solid during melting is
uniform throughout. Still, a magnifying glass
permits more certainty in the reading of the
scale, and slower heating would increase the uni-
formity of the sample temperature. With this
type of extra care, it is possible to tell that the
melting point of /raradichlorobenzene is 53.2°C,

Fig. 1-6. Every measurement involves some uncer-
tainty. The thermometer on the left can be
read to ±0.2°C. The one on the right can
be read to ±0.002°C.

not 53.0°C. Consider, however, the chance of re-
fining your thermometry sufficiently to distin-
guish between 53.200°C and 53.203°C. With the
equipment available to you, it just isn't possible.
We conclude, then, that a careful measurement
could establish the melting point to be 53.2°C.
Then the second statement (m.p. = 53.2°C)
would tell just what is known. The first state-
ment (m.p. = 53°C) does not tell all that is
known since only two figures are given, though
three were measured. The third statement
(m.p. = 53.203°C) tells far more than is known
since the last two figures were not learned ex-
perimentally. We see that the measurement gives
us three figures which are meaningful and sig-
nificant. The number 53.2°C is said to have three
significant figures.

oat

chemistry: an experimental SCIENCE I CHAP. 1

1-2.2 Uncertainty in a Derived Quantity:
Addition and Subtraction

Results of scientific observations are often com-
bined. For example, in Experiment 5 you will
determine the change of water temperature dur-
ing the combustion of a candle (or during the
solidification of candle wax). The change of tem-
perature, which we called At, is the result of two
measurements, not just one — it is a derived quan-
tity:

Temperature after heating = 38.5°C

Temperature before heating = 9.3°C

Difference (temperature change), At = 29.2°C

In accordance with good scientific practice, we
would like to express the temperature change so
as to include just what is known but no more.
To do this, we must investigate how the uncer-
tainties in the two temperature readings fix the
uncertainty in the difference, At. Suppose the
final temperature is measured by a second stu-
dent who finds the temperature to be 38.3°C.
Then a third student finds it to be 38.7°C. Ap-
parently, different students making the same
measurement may differ by a few tenths of a
degree. Apparently, the measurement recorded
by any one student could be in error by this
amount, about 0.2°C. Perhaps the temperature
recorded as 38.5°C could be as high as 38.7°C
(0.2° higher) or as low as 38.3°C (0.2° lower)!
This can be expressed briefly as follows:

Temperature after heating = 38.5 ± 0.2°C*

Presumably, the same uncertainty is present in
the first temperature measurement, so our cal-
culation becomes:

Temperature after heating
Temperature before heating

= 38.5 ± 0.2°C
= 9.3 ± 0.2°C

Difference (temperature change), At = 29.2 ± ??°C

To decide what uncertainty to ascribe to 29.2,
consider the worst possible combination of the
uncertainties. The first temperature could be as
low as 9.1°C and the final temperature could be
as high as 38.7°C. Then the difference would be
29.6°C. Thus the worst possible combination of
errors would place an error in the difference

* The symbol ± is read "plus or minus."

equal to the sum of the uncertainties in the parts,
0.2 + 0.2 = 0.4. Hence our derived result can
be written

Difference (temperature change) = 29.2 ± 0.4°C

We see that the uncertainty in a derived quantity
is fixed by the uncertainties in the measurements
that must be combined. For an addition or a
subtraction, the maximum uncertainty is simply
the sum of the uncertainties in the components:
0.2 + 0.2 = 0.4.

EXERCISE 1-1

In Experiment 5, the weight of a sample of water
is determined by subtracting the weight of the
empty can from the weight of the can containing
the water.

(wt. water) = (wt. can + water) — (wt. empty can)

Suppose the weight of the can is 61 ± 1 grams
and the weight of the can plus water is 406 ± 1
grams. Calculate the weight of the water and the
maximum uncertainty in the weight caused by
the uncertainties in each of the two weighings.

1-2.3 Uncertainty in a Derived Quantity:
Multiplication and Division

The temperature measurements made in Experi-
ment 5 enable you to calculate the quantity of
heat liberated when a known weight of candle is
burned. Heat is measured in units called calories;
one calorie is the heat needed to raise the tem-
perature of one gram of water one degree Centi-
grade.^ To raise the temperature of two grams
of water one degree would require two calories —
10 grams would require 10 calories. In general,

quantity of heat necessary to raise w grams
of water one degree

= w calories

But in the experiment, the temperature of the

t The heat necessary to raise the temperature of one
gram of water one degree is constant within ±0.2%
between 8 and 80°C. The calorie has been, in the past,
defined to be the heat necessary to raise the temperature
of one gram of water from 14.5 to 15.5°C.

SEC. 1-2 I UNCERTAINTY IN SCIENCE

water rises several degrees — we call the tempera-
ture change At. If it takes one calorie to warm
one gram of water one degree Centigrade, it
takes five calories to warm it five degrees. In
general,

quantity of heat necessary to raise w grams
of water At degrees

= w X At calories (/)

Again we are faced with the dual problem: first,
to calculate the quantity of heat, q; and, second,
to decide the uncertainty in q.

The quantity of heat, q, is calculated with the
aid of equation (7); it is simply the product of the
weight of water times the temperature difference.
Referring to our data, we find,

wt. of water = 345 ± 2 grams
At = 29.2 ± 0.4°C

The quantity of heat is equal to the product,

345
X 29.2
690
3105
690

10,074.0 calories
q = 10,074.0 ± ? calories

As in the last section, we can estimate the
uncertainty in q by considering the worst possible
combination of uncertainties. Suppose the weight
is actually 343 grams and At is actually 28.8°.
Then the product would be below 10,074.0 calo-
ries. But perhaps the weight is actually 347 grams
and At is actually 29.6°. Then the product would
be above 10,074.0 calories. These extremes deter-
mine the uncertainty in the product:

Minimum value

Maximum value

343

347

X 28.8

X29.6

2744

2082

2744

3123

686

694

9878.4 calories

10,271.2 calories

We see that the product 10,074.0 could be in
error by almost 200 calories. Now we can express
our result together with its uncertainty:

q = 10,074.0 ± about 196 calories

In view of this large uncertainty, we can round
off the answer sensibly:

q = 10,100 ± 200 calories

Once again the uncertainty in the product, a
derived quantity, is fixed by the uncertainties in
the measurements that must be combined.

EXERCISE 1-2

Calculate the uncertainty in the product w X At
caused by the temperature measurement alone
(assuming the uncertainty in the 345 gram weight
of water has been made negligible by more care-
ful weighings). Calculate the uncertainty caused
by the ±2 gram weighing uncertainty alone
(assuming the uncertainty in the temperature
change, 29.2°C, has been made negligible by use
of a more sensitive thermometer). Compare these
two contributions to the total uncertainty, about
200 calories.

The uncertainty in the product, ±200 calo-
ries, is not simply the sum of the uncertainties
in the factors, ±0.4°C and ±2 grams. Instead,
the sum of the percentage uncertainties in the
factors determines the uncertainty in a product
or a quotient.* Fortunately, there is an easy
method for estimating it roughly without cal-
culating percentages. This method, based upon
the number of figures written, is described in
Section 1-2.5.

1-2.4 The Absence off Certainty in Science

Each measuring device has limitations that fix its
accuracy. Hence every individual observation
has some uncertainty associated with it. Since
every regularity of nature is discovered through
observations, every regularity (law, rule, theory)
has uncertainty attached to it.

Every scientific statement involves some
uncertainty.

A corollary:

No scientific statement is absolutely certain.

* The calculation based upon percentage uncertainty is
presented in Appendix 4 of the Laboratory Manual.

chemistry: an experimental SCIENCE I CHAP. 1

1-2.5 How Uncertainty Is Indicated

We now have seen two methods for indicating
uncertainty in a number. The most informative
is to follow the number by the symbol db and
then the best estimate available of the uncer-
tainty. The less informative but more widely used
method is to indicate crudely the uncertainty by
the number of figures shown. The last figure
shown is generally the one in which there is some
uncertainty. Thus, the number 53.2°C indicates
there may be uncertainty in the figure 2, but
there is none in either of the figures 5 or 3. The
digits that are certain and one more are called
significant figures. The correct number of sig-
nificant figures should always be used and, wher-
ever possible, the more definite indication ±
should be added.

We need convenient rules for estimating the
maximum uncertainty in derived quantities. This
is rather easy for a sum or a difference. In either
case, merely add up the uncertainties in the
components. Fortunately there is an equally sim-

ple rule for estimating roughly the uncertainty
in a product or quotient. A product (a X b) or
quotient (a/b) has the same number of significant
figures as the less precise component (a or b).*

EXERCISE 1-3

In Section 1-2.3 we multiplied 345 X 29.2 to
obtain 10,074.0 calories.

(a) How many significant figures are there in the
factor 345? In the factor 29.2?

(b) How many significant figures should be re-
tained in the product, 10,074.0?

(c) Six figures are specified in the number
10,074.0 — more than are warranted. "Round
off" this number in accordance with your
answer to (b). Compare your answer with
the final result derived in Section 1-2.3,
q = 10,100 ± 200 calories.

1-3 COMMUNICATING SCIENTIFIC INFORMATION

One of the most important reasons for man's
progress in understanding and controlling his en-
vironment is his ability to communicate knowl-
edge to the next generation. It isn't necessary for
each twentieth century scientist to invent the
atomic description of matter. This was invented
by John Dalton in the nineteenth century, and
Dalton recorded his ideas in the scientific litera-
ture together with the observations that led him
to the model. By study of this and subsequent
literature a modern scientist can appraise the
nature of the description, the facts it will explain,
and the limitations. He is quickly able to ap-
proach the frontiers of knowledge — the frontiers
defined by the limitations in our accepted models
of the behavior of matter.

One can almost say that a scientific advance is
important only if it is communicated to others.
If Dalton had not told others of his ideas nor
tried to convince them (through his supporting

arguments), then someone else would have had
to do it all again.

Communication of knowledge is, then, an im-
portant part of scientific activity. The first re-
quirement is good use of language. If an idea is
not well expressed, whether orally or in writing,
it is not likely to be clearly understood. An argu-
ment loses its force if it is stated in an ambiguous
way. An essential thought can be lost in a maze
of excess words. Choose and use your language
carefully.

The manner in which you present an idea de-
pends to some extent upon the intended use, to
some extent upon the type of information avail-
able. Usually, the more precise the statement of
the regularity, the more valuable it is. In general,
there will be more than one way to express a
generalization, and judgment is needed in mak-

* The use of significant figures is discussed in Appendix
4 in the Laboratory Manual.

SEC. 1-3 I COMMUNICATING SCIENTIFIC INFORMATION

Fig. 1-7. A regularity in the behavior of a fixed
amount of gas. As pressure rises, volume
decreases.

ing a choice among them. This is best seen in
terms of an example.

Figure 1-7 shows a tire pump with a pressure
gauge attached to the hose so that the gauge
prevents escape of the gas in the pump. Pushing
on the handle of the pump causes the plunger to
go down, and reduces the volume occupied by
the gas. The gauge shows that the pressure is
higher. Pushing still harder on the handle in-
creases the pressure still more. Again the increase
in pressure causes a reduction in volume. We see
that as the pressure rises, the volume decreases.
This qualitative statement describes a regularity
in the behavior of a fixed amount of gas. Such a
qualitative statement is the crudest form in which
a regularity can be expressed.

A curious person, attempting to understand
this regularity, might see a necessity for more
careful measurements. He might build a new
piece of apparatus, one that would be more suit-
able for measuring volumes and pressures over
a wide range. After carrying out a series of

measurements, he would conclude with a table
of data, such as Table 1 -II. A table of data is a
second way in which the regularity can be ex-
pressed. In a third mode of expression, the meas-
urements can be presented graphically in a plot
of pressure against volume, as in Figure 1-8.

With these careful measurements, we might
also seek a mathematical statement of the behav-
ior. Sometimes inspection of the data suggests a
relationship. Sometimes the appearance of a
plot, such as Figure 1-8, reveals a mathematical
expression. In the example we're studying, the
curve resembles a hyperbola, a curve described
by the simple equation, xy = a constant. This
similarity leads us to multiply each pressure and
volume pair, as shown in the third column of
Table l-II. We find that during a ten-fold in-
crease in pressure, the product P X V is fairly
constant. There are some variations in the prod-
uct, as seen both in Table l-II and, as well, in
the scatter of the points around the smooth curve
in Figure 1-8. The random nature of the devia-
tions from constancy suggests that they measure
the uncertainty due to experimental technique.
We can use these deviations to provide an esti-
mate of the uncertainty in the average, ±0.6.
(How this is done is shown in Exercise 1-4.)

chemistry: an experimental SCIENCE I CHAP. 1

Hence, with reasonable confidence we can state
the regularity mathematically:

PXV= 22.4 ± 0.6

Thus we have found four ways to express the
regularity between the pressure and volume of
oxygen gas:

(a) Qualitatively: As the pressure rises, the vol-
ume decreases.

(b) Quantitatively: List the original data that
show how pressure and volume are related,
as in Table 1 -II.

(d) Mathematically:

PXV= 22.4 ± 0.6

P = pressure (in atmospheres)
V = volume (in liters) of 32.0 grams
of oxygen gas at 0°C

Obviously the regularity expressed in the qual-
itative form (a) is far less informative than any
one of the quantitative presentations, (b), (c), or
(d). The relative merits of the expressions (b),
(c), and (d) depend upon the use. Table l-II tells
in most detail exactly how much is known about
the pressure-volume behavior of oxygen gas
(from this experiment). In the graphical presen-
tation of Figure 1-8 the trend of the data is
shown by the smooth curve drawn to pass near
as many points as possible. Uncertainties caused

zoo

150

100

2 4 .6 .8

Pressure, atmospheres

1.0

Table l-II.

PRESSURE AND VOLUME OF 32.0 GRAMS
OF OXYGEN GAS t = 0 C

PRESSURE

(in units called
atmospheres)

VOLUME

(in units called
liters)

PXV

0.100
0.200
0.400
0.600
0.800
1.00

224

109
60.0
35.7
27.7
22.4

22.4
21.8
24.0
21.4
22.2
22.4

Average 22.4 ± 0.6

Fig. 1-8. A plot of the pressure versus volume of

32.0 grams of oxygen gas; t = 0°C.

by experimental errors cause the data points to
fall above and below this smooth curve. Hence
the graphical presentation reveals reliability of
the measurements. The smooth curve "irons
out" these uncertainties and provides a con-
venient basis for predicting volumes at inter-
mediate pressures (that is, for interpolating).
However, from the standpoint of usefulness, the
mathematical expression (d) is often the best.
It is the most compact way of stating the regu-
larity together with its uncertainty. Mathematics
is one of the most important tools of chemistry.
No matter how expressed, all scientific "rules,"
"laws," and "theories" are statements of regu-
larities of nature. Their usefulness depends upon
the amount of experimental evidence that shows
that the "rule," "law," or "theory" corresponds
to experimental reality. Within the bounds that
it is known to correspond to experimental reality,
the relation can be used for prediction.

EXERCISE 1-4

(a) Add the six values of P X V in Table l-II
and divide by 6 to obtain the average.
(P X V)„ or, {PV)„.

SEC. 1-4 I REVIEW

(b) Now add a fourth column to Table I —I I
showing the deviation of each P X V prod-
uct from (PV)av. Head this column with the
word "Deviation," and calculate each entry
by subtracting (PV)av from the measured
value. For example, the second entry will be
-0.6 (since, 21.8 - 22.4 = -0.6).

deviations, add the column (disregarding al-
gebraic signs) and divide by 6 to obtain an
average deviation,
(d) Compare your calculations in (a) and (c)
with the result given in Table I -II,

Average = 22.4 db 0.6

1-4 REVIEW

This chapter began with the statement that we
would find through experience what science is
all about. Already you have had opportunities
to do so in the laboratory. We see that science
is man's systematic investigation of his en-
vironment. Chapter 1 has told how this in-
vestigation proceeds. The remainder of the book
is concerned with those parts of this investigation
that are carried out by chemists. Before going
on to see what chemistry is, let us review your
laboratory accomplishments so far with empha-
sis on the activities of science.

1-4.1 Accumulating Information Through
Observation

Observation of a burning candle reveals an as-
tonishing complexity. It also reveals the impor-

Fig. 1-9. A good experimentalist is a good observer.

Record in your notebook at the moment of
observation. Prepare tables in advance.

tance and value of careful study and attention
to detail.

In your experimentation, be alert and ready
for unexpected developments. Record in your
notebook at the moment of observation a de-
scription of everything you see. The time of the
observation frequently has importance. Com-
pleteness is, by far, the most important property
of a good notebook. Next in importance, legi-
bility, neatness, and organization make your
notebook a more valuable record. Whenever
possible, prepare tables in advance for the results
of measurements you can anticipate. This guar-
antees that you won't forget to note important
information, and it frees you from clerical work
during an experiment.

Remember, chemistry is built upon the results
of experiments. An experiment is a controlled
sequence of observations. A good experimentalist
is a good observer.

1-4.2 Organizing Information and Seeking
Regularities in It

The mere cataloging of observations is not sci-
ence. In fact, the advance of our knowledge of
nature would long ago have ground to a halt if
we merely made observations. The multitude of
known facts can be dealt with only if it is pack-
aged efficiently. This packaging we have called
"organizing the information" and "seeking regu-
larities."

There is no single recipe for seeking regulari-
ties. That is probably why the search is so
interesting and why the scientist receives so much
personal satisfaction from his work. Here is op-

chemistry: an experimental SCIENCE I CHAP. 1

portunity for originality, opportunity for testing
one's wit and cleverness. You can experience the
pleasure a scientist derives from clarifying a pre-
viously mystifying behavior by careful experi-
ments of your own.

In our study of the candle the presence of
liquid at the top of the candle caught our atten-
tion. It led us to wonder about the behavior of
other familiar solids given a similar treatment.
In this case, we went looking for regularity— we
sought, through experiment, to discover how
other solids behave on heating. Our first studies,
when organized, led us to the generalization that
solids melt at a characteristic temperature when
heated. We made two gains thereby: we found
an efficient expression of the results of a number
of experiments, and we provided a basis for ex-
pectation on the effect of heating solids which
we have not studied before. The confidence this
expectation deserves is fixed by the amount of
evidence supporting the generalization.

1-4.3 Wondering Why

The culmination of the investigation of our en-

vironment we have dubbed "wondering why."
We seek explanations. Through an example, we
have seen that an explanation is the discovery of
likenesses connecting a process we do not under-
stand with processes we do understand. This is
the most rewarding activity of science. It leads
to exploration. Learn to ask yourself questions
beginning with "Why" when you observe — both
in and out of the chemistry laboratory. It is a
good habit to have, and it frequently makes life
more interesting.

You have had opportunities to ask many
"Why" questions already from your work in the
laboratory. In fact, there are enough such ques-
tions to provide the basis for the rest of the
course. Some of the questions that have been
raised in your experiments are listed at the end
of the chapter. Can you add to this list? How
many of the questions can you answer now? We
will find the answers to many of them in our
subsequent study. Some may not yet have satis-
factory answers. These are the most interesting
questions because they point into the future —
your future.

SOME QUESTIONS RAISED DURING THE STUDY OF A BURNING CANDLE

Why does a solid absorb heat when it melts?

Why is heat liberated in the burning of a candle?

Why is the heat effect so much larger in the chemical reaction than in the phase change you studied ?

Why does the candle react with air to give carbon dioxide and water rather than the reverse, carbon dioxide and

water reacting to give candle and air?
Why didn't the candle react with air (that is, burn) while the candle was stored in your desk drawer ? Why did it

wait until you wanted it to burn ? What is the role of the match you used to light the candle ?
Why does a candle burn slowly when you light the wick, in contrast to what happens when you light the "wick"

of a firecracker ?
What is the role of the wick of the candle?

How much water and carbon dioxide are produced from a burning candle?
Why does carbon dioxide cause limewater to become cloudy ?

Why does the burning of sulfur produce a bad smell while the burning of steel wool produces sparks?
Why does a flame emit colored light?
Why is the base of the flame blue?
What is the dark zone in the candle flame?
Why does the candle flame smoke more in a breeze ?
Why don't we run out of questions?

CHAPTER

A Scientific Model:
The Atomic Theory

• • • hypotheses ought to be fitted merely to explain the properties of
things and not attempt to predetermine them except in so far as they
can be an aid to experiments.

ISAAC NEWTON, 1689

One of the activities of science is the search for
regularity. Regularities that directly correlate ex-
perimental results are generally called rules or
laws. A more abstract regularity, expressing a
hidden likeness, is generally called a model,
theory, or principle. Thus, the behavior of oxy-
gen gas summarized in the equation P X V = a
constant is called a law.* The explanation of this
same regular gas behavior in terms of the motion
of particles is called a theory. It is a greater
abstraction to connect the PV product with the
mathematical equations that describe rebound-
ing billiard balls. Nevertheless, rules, laws, mod-
els, theories, and principles all have a common
aim — they all systematize our experimental
knowledge. They all state regularities among
known facts.

The more abstract regularities come from the
discovery of a hidden likeness. When the likeness

* It is called Boyle's Law, after Robert Boyle, the
scientist who first discovered this particular regularity.

involves a real physical system (such as rebound-
ing billiard balls), the explanation is usually
called a model. When the likeness involves an
abstract idea (such as a mathematical equation),
the explanation is usually called a theory. There
is no real distinction to be made, however, and
we shall use the words model and theory inter-
changeably.

When seeking an explanation, we sometimes
find more than one explanation. When this hap-
pens the model (or theory) that is most useful
is most used. A model that proves to be useful
generally points to new directions of thought.
The new directions guide us to new experiments
and, thereby, new facts come to light. Often, the
new facts will require growth of the model. Oc-
casionally the new facts forcibly contradict the
model and it must be abandoned in favor of
another. Both the growth and abandonment of
models or theories reflect increase in our under-
standing of the environment.

Let us see how a model grows.

A SCIENTIFIC MODEL: THE ATOMIC THEORY | CHAP. 2

2-1 IMPLICATIONS AND GROWTH OF A SCIENTIFIC MODEL

As an example, we can explore the implications
of our explanation of the behavior of gases.

Question: Why does a balloon expand as it is
inflated?

This model is useful, first, because we can
calculate in mathematical detail just how much
push a billiard ball exerts on a cushion at each
rebound, and, second, because exactly the same
mathematics describes the pressure behavior of
gas in a balloon. The success of the model leads
to new directions of thought. For example, we
might now wonder whether the pressure-volume
behavior of oxygen, as shown in Table l-II
(p. 14), can be explained in terms of the particle
model of a gas.

2-1.1 The Pressure-Volume Behavior
of Oxygen Gas

The experimental data in Table l-II show that
decreasing the volume by one-half doubles the
pressure (within the uncertainty of the measure-
ments). How does the particle model correlate
with this observation? We picture particles of
oxygen bounding back and forth between the
walls of the container. The pressure is deter-
mined by the push each collision gives to the
wall and by the frequency of collisions. If the
volume is halved without changing the number
of particles, then there must be twice as many
particles per liter. With twice as many particles
per liter, the frequency of wall collisions will be
doubled. Doubling the wall collisions will double
the pressure. Hence, our model is consistent with
observation: Halving the volume doubles the
pressure.

Fig. 2-1. In the particle model, wall collisions deter-
mine pressure. Halving the volume doubles
the pressure.

SEC. 2-1 I IMPLICATIONS AND GROWTH OF A SCIENTIFIC MODEL

2-1.2 The Pressure-Volume Behavior
of Other Gases

Having gained this understanding of the pres-
sure-volume behavior of oxygen, it is natural to
wonder whether the same model is applicable
to other gases. Thus, the development of the
theory leads us to perform new experiments.
Such experiments provide a systematic growth of
our knowledge of our environment. They are
generally much more effective than random,
"shot-in-the-dark" experiments.

Two other gases on the chemist's shelf are
ammonia and hydrogen chloride. Is the particle
model applicable to them as well as to oxygen?
To find out, we must perform experiments that
duplicate the conditions used in the study of
oxygen. Table 2-1 shows pressure-volume meas-
urements for 32.0 grams of gaseous ammonia at
0°C. Table 2-II shows the same type of data for
32.0 grams of gaseous hydrogen chloride at this
same temperature.

Table 2-1

PRESSURE AND VOLUME OF 32.0 GRAMS
OF AMMONIA GAS t = 0 C

PRESSURE

(atmospheres)

VOLUME

(liters) P X V

0.100

421

42.1

0.500

84.2

42.1

1.00

42.1

We see that these two gases also show the
behavior at a fixed temperature, PV = a con-
stant. The particle model should be useful for
these gases as well as for oxygen. On the other
hand, the numerical value of the constant varies
from one gas to another, if the same weight of
gas is considered. Thus 32.0 grams of oxygen at
0°C and one atmosphere occupy 22.4 liters. The
same weight of ammonia at this temperature and
pressure occupies 42.1 liters. The same weight of
hydrogen chloride occupies only 19.6 liters. The
particle model of gases must be modified to ex-
plain these differences.

Table 2-11

PRESSURE AND VOLUME OF 32.0 GRAMS
OF HYDROGEN CHLORIDE GAS t = 0°C

PRESSUR£

(atmospheres)

VOLUME

(liters) P X V

0.100

196

19.6

0.500

39.2

19.6

1.00

19.6

To explain this behavior, chemists have found
it convenient to consider a different weight of
each gas; they select that amount that gives the
same PV product as 32.0 grams of oxygen gas.
Consider, first, ammonia gas. At 0°C and a pres-
sure of one atmosphere, 32.0 grams of ammonia
occupies 42.1 liters. We have taken too large a
weight of ammonia. The weight of ammonia
needed to occupy only 22.4 liters at this pressure
is smaller by a factor 22.4/42.1 :

22.4
wt. ammonia = 32.0 g X ttt = 17.0 g

Pressure-volume data for this weight of ammonia
are shown in Table 2-III.

Table 2-III

PRESSURE AND VOLUME OF 17.0 GRAMS
OF AMMONIA GAS

PRESSURE

(atmospheres)

VOLUME

Giters) PX V

0.100

224

22.4

0.500

44.8

22.4

1.00

22.4

EXERCISE 2-1

If 32.00 grams of hydrogen chloride gas (at 0°C
and one atmosphere) occupy 19.65 liters, then a
larger weight of hydrogen chloride is needed to
occupy the larger volume, 22.4 liters. Show that
the weight needed is 36.5 grams.

A SCIENTIFIC MODEL: THE ATOMIC THEORY I CHAP. 2

Now the regularity between pressure and vol-
ume of these three gases can be expressed as
follows:

For 32.0 grams of oxygen at 0°C,

P X V = 22.4;
for 17.0 grams of ammonia at 0°C,

PX V = 22.4;
for 36.5 grams of hydrogen chloride at 0°C,

P X V = 22.4.

Each of the gases has a behavior consistent
with the particle model of a gas (PV = a con-
stant). However, the particles of the gas called
oxygen must differ from the particles of gas
called ammonia. These, in turn, must differ from
the particles of the gas called hydrogen chloride.
How do the particles differ? Why is it that 32.0
grams of oxygen give the same PV product as
17.0 grams of ammonia and 36.5 grams of hy-
drogen chloride (all at 0°Q? Do the particles
have different weights? Again, we are led to new
questions and new questions lead to new experi-
ments.

2-1.3 Some Properties of Gases

What gases do we find in the chemical stock-
room? How do they compare in other proper-
ties? Looking down the row of tanks we find the
names ammonia, chlorine, hydrogen, hydrogen
chloride, nitric oxide, nitrogen dioxide, and oxy-
gen, among others. Two of these gases are
colored — chlorine is yellow-green and nitrogen
dioxide is reddish-brown — and the other five are
colorless. The colorless gases can be further
sorted according to their solubilities in water.
Figure 2-2 shows what happens if a stoppered
test tube full of each gas is opened with the
mouth of the test tube under water. In the tubes
containing ammonia and hydrogen chloride the
water rises rapidly, filling the tubes. These two
gases dissolve readily in water. In each of the
other three test tubes the liquid level rises very
little, showing that little gas dissolves.

Fig. 2-2. Gases have different solubilities in water.

Immediately
after gases
placed in
contact
with water

Gases with high solubility in water
Ammonia Hydrogen chloride

Gases with low solubility in water
Nitric oxide Oxygen Hydrogen

After 30
minutes
in contact
with water

SBC. 2-2 I MOLECULES AND ATOMS

Though ammonia and hydrogen chloride both
dissolve in water, these two gases are very- dif-
ferent in other properties. For example, they
behave differently when placed in contact with
the dye, litmus. This dye, when moistened, turns
red if it is placed in hydrogen chloride. However,
if it is placed in ammonia, it turns blue.

We have not yet distinguished the gases nitric
oxide, hydrogen, and oxygen. Nitric oxide has
its own personality. Immediately upon exposure
to air, the colorless nitric oxide becomes reddish-
brown — exactly the color of nitrogen dioxide.
Neither oxygen nor hydrogen behaves this way.

The gases oxygen and hydrogen are readily
distinguished by their combustion properties.
When a glowing splint is plunged into oxygen,

the splint bursts into flame. When a brightly
glowing splint is plunged into hydrogen, the glow
is either extinguished or, if air has mixed with
hydrogen, it produces a small explosion.

Thus we find that each of these gases has dis-
tinctive properties. If these gases are made up of
particles, then the particles must be distinctive.
The particles that are present in ammonia cannot
be like the particles in hydrogen chloride, or like
those in the other gases. The nature of the am-
monia particles, then, is the key to the properties
of ammonia. The particles that make up a gas
determine its chemistry. They are so important
to the chemist that they are given a special name.
A gas is described as a collection of particles
called molecules.

2-2 MOLECULES AND ATOMS

The particles, or molecules, of the gas nitric
oxide cannot be exactly like those of nitrogen
dioxide. There must be differences that account
for the fact that one gas is colorless and the other
reddish-brown. Yet, when nitric oxide and air
are mixed, color appears, suggesting that nitro-
gen dioxide has been formed. Apparently mole-
cules present in air somehow combine with the
molecules of nitric oxide to form molecules of
nitrogen dioxide. We would like to develop our
picture of molecules so it will aid us in discussing
these changes.

To explain how molecules can rearrange and
change, we assume they must be built of smaller
fragments. These smaller fragments, or building

blocks, are called atoms. With this assumption,
we can explain differences between two mole-
cules in terms of the atoms present in each
molecule. Nitric oxide is different from hydrogen
chloride because it contains different atoms.
Nitric oxide exposed to air forms nitrogen di-
oxide by some rearrangement of the available
atoms. In general, the properties of a gas are
fixed by the number and types of atoms it con-
tains.

Fig. 2-3. Models of molecules. The properties of a
molecule are fixed by the number and types
of atoms it contains.

A model of
AMMONIA

A model of
HYDROGEN
CHLORIDE

A SCIENTIFIC MODEL: THE ATOMIC THEORY I CHAP. 2

SEARCH FOR AN EXPLANATION :

Why do equal volumes of ammonia and hydrogen chloride have different weights?

At room temperature

and one
atmosphere pressure.

and

1 2.45 liters of
hydrogen chloride
weigh 3. 65 grams

Chemists construct models of molecules to
show how many atoms they contain. Figure 2-3
shows some examples. The model of ammonia
represents three atoms of one kind attached to
one atom of another kind. The model of hydro-
gen chloride contains only two atoms, and of
different kind. We shall spend the entire year
discussing why these and other models are con-
structed as they are. We shall see that a molecule
of ammonia is pictured as shown in Figure 2-3
because this model helps us explain the proper-
ties of ammonia. Throughout the course, we
shall investigate properties of substances found
in nature or prepared in the laboratory and then
we will seek explanations in terms of the num-
bers, types, and arrangements of the atoms pres-
ent. These explanations are called the atomic
theory. The atomic theory is regarded as the
cornerstone of chemistry.

2-2.1 The Weights of Molecules

We have discovered that 32.0 grams of oxygen,
17.0 grams of ammonia, and 36.5 grams of hy-
drogen chloride each exhibits the regular be-
havior,

PV = 22.4 at 0°C (/)

For any of the three gases, if the pressure is
given, the volume occupied can be calculated.
At one atmosphere, each of the specified weights
of gas occupies 22.4 liters. At two atmospheres,
each gas is compressed into a smaller volume,
11.2 liters. We wonder, Why do 22.4 liters of

WHY?

Fig. 2-4. Why do equal volume of ammonia and hy-
drogen chloride have different weights?

ammonia weigh 17.0 grams when the same vol-
ume of hydrogen chloride weighs 36.5 grams?
There are two factors to consider. These are
the same two factors we would be concerned with
if we were to ask why a bag full of beans weighs
17.0 grams whereas the same bag full of marbles
weighs 36.5 grams. The explanation would be
found by comparing the number of beans in the
bag and the weight per bean to the number and
individual weights of marbles in the same bag.
In our gas problem we make the same kind of
comparison; the weight per molecule and num-
ber of ammonia molecules in 22.4 liters must be
compared to the weight per molecule and num-
ber of hydrogen chloride molecules in 22.4 liters.
There are two particularly simple possibilities:

(A) Perhaps:

(1) Equal volumes of these two gases con-
tain the same number of molecules, ai.d

(2) ammonia molecules weigh less, per mol-
ecule, than hydrogen chloride molecules
by the factor 17.0/36.5.

(B) Perhaps:

(1) Equal volumes of these two gases con-
tain different numbers of molecules, am-
monia containing fewer by the factor
17.0/36.5, and

(2) ammonia molecules weigh exactly the
same as hydrogen chloride molecules.

SEC. 2-2 I MOLECULES AND ATOMS

SEARCH FOR AN EXPLANATION 7

Why do equal volumes of ammonia and hydrogen chloride have different weights?

At room temperature

and one
atmosphere pressure.

' ^2.45 titers of
<l III

ammonia, weigh
ii
/. 70 grams

and

v..

n in
hydrogen chloride

WHY?

MODEL A

Perhaps:

1. 2. 45 liters of ammonia
contain the same
number of molecules
as 2.45 titers of
hydrogen, chloride.

and

2. Afnmonia molecules
weigh less than,
hydrogen chloride
molecules, (less by a
factor, J. 70/3. 65).

MODEL B

Perhaps:

1. 2.45 titers of ammonia
contain fewer molecules
than do 2.45 liters of
hydrogen, chloride,
(lest by a factor,
1.70/3.65).

and

2. Ammonia molecules
■weigh the same as
hydrogen chloride
molecules

Fig. 2-5. Two simple models to explain the weights of equal volumes of gases.

A SCIENTIFIC MODEL: THE ATOMIC THEORY I CHAP. 2

30 ml
ammonia

Fig. 2-6. An apparatus suited to the measurement of

volumes of gases.

These two possibilities are attractive because
they are simple — one factor alone is held respon-
sible for the weight difference. We must be
prepared, however, for disappointment. There is
the third possibility that neither of these pro-
posals, A or B, accounts for the properties of
gases. After all, neither A nor B applies to the
beans and marbles example. The bag probably
wouldn't contain the same number of beans as
marbles (as in B) but, in addition, beans and
marbles don't weigh the same (as in A). We need
more information to decide if either proposal A
or B applies to gases. More information is ob-
tained by observing how some gases behave
when mixed.

2-2.2 Mixtures off Ammonia and
Hydrogen Chloride

When hydrogen chloride and ammonia gases are
mixed, a white powder is formed. When hydro-
gen chloride molecules and ammonia molecules
are mixed, the atoms are rearranged and an en-

tirely different substance, a solid, results. A quan-
titative study of this process is informative.

Figure 2-6 shows an apparatus suited to the
measurement of volumes of gases. Thirty milli-
liters of ammonia have been admitted to the left
tube from the ammonia storage tank. Next, 50
ml of hydrogen chloride were admitted to the
right tube. The leveling bulbs were used to adjust
the pressure of each gas to one atmosphere. The
apparatus is ready. The hydrogen chloride sam-
ple can be transferred slowly into the tube con-
taining the ammonia. Figure 2-7 shows the prog-
ress of the experiment.

In Figure 2-7A we see the situation after 20 ml
of hydrogen chloride gas have been transferred.
A cloud of the white solid fills the left tube where
the gases mix and, after the leveling bulbs are
adjusted, we find that just 10 ml of ammonia
remain. Twenty milliliters of hydrogen chloride
combined with just 20 ml of ammonia, forming
the white solid. Ten milliliters more hydrogen
chloride are just enough to consume the last of
the ammonia, forming the solid of insignificant
volume, as shown in Figure 2-7B. Continued
addition of hydrogen chloride causes no further
solid formation but merely leaves an excess of
hydrogen chloride in the left tube (Figure 2-7C).

This is a significant and simple result. Thirty
milliliters of hydrogen chloride combine with
just 30 ml of ammonia, measured at the same
temperature and pressure. Therefore, one liter of
hydrogen chloride would combine with just one
liter of ammonia. Though a given volume of
ammonia weighs less than the same volume of

Fig. 2-7. Mixing measured volumes of hydrogen chlo-
ride gas and ammonia gas.

SEC. 2-2 I MOLECULES AND ATOMS

hydrogen chloride (less by the factor 17.0/36.5),
these equal volumes combine. This simple situa-
tion suggests that we should seek a simple ex-
planation. Our proposal A in Section 2-2.1 fits
nicely. If we propose that equal volumes contain
equal numbers of molecules, then 30 ml of am-
monia contain the same number of molecules as
do 30 ml of hydrogen chloride. Proposal A leads
us to conclude that one molecule of ammonia
combines with one molecule of hydrogen chlo-
ride to form the white solid. Through proposal
A, the combining volumes tell us the numbers
of molecules that combine. In contrast, there is
no correspondingly simple way to explain the
new data with proposal B.

Of course, a single example hardly furnishes
compelling evidence that equal volumes of any
pair of gases (at the same temperature and pres-
sure) contain equal numbers of molecules. Nei-
ther can we be convinced on the basis of
simplicity when we have but one example. How-
ever, many gases behave as simply as a mixture
of hydrogen chloride and ammonia. For ex-
ample,

2-2.3 The Relative Weights of Molecules

The importance of Avogadro's Hypothesis is
that it furnishes a basis for weighing molecules.
Two equal volumes of gas (at the same tempera-
ture and pressure) are weighed. If we assume
these two volumes contain identical numbers of
molecules, then we must also conclude that the
gas that weighs more must have heavier mole-
cules. Furthermore, the ratio of the weights of
the molecules must be exactly the ratio of the
weights of the two gas samples.

For example, in Table 1 -I I (p. 14) data were
given that show that 32.0 grams of oxygen at 0°C
and one atmosphere pressure occupy 22.4 liters.
This same volume of ammonia (also at 0°C and
one atmosphere pressure) weighs 17.0 grams. By
Avogadro's Hypothesis, these two volumes con-
tain equal numbers of molecules. Hence each
ammonia molecule must weigh less than an oxy-
gen molecule by the factor 17.0/32.0. By the
same argument, each hydrogen chloride mole-
cule must weigh more than an oxygen molecule
by the factor 36.5/32.0. We say "must weigh"
but, of course, this is a valid statement only if

two liters of
the gas
nitric oxide

combine
with

one liter of
the gas
oxygen

to form

nitrogen
dioxide

one liter of
the gas
oxygen

will burn
coal to
form

one liter of the
gas carbon
dioxide

two liters of
the gas
hydrogen

combine
with

one liter of
the gas
oxygen

These simple, integer volume ratios confirm
the usefulness of the interpretation that equal
volumes contain equal numbers of molecules.
This proposal was first made in 1811 by an
Italian scientist, Amadeo Avogadro; hence it is
called Avogadro's Hypothesis. It has been used
successfully in explaining the properties of gases
for a century and a half.

Avogadro's Hypothesis: Equal volumes of
gases, measured at the same temperature and
pressure, contain equal numbers of molecules.

to form

water

Avogadro's Hypothesis is applicable.

By many such weighings, scientists have
learned the relative weights of many gases. The
experiment is fairly simple. A carefully meas-
ured volume of oxygen is weighed at a fixed
pressure and temperature. Then the same volume
of another gas is weighed at this same pressure
and temperature. The relative weights of the
gases indicate the relative weights of the mole-
cules, provided Avogadro's Hypothesis is ap-
plicable. Neither the pressure nor the tempera-

A SCIENTIFIC MODEL! THE ATOMIC THEORY I CHAP. 2

ture need be measured, provided they are held
constant.

Table 2-IV

WEIGHTS OF EQUAL VOLUMES OF
GASES UNDER FIXED TEMPERATURE
AND PRESSURE (BASED ON THE VOLUME
OCCUPIED BY 32.0 GRAMS OF OXYGEN)

WEIGHT

gas (grams)

oxygen
ammonia
chlorine
hydrogen
hydrogen chloride
nitric oxide
nitrogen dioxide

(32.0)
17.0
71.0
2.02
36.5
30.0
46.0

Table 2-IV shows for some other gases the
weights that have the same volume as 32.0 grams

been formed. How can this change be discussed
in terms of our molecular model of a gas?

First, we explain the differences between nitric
oxide, oxygen, and nitrogen dioxide by asserting
that the molecules of nitric oxide, oxygen, and
nitrogen dioxide are somehow different. They
must be composed of smaller components that
we call atoms. The numbers and kinds of atoms
in a molecule of nitric oxide must be different
from the numbers and kinds of atoms in a mole-
cule of oxygen.

Now we find that nitrogen dioxide can be
formed from a mixture of nitric oxide and oxy-
gen. This means that the atoms in nitrogen di-
oxide must have come from those in nitric oxide
together with those in oxygen.

Finally, we discover that exactly two volumes
of nitric oxide combine with one volume of oxy-
gen and that exactly two volumes of nitrogen
dioxide are formed. According to Avogadro's
Hypothesis, this indicates that

two molecules
of nitric
oxide

combine
with

one molecule
of oxygen

to form

two molecules
of nitrogen
dioxide

of oxygen (at the same pressure and tempera-
ture).

2-2.4 The Number of Atoms in a Molecule

Figure 2-3 shows a model of an ammonia mole-
cule and a model of a hydrogen chloride mole-
cule. These models show how chemists picture
the molecule of ammonia: it contains four
atoms. A hydrogen chloride molecule contains
only two atoms. Chemists decide how to con-
struct these molecules from the same type of
information described in Section 2-2.2, by the
volumes of gases that combine.

Consider the combination of nitric oxide and
oxygen. Nitric oxide (a colorless gas) when mixed
with oxygen gas (also colorless) becomes reddish-
brown. The color is identical to that of another
gas, nitrogen dioxide. All the properties of the
nitric oxide-oxygen mixture are consistent with
the conclusion that the gas nitrogen dioxide has

All of the atoms in the two molecules of nitrogen
dioxide came from two molecules of nitric oxide
and one molecule of oxygen. Of course, the two
molecules of nitrogen dioxide have twice as many
atoms as does a single molecule of nitrogen di-
oxide. Hence, no matter how many atoms one
molecule of nitrogen dioxide might contain (for
example, one, two, three, four, . . .), two mole-
cules of nitrogen dioxide must contain an even
number of atoms (for example, two, four, six,
eight, . . .). The same statement is applicable to
the two molecules of nitric oxide that were com-
bined. No matter how many atoms one molecule
of nitric oxide contains, two molecules must con-
tain an even number of atoms.

Thus we see that after the even number of
atoms in two molecules of nitric oxide have
combined with the atoms in one molecule of
oxygen, there is still an even number of atoms.
This can be so only if a molecule of oxygen also
contains an even number of atoms. We are led

SEC. 2-2 I MOLECULES AND ATOMS

to the conclusion that a molecule of oxygen con-
tains an even number of atoms.

This can be demonstrated clearly in algebraic
language.

Suppose:
one molecule of nitric oxide contains

X atoms,
one molecule of oxygen contains

Y atoms, and
one molecule of nitrogen dioxide contains

Z atoms,
where X, Y, and Z are integers.

Then:
two molecules ot nitric oxide contain

2X atoms, and
two molecules of nitrogen dioxide contain

2Z atoms.

Also:
all of the atoms in two molecules of nitrogen
dioxide (2Z atoms) came from two mole-
cules of nitric oxide (2X atoms) plus one
molecule of oxygen ( Y atoms), or

IX + Y = 2Z

So:

we can solve for Y by subtracting 2X from
each side of this equation:

Y =2Z-2X (3)

Y=2(Z- X) (4)

Thus no matter what the integer values of Z
and X are, their difference (Z — X)\s an integer.
Since doubling any integer produces an even
number, Y = 2(Z — X) must be an even num-
ber. We have proved that a molecule of oxygen
must contain an even number of atoms.

The simplest acceptable structure we can pic-
ture for oxygen is that it contains two atoms.
More experiments are needed before we can
eliminate the possibility that oxygen contains
four, six, or a higher (but even) number of atoms.

EXERCISE 2-2

In Section 2-2.2 (p. 24) it was noted that two
volumes of hydrogen gas combine with one vol-
ume of oxygen gas and two volumes of gaseous

water are produced. According to Avogadro's
Hypothesis, this means that two molecules of
hydrogen combine with one molecule of oxygen
to form two molecules of water. If we define

X = the number of atoms in a molecule of

hydrogen,
Y = the number of atoms in a molecule of

oxygen,
Z = the number of atoms in a molecule of

water,

then we have the algebraic relationship
2X + Y = 2Z

(a) Convince yourself that from these data alone
we can conclude that Y must be even (that is,
oxygen molecules must contain an even num-
ber of atoms).

(b) By solving for X in terms of Y and Z, con-
vince yourself that from the above data
alone, X could be odd or even.

C?) 2-2.5 Atoms in Liquids and Solids

When a candle is burned, a gas is produced — a
gas containing carbon dioxide and water vapor.
It is useful to describe such a gas as a collection
of molecules, each molecule containing smaller
units called atoms. Each carbon dioxide mole-
cule contains one carbon atom and two oxygen
atoms. Each water molecule contains one oxygen
atom and two hydrogen atoms. Where did these
atoms come from? Were they present in the
candle before it burned?

Similar questions are raised if we consider the
effect of cooling the gases produced from the
candle. Cooling these gases results in condensa-
tion— drops of liquid water appear. If the water
vapor contains molecules, made up of atoms,
what happens to these molecules (and atoms)
when the gas condenses? Are they still present
in the liquid?

Scientists always seek the simplest explanation
that fits the known facts. Since we find it con-
venient to describe water vapor as a collection
of groups of atoms (called molecules), the sim-
plest assumption we can make about the con-

A SCIENTIFIC MODEL: THE ATOMIC THEORY | CHAP. 2

5-» ' .<—,--

Solid carbon dioxide

Liquid carbon dioxi

G-aseous carbon dioxide

Fig. 2-8. All matter consists of particles. In a gas, the
particles are far apart; in a liquid or solid,
they are close together.

densation of this vapor is that the liquid still
contains the atoms. Since a candle burns to
produce gases — collections of molecules — the
simplest assumption we can make is that the
candle already contained the atoms that formed
the gaseous molecules during combustion. These
simplifying assumptions — that liquids and solids
are made up of atoms — are acceptable and con-
venient as long as they prove to be consistent
with all that is known about liquids and solids.
Thus, we are led to the view that all matter

consists of particles. We can state this as a pro-
posal.

Proposal: All matter, whether solid, liquid, or gas,
consists of particles. In a gas these particles are
far apart; in a liquid or a solid, they are packed
close together.

This proposal is called the atomic theory.
As with any theory, its value depends upon its
ability to aid us in explaining facts of nature.
There is no more valuable theory in science than
the atomic theory. We shall use it throughout
this course. Later, in Chapter 14, we shall review
many of the types of experiments which cause
chemists to regard the atomic theory as the
cornerstone of their science.

2-3 SUBSTANCES: ELEMENTS AND COMPOUNDS

Molecules are clusters of atoms. Two types of
molecules are possible. Some molecules are clus-
ters of atoms in which all the atoms in a cluster
are identical; some molecules contain two or
more different kinds of atoms. These two kinds
of molecules are given different names.

An element or elementary substance contains
only one kind of atom.

A compound or compound substance contains
two or more kinds of atoms.

Usually a good deal of experimentation is
needed before a substance can be considered to
be pure. Even then, much more work and study
are needed before one can decide with confidence
that a given pure substance is an element or a
compound. Consider the substance water. Water
is probably the most familiar substance in our
environment and all of us recognize it easily. We
are familiar with its appearance and feel, its
density (weight per unit volume), the way in

sec. 2-3 I substances: elements and compounds

which it flows, the temperature at which it
freezes and boils, and the way in which it dis-
solves sugar and salt. Because water is identified
by constant and characteristic properties, it is a
pure substance. In a later experiment you will
see how we can change the pure substance water
into two other substances, hydrogen gas and
oxygen gas. The hydrogen and oxygen are pro-
duced in definite amounts. Since water can be
decomposed into two other substances, it must
contain at least two kinds of atoms. Hence water
is a compound.

Notice the pattern here. First, we established
the characteristic properties of water that cause
us to identify it as a pure substance. Second, we
found a change in which two other substances
were formed in definite amounts from water
alone. This second piece of information shows
that water contains more than one kind of atom
and that, hence, water is a compound.

Common sugar is another example of a sub-
stance. Most commercial samples of white sugar
are rather pure; that is, they contain only very
small amounts of substances other than sugar.
One characteristic property of sugar is its sweet-
ness. Another characteristic property is the way
it dissolves in water. Still another is the way it
behaves when heated. At a definite temperature
sugar not only begins to melt to a liquid but it
also begins to decompose. The liquid darkens
and gaseous water bubbles off. Finally, a black
solid (charcoal) remains in the container. We
recognize the black solid as a form of carbon.
Thus, pure sugar, identified by its characteristic
properties, can be decomposed to form water
and charcoal in definite amounts. Sugar is a
compound.

Water and sugar are compounds. What about
hydrogen and oxygen? Hydrogen, for example,
is a gas at normal conditions. It can be liquefied
at a characteristic temperature by cooling. By
further cooling it can be solidified at a second
characteristic temperature. It is a pure substance.
No treatment, however, causes it to form two
other substances. Hydrogen must contain only
one kind of atom, hence hydrogen is an element.
We call this kind of atom, the hydrogen atom.
Oxygen, too, has characteristic properties but

cannot be caused to form two other substances.
Oxygen, then, is an element — it contains only one
kind of atom, called the oxygen atom.

Now we can return to the decomposition of
water. Water can be decomposed to give hydro-
gen and oxygen. Since hydrogen contains only
hydrogen atoms and oxygen contains only oxy-
gen atoms, water molecules must contain some
hydrogen atoms and some oxygen atoms, but no
other kind of atom.

This type of problem is one of the most im-
portant in chemistry — deciding what atoms are
present in a given substance. How important this
is can be seen by comparing the three substances
water, oxygen, and hydrogen. Both water and
oxygen contain oxygen atoms but these sub-
stances are very different in their properties. Both
water and hydrogen contain hydrogen atoms but
these substances are no more alike than are
water and oxygen. The properties of water are
fixed by the combination of the two kinds of
atoms and these properties are distinctive.

EXERCISE 2-3

What differences between water, oxygen, and
hydrogen can you point out from your own ex-
perience? For example, you might consider

(a) boiling and melting points,

(b) role in combustion,

Sugar is another substance that contains both
oxygen and hydrogen atoms but it contains car-
bon atoms as well. Sugar does not resemble
water, oxygen, or hydrogen. The presence of the
various types of atoms and their arrangement
accounts for the distinctive properties which
identify sugar. In any substance, the atoms pres-
ent, their numbers, and their arrangement fix the
properties of that substance.

2-3.1 The Elements

An element is a pure substance that contains
only one kind of atom. There are about one hun-

A SCIENTIFIC MODEL: THE ATOMIC THEORY I CHAP. 2

dred different elements known today — hence
there are about one hundred kinds of atoms that
are chemically different. Some of these elements
occur pure in nature and hence have been known
for thousands of years. Such elements as iron,
silver, gold, mercury, and sulfur were known to
the ancients and were given Latin names by the
alchemists. For example, iron was called ferrum,
silver was called argentum, and gold was called
aurum.

During the nineteenth century the discovery of
elements increased as chemists began to adopt
quantitative methods. At the beginning of the
nineteenth century, perhaps 26 elements were
known. One hundred years later, at the begin-
ning of the present century, over 81 elements
were known. Over twice as many elements were
discovered in that one century as were discovered
in all of time before.

Figure 2-9 shows, as a function of time, our

Fig. 2-9. The discovery of the elements. A. The total
number of elements known as a function
of time. B. The number of elements discov-
ered in each half-century since 1700.

a *

100

80-

20-

1700

H3I

1800
Year

1900

m
%

13 before
1700

■

■,^wii

1700

1800
Year

1900

knowledge of the elements. In Figure 2-9A we
see how the total number of elements known has
increased since 1700. Figure 2-9B re-expresses
this same information in terms of the number of
elements discovered in each half-century since
1700. Both graphs show that the rate of discovery
of new elements is declining. The plots suggest
that there is a limited number of elements to be
found in nature.

Each element has been named and, for con-
venience, has been given a nickname — a short-
hand symbol of one or two letters. Thus the
element carbon is symbolized by the letter C, the
element neon by the letters Ne. The symbols are
adopted by international agreement among
chemists. Eleven of the elements have names
derived from the capitalized first letter of the
Latin name of the element and, if necessary, by
a second letter (uncapitalized).* These eleven in-
clude seven common metals known to the an-
cients. (See Table 2-V.)

The elements discovered more recently have
the same names in all languages, again by inter-
national agreement. Except for the eleven ele-
ments listed in Table 2-V, all of the elements
have symbols that can be derived from their
English names. For example, the symbols for
hydrogen (H), helium (He), carbon (C), nitrogen
(N), oxygen (O), calcium (Ca), and chlorine (CI)
are easily obtained from the names. Notice that
He is used for helium to distinguish it from H
for hydrogen. Again, since C is used for carbon,
the symbols for calcium and chlorine each have
a second letter added to the first. The table of
elements (inside the back cover of the book) con-
tains a complete list of the chemical symbols.

2-3.2 Chemical Formulas

Molecules are made up of atoms in definite num-
bers and definite arrangements. Models and sym-
bols for the elements aid us in showing the

* These symbols were adopted as a result of the urging
of an outstanding Swedish chemist, Jons Berzelius, during
the first half of the nineteenth century. Until Berzelius
lent his prestige as a chemist to this system, many of the
elements had several symbols in accordance with their
different names in different languages.

sec. 2-3 I substances: elements and compounds

A model of
WATtR

Fig. 2-10. A model of a molecule of water.

composition of a molecule. Figure 2-10 shows a
model of a water molecule. Experiments have
shown that the model should contain two atoms
of hydrogen and one atom of oxygen. The ad-
vantage of such a model is that it shows also
the spatial arrangement of the atoms. In a mole-
cule of water, each of the two hydrogen atoms
is connected to the oxygen atom in a triangular
arrangement. How the shape is determined and
how important it is in chemistry will be treated
later in the course.

The number and kinds of atoms in a molecule
can also be shown in a molecular formula. For
example, the water molecule is symbolized
"H20." In this molecular formula, "H" means
"hydrogen atom," "O" means "oxygen atom,"
and the subscript "2" following "H" indicates
there are two hydrogen atoms bound to the
single oxygen atom. The molecular formula of
ammonia, NH3, indicates that one molecule of
ammonia contains one atom of nitrogen (N) and

three atoms of hydrogen (H). Experiments show
that oxygen is diatomic (each molecule contains
two atoms), hence its molecular formula is 02.
Hydrogen gas is diatomic; its formula is H2.

Both the numbers and the arrangement of the
atoms in the molecule are shown by a structural
formula. The structural formulas, like the models
we have seen, show which atoms are attached to
each other. Thus, H20 has the structural formula

not

H— O— H
H— H— O

H H

In the structural formula H— O— H, the dashes
indicate the connections between the atoms. The
connections between atoms are called chemical
bonds. We see that each of the two hydrogen
atoms is bound to the oxygen atom. Both of
the alternate arrangements,

and

H— H— O

H H

agree with the molecular formula H20 but the
properties of water show that the atoms are not
so bonded.

No written formula is quite as effective as a
molecular model to help us visualize molecular
shape. Since chemists find that the shape of a
molecule strongly influences its chemical behav-
ior, pictures and models of molecules are im-
portant aids. A variety of types of models are

Table 2-V. chemical symbols that are not derivable from the common

ENGLISH NAME OF THE ELEMENT

COMMON NAME

SYMBOL

SYMBOL SOURCE

COMMON NAME

SYMBOL

SYMBOL SOURCE

antimony

stibnum

potassium

kalium

copper

cuprum

silver

argentum

gold

aurum

sodium

natrium

iron

ferrum

tin

stannum

lead

plumbum

tungsten

wolfram

mercury

hydrargyrum

A SCIENTIFIC MODEL: THE ATOMIC THEORY I CHAP. 2

Name Molecular Structural
Formula Formula

Models

Ball- and- Stick Ball- and- Spring Space-Filling

Hydrogen H2

Water H20

-H

-AT-

Ammonia NH*

" /

-H

Fig. 2-11. Different representations of molecules Hi,
H,0, and NHS.

commonly used, depending upon the emphasis
needed. Figure 2-1 1 shows some representations
of molecules of hydrogen, water, and ammonia.
The ball-and-stick and ball-and-spring models
display clearly the bonds and their orientations.
The ball-and-spring models indicate molecular
flexibility. The space-filling models provide a
more realistic view of the spatial relationships
and crowding among nonbonded atoms.

EXERCISE 2-4

Carbon dioxide has the formula C02. Remem-
bering that the prefix "di" means two, and "tri"
means three, write the molecular formula for
each of the following substances: carbon disul-
fide, sulfur dioxide, sulfur trioxide. (If you don't
know the symbol for an element, use the table
inside the back cover of the book.)

2-3.3 The Mole

Any sample of matter we examine contains a
very large number of atoms. We never work with

individual atoms or molecules but always with
collections of these particles. Chemists, there-
fore, have selected a unit larger than a single
atom or molecule for comparing amounts of
different materials. This unit, called the mole,
contains a very large number of particles,
6.02 X 1023. A mole of oxygen atoms, or a mole
of hydrogen atoms, or a mole of copper atoms
contains 6.02 X 1023 atoms of the specified kind.
A mole of oxygen molecules, 02, contains 2
moles of oxygen atoms (2 X 6.02 X 1023 oxygen
atoms) because each oxygen molecule contains
two atoms. A mole of P4 molecules contains
4 X 6.02 X 1023 phosphorus atoms, that is, four
moles of phosphorus atoms.*

A baker counts biscuits in dozens — a con-
venient number. Money is counted in dollars —
one hundred cents is, again, a convenient num-
ber. How did chemists choose to count in terms
of moles — the number 6.02 X 1023 seems an odd
choice. Why, for instance, didn't they settle on
some simpler number, such as exactly one billion
particles? There is a reason. Chemists prefer a
definition in terms of a quantity that can be
measured readily and with high accuracy. Weigh-

* If you have difficulty expressing numbers in terms of
powers of ten, refer to Appendix 5 in the Laboratory
Manual.

sec. 2-3 I substances: elements and compounds

ing is easier than counting when the number of
particles to be counted is so very large. Conse-
quently, chemists based the definition of the mole
upon a chosen weight rather than a chosen num-
ber of particles. During the nineteenth century,
chemists decided that the number of molecules
in a sample of oxygen weighing exactly 32 grams
would be taken as a standard number. Thus de-
fined, a mole is the number of oxygen mole-
cules in exactly thirty-two grams of oxygen*
The significance of a mole is most usefully con-
nected with this number of particles, rather than
the weight. The number, found later to be
6.02 X W23, is called Avogadro's number.
(Avogadro was the first to propose how to ob-
tain equal numbers of molecules of different sub-
stances.)

weighed on an ordinary balance. For practical
purposes, the weight of a mole of atoms is a
valuable number. This weight is called the atomic
weight. The atomic weight of an element is the
weight in grams of A vogadro's number of atoms.

Now consider compounds. Again, a useful
number to the chemist is the weight of a mole of
molecules. This weight is called the molecular
weight (or the molar weight). The molecular
aeight of a compound is the weight in grams oj
A vogadro's number of molecules.

Consider the substance hydrogen chloride.
This compound has the molecular formula HC1.
A chemist working with hydrogen chloride, HC1,
must often know the weight of a mole of mole-
cules (the molecular weight). This weight is
readily calculated from the atomic weights of
the two kinds of atoms, H and CI:

one molecule
of HC1

contains

one atom of
hydrogen

and

one atom of
chlorine

one mole of HC1
molecules

contains

one mole of
hydrogen
atoms

and

one mole of
chlorine
atoms

the weight of
one mole of
HC1 molecules

is the
same as

the weight of
one mole of
hydrogen atoms

the weight of
one mole of
chlorine atoms

molecular
weight of HC1

atomic weight
of hydrogen

1.01 g

atomic weight
of chlorine

35.5 g

molecular
weight of HC1

36.5 g

2-3.4 Atomic and Molecular Weights

Chemists deal with amounts of substances that
are readily measured. Although a chemist is
aware that the mass of a single oxygen atom is
2.656 X 10-23 gram, he finds it much more useful
to know that a mole of oxygen atoms weighs
16.00 grams. This is an amount that can be

* Recently the definition of a mole has been altered to
put it in terms of measurements made with higher accu-
racy. The change implied, — .0045 %, is unimportant from
a chemist's point of view.

In a similar way, the weight of a mole of H20
molecules is the weight of two moles of hydrogen
atoms plus the weight of one mole of oxygen
atoms. Hence

molecular wt. of H^O = 2 X (atomic wt. of H)

+ 1 X (atomic wt. of O)
- 2(1.01) + (16.00) g
= 18.02 g

Atomic weights have everyday importance to
a chemist. Therefore, the atomic weights must
be readily available. They are listed both in the

A SCIENTIFIC MODEL: THE ATOMIC THEORY I CHAP. 2

periodic table (inside the front cover) and in the
table of atomic weights (inside the back cover).

EXERCISE 2-5

Show that the weight of a mole (the molecular
weight) of COi is 44.0 grams and that the weight
of a mole of S02 is 64. 1 grams.

EXERCISE 2-€

In Experiment 6 you calculated the ratio of the
weight of carbon dioxide to the weight of the
same volume of oxygen. Oxygen has been as-
signed a molecular weight of 32.0. From the
molecular weight of oxygen and your measured
ratio, calculate the molecular weight of carbon

dioxide. Estimate the uncertainty in your result.
Compare to the value obtained in Exercise 2-5.

EXERCISE 2-7

What is the molecular weight of each of the sub-
stances sulfur (formula, S8), ammonia (formula,
NH3), and nitrogen (a diatomic molecule)?

EXERCISE 2-8

Calculate the weight of 6.02 X 1023 molecules of
carbon monoxide, CO.

EXERCISE 2-9

How many moles of iron atoms are present in
1.73 grams of iron?

2-4 REVIEW: THE ATOMIC THEORY

"Gases are composed of particles" was proposed
in Chapter 1 as a useful model to aid us in dis-
cussing certain properties of oxygen gas. We
have continued to use this model in discussing
other types of information. First it was tested
for applicability to other gases, ammonia and
hydrogen chloride. The assumption of molecular
particles turned out to fit the properties of these
two gases as well, but it was discovered that the
weights of the particles of different gases must
be different. Further studies of a variety of prop-
erties of gases (color, solubility in water, behav-
ior toward litmus, etc.) caused us to propose that
the molecules differ in their structures. We were
thus led to the view that molecules contain
smaller building blocks, called atoms.

Having proposed the existence of atoms, we
began representing the structure of molecules
through molecular models and molecular for-
mulas. These models and formulas picture what

is known about the numbers, types, and arrange-
ments of atoms in the molecules represented.
Our success in treating gases, using this atomic
theory, led us to propose that the atoms are
present, as well, in solids and liquids. Now we
postulate that atoms are present in all matter.
Thus a model (a theory) grows. As it is tested
in an ever widening range of experience, the
model often grows more complex. This is offset
by the advantage of developing interrelationships
among diverse phenomena (that is, by discover-
ing hidden likenesses). The atomic theory, as
developed to correlate chemical behavior, is
much more complicated than is needed to ex-
plain the simple gas behavior mentioned in
Chapter 1. Nevertheless, connections developed
between billiard balls rebounding from a cushion
and the pressure in a balloon have provided us
with a substantial start in understanding chem-
istry.

QUESTIONS AND PROBLEMS

1. Hydrogen, helium, and carbon dioxide are all
gases at normal temperatures. What differences
among the properties of these gases account for
the following?

(a) Hydrogen and helium are used in balloons
whereas carbon dioxide is not.

(b) Helium is less dangerous to use in balloons
than is hydrogen.

2. Four differences between helium gas and nitro-
gen gas are listed below.

(a) Dry air contains 80% nitrogen but only
0.0005 % helium (by volume).

(b) Helium is much less dense than nitrogen.

(d) Helium is much more expensive than nitro-
gen.

Which difference could account for the fact that
a diver is much less likely to suffer from the
bends if he breathes a mixture of 80% helium
and 20% oxygen than if he breathes air? (The
bends is a painful, sometimes fatal, disease
caused by the formation of gas bubbles in the
veins and consequent interruption of blood flow.
The bubbles form from gas dissolved in the
blood at high pressure.)

3. The most important step in the process for the
conversion of atmospheric nitrogen into impor-
tant commercial compounds such as fertilizers
and explosives involves the combination of one
volume of nitrogen gas with three volumes of
hydrogen gas to form two volumes of ammonia
gas.

From these data alone and Avogadro's Hy-
pothesis, how many molecules of hydrogen com-
bine with one molecule of nitrogen ? How many
molecules of ammonia are produced from one
molecule of nitrogen ?

4. Gaseous uranium hexafluoride is important in
the preparation of uranium as a source of
"atomic energy."

A flask filled with this gas is weighed under
certain laboratory conditions (temperature and
pressure), and the weight of the gas is found to
be 3.52 grams. The same flask is filled with oxy-
gen gas and is weighed under the same labora
tory conditions. The weight of the oxygen in the
flask is found to be 0.32 gram.

What is the ratio of the weight of one uranium
hexafluoride molecule to the weight of an oxygen
molecule? State any guiding principles needed
in answering this question.

5. Two volumes of hydrogen fluoride gas combine
with one volume of the gas dinitrogen difluoride
to form two volumes of a gas G.

(a) According to Avogadro's Hypothesis, how
many molecules of G are produced from one
molecule of dinitrogen difluoride?

(b) If X = number of atoms in a molecule of

hydrogen fluoride,
Y = number of atoms in a molecule of

dinitrogen difluoride,
Z = number of atoms in a molecule of G,
write the relation among X, Y, and Z appro-
priate to the combining volumes given.

If X is and Y is then Z must be

(d) No odd value of Y is suggested in question
(c). Prove that Y must be an even integer.

one volume of

combines

Ill 111C

one volume of

gdlClclI JJ<

3UCI 11

two volumes of

gas A

with

gas B

form

gas C

For example,

one volume of

combines

one volume of

two volumes of

carbon dioxide

with

carbon disulfide

form

carbon oxysulfide

one volume of

combines

one volume of

two volumes of

hydrogen

with

chlorine

form

hydrogen chloride

A SCIENTIFIC MODEL: THE ATOMIC THEORY I CHAP. 2

If X = number of atoms in a molecule of A,
Y = number of atoms in a molecule of B, and
Z = number of atoms in a molecule of C,

(a) Show that if X is even, Y must be even.

(b) Show that if X is odd, Y must be odd.

7. A pure white substance, on heating, forms a
colorless gas and a purple solid. Is the substance
an element or a compound?

8. What do the following symbols represent? K,
Ca, Co, CO, Pb.

9. Write formulas for

silicon dioxide (common sand, or silica),
sulfur dichloride,
nitrogen trifluoride,
aluminum trifluoride,
dinitrogen difluoride.

10. (a) Write formulas for

hydrogen chloride,
hydrogen bromide,
hydrogen iodide,
boron trichloride,
carbon tetrachloride,
nitrogen trichloride,
oxygen dichloride.
(b) Locate in the table inside the front cover the
symbol for each element involved in these,
compounds.

11.

For each of the following substances give the
name of each kind of atom present and the total
number of atoms represented in the formula
shown.

Name

Formula

(a) graphite (pencil lead)

(b) diamond

NaCl

(d) sodium hydroxide

NaOH

(e) calcium hydroxide

Ca(OH)>

(f) potassium nitrate

KNOs

(g) magnesium nitrate

Mg(N03)2

(h) sodium sulfate

Na2S04

(i) calcium sulfate

CaS04

12. All of the following substances are called
"acids." What element do they have in com-
mon?

(b) hydrochloric acid (or,
hydrogen chloride)

(d) sulfuric acid

(e) phosphoric acid

HC1

H2S04

H3P04

(a) nitric acid

HNO3

13. Here are the names of some common chemicals
and their formulas. What elements does each
contain?

(a) hydrogen peroxide H202

(b) jeweler's rouge Fe203

(d) tetraethyl lead Pb(C2H5)4

(e) baking soda NaHC03
(0 octane C8Hi8
(g) household gas CH4

14. (a) What does the molecular formula CBr4

mean?
(b) What information is added by the following
structural formula?

I
Br— C— Br

I
Br

15. How many particles are there in a mole?

16. A stone about the size of a softbail weighs
roughly a kilogram. How many moles of such
stones would be needed to account for the entire
mass of the Earth, about 6X10" grams?

17. If we had one mole of dollars to divide among
all the people in the world, how much would
each of the three billion inhabitants receive?

18. How many moles of atoms are in

(a) 9.0 grams of aluminum

(b) 0.83 gram of iron

Answer, (a) ^ mole.

19. The most delicate balance can detect a change
of about 10-8 gram. How many atoms of gold
would be in a sample of that weight ?

20. How many moles of oxygen atoms are in one
mole of nitric acid molecules? Of sulfuric acid
molecules ?

21. Determine the weight, in grams, of one silver
atom.

Answer. 1.79 X 10-22 gram.

QUESTIONS AND PROBLEMS

22. Write the formulas for the following compounds
and give the weight of one mole of each: carbon
disulfide, sulfur hexafluoride, nitrogen trichlo-
ride, osmium tetroxide.

23. Consider the following data

Element Atomic Weight

12.01

35.5

A and B combine to form a new substance, X.
If four moles of B atoms combine with one mole
of A to give one mole of X, then the weight of
one mole of X is

(a) 47.5 grams

(b) 74.0 grams

(d) 154.0 grams

(e) 166.0 grams

24. Calculate the molecular weight for each of the
following: SiF4, HF, Cl2, Xe, N02.

25. If 1^ moles of hydrogen gas (H2) react in a given
experiment, how many grams of H2 does this
represent ?

26. How many moles are contained in 49 grams of
pure H2S04?

Answer. 0.50 mole.

27. A chemist weighs out 10.0 grams of water, 10.0
grams of ammonia, and 10.0 grams of hydrogen
chloride (hydrochloric acid). How many moles
of each substance does he have?

28. (a) The ratio of the weight of a liter of chlorine

gas to the weight of a liter of oxygen gas,
both measured at room temperature and
pressure, is 2.22. Calculate the molecular
weight of chlorine,
(b) How does this value compare with the
atomic weight of chlorine found in the chart
of atomic weights? What is the formula of a
molecule of chlorine?

29. A flask of gaseous CCL, was weighed at a meas-
ured temperature and pressure. The flask was
flushed and then filled with oxygen at the same
temperature and pressure. The weight of the
CCI4 vapor will be about

(a) the same as that of oxygen,

(b) one-fifth as heavy as the oxygen,

(d) twice as heavy as the oxygen,

(e) one-half as heavy as the oxygen.

30. Suppose chemists had chosen a billion billion
(1018) as the number of particles in one mole.
What would the molecular weight of oxygen
gas be?

31. One volume of hydrogen gas combines with one
volume of chlorine gas to produce two volumes
of hydrogen chloride gas (all measured at the
same temperature and pressure). A variety of
other types of evidence suggests that hydrogen
is an element and that its molecules are diatomic.

(a) Which one of the following possible molecu-
lar formulas for the substance chlorine is not
consistent with the volumes that combine?
(Use only the data given here; do not pre-
sume the molecular formula of hydrogen
chloride.)

(i) CI,

(ii) CU

(iii) H2C12

(iv) H3C12

(v) H4C12

(b) For each formula in part (a) that is con-
sistent with the combining volumes data (and
the formula H2 for hydrogen), calculate the
molecular weight indicated by that formula.

CHAPTER

Chemical
Reactions

We might as well attempt to introduce a new planet into the solar sys-
tem, or to annihilate one already in existence, as to create or destroy a
particle of hydrogen. All the changes we can produce consist in sep-
arating particles that are in a state of • • • combination, and joining
those that were previously at a distance.

JOHN DAL TON, 1810

In Experiment 5 you compared the magnitude of
the heat released when melted wax solidifies to
the heat released when the same amount of wax
is burned. In each case an obvious change occurs
(liquid changes to solid or solid burns to a gas)
and it is accompanied by a measurable heat re-
lease. These are likenesses between solidification
and combustion. More apparent, however, are
striking differences. Through experiments on
warming and cooling wax, you know that after
wax melts to a liquid the solid wax can be re-
covered merely by recooling. After combustion
of the wax, however, cooling the gases produced
does not restore the wax. Instead, the products,

carbon dioxide and liquid water (and perhaps
some soot), bear no resemblance to wax. Equally
dramatic is the contrast of the heat effects. The
heat released during combustion is well over two
hundred times greater than the heat released
during solidification.

Because of these differences, chemists differ-
entiate these two kinds of change. We have
already named the solidification of wax — in Sec-
tion 1-1.2 we called this type of change a phase
change. A change like combustion, with its much
larger heat effects, is called a chemical change or
a chemical reaction.

3-1 PRINCIPLES OF CHEMICAL REACTIONS

The central activity of the chemist is to explore
and exploit chemical changes. Sometimes his
38

wish is to cause a change, sometimes to prevent
it, in a system of interest. Always he wishes to

SEC. 3-1 I PRINCIPLES OF CHEMICAL REACTIONS

understand and control the chemical changes
that might occur.

We shall begin our study of chemical changes
with a simple chemical reaction that forms a
familiar substance — water.

3-1.1 Formation of Water from Hydrogen
and Oxygen

In Section 2-2.2 it was mentioned that two vol-
umes of hydrogen combine with one volume of
oxygen. Water is produced. The reaction gives
off heat — a large amount of heat, as in the com-
bustion of a candle. The product, water, is not
at all like the starting materials, hydrogen and
oxygen. Hence, the change that occurs when
hydrogen and oxygen combine should be clas-
sified as a chemical reaction.

In terms of the atomic theory, we begin with
molecules of hydrogen and molecules of oxygen.
After reaction, we find molecules of water. The
bonds between atoms in the reacting substances
are broken, and the atoms rearrange to form the
new bonds in the product molecules. These
changes are readily pictured with the aid of our
molecular models. In Figure 3-1 two hydrogen
molecules (four atoms) and one oxygen molecule
(two atoms) are represented on the left. If these
molecules are to react to form water, the bonds
between the atoms in the oxygen molecule and
in the hydrogen molecules must be broken. Then,
the atoms can rearrange themselves to form two
water molecules. Notice that the atoms are re-
arranged as a result of the reaction but that the
total number of atoms does not change.

water could be formed? What would be left

over

EXERCISE 3-2

One million oxygen molecules react with suffi-
cient hydrogen molecules to form water mole-
cules. How many water molecules are formed?
How many hydrogen molecules are consumed?

Extending Figure 3-1, you can reason that the
production of 100 molecules of water requires
100 molecules of hydrogen and 50 molecules of
oxygen. Also, to produce one mole of water
(6.02 X 1023 molecules) you need one mole of
hydrogen gas (6.02 X 1023 molecules) and one-
half mole of oxygen (3.01 X 1023 molecules).
These results are summarized in Table 3-1.

Table 3-1.

REACTING AMOUNTS OF HYDROGEN
AND OXYGEN TO FORM WATER

Amounts Amounts

of Reacting of

Substances Product

HYDROGEN OXYGEN WATER

(a) In numbers

of molecules

100

6.02 X 10»

3.01 X 10a

6.02 X 10a

(b) In numbers

of moles

EXERCISE 3-1

Suppose ten hydrogen molecules and ten oxygen
molecules are mixed. How many molecules of

Fig. 3- J. Formation of water molecules from hydro-
gen molecules and oxygen molecules.

reacts with

The reaction between hydrogen gas and oxy-
gen gas proceeds more quickly if we mix the
gases and then ignite the mixture with a spark.
A violent explosion results. Even so, the quan-
tity of product, water, and the heat evolved are
the same per mole of hydrogen reacting as in
controlled burning.

to form

CHEMICAL REACTIONS I CHAP. 3

If we react one mole of pure hydrogen and
one-half mole of pure oxygen, one mole of water
is produced. The quantity of heat produced when
one mole of water is formed is 68,000 calories.
If we mix only 0.025 mole of pure hydrogen, only
(2) X (0.025) mole of oxygen is needed. The
amount of water produced is 0.025 mole. If only
0.025 mole of water is produced, only (0.025) X
(68,000) calories or about 1700 calories of heat
are released.

The source of this heat energy must be the re-
actants (hydrogen and oxygen) themselves since
no heat was supplied externally other than that
needed to ignite the mixture. We may conclude
that the water has less energy than did the re-
actants used to make it. Such a reaction in which
energy is released is called an exothermic re-
action. The quantity of energy produced when
one mole of hydrogen is burned, 68,000 calories
(68 kcal*), is called the molar heat of com-
bustion of hydrogen.

EXERCISE 3-3

How much heat is released when two moles of
hydrogen burn? One-half mole?

3-1.2 Decomposition of Water

We can decompose the water in a solution of
water and sulfuric acid by passing an electric
current through the solution in an electrolysis
apparatus, as shown in Figure 3-2. In this ap-
paratus, two conductors (called electrodes) are
immersed in the liquid. When the electrodes are
connected to a source of electrical energy hydro-
gen gas appears at one electrode and oxygen gas
appears at the other. If we operate the apparatus
until one mole of water has decomposed, one
mole of hydrogen gas and one-half mole of oxy-
gen gas are produced. We observe also that
energy (electrical energy in this case) is needed
to cause the water to decompose. If energy is

* 1 kilocalorie (1 kcal) = 1000 calories. The prefix
"kilo-" always means 1000. Thus, 1 kilogram = 1000
grams; 1 kilometer = 1000 meters.

Fig. 3-2. Electrolytic decomposition of water.

absorbed in a reaction, the reaction is called
endothermic.

Now we can compare the formation and de-
composition of water. As shown graphically in
Figure 3-3, the chemical change involved in the
formation of water is exactly the reverse of that
involved in the decomposition of water. These
changes are governed by simple rules. On the
left we find two atoms of oxygen, and on the
right we find two atoms of oxygen. On the left
we find four atoms of hydrogen, and on the
right we find four atoms of hydrogen. We see
that atoms are neither gained, nor are they lost.
In chemical reactions, atoms are conserved.

The number of molecules of H2 needed to
react with one molecule of 02 is the number
needed to produce two molecules of H20. If two
molecules of H20 are formed, four atoms of hy-
drogen are needed. Two molecules of H2 contain
four atoms of hydrogen. Remember, in chemical
reactions, atoms are conserved.

3-1.3 Conservation of Mass

Belief in the conservation of atoms is based upon
a generalization that has stood the test of many

SEC. 3-2 | EQUATIONS FOR CHEMICAL REACTIONS

decades. Matter can be neither created nor de-
stroyed. Since we often measure a quantity of
matter in terms of its mass (by weighing, for
example) we may say that mass is conserved.
Thus, one mole of liquid water weighs 18.0
grams; in the decomposition of one mole of
water, 2.0 grams of hydrogen and 16.0 grams of
oxygen are produced:

Fig. 3-3. The reactions of formation and decomposi-
tion of water shown with molecular models.

1 mole of

liquid water,

H20

18.0 g
18.0 g

1 mole of , 5 mole of
hydrogen, oxygen,

rO.

2.0 g + 16.0 g
18.0 g

Since the mass of a mole of water is the sum
of the masses of the atoms in the mole of water,
the conservation of mass implies conservation of
atoms.

reacts with

to form

is decomposed to

and

3-2 EQUATIONS FOR CHEMICAL REACTIONS

The graphic representations we have used for
reactions help us visualize the rearrangement of
atoms in the reactions. By a slight change we can
show the results in a less detailed but simpler
way. Chemical formulas can be used rather than
drawings of the atoms and molecules. Thus, the
formula of elementary hydrogen is H2, the for-
mula of elementary oxygen is 02, and the formula
of water is H20. By using the formulas to repre-
sent the molecules, we can replace the diagram
of Figure 3-3 by the following expressions:*

2 H2 + 1 02

4 H2 + 2 02
2H20

2H20
4H.0
1 02 + 2 H2

(2)
(i)
(4)

Such expressions are called chemical equations.
Notice that we show two molecules of a sub-
stance by writing the coefficient 2 before the

* Many chemists use an "equals" sign in place of the
arrow. Thus, equation (2) would be written:
2 H2 + 1 02 = 2 HjO

formula. The coefficient 2 before the formula
H20 means two molecules — it applies to every
symbol in the formula. In two molecules of
water, there are six atoms, four of hydrogen and
two of oxygen. Notice also that we can change
equation (2) to equation (5) merely by doubling
all the coefficients. We can change equation (2)
to equation (4) by reversing it. Equation (2) rep-
resents the formation of water; equation (4) rep-
resents the decomposition of water.

All equations are based upon the conservation
of atoms. Every symbol, when multiplied by the
subscript after it and the coefficient before the
formula, must appear as often on the left side of
the equation as on the right.

Natural gas contains mainly the substance
methane, with the formula CH4. The chemical
equation for the burning of methane is

1 CH4 + 2 02 — >- 1 C02 + 2 H20 (5)

The number of atoms in the reactants equals the

CHEMICAL REACTIONS I CHAP. 3

number of atoms in the products. One carbon
atom, four hydrogen atoms, and four oxygen
atoms are represented on each side of the equa-
tion.

Because of the relation between a molecule
and a mole, we can read equation (5) in either
of two ways: (1) "one molecule of methane reacts
with two molecules of oxygen to form one mole-
cule of carbon dioxide and two molecules of
water," or (2) "one mole of methane reacts with
two moles of oxygen to form one mole of carbon
dioxide and two moles of water."

EXERCISE 3-4

Write an equation containing the information
expressed in your answer to Question 1 of Ex-
periment 7.

3-2.1 Writing Equations for Reactions

How can you write the equations for a reaction
without first drawing a picture? Remember that
to do either you must:

(1) know what reactants are consumed and what
products form;

(2) know the correct formula of each reactant
and each product;

(3) satisfy the law of conservation of atoms.

Consider the reaction for the burning of mag-
nesium to form magnesium oxide. Magnesium
metal and magnesium oxide are solids. They
have the formulas Mg and MgO, respectively.
In preparation for writing the equation, we write
the formulas for the reactants on the left and the
formula for the product on the right:

Mg + 02 — >- MgO (a)

Expression (a) does not yet conserve atoms. We
must find numerical coefficients to place before
each formula so that there are the same number
of atoms of each element on the left side of the
equation as there are on the right. The process of
finding these coefficients is called balancing the
equation. For simple reactions, it is an easy and
logical process.

First, we may begin by choosing one mole of
oxygen as the amount of this reactant consumed
in (a):

Mg + 102 — >- MgO (b)

But one mole of 02, with its two moles of oxygen
atoms, will form two moles of magnesium oxide:

Mg + 102 — >- 2MgO (c)

Two moles of magnesium oxide require two
moles of magnesium metal. Thus,

2Mg + 102 — >- 2MgO (d) (<5)

Equation (6) is a chemical equation — since atoms
are conserved, it is said to be balanced.

We could have decided to begin by choosing
one mole of magnesium metal as the amount of
reactant consumed in (a):

lMg + 02 — >- MgO (bf)

One mole of magnesium contains a mole of
atoms, hence will form one mole of magnesium
oxide:

lMg + 02 — >- lMgO (c')

One mole of magnesium oxide contains one mole
of oxygen atoms, the number contained in one-
half mole of oxygen molecules. Thus,

lMg + i02 — ^ MgO {d') (7)

Equation (7) is also a chemical equation — again
atoms are conserved. It is just as correct an ex-
pression for the burning of magnesium as is (5).
To show this, we can multiply (7) by 2 to obtain
equation (<5). We can always multiply all the co-
efficients by a common factor or divide by a
common factor and obtain equally valid equa-
tions.

In equation (<5) the coefficient 1 may be
dropped but it is never wrong to retain it.

3-2.2 Other Examples of Chemical Equations

Hydrogen gas, H2, and chlorine gas, Cl2, react
to form hydrogen chloride gas, HC1.* The re-
actants are H2 and Cl2; the product is HC1:

1H2 + Cl2 — +■ HC1

* Hydrogen chloride, HC1, dissolved in water, is com-
monly called hydrochloric acid.

SEC. 3-2 I EQUATIONS FOR CHEMICAL REACTIONS

If we base the reaction upon one mole of H2, we
see that conservation of hydrogen atoms requires
that two moles of HC1 be produced:

1H2 + C12

2HC1

Now the product, 2HC1, contains two moles of
chlorine atoms. This is just the number of chlo-
rine atoms in one mole of chlorine. The reaction
is balanced. We may write

1H2 + 1C12 — ►- 2HC1
H2 + Cl2 — ■+- 2HC1
H2+ Cl2 = 2HC1

(5)

EXERCISE 3-5

Ammonia gas, NH3, can be burned with oxygen
gas, 02, to give nitrogen gas, N2, and water, H20.
See if you can follow the logic of the following
steps in balancing this reaction.

NH3+ 02

2NH3+ 02

2NH3 + |02

N2+ H20
1N2+ H20
1N2+ H20
1N2 + 3H20
1N2 + 3H20

(9)

State briefly what was done in each step.

The molecular formula for the substance form-
aldehyde is H2CO. Formaldehyde burns to form
carbon dioxide and water. What equation rep-
resents this reaction?

Again we begin by writing the formulas for
reactants and products:

lH2CO + 02

C02 + H20

Suppose we burn one molecule of formalde-
hyde. The one carbon atom, the two hydrogen
atoms, and one oxygen atom in the H2CO mole-
cule must appear in the products. Since, among
the products, carbon atoms appear only in car-
bon dioxide, there must be one molecule of C02:

1H2C0 + 02 — +- 1C02 + H20

Since hydrogen atoms appear in only one of the
products, water, there must be one molecule of

water to accommodate the two atoms of hydro-
gen. Now we have

lH2CO + y02 — *- 1C02 + 1H20

Notice that we have not yet determined the
coefficient of 02; we have designated it as y to
remind ourselves of this. Since the oxygen atoms
must also be conserved and three are required for
the products, three oxygen atoms must have been
present in the reactants. One oxygen atom was
present in the molecule of formaldehyde, so two
more are required. It follows that y must be 1.

We now have the balanced equation

lH2CO + 102

1C02 + 1H20 (70)

EXERCISE 3-6

A paraffin candle burns in air to form water and
carbon dioxide. Paraffin is made up of molecules
of several sizes. We shall use the molecular for-
mula C28H52 as representative of the molecules
present. One mole of candle contains the Avo-
gadro number of these molecules.

Formulas of Reactants
Q5Hs2 -+- 02 —

Formulas of Products
H20 + C02

Suppose one mole of paraffin (which weighs 353
grams) is burned. Using the method shown in
the preceding example, we obtain

lQsHs;. + v02

26H20 + 25C02

We still have not determined the coefficient
for 02. Since 76 O's are required for the products
[26 + (2 X 25) = 76], they must have been pres-
ent in the reactants. Show that it follows that y
must be 38:

C25H52 + 3802

26H20 + 25C02 (//)

Usually it is more useful to think of equations
in terms of moles rather than molecules since a
mole is a weighable amount. In equation (6) two
moles of magnesium weigh 48.6 grams; one
mole of oxygen weighs 32.0 grams; two moles
of MgO weigh 80.6 grams. Mass is conserved:
48.6 grams + 32.0 grams = 80.6 grams.

In equation (77) the 3802 is usually read as

CHEMICAL REACTIONS I CHAP. 3

38 moles, not 38 molecules; "38 moles of oxygen
gas" has experimental meaning. They weigh
38 X 32 grams =1216 grams.

3-2.3 Calculations Based upon Chemical
Equations

Equations give us all the information we need
for computing the weights of the substances con-
sumed or produced in chemical reactions. Sup-
pose we wish to know how many moles of water
are produced when 68 grams of ammonia are
burned. Equation (9) represents the reaction:

2NH3 + f02 — >- 1N2 + 3H20 (9)

One mole of ammonia weighs 17 grams. Two
moles of ammonia, weighing 34 grams, produces
three moles of water. We wish to burn 68 grams
of ammonia. How many moles is this?

68 grams

17 grams/mole
Hence we can write
2NH3 + f02

two moles
of ammonia

SO,

four moles
of ammonia

= 4.0 moles of ammonia

produce

IN, + 3H20

three moles
of water

produce

six moles
of water

We see that 68 grams (four moles) of ammonia
produce six moles of water.

Suppose we wish to know how many grams
of water are produced from the burning of one-
half mole of paraffin. Equation (77) shows the
reaction:

IQ5H5, + 3802

1 mole of
C25H52

\ mole of
Q25H52

produces

26H20 + 25C02

26 moles
of H20

Pr0duces tfSSo8

Since one mole of water weighs 18 grams, the
weight of 13 moles of water is (13 moles) X
(18g/mole) = 234 g.

EXERCISE 3-7

Show that 3.80 moles of oxygen are needed to
burn 35.3 g of paraffin by reaction (77).

EXERCISE 3-8

How many moles of oxygen, 02, are required to
produce 242 grams of magnesium oxide by equa-
tion (<*)?

2Mg + 102 — ■*- 2MgO (5)

EXERCISE 3-9

Write the equation for the reaction which took
place in Experiment 8, Part II. What was the
residue you obtained on evaporation of the solu-
tion in beaker number 2?

EXERCISE 3-10

In Experiment 8 you determined the number of
moles of silver chloride formed in the reaction
of some sodium chloride with a known amount
of silver nitrate. How many moles of sodium
chloride reacted with the silver nitrate? Compare
this with the number of moles of sodium chlo-
ride you added.

QUESTIONS AND PROBLEMS

One volume of hydrogen gas combines with one
volume of chlorine gas to give two volumes of
hydrogen chloride gas. On the basis of many re-
actions, we have learned that the molecular
formulas are, for hydrogen, H2, for chlorine, Cl2,
and for hydrogen chloride, HC1. The reaction,
in symbols, is

H2 + Cl2 — >- 2HC1

(a) According to this reaction, how many mole-
cules of hydrogen chloride, HC1, can be
formed from one molecule of hydrogen, H2?

(b) How many moles of hydrogen chloride, HC1,
can be formed from one mole of hydro-
gen, H2?

QUESTIONS AND PROBLEMS

(d) Eight moles of hydrogen chloride are formed
from how many moles of Cl2?

2. The reaction between nitric oxide, NO, and oxy-
gen, 02, is written

2NO + 02 — *■ 2N02

(a) Two molecules of nitric oxide give how many
molecules of nitrogen dioxide, N02 ?

(b) Two moles of NO give how many moles of
N02?

(d) How many moles of oxygen atoms are there
in one mole of 02?

(e) How many moles of oxygen atoms are there
in two moles of N02 ?

(0 Use the answers to parts (c), (d), and (e) to
verify that the reaction is written so as to
conserve oxygen atoms.

3. (a) Write the equation for the reaction between

nitrogen and hydrogen to give ammonia on
the basis of your answer to Problem 3 of
Chapter 2, and assuming the following mo-
lecular formulas: nitrogen, N2; hydrogen,
H2; ammonia, NH3.

(b) Verify that your equation conserves nitrogen
atoms.

4. When ammonia is decomposed into nitrogen and
hydrogen, the reaction absorbs heat. Written in
terms of moles, the equation is

2NH3 + 22 kcal — »- N2 + 3H2

(a) Two moles of ammonia produce how many
moles of nitrogen ?

(b) The production of one mole of nitrogen ab-
sorbs how much heat?

(d) Calculate the weight of two moles of am-
monia and compare it to the sum of the
weights of one mole of nitrogen, N2, plus
three moles of hydrogen, H2.

5. In the manufacture of nitric acid, HN03, nitro-
gen dioxide reacts with water to form HN03 and
nitric oxide, NO:

3N02 + H20 — >■ 2HN03 + NO

(a) Verify that the equation conserves oxygen
atoms.

(b) How many molecules of nitrogen dioxide are
required to form 25 molecules of nitric
oxide ?

6. If 3 grams of substance A combine with 4 grams
of substance B to make 5 grams of substance C
and some D, how many grams of D would you
expect ?

7. One step in the manufacture of sulfuric acid is
to burn sulfur (formula, S8) in air to form a
colorless gas with a choking odor. The name of
the gas is sulfur dioxide and it has the molecular
formula S02. On the basis of this information :

(a) Write the balanced equation for this reac-
tion.

(b) Interpret the equation in terms of molecules.

(d) Two moles of sulfur, S8, would produce how
many moles of sulfur dioxide, SO,?

8. When iron rusts, it combines with oxygen of the
air to form iron oxide, Fe203. Which of the fol-
lowing is FALSE?

(a) The equation is

302 + 4Fe

2Fe203

(b) There are five atoms represented by the for-
mula, Fe203.

(d) The mass of the reactants equals the mass
of the products.

(e) Atoms are conserved.

Balance the equations for each of the following
reactions. Begin on the basis of one mole of the
substance underscored.

(a) Li + Clo

(b) Na + Cl2

(d) Na + Br2

(e) Q2 + Cl2

(f) o2 + CJ2

LiCl

NaCl

NaF

NaBr

ci2o

CIO

10.

Show that your answers to parts (e) and (0 con-
tain the same information.

Balance the equations for each of the following
reactions involving oxygen. Begin on the basis
of one mole of the substance underscored.

CHEMICAL REACTIONS I CHAP. 3

(a) With metallic nickel:

Ni + Q2 — >- NiO

(b) With metallic nickel:

Ni + 02 — ►■ NiO

Li + Q2 — >■ Li20

(d) With the rocket fuel hydrazine, N2H4:

NFL. + 02 — ►- N2 + H-O

(e) With acetylene, QH2, in an acetylene torch
flame:

CM* + 02 — »- C02 + H.O

Answer. C2H2 + f 02 — >■ 2C02 + H20

(f) With the important copper ore, chalcocite,
Cu2S (the process called "roasting" the ore):

Cu2S + 02 — >- Cu .0 + SO,

(g) With the important iron ore, iron pyrites,
FeS2 (again, "roasting" the ore):

FeS2 + 02 — >■ Fe203 + S02

11. (a) Balance the equations for the decomposition
(to elements) of ammonia, NH3, nitrogen
trifluoride, NF3, and nitrogen trichloride,
NC13. Base each equation upon the produc-
tion of one mole of N2.

NH3

NF3
NCI3

1N2 + H2

1N2+F,
1N2 + Cl2

(b) Rewrite the equations to include the infor-
mation that the decomposition of ammonia
is endothermic, absorbing 22.08 kcal/mole
N2, the decomposition of NF3 is endother-
mic, absorbing 54.4 kcal/mole N2, and the
decomposition of NC13 is exothermic, re-
leasing 109.4 kcal/mole N2.

(c) One of the three compounds NH3, NF3, and
NCI3 is dangerously explosive. Which would
you expect to be the explosive substance?
Why?

12. Graphite, a form of carbon, C, burns in air to
produce the colorless gas, carbon dioxide. On
the basis of this information :

(a) Write the equation for the reaction.

(b) If one mole of graphite is burned, how many
moles of carbon dioxide are produced? What
is the weight in grams of this amount of
carbon dioxide?

(c) If two moles of graphite were burned, how
many moles of carbon dioxide would be
produced? What is the weight in grams?

(d) If five moles of graphite were burned in a
vessel containing 10 moles of oxygen gas,
what is the maximum number of moles of
carbon dioxide that could be produced ?

13. If a piece of sodium metal is lowered into a bottle
of chlorine gas, a reaction takes place. Table
salt, NaCl, is formed.

(a) Write the equation for the reaction.

(b) How many moles of NaCl could be formed
from one mole of Na ?

14. Methane, the main constituent of natural gas,
has the formula CH4. Its combustion products
are carbon dioxide and water.

(a) Write the equation for the combustion of
methane. Compare your answer with equa-
tion (5), p. 41.

(b) One mole of methane produces how many
moles of water vapor?

(d) How many moles of water vapor would be
produced by 4.0 grams of methane?

15. If potassium chlorate, KCIO3, is heated gently,
the crystals will melt. Further heating will de-
compose it to give oxygen gas and potassium
chloride, K.C1.

(a) Write the equation for the decomposition.

(b) How many moles of K.CIO3 are needed to
give 1.5 moles of oxygen gas?

(d) How many moles of oxygen gas would be
produced by 122.6 grams of K.CIO3?

16. One gallon of gasoline can be considered to be
about 25 moles of octane, CsHl8.

(a) How many moles of oxygen must be used to
burn this gasoline, assuming the only prod-
ucts are carbon dioxide and water?

(b) How many moles of carbon dioxide are
formed ?

QUESTIONS AND PROBLEMS

(d) What weight of carbon dioxide is released
into the atmosphere when your automobile
consumes 10 gallons of gasoline? Express
this answer in pounds (1 kg = 2.2 pounds).

17. Iron (Fe) burns in air to form a black, solid
oxide (Fe304).

(a) Write the equation for the reaction.

(b) How many moles of oxygen gas are needed
to burn one mole of iron ?

(d) Can a piece of iron weighing 5.6 grams burn
completely to Fe304 in a vessel containing
0.05 mole of 02?

18. Problem 5 relates to the manufacture of nitric
acid.

(a) According to the equation given in that
problem, how many grams of nitric acid are
formed from one mole of nitrogen dioxide ?

(b) How many more grams of nitric acid could
be formed if the nitric oxide formed could
be completely converted into nitric acid (as-
sume one mole of nitric oxide gives one mole
of nitric acid)?

19. Hydrazine, N2H4, can be burned with oxygen to
provide energy for rocket propulsion. The energy
released is 150 kcal per mole of hydrazine
burned.

(a) How much energy is released if 10.0 kg of
hydrazine fuel are burned?

(b) Compare the energy that would be released
if the same weight of hydrogen, 10.0 kg, were
burned as a fuel instead (see Section 3-1.1).

Gilbert Newton Lewis, one of the greatest American chem-
ists of the twentieth century, began his career teaching high
school chemistry. Born near Boston and reared in Ne-
braska, young Lewis returned to the East to study and
graduated from Harvard University. After a year teaching
high school, he returned to Harvard and received the Ph.D.
in 1899. There followed a year at universities in Germany,
another as Superintendent of Weights and Measures in the
Philippine Islands. Then, in seven years at the Massachu-
setts Institute of Technology he rose to the rank of Profes-
sor. Finally, in 1912 he accepted the position of Chairman
of what was then a little known chemistry department far
from the recognized scientific centers. He moved to the
University of California and spent the remainder of his
career at Berkeley, building one of the most powerful
chemistry departments in the world.

Lewis devoted most of his career to the understanding of
the structures of molecules and of thermodynamics, the
energy relations in chemical changes. His thinking was far
ahead of his lime and his theories have had profound in-
fluence on chemistry. His understanding of chemical bond-
ing has strongly influenced modern thinking on this subject.
Lewis was one of the first to recognize that energy effects
provide a basis for predicting what chemical reactions can
occur. Thus he awakened chemists to the crucial importance
of thermodynamics. His book on this subject, published in
1923, became a classic of the chemical literature. He pub-
lished over 150 research publications on topics extending
from the phases of sulfur to quantum mechanics.

G. N. Lewis enjoyed chemistry. Throughout his distin-
guished career he remained active in the laboratory and
never tired of the thrill of discovery. He favored simple
and direct experiments — many of his important discoveries
were performed with a few test tubes and simple chemicals.
His enthusiasm and burning interest in chemistry were
contagious — many of his students became great scientists.
Lewis virtually eliminated graduate courses, relying in-
stead upon the open debate of research seminars. He en-
couraged his colleagues to think critically, to challenge his
ideas, and to welcome challenge of their own.

G. N. Lewis died March 23, 1946, in the laboratory he
loved, surrounded by the beakers and books that were the
tools of his trade. He is remembered and respected by
chemists the world over.

CHAPTER

The Gas Phase:
Kinetic Theory

• • • it is my intention to make known some new properties in gases,
the effects of which are regular, by showing that these substances com-
bine amongst themselves in very simple proportions ■ • •

JOSEPH L. GAY-LUSSAC, 1808

We have already seen that the behavior of gases
is important to a chemist. The pressure-volume
behavior leads to the particle model of a gas.
Differences among gases (in properties such as
color, odor, and solubility) show that the par-
ticles of one gas differ from the particles of an-
other gas. In chemical reactions, the simple
combining volume relationships support Avo-
gadro's Hypothesis and, hence, give us a way to
measure molecular weights.

Thus we see that the properties of gases pro-
vide a substantial basis for developing the atomic
theory. The gaseous state is, in many ways, the
simplest state of matter for us to understand.
The regularities we discover are susceptible to
detailed mathematical interpretation. We shall
examine these regularities in this chapter. We
shall find that their interpretation, called the
kinetic theory, provides an understanding of the
meaning of temperature on the molecular level.

4-1 THE VOLUME OCCUPIED BY ONE MOLE OF GAS

To a chemist, one of the most important regu-
larities displayed by a gas relates to the volume
occupied by one mole of a gas. We shall begin
investigating this subject by comparing the sizes
of gaseous particles with the^average spacing
between them under normal conditions of tem-
perature and pressure. The comparison can be
based upon the volumes occupied by a mole of

nitrogen first as a solid, then as a liquid, and
finally as a gas.

4-1.1 The Volume Occupied by a Mole
off Nitrogen, N,

The molecular formula of nitrogen is N2; the
nitrogen molecule is diatomic. One mole of N2

THE GAS PHASB: KINETIC THEORY | CHAP. 4

molecules contains, then, two moles of nitrogen
atoms. The weight of one mole of nitrogen mole-
cules is 28.0 grams.

At a sufficiently low temperature, below
-210°C, nitrogen is a solid with a density of
1 03 grams per milliliter. The volume occupied
by one mole of the solid, called the molar
volume, is
Molar volume, solid. 2-ff^f = *" ■"***

Now if we warm the solid to -210°C, the
soUd melts to form liquid nitrogen. The density
of this liquid is 0.81 grams per milliliter. Now the
volume of a mole is
Molarvolume,/^: 2-|ff^ = 34.6 ml/mole

If we raise the temperature still further, the
liquid vaporizes to form nitrogen gas, taking
whatever density is necessary to fill the container.
The density now depends upon the volume of the
container and the temperature. For the sake of
comparison, suppose the gaseous nitrogen is
placed in that volume that gives a pressure of
one atmosphere when the container is placed m
an ice bath at 0°C. Then the density is found to
be only 0.00125 gram per milliliter. This means
that the volume required for one mole of gas is
Molar volume, gas at 0°C, 1 atm:

28.0 g/mole = 22 4 X 10s ml/mole
0.00125 g/ml

= 22.4 liters/mole

The volume of this gas is almost 1000 times as
great as the volume of the same weight of solid.
Experiments with other gases lead to similar
results. If the size of a single molecule is assumed
to be the same in the solid and gas, then the
molecules must have separated from each other
in the gas. The free space between gaseous mole-
cules is on the order of 1000 times the volume
a molecule occupies in the solid.

EXERCISE 4-2

(a) Calculate the volume (in milliliters) occupied
by one nitrogen molecule in the solid phase.

(b) Recognizing that one milliliter is 1.00 cubic
centimeter, estimate the size (in centimeters)
of a cube that has the volume calculated
in part (a). Use one significant figure. Now
express your answer in Angstroms (1 A =
10"8 cm).

EXERCISE 4-1

How many molecules of nitrogen are present in
one liter of the gas at 0°C and one atmosphere
pressure?

4-1.2 A Comparison of Molar Volumes of Gases

The volume calculated above, 22.4 liters, we
have seen before. In Table l-II the pressure-
volume product of 32.0 grams of oxygen, 02, was
found to be 22.4 at 0°C. (Notice that 32.0 grams
of Oa is the weight of one mole of oxygen.) So
we can use this relation,

liters X atmospheres (ftt 0oQ
r A r mole

to solve for the volume of a mole of O, at one
atmosphere pressure:

1 atm X V = 22.4 liters X atm/mole
liters X atm/mole
V " 2ZA 1 atm
= 22.4 liters/mole

This is the same volume as that just calculated
for a mole of nitrogen at 0°C and one atmos-
phere pressure (in Section 4-1.1). Furthermore,
it is the same volume occupied by 17.0 grams of
ammonia at 0°C and one atmosphere pressure
(See Table 2-III, p. 19). Since the molecular
formula of ammonia is NH3, its molecular weight
is (14.0 + 3 X 1.0) = 17.0 grams. Thus one mole
of ammonia, 17.0 grams, also occupies 22.4 liters
at 0°C and one atmosphere pressure. Experi-
ments on many other gases are in agreement and
lead to the generalization:

A mole of gas occupies 22.4 liters at PC and
one atmosphere pressure. v >

What happens to a gas as the temperature is
changed? An experiment provides the answer.
Table 4-1 shows some pressure-volume measure-

SEC. 4-1 I THE VOLUME OCCUPIED BY ONE MOLE OF GAS

ments for ammonia gas at 25°C (approximately
room temperature). Although the data shown
contain some experimental uncertainty, we find
again the regularity, PV = a constant.

Table 4-1

PRESSURE AND VOLUME OF 17.0 GRAMS
OF AMMONIA GAS, NHi t - 25°C

PRESSURE

VOLUME

(atmospheres)

(liters)

PX V

0.200

123

24.6

0.400

60.0

24.0

0.600

43.0

25.8

0.800

29.3

23.4

1.00

25.7

1.50

15.9

23.9

2.00

12.1

24.2
Average 24.5 ±0.7

This time, however, the pressure-volume prod-
uct for a mole of ammonia is 24.5 ± 0.7. We

Fig. 4-1. A mole of gas occupies 22.4 liters at 0'C,
1 atmosphere.

A mole of gas occupies 24.5 liters at 25' C,
1 atmosphere.

can compare this with our earlier result:

For one mole of ammonia at 0°C,

PXV = 22.4 liter-atm (2)

For one mole of ammonia at 25°C,

p x V = 24.5 liter-atm (5)

From (3) we see that the molar volume of
ammonia at 25°C and one atmosphere pressure
is 24.5 liters, whereas it is 22.4 liters at 0°C. The
molar volume of ammonia depends upon the
temperature. This result is no surprise —a sample
of gas expands when heated at constant pressure.
So when we compare the molar volumes of dif-
ferent gases, they should be at the same tem-
perature (and, by the same sort of argument, at
the same pressure).

Consider the following experiment. The air is
removed from a one-liter flask and it is weighed

Table 4-II THE VOLUME ° F A mole of gas at 25 c and one atmosphere

PRESSU RE

GAS

WT. OF WT. OF WT. OF MOLECULAR VOLUME

FLASK EMPTY FLASK + GAS 1 LITER OF GAS WEIGHT (liter/mole)

Wi(g) W»(g) Wx - W , (g/liter) MW(g/mole) MW/(Wt-Wx)

oxygen, Os

157.35

158.66

1.31

32.0

24.5

nitrogen, Nj

157.35

158.50

1.15

28.0

24.3

carbon

monoxide, CO

157.35

158.50

1.15

28.0

24.4

carbon

dioxide, COj

157.35

159.16

1.81

44.0

24.3

THE GAS PHASE: KINETIC THEORY I CHAP. 4

empty. Then the flask is weighed again filled with
a gas at one atmosphere pressure and at 25°C.
The difference in weight is the weight of one liter
of the gas. From this, we can calculate the vol-
ume of a mole of that gas. Table 4-II shows the
results. We find that all the gases have about the
same molar volume at 25°C and one atmosphere.
Whether the gas is 02, N2, CO, or C02, the same
volume, 24.5 ± 0.2 liters, contains 6.02 X 1023
molecules (at 25°C, one atmosphere). Whether
the gas is N2 or CO, a volume of 22.4 ± 0.1 liter
contains 6.02 X 1023 molecules at 0°C and one
atmosphere pressure.

4-1.3 Avogadro's Hypothesis

When different gases are compared at the same
temperature and pressure, they have the same
volume per mole. This is true at 0°C and one
atmosphere but, more important, it is true at
other temperatures and pressures as well.

When the two gases ammonia, NH3, and hy-
drogen chloride, HC1, react, one liter of ammo-
nia reacts with one liter of hydrogen chloride if
the two volumes are measured at the same tem-
perature and pressure. This simple one-to-one
volume ratio is observed at 0°C and one atmos-
phere pressure but, more important, this simple
ratio is observed at other temperatures and
pressures as well.

These results and many similar ones led Avo-
gadro to propose his famous hypothesis, as dis-
cussed in Section 2-2.3. The hypothesis states
that equal volumes of gases contain equal numbers
of molecules (at the same temperature and pres-
sure). Therefore, the molecular weight of a gas

can be determined by comparing the weight
of a known volume of the gas with the weight
of the same volume of another gas of known
molecular weight. It does not matter what t and
P are as long as they are the same for the
two gases.

"Avogadro's Hypothesis" is often called
"Avogadro's Law" because it has such wide ap-
plicability. It is one of the important generaliza-
tions of chemistry. It is important, not because
it is exact but because it applies to all gases,
regardless of whether their molecules are large
or small. The molecules of different gases actu-
ally have different sizes and slightly different
attractions for one another. As a result, different
gases do not have exactly the same number of
molecules in a given volume. Such variations are
small (usually less than 1 %) and do not impair
the usefulness of Avogadro's Hypothesis as a
method for determining the molecular weight of
a gas.

It is an interesting commentary on the progress
of science that this important regularity, now
often called a "Law," was not generally accepted
for half a century after it was proposed. Though
Avogadro published his idea in 181 1, its validity
was not widely recognized until the proposal was
reintroduced at an international conference of
chemists at Karlsruhe, Germany, in 1858. Today,
we find it easy to "discover" or "confirm" Avo-
gadro's Hypothesis because we can draw upon
a wealth of accumulated quantitative weight and
volume relations to develop a tightly knit pattern
of self-consistency. In contrast, even atomic
weights were in doubt early in the nineteenth
century and quantitative methods were rela-
tively crude.

4-2 THE KINETIC THEORY

Avogadro's Hypothesis provides a method for
identifying the molecules present in a gas. Also,
it explains why the volumes of gases that react
with each other are in the same simple ratio as
are the moles in the balanced equation. The
importance of these results makes the explana-

tion of the properties of gases important to a
chemist.

We have already observed that there are many
and close similarities between a gas and a col-
lection of particles in endless motion. It is essen-
tial in the particle model that each particle

SEC. 4-2 I THE KINETIC THEORY

possesses energy of motion, called kinetic
energy. Hence, the mathematical expression
that describes this model is called the kinetic
theory of gases. According to this theory, the
molecules of a gas are in rapid motion. They
travel in straight lines until they meet other mole-
cules of the gas or the atoms in the walls of the
container. They are then deflected and scattered.
The net result is a helter-skelter movement of
molecules traveling in all directions and at dif-
ferent speeds.

At room temperature, the average speed of a
nitrogen molecule is found to be about one-
quarter mile per second. In one second, however,
the nitrogen molecule has collided with many
other molecules, so its motion follows a zig-zag
path. Although the average distance between
molecules is small, the molecule passes by many
other molecules without hitting them, so the dis-
tance it travels between collisions is about fifteen
times the average distance between the molecules
(at room pressure and temperature).

4-2.1 Gas Pressure

Pressure is an important quantity in a discussion
of gas behavior. The applicability of the kinetic
theory to an understanding of gas pressure is,
then, an important success (see Section 2-1.1).
We shall investigate this success in more detail,
but first we should investigate how pressure is
measured.

MEASURING THE PRESSURE OF A GAS

A gas exerts pressure equally on all the walls of
its container. The standard method of measuring
this pressure is by measuring the height of a
mercury column supported by the gas. An in-
strument for measuring the pressure of the air is
illustrated in Figure 4-2A; it is called a barome-
ter. We can make a barometer by filling a long
tube (closed at one end) with mercury and in-
verting it in a dish of mercury. Mercury will flow
from the tube into the dish until the column of
mercury exerts a downward pressure which is
exactly balanced by the pressure of the air. In
the illustration, the pressure of the air is ex-
pressed as "755 millimeters of mercury" (written

mm Hg or mm). This is the height of the mercury
column. (Notice that only mercury vapor is pres-
ent to exert pressure in the space at the top of
the column. At room temperature, this vapor
pressure is negligible, about 10-3 mm.)

The pressure of a gas sample can be measured
in a device similar to a barometer, called a
manometer. Figures 4-2B and 4-2C show two
types. Figure 4-2B shows a closed-end manome-
ter. Here the downward pressure exerted by the
column of mercury is balanced by the pressure
of the gas sample placed in the flask. The gas
pressure is, in the example shown, 105 mm. As in
the barometer, only mercury vapor is present in
the right-hand tube.

The apparatus shown in Figure 4-2C differs in
that the right-hand tube is open. In this type of
manometer, atmospheric pressure is exerted on
the right-hand mercury column. Hence the pres-
sure in the flask plus the height of the mercury
column equals atmospheric pressure. In the ex-
ample shown, the pressure is 755 — 650 = 105
mm, the same as pictured in the closed-end
manometer, Figure 4-2B.

STP

Two conditions that are often important in
chemical experiments are temperature and pres-
sure. Consequently, chemists usually control and
measure these conditions during experiments. In
addition, it is useful to refer many experimental
results to a standard and generally accepted set
of temperature and pressure conditions. This
facilitates comparison of results of different
types and from different laboratories.

The temperature 0°C is readily obtained and
maintained with an ice water bath. The tem-
perature is one at which thermometers are cali-
brated; this aids measurement. A temperature
that is easy to maintain and easy to measure
makes a good standard temperature.

Air pressure varies somewhat from day to day
and from place to place. Nevertheless, air pres-
sure is always reasonably near 760 mm Hg, so
atmospheric pressure furnishes a convenient,
though approximate, reference pressure. How-
ever, it is not sufficiently constant for many
purposes. So, by international agreement, a

THE GAS PHASE: KINETIC THEORY I CHAP. 4

Air pressure,
755 mm

fli

Valve

■ Vacuum.

Gas inle-t

Valve

755

Fig. 4-2. Pressure measurement. A. Barometer: pres-
sure = 755 mm. B. Closed-end manometer:
pressure = 105 mm. C. Open-end manom-
eter: pressure = 755 — 650 = 105 mm.

standard pressure for gases is represented by a
height of 760 mm of mercury. This standard pres-
sure is often expressed merely as one atmos-
phere (1 atm).

Thus chemists have accepted 0°C and one
atmosphere as convenient standard conditions.
These conditions, 0°C and 760 mm pressure,
are called standard temperature and pres-
sure and are abbreviated STP.

The standard pressure is defined in terms of a pressure
reading on a standard barometer. A standard barometer

7.55-650 = 105 mm

takes account of the fact that the gravitational attraction
of the earth on the mercury varies slightly from place to
place, and the fact that mercury expands and becomes less
dense when it is heated. Thus, the mercury column of a
barometer is several millimeters longer at 20°C than at
0°C. In the standard barometer, the mercury is at 0°C.
You can find in published tables how much to subtract
from (or add to) your barometer reading to obtain the
same pressure value a standard barometer would give.
The correction is seldom more than 1 or 2 mm, and is
often negligible compared to other possible errors. Unless
your other experimental measurements are rather precise,
you need not correct your readings in this way.

THE CAUSE OF GAS PRESSURE

In Chapter 1 we explained how gases exert pres-
sure in terms of collisions of particles with the

SEC. 4-2 I THE KINETIC THEORY

container walls: this model of gas pressure is
part of the kinetic theory. Each time a gas mole-
cule strikes a wall, or a mercury surface, it exerts
a small push or force, just as a ball thrown at a
wall exerts a force upon it. The force per unit
area, called the pressure, depends directly upon
the number of molecules that strike the unit area
of the surface. Twice as many molecules in a
given volume result in twice as many collisions
per unit area, hence twice the original pressure.
Thus we explain why the pressure goes up as we
pump air into an automobile tire. If the volume
and temperature of the tire remain unchanged,
the pressure goes up in direct proportion to the
number of moles of air pumped in.

EXERCISE 4-3

A container of fixed volume contains two moles
of gas at room temperature. The pressure in the
container is four atmospheres. Three more moles
of gas are added to the container at the same
temperature. Use the result just stated to show
that the pressure is now 10 atmospheres.

(93 mm). The second bulb contains 0.001 1 mole
of water vapor. The pressure in this bulb is 20
mm Hg. The third bulb contains 0.0050 mole of
air and also 0.0011 mole of water vapor. The
third manometer shows that the pressure in the
last bulb is 113 mm Hg.

This experiment shows that the pressure ex-
erted by the mixture of gases is just the sum of
the pressure the air exerts when alone in the
flask and the pressure the water vapor exerts
when alone in the flask:

1 1 3 mm = 93 mm + 20 mm

(4)

The total pressure can be regarded as a sum of
the parts furnished by the individual pressures
exerted by each of the components of the gas
mixtures. The pressure exerted by each of the
gases in a gas mixture is called the partial
pressure of that gas. The partial pressure is the
pressure that the gas would exert if it were alone
in the container. In the example of Figure 4-3,
the total pressure in the third bulb is 113 mm.
The partial pressure of water vapor in this bulb
is 20 mm and the partial pressure of air is 93 mm.

4-2.2 Partial Pressure

Figure 4-3 shows three one-liter bulbs at 25°C.
The first bulb contains 0.0050 mole of air. The
manometer shows that the pressure is 93 mm Hg

Fie. 4-3. Pressure of a mixture of gases.

EXERCISE 4-4

Assume that 0.0050 mole of air contains 0.0040
mole of nitrogen, N2, and 0.0010 mole of oxygen,
02. What is the partial pressure of oxygen in the
first bulb in Figure 4-3? What is the partial pres-
sure of oxygen in the third bulb? Use three sig-
nificant figures.

93 mm

.0050 mole air-
in one liter

ir= Z5°C

20 mm

113 mm

.0011 mole yvater vapor
in one liter

t=ZS°C

.0050 mote air
-r .0011 mole water vapor
in one liter

t^ 25 °C

THE GAS PHASE: KINETIC THEORY | CHAP. 4

The pressure behavior shown in Figure 4-3 is
readily explained in terms of the kinetic theory
of gases. There is so much space between the
molecules that each behaves independently, con-
tributing its share to the total pressure through
its occasional collisions with the container walls.
The water molecules in the third bulb are seldom
close to each other or to molecules provided by
the air. Consequently, they contribute to the
pressure exactly the same amount they do in the
second bulb — the pressure they would exert if
the air were not present. The 0.0011 mole of
water vapor contributes 20 mm of pressure
whether the air is there or not. The 0.0050 mole
of air contributes 93 mm of pressure whether the
water vapor is there or not. Together, the two
partial pressures, 20 mm and 93 mm, determine
the measured total pressure.

4-2.3 Temperature and Kinetic Energy

If the kinetic theory is applicable to gases, we
should expect pressure to be affected by other
factors than the number of moles per unit vol-
ume. For example, the mass of the molecules and
their velocities should be important, as well.
After all, a baseball exerts more "push" on a
catcher's mitt than would a ping-pong ball
thrown with the same velocity. Also, a baseball
exerts more "push" on the mitt if a "fast ball"
is thrown rather than a "slow ball." To see how
the mass of the molecules and their velocities are
dealt with in the kinetic theory, we must consider
temperature.

To measure the temperature of a gas we im-
merse some kind of thermometer in it. If the
thermometer is colder than the system, heat flows
into the thermometer until the gas and the ther-
mometer are at the same temperature. Then we
read the thermometer to get a numerical value
for the temperature. If the thermometer were
hotter than the gas, heat would flow from the
thermometer. When there is no net flow of heat,
the thermometer is said to be in thermal
equilibrium with the gas.

There are many kinds of thermometers. Any
substance can be fashioned into a thermometer
if it has a readily measured property that is sen-

sitive to a change in temperature. The familiar
mercury thermometer depends upon the expan-
sion of the liquid as temperature is raised. Solids
and gases also change volume with temperature
change. Hence either can be (and both are) used
as a basis for a thermometer. A gas held at con-
stant volume also responds to a change in tem-
perature, the pressure rising with rising tem-
perature. This is the more common way in which
a gas is used in a thermometer: the volume is
fixed and the pressure varies with temperature.

So let us measure the temperature of a sample
of gas A by placing it in thermal contact with a
sample of gas B (our thermometer). There will
be heat flow between the two gas samples if they
are initially at different temperatures. Energy is
transferred from the hotter gas to the cooler gas.
When heat flow ceases, the gases have reached
thermal equilibrium. Then the gases have the
same temperature.

We can visualize what is going on with the aid
of the kinetic theory of gases. Suppose sample A
is initially at a high temperature relative to the
thermometer gas B. We interpret this to mean
that the molecules in gas A have more energy of
motion than those of gas B — the molecules of gas
A have higher kinetic energies (on the average).
When the samples are brought into thermal con-
tact, collisions permit the rapidly moving A
molecules to transfer kinetic energy through the
thermal connection to the slowly moving B mole-
cules. This transfer of kinetic energy from gas A
to gas B is the process that raises the temperature
of gas B and lowers the temperature of gas A.
When the thermal contact between molecules of
A and B no longer results in a net transfer of
kinetic energy from one gas to the other, then
gases A and B are in thermal equilibrium: they
have the same temperatures.

Thus, we picture heat flow between two gas
samples as a transfer of kinetic energy. The proc-
ess continues until the molecules of both gases
have the same average kinetic energy. Then the
gases are at the same temperature. This is a
basic premise of the kinetic theory: When gases
are at the same temperature, the molecules of the
gases have the same kinetic energy (on the aver-
age).

SEC. 4-2 I THE KINETIC THEORY

4-2.4 Absolute Temperature

The quantitative effects of temperature on gases
were first studied by Jacques Charles, a French
scientist, in 1787. He found that all gases expand
by the same fraction of their original volumes
when they are heated over the same temperature
range. (In these experiments the pressure re-
mained the same.) A simple experiment shows
the relations. Into a small-bore glass tube, one-
half meter in length and closed at one end, we
place a drop of mercury. This mercury falls and
finally traps a sample of air in the bottom of the
tube. (See Figure 4-4.) Since the tube has a uni-
form bore, .we can use the length of the air
sample as a measure of its volume. The mercury
plug moves up or down and maintains a con-
stant pressure.

Fig. 4-4. Apparatus for demonstrating the effect of
temperature on the volume of a gas.

Table 4-/II.

CHANGE OF VOLUME OF A GAS
WITH CHANGE IN TEMPERATURE

RELATIVE VOLUME

TEMPERATURE

(as measured by

(°C)

length of sample)

200

1.73

100

1.37

1.18

1.00

When we plot these results with relative vol-
umes on the ordinate (vertical axis) and tem-
peratures on the abscissa (horizontal axis), we
obtain the graph shown in Figure 4-5. The
straight line passes through the experimental
points. When extrapolated upward, it shows that
the volume at 273°C is double that at 0°C.
Extrapolated downward, the line shows that the

2.0

1.0

-273 "C

1
1

-200

-100 0 +100

+200

+300 °C

173 273 373
Te mpe ra tttre

■473

573 °K

Fig. 4-5. An absolute temperature scale from the
change of volume of a gas with temperature.

We may place the tube in ice water (0°C) and
measure the relative volume of the air sample.
If the tube is immersed in water boiling at one
atmosphere pressure (100°C), the relative vol-
ume has a higher value. From these data and
from similar measurements at other tempera-
tures, we collect data such as those in Table 4-III.

volume would become zero at — 273°C. The vol-
ume change per degree centigrade is yfg of the
volume at 0°C. Actually, all gases liquefy before
their temperature reaches — 273°C.

If gases are heated or cooled at constant vol-
ume, the pressure changes, also at the rate of ^
of its value at 0°C. Then the pressure of a gas

THB GAS PHASE: KINETIC THEORY I CHAP. 4

would become zero at — 273°C. In terms of the
kinetic theory, the motion of the molecules
ceases at this temperature. The kinetic energy
has become zero.

There are great advantages to an absolute tem-
perature scale that has its zero point at — 273°C.
Whereas the "zero" of temperature in the Centi-
grade scale is based upon an arbitrary tempera-
ture, selected because it is easily measured, the
zero point of the absolute scale has inherent
significance in the kinetic theory. If we express
temperatures on an absolute temperature scale,
we find that the volume of a fixed amount of gas
(at constant pressure) varies directly with tem-
perature* Also, the pressure of a fixed amount
of ?ns (at constant volume) varies directly with
temperature. And, according to the kinetic the-
ory, the kinetic energy of the molecules varies
directly with the absolute temperature. For these
reasons, in dealing with gas relations, we shall
usually express temperature on an absolute tem-
perature scale.

This temperature scale, with the same size
degrees as the Centigrade scale, is called the
Kelvin scale and values on this scale are ex-
pressed in degrees Kelvin (°K). Both Kelvin and
Centigrade temperatures are shown in Figure
4-5. Notice that all numerical values on the
Kelvin scale are 273 degrees higher than the
corresponding temperatures on the Centigrade
scale.

EXERCISE 4-5

(a) Express the following temperatures in de-
grees Kelvin:

Boiling point of water : 1 00°C

Freezing point of mercury: — 38.9°C

Boiling point of liquid nitrogen: — 196°C

(b) Express the following temperatures in de-
grees Centigrade:

Melting point of lead :

A normal room temperature:

Boiling point of liquid helium:

600°K

298°K

4°K

* This direct relation between volume and temperature
(at constant pressure) is called Charles' Law.

EXERCISE 4-6

In Experiment 9 a student obtained the result
that 2.00 X 10~3 mole of magnesium produced
a volume of hydrogen that would occupy 49.0
ml at 25°C and one atmosphere pressure.

(a) If one mole of magnesium produces one mole
of hydrogen, use these data to calculate the
volume of one mole of hydrogen at 25°C
(298°K) and one atmosphere.

(b) Calculate the volume one mole of hydrogen
would occupy at 0°C (273°K) and one at-
mosphere.

We have remarked that a temperature of zero
on the absolute temperature scale would cor-
respond to the absence of all motion. The kinetic
energy would become zero. Very interesting phe-
nomena occur at temperatures near 0°K (the
superconductivity of many metals and the super-
fluidity of liquid helium are two examples).
Hence, scientists are extremely interested in
methods of reaching temperatures as close to
absolute zero as possible. Two low temperature
coolants commonly used are liquid hydrogen
(which boils at 20°K) and liquid helium (which
boils at 4°K). Helium, under reduced pressure,
boils at even lower temperatures and provides a
means of reaching temperatures near 1°K. More
exotic techniques have been developed to pro-
duce still lower temperatures (as low as 0.00 1°K)
but even thermometry becomes a severe problem
at such temperatures.

4-2.5 Avogadro's Hypothesis and
the Kinetic Theory

The kinetic theory is based upon the premise
that if two gases are at the same temperature, the
molecules of the gases have the same average
kinetic energy. The ability of this kinetic theory
to explain Avogadro's Hypothesis is one of its
most important successes.

We may state Avogadro's Hypothesis in this
form: If two gases at the same temperature have
the same number of particles in a given volume,
they must exert the same pressure. Yet, as re-

SEC. 4-2 I THE KINETIC THEORY

marked in Section 4-2.3, the mass of a molecule,
as well as its velocity, should influence the pres-
sure exerted. If the molecules of our two gas
samples have different masses, they must have
different speeds in order to have the same kinetic
energies. The lighter molecules must travel faster,
so they will strike the container walls more times
per second. The effect of the more frequent col-
lisions exactly counteracts the lower "push" per
collision from these lower-mass molecules. The
result is in perfect accord with Avogadro's Hy-
pothesis: Two gases at the same concentration
and at the same temperature exert the same
pressure even though their molecules have dif-
ferent masses.

Avogadro's Hypothesis can be shown quite readily in
an approximate way. The kinetic energy of a moving
particle is expressed by the equation

KE = *mv2 (5)

where m is the mass of the particle and v is the velocity.
Therefore, for gas A and gas B at the same temperature,
we have

(KE)A = (KE)B (6)

IntAVA2 = jmaVfl1
or

rriAVA1 = mBVBl (7)

Now suppose we place n molecules in a cubical box of
dimension d. The pressure is fixed by the number of wall
collisions per second on each square centimeter times the
momentum transferred per collision:

_ _ /collisionsN/ 1 \/ momentum \ .»

\ second /\area/\ collision /

Momentum depends upon mass and velocity. The par-
ticle approaches the wall with momentum mv and leaves
with this same momentum in the opposite direction. The
momentum transferred to the wall is, then

Momentum = 2mv (9)

The collisions per second with the wall, on the other
hand, depend upon the container dimension and the
velocity (as the molecule bounces back and forth between
the walls). We can assume that one-third of the molecules
bounce back and forth in a given direction between two
opposite walls. Then if there are n molecules in the con-
tainer, there are «/3 hitting these two walls. One of these
walls receives a collision each time one of the molecules
travels the box dimension d and back, a distance of 2d.

collisions _ /no. particles bouncing back and forth \
second ~ \ time for a particle to travel distance 2d)

collisions _ / w/3 \ _ (n\( v_\ _ nv ....

second " \2d/v) ~ \3j\2d) ~ 6d ( }

Combining (8), (9), and (10), we find

_ /collisions\/ 1 N/momentumX
— \ second /\area/\ collision /

Applying equation (7/) to each of the gases A and B,

PA = \(^)(mAVA') U2)

Pb = \ (^)(»W) (13)

If the gases have the same pressure, Pa = Pb, we can
equate (12) and (13) so that

If the temperatures of the gases are the same, equation
(7) is applicable and equation (14) becomes

tiA riB
<P ~ d*

(15)

Thus we see that at the same temperature and pressure,
the two gases have the same number of molecules per
unit volume. This is Avogadro's Hypothesis.

4-2.6 The Perfect Gas

We have examined experimental pressure-volume data
for oxygen gas (Table l-II), ammonia gas (Table 4-1),
and hydrogen chloride gas (Table 2-II). In each case,
within the experimental uncertainty of the data shown,
the gases have the regular behavior, PV = a constant.
We find from many such experiments that many gases
follow this simple behavior. Of course such a generaliza-
tion is subject to uncertainty, as is any other scientific
statement. The generalization was derived from a set of
measurements, each of which involves some uncertainty,
and, hence, the constancy of the PV product is established
only within corresponding bounds of uncertainty. What's
more, there are limits to the pressure range over which
the behavior has been tested.

To be specific, consider the data for 17.0 grams of
ammonia gas at 25°C, as presented in Table 4-1 (p. 51).
These data show that PV = 24.5, but a complete state-
ment should include both the uncertainty and the range
over which the data are known to apply. In this case the
uncertainty is ±0.7 and the range is 0.2-2 atmospheres
pressure. It would not be safe, from these data alone, to
assume that the pressure-volume product is constant to
four significant figures, PV = 24.50. Neither would it be
safe to assume that the pressure-volume product is con-
stant outside the range of pressure studied, 0.2-2.0 at-
mospheres. Remember, a generalization is reliable within

THE GAS PHASE: KINETIC THEORY I CHAP. 4

the bounds defined by the experiments that led to the rule.
If we need four significant figures, or wish to know the
behavior at a higher pressure, more experiments are
needed.

More accurate pressure-volume measurements extend-
ing to much higher pressures have been performed. Table
4-IV shows the results of such experiments.

Table 4-IV

ACCURATE PRES
M EASU REM ENT
OF AMMONIA G

PRESSURE

(atmospheres)

SURE-VO
S FOR 17
AS AT 25

VOLUME

(liters)

LUME

00 GRAMS

PXV

0.1000

244.5

24.45

0.2000

122.2

24.44

0.4000

61.02

24.41

0.8000

30.44

24.35

2.000

12.17

24.34

4.000

5.975

23.90

8.000

2.925

23.40

9.800

2.360

23.10

condensation
beginning

9.800

0.020

0.20

no gas left;
liquid only

20.00

0.020

0.40

liquid only
present

50.00

0.020

1.0

liquid only
present

The most startling fact revealed in Table 4-IV is the
drastic deviation from PV = 24.5 that occurs when the
pressure is raised above 9.800 atmospheres. Suddenly the
relation PV = a constant is no longer applicable. Here

is dramatic evidence of the danger lurking in careless
extrapolation beyond the range of experience.

Even below the condensation pressure the pressure-
volume product was not perfectly constant. With meas-
urements of sufficient accuracy and precision, we can see
that the PV product of ammonia at 25°C is not really
constant after all. It varies systematically from 24.45 at
0.1000 atmospheres to 23.10 at 9.800 atmospheres, just
before condensation begins. Similar measurements on
28.0 grams of carbon monoxide at 0°C show that the PV
product is 22.410 at 0.2500 atmospheres pressure, but if
the pressure is raised to 4.000 atmospheres, the PV prod-
uct becomes 22.308. This type of deviation is common.
Careful measurements reveal the fact that no gas follows
perfectly the generalization PV = a constant at all pres-
sures. On the other hand, every gas follows this rule
approximately, and the fit becomes better and better as
the pressure is lowered. So we find that every gas ap-
proaches the behavior PV = a constant as pressure is
lowered.

There is a reasonable explanation for this type of
deviation. The kinetic theory, which "explains" the
pressure-volume behavior, is based upon the assumption
that the particles exert no force on each other. But real
molecules do exert force on each other! The condensation
of every gas on cooling shows that there are always
attractive forces. These forces are not very important
when the molecules are far apart (that is, at low pressures)
but they become noticeable at higher pressures. With this
explanation, we see that the kinetic theory is based on an
"idealized" gas — one for which the molecules exert no
force on each other whatsoever. Every gas approaches
such ideal behavior if the pressure is low enough. Then
the molecules are, on the average, so far apart that their
attractive forces are negligible. A gas that behaves as
though the molecules exert no force on each other is
called an ideal gas or a perfect gas.

Table 4- V. molar volumes of some gases

FORMULA

MOLAR WEIGHT

(grams)

MOLAR VOLUME AT
0°C AND 1 ATM

(liters)

hydrogen

2.0160

22.430

helium

4.003

22.426

("perfect" gas)

—

(22.414)

nitrogen

28.016

22.402

carbon monoxide

28.011

22.402

oxygen

32.000

22.393

methane

CH4

16.043

22.360

carbon dioxide

co2

44.011

22.262

hydrogen chloride

HC1

36.465

22.248

ammonia

NH3

17.032

22.094

chlorine

Cl2

70.914

22.063

sulfur dioxide

so2

64.066

21.888

QUBSTIONS AND PROBLEMS

Avogadro's Hypothesis is consistent with the kinetic
theory. Therefore a perfect gas follows Avogadro's Hy-
pothesis. At one atmosphere pressure and 0°C, one mole
(6.02 X 10" molecules) of a perfect gas occupies 22.414
liters. How closely real gases approximate a perfect gas
at one atmosphere pressure and 0°C is shown by measur-

ing the volume occupied by one mole of that gas, the molar
volume. Table 4-V shows the molar volumes of a num-
ber of gases. We see that real gases do approximate
closely (to three significant figures) the perfect gas be-
havior at one atmosphere and 0°C. Every gas becomes a
perfect gas as the pressure is reduced toward zero.

4-3 REVIEW

Regularities observed in the behavior of gases
have contributed much to our understanding of
the structure of matter. One of the most im-
portant regularities is Avogadro's Hypothesis:
Equal volumes of gases contain equal numbers
of particles (at the same pressure and tempera-
ture). This relationship is valuable in the deter-
mination of molecular formulas — these formulas
must be known before we can understand chemi-
cal bonding.

We have explored the meaning of temperature.
According to the kinetic theory, when two gases
are at the same temperature, the molecules of the
two gases have the same average kinetic energies.
Changing the temperature of a sample of gas at
constant pressure reveals that the volume is di-

rectly proportional to the temperature if the
temperature is expressed in terms' of a new,
absolute scale. The melting point of ice (0°C) on
this new scale (called the Kelvin scale) is 273°K.
The boiling point of water at one atmosphere
(100°C) is 373°K. The zero temperature on the
Kelvin scale corresponds to the hypothetical loss
of all molecular motion.

This progress gives us substantial basis for
confidence in the usefulness of the atomic theory
and it encourages us to develop the model fur-
ther. We shall see that the concepts we have
developed in our consideration of gases are also
useful when we consider the behavior of con-
densed phases— liquids and solids.

QUESTIONS AND PROBLEMS

1. How many molecules are there in a molar vol-
ume of a gas at 100°C? At 0°C?

2. What is the molar volume of water under each
of the following conditions?

(a) Solid, 0°C;

density of ice = 0.915 g/ml.

(b) Liquid, 0°C;

density of water (liquid, 0°C) = 1.000 g/ml.

density of water vapor (100°C, 1 atm) =
5.88 X 10~4 g/ml.

3. What is the molecular weight of a gas if at 0°C

and one atmosphere pressure, 1.00 liter of the

gas weighs 2.00 grams?

Answer. 44.8 g/mole

4. The gas sulfur dioxide combines with oxygen to
form the gas sulfur trioxide:

What ratio would you expect for the following?
number of SQ3 molecules produced
number of 02 molecules consumed
volume of SQ3 gas produced
volume of 02 gas consumed

(a)
(b)

5. A glass bulb weighs 108.1 1 grams after all of the
gas has been removed from it. When filled with
oxygen gas at atmospheric pressure and room
temperature, the bulb weighs 109.56 grams.
When filled at atmospheric pressure and room
temperature with a gas sample obtained from
the mouth of a volcano, the bulb weighs 111.01
grams. Which of the following molecular for-
mulas for the volcano gas could account for the
data?

2SO,(gas) + O-Xgas)

2SO,(gas)

co2
ocs

Si2H6

so2

NF3

S03

A gas mixture, half C02,
half Kr

THE GAS PHASE! KINETIC THEORY I CHAP. 4

6. Compressed oxygen gas is sold at a pressure of
130 atm in steel cylinders of 40 liters volume.

(a) How many moles of oxygen does such a
filled cylinder contain ?

(b) How many kilograms of oxygen are in the

cylinder?

Answer. 6.7 kg.

7. A carbon dioxide fire extinguisher of 3 liters
volume contains about 10 pounds (4.4 kg) of
C02. What volume of gas could this extinguisher
deliver at room conditions?

8. Hydrogen for weather balloons is often supplied
by the reaction between solid calcium hydride,
CaH2, and water to form solid calcium hydrox-
ide, Ca(OH)2, and hydrogen gas, H2.

(a) Balance the equation for the reaction and
decide how many moles of CaH2 would be
required to fill a weather balloon with 250
liters of hydrogen gas at normal conditions.

(b) What weight of water would be consumed in

forming the hydrogen ?

Answer. 0.18 kg.

9. Gas is slowly added to the empty chamber of a
closed-end manometer (see Figure 4-2B). Draw
a picture of the manometer mercury levels, show-
ing in millimeters the difference in heights of the
two mercury levels:

(a) before any gas has been added to the empty
gas chamber;

(b) when the gas pressure in the chamber is
300 mm;

(d) when the gas pressure in the chamber is
865 mm.

10. Repeat Problem 9 but with an open-end ma-
nometer (see Figure 4-2C). Atmospheric pres-
sure is 760 mm.

1 1 . The balloons that are used for weather study are
quite large. When they are released at the surface
of the earth they contain a relatively small vol-
ume of gas compared to the volume they acquire
when aloft. Explain.

12. A 1.50 liter sample of dry air in a cylinder exerts
a pressure of 3.00 atm at a temperature of 25°C.
Without change in temperature, a piston is
moved in the cylinder until the pressure in the

cylinder is reduced to 1.00 atm. What is the
volume of the gas in the cylinder now ?

13. Suppose the total pressure in an automobile tire
is 30 pounds/inch2 and we want to increase the
pressure to 40 pounds/inch2. What change in the
amount of air in the tire must take place ? As-
sume that the temperature and volume of the tire
remain constant.

14. The density of liquid carbon dioxide at room
temperature is 0.80 grams/ml. How large a car-
tridge of liquid C02 must be provided to inflate
a life jacket of 4.0 liters capacity at STP?

15. A student collects a volume of hydrogen over
water. He determines that there is 2.00 X 10~3
mole of hydrogen and 6.0 X 10~6 mole of water
vapor present. If the total pressure inside the
collecting tube is 760 mm, what is the partial
pressure of each gas?

Answer. Partial pressure H2 = 738 mm.
Partial pressure H20 = 22 mm.

16. A sample of nitrogen is collected over water at
18.5°C. The vapor pressure of water at 18.5°C
is 16 mm. When the pressure on the sample has
been equalized against atmospheric pressure, 756
mm, what is the partial pressure of nitrogen?
What will be the partial pressure of nitrogen if
the volume is reduced by a factor 740/760?

17. A candle is burned under a beaker until it ex-
tinguishes itself. A sample of the gaseous mix-
ture in the beaker contains 6.08 X 1020 molecules
of nitrogen, 0.76 X 1020 molecules of oxygen,
and 0.50 X 1020 molecules of carbon dioxide.
The total pressure is 764 mm. What is the partial
pressure of each gas?

18. A cylinder contains nitrogen gas and a small
amount of liquid water at a temperature of 25°C
(the vapor pressure of water at 25°C is 23.8 mm).
The total pressure is 600.0 mm Hg. A piston is
pushed into the cylinder until the volume is
halved. What is the final total pressure ?

Answer. 1176 mm.

19. Consider two closed glass containers of the same
volume. One is filled with hydrogen gas, the
other with carbon dioxide gas, both at room
temperature and pressure.

(a) How do the number of moles of the two
gases compare?

QUESTIONS AND PROBLEMS

(b) How do the number of molecules of the two
gases compare?

(d) If the temperature of the hydrogen container
is now raised, how do the two gases now
compare in:

(i) pressure,

(ii) volume,
(iii) number of moles,
(iv) average molecular kinetic energy.

20. The boiling points and freezing points in degrees
Centigrade of certain liquids are listed below.
Express these temperatures on the absolute tem-
perature (degree Kelvin) scale.

Liquid helium, boiling point = —269
Liquid hydrogen, freezing point = —259

Answer. 14°K.
Liquid hydrogen, boiling point = —253

Answer. 20°K.

Liquid nitrogen, freezing point = —210
Liquid nitrogen, boiling point = —196

Liquid oxygen, freezing point = — 219
Liquid oxygen, boiling point = — 183

21. If exactly 100 ml of a gas at 10°C are heated to
20°C (pressure and number of molecules remain-
ing constant), the resulting volume of the gas
will be which of the following?

(a) 50 ml,

(b) 1000 ml,

(d) 375 ml,

(e) 103 ml.

(a) Will the final pressure be greater or lower
than the original pressure?

(b) By what factor does the pressure change if
one mole of methane and one mole of oxygen
are mixed and reacted (with the temperature
changing from 25°C to 200°Q?

Answer. 2.38.

24. Automobiles are propelled by burning gasoline,
typical formula C8Hi8, inside a container (the
cylinder) that can change volume and drive the
wheels. Oxygen reacts with the gasoline to form
carbon dioxide and water, releasing enough en-
ergy to heat the gas from about 300°K to about
1500°K.

Balance the equation for the reaction and
decide whether the work done by the gas in the
cylinder is mainly due to pressure rise caused by
change in number of moles of gas or due to
pressure rise resulting from heating.

25. Why does the pressure build up in a tire on a
hot day? Answer in terms of the kinetic theory.

26. A vessel contains equal numbers of oxygen and
of hydrogen molecules. The pressure is 760 mm
Hg when the volume is 50 liters. Which of the
following statements is FALSE?

(a) On the average, the hydrogen molecules are
traveling faster than the oxygen molecules.

(b) On the average, more hydrogen molecules
strike the walls per second than oxygen
molecules.

(d) Equal numbers of moles of each gas are
present.

(e) The average kinetic energies of oxygen and
hydrogen are the same.

22. Why is it desirable to express all temperatures
in degrees Kelvin when working with problems
dealing with gas relationships?

23. A gaseous reaction between methane, CH4, and
oxygen, 02, is carried out in a sealed container.
Under the conditions used, the products are
hydrogen, H2, and carbon dioxide, C02. Energy
is released, so the temperature rises during the
reaction.

27. The vapor pressure of a molten metal can be
measured with a device called a Knudsen cell.
This is a container closed across the top by a
thin foil pierced by a small, measured hole. The
cell is heated in a vacuum, until the vapor above
the melt streams from the small hole (it effuses).
The weight of the material escaping per second
tells the rate at which gaseous atoms leave.

Two identical Knudsen cells are heated at
1000°C, one containing lead and the other con-
taining magnesium.

THE GAS PHASE: KINETIC THEORY I CHAP. 4

(a) Contrast the average kinetic energies of the
lead and magnesium atoms within each cell.

(b) Contrast the average velocities of the lead
and magnesium atoms leaving each cell.

(c) At this fixed temperature, the rate at which
atoms leave is determined by two factors,
the vapor pressure and the mass of the
gaseous particles. Explain.

28. The following table indicates the boiling points
and the molar volumes (0°C and 1 atm) of some
common gases.

(a) What regularity is suggested in the relation-
ship between the boiling points and molar
volumes?

(b) Account for this regularity.

GAS

FORMULA

BOILING
POINT
(°C)

MOLAR
VOLUME

(liters)

helium

-269

22.426

nitrogen

-196

22.402

carbon

monoxide

-190

22.402

oxygen

-183

22.393

methane

CH4

-161

22.360

hydrogen

chloride

HC1

-84.0

22.248

ammonia

NH3

-33.3

22.094

chlorine

-34.6

22.063

sulfur

dioxide

SO:

-10.0

21.888

CHAPTER

Liquids and Solids:
Condensed Phases
of Matter

Almost all the chemical processes which occur in nature, whether in
animal or vegetable organisms, or in the non-living surface of the earth,
• • • take place between substances in solution.

w. ostwald, 1890

Only a handful of substances are gases under
normal conditions of temperature and pressure.
Of the hundred or so elements, most are nor-
mally solids; two or three are liquids. As for
compound substances, more than a million have
been prepared by chemists, yet, more than 99%
of these are liquids or solids, each with distinc-
tive and characteristic properties. It is no sur-
prise, then, that there is great variety among all
of these substances. Rather, it is remarkable that
they can be classified into a small number of
types and that the wealth of information repre-
sented by the diversity of all of these compounds

can be treated within a simple framework. We
shall begin our study of this framework by con-
sidering the properties of pure substances in their
liquid and solid phases.

EXERCISE 5-1

The dozen or so elements that are normally
found as gases include nitrogen, oxygen, fluo-
rine, helium, neon, argon, krypton, xenon, and
chlorine. Where are these placed in the periodic
table (see inside front cover)?

5-1 PURE SUBSTANCES

A gas, when cooled, condenses to a liquid. Fur-
ther cooling causes solidification. The condensa-
tion of ammonia under pressure, the condensa-
tion of water vapor (steam) on cooling, and the

freezing of liquid water to ice are familiar cases.
These changes are called phase changes. We shall
consider liquid-gas changes first and then, solid-
liquid phase changes.

LIQUIDS AND SOLIDS: CONDENSED PHASES OF MATTER | CHAP. 5

5-1.1 Liquid-Gas Phase Changes

When a pan of water is warmed, the input of
heat causes the water temperature to rise. At a
certain point, however, the water begins to boil.
Then the temperature is constant as long as
liquid water remains, and continued heating
causes the formation of water vapor. Water
changes from the liquid phase to the gas phase,
absorbing energy though the temperature re-
mains constant. The energy of the liquid is less
than the energy of the same weight of gas.

Let us consider how much energy is needed
for this particular phase change.

H20(liquid) — ►- H20(gas) (7)

or, in abbreviation,

H2OfU — ►- H20(g) (7)

Suppose we wish to evaporate one mole of water,
as expressed in equation (7). One mole contains
the Avogadro number of molecules (6.02 X 1023)
and has a weight of 18.0 grams. Using a calo-
rimeter, as you did in Experiment 5, you could
measure the quantity of heat required to evapo-
rate one mole of water. It is 10 kilocalories per
mole. This value is called the molar heat of
vaporization of water. This is the energy re-
quired to separate 6.02 X 1023 molecules of water
from one another, as pictured in Figure 5-1.

Fig. 5-1. Evaporation of liquid water.

EXERCISE 5-2

+ 10

kcal
mole

One mole.
liquid water + 10 kcal

HzO(t)

6.02 * 1023 molecules
weight 18 g

When two moles of water are evaporated, how
much heat is required? One-half mole of water?

When water vapor condenses to liquid water,
the molecules release the energy it took to sepa-
rate them. A mole of gaseous water, therefore,
will release 10 kilocalories of heat when con-
densed to liquid water at the same temperature.
The amount of heat released is numerically equal
to the molar heat of vaporization.

Other liquid-gas phase changes are similar,
though boiling points vary over a wide range.
Table 5-1 shows the boiling points and heats of
vaporization of a variety of liquids. In each case,
energy is absorbed as the particles that make up
the liquid are separated into the molecules of the
gas. We shall see in Chapter 17 that the extreme
range of heats of vaporization shown in this
table can be explained using rather simple prin-
ciples. These principles provide a basis for quali-
tative predictions of boiling point, heat of
vaporization, and other properties.

5-1.2 Liquid-Gas Equilibrium: Vapor Pressure

Our knowledge of gas behavior helps us interpret
the evaporation of liquids. We have considered,
thus far, vaporization of a liquid at its usual

one mole water vapor

H20($)
6.02 x 1023 molecules
"weight 18 g

SEC. 5-1 ! PURE SUBSTANCES

Table 5-1. the normal boiling points and molar heats of vaporization

OF SOME PURE SUBSTANCES

SUBSTANCE

PHASE CHANGE

(liquid) — >- (gas)

BOILING POINT

°K °C

MOLAR HEAT
OF VAPORIZATION

(kcal/mole)

boiling point. But liquids vaporize at all tempera-
tures. Let us consider this process, beginning
with liquid water again.

If we place some liquid water in a flask at
20°C and seal the flask, some water molecules
leave the liquid and enter the gas phase. The
partial pressure of water vapor in the flask rises,
but when it reaches 17.5 mm no more change
can be observed. The amount of excess liquid
remains constant thereafter, and the partial
pressure of water vapor in the flask remains at
17.5 mm, as long as the temperature is main-
tained at 20°C. This partial pressure is called the
vapor pressure of water at 20°C. At this vapor
pressure, liquid and gaseous water can coexist
indefinitely at 20°C. This vapor pressure, 17.5
mm, is the same whether air is present or not;
it is a property of water. If the flask were origi-
nally evacuated, liquid would evaporate until the
pressure rose from 0 mm to 17.5 mm. If the flask
originally contained dry air at a pressure of 750
mm, liquid would evaporate until the pressure
rose from 750 mm to 767.5 mm (the partial pres-
sure of water vapor changing from 0 mm to 17.5
mm). When a liquid is in contact with its vapor
at the vapor pressure, the liquid and gas are said
to be in equilibrium. At equilibrium, no measur-
able changes are taking place.

EFFECT OF TEMPERATURE

The vapor pressure of water at 20°C is 17.5 mm.
At 40°C, the vapor pressure is 55.3 mm; at 60°C,
it is 149.4 mm. The vapor pressure of water in-
creases with increasing temperature.

Ethyl alcohol is also a liquid at room tempera-
ture. Its vapor pressure at 20°C is 44 mm, higher
than the vapor pressure of water at this same
temperature. At 40°C, ethyl alcohol has a vapor
pressure of 134 mm; at 60°C, the vapor pressure
is 352 mm. Again we find that the vapor pressure
increases rapidly with increasing temperature.
This is always so. The vapor pressure of every
liquid increases as the temperature is raised.

THE BOILING POINT

At any temperature, molecules can escape from
the surface of a liquid (vaporizing or evaporat-
ing) to enter the gas phase as vapor. At the spe-
cial temperature at which the vapor pressure just
equals the atmospheric pressure, a new phe-
nomenon occurs. There, bubbles of vapor can
form anywhere within the liquid. At this tempera-
ture, the liquid boils.

We see that the boiling point is fixed by the
surrounding pressure. For example, if the sur-
rounding pressure is 760 mm, water boils at
100°C. This is the temperature at which the
vapor pressure of water is just 760 mm. Ethyl
alcohol, having a higher vapor pressure, achieves
a vapor pressure of 760 mm at 78.5°C. Ethyl
alcohol boils at 78.5°C with this surrounding
pressure. Suppose, however, that the atmos-
pheric pressure drops to 750 mm (as it might just
before a storm). Then bubbles of vapor could
form anywhere in liquid water at a temperature
of 99.6°C since the vapor pressure of water is
750 mm at 99.6°C. Water boils at 99.6°C when
the surrounding pressure is 750 mm.

LIQUIDS AND SOLIDS: CONDENSED PH-ASES OF MATTER I CHAP. 5

The normal boiling point of a liquid is de- exercise 5-3

fined as the temperature at which the vapor w^ is the normal boiling point of ethyl al-

pressure of that liquid is exactly one standard cohol?
atmosphere, 760 mm Hg.

EXERCISE 5-4

Suppose a closed flask containing liquid water is
Fig. 5-2. Vapor pressure increases as temperature is connected t0 a vacuum pump and the pressure
raised

over the liquid is gradually lowered. If the water
temperature is kept at 20°C, at what pressure
will the water boil?

EXERCISE 5-5

Answer Exercise 5-4, substituting ethyl alcohol
for water.

5-1.3 Solid-Liquid Phase Changes

Solids and liquids are called condensed phases.
The attractive forces in a condensed phase, either
a solid or a liquid, tend to hold the molecules
close together. In liquids, molecules are irregu-
larly spaced and randomly oriented. In a crystal-
line solid, the molecules occupy regular posi-
tions, resulting in additional stability (relative to
the liquid).

The difference between the energy of a sub-
stance in liquid form and its energy in solid form
is usually much smaller than the difference be-
tween the energies of the liquid and gaseous
forms. For example, consider the heat of melting
a mole of ice,

H20(solid)
or, in abbreviation,

H20(liquid)

U20(l)

(2)

Liquid

in ice. hath., O C

The heat accompanying the phase change (2)
is 1.44 kcal/mole. This is much less than the mo-
lar heat of vaporization of water, 10 kcal/mole.
Table 5-II contrasts the melting points and the
heats of melting per mole (the molar heat of
melting, or the molar heat of fusion) of the same
pure substances listed in Table 5-1.

Once again, we find an extreme range among
the properties of these substances. The molar
heats of melting vary from 0.080 kcal/mole for

SEC. 5-2 I SOLUTIONS

Table 5-1 1, the melting points and heats of melting of some pure

SUBSTANCES

SUBSTANCE

neon

chlorine

water

sodium

sodium chloride

copper

PHASE CHANGE

(solid) — >■ (liquid)

MELTING POINT
°K °C

MOLAR HEAT
OF MELTING

(kcal/mole)

0.080

1.53

1.44

0.63

6.8

3.11

neon to 6.8 kcal/mole for the substance sodium very great differences in the forces that bind

chloride— a change by a factor of 85. There are these solids. Since these differences affect prop-

erties other than melting point and heat of melt-

Fig. 5-3. Melting of ice. ~ "»*' ^ are imPortant to a chemist.

+ i.+

5-2 SOLUTIONS

Sodium chloride, sugar, ethyl alcohol, and water
are four pure substances. Each is characterized
by definite properties, such as vapor pressure,
melting point, boiling point, density. Suppose we
mix some of these pure substances. Sodium chlo-
ride dissolves when placed in contact with water.
The solid disappears, becoming part of the liq-
uid. Likewise, sugar in contact with water dis-
solves. When ethyl alcohol is added to water, the
two pure substances mix to give a liquid similar
in appearance to the original liquids. The salt-
water mixture, the sugar-water mixture, and the

alcohol-water mixtures are called solutions. Solu-
tions differ from pure substances in that their
properties vary, depending upon the relative
amounts of the constituents. The behavior of
solutions during phase changes is dramatically
different from that just described for pure sub-
stances. These differences provide, at once, rea-
son for making a distinction between pure
substances and solutions and, as well, a basis for
deciding whether a given material is a pure sub-
stance or a solution.

LIQUIDS AND SOLIDS: CONDENSED PHASES OF MATTER I CHAP. 5

5-2.1 Differentiating Between Pure
Substances and Solutions

The earth has many unlike parts — it is heteroge-
neous. Some of the parts are uniform through-
out; that is, they are homogeneous. Familiar
examples of heterogeneous materials are granite
(which consists of various minerals suspended in
another mineral), oil and vinegar salad dressing
(which consists of droplets of oil suspended in
aqueous acetic acid), and black smoke (which
consists of particles of soot suspended in air).
Examples of homogeneous materials are dia-
mond, fresh water, salt water, and clear air.
Heterogeneous materials are hard to describe
and classify but we can describe homogeneous
materials rather precisely.

Both pure substances and solutions are ho-
mogenous. A homogeneous material that contains
only one substance is called a pure substance.
A solution is a homogeneous material that con-
tains more than one substance.

We have used the terms gas phase, liquid
phase, and solid phase. A phase is a homogene-
ous part of a system — a part which is uniform
and alike throughout. A system in turn, is any
region and the material in i* that we wish to
consider. It may include oniy one, or more than
one, phase.

Suppose we compare two liquid samples, one
of distilled water, and one of salt water. Each
sample is a homogeneous system consisting of a
single phase. However, one of the liquids is a
pure substance whereas the other is a solution.
We cannot tell, merely by visual observation,
which of these clear liquids is the pure substance
and which is the solution. True, there are dif-
ferences— for example, the salt water has a
greater density than the pure water — but even
this property does not indicate which is the pure
substance.

Let us compare the behavior of these two sys-
tems during a phase change. Consider, first, how
water acts when it is frozen or vaporized. Pure
water freezes at a fixed temperature, 0°C. If we
freeze half of a water sample to ice, remove the
ice, melt it in another container, and compare
the separate samples, we find that the two frac-
tions of the original sample are indistinguishable.

In a similar way, if we boil a sample of water
until half of it has changed to steam, condense
the steam to water in a different vessel, and then
compare the separate samples, we find that the
fractions of the original sample are indistin-
guishable. Such behavior on boiling (condensing)
or freezing (melting) characterizes pure sub-
stances. Solutions behave differently.

Suppose we boil off part of a sample of salt
water. The temperature of the liquid rises, as
shown in curve b of Figure 5-4, until boiling

Boiling begins

x — * — x — x — x — x — x — x — x — x — x — x — x — x — x-

Boiling begins

• Salt water
x Pure water

Time

Fig. 5-4. Behavior on boiling, (a) A pure substance,
(b) A solution.

begins. Already, a difference from the behavior
of pure water can be noted, shown in curve a
of Figure 5-4: the boiling point of the salt solu-
tion is higher. As boiling continues, the tempera-
ture of the pure water remains constant whereas
the temperature of the salt solution rises. As the
boiling point goes up, the remaining liquid be-
comes saltier. If we collect the steam from the
salt solution and condense it in a separate vessel,
we find that the resulting liquid behaves like pure
water rather than like the solution from which it
came. If we boil off all the water, solid salt
remains behind. Thus, by distilling — that is,
evaporating and recondensing in a separate vessel —
we can separate a pure liquid from a solution;
and, by crystallizing — that is, forming a crystal-
line solid — we can obtain a pure solid from a
solution. Chemists call the pure liquid obtained
by distilling and the pure solid obtained by crys-
tallizing, the components of the solution. In our

SEC. 5-2 I SOLUTIONS

Condenser

Cold water inlet-

Pure
■water

Fig. 5-5. A simple distillation apparatus.

experiment shown in Figure 5-4, the components
are salt and water.

Pure sodium chloride, like pure water, has a
definite melting (freezing) temperature (at a
given pressure). Separating operations — such as
distilling or freezing — do not separate the salt
into components. The composition of the salt,
whether expressed in relative numbers of sodium
and chlorine atoms or in the relative weights of
these atoms, is fixed and is represented by the
formula NaCl. Sodium chloride, like water, is an
example of a pure substance.

On the other hand, operations such as distill-
ing or freezing usually tend to separate solutions
into the pure substances that were the compo-
nents of the solution. The nearer alike the com-
ponents are, the harder it is to separate them
from the solution, but even in difficult cases, a
variety of methods in succession usually brings
about a separation. In nature, solutions are much
more common than pure substances, and hetero-
geneous systems are more common than solu-
tions. When we want pure substances, we often
must prepare them from solutions through suc-
cessive phase changes.

We are all familiar with liquid solutions. Gas
and solid solutions also exist. We shall consider
them briefly and then return to liquid solutions,
the most important from a chemist's point of
view.

5-2.2 Gaseous Solutions

All gas mixtures are homogeneous; hence all gas
mixtures are solutions. Air is an example. There
is only one phase — the gas phase — and all the
molecules, regardless of the source, behave as
gas molecules. The molecules themselves may
have come from gaseous substances, liquid sub-
stances, or solid substances. Whatever the source
of the constituents, this gaseous solution, air, is
a single, homogeneous phase. As with other
solutions, the constituents of air are separated by
phase changes.

5-2.3 Solid Solutions

Solid solutions are more rare. Crystals are stable
because of the regularity of the positioning of
the atoms. A foreign atom interferes with this
regularity and hence with the crystal stability.
Therefore, as a crystal forms, it tends to exclude
foreign atoms. That is why crystallization pro-
vides a good method for purification.

But in metals it is relatively common for solid
solutions to form. The atoms of one element may
enter the crystal of another element if their atoms
are of similar size. Gold and copper form such
solid solutions. The gold atoms can replace cop-
per atoms in the copper crystal and, in the same
way, copper atoms can replace gold atoms in the
gold crystal. Such solid solutions are called al-
loys. Some solid metals dissolve hydrogen or
carbon atoms — steel is iron containing a small
amount of dissolved carbon.

Solid solutions, alloys in particular, will be
considered again in Chapter 17.

5-2.4 Liquid Solutions

In your laboratory work you will deal mostly
with liquid solutions. Liquid solutions can be
made by mixing two liquids (for example, alcohol
and water), by dissolving a gas in a liquid (for
example, carbon dioxide and water), or by dis-
solving a solid in a liquid (for example, sugar and
water). The result is a homogeneous system con-
taining more than one substance — a solution.
In such a liquid, each component is diluted by
the other component. In salt water, the salt

LIQUIDS AND SOLIDS: CONDENSED PHASES OF MATTER I CHAP. 5

dilutes the water and, of course, the water dilutes
the salt. This solution is only partly made up of
water molecules and it is found that the vapor
pressure of the solution is correspondingly lower
than the vapor pressure of pure water. Whereas
water must be heated to 100°C to raise the vapor
pressure to 760 mm, it is necessary to heat a salt
solution above 100°C to reach this vapor pres-
sure. Therefore, the boiling point of salt water is
above the boiling point of pure water. The
amount the boiling point is raised depends upon
the relative amounts of water and salt. The more
salt that is added, the higher is the boiling point.

In a similar way, a lower temperature is re-
quired to crystallize ice from salt water or from
an alcohol-water solution than from pure water.
"Antifreeze" substances added to an automobile
radiator act on this principle. They dilute the
water in the radiator and lower the temperature
at which ice can crystallize from the solution.
Again, the amount the freezing temperature is
lowered depends upon the relative amounts of
water and antifreeze compound.

In general, the properties of a solution depend
upon the relative amounts of the components.
It is important to be able to specify quantita-
tively what is present in a solution, that is, to
specify its composition. There are many ways to
do this, but one method will suffice for our
purposes.

for various purposes. We shall use only one in
this course.

Chemists often indicate the concentration of
a substance in water solution in terms of the
number of moles of the substance dissolved per
liter of solution. This is called the molar con-
centration. A one-molar solution (1 M) contains
one mole of the solute per liter of total solution,
a two-molar solution (2 M) contains two moles
of solute per liter, and a 0.1 -molar solution
(0.1 M) contains one-tenth mole of solute per
liter. Notice that the concentration of water is
not specified, though we must add definite
amounts of water to make the solutions.

We can make a 1 M solution of sodium chlo-
ride by weighing out one mole of the salt. From
the formula, NaCl, we know that one mole
weighs 58.5 grams (23.0 grams + 35.5 grams).
We dissolve this salt in some water in a 1 liter
volumetric flask — a flask holding just 1 liter
when filled exactly to an etched mark. After the
salt dissolves, more water is added until the water
level reaches the etched mark to make the vol-
ume exactly 1 liter. Equally well, we can prepare
a 1 M sodium chloride solution using a 100 ml
volumetric flask. Then the final volume of the
solution will be 0.100 liter and we need only one-
tenth mole of salt. In this case, we weigh out
5.85 grams of salt, place it in the flask, dissolve
it, and add water to the 100 ml mark.

5-2.5 Expressing the Composition off Solutions 5-2.6 Solubility

The components of a solution are the pure sub-
stances that are mixed to form the solution. If
there are two components, one is sometimes
called the solvent and the other the solute. These
are merely terms of convenience. Since both must
intermingle to form the final solution, we cannot
make any important distinction between them.
When chemists make a liquid solution from a
pure liquid and a solid, they usually call the
liquid component the solvent.

To indicate the composition of a particular
solution we must show the relative amounts as
well as the kind of componenis. These relative
amounts chemists call concentrations. Chemists
use different ways of expressing concentration

When solid is added to a liquid, solid begins to
dissolve and the concentration of dissolved ma-
terial begins to rise. After all of the solid has
dissolved, the concentration remains constant,
fixed by the amount of dissolved solid and the
volume of the solution If more solid is now
added, the concentration will rise further. Fi-
nally, however, the addition of more solid no
longer raises the concentration of dissolved ma-
terial. When a fixed amount of liquid has dis-
solved all of the solid that it can, the concentra-
tion reached is called the solubility of that solid.
A solution in contact with excess solid is said to
be saturated.
The solubilities of solids in liquids vary widely.

SEC. 5-2 I SOLUTIONS

For example, sodium chloride continues to dis-
solve in water at 20°C until the concentration
is about six moles per liter. The solubility of
NaCl in water is 6 M at 20°C. In contrast, only
a minute amount of sodium chloride dissolves in
ethyl alcohol at 20°C. This solubility is 0.009 M.
Even in a single liquid, solubilities differ over
wide limits. The solids calcium chloride, CaCl2,
and silver nitrate, AgN03, have solubilities in
water exceeding one mole per liter. The solid
called silver chloride, AgCl, has a solubility in
water of only 10-5 mole per liter.

Because of this range of solubilities, the word
soluble does not have a precise meaning. There
is an upper limit to the solubility of even the
most soluble solid, and even the least soluble
solid furnishes a few dissolved particles per liter
of solution. If a compound has a solubility of
more than one-tenth mole per liter (0.1 M),
chemists usually say it is soluble. When the solu-
bility lies below 0.1 M (10-1 M), chemists usu-
ally say the compound is slightly soluble. Com-
pounds with solubility below about 10-3 M are
sometimes said to be very slightly soluble, and
if the solubility is so low as to be of no interest,
the solid is said to have negligible solubility. We
use glass containers for pure water because glass
has a negligible solubility in water.

5-2.7 Variations Among the Properties
off Solutions

Though many solutions are colorless and closely
resemble pure water in appearance, the differ-
ences among solutions are great. This can be
demonstrated with the five pure substances, so-
dium chloride (salt), iodine, sugar, ethyl alcohol,
and water. Two of these substances, ethyl alcohol
and water, are liquids at room temperature.
Let's investigate the properties of the solutions
these two substances form.

First we can investigate, qualitatively, the ex-
tent to which the solids dissolve in the liquids.
By adding a small piece of each solid to a milli-
liter of liquid, we easily discover that sugar dis-
solves both in water and ethyl alcohol, sodium
chloride dissolves readily in water but not in
ethyl alcohol, and iodine does not dissolve much

Fig. 5-6. Salt water readily conducts electricity; sugar
solution does not.

in water but dissolves readily in ethyl alcohol.
Thus we see that the solvent properties of the
two liquids are quite distinctive, at least as far
as sugar, salt, and iodine are concerned.

The experiment just described gives us four
solutions containing a substantial amount of
solute:

LIQUIDS AND SOLIDS: CONDENSED PHASES OF MATTER I CHAP. 5

Sugar in
water

Sugar in
ethyl
alcohol

III

Sodium
chloride
in water

Iodine in
ethyl
alcohol

Of these four solutions, IV is readily distin-
guished. This solution has a dark brown color.
The other three, I, II, and III, are colorless. They
can be easily distinguished by taste but chemists
have safer and more meaningful ways of distin-
guishing them. These solutions differ markedly
in their ability to conduct an electric current. The
two sugar solutions, I and II, have virtually the
same conductivity properties as the pure liquids
— they do not conduct electric current readily.

Solution III conducts electric current much more
readily than does pure water.

Thus we find great variation among solutions.
Iodine dissolves in ethyl alcohol, coloring the
liquid brown, but does not dissolve readily in
water. Sodium chloride does not dissolve readily
in ethyl alcohol but does dissolve in water, form-
ing a solution that conducts electric current.
Sugar dissolves readily both in ethyl alcohol and
in water, but neither solution conducts electric
current. These differences are very important to
the chemist, and variations in electrical con-
ductivity are among the most important. We
shall investigate electrical conductivity further
but, first, we need to explore the electrical nature
of matter.

5-3 ELECTRICAL NATURE OF MATTER

We have mentioned the electrical conductivity of
solutions as a means of distinguishing solutions.
The interest of a chemist in the electrical nature
of matter goes far deeper than this. We shall find
that an understanding of electrical behavior fur-
nishes a key to the explanation of chemical prop-
erties. We shall find that electrical effects aid us
in predicting molecular formulas, in explaining
chemical reactions, and in understanding energy
changes that accompany them.

5-3.1 Electrical Phenomena

Try to name several electrical phenomena that
you have often observed. Before you read on,
see if you can name five. Does your list include
the following?

(1) The attraction of a comb for your hair on a
dry day.

(2) The flash of a bolt of lightning.

(3) The shock you get if you touch a bare wire
in a radio set.

(4) The heat generated by an electric current
passing through the heating element of an
electric stove.

(5) The light emitted by the filament of a light
bulb as electric current is passed through it.

(6) The magnetic field generated by a current
passing through a coil of wire.

(7) The work done by an electric motor when
electric current passes through its coils.

(8) The emission of "radio waves" by the an-
tenna of a radio or television station.

We see electrical devices all around us — fur-
nishing power, light, means of communication —
influencing every facet of life. What does it mean
to say that an electric current "passes through"
a coil of wire? What is an electric current? To
begin answering these questions, we must explore
an electrometer, a device for detecting and meas-
uring electric charge.

5-3.2 Detection of Electric Charge

Figure 5-7 shows a simple electrometer. It con-
sists of two spheres of very light weight, each
coated with a thin film of metal. The spheres are
suspended near each other by fine metal threads
in a closed box to exclude air drafts. Each sus-
pending thread is connected to a brass terminal.
Next to the box is a "battery"— a collection of
electrochemical cells. There are two terminal
posts on the battery. We shall call these posts Pi
and Pi. If post Pi is connected by a copper wire

SEC. 5-3 ELECTRICAL NATURE OF MATTER

Fig. 5-7. A simple electrometer.

to the left terminal of the electrometer and post
Pi is connected to the right terminal, we observe
that the two spheres move toward each other.
Evidently the wires have transmitted to the
spheres the property of exerting force on each
other — an attractive force. The force is still pres-
ent when the air in the electrometer is removed
with a vacuum pump. The spheres react to each
other "across space." They feel "force at a dis-
tance."

If now the wires are disconnected, the attrac-
tive force remains. However, if the two elec-
trometer terminals are connected by a copper
wire the spheres return to their original positions
and hang vertically again. The attraction is lost.

We see that the battery transfers to the elec-
trometer spheres the property of attracting each
other. It is natural to imagine that something has
been transferred from the battery to the spheres.
This "something" is called electric charge. The
movement of this electric charge from the battery
through the metal threads to the spheres is called
an electric current. This electric charge is lost
when the two electrometer terminals are con-
nected by a copper wire.

We can learn more about electric charge by
another use of the electrometer. Connect one
wire from the battery (say, post P{) to the base
of the electrometer and the other wire (from post
P2) to both terminals, as in Figure 5-8.

This time the two spheres move apart — they
repel each other! When both spheres are given
electric charge from the battery post labeled P2,
they repel instead of attract. In Figure 5-7 we

saw that when one sphere received charge from
post A and the second sphere received charge
from post Pi, the two spheres attracted. There
must be at least two kinds of charge!

Now let us reverse the wires so that both
spheres are charged from battery post Pi. This
time, Pi is connected to the base of the electrome-
ter. Again we observe that the spheres move
apart. Whenever both spheres are connected to
the same battery post, the two spheres repel each
other.

This represents a large gain in our knowledge
of electric charge. One kind of charge comes
from battery post Pi. Until we have reason to do
otherwise, we shall call this kind of charge C\.
The other kind of charge comes from battery
post P2; we shall call this kind of charge d. The
spheres attract or repel each other when they
carry charge according to the following pattern:

C\ attracts d
C\ repels G
C2 repels C2

unlike charges attract
like charges repel
like charges repel

(i)

Fig. 5-8. The electrometer with both spheres con-
nected to battery post Pt.

LIQUIDS AND SOLIDS! CONDENSED PHASES OF MATTER | CHAP. 5

We have one more observation to symbolize.
When the spheres were given different charges
(as in Figure 5-7), the charges could be removed
by connecting the two terminals by a copper
wire. Then the spheres lost all attraction for each
other. We interpret this behavior to mean that
either G or C2 (or both) has moved through the
wire so as to join the other kind of charge. When
C\ and C2 are united, no charge remains. Sym-
bolically we can say,

Ci + Ci = no charge (4)

Our accumulated evidence shows that there
are at least two kinds of electric charge, which
we have symbolized G and C2. These two kinds
of charge possess the properties (3) and (4). We
may wonder if there are other kinds of charge.
This is answered by looking into other ways of
producing electric charges. Chemists can assem-
ble a variety of types of electrochemical cells
which show the same electrometer behavior
just described. Some frictional processes leave
charges on the two surfaces rubbed together. The
attraction of a comb for your hair is caused by
charges left on the comb as it rubbed against
your hair. Many decades ago the properties of
electric charge were investigated as they are pro-
duced by rubbing a hard rubber rod with cat fur.
The hard rubber is found to carry charge G, and
the cat fur carries charge C2. If a glass rod is
rubbed with silk, the glass rod is left with charge
C2 and the silk with charge G.

No matter how electric charge is produced, we
always find these same two types, G and C2, and

only these two. Any method of producing G also
produces an equivalent amount of C2. We con-
clude there are two and only two types of electric
charge.

5-3.3 The Effect of Distance

Figure 5-9 shows two electrometers which differ
in the spacing of the spheres. Though the charges
on the spheres come from the same battery, there
is more deflection of the spheres when they are
closely spaced (left) than when they are widely
spaced. When the spheres are closer together, the
deflection is larger. Hence, we conclude that the
force of attraction varies with distance and is
stronger when the charges are close to each
other. Careful quantitative studies show that the
force is inversely proportional to the square of
the distance r between the two spheres :

(Electric force) is proportional to —

(5)

where

r = distance between centers of the two spheres.

5-3.4 The Electron-Proton Model

These new facts about electrical phenomena can
be incorporated into our particle model of the
structure of matter if we again allow some

Fig.

5-9. Contrast of deflections in two electrometers
with different distances between spheres.

SEC. 5-3 I ELECTRICAL NATURE OF MATTER

growth of the model. The new idea is that matter
is made up of particles which carry the property
called electric charge. To be specific, we propose
that in atoms there are two kinds of particles
that carry unit charge, one which carries one
portion, or unit, of charge G and one which
carries one portion, or unit, of charge C2. These
particles are called electrons and protons.

Proposal: Matter includes particles, each of

which carries a unit of electric

charge.
Electrons: Each electron carries one

unit of charge G-
Protons: Each proton carries one unit

of charge G-

These particles exert force at a distance on
each other in accordance with the electrical be-
havior we have observed.

Since:

G repels G, electrons repel electrons;

G repels G, protons repel protons ;

G attracts C2, electrons attract protons;

G + C2 = no charge, one electron + one proton

= no charge;
or, one unit G + one unit C,

= no charge.

The atomic model now can cope with the facts
we have learned about electrical behavior. If a
piece of matter (such as one of the electrometer
spheres) has the same number of electrons and
protons, there are just as many units of charge
of type G as of type C2. Since G + C2 = no
charge, the sphere will have no charge. A body
with no net charge (with equal numbers of pro-
tons and electrons) is said to be electrically
neutral. If we remove some of the electrons from
the sphere, it will then have an excess of protons,
hence a net charge of type C2. If we add an excess
of electrons to the sphere, it will have a net
charge of type Q. The amount of net charge is the
difference between the amount of charge G and
charge C2.

It is a mathematical convenience if we express
the net charge in terms of algebraic symbols.
Henceforth we shall identify the type of charge
called G as "negative charge" and the type

called G as "positive charge." Notice the ad-
vantages.

The combination of 5 units of G and 3 units
of G leaves a net of 2 units of G- This now can
be expressed

5(-l) + 3( + l) = -5 + 3 = -2

EXERCISE 5-6

Suppose ten protons and eleven electrons are
brought together. These charges, grouped to-
gether, have the same net charge as how many
electrons? Remember that one proton plus one
electron gives no charge.

EXERCISE 5-7

Write an algebraic expression to obtain the re-
sult of Exercise 5-6, using numbers with algebraic
signs to represent charges.

5-3.5 Electric Force: A Fundamental
Property of Matter

We have learned that a battery can transfer to
the spheres of an electrometer a property called
electric charge. When this happens, the spheres
exert force on each other. The discussion brings
up two "wondering why" questions. The first is,
"Why do the electric charges appear?" What
caused the battery to transfer to the electrometer
the property called electric charge? We shall
examine this question carefully later in the course
because the subject of the operation of an elec-
trochemical cell is extremely important in chem-
istry. It is the topic of an entire chapter in this
book (Chapter 12). For the moment, all we can
say is that the electric charge did come from the
battery of electrochemical cells, thus indicating
that the matter within the cells contains electric
charges.

The second question probes deeper: "Why do
the two electrometer spheres, when charged, ex-
ert force on each other?" What is our explana-
tion of this phenomenon? We say that the
spheres have an excess of electrons (or protons)
and these electrons (or protons) exert force on

LIQUIDS AND SOLIDS: CONDENSED PHASES OF MATTER I CHAP. 5

each other. This does not really explain electric
force at a distance. We are left with the equiva-
lent questions, "Why do two electrons (or two
protons) repel each other? Why does an electron
attract a proton?" Without an answer, we say,
"It is a fundamental property of matter that it
can acquire electric charge and, when it does so,
it exerts force on other charged bodies." Such a
statement may be taken as a definition of a
"fundamental property" — a property which is
generally observed but for which diligent search
has failed to yield a useful model. Without an
explanation of a property, we call the property
"fundamental." It is a curious fact that after a
property has resisted explanation for quite a time
and it becomes classified as a fundamental prop-

erty, an explanation no longer seems to be neces-
sary.

EXERCISE 5-8

There was a time when atoms were said to be
fundamental particles of which matter is com-
posed. Now we describe the structure of the
atom in terms of the fundamental particles we
have just named, protons and electrons, plus
another kind of particle called a neutron. Why
are atoms no longer said to be fundamental par-
ticles? Do you expect neutrons, protons, and
electrons always to be called fundamental par-
ticles?

5-4 ELECTRICAL PROPERTIES OF CONDENSED PHASES

Now we are ready to investigate behavior of
condensed phases that shows evidence of the
presence and movement of electric charge. We
have already referred to one of the most impor-
tant examples — the movement of electric current
through water solutions.

5-4.1 The Electrical Conductivity
off Water Solutions

The movement of electric charge is called an
electric current. Hence when we say electric cur-
rent flows through a salt solution, we mean there
is a movement of electric charge through the
solution. We shall be concerned here with the
manner in which this charge moves.

Water is a very poor conductor of electricity.
Yet when sodium chloride dissolves in water, the
solution conducts readily. The dissolved sodium
chloride must be responsible. How does the dis-
solved salt permit charge to move through the
liquid? One possibility is that when salt dissolves
in water, particles with electric charge are pro-
duced. The movement of these charged particles
through the solution accounts for the current.
Salt has the formula, NaCl — for every sodium
atom there is one chlorine atom. Chemists have

decided that when sodium chloride dissolves in
water, the charged species present are chlorine
atoms, each carrying the negative charge of one
electron, and sodium atoms, each carrying the
positive charge of one proton. We symbolize a
chlorine atom with a negative charge as Cl~.
A sodium atom with a positive charge is sym-
bolized Na+. Atoms or molecules that carry elec-
tric charge are called ions.

With these symbols, we can write the equation
for the reaction that occurs when sodium chlo-
ride dissolves in water

NaCl(solid) + water — ►-

Na^(in water) + Cl~(in water) (6)

Equation (6) shows that when water dissolves
solid sodium chloride, Na+ ions (sodium ions)
and Cl~ ions (chloride ions) are present in the
solution. Chemists usually abbreviate this equa-
tion as much as is consistent with retaining es-
sential information. On the left of equation (<5),
the term "water" is usually not written since its
presence is implied by the symbols on the right.

NaCl(solid) — >■

Na+(in water) + Cl_(in water) (6)

We have already seen that NaCl(solid) is usually
written NaClfsj. There is a similar abbreviation

SEC. 5-4 | ELECTRICAL PROPERTIES OF CONDENSED PHASES

for "in water." To represent this, the expression
"in water" is replaced by the term "aqueous,"
commonly abbreviated "aq". * Thus equation
(6), showing the reaction of sodium chloride dis-
solving in water to form a conducting solution,
is usually written in the form:

NaClfsj — ►- Na+(aq) + C\~(aq)

(6)

Now we have a model of a salt solution that
aids us in discussing electrical conductivity. The
solid dissolves, forming the charged particles
Na.+(aq) and C\~(aq), which can move about
in the solution independently. An electric current
can pass through the solution by means of the
movement of these ions. The C\~(aq) ions move
in one direction, causing negative charge to move
that way. The Na+(aq) ions move in the oppo-
site direction, causing positive charge to move
this way. These movements carry charge through
the solution and current flows.

Sugar dissolves in water, but the resulting solu-
tion conducts electric current no better than does
pure water. We conclude that when sugar dis-
solves, no charged particles result: no ions are
formed. Sugar must be quite different from so-
dium chloride.

Calcium chloride, CaCl2, is another crystalline
solid that dissolves readily in water. The resulting
solution conducts electric current, as does the
sodium chloride solution. Calcium chloride is, in
this regard, like sodium chloride and unlike
sugar. The equation for the reaction is

CaCVsj

Ca+*(aq) + 2C\-(aq) (7)

Equation (7) shows that when calcium chlo-
ride dissolves, ions are present — Ca+2(aq) and
C\~(aq) ions. In this case, each calcium ion has
the positive charge of two protons. Therefore it
has twice the positive charge held by a sodium
ion, Na.+(aq). The chloride ion that forms,
C\~(aq), is the same negative ion that is present
in the sodium chloride solution, though it comes
from the calcium chloride solid instead of the
sodium chloride. Because both CaCh(s) and

NaClfsJ dissolve in water to form aqueous ions,
they are considered to be similar.

Silver nitrate, AgN03, is a third solid sub-
stance that dissolves in water to give a conduct-
ing solution. The reaction is

AgN03(s) — >- Ag+(aq) + NOs- (aq) (8)

This time the ions formed are silver ions,
Ag+(aq), and nitrate ions, N03~(aqj. The aque-
ous silver ion is a silver atom with the positive
charge of a proton ; it carries the same charge as
does an aqueous sodium ion. The aqueous ni-
trate ion carries the negative charge of an elec-
tron— the same charge carried by the aqueous
chloride ion. This time, however, the negative
charge is carried by four atoms, a nitrogen and
three oxygen atoms, that remain together. Since
this group, N03~, remains together and acts as
a unit, it has a distinctive name, nitrate ion.

These three solids, sodium chloride, calcium
chloride, and silver nitrate are similar, hence
they are classified together. They all dissolve in
water to form aqueous ions and give conducting
solutions. These solids are called ionic solids.

The ease with which an aqueous salt solution
conducts electric current is determined by how
much salt is dissolved in the water, as well as by
the fact that ions are formed. A solution con-
taining 0.1 moles per liter conducts much more
readily than a solution containing 0.01 moles per
liter. Thus the conductivity is determined by the
concentration of ions, as well as by their pres-
ence.

Silver chloride is a solid that shows this effect.
This solid does not dissolve readily in water.
When solid silver chloride is placed in water,
very little solid enters the solution and there is
only a very slight increase in the conductivity of
the solution. Yet there is a real and measurable
increase — ions are formed. Careful measure-
ments show that even though silver chloride is
much less soluble in water than sodium chloride,
it is like sodium chloride in that all the solid that
does dissolve forms aqueous ions. The reaction is

AgC\(s)

Ag+(aq) + Cl-(aq)

(9)

* The adjective, aqueous, comes from the Latin name
for water, aqua.

Silver chloride, like sodium chloride, is an ionic
solid.

LIQUIDS AND SOLIDS: CONDENSED PHASES OF MATTER I CHAP. 5

5-4.2 Precipitation Reactions
in Aqueous Solutions

Though both silver nitrate and sodium chloride
have high solubility in water, silver chloride is
very slightly soluble. What will happen if we mix
a solution of silver nitrate and sodium chloride?
Then, we will have a solution that includes the
species present in a solution of silver chloride,
Ag+(aq) and C\~(aq), but now they are present
at high concentration! The Ag+( aq) came from
reaction (8) and the C\~(aq) came from reaction
(6) and their concentrations far exceed the solu-
bility of silver chloride. The result is that solid
will be formed. The formation of solid from a
solution is called precipitation:

Ag+(aq) + C\-(aq) — > AgClfsJ (70)

Notice that reaction (70) indicates the change
that takes place when silver nitrate solutions and
sodium chloride solutions are mixed. We could
have written a more complete equation:

Ag+(aq) + NOz(aq) + Na+(aq) + C\~(aq)

—^ AgC\(s) + NtV (aq) + Na+(a<jj

However, the two ions N03~fagj and Na+(aq)
do not play an active role in the reaction, nor do
they influence the reaction that does occur [re-
action (70)]. Consequently, they are not included
in the equation for the reaction. The balanced
chemical equation should show only species which
actually participate in the reaction. These species
are called the predominant reacting species.

Equations (6), (7), (8), (9), and (70) involve
charged species, ions. When we considered how
to balance equations for chemical reactions (Sec-
tion 3-2.1), we dealt with reactions involving elec-
trically neutral particles. We were guided by the
rule that atoms are conserved. This principle is
still applicable to reactions involving ions. In
addition, we must consider the charge balance.
A chemical reaction does not produce or con-
sume electric charge. Consequently, the sum of
the electric charges among the reactants must be
the same as the sum of the electric charges among
the products. In reaction (7), calcium chloride
dissolves to give aqueous Ca+2 and Cl~ ions. The
balanced equation tells us that the neutral solid
calcium chloride dissolves to give one Ca^2 ion

for every two CI- ions. Summing these electric
charges,

(2+) +2 (1-)

0 (77)

In a balanced equation for a chemical reaction,
charge is conserved.

EXERCISE 5-9

Balance the equations for the reactions given
below. For each balanced equation, sum up the
charges of the reactants and compare to the sum
of the charges of the products.

(a) PbCUfs) —

(b) K2Cr207fsJ

?b+*(aq) + C\-(aq)

K+(aq) + Cr2Or(aq)

CTO;2(aq) + H+(aq)

5-4.3 The Electrical Conductivity of Solids

We have, in this chapter, encountered a number
of properties of solids. In Table 5-II, we found
that melting points and heats of melting of dif-
ferent solids vary widely. To melt a mole of solid
neon requires only 80 calories of heat, whereas
a mole of solid copper requires over 3000 calo-
ries. Some solids dissolve in water to form con-
ducting solutions (as does sodium chloride),
others dissolve in water but no conductivity
results (as with sugar). Some solids dissolve in
ethyl alcohol but not in water (iodine, for ex-
ample). Solids also range in appearance. There
is little resemblance between a transparent piece
of glass and a lustrous piece of aluminum foil,
nor between a lump of coal and a clear crystal
of sodium chloride.

The great variations among solids make it
desirable to find useful classification schemes.
Though this topic is taken up much later in the
course (Chapter 17), a beginning is provided by
a look at the electrical conductivity of solids.

The high electrical conductivity of a substance
like copper or silver is familiar to all. Conduc-

SEC. 5-4 I ELECTRICAL PROPERTIES OF CONDENSED PHASES

tivity measurements on many other solids show
that all of the substances that conduct electricity
as readily as copper and silver are of similar
appearance. These good conductors could al-
most all be classified visually as metals. The most
distinctive property of metallic substances is high
electrical conductivity.

When we study a solid that does not have the
characteristic lustrous appearance of a metal, we
find that the conductivity is extremely low. This
includes the solids we have called ionic solids:
sodium chloride, sodium nitrate, silver nitrate,
and silver chloride. It includes, as well, the
molecular crystals, such as ice. This solid, shown
in Figure 5-3, is made up of molecules (such as
exist in the gas phase) regularly packed in an
orderly array. These poor conductors differ
widely from the metals in almost every property.
Thus electrical conductivity furnishes the key to
one of the most fundamental classification
schemes for substances.

5-4.4 Ionic Solids

Referring to Tables 5-1 and 5-II, we find that
both sodium chloride and copper have extremely
high melting and boiling points. These two solids
have little else in common. Sodium chloride has
none of the other properties that identify a metal.
It has no luster, rather, it forms a transparent
crystal. It does not conduct electricity nor is it a
good heat conductor. The kind of forces holding
this crystal together must be quite different from
those in metals.

The sodium chloride crystal contains an equal
number of sodium atoms and chlorine atoms,

but they are not present as molecules. On the
basis of much experimental evidence, chemists
have concluded that sodium chloride crystals are
built up of sodium ions, Na+, and chloride ions,
Cl~, rather than of neutral atoms or molecules.
The numbers of Na+ and Cl~ ions must be equal
because the entire crystal is electrically neutral.
Nevertheless, there is electrical attraction be-
tween these oppositely charged particles. This
attraction between positive and negative ions
accounts for the binding in an ionic solid.

To represent the composition of such a solid,
we use the formula NaCl. However, this formula
does not indicate that molecules of sodium chlo-
ride are present — it is not a molecular formula.
It shows the composition by giving the relative
number of each kind of atom present and is
called an empirical formula.

Figure 5-10 shows a representation of the ar-
rangement of the ions in the sodium chloride
crystal. The ions are arranged in layers. A layer
in the interior of a crystal has a similar layer
lying in front of it and a similar layer lying
behind it. These layers are displaced so that a
Cl~ ion lies in front of each Na+ ion and a Cl~
ion lies behind each Na+ ion. Thus, each ion is
surrounded by six oppositely charged ions. We
call this arrangement the sodium chloride ar-
rangement or sodium chloride lattice. Because of
the proximity of the oppositely charged ions in
this arrangement, it is strongly bonded and the
melting point of such a crystal is high.

Fig. 5-10. The packing of ions in an ionic crystal:

sodium chloride.

LIQUIDS AND SOLIDS: CONDENSED PHASES OF MATTER I CHAP. 5

When an ionic solid like sodium chloride is
melted, the molten salt conducts electric current.
The conductivity is like that of an aqueous salt
solution: Na+ and C\~ ions are present. The ex-
tremely high melting temperature (808°C) shows
that a large amount of energy is needed to tear
apart the regular NaCl crystalline arrangement
to free the ions so they can move.

In contrast, solid sodium chloride dissolves
readily in water at room temperature and with-
out a large heat effect. This can only mean that
the water interacts strongly with the ions — so
strongly that aqueous ions are about as stable
as are ions in the crystal. In fact, water interacts

so strongly with ions that some molecular crys-
tals dissolve in water to form conducting solu-
tions. For example, solid hydrogen chloride,
HC\(s), is a molecular crystal similar to the ice
crystal. The solid is made up of HC1 molecules,
not of ions like the ionic sodium chloride. Yet
UC\(s) dissolves in water to form a conducting
solution containing hydrogen ions, H+(aq), and
chloride ions, Cl~(aq). Thus we cannot safely
interpret the conductivity of an aqueous solution
to mean that the solid dissolved was an ionic
solid. We can, however, state the opposite: When
an ionic solid dissolves in water, a conducting
solution is obtained.

QUESTIONS AND PROBLEMS

1. A liquid is heated at its boiling point. Although
energy is used to heat the liquid, its temperature
does not rise. Explain.

2. What is the maximum amount of heat that you
can lose as one gram of water evaporates from
your skin?

3. Note in Table 5-1 the correlation between the
normal boiling point and heat of vaporization
of a number of liquids. Suggest possible reasons
for this regularity.

4. Which would likely cause the more severe burn,
one gram of H20(g) at 100°C or one gram of
H,0(7j at 100°C?

5. Liquids used in rocket fuels are passed over the
outer wall of the combustion chamber before
being fed into the chamber itself. What advan-
tages does this system offer?

6. Which of the following will require more energy ?

(a) Changing a mole of liquid water into gaseous
water.

(b) Decomposing, by electrolysis, one mole of
water.

Explain.

7. Pick the liquid having the higher vapor pressure
from each of the following pairs. Assume all
substances are at room temperature.

10.

(a) Mercury, water.

(b) Gasoline, motor oil.

Explain why the boiling point of water is lower
in Denver, Colorado (altitude, 5,280 feet), than
in Boston, Massachusetts (at sea level).

Both carbon tetrachloride, CC14 (used in dry
cleaning and in some fire extinguishers) and mer-
cury, Hg, are liquids whose vapors are poisonous
to breathe. If CC14 is spilled, the danger can be
removed merely by airing the room overnight
but if mercury is spilled, it is necessary to pick
up the liquid droplets with a "vacuum cleaner"
device. Explain.

Because of its excellent heat conductivity, liquid
sodium has been proposed as a cooling liquid
for use in nuclear power plants.

(a) Over what temperature range could sodium
be used in a cooling system built to operate
at one atmosphere pressure or lower?

(b) How much heat would be absorbed per kilo-
gram of sodium to melt the solid when the
cooling system is put in operation?

Use the data in Tables 5-1 (p. 67) and 5-II (p. 69).

QUESTIONS AND PROBLEMS

11. Water is a commonly used cooling agent in
power plants. Repeat Problem 10 considering
one kilogram of water instead of sodium. Con-
trast the results for these two coolants.

12. How much heat must be removed to freeze an
ice tray full of water at 0°C if the ice tray holds
500 grams of water?

13. List three heterogeneous materials not given in
Section 5-2.1.

14. List three homogeneous materials not given in
Section 5-2.1.

15. Which of the following statements about sea
water is FALSE?

(a) It boils at a higher temperature than pure
water.

(b) It melts at a lower temperature than pure
water.

(d) The melting point falls as the liquid freezes.

(e) The density is the same as that of pure water.

16. How many grams of methanol, CH3OH, must
be added to 2.00 moles of HjO to make a solu-
tion containing equal numbers of H.O and
CH3OH molecules? How many molecules (of all
kinds) does the resulting solution contain?

17. How many grams of ammonium chloride,

NH4C1, are present in 0.30 liter of a 0.40 M

NH4C1 solution?

Answer. 6.4 grams.

18. Write directions for preparing the following
aqueous solutions:

(a) 1.0 liter of 1.0 M lead nitrate, Pb(N03)2,
solution.

(b) 2.0 liters of 0.50 M ammonium chloride,
NH4C1, solution.

19. How many liters of a 0.250 M K2Cr04 solution
contain 38.8 grams of K2Cr04?

20. List three properties of a solution you would
expect to vary as the concentration of the solute
varies.

21. Give two forces other than electric that are felt
at a distance.

22. What would you expect to observe if one elec-
trometer sphere were charged by your hair and
the other by the comb used to comb your hair?

23. Why do scientists claim there are only two kinds
of electric charge?

24. It is known that electric charges attract or repel
each other with a force that is inversely propor-
tional to the square of the distance between
them. If two spheres like those in the electrome-
ter (Figure 5-7) are negatively charged, what
would be the change in the force of repulsion if
the distance between them were increased to four
times the original distance?

25. Why do two electrically neutral objects with
mass attract each other?

26. Each of the following ionic solids dissolves in
water to form conducting solutions. Write equa-
tions for each reaction.

(a) potassium chloride, KC1

(b) sodium nitrate, NaN03

Answer. CaBr/sj — >- Ca+*~(aq) + lBr~(aq).

(d) lithium iodide, Lil

27. A chloride of iron called ferric chloride, FeCl3,
dissolves in water to form a conducting solution
containing ferric ions, Fe+3, and chloride ions,
CI".

(a) Write the equation for this reaction.

(b) If 0.10 mole of FeCl3 is dissolved in water
and diluted to 1.0 liter, what is the concen-
tration of ferric ion and of chloride ion?

Answer. Concentration of Fe+3 = 0.10 M.
Concentration of CI" = 0.30 M.

28. The salt ammonium sulfate, (NH4)2S04, dissolves
in water to form a conducting solution contain-
ing ammonium ions, NH^, and sulfate ions,
SO4"2.

(a) Write the balanced equation for the reaction
when this ionic solid dissolves in water.

(b) Verify the conservation of charge by com-
paring the charge of the reactant to the sum
of the charges of the products.

(c) Suppose 1.32 grams of ammonium sulfate is
dissolved in water and diluted to 0.500 liter.
Calculate the concentrations of NH^" (aq)
and S04" 2(aq).

LIQUIDS AND SOLIDS: CONDENSED PHASES OF MATTER I CHAP. 5

29. 1.00 liter of solution contains 0.100 mole of
ferric chloride, FeCl3, and 0.100 mole of ammo-
nium chloride, NH4C1. Calculate the concentra-
tions of Fe+3, C\~, and NH4+ ions.

Answer. Concentration Fe+3 = 0.100 M,
Concentration NH4+ =0.100 M,
Concentration CI" = 0.400 M.

30. In Experiment 10 you mixed lead nitrate and
sodium iodide. Write an equation for the reac-
tion that occurred. Show only the predominant
reacting species.

31. Write equations for the reactions between aque-
ous bromide ions and:

(a) aqueous lead ions,

(b) aqueous silver ions.

Both lead bromide, PbBr2, and silver bromide,
AgBr, are only slightly soluble.

32. When solutions of barium chloride, BaCl2, and
potassium chromate, K2Cr04, are mixed, the
following reaction occurs:

2K+ (aq) + CxOl2 (aq) + Ba+YaqJ + 2C\~(aq)
— >■ BaCr04fs; + 2K+(aq) + 2Cl-(aq;

(a) Show how charge is conserved.

(b) Rewrite the equation showing predominant
reacting species only.

(c) Suppose 1.00 liter of 0.500 M BaCl2 is mixed
with 1.00 liter of 0.200 A/K2Cr04. Assuming
BaCr04 has negligible solubility, calculate
the concentrations of all ions present when
precipitation stops.

Answer. Concentration K+ = 0.200 M,
Concentration Or = 0.500 M,
Concentration CrO^2 = negligible,
Concentration Ba+2 = 0.150 M.

CHAPTER

Structure of
the Atom and
the Periodic Table

The eighth element, starting from a given one, is a kind of repetition of
the first, like the eighth note of an octave in music.

J. A. R. NEWLANDS, 1864

We have already learned that nature has great
variety. Around us we find gases, liquids, and
solids. To liquefy air, we must cool it to about
— 180°C, far colder than the coldest winter. To
liquefy rock, we must heat it to temperatures
above 1000°C, the climate found in an active
volcano. When we examined chemical reactivity,
we found even more variety. A candle burns
quietly and slowly, once lit, though it does not
react appreciably until lit. Iron also reacts with
oxygen very slowly (it rusts), though not as
slowly as we might like. Hydrogen, by contrast,
reacts explosively with oxygen when it is ignited.
In contrast to the slow reactions of paraffin wax
and iron with oxygen, and the instantaneous re-
action of hydrogen with oxygen, helium gas will
never react with oxygen.

Turning to the atomic view of matter, we find
more than a hundred different elements. Each of
these elements has a kind of atom that is some-
how different from all of the others. With these
100 elements, chemists have prepared about one

and one-half million different compounds, each
having its own special properties. Each year
about 100,000 new compounds are reported.
Again we must deal with great variety.

We have already remarked in Chapter 1 that
"the mere cataloguing of observations is not
science." We could never cope with this great
variety in nature if we did not make use of its
regularities in organizing our knowledge. The
fact that chemists have been able to synthesize
more than a million compounds shows that they
have been successful in this organization. Their
success stems in large part from the regularities
embodied in the periodic table.

The periodic table groups elements with simi-
lar chemistry. It is of great value just as a
correlating device. It is even more powerful when
coupled with an understanding of the structure
of atoms. So, it is appropriate to consider this
topic before examining the relationships that es-
tablish the periodic table.

STRUCTURE OF THE ATOM AND THE PERIODIC TABLE I CHAP. 6

6-1 STRUCTURE OF THE ATOM

Scientists have developed a highly sophisticated
view of the structure of the atom. The currently
accepted model is called "the nuclear atom." We
shall present it without trying to show immedi-
ately all of the experimental evidence that led to
this particular model. Rest assured, though, that
every feature of the nuclear atom picture rests
upon experimental evidence, as we shall see in
Chapter 14.

6-1.1 A Model: The Nuclear Atom

An atom contains electrons and protons. Since
mass is associated with all matter, it is natural
to assume that atoms, which form matter, have
mass. And since any sample of matter occupies
a certain volume, we can also assume that each
atom has volume. Almost all the mass of the
atom is concentrated in a region that is much
smaller than the total volume of the atom. This
region is called the nucleus of the atom. The rest
of the volume of the atom is occupied by electrons.
The nucleus carries a positive electric charge.
The element hydrogen has the lightest atoms,
and the nuclei of these atoms have the smallest
positive charge anyone has observed. Every atom
of hydrogen has one proton in its nucleus. The
charge on the nucleus of an atom of hydrogen

is that of a single proton, 1+ unit of charge.
All other nuclei have positive charges that are
exactly an integer times the proton charge; a
nucleus may have 2+ units, 3+ units, 4+ units,
and so on. Each nucleus contains a definite num-
ber of protons and the charge on the nucleus is
fixed by this number. All atoms of a particular
element have the same nuclear charge. All hydro-
gen atoms have a nuclear charge of 1 + ; all
helium atoms have a nuclear charge of 2+ ; all
lithium atoms have a nuclear charge of 3+ ; and
so on. We shall see that the nuclear charge deter-
mines the chemistry of an atom.

Since the nucleus has positive charge, it at-
tracts electrons (each with negative charge). If a
nucleus attracts the number of electrons just
equal to the nuclear charge, an electrically neu-
tral atom is formed. Consider a nucleus contain-
ing two protons, a helium nucleus. When the
helium atom has two electrons as well (2—
charge), an electrically neutral helium atom
results:

2 protons + 2 electrons = no charge

(2+) + (2-) = 0

Electrons can be removed from or added to a
neutral atom, giving it a net charge. This is how
ions are formed. Thus,

'neutral
helium atom,

2 protons
2 electrons

+ energy

'positive
helium ion,

2 protons
.1 electron

He+

+ (electron)

(/)

and, as another example,

"neutral
fluorine atom,

9 protons
_9 electrons

+ energy

"positive
fluorine ion,

9 protons
.8 electrons _

+ (electron)

(2)

+ energy

F+

(2)

SEC. 6-1 I STRUCTURE OF THE ATOM

Since a positive charge attracts a negative charge,
it is difficult to take the electron away from the
positive helium or fluorine nucleus. Scientists
say "work must be done" or "energy is required"
to form a positive ion from a neutral atom, as in
(7 or (2). "Work" and "energy" are synony-
mous here and they indicate that an external
agent must exert force on the electron to make
it leave the neutral atom. In analogy, the attrac-
tion between electron and nucleus is like a
stretched rubber band connecting the two par-
ticles. Continued force applied to the two parti-
cles can result in the rubber band being stretched
until it breaks, releasing the two particles, but at
the expense of work.

Some neutral atoms can gain electrons, form-
ing negative ions. Thus a neutral fluorine atom
can add an electron to form a negative ion, F~.
This change, for fluorine atoms, does not require
the input of energy; it releases energy:

discovered. The neutron carries no charge; it is
a neutral particle. Its mass is almost identical to
the mass of the proton. Thus the nucleus of the
helium atom must consist of two neutrons and
two protons. Then its charge will be 2+ but its
mass will be four times the mass of the hydrogen
atom.

Now our nuclear model suffices. We can build
up the atoms for all elements. Each atom has a
nucleus consisting of protons and neutrons. The
protons are responsible for all of the nuclear
charge and part of the mass. The neutrons are
responsible for the rest of the mass of the nu-
cleus. The neutron plays a role in binding the
nucleus together, apparently adding attractive
forces which predominate over the electrical re-
pulsions among the protons.*

Around the nucleus are enough electrons to
make the atom as a whole, electrically neutral.

"neutral
fluorine atom,

9 protons
9 electrons

-+- (electron)

"negative
fluorine ion,

9 protons
.10 electrons.

+ energy

(J)

(i)

6-1.2 The Mass of an Atom and Its Parts

Protons are in the nucleus and electrons sur-
round it. Most of the mass of he atom is in the
nucleus. These two statements imply that an
electron weighs far less than a proton; this is the
case. Experiments have been performed in which
individual electrons and protons have been
weighed (they are described in Chapter 14).
These experiments show that the mass of the
electron is smaller than that of a proton by a
factor of -n^.

This means that most of the mass of the atom
must be furnished by the nucleus. However, the
mass of the nucleus is not determined by the
number of protons alone. For example, a helium
nucleus has two protons and a hydrogen nucleus
has one proton. Yet a helium atom is measured
to be four times heavier than a hydrogen atom.
What can be the composition of the helium nu-
cleus? A partial answer to this problem was
obtained when a third particle, the neulroD, was

Table 6-1

CHARGE AND MASS OF SOME
FUNDAMENTAL PARTICLES

PARTICLE

CHARGE

(relative to the
electron charge)

APPROXIMATE MASS

(relative to the
mass of a proton)

The charge and mass of each of the three funda-
mental particles we have discussed are shown in
Table 6-1.

* This model still does not explain what forces hold
the nucleus together in spite of the proton repulsions.
We do know that the helium nucleus is stable — it can
exist indefinitely — but the model does not explain why it
is stable. Thus we use models because they help us to
explain many important facts, even though they do not
explain aLl the facts.

STRUCTURE OF THE ATOM AND THE PERIODIC TABLE I CHAP. 6

6-1.3 The Sizes of Atoms

How large is an atom? We cannot answer this
question for an isolated atom. We can, however,
devise experiments in which we can find how
closely the nucleus of one atom can approach
the nucleus of another atom. As atoms approach,
they are held apart by the repulsion of the posi-
tively charged nuclei. The electrons of the two
atoms also repel one another but they are at-
tracted by the nuclei. The closeness of approach
of two nuclei will depend upon a balance between
the repulsive and attractive forces. It also de-
pends upon the energy of motion of the atoms
as they approach one another. If we think of
atoms as spheres, we find that their diameters
vary from 0.000 000 01 to 0.000 000 05 cm (from
1 X 10~8 to 5 X lO-8 cm). Nuclei are much
smaller. A typical nuclear diameter is 10-13 cm,
about 1/100,000 the atom diameter.

Suppose we take Yankee Stadium (seating
capacity, 67,000) as a model for the atom. To
keep the proper scale, the nucleus would be
about the size of a flea! For the hydrogen atom,
the flea would represent one proton. He would
be located at the center of the stadium, some-

Fig. 6-1. The nucleus is much smaller than the atom.

The nucleus occupies about the same rela-
tive volume in the atom as does a flea in
Yankee Stadium.

where behind second base. The one electron
present in the neutral atom would wander hither
and yon, occupying all the rest of the stadium.
For the helium atom, the nucleus would be re-
placed by a cluster of four fleas — representing
two protons and two neutrons. The two electrons
of the neutral helium atom would now share
the ample space of the huge stadium. This situa-
tion is made even more difficult to picture by the
fact that the fleas, occupying a minute volume in
center field, represent almost all of the mass that
is carried in the stadium.

6-1.4 Atomic Number

What do the atoms of one element have in com-
mon to distinguish them from the atoms of all
other elements? Each hydrogen nucleus bears a
charge of 1 + . Each neutral hydrogen atom has
one electron, charge 1 — , situated in the rela-
tively large volume of the atom outside the nu-
cleus. Each helium atom has a nucleus whose
charge is 2+ and each neutral atom has two
electrons around the nucleus. The element lith-
ium has atoms heavier than hydrogen or helium,
and each lithium atom has a nucleus whose
charge is 3+. Three electrons around the nucleus
are required to form a neutral lithium atom.

Thus, each of the chemical elements consists
of atoms whose nuclei contain a particular num-
ber of protons, hence a particular nuclear charge.
The number of protons in the nucleus is called the

SEC. 6-1 I STRUCTURE OF THE ATOM

atomic number. Of course, all atomic num-
bers are whole numbers. Thus, oxygen with
atomic number 8, has eight protons in the nu-
cleus (nuclear charge 8+). A neutral oxygen
atom must have eight electrons (each with charge
1 — ) as well.

The atomic number of each of the elements is
listed in the table on the inside of the back cover
of this book. You will find there that each ele-
ment has a distinctive name, symbol, and atomic
number. A given element can be identified by
any of these. For example, helium can be called
by its name, helium, by its symbol, He, or by its
atomic number, the element of atomic number 2.

In the periodic table we shall see that the ele-
ments have been listed in the order of increasing
atomic number. An example of the periodic table
is on the wall of your classroom and there is a
copy on the inside of the cover at the front of
this book. In each box in the table the number

above the chemical symbol for the element is the
atomic number.

6-1.5 Mass Number and Isotopes

All of the atoms of an element have the same
nuclear charge. Do all of the atoms of an element
have the same mass? Almost all hydrogen atoms
do have the same mass — the sum of the mass of
one proton and the mass of one electron. For
these atoms the nucleus consists of a single pro-
ton. However, a small fraction of the hydrogen
atoms 0.016% of them) have nuclei whose mass
is approximately twice as great as the mass of
the proton. (Compare with the helium nucleus.)
To explain the mass of these hydrogen atoms,
we conclude that each of their nuclei consists of a
neutron (charge zero, mass 1) and a proton
(charge 1 + , mass 1). This kind of hydrogen atom
is called hydrogen-2; another name commonly

Table 6-II. vital statistics of some common isotopes

NAME

ISOTOPE

ABUNDANCE

NATURE

ATOMIC
NUMBER

MASS
NUMBER

COMPOSITION'

NUCLEUS NUMBER OF

ELECTRONS IN

MASS CHARGE NEUTRAL ATOM

hydrogen- 1
hydrogen-2

99.984%
0.016

1
1

1
2

lp,

1/7

1
2

1 +
1 +

1
1

helium-3
helium-4

1.34 X 10-4
100

3
4

2p,
2p,

1/7
In

3
4

2 +
2 +

2
2

lithium-6
lithium-7

7.40
92.6

3
3

6
7

3p,
3/7,

3/7

4/7

3 +
3 +

beryllium-9

100

4/>,

5/1

4 +

boron-10
boron- 11

18.83
81.17

5
5

10
11

Sp,
Sp,

5/7
6/7

10
11

5 +
5 +

5
5

carbon-12
carbon-13

98.892
1.108

6
6

12
13

6p,

6/7,

6/7
In

12
13

6+
6+

6
6

nitrogen-14
nitrogen- 15

99.64
0.36

14
15

lp,
lp,

In
8/:

14
15

7 +

7
7

oxygen- 16
oxygen-17
oxygen- 18

99.76
0.04
0.20

8
8
8

16
17
18

Sp,
Sp,
Sp,

8/7

9/7

10/7

16
17
18

8 +
8 +
8 +

8
8
8

fluorine-19

100

9p,

10/7

9 +

chlorine-35
chlorine-37

75.4
24.6

35
37

lip,
I7p,

18/7

20/7

35
37

17 +
17 +

17
17

gold- 197

100

197

79/7,

118/7

197

79 +

uranium-235
uranium-238

0.71

99.28

92
92

235
238

92/7,
92p,

143/7
146/7

235
238

92 +

92
92

1 p = proton, n = neutron.

STRUCTURE OF THE ATOM AND THE PERIODIC TABLE I CHAP. 6

used is deuterium. The two kinds of hydrogen
atoms (having the same atomic number but dif-
ferent masses) are called isotopes. An isotope is
identified by specifying, first, which element it is
(usually by the symbol or name of the element)
and, second, the sum of the number of protons
and the number of neutrons. The number of pro-
tons plus the number of neutrons in a nucleus is
called the mass number of the nucleus. The mass
number is, of course, always an integer.
Reviewing:

Atomic number = number of protons in the
nucleus (fixes nuclear charge)

Mass number = number of protons and neu-
trons in the nucleus (fixes
nuclear mass)

Most of the chemical elements consist of mix-
tures of isotopes. Oxygen, atomic number 8, has
three stable isotopes. The kind having mass num-
ber 16 is most abundant. About 99.76% of the
oxygen atoms consist of this isotope. Only 0.04%

of the oxygen atoms have mass number 17 and
about 0.20% have mass number 18. The nucleus
of an oxygen- 16 atom consists of eight protons
and eight neutrons — charge 8+, mass 16. The
nucleus of an oxygen- 17 atom consists of eight
protons and nine neutrons — charge 8+, mass
17. The nucleus of an oxygen- 18 atom consists
of eight protons and ten neutrons — charge 8+,
mass 18. Table 6-II summarizes the data on the
atomic structure of a few common isotopes.

The nuclear charge and the electrons it attracts
primarily determine the ways in which atoms
behave toward other atoms. Mass differences
cause only minor chemical effects. Since the iso-
topes of an element have the same nuclear charge
and the same number of electrons per neutral
atom, they react in the same ways. Thus we can
speak of the chemistry of oxygen without speci-
fying which one of the three stable isotopes is
reacting. Only the most precise measurements
will indicate the very slight chemical differences
among them.

6-2 THE SIMPLEST CHEMICAL FAMILY— THE INERT GASES

We are now ready to consider why the elements
are arranged as they are in the periodic table. We
shall examine the known elements to discover
the significance and usefulness of this table — a
table so important it is printed on the inside
cover of this and almost every other general
chemistry textbook.

When the elements are arranged in order of
increasing atomic number, each successive ele-
ment has different chemistry from its neighbors.
On the other hand, there are striking likenesses
among some of the elements. For example, of the
one-hundred or so known elements, only about
a dozen are gases at normal conditions of tem-
perature and pressure. Of these, six elements are
so remarkably similar in chemistry that they are
conveniently considered together. Further, they
are identified by a family name, "the inert gases."

If we look at the properties of other elements,
we find that this recurrence of properties is usual.
It is convenient to group all of the elements

according to their chemistry into families or
groups. The elements in a particular group have
similar chemistry. Knowledge about one element
in a group then aids in understanding the chem-
istry of other elements in that group. In the
periodic table each group appears in a vertical

Fig. 6-2. The inert gas elements — a chemical family.

helium

SEC. 6-2 I THE SIMPLEST CHEMICAL FAMILY -THE INERT GASES

column. The key to this arrangement lies in the
elements called the inert gases. It is one of
nature's quirks that these gases provide the key
to the organization of our chemical knowledge
because these elements are distinctive in their
almost total absence of chemical reactivity. The
first of the inert gases is helium.

6-2.1 Helium

Helium, the second element in the periodic table,
has atomic number 2. This means its nucleus
contains two protons and has a 2+ charge. The
neutral atom, then, contains two electrons. There
are two stable isotopes, helium-4 and helium-3,
but the helium found in nature is almost pure
helium-4. Helium is found in certain natural gas
fields and is separated as a by-product. Sources
of helium are rare and most of the world supply
is produced in the United States, mainly in
Texas and Kansas.

Helium is a monatomic gas and, as yet, no
stable compounds of helium have been found.
The attractive forces between the atoms of he-
lium are unusually weak, as shown by the normal
boiling point. To liquefy helium, it must be
cooled to -268.9°C or 4.2°K. No other element
or compound has a boiling point as low. Helium
has another distinction which reflects these weak
forces: it is the only substance known which
cannot be solidified at any temperature unless it
is subjected to pressure. Helium becomes solid
at 1.1 °K at a pressure of 26 atmospheres.

Chemical reactions tend to occur among at-
oms to form more stable arrangements of the
atoms. Helium takes part in no chemical reac-
tions. Helium atoms must be particularly stable.

6-2.2 Neon, Argon, Krypton, Xenon, Radon

Among all of the hundred or so known elements,
there are only five with chemistry resembling that
of helium. We have remarked that these ele-
ments; neon, argon, krypton, xenon, and radon,
together with helium, are known as the inert
gases. It is only since 1962 that chemists have
been able to prepare any compounds at all in-
volving these elements. The few compounds that
have been prepared are extremely reactive, de-
composing readily to restore the inert gas to the
elementary state. Everything that is known of
the chemistry of the inert gas elements indicates
that the atoms of the inert gases must be par-
ticularly stable.

The physical properties of the inert gases are
shown in Table 6-III. There is much information
contained in this table, and we shall examine it
in parts.

BOILING POINT

The inert gas elements are all gases at room
temperature. Helium has the lowest known boil-
ing point, 4.2°K. Neon has the third lowest
known boiling point, 27.2°K. (Hydrogen, H2, has
the second lowest known boiling point, 20.4°K.)
The boiling point of argon is still quite low,
87.3°K, but not so low as to distinguish it from
a number of diatomic molecules. (Compare the
boiling points of nitrogen, N2, b.p. = 77.4°K;
fluorine, F2, b.p. = 85.2°K; oxygen, 02, b.p. =
90.2°K.) Krypton, xenon, and radon have suc-
cessively higher boiling points. Apparently, as
the atomic number goes up, the boiling point
goes up. Figure 6-3 shows the boiling point trend
of the inert gases in a plot of boiling point against
atomic number. Since the atomic number gives

Table 6-III. some properties of the inert gas elements

property He Ne Ar Kr

Atomic number

Atomic weight

4.00

20.2

39.9

83.7

131

222

Boiling point (°K)

4.2

27.2

87.3

120

165

211

Melling point (°K)

—

24.6

83.9

116

161

202

Atomic volume, liquid

(ml /mole of atoms)

31.8

16.8

28.5

32.2

42.9

50.5

STRUCTURE OF THE ATOM AND THE PERIODIC TABLE I CHAP. 6

200

■

*C 150

♦•

$100

rii

~f/

20 40 60 80 IOC

Atomic number,
Number of electrrorts

Fig. 6-3. The correlation of boiling point and number
of electrons per atom for the inert gases.

the number of protons in the nucleus, it also
gives the number of electrons held by each atom.
We can interpret a higher boiling point to mean
that more energy must be added to disrupt the
liquid state. Hence, the weak attractive forces
which cause the inert gases to liquefy increase as
the number of electrons per atom increases.

MELTING POINT

At temperatures only slightly below the liquefac-
tion temperatures, the liquids freeze. The solids
are all simple crystals in which the atoms are
close-packed in a regular lattice arrangement.
The narrow temperature range over which any
one of these liquids can exist suggests that the
forces holding the crystal together are very much
like the forces in the liquid.

ATOMIC VOLUME

We picture the atoms in a liquid and in a solid
as being packed rather tightly. The packing is
random in the liquid and regular in the solid.
With this picture, we can deduce from the vol-
ume per mole of atoms the volume to be assigned
to a single atom. Consulting Table 6-1II, we find
that helium is distinctive in its atomic volume

(indeed, as it is in every other property listed in
this table). Otherwise, we see that volume per
mole of atoms increases with atomic number,
that is, with the number of electrons around the
nucleus. Notice that the volume per mole of
atoms of neon (16.8 ml) is only slightly less than
that of water (18 ml). The water molecule occu-
pies slightly more space than does the neon
atom.

PHYSICAL PROPERTIES: SUMMARY

We have compared the physical properties of the
inert gases among themselves and with the cor-
responding properties of a few simple diatomic
molecules. We find that the forces between the
atoms are indeed weak though comparable to
the forces between some stable molecules. The
physical properties, however, do not distinguish
this group of elements. The uniqueness of the
inert gases relates to compound formation; when
compared with the other elements, the inert gases
are seen to have a specially small tendency to
form stable compounds.

6-2.3 Number of Electrons and Stability

The inert character of stable compounds makes
them of special interest to us. We might make a

Fig. 6-4. Trends in (he physical properties of the inert
gases.

200°K- (a) mpl' °K

100 "K- _ 1' """"

200*- (»*-P:°K

100'K-

40 r (c) Atomic volume,

ml /mole
30

SEC. 6-3 I THE ALKALIES

list of the number of electrons possessed by each
inert gas. (Remember the lost child organizing
his information?) We shall find these numbers
to be of particular value in all of our future study
of chemistry.

In Table 6-IV there is much food for thought.
First, each of the especially stable atoms has an
even number of electrons. Next we see that there
seems to be some regularity in the differences
between the number of electrons possessed by a
given inert gas and that of its predecessor. The
first two differences are eight and the second two
differences are eighteen. If there were another
inert gas, would it have 86 + 32 = 118 elec-
trons? What is the special significance of the
numbers 8, 18, and 32? Is it true for any other
element that the electron arrangements of he-
lium, neon, argon, and so on are especially
stable? We shall see that this is true, not just for
some elements, but for all elements.

Table 6-IV

THE ELECTRON POPULATIONS
OF THE INERT GASES

INERT TOTAL NUMBER CHANGE IN NUMBER

GAS OF ELECTRONS OF ELECTRONS

helium

neon

10 - 2 = 8

argon

18 - 10 = 8

krypton

36 - 18 = 18

xenon

54 - 36 = 18

radon

86 - 54 = 32

6-2.4 Sodium Chloride — Atoms Trying
to Be Inert Gas Atoms

Sodium chloride is a compound of an element,
chlorine, which just precedes an inert gas, and

an element, sodium, which just follows an inert
gas. Sodium has atomic number (nuclear charge)
1 1 , one larger than neon, which has atomic num-
ber 10. Hence, the neutral atom of sodium has
one more electron than the number held by the
especially stable neon atom. Chlorine has atomic
number (nuclear charge) 17, one below that of
argon, 18. The neutral atom of chlorine has one
less than the number of electrons held by the
especially stable argon atom. We find sodium
and chlorine combined in a one-to-one ratio in
the very stable compound called sodium chlo-
ride, NaClfsJ.

We have already discussed the structure of
solid sodium chloride in Chapter 5. We said
there, "On the basis of much experimental evi-
dence, chemists have concluded that sodium
chloride crystals are built up of charged particles
rather than of neutral atoms." The discussion
went on to identify the ions in the lattice as Cl~
ions packed tightly around Na+ ions (see Figure
5-10 on page 81). But how many electrons does
a chlorine ion, Cl~, have? By gaining an electron,
the chlorine now has 18 electrons, exactly the
same number of electrons as does argon, the
adjacent inert gas. In a similar way, but by losing
an electron, the sodium has contrived to reach
the 10 electron population of its adjacent inert
gas, neon. The atoms reached these inert gas-like
electron arrangements through compound for-
mation and the resulting compound thereby
acquired some of the unique stability of the
inert gases.

With this in mind, let us explore the chemistry
of all of the elements immediately adjacent to
the inert gases. These two vertical columns of
the periodic table are called the alkalies and the
halogens.

6-3 THE ALKALIES

The six elements adjacent to and following the
six inert gases are lithium, sodium, potassium,
rubidium, cesium, and francium. These elements
have similar chemistries and are called the

alkalies (or, the alkali metals). Figure 6-5 shows
that these elements are neighbors to the inert
gases. Their chemistries can be understood in
terms of the special stabilities of the 1 + ions

STRUCTURE OF THE ATOM AND THE PERIODIC TABLE I CHAP. 6

z
He

Wk.r*&

18
At*

54
Xc

^■*J^H

lithium
Sodium
potassium
■rubidium
cesium,
francium

Fig. 6-5. The alkali metals.

which have the inert gas electron arrangements.

All the alkalies are metals in their elementary
states. When the metal surfaces are clean, they
have a bright, silvery luster. The metals are ex-
cellent conductors of electricity and heat. They
are soft and malleable, and have low melting
points (compared with almost all other elemen-
tary metals).

In Chapter 5 we identified metals by their high
electrical conductivity. Now we can explain why
they conduct electric current so well. It is because
there are some electrons present in the crystal
lattice that are extremely mobile. These "con-
duction electrons" move throughout the metallic
crystal without specific attachment to particular
atoms. The alkali elements form metals because
of the ease of freeing one electron per atom to
provide a reservoir of conduction electrons. The
ease of freeing these conduction electrons derives
from the stability of the residual, inert gas-like
atoms.

Fig. 6-6. Potassium is soft; it can be cut with a knife.

6-3.1 Physical Properties of the Alkali Elements

Table 6-V lists the same properties for the alkali
metals that were listed in Table 6-III for the
inert gases.

BOILING POINT AND MELTING POINT

All the alkali metals are solids at room tempera-
ture, though cesium melts just above room tem-
perature. Notice that both melting points and
boiling points decrease as atomic number in-
creases (the opposite of the inert gas behavior).
Figure 6-7a and 6-7b contrast the trends in the
alkali metal melting and boiling points to the
opposite trends in the inert gases. Notice also
the extremely wide temperature range over which
the alkali liquids are stable. Sodium, for ex-
ample, melts at 371°K and boils at 1162°K,

Table 6-V. some properties of the alkali elements

PROPERTY

LITHIUM

SODIUM

POTASSIUM

RUBIDIUM

CESIUM

Atomic number

Atomic weight

6.94

23.0

39.1

85.4

133

Boiling point (°K)

1599

1162

1030

952

963

(°C)

1326

889

757

679

690

Melting point (°K)

453

371

336.4

311.8

3C1.7

(°C)

180

63.4

38.8

28.7

Atomic volume, solid

(ml/mole of atoms)

13.0

23.7

45.4

55.8

70.0

Density of solid at 20°C

0.535

0.971

0.862

1.53

1.90

SEC. 6-3 I THE ALKALIES

SOO'K-

(a) m.p, "K

■--in

1.500°K-

1.000'K-

500"K

— fir—

He It Ne Ml Ar K Kr Rb Xe Cs

(b) b.P.. °K

60
40V

He Li Ne Ml Ar K Kr Rb Xe Cs

20 Y

He Li Ne Ma Ar K Kr Rb Xe Cs

Fig. 6-7. Trends in the physical properties of the
alkalies.

almost 800° higher. Contrast neon, which melts
at 24.6°K and boils only 2.6° higher, at 27.2°K.
How different are the alkali metals from the
inert gases, only one atomic number removed!

ATOMIC VOLUME

The atomic volumes of the alkali metals increase
with atomic number, as do those of the inert
gases. Notice, however, that the volume occupied
by an alkali atom is somewhat larger than that
of the adjacent inert gas (with the exception of
the lithium and helium — helium is the cause of
this anomaly). The sodium atom in sodium metal
occupies 30% more volume than does neon.
Cesium occupies close to twice the volume of
xenon.

6-3.2 Chemistry of the Alkali Metals

The alkali metals are exact opposites of the inert
gases in chemical reactivity. These metals react

vigorously when in contact with oxygen and
chlorine, and even with such a placid reagent as
water. Let us investigate some of these reactions.

REACTIONS OF THE ALKALI METALS
WITH CHLORINE

When chlorine gas is brought into contact with
sodium metal, sodium chloride is formed:

Nafsj + \C\,(g) — ►- NaClfsJ + energy (4)

Sodium chloride has a lattice built up of so-
dium ions, Na+, and chloride ions, Cl~. There-
fore, the reaction (4) involves the transfer of
electrons from sodium atoms to chlorine atoms.
The resulting ions attract each other because
they have opposite electric charges. Of course, it
takes some energy to remove an electron from a
sodium atom to form a sodium ion. However,
it takes only a moderate amount of energy be-
cause the Na+ ion that is formed has the electron
population of an inert gas, neon. This electron,
removed from a sodium atom, is then added to
a chlorine atom to form a chloride ion, Cl~. This
reaction releases a small amount of energy — the
Cl~ ion also has the electron population of an
inert gas, argon. At a rather low net cost of en-
ergy, an electron can be removed from a sodium
atom and transferred to a chlorine atom, giving
Na+ and Cl~ ions. Now these two ions can move
toward each other with a large reduction in total
energy. The stability of the sodium chloride
crystal can be said to depend upon the electrical
attraction of the oppositely charged ions. The
crystal is said to be held together by ionic bonds.

This chemistry is characteristic of all of the
alkali metals. Each of them reacts with chlorine
gas in a similar way:

Li(s) +hC\2(g)
Nafsj + iCUfcl
K(s) + \C\,(g)
Rb(s) + \CUg)
Cs(s) + iCUg)

LiClfs,) + energy (5)

NaClfsJ -f energy (6)

KClfsJ + energy (7)

RbClfsJ + energy (8)

CsC\(s) + energy (9)

In every case, the alkali metal reacts to form
a stable, ionic solid in which the alkali is present
as an inert gas-like ion. The product is, in each
case, a crystalline substance with high solubility
in water.

STRUCTURE OF THE ATOM AND THE PERIODIC TABLE I CHAP. 6

REACTIONS OF THE ALKALI METALS
WITH WATER

Sodium metal reacts vigorously with water to
form hydrogen gas and an aqueous solution of
sodium hydroxide, NaOH:

2Na(s) + 2H20 — *-

2Na+(aq) + 20H-(aq) + H2(g) + energy (10)

Energy is liberated and the reaction often takes
place so rapidly that the temperature rises and
the hydrogen, mixing with air, explodes. Thus,
sodium metal is dangerous and must be handled
with caution. This chemistry is also character-
istic of all of the alkali metals.

EXERCISE 6-1

Write the equations for the reactions between
water and: lithium, potassium, rubidium, ce-
sium.

Again, we see that the alkali metals display
likeness in their reactions with water. Further-
more, the reaction products always include an
aqueous ion of the alkali element in which one
electron has been removed, giving a 1+ ion.

SUMMARY OF CHEMISTRY
OF THE ALKALI ELEMENTS

The alkali metals are extremely reactive. Thus,
there is a dramatic change in chemistry as we
pass from the inert gases to the next column in
the periodic table. The chemistry of the alkali
metals is interesting and often spectacular. Thus,
these metals react with chlorine, water, and oxy-
gen, always forming a +1 ion that is stable in
contact with most substances. The chemistry of
these +1 ions, on the other hand, is drab, re-
flecting the stabilities of the inert gas electron
arrangements that they have acquired.

6-4 THE HALOGENS

Now let's slide to the left in the periodic table
and consider the column of elements fluorine,
chlorine, bromine, iodine, and astatine. Each of
these elements has one less electron than does
its neighboring inert gas. These elements are
called the halogens. (The discussion that follows
does not include astatine because this halogen is
very rare.)

Fig. 6-8. The halogens.

fluorine

10
Ne

11
Na

chlorine Kjfl

16
Ar

hromtne Wjj£

36
Kr

iodine ■!"■

astatine WWM

86
Kn

Ft-

6-4.1 Physical Properties of the Halogens

Table 6-VI lists some properties of the halogens.
In the elemental state, the halogens form stable
diatomic molecules. This stability is indicated by
the fact that it takes extremely high temperatures
to disrupt halogen molecules to form the mona-
tomic species. For example, it is known that the
chlorine near the surface of the sun, at a tem-
perature near 6000°C, is present as a gas consist-
ing of single chlorine atoms. At more normal
temperatures, chlorine atoms react with each
other to form molecules:

2C\(g) — *- Cl2(g) (11)

Then, no further reactions among chlorine mole-
cules occur.

Apparently the diatomic molecules of the hal-
ogens already have achieved some of the stability
characteristic of the inert gas electron arrange-
ment. How is this possible? How could one
chlorine atom satisfy its need for one more elec-
tron (so it can reach the argon stability) by

SEC. 6-4 I THE HALOGENS

Table 6- VI. some properties of the halogens

PROPERTY

FLUORINE

CHLORINE

BROMINE

IODINE

ASTATINE

Atomic number

Atomic weight

19.0

35.5

79.9

127

Molecular formula

Br2

Boiling point (°K)

238.9

331.8

457

(°C)

-188

-34.1

58.8

184

Melting point (°K)

172

265.7

387

(°C)

-218

-101

-7.3

114

Atomic volume, solid

(ml/mole of atoms)

14.6

18.7

23.5

25.7

combining with another chlorine atom, an atom
with a similar need? We answer this question by
suggesting that the two atoms share two elec-
trons, each atom contributing one electron. If
the two atoms huddle close together and place
this communal pair of electrons between them,
each atom acts more as though it had the special
stability of the inert gas. This results in the for-
mation of a stable aggregate of atoms, a mole-
cule. Its formula is Cl2. The same argument can-
be made to explain the diatomic molecular for-
mulas of the other halogens. Because each of
these molecules is bonded by a shared pair of
electrons, the bond is called a covalent bond.

BOILING POINT AND MELTING POINT

We have already made a comparison between
the physical properties of some of the halogens
and those of inert gases. The comparison sug-
gests that after the formation of a diatomic
molecule, the bonding capacity of the two halo-
gen atoms is "used up." The only attractive
interactions remaining between two such dia-
tomic molecules are of the extremely weak
variety that account for the liquefaction of the
inert gases. Thus, the melting points rise as
atomic number increases (remember that the
alkali metals did the reverse) and there is but a
narrow temperature range over which the liquid
is stable. Fluorine and chlorine are gases under
normal conditions, bromine is a liquid, and
iodine is a solid. These differences in physical
state result from the accident that normal condi-
tions fall where they do. On a planet with a
"normal" temperature of 25°K, all of the halo-

gens would be solids whereas neon would be a
liquid, helium a gas, and argon, krypton, and
xenon would be solids.

Fig. 6-9. Trends in the physical properties of the
halogens.

500°K

(a) m.f>., A"

F a

Ne Ar

Na K

1500°K

1,000°K

500°K

(b) b. p., °K

20 -

He
Li

STRUCTURE OF THE ATOM AND THE PERIODIC TABLE I CHAP. 6

ATOMIC VOLUMES

Here we find a continuation of the trend dis-
played by the inert gases and alkali metals. Com-
pare the atomic volumes of the three adjacent
elements in the solid state:

fluorine

14.6 ml

chlorine

18.7 ml

bromine
23.5 ml

neon
20.2 ml

argon
24.2 ml

krypton
41.9 ml

sodium

23.7 ml

potassium
45.4 ml

rubidium

55.8 ml

In each set, the atomic volumes increase going
from halogen to inert gas to alkali metal, as
shown graphically in Figure 6-9c. Figure 6-10
shows models constructed on the same scale to
show the relative sizes of atoms indicated by the
atomic volumes and by the packing of the ions
in the ionic solids.

Figure 6-10. Models (to scale) of halogen atoms,
inert gas atoms, and alkali atoms.

6-4.2 Chemistry off the Halogens

The reactions of the alkali metals with chlorine
were used to display the similarities of the alkali
metals. In a similar way, the reactions of the
halogens with one of the alkali metals, say so-
dium, show similarity within this group. The
reactions that occur are as follows.

Nafsj + \¥,(g)
Nafsj + \C\2(g)
NafsJ + hBr2(g)
NafsJ + \h(g)

NaFfsJ + energy (12)

NaClfsj + energy (13)

NaBrfsJ + energy (14)

Nal(s) + energy (15)

These reactions all proceed readily and they
produce ionic solids with the general empirical
formula, NaX Each of these solids has a crystal
structure made up of positively charged sodium
atoms and negatively charged halogen atoms.
These negative ions, F-, CI-, Br-, and I-, are
called halide ions. The stabilities of these ions
can be related to the stabilities of the correspond-
ing inert gas electron arrangement.

The halogens also react with hydrogen gas to
form the hydrogen halides:

Halogens
Molecules —1 Ions

Inert &ases
Atoms

Alkali Elements
+1 Ions Metallic Atoms

Ka (in metal)

ci.

K+ K (in metal)

Br,

Rb^ Rb (in metal)

SEC. 6-5 I HYDROGEN — A FAMILY BY ITSELF

Ht(g) + F2(g)
HJg) + C\,(g)
H2(g) + Brjg)
Wg) + h(g)

2UF(g)

hydrogen fluoride (76)
2HC\(g)

hydrogen chloride (17)

2HBr(g)

hydrogen bromide (18)
2Hl(g)

hydrogen iodide (19)

None of these reactions, (76), (17), (18), or
(19), proceeds readily at room temperatures.
This is because the bonds holding the atoms to-
gether in the hydrogen and in the halogen mole-
cules must be broken if new bonds are to form
between hydrogen atoms and halogen atoms.
The breaking of the bonds is favored by high
temperatures, however, and, once started, these
reactions tend to proceed rapidly or even ex-
plosively.

6-4.3 Chemistry off the Halide Ions

Because of the stabilities of halides, most ele-
ments form stable halide compounds. Thus cal-
cium forms the compounds CaF2, CaCl2, CaBr2,
and Cal2, all ionic solids. In each crystal, the
calcium ion carries a +2 charge, and each of the
halide ions carries a — 1 charge. The empirical
formulas are all of the type Ca.X2.

The alkali halides are relatively unreactive
substances. They all display high solubility in
water and quite low solubility in ethyl alcohol.

We have seen in Experiment 8 that silver chlo-
ride has low solubility in water. This is also true
for silver bromide and silver iodide. In fact, these
low solubilities provide a sensitive test for the
presence of chloride ions, bromide ions, and
iodide ions in aqueous solutions. If silver nitrate

solution, which contains silver ions, Ag+faqJ,
and nitrate ions, N03" (aq), is added to a solution
containing l~(aq) ions, a yellow precipitate of
Agl(s) forms:

Ag+(aq) + l-(aq)

AglfsJ
yellow

If Br- ions are present, the reaction is

Ag+faqJ + Br-(aq) — >- AgBrfsJ

light yellow

and if Cl~ ions are present, the reaction is

Ag+(aq) + C\-(aq)

AgClfsJ

white

(20)

(21)

(22)

Silver fluoride is soluble. Therefore, no precipi-
tate forms as Ag+ ions are added to a solution
of F~ ions.

All the hydrogen halides are gaseous at room
temperature but hydrogen fluoride liquefies at
19.9°C and 1 atmosphere pressure. The most
important chemistry of the hydrogen halides
relates to their aqueous solutions. All of the
hydrogen halides dissolve in water to give solu-
tions that conduct electric current, suggesting
that ions are present. The reactions may be
written:

HF(g) + water
HCl(g) + water
HBr(g) + water
Hl(g) + water

H+(aq) + F-(aq) (23)

H+(aq) + C\-(aq) (24)

H+(aq) + Bx-(aq) (25)

H+(aq) + l-(aq) (26)

These solutions have similar properties and are
called acid solutions. The common species in the
solutions is the aqueous hydrogen ion, H+(aq),
and the properties of aqueous acid solutions are
attributed to this ion. We shall investigate these
solutions in Chapter 11.

6-5 HYDROGEN— A FAMILY BY ITSELF

Perhaps you are wondering why the element
hydrogen was not included among the halogens.
It is, after all, an element with one less electron
than its neighboring inert gas, helium. On the
other hand, the hydrogen atom has but one elec-

tron and, in a sense, it is like an alkali metal.
The removal of one electron from an alkali metal
atom leaves a specially stable electron popula-
tion— that of an inert gas. The removal of one
electron from a hydrogen atom leaves it with no

100

STRUCTURE OF THE ATOM AND THE PERIODIC TABLE I CHAP. 6

electrons, which also turns out to be specially
stable. We shall see both of these influences in
the chemistry of hydrogen. This element forms
a family by itself, one having some similarities
to the halogens and some similarities to the
alkalies.

6-5.1 Physical Properties

Hydrogen is a diatomic gas at normal conditions.
Its melting point is 15.9°K and its normal boiling
point is 20.4°K. This is the second lowest boiling
point of any element. Table 6-VII lists these
properties.

Table 6-VII

SOME PROPERTIES OF HYDROGEN

Atomic number

Atomic weight

1.008

Molecular formula

Boiling point (°K)

20.4

(°Q

-252.8

Melting point (°K)

14.0

(°C)

-259.2

Atomic volume, solid

(ml/mole of atoms)

13.1

We see that the physical properties of hydro-
gen are like those of the halogens. Hydrogen is
a diatomic gas, like the halogens, rather than a
metal, like the alkalies. Its melting point is very
low and it has a narrow temperature range over
which the liquid is stable. However, the family
relationships among the elements are based upon
their chemistry, so we must investigate the reac-
tions of hydrogen before classifying this unique
element.

6-5.2 Some Chemistry of Hydrogen

One of the most distinctive reactions charac-
terizing both the alkalies and the halogens is their
reaction with each other. The example we have
discussed most is the reaction between sodium
and chlorine to sive sodium chloride. Sodium

chloride is an ionic solid that dissolves in water
to give positively charged sodium ions, Na+(aq),
and negatively charged chloride ions, C\~(aq):

NafsJ + \Ck(g)
NaC\(s) + water

NaClfs; (27)

Na+(aq) + Cl-(aq) (28)

Does hydrogen react like sodium or chlorine
in reaction (27)1 Experiments show that hydro-
gen can take either position in reaction (27):

Nafsj + \Wg)

\Wg) + hch(g)

HC\(g) + water

NaHfsJ (29)

HCl(g) (30)

H+(aq) + C\-(aq) (31)

The compound sodium hydride, formed in
reaction (29), is a crystalline compound with
physical properties similar to those of sodium
chloride. The chemical properties are very dif-
ferent, however. Whereas sodium burns readily
in chlorine, it reacts with hydrogen only on heat-
ing to about 300°C. While sodium chloride is a
stable substance that dissolves in water to form
Na+(aq) and C\~(aq), the alkali hydrides bum
in air and some of them ignite spontaneously.
In contact with water, a vigorous reaction oc-
curs, releasing hydrogen:

NaH(s) + H20 — *-

H2(g) + Ua+(aq) + OH-(aq) (32)

Thus, in reaction (29), hydrogen reacts with
sodium like a halogen [as in reaction (27)], but
the product, sodium hydride, is very different in
its chemistry from sodium chloride.

Reaction (30) shows hydrogen acting like an
alkali metal. Though the product, hydrogen
chloride, is not an ionic solid like sodium chlo-
ride, it does dissolve in water to give aqueous
ions. The formation of H+(aq) and C\~( aq) is
strikingly like the analogous formation of
Na+( aq) and C\~(aq) by sodium chloride. In
fact, the tendency of hydrogen to form a posi-
tively charged ion in water, H+(aq), and the
absence of any evidence for a negatively charged
ion in water, H~(aq), is one of the most signifi-
cant differences between hydrogen and the halo-
gens.

An overall view of the chemistry of hydrogen
requires that it be classified alone— as a separate

SEC. 6-6 | THE THIRD-ROW ELEMENTS

101

1
H

10
Ne

11 12

N*M9

13 14

At Si

15 16 17 18 1

P S CI Ar\

35
Br

37
Kb

53
I

87
Fr

Fig. 6-11. The placement of the third-row elements

in the periodic table.

chemical family. There are some important simi-
larities to halogens — as we have seen, it is a

stable diatomic gas — but its chemistry is more
like that of the alkalies. Therefore, hydrogen is
usually shown on the left side of the periodic
table with the alkalies but separated from them
to indicate its distinctive character.

6-6 THE THIRD-ROW ELEMENTS

The chemistry of the elements we have examined
thus far in this chapter is dominated by the
special stabilities of the inert gas electron popu-
lations. We can expect to see this same factor at
work in the chemistry of the elements in other
parts of the periodic table. We shall now take an

excursion across a horizontal row of the periodic
table to see the trend in chemistry as we pass
from element to element. We shall consider the
third row, which contains the elements sodium,
magnesium, aluminum, silicon, phosphorus, sulfur,
chlorine, and argon.

Table 6- VIII. some properties of the elements of the third row

OF THE PERIODIC TABLE

PROPERTY

Atomic number

Atomic weight

23.0

24.3

27.0

28.1

31.0

32.1

35.5

39.9

Molecular formula

metal

network
solid

Cl2

Boiling point (°K)

1162

1393

2600

2628

553

718

238.9

(°C)

889

1120

2327

2355

280

445

-34.1

-186

Melting point (°K)

371

923

933

1683

317.2

392

172

(°C)

650

660

1410

44.2

119

-101

-189

Atomic volume, solid

(ml/mole of atoms)

23.7

14.0

9.99

12.1

16.9

15.6

18.7

24.2

102

STRUCTURE OF THE ATOM AND THE, PERIODIC TABLE I CHAP. 6

6-6.1 Physical Properties of the
Third-row Elements

Table 6-VII1 presents some properties of the
elements we are considering. The first three,
sodium, magnesium, and aluminum, are metal-
lic. The melting points and boiling points are
high and increase as we go from element to
element. This trend reflects stronger and stronger
bonding and it is paralleled by a decrease in the
atomic volume.

The fourth element, silicon, forms a solid in
which each silicon atom is bonded to four neigh-
boring silicon atoms placed equidistant from
each other. (This arrangement places the four
neighbors at the four corners of a regular tetra-
hedron.) This arrangement generates a three-
dimensional network, hence is called a network
solid. Typically such a network solid has a high
melting point and a high boiling point.

The remaining four elements form molecular
solids. The atoms of white phosphorus, sulfur,
and chlorine are strongly bonded into small
molecules (formulas, P4, S8, and Cl2, respectively)
but only weak attractions exist between the
molecules. The properties are all appropriate to
this description. Of course there is no simple
trend in the properties since the molecular units
are so different.

6-6.2 Compounds of the Third-Row Elements

To see the trend in chemistry as we move across
the periodic table, we will consider three types

of compounds: the hydrides, the chlorides, and
the oxides.

HYDRIDES

The hydrides are the compounds formed with
hydrogen. Hot, molten sodium metal reacts
with hydrogen gas to form a solid, salt-like hy-
dride having the empirical formula, NaH. The
ions in the salt are thought to be the Na+ ion
and the H~ ion. The H~ ion may exist in the solid
or its melt but it does not exist in a water solu-
tion. Magnesium forms a similar salt-like hydride
having the empirical formula MgH2. Apparently
two electrons per neutral magnesium atom are
removed to form the Mg+2 ion (with the stable,
neon-like electron arrangement).

Aluminum forms a hydride which seems to be
more molecular in character than it is salt-like.
The empirical formula is, however, A1H3. The
remaining elements form hydrides which are
definitely molecular, gaseous compounds. The
formulas are, respectively, SiH4, PH3, H2S, and
HC1. These formulas are shown in Table 6-IX,
together with the ratio of number of atoms of
hydrogen per atom of the element (H/M). We
see the regularity of the trend in combining ca-
pacity of the elements (as reflected in the ratio,
H/M). On the left side, the elements sodium and
magnesium (and, to some extent, aluminum)
achieve the neon-like electron arrangement by
surrendering electrons to hydrogen atoms, one
electron to each hydrogen atom. Note that each
hydrogen atom thereby achieves the helium-like
electron structure. From silicon onward, the

Table 6-IX. the formulas of some compounds of the third-row elements

Na Mg Al Si P S CI Ar

Hydrides

Formula

NaH

MgH2

A1H3

SiH4

PH,

H2S

HC1

—

H/M

Chlorides

Formula

NaCl

MgCl2

A12CU

SiCU

PCU, PCI,

SjCl,

Cl2

—

Cl/M

5, 3

Oxides

Formula

Na20

MgO

A120,

Si02

P4OU

SO,

Cl207, ci2o

—

2(0/M)

7. 1

SEC. 6-7 I THE PERIODIC TABLE

103

molecular compounds indicate that the elements
are achieving the next higher inert gas electron
arrangement (argon) by sharing electrons with
hydrogen atoms. For these compounds, the
bonding to the hydrogen atoms resembles more
the covalent bonding we find in Cl2 than the
ionic bonding we find in sodium chloride. What-
ever the nature of the bond, however, the com-
bining capacity of each element is influenced by
the tendency to form the next lower or next
higher inert gas electron population.

CHLORIDES AND OXIDES

We can find the same sort of trends in the com-
bining capacity by examining the chlorides and
the oxides.* Again, we find the bonding can be
understood in terms of electron transfer or elec-
tron sharing, in either case, to reach an inert gas
electron arrangement. There is only one apparent
anomaly — the combining capacity of sulfur in
the compound SjCl2. This is not a real anomaly,
however, for the structure of this compound
reveals the expected combining power of 2. The

Fig. 6-12. The structure of S2C1>.

structure is shown in Figure 6-12. The atoms in
this molecule are arranged so that each sulfur
atom reaches the argon-like electron arrange-
ment by sharing one pair of electrons with a
chlorine atom and one pair of electrons with
another sulfur atom.

SUMMARY

The simple trend in the formulas shown by
the third-row elements demonstrates the impor-
tance of the inert gas electron populations. The
usefulness of the regularities is evident. Merely
from the positions of two atoms in the periodic
table, it is possible to predict the most likely em-
pirical and molecular formulas. In Chapters 16
and 17 we shall see that the properties of a
substance can often be predicted from its molec-
ular formula. Thus, we shall use the periodic
table continuously throughout the course as an
aid in correlating and in predicting the properties
of substances.

6-7 THE PERIODIC TABLE

The power of the periodic table is evident in the
chemistry we have viewed. By arranging the
elements in the array shown on the inside of the
front cover, we simplify the problem of under-
standing the variety of chemistry found in na-

* It must be remarked that some stable compounds
have been omitted (for example, Na202, S02). The com-
pounds listed are stable and display the trends in bonding
capacity.

ture. The elements grouped in a vertical column
have pronounced similarities. General state-
ments can be made about their chemistries and
the compounds they form. Furthermore, the
formulas of these compounds and the nature of
the bonds that hold them together can be under-
stood in terms of the special stabilities of the
inert gases.
The periodicity of chemical properties was dis-

104

STRUCTURE OF THE ATOM AND THE PERIODIC TABLE I CHAP. 6

covered about one-hundred years ago. J. W.
Dobereiner (a German chemist) in 1828 recog-
nized similarities among certain elements (chlo-
rine, bromine, and iodine; lithium, sodium, and
potassium; etc.) and he grouped them as "tri-
ads." (Remember, "Cylindrical objects burn?")
J. A. R. Newlands (an English chemist) in 1864
was ridiculed for proposing a "law of octaves"
which foresaw the differences of eight that we
noted in Table 6-IV. Simultaneously, Lothar
Meyer (a German chemist and physicist) pro-
posed a periodic table similar to that of New-

lands. Independently and in this same year (the
time was ripe for the next step, "Wooden objects
burn") D. I. Mendeleev (a Russian) framed the
periodic table in more complete form. He even
predicted both the existence and properties of
elements not then known. The subsequent dis-
covery of these elements and corroboration of
their properties solidified the acceptance of the
periodic table. It remains, one-hundred years
later, the most important single correlation of
chemistry. It permits us to deal with the great
variety we find in nature.

QUESTIONS AND PROBLEMS

1. For which of the following processes will energy
be absorbed?

(a) Separating an electron from an electron.

(b) Separating an electron from a proton.

(d) Removing an electron from a neutral atom.

2. Which of the following statements is FALSE?
The atoms of oxygen differ from the atoms of
every other element in the following ways:

(a) the nuclei of oxygen atoms have a different
number of protons than the nuclei of any
other element;

(b) atoms of oxygen have a higher ratio of neu-
trons to protons than the atoms of any other
element;

(d) atoms of oxygen have different chemical be-
havior than do atoms of any other element.

3. For every atom, less energy is needed to remove
one electron from the neutral atom than is
needed to remove another electron from the
resulting ion. Explain.

4. List the number and kind of fundamental par-
ticles found in a neutral lithium atom that has a
nucleus with a nuclear charge three times that
of a hydrogen nucleus and with seven times the
mass.

5. The nucleus of an aluminum atom has a diame-
ter of about 2 X 10-13 cm. The atom has an
average diameter of about 3 X 10~8 cm. Calcu-
late the ratio of the diameters.

6. Suppose a copper atom is thought of as occupy-
ing a sphere 2.6 X 10~8 cm in diameter. If a
spherical model of the copper atom is made with
a 5.2 cm diameter, how much of an enlargement
is this?

7. Suppose an atom is likened to bees flying around
their beehive. The beehive would be compared
to the nucleus and the bees roving about the
countryside would be compared to the electrons
of the atom.

(a) If the radius of the beehive is 25 cm, what
would be the average radius of the flight of
the bees to maintain proper scale with the
atom? Express your answer in kilometers.

(b) At any instant, where is the concentration of
bees apt to be highest?

8. Helium, as found in nature, consists of two iso-
topes. Most of the atoms have a mass number 4
but a few have a mass number 3. For each
isotope, indicate the:

(a) atomic number;

(b) number of protons;

(d) mass number;

(e) nuclear charge.

QUESTIONS AND PROBLEMS

105

9. Fill in the blanks of the following table.

aluminum (Al)
beryllium (Be)
bismuth (Bi)
calcium (Ca)
carbon (C)
fluorine (F)
phosphorus (P)
iodine (I)

ATOMIC
NO.

PARTICLES PER ATOM

13
83

HUH

III II

II III

MASS

PROTONS ELECTRONS NEUTRONS NUMBER

10. How do isotopes of one element differ from each
other? How are they the same?

11. How much would 0.754 mole of chlorine-35
atoms weigh? How much would 0.246 mole of
chlorine-37 atoms weigh? What is the weight of
a mole of "average" atoms in a mixture of the
above samples? What is the atomic weight of
the naturally occurring mixture of these two
isotopes of chlorine?

12. What is the significance of the trends in the boil-
ing points and melting points of the inert gases
in terms of attractions among the atoms?

13. Why is argon used in many electric light lamps?

14. Calculate the ratio of the number of electrons
in a neutral xenon atom to the number in a
neutral neon atom. Compare this number to the
ratio of the atomic volumes of these two ele-
ments. On the basis of these two ratios, discuss
the effects of electron-electron repulsions and
electron-nuclear attractions on atomic size.

15. The molar heats of vaporization of the inert
gases (in kcal/mole) are: He, 0.020; Ne, 0.405;
Ar, 1.59; Kr, 2.16; Xe, 3.02; Rn, 3.92. Using the
data in Table 6-III, (p. 91), plot the boiling
points (vertical axis) against the heats of vapori-
zation (horizontal axis). Suggest a generalization
based upon a simple curve passing near the
plotted points. Write an equation for the
straight line passing through the origin (that is,
through zero) and through the point for radon.

16. Lithium forms the following compounds: lith-
ium oxide, Li20; lithium hydroxide, LiOH;
lithium sulfide, Li2S. Name and write the for-
mulas of the corresponding sodium and potas-
sium compounds.

17. An alkali element produces ions having the same
electron population as atoms of the preceding
inert gas. In what ways do these ions differ from
the inert gases? In what ways are they alike?

18. There is a large difference between the energy
needed to remove an electron from a neutral,
gaseous sodium atom and a neutral, gaseous
neon atom:

NafgJ+ 118.4 kcal
Ne(g) + 497.0 kcal

Na+(gJ + e~
Ne+(g) + e~

19.

Explain how these energies are consistent with
the proposal that the electron arrangements of
the inert gases are specially stable.

Refer to the halogen column in the periodic
table. How many electrons must each halogen
atom gain to have an electron population equal
that of an atom of the adjacent inert gas? What
property does this population impart to each
ion?

20. How do the trends in physical properties for the
halogens compare with those for the inert gases ?
Compare boiling points, melting points, and
atomic volumes.

106

STRUCTURE OF THE ATOM AND THE PERIODIC TABLE I CHAP. 6

21. Use your knowledge of the usefulness of the
periodic table to fill in the blank spaces in Table
6-VI, p. 97, under Astatine. List some chemical
reactions expected for astatine.

22. Chlorine is commonly used as a germicide in
swimming pools. When chlorine dissolves in
water, it reacts to form hypochlorous acid,
HOC1, as follows:

Cl2 + H20 +=»=

HOCl(aq) + H+(aq) + C\-(aq)

Predict what happens when bromine, Br2, dis-
solves in water. Write the equation for the re-
action.

23. Zinc metal dissolves in a solution of gaseous
chlorine in water as follows:

Zn(s) + G2(aq) — >- Zn+2(aq) + 2Q-(aq)

Zinc does not dissolve in a solution of gaseous
hydrogen in water but it does dissolve in an
aqueous solution of hydrogen chloride :

Zn(s) + 2H+(aq) + 2C\-(aq) — >■

Zn+2(aq) + H2(g) + 2Cl-(aq)

Recognizing that zinc metal must release elec-
trons to form Zn+2(aq), explain how these re-
actions demonstrate that gaseous hydrogen does
not behave like a halogen.

24. Write the molecular formulas of the hydrogen
compounds of the second-row elements, Li, Be,
B, C, N, O, F, Ne. Indicate, for each compound,
the H/M ratio.

25. Indicate the electron rearrangement (gain or
loss) in each kind of atom assuming it attains
inert gas-like electron structure in the following
reactions.

2RbBr
2CsI
MgS
2BaO

(a) 2Rb + Br2

(b) 2Cs + I2

(d) 2Ba + 02

26. Which of the following is NOT a correct formula
for a substance at normal laboratory conditions ?

(a) H2S(g) (d) NaNefsj

(b) CaCl2(s) (e) Al203(sj

27. Magnesium metal burns in air, emitting enough
light to be useful as a flare, and forming clouds
of white smoke. Write the equation for the re-
action. What is the composition of the smoke?

28. Use the formulas for magnesium oxide, MgO,
and magnesium chloride, MgCl2, together with
the periodic table to decide that magnesium ions
have the same number of electrons as each of
the following, EXCEPT

(a) neon atoms, Ne;

(b) sodium ions, Na+;

(d) oxide ions, O-2;

(e) calcium ions, Ca+2.

29. Sodium metal reacts with water to form sodium
ions, Na+, hydroxide ions, OH~, and hydrogen
gas, H2, as follows:

2Na(sJ + 2H20 — ►■

2Na+(aq) + 20H-(aqj + H2(g)

Assuming calcium metal reacts in a similar way,
write the equation for the analogous reaction
between calcium and water. Remember that
calcium is in the second column of the periodic
table and sodium is in the first.

30. Use Table 6-IX, p. 102, and the periodic table
to write possible formulas for the following com-
pounds:

(a) a hydride of barium, element 56;

(b) a chloride of germanium, element 32;

(d) an oxide of cesium, element 55;

(e) a fluoride of tin, element 50.

31. All of the isotopes of the element with atomic
number 87 are radioactive. Hence, it is not found
in nature. Yet, prior to its preparation by nuclear
bombardment, chemists were confident they
knew the chemical reactions this element would
show. Explain. What predictions about this ele-
ment would you make?

DMITRI MENDELEEV, 1834-1907

Element 101 is named Mendelevium in honor of the great
Russian chemist, Dmitri Mendeleev. The youngest of
seventeen children, he was born in Tobolska where his
grandfather published the first newspaper in Siberia and
his father was the high school principal. Dmitri received his
early education from a political exile, but when his father
died, his mother traveled west in search of better educa-
tional opportunities for Dmitri.

At the University of St. Petersburg {now Leningrad), he
distinguished himself in science and mathematics and earned
the doctorate with a thesis on a subject that remains of
current interest, "The Union of Alcohol and Water.'1'' Sub-
sequent studies in France and Germany permitted him to
attend the 1858 Karlsruhe (Germany) conference at which
Avogadro's Hypothesis was heatedly debated. Later, he
visited the oilfields of Pennsylvania to see the first oil well.
Upon his return to Russia, he developed a new commercial
distillation process.

He became a professor of chemistry at St. Petersburg

when only 32. Searching for regularities, he arranged the
elements by their properties. This organization led him to
propose the periodic table and use it to predict the existence
and properties of a number of additional elements. When
some of those that were foretold in 1869 were actually
discovered a few years later, Mendeleev was hailed as a
prophet.

This inspiring teacher and tireless experimenter was so
deeply concerned over social issues that he resigned his
professorship rather than obey an order to cease interfering
with affairs of government. He made enemies by supporting
liberal movements and even defied the Czar's wishes by
refusing to cut his hair. Nevertheless, he won the appoint-
ment as Director of the Bureau of Weights and Measures.

When Mendeleev first published his chart, there were
63 elements known. One year after his death, there were 86.
The rapidity of this increase was made possible by the
most important generalization of chemistry, the periodic
table.

CHAPTER

Energy Effects in
Chemical Reactions

Although a typical chemical reaction • • • may appear far removed
from the working of an engine, the same fundamental principles of heat
and work apply to both.

f. T. wall, 1958

Chemical reactions form the heart of chemistry.
And there is no more important aspect of chemi-
cal reactions than the energy effects that are
caused. You will realize this if you let your
thoughts wander between the warmth the little
child in the fable derived from the combustion

of wood and the celestial joy ride an astronaut
receives from the reactions of his rocket fuels.
How much energy is involved in a chemical re-
action? How do we find this out? Where does
this energy come from? We shall investigate
these questions in this chapter.

7-1 HEAT AND CHEMICAL REACTIONS

At a temperature of 600°C, steam passed over
hot coal (coal is mostly carbon) reacts to give
carbon monoxide and hydrogen:

H20(g) + C(s) — >- CO(g) + H2(g) (7)

This reaction is quite useful because the mixture
of product gases, called "water gas," is an ex-
cellent industrial fuel. In the commercial prepa-
ration of water gas, the chemical engineer must
allow for the absorption of heat during the reac-
tion. In fact, he must periodically turn off the
steam and reheat the coal to keep the reaction
going. To aid the engineer, we might measure
108

the amount of heat absorbed by the system and
write it into the chemical reaction. Such a meas-
urement shows that 3 1 .4 kcal of heat are absorbed
per mole of carbon reacted. Since the heat is
used up (as are the reactants), we can rewrite
reaction (7) and show the heat on the left side
of the equation:

H20(g) + C(s) + 31.4 kcal — >■

CO(g) + H2(g) (la)

Now we might think of the mechanical engi-
neer who is designing a boiler to be heated with
water gas fuel. He is interested in burning the

SEC. 7-1 I HEAT AND CHEMICAL REACTIONS

109

water gas. This involves two chemical reactions
of combustion:

and

co(g) + \o,(g) — >■ co2(g) (2)

Wg) + \02(g) —> H2Q(g) (3)

These reactions release heat, and our mechanical
engineer wishes to know how much. Again, we
might help by measuring these amounts of heat
and adding the information to reactions (2) and
(3). Since heat is produced by the reaction (as is a
chemical product), we should place it on the
right side of the equation. Experiments show:

CO(g) + \02(g)
H2(g) + \Q2(g)

C02(g) + 67.6 kcal (2a)
H2Q(g) + 57.8 kcal (3a)

Now let us talk with the business manager. He
thinks in terms of gains and losses. He is likely
to observe that the consumption of coal and
water to generate water gas is followed by the
combustion of the water gas to form carbon
dioxide and water. Without knowing much
chemistry, he can see that what is finally accom-
plished is the combustion of coal to form carbon
dioxide. The overall reaction is obtained by add-
ing reactions (7), (2), and (3):

H20(g) + C(s)

CO(g) + \o2(g)

Wg) + \02(g)

CO(g) + H2(g) (1)
C02(g) (2)

H2OteJ (3)

Overall
reaction :

C(s) + 02(g)-+C02(g)

(4)

The business manager is frugal so he asks, "Why
not burn the coal directly and save the cost ol
manufacturing the water gas?" The mechanical
engineer is practical so he asks, "How much heat
will the boiler receive if I use coal instead of
water gas?" The chemical engineer goes to the
laboratory to find the answers by measuring the
heat released per mole of carbon burned in reac-
tion (4). The laboratory result shows that re-
action (4) releases 94.0 kcal/mole:

C(s) + 02(g) -+ C02(g) + 94.0 kcal (4a)

The chemical engineer now can answer all of the
questions. If one mole of carbon is burned di-
rectly, 94.0 kcal of heat are released for the

mechanical engineer. The same amount of coal
converted into water gas releases the sum of the
heats evolved in reactions (2a) and (5a):

67.6 + 57.8 = 125.4 kcal

The mechanical engineer has a better fuel in
water gas than in coal.

With new respect, the business manager might
now ask the chemical engineer, "Where did this
extra heat come from?" "Did we get something
for nothing?" The answer is, "No." The water
gas releases more heat per mole of carbon be-
cause the chemical engineer put that amount of
heat in during reaction (7a). The business man-
ager's ledger is shown in Table 7-1.

Table 7-1

HEAT EFFECTS IN THE MANUFACTURE
AND USE OF WATER GAS

DEBIT CREDIT

Reaction (7a): heat absorbed

31.4 kcal

—

Reaction (2a): heat released

—

67.6 kcal

Reaction (3a) : heat released

—

57.8 kcal

Overall reaction:
(la) + (2a) + (3a) = (4a)

31.4 kcal
absorbed

125.4 kcal
released

125.4

- 31.4

94.0 net

Experiment reaction (4a):
Heat released

94.0 kcal

7-1.1 Heat Content of a Substance

The example just given shows that the 31.4 kcal
absorbed during reaction (7a) was "stored" in
the water gas. Furthermore, the amount of en-
ergy "stored" is definite, not alterable at the
demand of the business manager or the whim
of the chemical engineer. How much energy is
stored depends upon the reactants and products
in reaction. We must add a fixed amount of
energy (as heat) to coal and steam to make a
specified amount of carbon monoxide and hy-
drogen. This heat is retained by the CO and H2
molecules, as shown in Table 7-1. We might say
that reaction (7a) increases the "heat content" of

110

ENERGY EFFECTS IN CHEMICAL REACTIONS I CHAP. 7

the atoms of the reactants by rearranging them
to form the products. Apparently a mole of each
molecular substance has a characteristic "heat
content" just as it has a characteristic mass. This
heat content measures the energy stored in a
substance during its formation. The heat effect in
a chemical reaction measures the difference be-
tween the heat contents of the products and the
heat contents of the reactants. If more energy is
stored in the reactants than in the products, then
heat will be released during reaction. Conversely,
heat will be absorbed if more energy is stored
in the products than in the reactants.

This idea — that each molecular substance has
a characteristic heat content — provides a good
explanation of the heat effects found in chemical
reactions. Chemists symbolize heat content by H.
Since the heat effect in a reaction is the difference
between the 7/'s of the products and the 7/'s of
the reactants, the heat of reaction is called AH,
the Greek letter A (delta) signifying "difference."

We can see what AH means in terms of an
example. Consider reaction (7):

H20(g) + C(s) — >- CO(g) + H2(g) (1)

A heat content, H, is assigned to each substance.
Then AH for reaction (7) is the difference :

Afr _ I heat content \ / heat content \

\ of products / \ of reactants /

= \Hco + 7/h«] — [7/hjO + He]

= Hco 4" 7/hi — Hhio — He

Products

Heat
content

AH
positive

Reactants

Since the reaction consumes heat, the heat con-
tent of the products is higher and AH will be
positive. We can express this by writing

H20(g) + C(s) — * CO(g) + H2(g)

AH = +31.4 kcal (lb)

Reaction (lb) is exactly equivalent to reaction
Ua):

UiO(g) + C(s) + 31.4 kcal — >-

CO(g) + H2(g) (la)

Let's try this on reaction (2):

.„ _ /heat content \ _ /heat content \
\of products / \of reactants /

= Hco, — [Hco + T/iOi]
= Hcch — Hco — Hio,

In this reaction, heat is evolved ; hence, the heat
content of the products is below that of the re-
actants. Therefore, AH must be negative:

CO(g) + ±02(g) -+ C02(g)

AH = -67.6 kcal (2b)

This has exactly the same meaning as

CO(g) + \02(g) — >- C02(g) + 67.6 kcal (2a)

We see that the sign of AH is sensible. It is
positive when heat content is rising (by heat
absorption) and it is negative when heat content
is dropping (by heat evolution). This is shown
diagrammatically in Figure 7-1.

Fig. 7-1. Heat content change during a reaction.

'Rea.c1ra.nJrs

Heatr
Con.iren.tr

H
ative

Products

Reaction, proceeding

Heair is ahsorhed
A H is positive

Reaction, proceeding

Ueatr is evotved
AH is negative

SEC. 7-1 I HEAT AND CHEMICAL REACTIONS

111

7-1.2 Additivity of Reaction Heats

Let us return to the debit and credit balance we
found in our water gas fuel problem. In terms
of AH, the heat effects are as follows:

Source of
direct current

H20(g) + C(s)

CO(g) + hojg)
H,(g) + \Q2(g)

CO(g) + H2(g)

AH, = +31.4 kcal (lb)

C02(g)

AH, = -67.6 kcal (2b)

H20(g)

AHS = -57.8 kcal (3b)

Overall reaction (/) + (2) + (3) = (4):

Qs) + cygj — ►- co,(g)

A//, = -94.0 kcal (4b)

We discover that not only is reaction (4) equal
to the sum of reactions (lb) + (2b) + (3b) in
terms of atoms, but also that

AH4 = AH, + AHt + AH,

= 31.4 + (-67.6) + (-57.8)

= 31.4 - 67.6 - 57.8

= -94.0 kcal

We see that when a reaction can be expressed as
the algebraic sum of a sequence of two or more
other reactions, then the heat of the reaction is the
algebraic sum of the heats of these reactions. This
generalization has been found to be applicable
to every reaction that has been tested. Because
the generalization has been so widely tested, it
is called a law — the Law of Additivity of
Reaction Heats.*

7-1.3 The Measurement off Reaction Heat

The measurement of reaction heats is called
calorimetry — a name obviously related to the
unit of heat, the calorie. You already have some
experience in calorimetry. In Experiment 5 you
measured the heat of combustion of a candle and
the heat of solidification of paraffin. Then in
Experiment 13 you measured the heat evolved
when NaOH reacted with HC1. The device you
used was a simple calorimeter.

* This generalization was first proposed in the year
1840 by G. H. Hess on the basis of his experimental
measurements of reaction heats. It is sometimes called
Hess's Law of Constant Heat Summation.

Thermometer

Insulation.

Water
Reaction chamber

Resistance wire
for
y< pS ianitiru/ charge

Fig. 7-2. General plan of a calorimeter.

Calorimeters vary in details and are adapted
to the particular reaction being studied. Figure
7-2 shows the general plan of a calorimeter that
might be used in measuring the heat evolved
during a combustion reaction. It might be ap-
plied to the combustion of a candle to yield a
much more reliable answer than can be obtained
by the crude technique of Experiment 5.

Essentially, the device consists of an insulated
vessel containing a known weight of water. A
weighed amount of the substance to be burned
and an excess of oxygen are introduced under
pressure into a reaction chamber placed in the
vessel. The reaction mixture is ignited by means
of an electrical resistance wire sealed into the
reaction chamber. The heat produced by the
reaction changes the temperature of the water,
which is stirred to keep its temperature uniform.
From this temperature change and from the
amount of heat required to raise the temperature
of the calorimeter and its contents by one degree,
the heat of combustion per mole of substance
burned can be calculated. By using a large
amount of water, the actual change in tempera-
ture is kept small. This is desirable in that it
keeps the final temperature of the products of
reaction fairly close to the initial temperature of
the reactants.

112

ENERGY EFFECTS IN CHEMICAL RBACTIONS I CHAP. 7

EXERCISE 7-1

Suppose reactants are mixed in a calorimeter at
25°C and the reaction heat causes the tempera-
ture of the products and calorimeter to rise to
35°C. The resultant determination of AH applies
to what temperature? Explain why it is desirable
to keep the final temperature close to the initial
temperature in a calorimetric measurement.

In the combustion reaction as carried out in the
calorimeter of Figure 7-2, the volume of the system is kept
constant and pressure may change because the reaction
chamber is sealed. In the laboratory experiments you
have conducted, you kept the pressure constant by leaving
the system open to the surroundings. In such an experi-
ment, the volume may change. There is a small difference
between these two types of measurements. The difference
arises from the energy used when a system expands
against the pressure of the atmosphere. In a constant
volume calorimeter, there is no such expansion; hence,
this contribution to the reaction heat is not present. Ex-
periments show that this difference is usually small. How-
ever, the symbol AH represents the heat effect that
accompanies a chemical reaction carried out at constant
pressure — the condition we usually have when the reac-
tion occurs in an open beaker.

7-1.4 Predicting the Heat of a Reaction

Chemists have measured the heats of many re-
actions. With the measured values, many un-

measured reaction heats can be predicted by
applying the Law of Additivity of Reaction
Heats. Consequently, a compilation of known re-
action heats is extremely useful. Table 7-II is
such a compilation.

Suppose we are interested in the heat of com-
bustion of nitric oxide, NO:

NOfej + \02(g) — ►- N02fg; AH5 (5)

Since reaction (5) can be obtained by combin-
ing two reactions in Table 7-II, we can predict
AH5. In Table 7-II we find

\K2(g) + \02(g) — > NO(g)

AH6 = +21.6 kcal/mole NO (6)

hK2(g) + 02(g) -^N02(g)

AH7 = +8.1 kcal/mole N02 (7)

Now we wish to obtain reaction (5) by combin-
ing reactions (6) and (7). Since NO is a reactant
in reaction (5), we need the reverse of reaction
(<5). We obtain the heat of the reverse reaction
merely by changing the algebraic sign of AH6. If
21.6 kcal of heat are absorbed when one mole
of NO is formed, then 21.6 kcal of heat will be
released when one mole of NO is decomposed
in the reverse reaction:

NO(g) — >■ ±N2(g) + \02(g)

AH8 = -21.6 kcal/mole NO (8)

Now we can add reactions (7) and (8) to obtain
reaction (5):

Table 7 -II. heats of reaction between elements, t = 25°C, p = 1 atm

ELEMENTS

COMPOUND

FORMULA

NAME

HEAT OF REACTION

(kcal/mole of product)

Wg) + hOt(g) — ►■ H20(gj water vapor

H2(gJ + \02(g) — >- H20(7J water

S(s) + 02(g) — ►■ S02(g) sulfur dioxide

H2(g) + S(s) + 202(g) — >- H2SOi(l) sulfuric acid

hWg) + k02(g) — ►■ NOteJ nitric oxide

^N2fgJ + 02(g) — >- N02(gj nitrogen dioxide

h^2(g) + fHrf*; — ► NH3fej ammonia

C(s) + \02(g) — ►■ CO(g) carbon monoxide

C(s) + 02(g) — >- C02(g) carbon dioxide

2C(s) + 3H2(g) — )- dH,(g) ethane

3C(s) + 4U2(g) — >■ CsHs(g) propane

*H2(gJ + \\t(g) — >■ Hl(g) hydrogen iodide

-57.8
-68.3
-71.0
-194
+21.6

+8.1
-11.0
-26.4
-94.0
-20.2
-24.8

+6.2

SEC. 7-2 I THE LAW OF CONSERVATION OF ENERGY

113

NO(g) — > $Wg) + \Oi(g)

hWg) + o2(g) — ►■ no^j

Overall reaction:

NO(g) + 02(g) + h*i2(g) -+ N02(g) + \02(g) + iN./gj

NO(g) + \Q2(g) —*■ NO/gJ

AH, = -21.6 kcal/mole NO (5)

AH7 = +8.1 kcal/mole N02 (7)

AH6 = -21.6 + 8.1

AHS = -13.5 kcal/mole NO (5)

EXERCISE 7-2

Predict the heat of the reaction

CO(g) + ho2(g) — + C02(g)
from two reactions listed in Table 7-II. Compare
your result with AHeb given in Section 7-1.2.

The usefulness of Table 7-II is obvious from
these examples. Many, many reaction heats can
be predicted— in fact, the heat of any reaction
that can be obtained by adding two or more of
the reactions in the table. Furthermore, there is
an easy way to decide whether the table contains
the necessary information for a particular ex-
ample. A given reaction can be obtained by add-
ing reactions in Table 7-II, provided every com-
pound in the reaction is included in the table.
The elements participating in the reactions auto-
matically will appear in proper amount.

Consider a more complicated example — the
oxidation of ammonia, NH3:

KH3(g) + i02(g) — >-

K02(g) + %H20(g) AH9 (9)

In reaction (9) we find three compounds, NH3(gJ,
N02(g,), and H20(g). These are all found in
Table 7-11. Consequently, we are able to calcu-
late AH9.

EXERCISE 7-3

Convince yourself that reaction 9) and also
AH9 = —67.6 kcal can be obtained by carrying
out the indicated summation:

From

Subtract

\N2(g) + 02(g) —+ N02(gj
AH = +8.1 kcal

hn2(g) + § H2(g) — > Ntug)

Subtract AH = -11.0 kcal

And now add § times H2(g) + \02(g) — >■ H20(g)
Add | times AH = -57.8 kcal

Thus, when we wish to predict the heat of
some reaction, it takes but a moment to decide
whether the compilation of Table 7-11 includes
the necessary reactions. If every compound in the
reaction of interest is in the "Compound" col-
umn of Table 7-II, then the prediction can be
made. This is a convenience provided by a com-
pilation of heats of reaction between elements
and explains why these reaction heats are the
ones chemists tabulate. Of course, the list in
Table 7-II includes only a small fraction of the
known values. Many more reaction heats are
tabulated in handbooks; they are listed in in-
dexes under Heat of Formation.

7-2 THE LAW OF CONSERVATION OF ENERGY

We have seen how chemists measure the heat of
a reaction. Using a compilation of measured
values, we can predict the energy changes of
many reactions that have not been measured.
Thus, the rule of Additivity of Reaction Heats

is a very useful and reliable generalization. It
makes us wonder "Why should it be so?" The
explanation, as usual, is found by connecting the
behavior of a chemical system to the behavior
of other systems that are better understood.

114

ENERGY EFFECTS IN CHEMICAL REACTIONS I CHAP. 7

Fig. 7-3. Conservation of energy in a collision of
billiard balls.

7-2.1 Conservation off Energy in a Billiard
Ball Collision

Figure 7-3 shows an experimental study of the
collision of hard spheres. The experimenter im-
parts energy of motion to the white ball (see
Figure 7-3A, 7-3B). He does so by doing work
by striking the ball with the end of a cylindrical
stick (a cue). The amount of energy of motion
(kinetic energy) received by the ball is fixed by
the amount of work done. If the ball is struck
softly (little work being done), it moves slowly.
If the ball is struck hard (much work being done),
it moves rapidly. The kinetic energy of the white
ball appears because work was done — the
amount of work, Wu determines and equals the
amount of kinetic energy, (KE)X. In symbols,

Wx = (KE)V

UO)

Suppose the direction of motion of the white
ball causes it to contact the motionless red ball.
A collision occurs. Figure 7-3C shows the result.
The white ball has a lower kinetic energy, (KE}i,
but now the red ball is moving! The red ball now
has energy of motion — let's call it (KE)3. Meas-
urements show that velocities are such that the

kinetic energy gained by the red ball is equal to
the kinetic energy lost by the white ball. In sym-
bols,

(KE)X = (KE)2 + (KE)3 (11)

Let us review this experiment. First, an
amount of work was performed, Wx, with the
cylindrical stick. The amount of kinetic energy
received by the white ball, (KE)X, was exactly
fixed by Wx. After the white ball collided with
the red ball, the sum of the energies of the two
balls, (KE}t + (KE)3, is exactly equal to (#£),.
We can write

Wx = (KE), = (KEh + (KE)3 (12)

If we recognize work as a form of energy,
we may say that energy was conserved during the
experiment. Every quantity in (12) is known —
we can measure the work done as we do it and
we can learn the magnitudes oi(KE)u (KE)?, and
(KE)3 through velocity and mass measurements.
Many such experiments are performed every day
and the results are always in agreement. In a
billiard parlor, energy is conserved.

7-2.2 Conservation off Energy in a
Stretched Rubber Band

Figure 7-4 shows an experimental study of the
stretching of a rubber band. The rubber band is

Fig. 7-4. Conservation of energy in a stretched rubber
band.

SBC. 7-2 I THB LAW OF CONSERVATION OF ENERGY

115

stretched and hooked over the end of the testing
device, as shown in Figure 7-4A. Work must be
done to stretch it— let's call it Wx. In Figure 7-4C
the rubber band has been released and has re-
turned to its initial length, but it now has energy
of motion, (KE)3. How much energy? The kinetic
energy the rubber band has depei. jts upon how
much work wa' done in stretching it. Wx fixes
(KE)z. We may write

Wx = (KE)»

U3)

Again we recognize work as a form of energy
and, since Wx = (KE)3, the overall result is that
energy was conserved. But was energy conserved
in Figure 7-4B? If so, where is the energy? The
work Wx has already been done. Though the
rubber band is stretched, there is no outward
manifestation of energy. The rubber band is mo-
tionless, so it has received no kinetic energy.
Yet, we know (from previous experiments) that
the rubber band will receive the energy (KE)3 at
any instant that we choose to release it. The
initial and final result are reminiscent of the
billiard balls: energy is conserved. It would be
convenient to say that energy is conserved in
Figure 7-4B, as well. So, we invent a new form
of energy — energy of position, or potential en-
ergy. We say that as the work Wx was performed
it was stored in the rubber band as potential
energy.

Now we can review this experiment. In sym-
bols we have

Wx = (PE)2 = (,KE)t

work potential kinetic
energy energy

U4)

The amount of work performed fixed Wx. Meas-
urements of mass and velocity of the rubber band
tell us, experimentally, the magnitude of (KE)Z.
How do we know (PE^l How are we sure that
(P£>2 is equal to Wx and to (KE^l The evidence
we have is that we put an amount of energy into
the system and can recover all of it later at will.
It is natural to say the energy is stored in the
meantime. Then we can say that the rubber band
is just like the billiard ball collision: energy is
conserved at all times.

Electrodes

Source

electrical

energy,

direct

current-

Fig. 7-5. Electrolytic decomposition of water.

7-2.3 Conservation of Energy in a
Chemical Reaction

Figure 7-5 shows an apparatus in which an elec-
tric current can be passed through water. As
remarked in Section 3-1.2, the electric current
causes a decomposition of water. As work is
done (electrical work), hydrogen gas and oxygen
gas are produced. Measurements of the electric
current and voltage show that 68.3 kcal of elec-
trical work, Wx, must be done to decompose one
mole of water. The equation for the reaction is

68.3 kcal + HMD —+ Wg) + Wg) (75)

electrical
work

Now suppose we measure the heat of reaction
of hydrogen and oxygen in a calorimeter like
that shown in Figure 7-2. This experiment has
been performed many times; 68.3 kcal of heat,
Qz, are produced for every mole of water formed.
The equation for this reaction is

Wg) + \Oi(g) — >- H20(l) + 68.3 kcal heat (76)

We have a situation just like that of the rubber
band. We put a readily measured amount of
energy, Wx, into the system and, at any time

116

ENERGY EFFECTS IN CHEMICAL REACTIONS I CHAP. 7

later that we choose, this energy can be recovered
as heat, Q%:

W, = Q, (17)

The overall result is that energy is conserved.

Figure 7-7 shows this schematically in a dia-
gram like that of Figure 7-1. If the heat content
of two moles of water is represented by a line
on this diagram, then the energy of two moles
of hydrogen plus one mole of oxygen should be
represented by a line 136.6 kcal higher. Now the
diagram indicates that when water is decom-
posed, energy must be put in to raise the heat
content enough to form the products. When hy-

Fig. 7-6. Conservation of energy in a chemical
reaction.

drogen burns in oxygen, the heat content drops
and energy is released.

Again we may ask : Where was the energy put
in reaction (15) before we carry out reaction
(16)1 The rubber band experiment guides us.

It is easier to explain why Wx = Q3 if we say
that the energy W\ was stored in the chemical
substances U2(g) and O-rfg). We assign to these
(and all other) substances the capacity to store
energy and we call it heat content. This permits
us to say that energy is conserved at all times
during a chemical reaction as it is in billiard ball
collisions and in stretched rubber bands.

7-2.4 The Basis for the Law

off Conservation of Energy

We see that there are many forms of energy. We
have talked about work as a form of energy and

Two moles
liqti id wcvte r>

2H20 (I)

-h 2(68.3 kcal)

, 136. 6 kcal

electrical work

+■

tifcl

two moles
hydroge-n gas

2Kz(g)

one mole
oxygen, qcls

+ 02 (f)

+ Z [68.3 kcal)

Two mx>les

+■

one m-ole
oxygen. ga\s

two moles

136.6 kcal

hydrogen gas

liquid wafer

heat

2H2 (9)

+■

n / l

°z (<fJ

*■ 2J120 ( IJ

= 0s

SEC. 7-3 | THE ENERGY STORED IN A MOLECULE

117

Heat
content
(potential
energy)

2H2(9)+ Q2(g)

AH =136.6 kcal

Decomposition of water-
Formation of water-

Fig. 7-7 . The energy change in a chemical reaction.

referred to two kinds of work, muscular work
and electrical work. Kinetic energy and heat are
other readily measured forms of energy. We have
added two additional forms that are only in-
directly detectable — the potential energy of the
stretched rubber band and the heat content of
chemical substances. With these two added
forms, we can write:

Energy is always conserved.

This is called the Law of Conservation of
Energy. It says that in every experiment so far
performed, energy was conserved provided all of
the different forms of energy are taken into ac-
count. Because the number of such experiments
is extremely large and varied in type, the law
gives a reliable basis for predicting. Don't be
upset that the law is true only because we added
forms of energy to account for energy not di-
rectly measurable. This law is like any other law
— its usefulness justifies it. As long as the several
forms of energy give us a model that is always
consistent with experiments, the law remains
useful.

In specific reference to the heat effects in chem-
ical reactions, hundreds of different reactions
have been studied calorimetrically. The results
are always in accord with the Law of Additivity
of Reaction Heats. If we assign a characteristic
heat content to each chemical substance, then
all of these experiments support the Law of Con-
servation of Energy. Since the Law of Conserva-
tion of Energy is consistent with so many dif-
ferent reactions, it can be safely assumed to apply
to a reaction which hasn't been studied before.

The term "law" is usually applied to the older scientific
generalizations. Modern scientists do not apply the term
to new generalizations because they realize that all "laws
of nature" are human statements — human generalizations
— and are subject to revision. For example, later in this
chapter you will find that matter and energy are one and
the same. At that time, you will see that the two conserva-
tion laws, the Conservation of Matter and the Conser-
vation of Energy, are really but one law, the Conservation
of Matter (which is Energy). Yet in any chemical process
the mass equivalent of the reaction heat is negligible.
Under these conditions, the Law of Conservation of
Matter and the Law of Conservation of Energy can be
considered as independent statements. In this form the
conservation laws are very useful, even though the state-
ments we have made about them do not apply under all
conditions.

7-3 THE ENERGY STORED IN A MOLECULE

In the discussion of Sections 7-2.1 to 7-2.4, we
found it useful to talk about different "forms"
of energy. Two of these are heat and heat con-

tent. Heat content is sometimes called "chemical
energy" because its magnitude is intimately tied
up with chemical composition. These are macro-

118

ENERGY EFFECTS IN CHEMICAL REACTIONS I CHAP. 7

scopic* manifestations of energy. Two other
macroscopic manifestations of energy are kinetic
energy (of a thrown baseball, for example) and
potential energy (of a baseball at the top of the
flight of a high foul ball, for example). Thus, we
need to identify several "forms" of energy when
discussing the macroscopic properties of sub-
stances: heat, heat content (chemical energy),
kinetic energy, potential energy, electrical en-
ergy, and mechanical work. The presence and
amount of each energy form is determined by
methods uniquely applicable to that form. We
determine the quantity of heat released in a
calorimeter by measuring a temperature rise with
a thermometer. We measure the kinetic energy
of a baseball with a watch and a meter stick.
You would learn little about the kinetic energy
of a baseball by throwing it at a thermometer
and nothing about water temperature by wearing
your wrist watch in the shower.

When we turn to the molecular scale, however,
we discover that all of these macroscopic forms
of energy can be discussed in terms of the two
kinds of energy we assigned to the baseball,
kinetic and potential energies. We can "explain"
all macroscopic forms of energy with a micro-
scopic model involving only the energy of motion
and energy of position of the atomic and molecu-
lar particles. The explanation has the special
advantage given in Section 1-1.3 (pp. 5-8).

* Macroscopic means on a large scale — the opposite
of microscopic. In general, it is used to indicate weighable
and visible amounts.

/'■■'.' \

' J>

^•0--£J

?>-i

i~iu& v^

»w

))»■

All of our experience and knowledge about the
properties of moving baseballs (and billiard balls
and rubber bands and automobiles and pendu-
lums and gyroscopes) can be used in clarifying
the nature of heat, heat content, electrical energy,
etc. To see this, we must consider how chemists
discuss the energy held by a molecule.

7-3.1 The Energy off a Molecule

Let us picture a molecule in terms of a model
consisting of balls of proper relative masses
hooked together by springs. The springs repre-
sent the bonds between the atoms. We can start
the springs vibrating and then toss the entire
assembly through space in an end-over-end mo-
tion. There are now three kinds of kinetic energy
associated with our model, as pictured in Fig-
ure 7-8.

This model applies quite well to a molecule
in the gaseous state, but in the liquid state, and
(even more so) in the solid state, all these mo-
tions are restricted. In these phases the chief
kinetic energy manifestation is a back-and-forth
motion of the molecule about a fixed point.

Fig. 7-8. Types of motion of a molecule of carbon
dioxide, CO2. A. Translational motion; the
molecule moves from place to place. B. Ro-
tational motion; the molecule rotates about
its center of mass. C. Vibrational motion;
the atoms move alternately toward and away
from the center of mass.

\ I

)))) > .

5 \m>

SEC. 7-3 I THE ENERGY STORED IN A MOLECULE

119

In addition to the various kinds of kinetic
energy listed in Figure 7-8, there is potential en-
ergy related to the forces which act between
molecules. These forces are attractive, having a
very small average value in the gaseous state in
which the molecules are far apart. The forces are,
on the average, larger in the liquid state and are
still larger in the solid state.

Next, there is present, within the molecule,
chemical energy which is related to the forces
which hold the atoms together in the molecule.
This is referred to as chemical bond energy.

In addition, each atom has energy, some as-
sociated with the electrons and some with the
nucleus. The electrons in the atom possess kinetic
energy and, because of their attraction to the
nucleus and repulsion from each other, they also
possess potential energy. The algebraic sum of
these kinetic and potential energies represents
the energy necessary to pull an electron away
from an atom.

Finally, there is present within the nucleus of
each atom a store of energy. This energy is re-
lated to the forces holding the nuclear particles
together. Since each nucleus remains intact and
apparently uninfluenced through chemical reac-
tions, this nuclear energy does not change.
Hence, the nuclear contribution to the molecular
heat content does not usually concern a chemist.

The sum of all of these forms of molecular
energy makes up the molecular heat content. If
we add together the molecular heat content of
6.02 X 1023 molecules of a given kind, we obtain
the molar heat content of that substance.

7-3.2 Energy Changes on Warming

Having this view of the make-up of the heat con-
tent of a substance, we can now visualize the
effects brought on by warming the substance. If
the temperature is low at first, the substance will
be a solid. Warming the solid increases the ki-
netic energy of the back-and-forth motions of
the molecules about their regular crystal posi-
tions. As the temperature rises, these motions
disturb the regularity of the crystal more and
more. Too much of this random movement de-
stroys the lattice completely. At the temperature

above which the kinetic energy of the particles
causes so much random movement that the lat-
tice is no longer stable, a phase change occurs:
the solid melts.

In the liquid each molecule has considerably
more freedom of movement, particularly for
translation and rotation. Warming the liquid
enhances the amount of molecular movement.
As always, kinetic energy provides a randomiz-
ing effect, tending to carry the molecules every-
where in the container. As the energy of motion
rises (with the rising temperature), more of the
molecules are able to move away from the liquid
region where the potential energy is a minimum.
Another phase change occurs: the liquid va-
porizes.

If, now, we continue warming the substance
sufficiently, we will reach a point at which the
kinetic energies in vibration, rotation, and trans-
lation become comparable to chemical bond en-
ergies. Then molecules begin to disintegrate.
This is the reason that only the very simplest
molecules — diatomic molecules — are found in
the Sun. There the temperature is so high
(6000°K at the surface) that more complex mole-
cules cannot survive.

Finally, if we continue the heating still further,
we will ultimately reach a temperature at which
the kinetic energies are large enough to disrupt
the nuclei. Then, "nuclear reactions" begin. The
conditions in some stars are considered to be
suitable for rapid nuclear reactions.

To conclude this study, let's consider the mag-
nitudes of the energy effects. Phase changes usu-
ally involve energies of several kilocalories per
mole. Chemical reactions usually involve ener-
gies of 50 to several hundred kilocalories per
mole. Thus, we see that the energies involved in
chemical reactions are usually 10 to 100 times
larger than those involved in phase chages.

EXERCISE 7-4

Show that the ratio of the molar heat of for-
mation of gaseous water from the elements (a
chemical reaction) to the molar heat of the fusion
of water (a phase change) is of the order of 50.

120

ENERGY EFFECTS IN CHEMICAL REACTIONS I CHAP. 7

7-4 THE ENERGY STORED IN A NUCLEUS

Now let us consider nuclear changes. The fact
that nuclei do remain intact during chemical re-
actions suggests that much larger energies are
required for nuclear changes. Experimentally,
this proves to be true. Nuclear reactions usually
involve energy changes over a million times
greater than those we find in chemical reactions.
This enormous factor accounts for the current
interest in nuclear reactions as a source of en-
ergy.

One such nuclear reaction is represented by
the equation

2giU + In — >- 'tlBa + llKr + Ion + energy (18)

Before we examine the details of this rather
strange looking equation, let us focus our atten-
tion on the "+ energy" term. The numerical
value is of the order of 4.5 X 109 kcal/mole of
uranium. Look at that figure again and compare
it to the molar heat of combustion of carbon.
Roughly, what is the ratio of these two energies?
It is 107, or 10 million!

Now let us examine the reaction in more de-
tail. Forget momentarily the subscripts and su-
perscripts. Recall from Chapter 6 that the
neutron («) is one of the fundamental particles
visualized as present in nuclei. What has hap-
pened?

U + n — >- Ba + Kr + 3/7 (19)

Instead of producing new kinds of substances by
combination of atoms, the element uranium has
combined with a neutron and as a result has
split into two other elements — barium and kryp-
ton— plus three more neutrons. Atoms of a given
element are characterized by their atomic num-
ber, the number of units of positive charge on
the nucleus. For one element to change into
another element the nucleus must be altered. In
our example the uranium nucleus, as a result of
reacting with a neutron, splits or fissions into two
other nuclei and releases, in addition, neutrons.*

How we get neutrons and how we get them to
react with uranium nuclei is not essential to our
present discussion.

A glance at the periodic table will show that
the subscripts we have attached to our symbols
are the atomic numbers of the elements desig-
nated by the symbols — 92 for U, 56 for Ba, 36
for Kr. The zero subscript attached to the neu-
tron denotes the lack of charge on this particle.
If we look at the subscripts,

92U + 0n — >■ MBa + seKr + 3o/i (20)

we notice that their sum on each side of the
equation is identical:

92 + 0 = 56 + 36 + (3 X 0)

This identity is but another way of expressing
the law of conservation of charge.

In the model of nuclear structure you were
given in Chapter 6, the nucleus was pictured as
being built up of protons and neutrons. These
two kinds of particles are given the general name
nucleon. The mass number of a nucleus is
equal to the number of nucleons present. The
superscripts in our equation are mass numbers:

235IJ + 1„

'Ba + wKr + 3'/i

(21)

Apparently the mass numbers are also con-
served :

235 + 1 = 141 + 92 + (3 X 1)

We may rephrase this in the form of a rule: The
total number of nucleons is unchanged during nu-
clear reactions.

EXERCISE 7-5

According to the model of Chapter 6, how many
nucleons would be present in a uranium nucleus
of mass number 235? How many protons are
pictured as being present? How many neutrons?

* Perhaps you have already recognized our nuclear
reaction as a fission reaction. It is of the type of reaction
used in an atomic pile, the energy source of a nuclear

power plant. The example we have selected is only one
of the ways the uranium nucleus can divide. Lanthanum
and bromine nuclei are also produced, cerium and sele-
nium, and so on, each pair of fission products being such
that the sum of their atomic numbers is always 92.

SEC. 7-4 I THE ENERGY STORED IN A NUCLEUS

121

Actually, then, by our symbol "iJU we are
representing not an atom, but a nucleus. Our
equation is written in terms of nuclei and par-
ticles associated with them. This nuclear equa-
tion tells us nothing about what compound ot
uranium was bombarded with neutrons or what
compound of barium is formed. We are sum-
marizing only the nuclear changes. During the
nuclear change there is much disruption of other
atoms because of the tremendous amounts of
energy liberated. We do not know in detail what
happens but eventually we return to electrically
neutral substances (chemical compounds) and
the neutrons are consumed by other nuclei.

It is easy to determine that there is an electron
balance between the reactants and products:

Nucleus: »2U on wBa 36Kr on

Electrons associated
with nucleus:

92 0

92 = 56 + 36

7-4.1 Exact Mass Relationships

Although the mass numbers of the proton and neutron
are both one, the masses of these fundamental particles
are not identical. The mass of one mole of protons is
1.00762 grams and that of one mole of neutrons is
1.00893 grams. Further investigation would show that
the experimentally measured mass of the nucleus of any
given isotope is not the exact sum of the masses of
protons and neutrons confined in the nucleus according
to our model. For example, the mass of the nucleus of
the uranium isotope of mass number 235 is less than the
exact sum of the masses of 92 protons and 143 neutrons.

One of the consequences of the special theory of rela-
tivity formulated in 1905 by the great German theoretical
physicist, Albert Einstein, was that we came to realize
that mass and energy are one and the same. Although
this was a very radical notion at the time Einstein first
presented his theory, the equation relating mass to energy
is probably already familiar to you. The formula E = mc*
has become almost a part of common idiom since the
successful application of nuclear energy became a part
of modern technology in the mid 1940's. In this equation
c is the speed of light, 3.00 X 10'" cm/second. Apparently
a small value of mass (m) is equivalent to a tremendous
amount of energy since the proportionality constant (c2)
relating mass to energy is numerically 9.00 X 10'°.

We can use this idea of the relation of mass to energy
in several ways. The mass of a " U nucleus is less than
the sum of the masses of the 92 protons and 143 neutrons
postulated to be in it. The difference in mass represents
the binding energy which holds the nucleons together in

the nucleus. Here we have used the concept of expressing
the nuclear binding energy in terms of the implied de-
crease in mass. We can do the same, if we wish, for a
chemical reaction. Again let us return to the molar heat
of combustion of carbon, roughly 102 kcal:

Qs) + Ot(g)

CO>(g) AH = -94 kcal (22)

The mass change associated with an energy change of
102 kcal* is of the order of 5 X 10~9 grams. This is a
quantity far too small to be detected on any balance
capable of weighing the 12 grams of carbon and 32 grams
of oxygen consumed in the reaction. Since the chemical
"mass defects" are too small to measure, we do not use
this terminology in chemistry.

If we wish to gain some idea of the alteration of mass
in a nuclear change, we cannot use the fission reaction
because the exact masses of the nuclei involved are not
known. Let us look at another type of reaction of pos-
sible importance in the production of nuclear energy:

iH +

•'He + lQn

(23)

This reaction is called fusion since nuclei are combining
to form a heavier nucleus. The energy associated with
this change is 4.05 X 108 kcal/mole of ?H nuclei.

Let us do a little bookkeeping with the exact masses
of these nuclei. Actually we will simplify a bit and use
the exact masses of the atoms. This will make no dif-
ference. The masses of the atoms differ from the nuclear
masses by the masses of the number of electrons in each
atom. We have shown that electrons are conserved in
nuclear changes. Exact masses of atoms (that is, exact
masses of each isotopic species and not the chemical
atomic weights shown on the inside back cover) are
readily available. For our hydrogen-helium reaction we
have

Reactants: 2H 2.01471 g/mole

Products:

3H
4He

3.01707
5.03178
4.00390
1.00893
5.01283

Reactants:
Products:
Mass Difference:

5.03178
5.01283
0.01895 g/mole

Compare this mass difference of about 0.02 g/mole with
one of about 5 X 10-9 g/mole for the combustion of
carbon.

In closing, let us remind ourselves of the difference
between nuclear and chemical reactions. In nuclear re-
actions, changes in the nuclei take place. In chemical
reactions, the nuclei remain intact and the changes are
explainable in terms of the electrons outside the nucleus.

* Before using the E = mc- relation to calculate the
amount of mass associated with this energy change, you
must pay attention to the relation of various energy units.

122

ENERGY EFFECTS IN CHEMICAL REACTIONS I CHAP. 7

QUESTIONS AND PROBLEMS

1. Given

3C(s) + 2Fe203fsj +110.8 kcal — *-

4Fe(s) + ZCO,(g)

Rewrite the equation using one mole of carbon
and use the AH notation.

Given

\Wg) + \*r2(l)

— >■ HBr(g)

AH = -8.60 kcal/mole HBr

Rewrite the equation for one mole of hydrogen
gas and include the heat effect as a term in the
equation.

Which of the following reactions are endother-
mic?

(a) H2(g) + \02(g) -

(b) h^g) + \Oi(g)

U*0(g)

AH = -57.8 kcal

NOfgJ

AH = +21.6 kcal

(d) hK2(g) + W*(g) — >■ NH3(gj +11.0 kcal

(e) NH3(gj — ►- ±N2(g) + |H2(gj

AH = +11.0 kcal

4. What is the minimum energy required to syn-
thesize one mole of nitric oxide, NO, from the
elements ?

5. How much energy is liberated when 0.100 mole
of H2 (at 25°C and 1 atmosphere) is combined
with enough 02(g) to make liquid water at 25°C
and 1 atmosphere?

6. How much energy is consumed in the decom-
position of 5.0 grams of H20(l) at 25°C and
1 atmosphere into its gaseous elements at 25°C
and 1 atmosphere?

7. Using Table 7-II, calculate the heat of burning
ethane in oxygen to give C02 and water vapor.

Answer. AH = —341 kcal/mole C2H6.

8. Given

C(diamond) + 02(g)
C(graphite) + 02(g)

C02(g)
AH = -94.50 kcal

C02(g)
AH = -94.05 kcal

Find AH for the manufacture of diamond from
graphite.

C(graphite)

C(diamond)

Is heat absorbed or evolved as graphite is con-
verted to diamond?

9. To change the temperature of a particular calo-
rimeter and the water it contains by one degree
requires 1 550 calories. The complete combustion
of 1.40 grams of ethylene gas, C2H4(g), in the
calorimeter causes a temperature rise of 10.7
degrees. Find the heat of combustion per mole
of ethylene.

10. The "thermite reaction" is spectacular and
highly exothermic. It involves the reaction be-
tween Fe203, ferric oxide, and metallic alumi-
num. The reaction produces white-hot, molten
iron in a few seconds. Given :

2A1 + f 02 — >- A1203 AHx = -400 kcal/mole

2Fe + f02 — ►- Fe2Os AH2 = -200 kcal/mole

Determine the amount of heat liberated in the
reaction of 1 mole of Fe203 with Al.

Answer. AH = —200 kcal/mole Fe203.

11. How much energy is released in the manufacture
of 1.00 kg of iron by the "thermite reaction"
mentioned in Problem 10?

12. How many grams of water could be heated from
0°C to 100°C by the heat liberated per mole of
aluminum oxide formed by the "thermite reac-
tion," as described in Problem 10?

13. Which would be the better fuel on the basis of
the heat released per mole burned, nitric oxide,
NO, or ammonia, NH3? Assume the products
are N02(gj and H20(g).

14. What is the minimum energy required to syn-
thesize sulfur dioxide from sulfuric acid?

H2S04f/J — > SO/gJ + H20(g) + \02(g)

Answer. AH = +65 kcal/mole S02(g).

15. Why is the Law of Conservation of Energy con-
sidered to be valid?

16. What do you think would happen in scientific
circles if a clearcut, well-verified exception was

QUESTIONS AND PROBLEMS

123

found to the Law of Conservation of Energy as
stated in the text?

17. Is energy conserved when a ball of mud is
dropped from your hand to the ground? Explain
your answer.

18. What becomes of the energy supplied to water
molecules as they are heated in a closed con-
tainer from 25°C to 35°C?

19. Outline the events and associated energy changes
that occur on the molecular level when steam at
150°C and 1 atmosphere pressure loses energy
continually until it finally becomes ice at - 10°C.

20. The heat of combustion of methane, CH4, is
—210 kcal/mole:

CHA(g) + 202

C02 + 2H20

AH = -210 kcal

Discuss why this fuel is better than water gas if
the comparison is based on one mole of carbon
atoms.

21. In a nuclear reaction of the type called "nuclear
fusion," two nuclei come together to form a
larger nucleus. For example, deuterium nuclei,

23.

iH, and tritium nuclei, jH, can "fuse" to form
helium nuclei, 2He, and a neutron:

aHe + In

AH = -4.05 X 108 kcal

?H + ?H

How many grams of hydrogen would have to be
burned (to gaseous water) to liberate the same
amount of heat as liberated by fusion of one
mole of iH nuclei? Express the answer in tons
(1 ton = 9.07 X 10s g).

22. Which of the following reactions is most likely
to have a heat effect of - 505 kcal? Which would
be -1.7 X 10« kcal? Which would be +7.2
kcal?

(a) UF,fi; — >■ UFt(g)

(b) U(s) + 3F2(g) -^ UF6

239, T
92 *->

AH = ?
AH = ?
AH = ?

Fission of uranium gives a variety of fission
products, including praseodymium, Pr. If the
process by which praseodymium is formed gives
^gPr and three neutrons, what is the other nu-
clear product?

235i t _i_ x„

^Pr + ? + 3j/i

CHAPTER

The Rates of
Chemical Reactions

• • • a molecular system • • • [passes] • • • from one state of
equilibrium to another • • - by means of all possible intermediate
paths, but the path most economical of energy will be more often travelled.

HENRY EYRING, 1945

A candle remains in contact with air indefinitely
without observable reaction but it reacts when
given a start with a lighted match. A mixture of
household gas and air in a closed room remains
indefinitely without reacting but it may explode
violently if so much as a glowing cigarette is
brought into the room. A piece of iron reacts
quite slowly with air (it rusts) but a piece of white
phosphorus bursts into flame when it is exposed
to air. These are all reactions with oxygen from
the air but they have extremely different time
behaviors. Reactions proceed at different rates.
We care about reaction rate because we must
understand how rapidly a reaction proceeds and
what factors determine its rate in order to bring
the reaction under control.*

Let us see what the expression "the rate of a
reaction" means in terms of an example — the
reaction between carbon monoxide gas, CO, and
nitrogen dioxide, N02. Chemical tests show that

* The study of reaction rates is called chemical
kinetics.
124

the products are carbon dioxide, C02, and nitric
oxide, NO. The equation for the reaction is

CO + N02

CO. + NO

(')

Suppose we prepare a mixture of carbon mon-
oxide and nitrogen dioxide and then heat it to
200°C. When the gas is heated, we observe a
gradual disappearance of the reddish brown
color of N02. Reaction is taking place. We can
find its time behavior by measuring the change
in color during a measured time interval. Since
the other gases are colorless, this color change
indicates the number of moles of N02 that have
reacted during the time interval. The quotient of
these two, moles reacted divided by the time
interval, is called the rate of the reaction:

Rate =

quantity NQ2 consumed

time interval
= quantity N02 consumed per unit time

We can express the rate of reaction (7) in terms
of the rate of consumption of either CO or N02.
Equally well, we can express the time behavior

SEC. 8-1 I FACTORS AFFECTING REACTION RATES

125

of the reaction in terms of the appearance of
either product, C02 or NO. Which is used de-
pends upon convenience of measurement. If the
experimenter prefers to measure the production
of carbon dioxide, he would express the rate in
the form

_ quantity CO? produced
time interval

= quantity COj produced per unit time

The quantity consumed or produced is conven-
iently expressed in partial pressure units if the
substance is a gas. Concentration units are con-
venient if the reactant or product is in solution.
The time measurement is also expressed in what-
ever units fit the reaction: microseconds for the
explosion of household gas and oxygen, seconds
or minutes for the burning of a candle, days for
the rusting of iron, months for the rotting of
wood.

8-1 FACTORS AFFECTING REACTION RATES

In the laboratory you have observed the reaction
of ferrous ion, Fe+2(aq), with permanganate ion,
Mn04~ (aq), and also the reaction of oxalate ion,
CiQi2(aq), with permanganate ion, Mn04 (aq).
These studies show that the rate of a reaction
depends upon the nature of the reacting substances.
In Experiment 14, the reaction between I03" and
HS03~ shows that the rate of a reaction depends
upon concentrations of reactants and on the tem-
perature. Let us examine these factors one at a
time.

8-1.1 The Nature off the Reactants

Compare the following three reactions, all of
which occur in water solutions:

5Q04- 2(aq) + 2Mn04- (aq) + \6H+(aq) — *•

\0CO,(g) + 2Mn+i(aq) + 8H,0 slow (2)

5Fe+2(aqJ + Mn04- (aq) + SH+(aq) — *-

5Fe+3(aq) + Mn+2(aqJ + 4H20 very fast (J)

Fe+2(aqJ + Ce+YaqJ — >■

Ft+*(aq) + Ce+Yaqj very fast (4)

Both ferrous ion, Fe**(aq) and oxalate ion,
C204~ 2(aq), have the capability of decolorizing
a solution containing permanganate ion at room
temperature. Yet, there is a great contrast in the
time required for the decoloration. The differ-
ence lies in specific characteristics of Ft+2(aq)
and Ci0^2(aq). On the other hand, Fe+2(aq) is
also changed to Fe+'6(aq) by reacting either with
Mn04" (aq) or with eerie ion, Ct+4(aq). One of
these reactions is simple and the other involves

many molecules. Yet, these are both rapid re-
actions.

Here are two reactions that take place in the
gas phase.

2NO + 0> — >- 2NO, moderate at 20°C (5)

CH4 + 202 — »- C02 + 2H>0

extremely slow at 20°C (<5)

The oxidation of nitric oxide, NO, is a reaction
involved in smog production. It is moderately
rapid at normal temperatures. The oxidation of
methane, CH4 (household gas), however, occurs
so slowly at room temperature that we may say
that, for all practical purposes it doesn't react at
all. Again, the difference in the reaction rates
must depend upon specific characteristics of the
reactants, NO and CH4.

The determination of the molecular character-
istics which are important in rate behavior is an
interesting frontier of chemistry. It seems that
chemical reactions which involve the breaking of
several chemical bonds and the formation of new
chemical bonds tend to proceed slowly at room
temperature. Reaction (2) is of this type— there
are many bonds which must be broken in the
five C204-2 ions and the two Mn04~ ions to form
the 10CO2 and 2Mn+2. This reaction might be
expected to proceed slowly (as it does). Reac-
tion (6) also involves breaking of bonds and
forming of new bonds and it is slow at room
temperature. In contrast, reaction (3) is very
rapid though it involves breaking of chemical
bonds and it might be expected to proceed slowly.

126

THE RATES OF CHEMICAL REACTIONS I CHAP. 8

We see that we cannot be certain of a prediction
that a reaction might be slow. Reaction (4) ap-
parently does not require bond breaking or bond
formation. It can be expected to be rapid (as it
is). A prediction of this type is usually reliable.
Reaction (5) requires breaking of but one bond
and the formation of two. It has a moderate
reaction rate, rapid at high pressures and slow
at low pressures.

These and other examples lead to the follow-
ing rules:

(1) Reactions that do not involve bond rear-
rangements are usually rapid at room tem-
perature.

(2) Reactions in which bonds are broken tend
to be slow at room temperature.

We can say little more about how the nature
of the reactants determines the reaction rate
until we consider in detail how some reactions
take place. For the time being, it will suffice to
observe that this is an active field of study and
much remains to be learned.

EXERCISE 8-1

Are any of the following three reactions likely
to be extremely rapid at room temperature? Are
any likely to be extremely slow at room tempera-
ture? Explain.

(a) C^(aq) + Fe+3(aq) — ►-

Ci+3(aq) + Fe+i(aq)

(b) 3Fe+2(agJ + ^O^(aq) + 4H+(aq) — ►-

3Fc+3(aq) + NOfeJ + 2H20

liquid
gasoline

8-1.2 Effect of Concentration:
Collision Theory

Henceforth we shall concentrate our attention on
one reaction at a time. The nature of the reac-
tants will be held constant while the other factors
that affect rates are considered. The first of these
factors is concentration.
Chemists have learned that, for many reac-

tions, raising the concentration of a reactant
increases the reaction rate. Not infrequently,
though, there will be no effect. In this section we
shall consider how a rate increase with rising
reactant concentration is explained. In Section
8-1.3 we shall explore why some reactions pro-
ceed at a rate independent of the concentration
of one or more reactants. Both explanations are
based upon a model of the way chemical reac-
tions take place on the molecular scale.

In the molecular view of matter, it is natural
to assume that two molecules must come close
together in order to react. Therefore, we postu-
late that chemical reactions depend upon colli-
sions between the reacting particles — atoms,
molecules, or ions. This model of reaction rate
behavior is called the collision theory. It pro-
vides a successful basis for understanding the
effect of concentration. Just as an increase of the
number of cars in motion on a highway leads
to a higher rate of formation of dented fenders,
increasing the number of particles in a given
volume gives more frequent molecular collisions.
The higher frequency of collisions results in a
higher rate of reaction.

Consider a homogeneous system — one in
which all components are in the same phase.
According to the collision theory, we can expect
that increasing the concentration of one or more
reactants will result in an increase in the rate of
the reaction. Lowering the concentration has the
opposite effect. This is exactly the behavior found
in the reaction between HS03" (aq) and I03" (aq)
when the concentrations are varied by adding or
removing reactants or solvent (Experiment 14).
In gases, also homogeneous systems, the con-
centration of an individual reactant can be
raised by admitting more of that substance into
the mixture. The concentrations of all gaseous
components can be raised simultaneously by de-
creasing the volume occupied by the mixture.
Decreasing the volume by compressing the gas
raises the concentration of all reactants, hence
increases the rates of reactions taking place. In-
creasing the volume by expanding the gas has
the opposite effect on concentrations, hence de-
creases reaction rates.

In a heterogeneous reaction system, the com-

SEC. 8-1 I FACTORS AFFECTING REACTION RATES

127

Fig. 8-1. The number of collisions per second depends
upon concentration.

ponents are in two or more different phases. As
an example, consider

burning

wood (solid) + oxygen (gas) >-

carbon dioxide (gas) + water (gas)

In a system of this sort, the rate of the reaction
depends upon the amount of interface between
the phases, or, in other words, the area of con-
tact between them. For example, a log burns in
air at a relatively slow rate. If the amount of
exposed surface of the wood is increased by re-
ducing the log to splinters, the burning is much
more rapid. If, further, the wood is reduced to
fine sawdust and the latter is suspended in a
current of air, the combustion takes place ex-
plosively. Where one of the reactants is a gas,
such as in the above example, the concentration
of the gas is also a factor. A piece of wood burns
much more rapidly in pure oxygen than it does
in ordinary air, in which the oxygen makes up
only about 20% of the mixture.

We see that the collision theory provides a
good explanation of reaction rate behavior. It is
quite reasonable that the reaction rate should
depend upon collisions among the reactant mole-
cules. In fact, it is so reasonable that we are left
wondering why the concentrations of some re-
actants in some reactions do not affect the rate.

The explanation is found in the detailed steps
by which the reaction takes place, the reaction
mechanism.

8-1.3 Reaction Mechanism

As has been proposed, in order for a chemical
reaction to occur, particles must collide. The
particles may be atoms, molecules, or ions. As
a result of collisions, there can be rearrangements
of atoms, electrons, and chemical bonds, with
the resultant production of new species. As an
example, let us take another look at the reaction
between Fe+2 and MnO*- in acid solution:

5Ft+2(aq) + Mn04" (aq) + SH+(aq) — >-

5Fe+3(aq) + Mn+2(aqJ + 4H20 (7)

The equation indicates that one Mn04" ion, five
Fe+2 ions, and eight H+ ions (a total of four-
teen ions) must react with each other. If this
reaction were to take place in a single step, these
fourteen ions would have to collide with each
other simultaneously. The probability of such an
event occurring is extremely small — so small that
a reaction which depended upon such a collision
would proceed at a rate immeasurably slow.
Since the reaction occurs at an easily measured
rate, it must proceed by some sequence of steps,
none of which involves such an improbable col-
lision.

As a matter of fact, chemists regard the colli-
sion of even four molecules as an extremely
improbable event if they are at low concentration
or if they are in the gas phase. We conclude that

128

THE RATES OF CHEMICAL REACTIONS I CHAP. 8

a complex chemical reaction which proceeds at
a measurable rate probably takes place in a
series of simpler steps. The series of reaction
steps is called the reaction mechanism.

Consider the oxidation of gaseous hydrogen
bromide, HBr, a reaction that is reasonably
rapid in the temperature range from 400 to
600°C:

4HBrfgJ + 02(g) — ► 2H20(g) + 2Br2(g) (8)

By the collision theory, we expect that increasing
the partial pressure (and thus, the concentration)
of either the HBr or 02 will speed up the reaction.
Experiments show this is the case. Quantitative
studies of the rate of reaction (8) at various pres-
sures and with various mixtures show that oxy-
gen and hydrogen bromide are equally effective
in changing the reaction rate. However, this re-
sult raises a question. Since reaction (8) requires
four molecules of HBr for every one molecule
of 02, why does a change in the HBr pressure
have just the same effect as an equal change in
the 02 pressure?

The explanation is found by considering the
details of the process by which reaction (8) oc-
curs. The overall reaction brings together five
molecules, four of HBr and one of 02. However,
the chance that five gaseous molecules will col-
lide simultaneously is practically zero. The reac-
tion must occur in a sequence of simpler steps.

All of the studies of reaction (8) are explained
by the following series of reactions:

HBr + 02

HOOBr + HBr

HOBr + HBr

HOOBr slow (9)
2HOBr fast (70)

H20 + Br2 fast (77)

First, observe that adding reactions (9) and (70)
plus twice reaction (11) gives the overall reac-
tion (8). Next, we see that each step in the se-
quence requires only two molecules to collide.
Finally, the proposal that reaction (9) is slow
whereas reactions (10) and (//) are fast explains
why HBr and 02 have the same effect on the
reaction rate.

Reaction (9) is a "bottle-neck" in the oxida-
tion of hydrogen bromide. As fast as HOOBr is
formed by this slow reaction it is consumed in
the rapid reaction (10). But no matter how rapid

reactions (70) and (77) are, they can produce
H20 and Br2 only as fast as the slowest reaction
in the sequence. Hence, the factors that deter-
mine the rate of reaction (9) determine the rate
of the overall process.

The sequence of reactions (9), (70), and (77)
is called the reaction mechanism of the overall
reaction (8). Because it is the slowest reaction in
the mechanism, reaction (9) is the step that fixes
the rate. The slowest reaction in a reaction mecha-
nism is called the rate determining step.

There are two features of this example that
are rather common. First, none of the steps in
the reaction mechanism requires the collision of
more than two particles. Most chemical reactions
proceed by sequences of steps, each involving only
two-particle collisions. Second, the overall or net
reaction does not show the mechanism. In gen-
eral, the mechanism of a reaction cannot be de-
duced from the net equation for the reaction; the
various steps by which atoms are rearranged and
recombined must be determined through experi-
ment.

EXERCISE 8-2

Imagine five people working together to wash a
stack of very greasy dishes. The first two clear
the table and hand the dishes to the third person
who washes them and hands them on. The last
two persons dry and stack them. Which step is
likely to be the rate determining step? In the
light of your answer, discuss how the rate of the
overall process would be affected if a sixth person
joined the group (a) as a table clearer; (b) as a
second dishwasher; (c) as a dish dryer.

8-1.4 The Quantitative Effect of Concentration

The reaction mechanism is deduced from quantitative
studies of the dependence of the rate upon the concentra-
tions or pressures of the various reactants. To interpret
such studies, we need to develop our collision theory
model.

Consider the reaction between gaseous hydrogen, H2,
and gaseous iodine, I»:

Ht(g) + U(g)

201(g)

(12)

SEC. 8-1 I FACTORS AFFECTING REACTION RATES

129

Each time a molecule of H2 collides with an iodine mole-
cule, reaction may occur. The frequency of these en-
counters, for a particular H2 molecule, is determined by
how many b molecules are present. Doubling the number
of I2 molecules per unit volume would just double the
collisions. Tripling the number of h molecules per unit
volume would triple the collisions. Since the iodine partial
pressure fixes the iodine concentration, the rate of the
reaction is proportional to the iodine partial pressure:

(rate) is proportional to

iodine

partial

_pressure.

(13)

In the same way, a particular iodine molecule must find
a hydrogen molecule to react. The rate of the reaction is
proportional to the partial pressure of the hydrogen:

(rate) is proportional to

hydrogen
partial
_ pressure

(14)

In view of (13) and (14), the rate must be proportional
to the product of the partial pressure of iodine and
hydrogen :

(rate) is proportional to

"hydrogen

partial
_ pressure

"iodine
partial
pressure

In symbols, we can write

rate =

= k[pHi] X

>lJ

U5)

(16)

8-1.5 Effect of Temperature: Collision Theory

In Experiments 12 and 14, you discovered that
temperature has a marked effect upon the rate
of chemical reactions. Thus, raising the tempera-
ture speeded up the reaction between 103" and
HSO3". That is the same effect, qualitatively, that
is observed in the reaction of a candle with air.
The match "lighted" the candle by raising its
temperature (at the wick). Once started, the re-
action of combustion releases enough heat to
keep the temperature high, thus keeping the re-
action going at a reasonable rate. Raising the
temperature speeded up the reaction. The same
type of explanation applies to the explosion of
a kitchen full of household gas and air when a
cigarette is brought into the room. Around the
glowing tip of the cigarette the gas temperature
is raised. At this locale, the reaction speeds up,
liberating heat. This heat warms the nearby re-
gion even more and the reaction goes somewhat

faster. This acceleration continues until finally
(in a millisecond or so) it reaches an ex-
plosive rate — the most rapid reaction permitted
by the collisional properties of the gas. Raising
the temperature started it all by speeding up the
reaction.

In all of these reactions (and in almost all
others), increasing the temperature has a very
pronounced effect, always speeding up the reac-
tion. Two questions come to mind. "Why does
a temperature rise speed up a reaction?" and
"Why does a temperature rise have such a large
effect?" To answer these questions, we return to
our collision theory.

From what we know about molecular sizes,
we can calculate that a particular CH4 molecule
collides with an oxygen molecule about once
every one-thousandth of a microsecond (10-9
seconds) in a mixture of household gas (methane,
formula CH4) and air under normal conditions.
This means that every second this methane mole-
cule encounters 109 oxygen molecules! Yet the
reaction does not proceed noticeably. We can
conclude either that most of the collisions are
ineffective or that the collision theory is not a
good explanation. We shall see that the former
is the case — we can understand why most colli-
sions might be ineffective in terms of ideas that
are consistent with the collision theory.

Chemists have learned that chemical reactions
occur when collisions occur but only when the
collision involves more than a certain amount of
energy. We can understand this by returning to
our analogy of cars bumping each other on a
highway. In a line of heavy traffic one frequently
receives gentle bumps from the car in front or
the car behind. No damage is done to the ears-
only to tempers. But occasionally a high speed
collision occurs. If this occurs with enough en-
ergy, a bumper may be knocked off a car and a
fender may be dented. It is the high energy col-
lisions which cause the auto damage and it is
high energy molecular collisions which cause the
"molecular damage" that we call a chemical re-
action. Just as a certain amdunt of energy is
required to break loose a bumper, a certair
amount of energy is required to cause a chemical
reaction. In either instance, if there is more than
this "threshold energy," the reaction can occur
and if there is less, it cannot.

130

THE RATES OF CHEMICAL REACTIONS I CHAP. 8

Collimator

Tin vapor
Molten ti

Oven, at

controlled

temperature

Fig. 8-2. Rotating disc jur measurement of atomic velocities.

SEC. 8-1 I FACTORS AFFECTING REACTION RATES

131

9 3 7 6 5

Pie slice numher

Disc D2 after many revolutions'

Increasing velocity of tin atoms

Fig. 8-3. Distribution of atomic (or molecular) veloc-
ities from the rotating disc.

DISTRIBUTION OF KINETIC ENERGIES

This discussion of threshold energy causes us to
wonder what energies are possessed by molecules
at a given temperature. We have already com-
pared the molecules of a gas with billiard balls
rebounding on a billiard table. When billiard
balls bounce around, colliding with each other,
some of them move rapidly and some slowly. Do
molecules behave this way? Experiment provides
the answer.

Figure 8-2 shows a device for measuring the
distribution of atomic or molecular velocities.
It consists of two discs, A and A, rotating
rapidly on a common axle. They rotate in a
vacuum chamber in front of an oven containing
molten tin and held at a controlled temperature.
Vapor streams out of the small opening in the
oven and strikes the rotating disc, A. When the
disc has rotated to the position shown in Figure
8-2B, a small amount of gas has passed through
the slot in disc A- A short time later, shown in
Figure 8-2C, the atoms of tin have traveled part
of the way toward the second rotating disc. The
fastest moving atoms have traveled farther than
the others — they are leading the way. The slowest
moving atoms are beginning to lag behind. Still
later, Figure 8-2D, the atoms have spread out in
space even more, and the fastest atoms have
already reached the second rotating disc. Now
as the atoms reach disc A, they condense on it.
The position where a given atom condenses on
disc A depends upon how long that atom took to

travel from A to A and how fast A is rotating.

As the slotted disc A lets through burst after
burst of tin atoms, a layer of tin builds up on the
surface of disc A- The pattern of this layer is
determined by the distribution of velocities of
the atoms escaping from the oven at tempera-
ture T. Figure 8-3 shows the disc A divided into
sections, like slices of a pie. The fastest moving
atoms are condensed on pie slices 3, 4, and 5.
The slowest moving atoms are condensed on the
pie slices 10 and 1 1 . If the disc is cut up and each
slice weighed, the amount of tin can be deter-
mined. A plot of the weight of tin against the
pie slice number indicates the distribution of
atomic velocities.

Clearly, the plot of Figure 8-3 contains infor-
mation about the distribution of kinetic energies.
From the rate of rotation of the discs and the
distance between them we can calculate the ve-
locity an atom must have to condense on a
particular pie slice. From the atomic mass and
its velocity, we learn the atom's kinetic energy.
Figure 8-4 shows the result. At a temperature Ti

Fig. 8-4. Effect of temperature on atomic {or molec-
ular) kinetic energy distribution.

Tj greater than 7}

Kinetic energy

132

THE RATES OF CHEMICAL REACTIONS CHAP.

a few atoms have very low kinetic energies and
some have very high kinetic energies. Most
of them have intermediate kinetic energies, as
shown by the solid curve. At a higher tempera-
ture, r2, the energy distribution is altered to that
shown as a dashed curve. As can be seen, increas-
ing the temperature causes a general shift of the
distribution toward one with more molecules
having high kinetic energies. Moreover, in going
from Tx to T2 there is a large increase in the
number of molecules having kinetic energies
above a certain value.

THRESHOLD ENERGY AND REACTION RATE

We can apply these curves to our reaction rate
problem. Suppose a reaction can proceed only if
two molecules collide with kinetic energy exceed-
ing a certain threshold energy, E. Figure 8-4
shows us a typical situation. At Tx the darkly
shaded area is proportional to the number of the
molecules which possess this energy or more.
Since only a small number of molecules have as

much as this energy, few collisions are effective,
and the reaction is slow. But if we raise the tem-
perature to T2, the number of molecules with
energy E or greater is raised in proportion to the
diagonally shaded area. Only a small tempera-
ture change is needed to make a large change in
the area out on the tail of the energy distribution
curve. Consequently, the reaction rate is very
sensitive to change in temperature.

This argument is based upon the "typical" situation in
which E is well out on the tail of the curve. Suppose it is
not; suppose E is near the maximum of the curve at 7*i,
or is even to the left of it. Then a large number of the
molecules have the requisite energy, even at the lower
temperature, Tt. Since collisions occur so rapidly (re-
member, one every 10~9 second or so), the reaction is
over in a blink of the eye. This reaction would be called
"instantaneous." The circumstances shown in Figure 8-4
are "typical" only of a slow reaction.

It should be remarked that raising the temperature also
increases the rate by increasing the frequency of the
collisions. This is, however, a very small effect compared
with that caused by the increase in the number of mole-
cules with sufficient energy to cause reaction.

8-2 THE ROLE OF ENERGY IN REACTION RATES

Now that we have this evidence about the mo-
lecular velocity distribution, we can see how
temperature changes the fraction of the molecu-
lar collisions involving an energy exceeding the
threshold energy E. Now our understanding ot
the role of energy in fixing reaction rates can be
expanded.

8-2.1 Activation Energy

Suppose you wished to make an automobile trip
from Los Angeles to San Francisco. This trip re-
quires that the Tehachapi Mountains be crossed.
As shown in Figure 8-5, this can be accomplished
by taking the highway through the Tejon Pass.
Of course, the high altitude of this pass, 4200
feet, makes this part of the trip the slowest and
the biggest test of your automobile. This is the
point in the trip where radiator fluids boil and
engine troubles develop. This is the point where
the older cars turn back and limp home. Yet,
this is the lowest pass in these mountains, hence
the most favorable route. If your car will sur-

mount a 4200 foot pass, it will undoubtedly
make the trip to San Francisco.

Chemical reactions are similar. As molecules
collide and reaction takes place, the atoms must
take up, momentarily, bonding arrangements
that are less stable than either reactants or prod-
ucts. These high energy molecular arrangements
are like the mountain pass — they place an energy
barrier between reactants and products. Only if
the colliding molecules have enough energy to
surmount the barrier imposed by the unstable
arrangements can reaction take place. This bar-
rier determines the "threshold energy" or mini-
mum energy necessary to permit a reaction to
occur. It is called the activation energy.

This barrier can be shown graphically by am-
plifying Figure 7-1 (p. 110), which showed the
relative energies of reactants and products. In
Figure 8-6 we see that the diagram becomes the
equivalent of the road map in our automobile
trip analogy. This diagram applies to the reac-
tion between carbon monoxide, CO, and nitro-
gen dioxide, N02, to produce carbon dioxide,

SEC. 8-2 I THE ROLE OF ENERGY IN REACTION RATES

133

SAN

LOS ANGELES

4200

Travel coordirvate
-»- Ter'on Pass

Fig. 8-5. Crossing the mountains between Los Angeles
and San Francisco.

C02, and nitric oxide, NO. The horizontal axis
of the diagram, called the reaction coordinate,
shows the progress of the reaction. Proceeding

from left to right along this reaction coordinate
signifies the CO and N02 molecules approaching
each other, colliding, and going through inter-
mediate processes of reaction which result in the
formation of C02 and NO, and, finally, the sepa-
ration of the latter two molecules. The vertical
axis represents the total potential energy of the

134

THE RATES OF CHEMICAL REACTIONS I CHAP. 8

Potential

energy,

kcaly/mole

+SO

j^ZcnWTED COMPLEX

Activation
energy

/S^

—

CO -r tfOz

Net energy released

-50

sv^cq2 * no

Reaction coordinate

Reactants

Activated complex.

-*- Products

Fig. 8-6. Potential energy diagram for the reaction
CO + N02 — >- C02 + NO.

system. Thus, the curve provides a history of the
potential energy change during a collision which
results in a reaction. The energy required to sur-
mount the potential energy "barrier" to reaction
usually is provided by the kinetic energies of the
colliding particles, as fixed by the temperature.
Let us move from left to right along this curve
and describe the events which occur. Along the
flat region at the left, CO and N02 are approach-
ing each other. In this region, they possess kinetic
energy and their total potential energy shows no
change. The beginning of the rise in the curve
signifies that the two molecules have come suffi-
ciently close to have an effect on each other.
During this approach, the molecules slow down
as their kinetic energies furnish the potential
energy to climb the curve. If they have sufficient
kinetic energy, they can ascend the left side of
the "barrier" all the way up to the summit. At-
taining this point is interpreted as follows: CO
and N02 had sufficient kinetic energy to over-
come the mutually repulsive forces of their nuclei
and negative electron clouds and thus come very
close to each other. Here at the summit the
molecular cluster is unstable with respect to
either the forward reaction (to give C02 and NO)
or the reverse reaction (to restore the molecules
of CO and N02). This transitory arrangement is
of key importance (its potential energy fixes the

activation energy); it is called the activated
complex.

Now there are two possibilities: (1) the acti-
vated complex may separate into the two original
CO and N02 molecules, which would then re-
trace their former path on the curve, or, (2) the
activated complex may separate into C02 and
NO molecules. The latter possibility is repre-
sented by moving down the right side of the
"barrier." In the flat region at the right, C02 and
NO have separated beyond the point of having
any effect on each other and the potential energy
of the activated complex has become kinetic en-
ergy again.

In the event that the CO and N02 molecules
do not have sufficient energy to attain the sum-
mit, they reach a point only part way up the left
side of the "barrier." Then, repelling one an-
other, they separate again, going downhill to the
left.

We have labeled the difference between the
high potential energy at the activated complex
and the lower energy of the reactants as the acti-
vation energy. The activation energy is the
energy necessary to transform the reactants into
the activated complex. This may involve weaken-
ing or breaking bonds, forcing reactants close
together in opposition to repulsive forces, or
storing energy in a vibrating molecule so that it
reacts on collision. Increasing the temperature
affects reaction rate by increasing the number of
molecular collisions that involve sufficient energy
to form this activated complex. The magnitude

SBC. 8-2 I THB ROLB OF ENERGY IN REACTION RATES

135

of ihe activation energy for a reaction can be
determined by measuring experimentally the
change in reaction rate with temperature.

8-2.2 Heat of Reaction

We can deduce the heat of reaction from Figure
8-6. In our example, the reactants are at a higher
total energy than are the products. This means
that in the course of the reaction there will be a
net release of energy. This reaction is exothermic.
Figure 8-6 shows that the reaction releases 54
kcal of heat per mole of carbon monoxide con-
sumed. Notice that the height of the energy
barrier between reactants and products has no
effect on the net heat release. We must put in an
amount of energy equal to the activation energy
to get to the top of the barrier but we get it all
back on the way down the other side.

Now let us consider the reverse reaction. We
need not draw another reaction diagram, since
Figure 8-6 will suffice. Now we are interested in
the reaction between C02 and NO to produce
CO and N02:

CO + N02 -<— C02 + NO (17)

This reaction begins at the lower energy appro-
priate to the chemical stability of C02 + NO
(at the right side of Figure 8-6) and ends at the
higher energy appropriate to the chemical sta-
bility of CO + N02 (at the left side of Figure
8-6). The difference in energy, the heat of this
reaction, is just equal to that of the reverse
reaction but is opposite in sign. This reaction
absorbs 54 kcal of heat per mole of carbon
monoxide produced. It is endothermic.

Figure 8-6 contains one other very interesting
piece of information concerning the rate of the
reverse reaction, (77). This reaction rate is con-
trolled by the energy barrier confronting the
colliding molecules of C02 and NO. We see from
the diagram that the activation energy for this
reaction is higher than that for the reaction we
studied earlier. Further, it is higher by exactly
the heat of reaction. We conclude that the reac-
tion between C02 and NO will be slower, at any
given temperature, than the reverse reaction be-
tween CO and N02, if the rates are compared
at the same partial pressures.

The relationship between activation energies
for the forward and reverse reactions can be ex-
pressed mathematically. The activation energy is
denoted by the symbol AHX (read "delta-//-
cross") and the heat of the reaction by AH.
Hence we may write:

AH}{ = AHl + AH (18)

AH\i = activation energy for reaction pro-
ceeding to right (always endothermic)
AHi = activation energy for reaction pro-
ceeding to left (always endothermic)
AH = heat absorbed during reaction pro-
ceeding left to right (either endother-
mic or exothermic).

The heat of reaction, AH, is positive if heat is
absorbed as the reaction proceeds, left to right.
It is negative if heat is evolved. In our example,
CO + N02 — >- C02 + NO,

A//fe = +32 kcal/mole
AHi = +86 kcal/mole
AH = — 54 kcal/mole

We see that

(32) = (86) + (-54)

which is in accordance with equation (18). This
relationship is important because it implies that
we need only two of the three quantities, AHlR,
AH[, and AH and then we can calculate the
third. For example, if we measure the heat of the
reaction, AH, and also measure the rate of the
reaction between CO and N02 in order to deter-
mine AH)t, then we can calculate AH[ by equa-
tion (18). From AHXL we learn something about
the rate of reaction of C02 + NO.

8-2.3 Action off Catalysts

Many reactions proceed quite slowly when the
reactants are mixed alone but can be made to
take place much more rapidly by the introduc-
tion of other substances. These latter substances,
called catalysts, are not used up in the reaction.
The process of increasing the rate of a reaction
through the use of a catalyst is referred to as
catalysis. You have seen at least one example of
catalytic action, the effect of Mn+2(aq) in speed-
ing up the reaction between CiO;2(aq) and
Mn04" (aq).

136

THE RATES OF CHEMICAL REACTIONS I CHAP.

SAM FRANCISCO

TEJON PASS
A ttitude 4, 2. OOft. I

LOS ANGELES

Travel coor-din,ate

Fig. 8-7. An alternate, easier route through the moun-
tains between Los Angeles and San Fran-
cisco.

The action of a catalyst can be explained in
terms of our mountain pass analogy. In Figure
8-5 we see a formidable mountain pass obstruct-

ing travel between Los Angeles and San Fran-
cisco. Many people who would like to take this
trip cannot because their automobile isn't up to
this much of a climb. For this reason, an alter-
nate route was improved. The coast route, shown
in Figure 8-7 is somewhat longer but it is easier
because the highest pass, Gaviota Pass, is only

SEC. 8-2 I THE ROLE OF ENERGY IN REACTION RATES

137

Potential
energy

Reaction coordinates

Fig. 8-8. Effect of a catalyst on a reaction and its
reverse.

900 feet above sea level. Some travelers still use
Tejon Pass but, in addition, the trip can be made
via the easier Gaviota Pass. The net result is that
more people per day are able to make the trip
from Los Angeles to San Francisco. Of course,
the lower pass serves to increase the rate of re-
turn travel as well.

Figure 8-8 shows the analogous situation for
a chemical reaction. The solid curve shows the
activation energy barrier which must be sur-
mounted for reaction to take place. When a
catalyst is added, a new reaction path is provided
with a different activation energy barrier, as sug-
gested by the dashed curve. This new reaction
path corresponds to a new reaction mechanism
that permits the reaction to occur via a different
activated complex. Hence, more particles can get
over the new, lower energy barrier and the rate
of the reaction is increased. Note that the activa-
tion energy for the reverse reaction is lowered
exactly the same amount as for the forward re-
action. This accounts for the experimental fact
that a catalyst for a reaction has an equal effect
on the reverse reaction; that is, both reactions
are speeded up by the same factor. If a catalyst
doubles the rate in one direction, it also doubles
the rate in the reverse direction.

8-2.4 Examples of Catalysts

In all cases of catalysis, the catalyst acts by inserting
intermediate steps in a reaction — steps that would not

occur without the catalyst. The catalyst itself must be
regenerated in a subsequent step. (An added substance
which is permanently used by reaction is a reactant, not
a catalyst.) An example is provided by the catalytic action
of acid on the decomposition of formic acid, HCOOH.
A model of formic acid is shown in Figure 8-9. The
carbon atom has attached to it a hydrogen atom, an
oxygen atom, and an OH group.

Figure 8-10 shows how this molecule might decompose.
If the hydrogen atom attached to carbon migrates over
to the OH group, the carbon-oxygen bond can break to
give a molecule of water and a molecule of carbon mon-
oxide. This migration, shown in the center drawing, re-
quires a large amount of energy. This means there is a
high activation energy. Hence, the reaction occurs very
slowly :

HCOOH — >■ H,0 + CO (19)

Fig. 8-9. A model of formic acid.

Ball -and- Spring Model

Space. -FiltiriQ Afodel

138

THE RATES OF CHEMICAL REACTIONS I CHAP. 8

"React-ion. coordinate, uncatalyzed decomposition

Fig.

8-10. Potential energy diagram for the uncata-
lyzed decomposition of formic acid.

If sulfuric acid, H2SO4, is added to an aqueous solution
of formic acid, carbon monoxide bubbles out rapidly.
This also occurs if phosphoric acid, H3PO,, is added in-
stead. The common factor is that both of these acids
release hydrogen ions, H+. Yet, careful analysis shows
that the concentration of hydrogen ion is constant during
the rapid decomposition of formic acid. Evidently, hy-
drogen ion acts as a catalyst in the decomposition of
formic acid.

Chemists have a rather clear picture of how H+ cata-
lyzes reaction (79). The availability of H+ in the solution
makes a new reaction path available. The new reaction
mechanism begins with the addition of a hydrogen ion to
formic acid, as shown in Figure 8-11. Thus, the catalyst
is consumed at first, forming a new species, (HCOOH2)+.
In this species one of the carbon-oxygen bonds is weak-
ened. With only a small expenditure of energy, the next
reaction shown in Figure 8-11 can occur, producing
(HCO)+ and H20. Finally, (HCO)+ decomposes to pro-
duce carbon monoxide, CO, and H+. This last reaction
of the sequence regenerates the catalyst, H+.

Each of the steps in this new reaction mechanism is
governed by the same principles that govern a simple re-
action. Each reaction has an activation energy. The over-
all reaction has a potential energy diagram that is merely
a composite of the simple energy curves of the succeeding
steps.

The highest energy required in this new reaction path
is only 18 kcal, much lower than the activation energy
shown in Figure 8-10 for the uncatalyzed reaction. Hence
the rate of decomposition is much faster when acid is
present.

Notice that catalytic action does not cause the reaction.
A catalyst speeds up a reaction that might take place in
its absence but at a much lower rate.

In some cases, the catalyst is a solid substance on whose
surface a reactant molecule can be held (adsorbed) in a
position favorable for reaction until a molecule of another
reactant reaches the same point on the solid. Metals such
as iron, nickel, platinum and palladium seem to act in
this way in reactions involving gases. There is evidence
that in some cases of surface adsorption, bonds of re-
actant particles are weakened or actually broken, thus
aiding reaction with another reactant particle.

A very large number of catalysts, called enzymes, are
found in living tissues. Among the best known examples
of these are the digestive enzymes, such as the ptyalin in
saliva and the pepsin in gastric juice. A common func-
tion of these two enzymes is to hasten the breakdown of
large molecules, such as starch and protein, into simpler
molecules which can be utilized by body cells. In addition
to the relatively small number of digestive enzymes, there
are many other enzymes involved in biochemical proc-
esses. Enzymes are considered again in Chapter 24.

The specific methods by which catalysts work are not
clearly understood in most cases. Finding a catalyst suit-
able for a given reaction usually requires a long period

QUESTIONS AND PROBLEMS

139

Reaction coordinat-e-, catatyzed decomjposi-Hon.

of laboratory experimentation. Yet, we can look forward
to the time when a catalyst can be tailor-made to fit a
particular need. This exciting prospect accounts for the
great activity on this chemical frontier.

Fig. 8-11. Potential energy diagram of the catalyzed
decomposition of formic acid.

QUESTIONS AND PROBLEMS

1 . The rate of movement of an automobile can be
expressed in the units miles per hour. In what
units would you discuss the rate of:

(a) movement of movie film through a pro-
jector;

(b) rotation of a motor shaft;

(d) consumption of milk by a family;

(e) production of automobiles by an auto as-
sembly plant.

2. Pick the member of each pair having the greater
reaction rate. Assume similar conditions within
each pair.

(a) Iron rusting or copper tarnishing.

(b) Wax burning or paper burning.

3. Describe two homogeneous and two heteroge-
neous systems that are not described in the text.

4. Explain why there is danger of explosion where
a large amount of dry, powdered, combustible
material is produced.

5. Explain (at the molecular level) why an increase
in concentration of a reactant may cause an in-
crease in rate of reaction.

6. Consider two gases A and B in a container at
room temperature. What effect will the following
changes have on the rate of the reaction between
these gases?

(a) The pressure is doubled.

(b) The number of molecules of gas A is
doubled.

140

THE RATES OF CHEMICAL REACTIONS CHAP. 8

7. In an important industrial process for producing
ammonia (the Haber Process) the overall re-
action is

N2fgJ + 3H2(g) — ►- 2NH3(g) + 24,000 calories

A yield of approximately 98 % can be obtained
at 200°C and 1000 atmospheres of pressure. The
process makes use of a catalyst which is usually
finely divided, mixed iron oxides containing
small amounts of potassium oxide, K20, and
aluminum oxide, A1203.

(a) Is this reaction exothermic or endothermic?

(b) Suggest a reason for the fact that this reac-
tion is generally carried out at a temperature
of 500°C and 350 atmospheres in spite of the
fact that the yield under these circumstances
is only about 30%.

(d) How many grams of hydrogen must react to
form 1.60 moles of ammonia?

8. Give several ways by which the rate of combus-
tion in a candle flame might be increased. State
why the rate would be increased.

9. State three methods by which the pressure of a
gaseous system can be increased.

10. Do you expect the reaction

CtlUg) + 302(g) — >- 2C02(g) + 2H-20(g)

to represent the mechanism by which ethylene,
C2H4, burns? Why?

11. A group of students is preparing a ten-page
directory. The pages have been printed and are
stacked in ten piles, page by page. The pages
must be: (1) assembled in order, (2) straightened,
and (3) stapled in sets. If three students work
together, each performing a different operation,
which might be the rate-controlling step? What
would be the effect on the overall rate if the first
step were changed by ten helpers joining the
individual assembling the sheets? What if these
ten helpers joined the student working on the
second step? The third step?

1 2. Describe the life and death of an ordinary, empty
water glass, Utilize the concept "threshold en-
ergy."

13. An increase in temperature of 10°C rarely dou-
bles the kinetic energy of particles and hence the
number of collisions is not doubled. Yet, this
temperature increase may be enough to double
the rate of a slow reaction. How can this be
explained?

14. In a collision of particles, what is the primary
factor that determines whether a reaction will
occur?

15. In Figure 8-6, why is kinetic energy decreasing as
NO> and CO go up the left side of the barrier
and why is kinetic energy increasing as C02 and
NO go down the right side? Explain in terms of
conservation of energy and also in terms of what
is occurring to the various particles in relation
to each other.

16. Phosphorus, P4, exposed to air burns spontane-
ously to give P4Oi0; the AH of this reaction is
— 712 kcal/mole P4.

(a) Draw an energy diagram for the net reaction,
explaining the critical parts of the curve.

(b) How much heat is produced when 12.4
grams of phosphorus burn?

17. Considering that so little energy is required to
convert graphite to diamond (recall Problem 8,
Chapter 7), how do you account for the great
difficulty found in the industrial process for ac-
complishing this?

18. Why does a burning match light a candle?

19. Draw an energy diagram for the reaction

C(s) + Q2(g)

CO-2(g)

(a) when the C is in large chunks of coal.

(b) Is the curve changed if very fine carbon
powder is used?

20. Sketch a potential energy diagram which might
represent an endothermic reaction. (Label parts
of curve representing activated complex, activa-
tion energy, net energy absorbed.)

21. Why is it difficult to "hardboil" an egg at the
top of Pike's Peak? Is it also difficult to cook
scrambled eggs there? Explain.

22. Give two factors that would increase the rate
of a reaction and explain why these do increase
the rate.

HENRY EYRING, 1901 -

Henry Eyring is one of the most active and honored chem-
ists of our time. His advancement of the theory of reaction
rates benefits practically every field of chemistry and chemi-
cal technology. His 300 published papers and five books
range through chemistry, physics, metallurgy, and biology.

Henry was born in Chihuahua, Mexico, on a cattle
ranch. The Eyring family was forced to abandon this home
eleven years later under threat by the revolutionist Salazar.
With other displaced American colonists, the family moved
to Texas, then to Arizona. By distinguishing himself in
high school, Henry Eyring earned a scholarship to the
University of Arizona. Years later, this university gave
him their Distinguished Alumnus Award, proud that he had
earned a B.S. in mining engineering and an M.S. in
metallurgy.

After graduation he was highly successful but not satis-
fied as an engineer in a flotation mill. He turned to the
University of California where he received the Ph.D. in
physical chemistry and where he felt the inspiration of the
great G. N. Lewis. Two years of teaching at the University
of Wisconsin, a year of study in Germany, and a year as
lecturer at the University of California won him a faculty
position at Princeton University. In 1946 he went to the
University of Utah as Chairman of the Chemistry Depart-
ment and Dean of the Graduate School, carrying with him
world recognition in chemistry.

Henry Eyring'' s research has been original and frequently
unorthodox. He was one of the first chemists to apply
quantum mechanics in chemistry. He unleashed a revolu-
tion in the treatment of reaction rates by use of detailed
thermodynamic reasoning. Having formulated the idea of
the activated complex, Eyring proc ceded to find a myriad
of fruitful applications — to viscous flow of liquids, to dif-
fusion in liquids, to conductance, to adsorption, to catalysis.

Anyone who hears Eyring speak on chemistry leaves
convinced that clarifying chemical behavior is exhilarating
fun. Ey ring's enjoyment of science is as obvious as is the
importance of his fundamental contributions. There remains
the noteworthy facet of this great scientist that he is deeply
religious and gives generously of his time and energy to his
church. He shows deep concern about the political, social,
and ethical implications of science and is always ready to
discuss them. Thus it can be said that, in the broadest
sense, Henry Eyring acts as a catalyst of men's minds.

CHAPTER

Equilibrium in
Chemical Reactions

• • • by • • • equilibrium, we mean a state in which the properties of
a system, as experimentally measured, would suffer no further observ-
able change even after the lapse of an indefinite period of time. It is not
intimated that the individual particles are unchanging.

G. n. lewis and m. randall, 1923

In Chapter 8 we discussed the rate of the reaction
between CO and N02,

CO(g) + NCVgj — >- C02(g) + HO(g) (7)

Then, later in that chapter, we turned to the
question of the rate of the reaction that is the
reverse of (7),

C02(g) + NO(g) — »- CO(g) + N02(g) (2)

Indeed! Can't this reaction make up its mind?
If we mix CO(g) and N02(g), reaction (7) be-
gins. But as soon as it does, C02(g) and NO(g)
are formed. As these products accumulate, re-
action (2) becomes possible, undoing reaction

(7). Which wins out?

By direct observation of the reddish-brown
color of N02 we can see the progress of reaction
(7). The N02 is consumed at first, but after a
time the color stops changing. When changes
no longer occur in a reacting chemical system,
we say the system has reached a state of equi-
librium. The equilibrium situation raises many
interesting questions. How do we recognize equi-
librium? What are the molecules doing at the
state of equilibrium? What factors change the
state of equilibrium? What is the composition of
the gas mixture at equilibrium? In this chapter
we shall seek answers to these questions.

9-1 QUALITATIVE ASPECTS OF EQUILIBRIUM

We have encountered equilibrium before — in our
consideration of phase changes. In Section 5-1.2
142

we considered the liquid-gas equilibrium that
fixes the vapor pressure of a liquid, and in Sec-

SEC. 9-1 I QUALITATIVE ASPECTS OF EQUILIBRIUM

143

Solid begins
to dissolve

Solid stilt dissolving,
color deepening

JVb -morQ changes ,
equilibrium exists

Fig.

9-1. Iodine dissolving in an alcohol-water mix-
ture. Equilibrium is recognized by constant
color of the solution.

tion 5-2.4 we considered the solid-liquid equilib-
rium that fixes the solubility of a solid in a liquid.
With this as background, let us consider the first
question about equilibrium: How do we recog-
nize it?

9-1.1 Recognizing Equilibrium

Figure 9-1 shows the addition of solid iodine to
a mixture of water and alcohol. At first the
liquid is colorless but very quickly a reddish
color appears near the solid. Stirring the liquid
causes swirls of the reddish color to move out —
solid iodine is dissolving to become part of the
liquid. Changes are evident: the liquid takes on
an increasing color and the pieces of solid iodine
diminish in size as time passes. Finally, however,
the color stops changing (see Figure 9-1). Solid
is still present but the pieces of iodine no longer
diminish in size. Since we can detect no more
evidence of change, we say that the system is at
equilibrium. Equilibrium is characterized by con-
stancy of macroscopic properties*

Calcium carbonate, CaC03, decomposes upon

* Remember that the word macroscopic was defined in
Chapter 7. It means a large amount of material — enough
to see and weigh.

heating to form carbon dioxide gas, C02, and
calcium oxide (lime), CaO:f

CaCOi(s) ^S- CaO(s) + C02(g) (5)

temp.

Figure 9-2 shows the result of heating solid
CaC03, initially under a vacuum, to 800°C (part
A). Decomposition begins according to reaction
(3) and the gas pressure rises (part B). The pres-
sure continues to rise until it reaches 190 mm
(part C). Thereafter, no further change is evident.
Since we can detect no more evidence of change,
we say that the system is at equilibrium. Equi-
librium is characterized by constancy of macro-
scopic properties.

Though a system at equilibrium is constant in
properties, constancy is not the only require-
ment. Consider a laboratory burner flame. There
is a well-defined structure to the flame — an inner
cone surrounded by a luminous region whose
appearance does not change. A temperature
measurement made at a particular place in the
flame shows that the temperature at that spot
is constant. At another place in the flame the
temperature might be different but, again, it
would be constant, not changing with time. A
measurement of the gas flow rate shows a con-
stant movement of gas into the flame. Yet a
laboratory burner flame is not at equilibrium be-

f Reaction (3) is used for the manufacture of millions
of tons of lime every year in the United States, for use,
principally, in plaster.

144

EQUILIBRIUM IN CHEMICAL REACTIONS I CHAP. 9

CaCO, (s)

Vacuum

Ca COj (S) -h CaO (S)

Solid firstr lieatad.
PresPitra ~ O

JP-K&ssura t-i&incj.

Calciurn ox.ide and

carhon. dioxide forming.

Pressure. —190 -vnm.
2>/o -more chances.
Equilibrium exis-ts.

Fig. 9-2. The thermal decomposition of calcium car-
bonate, CaCO:,(.y) *± CaO(i) + CO An).

cause chemical change is occurring. Methane,
CH4, and oxygen, Oj, are continuously fed into
the flame and carbon dioxide, C02, and water,
H20, are continuously leaving. Substances are
entering and leaving at all times. Such a system
is called an open system. Furthermore, the tem-
perature is not uniform throughout the system.
Equilibrium can exist only in a closed system —
a system containing a constant amount of matter
with all of this matter at the same temperature.
The laboratory burner flame is called a steady
state to indicate that some of its properties are
constant but equilibrium does not exist.

Now we can give a complete statement about
recognizing equilibrium: equilibrium is rec-
ognized by the constancy of macroscopic
properties in a closed system at a uniform
temperature.

EXERCISE 9-1

Which of the following systems constitute steady
state situations, and which are at equilibrium?
For each, a constant property is indicated.

(a) An open pan of water is boiling on a stove.
The temperature of the water is constant.

(b) A balloon contains air and a few drops of
water. The pressure in the balloon is con-
stant.

9-1.2 The Dynamic Nature off Equilibrium

The constancy of properties at equilibrium refers
to macroscopic measurements. Now we will con-
sider what the equilibrium is like on the molec-
ular level, as chemists picture it.

SOLUBILITY

Figure 9-1C shows a system at equilibrium.
Solid iodine has dissolved in an alcohol-water
mixture until the solution is saturated. Then no
more solid dissolves and the color of the solution
remains constant.

Among the molecules, however, business is
going on as usual. Iodine dissolves by the detach-
ment of surface layer molecules from the iodine
crystals. The rate at which this process occurs is
fixed by the stability of the crystal (tending to
hold the molecules in the surface layer) and the
temperature (the thermal agitation tending to
dislodge the molecules from their lattice posi-
tions). As the dissolving continues, the concen-
tration of iodine molecules in the solution in-
creases.

Occasionally a molecule moving about in the
solution encounters the surface of an iodine
crystal and lodges there. This addition to the
crystal is called precipitation, or crystallization,
and it occurs more and more often as the con-
centration of iodine in solution rises.

Here we have two opposing processes. At a
given temperature, molecules leave the surface
of the crystal at a constant rate, tending to in-
crease the concentration in solution. On the
other hand, dissolved molecules are continually
striking the surface and precipitating, tending to

SEC. 9-1 I QUALITATIVE ASPECTS OF EQUILIBRIUM

145

decrease the concentration of molecules in solu-
tion. When enough material has dissolved so that
the rate of return of molecules to the surface of
the solid is just equal to the rate at which they
are leaving the surface, no more net change will
occur. Even though molecules are continually
dissolving and others are precipitating, as long
as these two processes are in balance, the amount
of iodine dissolved per unit volume of solution
will be constant. This macroscopic property, the
solubility, is now constant: the system is in solu-
bility equilibrium. But chemists interpret this
constancy as a balance between two oppos-
ing processes which continue at equilibrium.
At equilibrium, microscopic processes con-
tinue but in a balance that yields no macro-
scopic changes.

VAPOR PRESSURE

Consideration of the dissolving of iodine in an
alcohol-water mixture on the molecular level
reveals the dynamic nature of the equilibrium
state. The same type of argument is applicable
to vapor pressure.

We have already noted that if we place liquid
water in a flask at 20°C and seal the flask, some
water molecules leave the liquid and enter the
gas phase. The partial pressure rises as more and
more water molecules become part of the gas.
Finally, however, the pressure stops rising and
the partial pressure of water becomes constant.
This partial pressure is the vapor pressure and
equilibrium now exists.

Yet, it is reasonable to suppose that water
molecules from the liquid are still evaporating,
even at equilibrium. Molecules in the liquid have
no way of "knowing" that the partial pressure
of the vapor is equal to the vapor pressure. In
the gas phase, the randomly moving molecules
continue to strike the surface of the liquid and
condense. Equilibrium corresponds to a perfect
balance between this continuing evaporation and
condensation. Then no net changes can be de-
tected.*

* When the partial pressure of the water equals the
vapor pressure, the gas above the liquid is said to be
saturated. The word "saturated" has the same meaning
as it did relative to solubility: the gas phase contains as
much water vapor as it can hold at equilibrium.

»1

OB nnw

Unsaturated

Saturated

Supers a tu rated

vapor"

vapor,
EQUILIBRIUM

Vapor

Fig. 9-3. Exchange of molecules between liquid and

gas (A) when the partial pressure is below
the vapor pressure; (B) at equilibrium; (C)
when the partial pressure is above the vapor
pressure.

Figure 9-3 shows this schematically. If the
partial pressure of the vapor is less than the
equilibrium value (as in Figure 9-3A), the rate of
evaporation exceeds the rate of condensation
until the partial pressure of the vapor equals the
equilibrium vapor pressure. If we inject an excess
of vapor into the bottle (as in Figure 9-3C), con-
densation will proceed faster than evaporation
until the excess of vapor has condensed. The
equilibrium vapor pressure corresponds to that
concentration of water vapor at which condensa-
tion and evaporation occur at exactly the same
rate (as in Figure 9-3 B). At equilibrium, micro-
scopic processes continue but in a balance that
yields no macroscopic changes.

CHEMICAL REACTIONS

Let us examine a chemical reaction to see if these
same conditions apply. Suppose we fill two iden-
tical bulbs to equal pressures of nitrogen dioxide.
Now immerse the first bulb (bulb A) in an ice
bath and the second bulb (bulb B) in boiling
water, as in Figure 9-4. The gas in bulb A at
0°C is almost colorless; the gas in bulb B at
100°C is reddish-brown. The predominant mo-
lecular species in the cold bulb must be different
from that in the hot bulb. A variety of experi-
ments shows that the cold bulb contains mostly
N2O4 molecules. These same experiments show
that the hot bulb contains mostly N02 molecules.
The N2O4 molecules absorb no visible light, so

EQUILIB

RIUM IN CHEMICAL REACTIONS | CHAP. 9

r - 0°C

t^2S°C

t=100°C

Fig 9-4. Nitrogen dioxide gas at different tempera-
tures. Bulb A: At 0°C: N20, {almost color-
less). Bulb B: At 100°C: N02 (reddish-
brown).

the cold gas is almost colorless. The N02 mole-
cules do absorb some visible light, so the hot gas
is reddish-brown.

Now let us transfer these two bulbs to a bath
at room temperature, as shown in Figure 9-5.
Immediately the color begins to intensify in bulb
A. The color shows that a chemical change has
occurred, forming N02 molecules from N204:
In bulb A N204(gj — ►- 2N02(g,> 00

At the same time, the color in bulb B begins to
pale, showing that a chemical change has oc-
curred in this bulb as well, forming N204 mole-
cules from N02:

In bulb B 2N02(gj — *■ WJg) (5)

In each bulb the colors continue to change, bulb
A becoming darker and bulb B becoming more

Fig. 9-5. Nitrogen dioxide gas at room temperature.

Bulb A and bulb B after transfer to water
bath at 25°C.

pale. Finally, as the two bulbs approach the same
temperature, the colors stop changing. A close
examination shows that the two colors are now

the same!

By direct visual observation we can watch the
contents of these two bulbs approach the con-
stancy of macroscopic properties (in this case,
color) that indicates equilibrium. In bulb A equi-
librium was approached by the dissociation of
N204, reaction (4); in bulb B it was approached
by the opposite reaction, reaction (5). Here it is
clear why the color of each bulb stopped chang-
ing at the particular hue characteristic of the
equilibrium state at 25°C. The reaction between
N02 and N204 can proceed in both directions:

N204(gj — *■ 2N02(gj 00

N204(gj ■<— 2N02(g>) (5)

Since N204 molecules can dissociate in bulb A,
they must also be able to dissociate in bulb B.
Surely an N204 molecule doesn't act differently
in bulb A (at 25°C) than it does in bulb B (at
25°C). The same sort of statements must apply
to the combination of two N02 molecules. If the
reaction occurs in bulb B, then it must also occur
in bulb A. The net change we see (by observing
the changing N02 color) represents, then, the
difference in the rate of production of N02 by
reaction (4) and the rate of loss of N02 by reac-
tion (5). Changes will cease when these two rates
are exactly equal. If we approach equilibrium
from a lower temperature (which favors N204),
then reaction (4) predominates at first. But as
more and more NOz is produced, reaction (5)
becomes faster and faster. When reaction (5) be-
comes just as fast as reaction (4), then equilib-
rium has been reached: macroscopic properties
no longer change even though both reactions still
proceed in a state of balance. At this time we
replace the single arrow in reaction (4) by a
double arrow (+±) or an equals sign (=) to
show that equilibrium prevails:*

NtOt(g) q=fc 2NOt(g) (*)

NjCVsJ = THOjg) «*)

*=25°C

♦These alternative notations, +±. and =, are used
by chemists interchangeably in equations for chemical
reactions. Both notations will be seen.

SEC. 9-1 I QUALITATIVE ASPECTS OF EQUILIBRIUM

147

In bulb B we approached equilibrium from a
higher temperature (which favors N02); then re-
action (5) predominated at first. Using the same
sort of argument we applied to bulb A, we see
that as time progresses, reaction (4) becomes
more and more rapid (as N204 is produced) and
reaction (5) becomes slower (as N02 is used up).
Finally, when the rates become equal, equilib-
rium is reached and the equilibrium expression
(6) is applicable in bulb B.

For chemical reactions, just as for phase
changes, at equilibrium, microscopic processes
continue but in a balance which gives no macro-
scopic changes.

9-1.3 The State of Equilibrium

It is most important to note that in our descrip-
tion of the equilibrium state we have not implied
that at equilibrium the number of moles of N204
remaining is the same as the number of moles of
N02 produced. Equation (<5) gives us no informa-
tion concerning the fraction of the nitrogen di-
oxide present as NOi at equilibrium. This is
easily verified by heating the water surrounding
bulbs A and B about 10°. The colors of the gases
in both bulbs change to a new equilibrium color
(one corresponding to the presence of more
N02). Yet the same expression is applicable:
N204(gj =«=*: 2HOJg) (6)

What does equation (<5) tell us, then? First, it
tells us that equilibrium prevails (ihe q=^ sign
tells us that). Next, it tells us that there are two
types of molecules present, N204 molecules and
N02 molecules. Finally, it tells us that during the
approach to equilibrium, two molecules of N02
are produced (or consumed) for every one mole-
cule of N204 dissociated (or formed). It does not
tell us whether at equilibrium there will be much
or little N02 compared with the amount of N204.
To emphasize this point, let us consider an-
other familiar reaction,

HMg) qj=fc H2(g) + \02(g) (7)

Until we are given the necessary information we
have no idea how complete the decomposition of
water is at equilibrium. All we know is that for
every mole of water which decomposes we will

obtain 1 mole of hydrogen and \ mole of oxygen.

It has been determined that in a closed vessel
at 2273°K and a total pressure equal to one
atmosphere at the time equilibrium is attained,
0.6% of the water has dissociated. If we started
with one mole of water, 0.6% = 0.6 X ^hs =
0.006 mole would decompose. There would be
left 1 - 0.006 = 0.994 mole of undecomposed
water. There would be formed 0.006 mole of H2
and 0.003 mole of oxygen. We can summarize
this as follows:

H&g ) q=b H2(g) + \02(g)

Initial moles 1 0 0

Moles present
at equilibrium

0.994

0.006 0.003

In other words, if we start with water, not much
of it has decomposed when the equilibrium state
is attained at 2273°K.

Now what about approaching the equilibrium
state by starting with hydrogen and oxygen? Let
us start with 1 mole of hydrogen and \ mole of
oxygen and allow the reaction to attain equilib-
rium at 2273°K and a total pressure equal to one
atmosphere. At equilibrium we find present 0.994
mole of water, 0.006 mole of Hj, and 0.003 mole
of 02. This can be summarized as follows:

Initial moles

Moles present
at equilibrium

Wg) + \0«(g)
1 0.5

0.006 0.003

U£>(g)
0

0.994

If we start with hydrogen and oxygen, equilib-
rium is attained after most of the hydrogen and
oxygen have united to form water. More im-
portant, though, the partial pressures at equilib-
rium are the same as those obtained beginning
with pure H20. The equilibrium pressures are
fixed by the temperature, the composition, and
the total pressure; they do not depend upon the
direction from which equilibrium is approached.
The balanced equation does not indicate the
concentrations (or partial pressures) at equilib-
rium.

9-1.4 Altering the State of Equilibrium

We have seen that, qualitatively, the state of
equilibrium for a system is characterized by the

148

EQUILIBRIUM IN CHEMICAL REACTIONS I CHAP. 9

relative amounts of products and reactants pres-
ent. With reference to the decomposition of
water, any change in conditions which would
cause more than 0.6% of the water to dissociate
at equilibrium would be said to change the state
of equilibrium for the reaction,

H,0(g) +± U2(g) + \Q2(g)

(7)

in favor of the formation of more hydrogen and
oxygen.

What conditions might alter the equilibrium
state? Concentration and temperature! These
are factors that affect the rate of reaction. Equi-
librium is attained when the rates of opposing
reactions become equal. Any condition that
changes the rate of one of the reactions involved
in the equilibrium may affect the conditions at
equilibrium.

CONCENTRATION

Consider the reaction you encountered in the
laboratory — that between ferric ion (Fe+3) and
thiocyanate ion (SCN-):

Fe+YaqJ + SCN-(aq)

FeSCN+2(aqj (5)

Again we have visual evidence of concentration
at equilibrium since the intensity of the color is
fixed by the concentration of the FeSCN+2 ion.
The addition of either more ferric ion [by adding

Fig. 9-6. Equilibrium conditions are affected by the
reactant concentrations.

a soluble salt such as ferric nitrate, Fe(N03)3] or
more thiocyanate ion (by adding, say, sodium
thiocyanate) changes the concentration of one of
the reactants in (8). Immediately the color of
the solution darkens, showing that there is an
increase in the amount of the colored ion,
FeSCN+2. The equilibrium concentrations are af-
fected if the concentrations of reactants {or prod-
ucts) are altered.

TEMPERATURE

We have already considered an example of the
change of equilibrium concentrations as the tem-
perature is altered. The relative amounts of N02
and N204 are readily and obviously affected by
a temperature change. The equilibrium concen-
trations are affected if the temperature is altered.

CATALYSTS

Catalysts increase the rate of reactions. It is
found experimentally that addition of a catalyst
to a system at equilibrium does not alter the
equilibrium state. Hence it must be true that any
catalyst has the same effect on the rates of the
forward and reverse reactions. You will recall
that the effect of a catalyst on reaction rates can
be discussed in terms of lowering the activation
energy. This lowering is effective in increasing
the rate in both directions, forward and reverse.
Thus, a catalyst produces no net change in the
equilibrium concentrations even though the sys-
tem may reach equilibrium much more rapidly
than it did without the catalyst.

9-1.5 Attainment of Equilibrium

The equilibrium state is not always attained in
chemical reactions. Consider reaction (7):

H20(g) q=fc H2(g) + \02(g)

AH = +57.8 kcal (7)

A large amount of heat is absorbed in this reac-
tion, 57.8 kcal/mole of water decomposed. If the
temperature is lowered, the state of equilibrium
is even more favorable to the production of
water at room temperature than it is at 2273°K.
Yet a mixture of hydrogen and oxygen can re-
main at room temperature for a long period
without apparent reaction. Equilibrium is not

SEC. 9-1 I QUALITATIVE ASPECTS OF EQUILIBRIUM

149

attained in this system because the rate of the
reaction between hydrogen and oxygen at room
temperature is too low. This explanation is easily
verified by speeding up the reaction slightly. If a
mixture of H2 and 02 is disturbed with a small
spark, reaction begins and it enthusiastically (and
explosively) continues until most of the gases
have been converted to water.

This distinction between the conditions in a
chemical system at equilibrium and the rate at
which these conditions are attained is very im-
portant in chemistry. By arguments that we shall
consider a chemist can decide with confidence
whether equilibrium favors reactants or products
or neither. He cannot predict, however, how
rapidly the system will approach the equilibrium
conditions. That is a matter of reaction rates,
and the chemist must perform separate experi-
ments to learn whether a given rate is rapid or
not.

9-1.6 Predicting New Equilibrium

Concentrations: Le C ha teller's Principle

We are not satisfied with the conclusion that this
change or that change affects the equilibrium
concentrations. We would also like to predict
the direction of the effect (does it favor products
or reactants?) and the magnitude of the effect
(how much does it favor products or reactants?).
The first desire, to know the qualitative effects,
is answered by a generalization first proposed by
a French chemist, Henry Louis Le Chatelier, and
now called Le Chatelier's Principle.

Le Chatelier sought regularities among a large
amount of experimental data concerning equi-
libria. To summarize the regularities he found,
he made this generalization: If an equilibrium
system is subjected to a change, processes
occur that tend to counteract partially the
imposed change. This generalization has been
found to be applicable to such a large number
of systems that it is now called a principle. Let
us see how it applies to our examples.

CONCENTRATION AND LE CHATELIER'S
PRINCIPLE

If a soluble thiocyanate salt is added to an equi-
librium solution containing both ¥e+3(aq) and

SCN-(aq), the color of the complex ion in-
creases :

Fe+3(aq) + SCN-faqj 3=fc FeSCN+2( aq) (8)

A new state of equilibrium is then attained in
which more FeSCN+2 is present than was there
before the addition of SCN-. Increasing the con-
centration of SCN- has increased the concentra-
tion of the FeSCN+2 ion. This is in accord with
Le Chatelier's Principle. The change imposed
on the system was an increase in the concentra-
tion of SCN-. This change can be counteracted
in part by some Fe+3 and SCN- ions reacting to
form more FeSCN+2. The same argument applies
to an addition of ferric ion from a soluble ferric
salt. In each case, the formation of FeSCN+2 uses
up a portion of the added reactant, partially
counteracting the change.

PRESSURE AND LE CHATELIER'S
PRINCIPLE

Instead of altering the concentration of one in-
dividual component in an equilibrium system,
we can alter the concentration of all gaseous
components by changing the pressure at which
the system is confined. Let us start with the sys-
tem represented by equation (7) and double the
total pressure. The system now occupies a much
smaller volume than it did previously. The total
number of moles present per unit volume is
greater than it was under the original equilibrium
conditions. This change can be counteracted in
part if some hydrogen and oxygen combine to
form gaseous water. Then the total number of
moles present is reduced (1$ moles unite to form
1 mole). Hence, we can predict that increasing
the concentration of all components by increas-
ing the pressure will shift the state of equilibrium
in favor of the formation of gaseous water. This
is in accord with experiment.

A change in total pressure does not always
shift equilibrium. The first reaction mentioned
in this chapter exemplifies this:

CO(g) + NCVgj +± CO,(g) + NO(g) (9)

If we increase the pressure on a mixture of these
four gases at equilibrium, the gases are com-
pressed to a smaller volume. Once again the con-

150

EQUILIBRIUM IN CHEMICAL REACTIONS I CHAP. 9

centrations are all increased. What does Le
Chatelier's Principle tell us here? If the equilib-
rium state is altered to favor products, some CO
and N02 molecules react to form an exactly
equal number of molecules of C02 and NO.
Since there is no change in the total number of
moles, the proposed change in the equilibrium
does not partially reduce the pressure change.
Le Chatelier's Principle tells us that processes
occur so as to "counteract partially the imposed
change." Here neither a change favoring the
reactants nor a change favoring products will
"counteract" the imposed pressure change.
Hence, Le Chatelier's Principle leads us to expect
no change of the equilibrium state for reaction
(9) when the pressure is altered. Experimentally
we find that no change is observed. The equilib-
rium state is not affected by a pressure change
for any equilibrium gas mixture where the num-
ber of reactant molecules is the same as the
number of product molecules in the balanced
reaction.

EXERCISE 9-2

Does Le Chatelier's Principle predict a change of
equilibrium concentrations for the following re-
actions if the gas mixture is compressed? If so,
does the change favor reactants or products?

(a) N20<fej q=b 2N02(gj

(b) H2(g) + Ug) +± 2Hl(g)

TEMPERATURE AND LE CHATELIER'S
PRINCIPLE

Let us add to reaction (4) the information that
the decomposition of N204 is endothermic:

WJg) +±: 2NOi(g) AH = +14.1 kcal (4)

Our experimental observations indicated that
warming a bulb containing N02 and N204 caused
a shift of the equilibrium state in favor of the
formation of N02 (the reddish-brown color
deepened). It is easy to see that this is in accord
with Le Chatelier's Principle. A rise in tempera-
ture is caused by an input of heat. At the higher

temperature, the equilibrium is changed to form
more N02. The formation of N02 absorbs a por-
tion of the heat that caused the temperature rise.
Raising the temperature of liquid water raises
its vapor pressure. This is in accord with Le
Chatelier's Principle since heat is absorbed as the
liquid vaporizes. This absorption of heat, which
accompanies the change to the new equilibrium
conditions, partially counteracts the temperature
rise which caused the change.

9-1.7 Application of Equilibrium Principles:
The Haber Process

Knowledge of chemical principles pays rewards
in technological progress. Control of chemical
reactions is the key. The large scale commercial
production of nitrogen compounds provides a
practical example of the beneficial application of
Le Chatelier's Principle.

The most difficult step in the process for the
conversion of the inert nitrogen of the atmos-
phere into important commercial compounds
such as fertilizers and explosives involves the
reaction

N2(g) + 3Hi(gj +±z 2NH3(g) + 22 kcal (10a)

NJg) + 3H2(g)

2NH3(g)

AH = -22 kcal (Wb)

Can we predict the optimum conditions for a
high yield of NH3? Should the system be allowed
to attain equilibrium at a low or a high tempera-
ture? Application of Le Chatelier's Principle sug-
gests that the lower the temperature the more
the equilibrium state will favor the production
of NH3. Should we use a low or a high pressure?
The production of NH3 represents a decrease
in total moles present from 4 to 2. Again Le
Chatelier's Principle suggests use of pressure to
increase concentration. But what about prac-
ticality? At low temperatures reaction rates are
slow. Therefore a compromise is necessary. Low
temperature is required for a desirable equilib-
rium state and high temperature is necessary for
a satisfactory rate. The compromise used in-
dustrially involves ?" intermediate temperature
around 500°C and even then the success of the

SEC. 9-2 I QUANTITATIVE ASPECTS OF EQUILIBRIUM

151

process depends upon the presence of a suitable
catalyst to achieve a reasonable reaction rate.
With regard to pressure, another compromise
is needed. It is expensive to build high pressure
equipment. A pressure of about 350 atmospheres
is actually used. Under these conditions, 350
atmospheres and 500°C, only about 30% of the
reactants are converted to NH3. The NH3 is re-
moved from the mixture by liquefying it under
conditions at which N2 and H2 remain as gases.

The unreacted N2 and H2 are then recycled until
the total percent conversion to ammonia is very
high.

Prior to World War I the principal sources of
nitrogen compounds were some nitrate deposits
in Chile. Fritz Haber, a German chemist, suc-
cessfully developed the process we have just
described, thus allowing chemists to use the
almost unlimited supply of nitrogen in the at-
mosphere as a source of nitrogen compounds.

9-2 QUANTITATIVE ASPECTS OF EQUILIBRIUM

Le Chatelier's Principle permits the chemist to
make qualitative predictions about the equilib-
rium state. Despite the usefulness of such predic-
tions, they represent far less than we wish to
know. It is a help to know that raising the pres-
sure will favor production of NH3 in reaction
(10a). But how much will the pressure change
favor NH3 production? Will the yield change by
a factor of ten or by one-tenth of a percent? To
control a reaction, we need quantitative informa-
tion about equilibrium. Experiments show that
quantitative predictions are possible and they
can be explained in terms of our view of equilib-
rium on the molecular level.

9-2.1 The Equilibrium Constant

By means of colorimetric determination in the
laboratory you measured the concentration of
FeSCN+2, which we shall designate [FeSCN+2],*
in solutions containing ferric and thiocyanate
ions, Fe+3 and SCN~. The reaction is

Fe+YaqJ + SCN-(aq) +± FeSCN+*faqj (8)

From [FeSCN+2] and the initial values of [Fe+3]
and [SCN-] you calculated the values of [Fe+3]
and [SCN-] at equilibrium. You then made cal-
culations for various combinations of these val-

* Hereafter, we shall regularly use the square brackets
notation, [ ], to indicate concentration. Thus, we read
[Fe+I] as "ferric ion concentration."

ues. Many experiments just like these show that
the ratio

[FeSCN+2]

[Fe+3][SCN-]

(//)

comes closest to being a fixed value. Note that
this ratio is the quotient of the equilibrium con-
centration of the single substance produced in
the reaction divided by the product of the equi-
librium concentrations of the reactants.

Colorimetric analysis based on visual estima-
tion is not very exact. Some more accurate data
on the H2, I2, HI system at equilibrium are
shown in Table 9-1. The reaction is

2Hl(g)

H2(g) + h(g)

(12)

The data have been expressed in concentrations,
although pressure units are more usual for a
reaction involving gases.

EXERCISE 9-3

For the last two experiments in Table 9-1, num-
bers 4 and 5, why is [H2] = [I2]? For experiment
1, what were the initial concentrations of H2 and
I2 before the reaction occurred to form HI ?

Let's work with these data. Heartened by your
results from your own laboratory data, let us
compute the value of the ratio

[H«][l«]
[HI]

(13)

152

EQUILIBRIUM IN CHEMICAL REACTIONS | CHAP. 9

Table 9-1. equilibrium concentration at 698. e k of hydrogen,

IODINE, AND HYDROGEN IODIDE

EXPT.
NO.

[H2]
(moles/liter)

(moles/liter)

[HI]

(moles/liter)

1.8313 X 10"3

3.1292 X 10"3

17.671 X 10"3

2.9070 X 10"3

1.7069 X 10-3

16.482 X 10"3

4.5647 X 10-3

0.7378 X 10-3

13.544 X 10-3

0.4789 X 10-3

3.531 X 10-3

1.1409 X 10"3

1.1409 X 10-3

8.410 X 10"»

* Values above the line were obtained by heating hydrogen and iodine
together; values below the line, by heating pure hydrogen iodide.

We obtain the numbers in Table 9-II. In view of
the precision of the data from which these ratios
are derived, the ratios are far from constant.
Now let us try the ratio

rH2][i2]

[HI]2

(14)

These calculations are summarized in Table
MIL

The results are most encouraging and imply
that with a fair degree of accuracy we can write

[H2HI2] f ,

1 rHI]2 J = a constant

= 1.835 X 10"2 at 698.6°K (75)
Look at this ratio in terms of reaction (12):

2Hl(g) ^± H2(g) + h(g) (12)

The ratio (75) is the product of the equilibrium
concentrations of the substances produced in the
reaction, [H2] X [h], divided by the square of

the concentration of the reacting substance,
[HI]2. In this ratio, the power to which we raise
the concentration of each substance is equal to
its coefficient in reaction (12).

9-2.2 The Law of Chemical Equilibrium

Let us summarize what we have learned. For the
reaction

Ft+'(aq) + SCN-(aq) +=h FeSCN+2(aqj (5)

we found that the concentrations of the mole-
cules involved have a simple relationship.

[FeSCN+2] , ,

— ■*— = a constant

(76)

[Fe+3][SCN-]

Then we considered precise equilibrium data for
the reaction

2Hl(g)

Wg) + h(g)

(12)

Table 9-11

VALUES OF
OF TABLE 9

[HtHlt]
[HI]

FOR DATA

EXPT.
NO.

Table 9-111

VALUES OF

l&ffi FOR DATA

OF TABLE 9

-1

[H.][l.l
[HI]

EXPT. TH2][I2]
NO. [HI]2

32.429 X 10-*

30.105 X 10-6

24.866 X 10"s

6.495 X 10~6

15.477 X 10-*

1.8351 X 10"2
1.8265 X 10-2
1.8359 X 10"2
1.8390 X 10"2
1.8403 X 10-2

Average 1.835 X lO"2

SEC. 9-2 | QUANTITATIVE ASPECTS OF EQUILIBRIUM

153

The concentrations of the molecules appearing
in reaction (12) were found to have a simple
relationship,

[H«J[I»3 =
[HI?

= a constant

(14)

In each of our simple relationships, (76) and (74),
the concentrations of the products appear in the
numerator. In each relationship the concentra-
tions of reactants appear in the denominator. In
reaction (12), two molecules of hydrogen iodide
react. This influences expression (14) because it
is necessary to square the concentration of hy-
drogen iodide, [HI], in order to obtain a con-
stant ratio.

These observations and many others like them
lead to the generalization known as the Law of
Chemical Equilibrium. For a reaction

aA + bB +±: eE + fF

(17)

when equilibrium exists, there will be a simple
relation between the concentrations of products,
[E] and [F], and the concentrations of reactants,

[A] and [B] :

\EY\FV

■ /-. rniK = K = a constant at

constant temperature

(18)

In this generalized equation, (18), we see that
again the numerator is the product of the equi-
librium concentrations of the substances formed,
each raised to the power equal to the number of
moles of that substance in the chemical equation.
The denominator is again the product of the
equilibrium concentrations of the reacting sub-
stances, each raised to a power equal to the
number of moles of the substance in the chemi-
cal equation. The quotient of these two remains
constant. The constant K is called the equilib-
rium constant. This generalization is one of the
most useful in all of chemistry. From the equa-
tion for any chemical reaction one can imme-
diately write an expression, in terms of the con-
centrations of reactants and products, that will
be constant at any given temperature. If this
constant is measured (by measuring all of the
concentrations in a particular equilibrium solu-
tion), then it can be used in calculations for any
other equilibrium solution at that same tempera-
ture.

In Table 9-IV are listed some reactions along
with the equilibrium law relation of concentra-
tions and the numerical values of the equilibrium
constants. First, let's verify the forms of the
equilibrium law relation among the concentra-

Table 9-IV. some equilibrium constants

Cu(s) + 2Ag+(aq) +± Cu+i(aq) + 2Ag(s)
Ag+(aq) + 2NH3(aq) +± Ag(UH3h+(aq)
N2Ot(g) +± 2N02(g)
2Hl(g) :<=£ U2(g) + \2(g)

HSOr(aq)

CH3COOHfa</J

AgC\(s)
H20

Ag\(s)

H+(aq) + SOrHaq)

H+(aq) + CH.COO- (aq)

Ag+(aq) + C\-(aq)
H+(aq) + OH-(aq)
Ag+(aq) + l-(aq)

EQUILIBRIUM LAW
RELATION

K =

[Cu+;
[Ag+]

[Ag(NFWl

[Ag+][NH3p

[N20<]

K =

[H8][l«]
[HI?

fH+irsor'1

[HSCV]

[H+][CH;,COO-l
[CHjCOOHJ

K = [Ag+][C1-]

A: = [H+][OH]

K = [Ag+][1-]

K AT STATED TEMP.

2 X 1015 at 25°C

1.7 X 107at 25°C
0.87 at 55°C
0.018 at 423°C
0.013 at 25°C

1.8 X 10"5at 25°C

1.7 X 10-'° at 25°C
10-" at 25°C
10-16 at 25°C

154

EQUILIBRIUM IN CHEMICAL REACTIONS | CHAP. 9

tions. The very first has an unexpected form. For
the reaction,

Cu(s) + 2Ag+(aq) = Cu+YaqJ + 2Ag(s) (19)

you do not find

rr\.+2irAoi2

(20)

rcu+'irAgi* K

[Ag+]'[Cu]

but rather, you find

[Cu^]
[Ag+?

(21)

This is because the concentrations of solid cop-
per and solid silver are incorporated into the
equilibrium constant. The concentration of solid
copper is fixed by the density of the metal — it
cannot be altered either by the chemist or by the
progress of the reaction. The same is true of the
concentration of solid silver. Since neither of
these concentrations varies, no matter how much
solid is added, there is no need to write them
each time an equilibrium calculation is made.
Equation (21) will suffice.

EXERCISE 9-4

If we assign the equilibrium constant K' to ex-
pression (20) and K to expression (21),

[Ag+ftCu] [Ag+?

show that

[Ag]2

Another K of unexpected form applies to the
reaction

H,0 +±: H+(aq) + OH(aq) (22)

For this reaction we might have written

[H20]

(23)

Instead, Table 9-1 V lists expression (24) as

[H+][OH] = K (24)

The concentration of water, [H20], does not
appear in the denominator of expression (24).
This is usually done in treating aqueous reactions

that consume or produce water. It is justified
because the variation in the concentration of
water during reaction is so slight in dilute aque-
ous solutions. We can treat [H2OJ as a concen-
tration that does not vary. Hence, it can be in-
corporated in the equilibrium constant.

EXERCISE 9-5

Water has a density of one gram per milliliter.
Calculate the concentration of water (expressed
in moles per liter) in pure water. Now calculate
the concentration of water in 0.10 M aqueous
solution of acetic acid, CHLCOOH, assuming
each molecule of CH3COOH occupies the same
volume as one molecule of H20.

In summary, the concentrations of solids and
the concentrations of solvent (usually water) can
be and usually are incorporated in the equilib-
rium constant, so they do not appear in the
equilibrium law relation.

Now look at the numerical values of the equi-
librium constants. The ICs listed range from 10+15
to 10-16, so we see there is a wide variation. We
want to acquire a sense of the relation between
the size of the equilibrium constant and the state
of equilibrium. A large value of K must mean
that at equilibrium there are much larger con-
centrations present of products than of reactants.
Remember that the numerator of our equilib-
rium expression contains the concentrations of
the products of the reaction. The value of
2 X 1015 for the K for reaction (19) certainly
indicates that if a reaction is initiated by placing
metallic copper in a solution containing Ag+
(for example, in silver nitrate solution), when
equilibrium is finally reached, the concentration
of Cu+2 ion, [Cu+2], is very much greater than
the square of the silver ion concentration, [Ag+]2.

A small value of K for a given reaction implies
that very little of the products have to be formed
from the reactants before equilibrium is attained.
The value of K = 10-16 for the reaction

Agl(s) +±: Ag+(aq) + \(aq) (25)

indicates that very little solid Agl can dissolve

SEC. 9-2 | QUANTITATIVE ASPECTS OF EQUILIBRIUM

155

before equilibrium is attained. Silver iodide has
extremely low solubility. Conversely, if 0.1 M
solutions of KI and AgN03 are mixed, the values
of [Ag+] and [I~] are large and the equilibrium
state cannot be reached until the [Ag+] and [I~]
have been greatly reduced by the precipitation
of Agl.

9-2.3 The Law of Chemical Equilibrium Derived
from Rates off Opposing Reactions

Chemists picture equilibrium as a dynamic balance be-
tween opposing reactions. An understanding of the Law
of Chemical Equilibrium can be built upon this basis.
Consider the oxidation of nitric oxide, NO, to nitrogen
dioxide, Nd:

2NOteJ + Oi(g) — »■ mQh(g) (26)

The reaction to the right, (/?), proceeds with a rate that
is found experimentally to depend upon the concentra-
tions of the reactants as follows :

(rate)* = **[NO]*[0,] (27)

The reverse reaction to the left, (L), has been studied as
well. The rate of this reaction

2NO(g) + Ot(g) ■«— 2NCVSJ (28)

depends upon the concentrations as follows:

(rateU = *L[NO,]» (29)

Expressions (27) and (29) show how the rates of reaction
(26) and its reverse, reaction (28), depend upon the con-
centrations. Now we can apply our microscopic view of
the equilibrium state. Chemical changes will cease (on
the macroscopic scale) when the rate of reaction (26) is
exactly equal to that of reaction (28). When this is so,
we can equate expressions (27) and (29):

(rate)* = (rate)*, (30)

**[NO]s[02] = fa[NO,]» (31)

By algebraic rearrangement, equation (31) can be written

k* [NO,]' 2

kL [NO]*[0,] v }

Since both kR and kL are constants at a given temperature,
their ratio is constant. Hence (32) is the equilibrium law
expression for the equilibrium

2KO(g) + Oz(g) q=b 2NO,(g)

and the equilibrium constant

(33)

(34)

Thus we see that the experimental rate laws for this
reaction and its reverse lead to the equilibrium law. In
every reaction that has been sufficiently studied, this same
result is obtained. We are led to have confidence in the

molecular view of equilibrium as a dynamic balance
between opposing reactions.

9-2.4 The Factors Which Determine
Equilibrium

We have learned much about equilibrium. It is
characterized by constancy of macroscopic prop-
erties but with molecular processes continuing
in a state of dynamic balance. At equilibrium we
can conclude that every reaction that takes place
does so at the same reaction rate as its reverse
reaction.

We have gone further and discovered that the
equilibrium conditions imply a constant relation-
ship among the concentrations of reactants and
products. This relationship is called the Law of
Chemical Equilibrium. Using this law, we can
express the conditions at equilibrium in terms of
a number K, called the equilibrium constant.

Despite this detailed familiarity with equilib-
rium, there is one facet we have not considered
at all. What determines the equilibrium con-
stant? Why does one reaction favor reactants
and another reaction favor products? What fac-
tors cause sodium chloride to have a large solu-
bility in water and silver chloride to have a low
solubility? Why does equilibrium favor the re-
action of oxygen with iron to form Fe203 (rust)
but not the reaction of oxygen with gold? As
scientists, we cannot resist wondering what fac-
tors determine the conditions at equilibrium.

This is the activity of science we called "Won-
dering why" (Section 1-1.3) — we are searching
for explanation. An explanation is a likeness
which connects the system under study with a
model system which is well understood. We
might begin by considering Figure 9-7. Here we
see a golf bag thrown into the rear of a station
wagon. Unfortunately, the ball pocket is open
and all of the golf balls have spilled out onto the
floor of the station wagon. Because the floor has
a step in it, the golf balls on the upper level pos-
sess some potential energy (energy of position).
The golf balls tend to roll to the lower level
spontaneously, as shown in Figure 9-8. As a golf
ball does this, its potential energy becomes ki-
netic energy (energy of motion). Finally, this

156

EQUILIBRIUM IN CHEMICAL REACTIONS ] CHAP. 9

Fig. 9-7. Golf balls rolling on the floor of a station
wagon.

kinetic energy is dissipated into heat. Now the
golf balls lie at rest at the lower floor level.

This situation has some similarities to the
chemical change in a spontaneous, exothermic
reaction. The reactants of high heat content re-
act spontaneously to form products of lower
heat content. As each molecular reaction occurs,
the excess heat content becomes kinetic energy.
The product molecules separate from each other
with high kinetic energy. As they collide with

other molecules, this energy is dissipated into
heat. Figure 9-8 shows this through a heat con-
tent diagram for the chemical reaction.

(1) There are two states of each system:

Initial State Final State

Golf balls: on upper level
Reaction: reactants

on lower level
products

Fig. 9-8. Comparison of a chemical reaction to golf
balls rolling downhill.

Ht3h

heat

content

Low

heat

content

SEC. 9-2 I QUANTITATIVE ASPECTS OF EQUILIBRIUM

157

(2) The potential energy of the initial state is
higher than the potential energy of the final
state:

Golf balls:

Reaction:

Initial State

high potential
energy

high heat
content

Final State

low potential
energy

low heat
content

(3) As the change from initial state to final state
proceeds, the form of the energy changes:

Initial State

Golf balls: potential
energy

Reaction: heat content

Final State

kinetic energy,
and then heat

molecular kinetic
energy, and
then heat

a liquid. Water evaporates spontaneously
but it absorbs heat as it does so. It is not
"rolling downhill" energetically. When am-
monium chloride dissolves in water, the solu-
tion becomes cooler. Again heat is absorbed
— yet the ammonium chloride goes ahead
and dissolves.
(2) Another difficulty is that spontaneous chemi-
cal reactions do not go to completion. Even
if a spontaneous reaction is exothermic, it
proceeds only till it reaches equilibrium. But
in our golf ball analogy, "equilibrium" is
reached when all of the golf balls are on the
lower level. Oui analogy would lead us to
expect that an exothermic reaction would
proceed until all of the reactants are con-
verted to products, not to a dynamic equi-
librium.

(4) The changes from initial to final state pro-
ceed spontaneously toward lowest potential
energy, the direction corresponding to "roll-
ing downhill":

Initial State

Final State

Golf balls:
Reaction:

spontaneous
spontaneous

Having established these similarities, we might
offer a possible generalization:

Since: golf balls always roll downhill spon-
taneously

Perhaps: reactions always proceed spontane-
ously in the direction toward minimum
energy.

This proposal leads us to expect that a reaction
will tend to proceed spontaneously if the prod-
ucts have lower energy than the reactants. This
expectation is in accord with experience with
many reactions, especially for those which re-
lease a large amount of heat.

There are two basic and serious difficulties
with our proposed explanation. (Remember
"Cylindrical objects burn"?)

(1) Some endothermic reactions proceed spon-
taneously. One example is the evaporation of

Because of these failures, we need to alter our
proposed explanation. We must seek a new anal-
ogy that gives a better correspondence with the
behavior of chemical reactions. How should we
alter our golf ball analogy to bring it into better
accord with experimental facts? Here is a pos-
sible view.

Consider how the golf ball situation shown in
Figure 9-8 will change when the station wagon
is driven over a bumpy road. Now the golf balls
are shaken and jostled about; they roll around
and collide with each other. Every now and then
one of the golf balls even accumulates enough
energy (through collisions) to return to the upper
level of the station wagon floor. Of course^ any
golf ball that is bounced up tends to roll back
down to the lower level a little later. As this
bumpy ride continues, a state is reached in which
golf balls are being jostled up to the higher level
at the same rate they are rolling back down to
the lower level. Then "equilibrium" exists. Some
of the golf balls are on the lower level and some
on the upper level. Since the rate of rolling up
equals the rate of rolling down, a dynamic bal-
ance exists.

This analogy solves the problems of the sim-
pler "golf balls roll downhill" picture. The
bumpy road model contains a new feature that
gives a basis for expecting "reaction" in the

158

EQUILIBRIUM IN CHEMICAL REACTIONS I CHAP. 9

High

heat

content-

heat
content

iiiiih

F/g. 9-9. Golf balls rolling on the floor of a station
wagon driving on a bumpy road.

endothermic direction. Some golf balls roll uphill
if they are shaken hard enough. The tendency to
roll back down will always keep them coming
back to the lower level and, finally, equilibrium
will be reached when the rate of rolling down
equals the rate of jostling up.

What happens if the road becomes smoother?
The "jostling up" reaction is less favored— the
equilibrium conditions change in favor of the
golf balls at the lower level.

Now turn to the chemical reaction. What fea-
ture in a reacting chemical system corresponds
to the jostling of the bumpy road in our analogy?
It is the temperature. At any temperature except
absolute zero there is a constant random jostling
of the molecules. Some molecules have low ki-
netic energies, some have high kinetic energies —
we looked at the distribution of the energies in
Chapter 8 (see Figure 8-4, page 131). Some of the
molecules will occasionally accumulate enough
energy to "roll uphill" to less stable molecular
forms. On the one hand, molecular changes will
take place in the direction of minimum energy.

On the other hand, the molecular changes will
finally reach a dynamic equilibrium when the
random jostlings or energy transfers at the tem-
perature of the system are restoring molecules to
the molecular forms of higher energy at the same
rate as they are "rolling downhill" to the lower
energy forms.

Now we have an analogy that does aid us in
understanding chemical reactions and equilib-
rium. We can see the following features of chem-
ical reactions:

(1) Chemical reactions proceed spontaneously
to approach the equilibrium state.

(2) One factor that fixes the equilibrium state is
the energy. Equilibrium tends to favor the
state of the lowest energy.

(3) The other factor that fixes the equilibrium
state is the randomness implied by the tem-
perature. Equilibrium tends to favor the
state of greatest randomness.

(4) The equilibrium state is a compromise
between these two factors, minimum en-
ergy and maximum randomness. At very
low temperatures, energy tends to be the
more important factor. Then equilibrium
favors the molecular substances with the

SEC. 9-2 | QUANTITATIVE ASPECTS OF EQUILIBRIUM

159

lowest heat content. At very high tempera-
tures, randomness becomes more important.
Then equilibrium favors a random distribu-
tion among reactants and products without
regard for energy differences.

Our analogy can be stretched one point fur-
ther. We might ask whether the relative area of
the upper floor level compared with that of the
lower floor level has any bearing on the distribu-
tion of golf balls. After all, if the upper level area
is small, as in Figure 9-7, few golf balls are likely
to remain there. Contrast Figure 9-10, in which
the golf bag has been removed. Now the upper
floor level has much more area. Golf balls which
reach the upper level will now have a great deal
of space. They can roll around longer before
returning to the lower level. The effect of extend-
ing the upper level will be to increase the fraction
of the golf balls that occupy the upper level at
"equilibrium."

This extension of the analogy increases its
value in considering chemical reactions. The
simplest example is probably the vaporization of
a liquid. It is true that the molecules have lower
energy when they cluster together tightly in the
liquid state. On the other hand, the gaseous state
provides a broad upper level. Every molecule
which vaporizes has an amount of space avail-

Fig. 9-10. Golf balls rolling on the floor of a station
wagon. The effect of extending the upper
level.

able to it much larger than it had in the crowded
liquid. This "available space" factor, accom-
panied by the random jostlings of temperature to
overcome the potential energy difference, aids
vaporization.

Now let us review what happens as we warm
a solid substance from a very low temperature
to a very high temperature. As the temperature
is raised, small energy differences become unim-
portant. Thus, if the temperature of the solid is
raised too much, the lower energy of the regular
solid becomes unimportant compared with the
random thermal energies. The solid melts, sur-
rendering this energy stability in return for the
randomness of the liquid state. If the tempera-
ture is raised still further, the energies of attrac-
tion among the molecules become unimportant
compared with the random thermal energies.
Then the liquid vaporizes, surrendering the lower
potential energy afforded by the molecules re-
maining close together in favor of the still higher
randomness of the gaseous state. If we raise the
temperature still further, the energies that hold
molecules together begin to become unimportant
compared with random thermal energies. Fi-
nally, at extremely high energies, molecules no
longer exist — all is chaos. This is the chemical
situation within the Sun. Since at such high en-
ergies chemical reactions become unimportant,
all chemists on the Sun are out of work. We'd
better return to room temperature to apply our
knowledge of equilibrium to chemical systems
within our interest.

160

EQUILIBRIUM IN CHEMICAL REACTIONS | CHAP. 9

QUESTIONS AND PROBLEMS

1. Which of the following are equilibrium situa-
tions and which are steady state situations?

(a) A playing basketball team and the bench of
reserves. The number of players "in the
game" and the number "on the bench" are
constant.

(b) The liquid mercury and mercury vapor in a
thermometer. Temperature is constant.

(d) A well-fed lion in a cage. The lion's weight
is constant.

Answer, (a) and (b) are equilibrium situations.

2. Which of the following are equilibrium situa-
tions and which are steady state situations?

(a) A block of wood floating on water.

(b) During the noon hour, the water fountain
constantly has a line of ten persons.

(c) When a capillary tube is dipped in water,
water rises in the capillary (because of sur-
face tension) to a height /; and remains con-
stant there.

(d) The capillary system of (c) considered over
such a long period that evaporation of water
out the end of the capillary cannot be neg-
lected.

(e) At a particular point in the reaction chamber
of a jet engine, the gas composition (fuel,
air, and products) is constant.

3. What, specifically, is "equal" in a chemical re-
action that has attained a state of equilibrium?

4. One drop of water may or may not establish a
state of vapor pressure equilibrium when placed
in a closed bottle. Explain.

5. Why are chemical equilibria referred to as
"dynamic"?

6. What do the following experiments (done at
25°C) show about the state of equilibrium?

(a) One liter of water is added a few milliliters
at a time, to a kilogram of salt which only
partly dissolves.

(b) A large salt shaker containing one kilogram
of salt is gradually emptied into one liter of
water. The same amount of solid dissolves
as in (a).

7. The following chemical equation represents the
reaction between hydrogen and chlorine to form
hydrogen chloride:

H,(g) + C\,(g) ^± mC\(g) + 44.0 kcal

(a) List four important pieces of information
conveyed by this equation.

(b) What are three important areas of interest
concerning this reaction for which no infor-
mation is indicated?

8. How does a catalyst affect the equilibrium con-
ditions of a chemical system?

9. In any discussion of chemical equilibrium why
are concentrations always expressed in moles,
rather than in grams, per unit volume?

10. If the phase change represented by

Heat + H,Of JJ +± HjO(g)
has reached equilibrium in a closed system:

(a) What will be the effect of a reduction of vol-
ume, thus increasing the pressure?

(b) What will be the effect of an increase in tem-
perature?

11. Methanol (methyl alcohol) is made according to
the following net equation:

CO(g) + 2H«(g)

CH3OH(gJ + heat

Predict the effect on equilibrium concentrations
of an increase in: (a) Temperature, (b) Pressure

Answer, (a) Decreases CH3OH
(b) Increases CH3OH.

12. Consider the reaction:

4HCl(g) + 02(g) +±

2H2Ofgj + 2Ch(g) + 27 kcal

QUESTIONS AND PROBLEMS

161

What effect would the following changes have on
the equilibrium concentration of Cl2? Give your
reasons for each answer.

(a) Increasing the temperature of the reaction
vessel.

(b) Decreasing the total pressure.

(d) Increasing the volume of the reaction cham-
ber.

(e) Adding a catalyst.

13. Write the equation for the dissociation of Hl(g)
into its elements.

(a) Will HI dissociate to a greater or a lesser
extent as the temperature is increased?
AH = -6.2 kcal/mole H\(g).

(b) How many grams of iodine will result if at
equilibrium 0.050 mole of HI has disso-
ciated ?

14. Consider two separate closed systems, each at
equilibrium:

(a) HI and the elements from which it is formed,

(b) H2S and the elements from which it is
formed.

What would happen in each if the total pressure
were increased? Assume conditions are such that
all reactants and products are gases.

15. Each of the following systems has come to equi-
librium. What would be the effect on the equilib-
rium concentration (increase, decrease, no
change) of each substance in the system when
the listed reagent is added?

REACTION

REAGENT

(a) CtfVf J +=±H/£J + CHt(g)

H,(g)

(b) Cu-wfaqj + 4NH3f£j q=>=

Cu(NH.dt2(«q)

CuSO/sj

AgClCsJ

AgC\(s)

(d) PbSO/sJ + H+(aq) 3=fc

Pb+2(aq) + USO^(aq)

Pb(NOa),

solution

(e) CO(g) + \0,(g) q=b

CO>(g) -f heat

heat

16. Nitric oxide, NO, releases 13.5 kcal/mole when
it reacts with oxygen to give nitrogen dioxide.
Write the equation for this reaction and predict
the effect of (i) raising the temperature, and of
(ii) increasing the concentration of NO (at a
fixed temperature) on:

(a) the equilibrium concentrations;

(b) the numerical value of the equilibrium con-
stant;

17. Given:

S02(g) + \Q2(g)

SQ3(g) + 23 kcal

(a) For this reaction discuss the conditions that
favor a high equilibrium concentration of
S03.

(b) How many grams of oxygen gas are needed
to form 1.00 gram of SQ3?

Answer. 0.200 gram of 02.

18. Given:

CaCO/sJ =?=fc CaO(s) + CO,(g)

(closed system)

At a fixed temperature, what effect would adding
more CaC03 have on the concentration of C02
in the region above the solid phase? Explain.

19. Given:

Wg) + h(g)

2Hl(g) (closed system)

Answer, (a) C2H6 (increase), H2 (increase),
QH4 (decrease).

At 450°C, A: = 50.0 for the above reaction.
What is the equilibrium constant of the reverse
reaction at 450°C?

20. Write the expression indicating the equilibrium
law relations for the following reactions.

(a) Wg) + 3H2f*J *± 2NH3(g)

(b) CO(g) + N02(g) ^^ CQ2(g) + NO(g)

(d) Pbh(s) +± Pb+*(aq) + 2l~(aq)

(e) CN-(aq) + H20(/; +=±

HCNfaqJ + OH-faqJ

21. Equilibrium constants are given for several sys-
tems below. In which case does the reaction as
written occur to the greatest extent ?

162

EQUILIBRIUM IN CHEMICAL REACTIONS I CHAP. 9

REACTION

(a) CHzCOOHdiq) +±

H+(aq) + CH3COO(aq)

(b) CdS(s) +±:

Cd^(aq) + S^(aq)

HS(aq)

1.8 X 10-&
7.1 X 10-28
1 X 107

22. In the reaction
2Hl(g)

Wg) + h(g)

at 448°C the partial pressures of the gas at
equilibrium are as follows:

[HI] = 4 X 10-3atm;
[H,] = 7.5 X 10-3atm;
[I,] = 4.3 X 10-B atm.

What is the equilibrium constant for this re-
action?

23. Reactants A and B are mixed, each at a concen-
tration of 0.80 mole/liter. They react slowly, pro-
ducing C and D: A + B +± C + D. When
equilibrium is reached, the concentration of C
is measured and found to be 0.60 mole/liter.
Calculate the value of the equilibrium constant.

Answer. K = 9.0.

24. The water gas reaction

C02(g) + H2(g) *=± CO(g) + H20(g)

was carried out at 900°C with the following
results.

PARTIAL PRESSURE,
ATM AT EQUILIBRIUM

TRIAL

NO.
1

2
3

0.352
0.266
0.186

H.Q

0.352
0.266
0.686

CO,

0.648
0.234
0.314

H-

0.148
0.234
0.314

(a) Write the equilibrium constant expression.

(b) Verify that the expression in (a) is a con-
stant, using the data given.

25. Select from each of the following pairs the more
random system.

(a) A brand new deck of cards arranged accord-
ing to suit and number.

(a') The same deck of cards after shuffling.

(b) A box full of sugar cubes.

(b') Sugar cubes thrown on the floor.

(c') Stacked fire wood.

26. For each of the following reactions, state: (i)
whether tendency toward minimum energy fa-
vors reactants or products, (ii) whether tendency
toward maximum randomness favors reactants
or products.

(a) H20(l) +± H20(s)

(b) H20(l) +±: H20(g)

(d) h(s) + 1.6 kcal +±

AH = -1.4 kcal
AH = +10 kcal

^± CaO(s) + C02(g)

I2 (in alcohol)

(e) 4Fe(s) + 3Q2(g) +± 2Fe-,03(s) + 400 kcal

CHAPTER

Solubility
Equilibria

• • • solubility • • • depends fundamentally upon the ease with which

• • • two molecular species are able to mix, and if the two species dis-
play a certain hostility toward mixing, • • • saturation [is] attained at
smaller concentration • • •

JOEL H. HILDEBRAND, 1936

The principles of equilibrium have wide applica-
bility and great utility. For example, they aid us
in understanding and controlling the solubility
of solids and gases in liquids. We shall consider,
first, the solubility of a molecular solid in a liq-

uid, then the solubility of a gas in a liquid. The
usefulness of equilibrium principles is even
greater in treating the solubilities of ionic solids
in water. Much of this chapter will be devoted to
aqueous solutions of ionic solids.

10-1 SOLUBILITY: A CASE OF EQUILIBRIUM

The starting point in any quantitative equilib-
rium calculation is the Equilibrium Law. For a
generalized reaction,

aA + bB +

eE + fF +

(/)

equilibrium exists when the concentrations sat
isfy the relation

K = a constant

(2)

[A]'[B? • • •
First, we shall apply expression (2) to the solu

tion system of solid iodine dissolving in liquid
ethyl alcohol.

10-1.1 The Solubility of Iodine
in Ethyl Alcohol

As a solid dissolves in a liquid, atoms or mole-
cules leave the solid and become part of the
liquid. These atoms or molecules may carry no
charge (then they are electrically neutral) or they
may be ions. The iodine-alcohol system is of the

163

164

SOLUBILITY EQUILIBRIA I CHAP. 10

former kind. As iodine dissolves, neutral mole-
cules of iodine, I2, leave the regular crystal lattice
and these molecules become part of the liquid
phase. At equilibrium, excess solid must remain
and a fixed concentration of iodine is present in
solution. This fixed concentration is called the
solubility.
For this system at equilibrium, the reaction is

I2(solid) = I2(alcohol solution) (3)

The Equilibrium Law applied to this reaction
gives

K = a constant = [concentration I2 in alcohol]

K = [I,] (4)

crystal than in the solution. The potential energy
must rise as a molecule leaves the crystal and the
principles that govern rates of reaction are op-
erative. Presumably there is an activated complex
for the process. The rate at which molecules leave
a square centimeter of surface, passing over the
energy barrier, is determined by the height of the
barrier and by the temperature. We can call this
rate kd. Changing the temperature does not affect
the activated complex, but the molecular energy
distribution is altered (see Figure 8-3, p. 131).
Hence, kd is a function of temperature. These
two factors determine completely the rate of dis-
solving:

/rate molecules leave \

/ surface \ /rate molecules leave \
(rate of dissolving) = ( area ) X I a square centimeter j

\ of crystal surface /

- (A)
(rate of dissolving) = A X kd

10-1.2 The Dynamic Nature of Solubility
Equilibrium

The simple form of the equilibrium expression
(4) follows directly from the dynamic nature of
the solubility equilibrium. There must be a dy-
namic balance between the rate that iodine mole-
cules leave the crystal and the rate that iodine
molecules return to the crystal. To understand
this dynamic balance, we must consider the fac-
tors that determine these two rates.

RATE OF DISSOLVING

One of the factors that influences the rate of dis-
solving of solid is the area, A, of the crystal
surface that contacts the liquid. If many crystals
(with large A) are dissolving simultaneously, the
rate of dissolving is faster than if only a few
crystals (with small A) are in the solvent. The
rate of dissolving is proportional to this liquid-
solid surface area, A.
A molecule of iodine is more stable in the

(*->

(5)

RATE OF PRECIPITATION

The rate of precipitation is the rate at which
molecules return to the surface and fit into the
crystal lattice. To do this, the molecules in solu-
tion must first strike the crystal surface. Again,
the more surface, the more frequently will dis-
solved molecules encounter a piece of crystal.
The rate of precipitation is proportional to A.
In addition, the rate that molecules strike the
surface depends upon how many molecules there
are per unit volume of solution. As the concen-
tration rises, more and more molecules strike
the surface per unit time. The rate of precipita-
tion is proportional to the iodine concentration,

[hi

The last factor is, again, the rate that mole-
cules can pass over the energy barrier — the acti-
vated complex for precipitation. Again there is
a rate constant, kp, that is determined by tem-
perature and the height of the energy barrier to
precipitation.

We have, then, three factors that determine
the rate of precipitation:

(rate of precipitation) = f surfaceVconcentrati°n ofWrate dissolved molecules pass\
\area /\dissolved I2 /\over activation energy barrier/

= (A) X [I2] X (*,)

(rate of precipitation) = A X kp X [I2]

(6)

SEC. 10-1 l solubility: a case of equilibrium

165

THE DYNAMIC NATURE
OF EQUILIBRIUM

At equilibrium, we can equate (5) and (6):

(rate of dissolving) = (rate of precipitation)

A X kd = A X kp X [I2] (7)

The area of contact, A, appears both on the left
and on the right of expression (7). Hence, it
cancels out. Dividing both sides of (7) by kp, we
obtain

^ = [I2] (8)

Since kd and kp each depend upon temperature,
their ratio depends upon temperature. Other-

Fig. 10-1. Solubility equilibrium is dynamic.

THE DISSOLVING PROCESS

wise, however, each is constant. We can write
K = [h] 00

where

K~ k„

Thus, by expressing the dynamic balance be-
tween the rates of dissolving and precipitation,
we obtain (4). The concentration of I2 at equi-
librium is a constant, fixed by the temperature.
This constant equals the solubility.

10-1.3 The Factors That Fix Solubility
of a Solid

All of the discussion we have just applied to the
dissolving of iodine in ethyl alcohol applies
equally well to the dissolving of iodine in carbon

THE PRECIPITATION PROCESS

Area A

Solid

Solution

So lut ion

Solid

fd*sfol£j = *d * ra~4 (/r^itLon) = *J> f~1> (concentration)

At equilibrium, {j**** °f ) = / Rate . °f )
I dissolving/ [precipitation/

166

SOLUBILITY EQUILIBRIA | CHAP. 10

tetrachloride, CC14. Iodine at room temperature
dissolves in carbon tetrachloride at a certain rate
that, at equilibrium, exactly equals the rate of
precipitation. Again we reach the simple equi-
librium expression

K=jf= M

(4)

Despite this qualitative similarity, the solu-
bility of iodine in CCU is very different from its
solubility in alcohol. One liter of alcohol dis-
solves 0.84 mole of iodine, whereas one liter of
CCU dissolves only 0.12 mole:

^alcohol = 0.84 mole/liter (9)

Kccu = 0.12 mole/liter (10)

Why are these constants so different? To see
why, we must turn to the two factors that control
every equilibrium, tendency toward minimum
energy and tendency toward maximum random-
ness.

THE EFFECT OF RANDOMNESS

In either solvent, alcohol or carbon tetrachloride,
the dissolving process destroys the regular crys-
tal lattice of iodine and forms the disordered
solution. The dissolving process increases ran-

Fig. 10-2. A large energy difference between solid
and solution lowers the solubility.

domness. The tendency toward maximum ran-
domness tends to cause solids to dissolve.

THE EFFECT OF ENERGY

Experiment shows that heat is absorbed as iodine
dissolves. The regular, ideally packed iodine
crystal gives an iodine molecule a lower potential
energy than does the random and loosely packed
solvent environment. We see that the second
factor, tendency toward minimum energy, favors
precipitation and growth of the crystal.

Now we see the opposing factors at equilib-
rium. To increase randomness, solid tends to dis-
solve. To lower energy, solid tends to precipitate.
Equilibrium is reached when the concentration
is such that these two tendencies are equal.

How much the energy factor favors the crystal
depends upon the change in heat content as a
mole of solid dissolves. This change is called the
heat of solution. The heats of solution of iodine
in these two solvents have been measured; they
are as follows:

h(s) + 1.6 kcal
h(s) + 5.8 kcal

I2(in alcohol)
I2(in CCU)

(11)
(12)

We see that there is a much greater energy rise
when a mole of I2 dissolves in CC14 than when
a mole of I2 dissolves in alcohol. Thus the energy
factor (favoring the crystal) that opposes the
randomness factor (favoring solution) is much

CCL

Solid

I- iri alcohol

Solid Iz

A large energy effect
opposing dissolving

.A small energy effect
opposing dissolving

SEC. 10-1 I solubility: a case of equilibrium

167

larger for CC14 than for alcohol. The solubility
of iodine in CC14 is not as high as it is in alcohol.
This energy rise establishes, for this case, the
"hostility" toward mixing referred to in the quo-
tation at the beginning of the chapter. The larger
the "hostility," as measured by heat absorbed
on mixing, the lower will be the solubility.

THE EFFECT OF TEMPERATURE

Raising the temperature always tends to favor
the more random state. For these solvents, this
means that more solid will dissolve, since the
liquid solution is more random than the solid.
The solubility of iodine increases as temperature
is raised, both in alcohol and in carbon tetra-
chloride.

EXERCISE 10-1

The heat of solution of iodine in benzene is +4.2
kcal/mole (heat is absorbed). Assuming the in-
crease in randomness is the same when iodine
dissolves in liquid benzene as it is in ethyl alco-
hol and in CC14, justify the prediction that the
solubility of I2 in benzene is higher than in CC14
but lower than in alcohol.

10-1.4 Solubility of a Gas in a Liquid

Gases, too, dissolve in liquids. Let us apply our
understanding of equilibrium to this type of sys-
tem.

THE EFFECT OF RANDOMNESS

The gaseous state is more random than the liquid
state since the molecules move freely through a
much larger space as a gas. Hence randomness
decreases as a gas dissolves in a liquid. In this
case, unlike solids, the tendency toward maxi-
mum randomness favors the gas phase and op-
poses the dissolving process.

THE EFFECT OF ENERGY

In a gas the molecules are far apart and they
interact very weakly. As a gas molecule enters

the liquid, it comes close to the solvent molecules
and they attract each other, lowering the poten-
tial energy. Again we find a contrast to the
behavior of solids. When a gas dissolves in a
liquid, heat is evolved. The tendency toward
minimum energy favors the dissolving process.

Thus we see that the equilibrium solubility of
a gas again involves a balance between random-
ness and energy as it does for a solid, but the
effects are opposite. For a gas, the tendency
toward maximum randomness favors the gas
phase, opposing dissolving. The tendency toward
minimum energy favors the liquid state, hence
favors dissolving.

As an example, consider the solubilities of the
two gases, oxygen, 02, and nitrous oxide, N20,
in water. The heats of solution have been meas-
ured and are as follows:

02(g) +± 02(aq) + 3.0 kcal/mole 02 (75)

KiO(g) +±l H20(aq) + 4.8 kcal/mole N20 (14)

Assuming the randomness factor is about the
same, the gas with the larger heat effect (favoring
dissolving) should have the higher solubility. The
measured solubilities at one atmosphere pressure
and 20°C of oxygen and nitrous oxide in water
are, respectively, 02, 1.4 X 10~3 mole/liter and
N20, 27 X 10~3 mole/liter, consistent with our
prediction.

THE EFFECT OF TEMPERATURE

Raising the temperature always tends to favor
the more random state. This means that less gas
will dissolve, since the gas is more random than
the liquid. The solubility of a gas decreases as
temperature is raised.

EXERCISE 10-2

From the heat of solution of chlorine in water,
—6.0 kcal/mole (heat evolved), how do you ex-
pect the solubility of chlorine at one atmosphere
pressure and 20°C to compare with that of oxy-
gen and of nitrous oxide, N20?

168

SOLUBILITY EQUILIBRIA I CHAP. 10

SOLIDS

Tendency to
minimum energy

Solid

Opposes dissolving

Te-n.de-n.cy -to
maximunt randomness

Favors dissolving

Raising T favors solution-
solubility rises as T rises.

GASES

Gas

Solution

Favors dissolving

Fig. 10-3. Maximum randomness versus minimum
energy — solubility of solids and gases.

Solution

Opposes dissolving

Raising T favors gas :
Solubility drops as T rises.

10-2 AQUEOUS SOLUTIONS

Expression (4) is applicable to the solubility equi-
libria of some substances in water, but not of all.
Contrast, for example, water solutions of sugar,
salt, and hydrochloric acid. Sugar forms a mo-
lecular solid and, as it dissolves in water, the
sugar molecules remain intact. These molecules

leave the crystal and become a part of the liquid.
This is exactly the situation we described for
iodine in alcohol, and expression (4) is applicable
to aqueous sugar solutions. But sodium chloride,
NaCl, behaves quite differently. As salt dissolves,
positively charged sodium ions and negatively

SEC. 10-2 | AQUEOUS SOLUTIONS

169

charged chloride ions enter the solution and
these ions behave quite independently:

NaClfsj — >- Na+(aq) + Cl~( aq) (75)

Hydrochloric acid, HC1, is similar. This sub-
stance is a gas at normal conditions. At very low
temperatures it condenses to a molecular solid.
When HC1 dissolves in water, positively charged
hydrogen ions and negatively charged chloride
ions are found in the solution. As with sodium
chloride, a conducting solution containing ions
is formed:

HCl(g) — >- H+(aq) + C\-(aq)

(76)

Substances like NaClfsj and HC\(g) that dis-
solve in water to form conducting solutions are
called electrolytes. The conduction involves
movement of the ions through the solution, posi-
tive ions moving in one direction and negative
ions in the other. This shows that the positive
and negative ions behave independently. In view
of this independence of the ions, solubility be-
havior in an electrolyte solution is more com-
plicated than that given by expression (4). We
shall find that equilibrium principles are corre-
spondingly more important.

10-2.1 Types off Compounds That
Are Electrolytes

The ions in an electrolyte solution can arise in
two major ways. They may already be present
in the pure compound, as in ionic solids. When
such a solid is placed in water, the ions separate
and move throughout the solution.* However,
some compounds that form ions in water are not
considered to contain ions when pure, whether in
the solid, liquid, or gas phase. Hydrochloric acid,
HC1, and sulfuric acid, H2S04, are good exam-
ples of the second type of compound. They form
molecular liquids (or solids, if cold enough). But
in water they form ions: HC1 gives hydrogen
ion, H+(aq), and chloride ion, C\-(aq); H2S04

* Liquids that form conducting solutions are called
ionizing solvents. A few other compounds (ammonia,
NH3, sulfur dioxide, S02, sulfuric acid. H2SO,, etc.) are
"ionizing solvents" but water is by far the most impor-
tant. We will discuss water exclusively but the same ideas
apply to the other solvents in which ions form.

gives hydrogen ion, H+(aq), hydrogen sulfate or
bisulfate ion, HS04~, and sulfate ion, S04~2:

HCl(g) + water — >- H+(aq) + C\-(aq) (76)

H2S04frJ + water — *- H+(aq) + HSCV (aq) (77)

HS04" (aq) ^± H+(aq) + SOf 2faqj (7S)

However they are formed, and from whatever
source, aqueous ions are individual species with
properties not possessed by the materials from
which they came. Furthermore, the properties of
a particular kind of ion are independent of the
source. Chloride ions from sodium chloride,
Na.C\(s), have the same properties as chloride
ions in an aqueous solution of hydrochloric acid,
HC1. In a mixture of the two, all of the chloride
ions act alike; none "remembers" whether it
entered the solution from an ionic NaCl lattice
or from a gaseous HC1 molecule.

Since the properties of an ionic solution (that
is, a solution containing ions) differ in important
ways from those of nonconducting solutions, it
is important to be able to predict which sub-
stances are likely to form ionic solutions in
water. The periodic table guides us.

In Chapter 6 we saw that the chemistry of
sodium can be understood in terms of the special
stability of the inert gas electron population of
neon. An electron can be pulled away from a
sodium atom relatively easily to form a sodium
ion, Na+. Chlorine, on the other hand, readily
accepts an electron to form chloride ion, Cl~,
achieving the inert gas population of argon.
When sodium and chlorine react, the product,
sodium chloride, is an ionic solid, made up of
Na+ ions and Cl~ ions packed in a regular lat-
tice. Sodium chloride dissolves in water to give
Na+(aq) and C\~(aq) ions. Sodium chloride is
an electrolyte; it forms a conducting solution in
water.

This example illustrates the guiding principles.
Sodium is a metal — electrons can be pulled away
from sodium relatively easily to form positive
ions. Chlorine is a nonmetal — it tends to accept
electrons readily to form negative ions. When a
metallic element reacts with a nonmetallic ele-
ment, the resulting compound usually forms a
conducting solution when dissolved in water.

170

SOLUBILITY EQUILIBRIA | CHAP. 10

Positive ions -t)

tair form, s-oluble
■A ALL anions

compounds yvii

Li +

Be*2

co;2

/no;

OH'

Not

M9»

At'3

po;3

/£o+

K +

Ca'2

5c'3

Cr+3

Mrt2

Fe'2/

Co+2

Ni'2

Cut/

/ttt2

Zn+Z

Oct*

As*3/

4**

Br~

Rb+

sS2

Y+3

A$+

Cd'z

if3

St/
/t+

5b'3/
/h's

Cs+

Bet2

It3

&R.E.

H<3z7
/Hg

n +

Pb'2

Bi+3

At~

Fr+

Rt2

The metals are found toward the left side of the
periodic table and the nonmetals are at the right
side. A compound containing elements from the
opposite sides of the periodic table can be expected
to form a conducting solution when dissolved in
water. Notice from our examples that hydrogen
reacts with nonmetals to form compounds that
give conducting solutions in water. In this sense,
hydrogen acts like a metallic element.

EXERCISE 10-3

Using the periodic table as a guide, predict which
of the following compounds form ionic solutions
in water: silicon carbide, SiC; magnesium bro-
mide, MgBr2; carbon tetrabromide, CBr4; chro-
mic chloride, CrCl3.

10-2.2 A Qualitative View of Aqueous
Solubilities

Hereafter in this chapter we shall be concerned
exclusively with substances that form ionic solu-
tions in water. Since each substance is electrically
neutral before it dissolves, it must form ions of
positive charge and, as well, ions of negative
charge. Ions with positive charges are called
cations. Ions with negative charges are called
anions. A conducting solution is electrically
neutral; it contains both anions and cations.

First, let us consider substances with high solu-
bility. As was stated on p. 73, chemists consider

Fig. 10-4. Almost all compounds of the alkalies, hy-
drogen ion, and ammonium ion are soluble
in water.

a substance to be soluble if it dissolves to a
concentration exceeding one-tenth of a mole per
liter (0.1 M) at room temperature. Using this
meaning of the word soluble, we can say that
some cations (positive ions) form soluble com-
pounds with all anions (negative ions). These
cations are the hydrogen ion, H+(aq), ammo-
nium ion, NH^, and the alkali ions, Li+, Na+,
K+, Rb+, Cs+, Fr+. Figure 10-4 shows the place-
ment of these ions in the periodic table.

The same sort of remark can be made about
two anions (negative ions). Almost all com-
pounds involving nitrate ion, N03_, and acetate
ion, CH3COO-, are soluble in water.*

Other anions (negative ions) form compounds
of high solubility in water with some metal
cations (positive ions) and compounds of low
solubilities with others. Figure 10-5 indicates, for
five anions, the metal ions that form compounds
of low solubilities. Figure 10-5 A refers to chlo-
rides, bromides, and iodides, CI-, Br-, and I-;
Figure 10-5B refers to sulfates, S04-2; Figure
10-5C refers to sulfides, S-2. Notice the difference
between Figure 10-4 and Figure 10-5. The cross-

* There are a few compounds of alkalies, nitrate, and
acetate that have low solubilities, but most of them are
quite complex in composition. For example, sodium
uranyl acetate, NaU02(CH3COO)3 has low solubility.
Silver acetate is an exception but its solubility is moderate.

SEC. 10-2 I AQUEOUS SOLUTIONS

171

hatching in Figure 10-4 identifies metal ions that
form soluble compounds. Figure 10-5 identifies
the ones with low solubility.

Figure 10-6 continues this pictorial presenta-
tion of solubilities. Figure 10-6A shows the posi-
tive ions that form hydroxides of low solubility.
Figure 10-6B shows the positive ions that have
low solubility when combined with phosphate
ion, P(V3, carbonate ion, C03-2, and sulfite ion,

so3-2.

These figures furnish a handy summary of
solubility behavior. We see from Figure 10-5 A
that few chlorides have low solubilities. The few
that do contain cations of metals clustered to-
ward the right side of the periodic table (silver
ion, Ag+, cuprous ion, Cu+, mercurous ion,
Hg2+2, and lead ion, Pb+2) but they do not fall
in a single column. This irregularity is not un-

usual in solubility behavior and is seen again in
Figures 10-5B and 10-6A. In these two figures,
the elements in the second column (the alkaline
earths) show a trend in behavior. Thus, beryllium
and magnesium ions (Be+2 and Mg+2) form solu-
ble sulfates. The others — calcium, strontium,
barium, and radium ions (Ca+2, Sr+2, Ba+2, and
Ra+2) — form sulfates with low solubilities. Just
the opposite behavior is seen in Figure 10-6A for
the compounds of these same elements with hy-
droxide ion, OH-. As for the elements in the
middle of the periodic table, we see that they
form compounds of low solubilities with the
ions: sulfide, S-2, hydroxide, OH~, phosphate,
P04-3, carbonate, C03"2, and sulfite, S03-2.

The information contained in Figures 10-4,
10-5, and 10-6 is collected for reference in Table
10-1.

Table 10-1. solubility of common compounds in water

NEGATIVE IONS

(Anions)

POSITIVE IONS

(Cations)

FORM

compounds with
solubility:

All

Alkali ions

Li+, Na+, K+, Rb+, Cs+, Fr+

Soluble

All

Hydrogen ion, H+(aq)

Soluble

All

Ammonium ion, NH4+

Soluble

Nitrate, N03_

All

Soluble

Acetate, CH3COO"

All

Soluble

Chloride, CI"]
Bromide, Br- >
Iodide, I" J

Ag+, Pb+2, Hg2+2, Cu+
All others

Low Solubility
Soluble

Sulfate, S04-*

Ag\ Ca+2, Sr+2, Ba+2, Pb+2
All others

Low Solubility
Soluble

Sulfide, S-*

Alkali ions, H+(aq), NH4\ Be+2,

Mg+2, Ca+2, Sr+2, Ba«
All others

Soluble
Soluble

Low Solubility

Hydroxide, OH"

Alkali ions, H+(aq), NH«\ Sr+2, Ba""
All others

Soluble

Low Solubility

Phosphate, PCV8 i
Carbonate, COa 2
Sulfite, SOr2 J

Alkali ions, H+(aq), NH4+
All others

Soluble

Low Solubility

172

SOLUBILITY EQUILIBRIA I CHAP. 10

Li*

Be*2

cor2

vuy

/no;

OH'

with

CIZ Br-T I'\

Na*

Mo*2

Al*3

po;3

s-y

/so*

Br~
I"

Ca*2

5c*3

Cr*3

Mn*2

Fe+V

/Pe+i

Co*2

*-«k

Ztt2

OS3

G-e

As*3/

A*5

Rb*

Sr*2

Y+3

Pd 1 1 Cd*2

In*3

5n y

/ +4
/on

sb*y
/b*s

Cs*

Ba*2

La*3
&R.E.

To.

Y^9

ti*\ 9 Br3

At~

Fr*

Ra*2

Positive ions

1 -that form.

Li*

Be*2

compounds oJ~ jLUW soLtLbilit-y
with p^S

CO'/

/no;

OH'

Na*

M9*2

Al*3

po;3

s'2A

Ca*2
Sr*2
Ba*2

5c*3

Cr*3

Mn*2

Fe*2/
/Fe+i

Co*2

Nt*2

Cu/
/tJ2

Zn*2

aa*3

As*3/
/s+S

Br~

Rb*

y+3

A«+

Cd*Z

In*3

Sn*V

sb*y
/b*s

la*3
&R.E.

Hsiy

Tl*\ \Bi*3

At~

Fr*

Ra*2

H +

■thxiir form

1 Positive ions

Li*

Be*2

compotinas oj~ jluw solubility
with | Q

CO'/

vuy

/no;

OH'

F-

Na*

Mo*2

M"m

po;3

Ca*2\

Sc*3 Ti V Cr*3 Mn*2 **1*3 Co*2 Ni+2 ** \, Z*t+Z

Fe*3 Cu2

Ga+3 &e A**J+s
As5

Br~

Rb*

Sr*2]

Rh Pd Ag* Cd*2

T +3 Sn*2 5b*3
ln rS+ si*5

Cs*

Ba+2%

TI* Pb*2 Bi*3

At~

Fr*

Ra*2

Fig. 10-5. Positive ions forming compounds of low solubilities with various anions.

SEC. 10-2 | AQUEOUS SOLUTIONS

173

Lim

Na+ti

K*\

sS3

Rb +

Sr+Z

y+3

Cs*

Bct2\

La*3

Fr +

Ra*2

Positive ions

that form
compounds of LOW solubility
with

hydroxtde ton, OH

+3M*?2Fe

,*2 Ni*2 Cu"„ Zn*2 Ga*3

Mo Tc Ru Rh Pd Ao* Cd*z In

Jr Pt At

Zn~ Go" Ge

*Z t+3 Sn

£J

t form

H Positive ions |

Lt +

Be*2\

1 PO/3, CO/2, and SO/2 ■

CO/2

my
/do;

OH'

No*

m 1*3

Mo*2m

po4j

so/

Ca*2

SS>

V Cr Mn

hmUmI ^■■t

As*3

A/s

Br~

Rb +

C +2

*r+3

Nb Mo Tc

In*3

Sn*2
in*4

Sb*3
Sb*5

or i

tin ra. j\y t-a

Cs*

**z til v

w +2
Ta W Re Os Ir Pt Au "** +2

TV*

Pb*z

Bi*3

At"

Fr*

Ra+Z

Fig. 10-6. More positive ions forming compounds of
low solubilities with various anions.

EXERCISE 10-4

Use Figures 10-4, 10-5, and 10-6 to decide the
solubility of each of the compounds listed below.
Write "sol" if the compound is soluble and
"low" if it has low solubility.

Mg(N03)2, MgCl2, MgS04, Mg(OH)2, MgC03,

Ca(N03)2, CaCl* CaS04, Ca(OH)2, CaC03,

Sr(N03)2, SrCl2, SrS04, Sr(OH)2, SrC03.

EXERCISE 10-5

Decide the formula of each of the following com-
pounds and indicate those with low solubility in

water. Silver carbonate; aluminum chloride;
aluminum hydroxide; cuprous chloride (the
chloride of Cu+); cupric chloride (the chloride
of Cu+2); ammonium bromide.

10-2.3 The Equilibrium Law

Table 10-1 and the schematic presentations of
Figures 10-4, 10-5, and 10-6 are useful for a
quick and qualitative view of solubilities. But
chemists are not satisfied with the statement that
a substance has low solubility. We must know
how much of the substance dissolves. We must be
able to treat solubility in a quantitative fashion.
Fortunately this can be done with the aid of the

174

SOLUBILITY EQUILIBRIA I CHAP. 10

principles of equilibrium, developed in Chap-
ter 9.

As mentioned earlier, the quantitative concen-
tration relationship that exists at equilibrium is
shown in the Equilibrium Law Relation:

EXERCISE 10-6

K =

\EV\F]>

(2)

Expression (2) applies to a solubility equilibrium,
provided we write the chemical reaction to show
the important molecular species present. In Sec-
tion 10-1 we considered the solubility of iodine
in alcohol. Since iodine dissolves to give a solu-
tion containing molecules of iodine, the con-
centration of iodine itself fixed the solubility.
The situation is quite different for substances
that dissolve to form ions. When silver chloride
dissolves in water, no molecules of silver chlo-
ride, AgCl, seem to be present. Instead, silver
ions, Ag+, and chloride ions, Cl~, are found in
the solution. The concentrations of these species,
Ag+ and CI-, are the ones which fix the equilib-
rium solubility. The counterpart of equation (7)
will be

AgC\(s) q=b Ag+(aq) + Or(aq) (19)

and equilibrium will exist when the concentra-
tions are in agreement with the expression

K = a constant = [Ag+] X [CI-]

(20)

Just as in expression (4), the "concentration" of
the solid (silver chloride) does not appear in the
equilibrium expression (20); it does not vary.
To consider a more complicated example, con-
sider the application of expression (2) to the
solubility of lead chloride, PbCl2:

PbC\2(s)

Pb+*(aq) + 2Cl-(aq) (21)

At equilibrium,

K = a constant = [Pb+2] X [CI"]2 (22)

Solubility equilibrium constants, such as (20)
and (22), are given a special name — the solubility
product. It is symbolized Ksp. A low value of
Ksp means the concentrations of ions are low at
equilibrium. Hence the solubility must be low.
Table 10-11 lists solubility products for some
common compounds.

Write the equation for the dissolving of calcium
sulfate, CaS04, and the solubility product ex-
pression.

EXERCISE 10-7

Write the equation for the dissolving of silver
chromate, Ag2Cr04, and the solubility product
expression. Silver chromate dissolves to give Ag+
and Cr04~2 ions.

10-2.4 Calculation off the Solubility
of Cuprous Chloride in Water

The solubility product is learned from measure-
ments of the solubility. In turn, it can be used as
a basis for calculations of solubility. Suppose we
wish to know how much cuprous chloride, CuCl,
will dissolve in one liter of water. We begin by
writing the balanced equation for the reaction:

CuClfsj +±: Cu+(aq) + C\~(aq) (23)

From this equation, we can write the equilibrium
expression:

K,p = [Cu+][C1-] (24)

Now the numerical value of Ktp is found in Table
10-11:

K,p = 3.2 X 10-7 - [Cu+][C1-] (25)

Table 10-11

SOME SOLUBILITY PRODUCTS AT
ROOM TEMPERATURE

TIC1

1.9 X 10-4

SrCr04

3.6 X 10-'

CuCl

3.2 X 10-7

BaCr04

8.5 X 10-"

AgCl

1.7 X 10-'°

PbCr04

2 X 10-18

TIBr

3.6 X 10-6

CaS04

2.4 X lO"4

CuBr

5.9 X 10-9

SrSO«

7.6 X 10"7

AgBr

5.0 X 10~18

PbSO,

1.3 X 10-»

BaS04

1.5 X lO"9

Til

8.9 X 10-8

RaSO«

4 X lO"11

Cul

1.1 X io-«

Agl

8.5 X 10-17

AgBr03

5.4 X 10"8

AglOj

3.1 X lO"8

SEC. 10-2 I AQUEOUS SOLUTIONS

175

Expression (25) indicates that cuprous chloride
dissolves, according to reaction (23), until the
molar concentrations of cuprous ion and chlo-
ride ion rise enough to make their product equal
to 3.2 X 10"7.

Now is the time to dust off the algebra and put
it to work. Suppose we designate the solubility
of cuprous chloride in water by a symbol, s. This
symbol s equals the number of moles of solid
cuprous chloride that dissolve in one liter of
water. Remembering equation (23), we see that s
moles of solid cuprous chloride will produce s
moles of cuprous ion, Cu+, and s moles of chlo-
ride ion, CI-. Hence these concentrations must
be equal, as shown below.

[Cu+] = [CI-] = s moles/liter

= moles CuCl dissolved (26)

Substituting (26) into (25), we have

K,p = 3.2 X 10-7 = (s) X (s) = a»

s1 = 3.2 X 10"7 = 32 X 10-8

s = V32 X 10-8 = 5.7 X 10-< mole/liter

= 0.00057 M (27)

EXERCISE 10-8

Calculate the solubility, in moles per liter, of
calcium sulfate in water, using the solubility
product given in Table 10-11.

10-2.5 Will a Precipitate Form?

When two solutions are mixed, a precipitate may
form. For example, suppose solutions of calcium
chloride, CaCl2, and sodium sulfate, Na2S04, are
mixed. The mixture contains both calcium ions,
Ca+2, and sulfate ions, S04-2, so solid calcium
sulfate may form. The solubility product per-
mits us to predict with confidence whether it will
or not.

Let us consider two cases to show ' ow the
prediction is made:

(1) Equal volumes of 0.02 M CaCl2 and 0.0004 M
Na2S04 are mixed.

(2) Equal volumes of 0.08 M CaCl2 and 0.02 M
Na2S04 are mixed.

Fig. 10-7. Will a precipitate form?

Will a precipitate form in either case? In both?

The first step is to write the balanced equation
for the reaction of calcium sulfate dissolving in
water and then use the Equilibrium Law:

CaSO/sj +±l Ca+i(aq) + SQZ*(aq) (28)

K.p = [Ca+2][S04-2] (29)

The next step is to find the concentration of each
ion in the final mixture. After we mix equal vol-
umes, each ion is present in twice as much solu-
tion so the concentration is only half as great as
before mixing. Therefore, in case 1 ,

[Ca+2] = 0.02M = Q Q1 Msslx 10-2 M

[SCV1] =

0.0004 M

= 0.0002 M = 2 X 10-« M

A trial value of the ion product must be com-
pared to K,v\

[Ca+J] X [StV2] = (1 X 10-') X (2 X 10"<)
= 2 X 10"«

The trial product, 2 X 10-6, is less than K,v =
2.4 X 10-4 so precipitation will not occur in the
mixture of case 1. In case 2,

176

SOLUBILITY EQUILIBRIA | CHAP. 10

[Ca*»] = ™±M = 0.04 M

4 X 10~2 M

[SCV2] = °-02M = 0.01 M = \ X 10~2 M

Again we calculate a trial value of the ion
product,

[Ca+2] X [SOf2] = (4 X 10-2) X (1 X 10~2)
= 4 X 10-"

This time the trial product, 4 X 10-4, is greater
than Ksp = 2.4 X 10-5 so a precipitate does form.
Solid calcium sulfate, CaS04, will continue to
form, lowering the concentrations [Ca+2] and
[S04-2] until they are low enough that the ion
product equals Ksp. Then equilibrium exists and
no more precipitation occurs.

EXERCISE 10-9

A 50 ml volume of 0.04 M Ca(N03)2 solution is
added to 150 ml of 0.008 M (NH4)2S04 solution.
Show that a trial value of the calcium sulfate ion
product is 6 X 10~5. Will a precipitate form?

10-2.6 Precipitations Used for Separations

A chemist is often interested in separating sub-
stances in a solution mixture. Such a problem is

solved by applying equilibrium considerations.
Suppose we have a solution known to contain
both lead nitrate, Pb(N03)2, and magnesium
nitrate, Mg(N03)2. The lead and magnesium can
be separated by removing from the solution all
of the lead ion, Pb+2, as a solid lead compound.
We must avoid, of course, precipitation of any
magnesium compound. Consulting Figure 10-5
or Table 10-1, we see that lead ion and sulfate
ion form a compound with low solubility. If
enough sodium sulfate, Na2S04, is added, lead
sulfate, PbS04, will precipitate. Since Figure
10-5B and Table 10-1 indicate that magnesium
sulfate, MgS04, is soluble, there will be no pre-
cipitate of MgS04. The solid can be removed
from the liquid by filtration and the desired
separation has been obtained.

EXERCISE 10-10

Use Figures 10-5 and 10-6 or Table 10-1 to decide
which of the following soluble salts would permit
a separation of magnesium and lead through a
precipitation reaction: sodium iodide, Nal; so-
dium sulfide, Na2S; sodium carbonate, Na2C03.

QUESTIONS AND PROBLEMS

1 . Sugar is added to a cup of coffee until no more
sugar will dissolve. Does addition of another
spoonful of sugar increase the rate at which the
sugar molecules leave the crystal phase and enter
the liquid phase? Will the sweetness of the liquid
be increased by this addition? Explain.

2. In view of the discussion of the factors that de-
termine the rate of dissolving (Section 10-1.2),
propose two methods for increasing the rate at
which sugar dissolves in water.

3. When a solid evaporates directly (without melt-
ing), the process is called sublimation. Evapo-
ration of "dry ice" (solid C02) is a familiar
example. Two other substances that sublime are
FCN and ICN:

FCN(s) +± FCN(g) AH = +5.7 kcal
lCN(s) +± lCN(g) AH = +14.2 kcal

(a) In sublimation, does the tendency toward
maximum randomness favor solid or gas?

(b) In sublimation, does the tendency toward
minimum energy favor solid or gas?

(c) The vapor pressure of solid FCN is 760 mm
at 201°K. In view of part b, would you expect
solid ICN to have a lower or higher vapor
pressure than solid FCN at this same tem-
perature, 201 °K?

4. Liquid chloroform, CHC13, and liquid acetone,
CH3COCH3, dissolve in each other in all pro-
portions (they are said to be miscible).
(a) When pure CHC13 is mixed with pure ace-
tone, is randomness increased or decreased ?

QUESTIONS AND PROBLEMS

177

.(b) Does the tendency toward maximum ran-
domness favor reactants or product in the
reaction :

CHCh(l) + CH3COCH3(l)

>- 1 : 1 solution
AH = -495 cal

(d) In view of your answers to parts b and c,
discuss the experimental fact that these two
liquids are miscible.

5. Which of the following substances can be ex-
pected to dissolve in the indicated solvent to
form, primarily, ions ? Which would form mole-
cules?

(a) sucrose in water.

(b) RbBr in water.

(d) CsN03 in water.

(e) HNO3 in water.

(f) S8 in carbon disul-
fide, CS2.

(g) IC1 in ethyl alcohol.

6. Which of the substances listed in Problem 5
would be called electrolytes?

7. Assume the following compounds dissolve in
water to form separate, mobile ions in solution.
Write the formulas and names for the ions that
can be expected.

(a) HI

(b) CaCl2

(d) Ba(OH)2

(e) KNO3

(f) NH4C1

8. Write the equation for the reaction that occurs
when each of the following electrolytes is dis-
solved in water.

(a) lithium hydroxide (solid).

(b) nitric acid (liquid).

(d) sodium nitrate (solid).

(e) ammonium iodide (solid).

(f) potassium carbonate (solid).

Answer, (a) LiOHfsJ — >■ U+(aq) + OH~ (aq).

9. What would you expect to happen if equal vol-
umes of 0.1 M MgS04 and 0.1 M ZnCl2 were
mixed together?

10. Predict what would happen if equal volumes of
0.2 M Na2S03 and 0.2 M MgS04 were mixed. If
a reaction takes place, write the net ionic equa-
tion.

11. Using Figures 10-4 to 10-6 (or Table 10-1), make
a statement about the solubilities of the com-
pounds containing the following ions.

Anion

(a) carbonate, C03~ 2

(b) carbonate, C03" 2

(d) hydroxide, OH~

(e) chloride, Cl"

Cations
alkali ions, Li+, Na+,

K+, Rb+, Cs+
alkaline earth ions, Be+2,

Mg+2, Ca+2, Sr+2, Ba+2
alkaline earth ions, Be+2,

Mg+2, Ca+2, Sr+2, Ba+2
the cations of the fourth

row of the periodic

table
the cations of the fifth

row of the periodic

table

12.

Answer, (a) All alkali carbonates are soluble,
(b) All alkaline earth carbonates
have low solubilities.

Write the empirical formulas for each of the
following compounds and indicate which have
low solubilities.

(a) silver sulfide.

(b) potassium sulfide.

(d) nickel sulfide.

(e) ferrous sulfide (Fe+2).

(f) ferric sulfide (Fe+3).

13. Write net ionic equations for any reactions that
will occur upon mixing equal volumes of 0.2 M
solutions of the following pairs of compounds.

(a) silver nitrate and ammonium bromide.

(b) SrBr2 and NaN03.

(d) Nal and Pb(N03)2.

(e) barium chloride and sodium sulfate.

Answer, (a) Ag+(aq) + Br~(aq) — >- AgBrfsJ.

14. What ions could be present in a solution if
samples of it gave:

(a) A precipitate when either C\~(aq) or
S04"%qj is added?

(b) A precipitate when C\~(aq) is added but
none when S04~ (aq) is added?

178

SOLUBILITY EQUILIBRIA I CHAP. 10

15. What cations from the fourth row of the periodic
table could be present in a solution with the
following behavior.

(a) No precipitate is formed with hydroxide ion.

(b) A precipitate forms with hydroxide ion and
with sulfate ion.

(d) A precipitate forms with carbonate ion, none
with sulfide ion.

16. The solubility of silver chloride is so low that all
but a negligible amount of it is precipitated when
excess sodium chloride solution is added to silver
nitrate solution. What would be the weight of
the precipitate formed when 100 ml of 0.5 M
NaCl is added to 50.0 ml of 0.100 M AgNOs?

Answer. 0.715 gram.

17. Write the solubility product expression for each
of the following reactions.

(a) BaS04fsJ *=■

(b) Zn(OH)/S/) :

Ba+S (aq) + SOr(aq)
± Zn+2(aq) + 20H-(aq)
=± 3Ca+2f aq) + 2P04"3(aqJ

18. Write the solubility product expression applica-
ble to the solubility of each of the following
substances in water.

(a) calcium carbonate.

(b) silver sulfide.

19. The solubility product of AgCl is 1.4 X 10~4 at
100°C. Calculate the solubility of silver chloride
in boiling water.

20. Experiments show that 0.0059 gram of SrC03
will dissolve in 1.0 liter of water at 25°C. What
is K,p for SrCOa?

Answer. 1.6 X 10"9.

21. How many milligrams of silver bromide dissolve
in 20 liters of water? (Use the data given in
Table 10-11.)

22. To one liter of 0.001 M H2S04 is added 0.002
mole of solid Pb(N03)2. As the lead nitrate dis-
solves, will lead sulfate precipitate?

23. Suppose 10 ml of 1.0 M AgN03 is diluted to one
liter with tap water. If the chloride concentration
in the tap water is about 10~6 M, will a precipi-
tate form?

24. The test described in Problem 23 does not give
a precipitate if the laboratory distilled water is
used. What is the maximum chloride concentra-
tion that could be present ?

25. Will a precipitate exist at equilibrium if \ liter
of a 2 X 10~3 M A1C13 solution and \ liter of a
4 X 10-2 M solution of sodium hydroxide are
mixed and diluted to 103 liters with water at
room temperature? (Ksp = 5 X 10~33.)

26. Use Figures 10-5 and 10-6 or Table 10-1 to decide
which of the following soluble substances would
permit a separation of aqueous magnesium and
barium ions. For those that are effective, write
the equation for the reaction that occurs.

(a) ammonium carbonate.

(b) sodium bromide.

(d) sodium hydroxide.

27. To a solution containing 0.1 M of each of the
ions Ag+, Cu+, Fe+2, and Ca+2 is added 2 M
NaBr solution, giving precipitate A. After filtra-
tion, a sulfide solution is added to the solution
and a black precipitate forms, precipitate B.
This precipitate is removed by filtration and 2 M
sodium carbonate solution is added, giving pre-
cipitate C. What is the composition of each
precipitate, A, B, and C?

CHAPTER

Aqueous Acids
and Bases

The acids, • • • are compounded of two substances • • • the one
constitutes acidity, and is common to all acids, • • • the other is pecul-
iar to each acid, and distinguishes it from the rest • • •

A. LAVOISIER, 1789

In Chapter 10 we used the principles of equilib-
rium to help us understand solubility in liquids.
In such a system constituents in solution reach
the dynamic balance of equilibrium with another
phase, a solid or a gas. Equilibrium can also
exist among two or more constituents present in
the same solution. One of the examples already
encountered (in Chapter 9 and in Experiment
15) is

Ft**(aq) + SCN-faqj +±: FeSCN^(aq) (1)

for which experiment showed

„ _ [FeSCN^I
[Fe+3][SCN-]

(2)

In reaction (/), all of the molecular species in-
volved in the equilibrium are in the solution as
dissolved species. Though the equilibrium rela-
tionship that exists among the concentrations
is a little more complicated than in the solu-
bility product expressions, the guiding princi-
ples are the same.

In this chapter we shall explore some more
equilibria like (/), in which all of the important
species are dissolved. We will consider mainly
those equilibria in which one of the ions is
H+(aq). This type of equilibrium furnishes one
of the most important classes of chemical reac-
tions of all those that occur in water.

11-1 ELECTROLYTES— STRONG OR WEAK

In Section 10-2 we considered electrolytes, sub-
stances that dissolve in water to give solutions
containing ions. Thus far, we have considered
electrolytes such as UC\(g) and NaOHfsJ:

HClf*; + water — >■ H+(aq) + C\-(aq) (5)

NaOHfsJ + water — ■+- Na+(aq) + OR-(aq) (4)

According to reaction (3), when HC1 gas dis-
solves in water, all of the HC1 molecules break
up, or dissociate, into ions, H+(aq) and C\~(aq).
There is no experimental evidence for the pres-

179

180

AQUEOUS ACIDS AND BASES I CHAP. 1 1

ence of molecules of HC1 in aqueous hydro-
chloric acid solutions. In a similar way, there is
no evidence for the presence of molecules of
NaOH in aqueous solution — apparently the so-
dium hydroxide crystal breaks up completely
into sodium ions, Na+(agJ and hydroxide ions,
OH-(aq). A substance that dissolves and exclu-
sively gives ions is called a strong electrolyte.
Not all substances that form conducting solu-
tions break up, or dissociate, so completely. For
example, vinegar is just an aqueous solution of
acetic acid. Such a solution conducts electric
current, showing that ions are present:

CH3COOH +±l H+(aq) + CHaCOO-faqj (5)

The conductivity of a 0.1 M acetic acid solu-
tion is much lower, however, than that of a
0. 1 M hydrogen chloride solution. This and other
experiments show that only a small fraction of
the dissolved acetic acid, CH3COOH, has formed
ions. Such a substance that dissolves and dis-
sociates to ions only to a limited extent is called a
weak electrolyte.

Fig. 11-1. A strong electrolyte solution conducts bet-
ter than a weak electrolyte solution.

Careful measurements show that water is,
itself, a weak electrolyte. We shall consider it
first.

11-1.1 Water: A Weak Electrolyte

Pure water does not conduct electric current
readily. Yet an extremely sensitive meter shows
that even the purest water has a tiny conduc-
tivity. To conduct electric current, water must
dissociate to a very small extent, forming ions.
The ions prove to be hydrogen ion, H+(aq), and
hydroxide ion, OH~( aq):

HiOd) +± H+(aq) + OH-(aq) (5)

This is an equilibrium involving three species in
the liquid phase, H20, H+(aq), and OH~(aq).
The equilibrium law can be written

[H+]fOH-]
[H20]

(7)

However, as we have remarked in Section 9-2.2,
the concentration of H20 in water is so large,
55.5 M, and so few ions are formed that its con-
centration is virtually constant. Consequently,
expression (7) is usually simplified by incorporat-

HCl : a strong electrolyte

CH3COOH ■' a weak elecrtrolylre

SEC. 11-1 | ELECTROLYTES — STRONG OR WEAK

181

ing the factor 55.5 in the constant. We shall do
this, and label the constant Kw to indicate that
it includes the factor [H2OJ:

Km = [H+][OH-] (8)

Kw = [H,0] X K = 55.5 X K (9)

The magnitude of Kw is given in Table 9-IV
at 25°C,

Kw= 1.00 X 10-" (70)

Having this value of Kw, we can calculate the ion
concentrations, using the same methods we ap-
plied to solubility.

In reaction (6) there is one H+(aq) ion formed
for every OH~(aq) ion. Hence, in pure water
where the only source of ions is reaction (6), the
concentrations of H+(aq) and OH~(aq) must
be equal. That is,

in pure water, [OH"] - [H+] (77)

Substituting (77) into the equilibrium expression,
we can calculate the concentrations of the two
types of ions:

K„ = [H+] X [OH-] = [H+] X [H+] = [H+]2

or,

[H+] = Vkw = Vim x io-14

[H+] = 1.00 X 10~7M

Because of (77), we can also write

[OH-] = [H+] = 1.00 X 10"7M

(72)

(7i)

Now we can explain the low conductivity of
pure water. Though water dissociates into ions,
H+(aq) and OH~(aq), it does so only to a very
slight extent. At equilibrium, the ion concentia-
tions are only 10-7 M. Water is a weak elec-
trolyte.

11-1.2 The Change off K, with Temperature

Experiments show that reaction (6) absorbs energy:

HiO(l) + 13.68 kcal — )- H+(aq) + OH(aq) (14)

Even though only a minute fraction of the water present
actually is dissociated at equilibrium, if we measure the
energy effect and divide by the number of moles, we find
that it takes 13.68 kcal per mole of water broken into
ions.

This heat effect can be used in predicting how Kv
changes with temperature. Le Chatelier's Principle indi-

cates that increased temperature will shift the equilibrium
condition toward larger concentrations of the ions (so as
to absorb heat). Hence, Ku is expected to increase. This
agrees with the experimental values for Kw, given for
various temperatures in Table 11-1.

Table II -I

VALUES OF K AT VARIOUS
TEMPERATURES

TEMPERATURE (°C)

0.114 X 10-"

0.295 X 10-'<

0.676 X 10-"

1.00 x io-»

9.55 X 10-"

11-1.3 The Special Roles off H aq and
OH (aqj in Water

The equilibrium reaction (6) gives the two ions
W+(aq) and OH~(aq) special roles in aqueous
solutions:

W(l)

H+(aq) + OH-(aq)

(6)

In pure water, where the only source of ions
is reaction (6), the concentrations of H+(aq) and
OH~(aq) must be equal. But what if we add
some HC1 to the solution? We have already
noted that HC1 is a strong electrolyte, dissolving
to give the ions W+(aq) and C\~(aq). Thus, hy-
drogen chloride adds H+(aq) but not OH~(aq)
to the solution. The concentrations [H+] and
[OH-] are no longer equal. However, they are
still found to be "tied together" by the equilib-
rium relationship

or,

Kw = [H+][OH-]

[OH-] =

[H+]

(5)
(75)

Expression (75) shows that as the concentration
of H+ rises (for example, as we add HC1 to
water), the concentration of OH" must de-
crease.

Suppose, on the other hand, that we add so-
dium hydroxide, NaOH, to pure water. Sodium
hydroxide is also a strong electrolyte, adding

182

AQUEOUS ACIDS AND BASES | CHAP. 11

OH- ions to the solution. Now we can rewrite
(8) in the form

[H+] =

[OH-]

(16)

and we see that raising the concentration [OH-]
lowers the concentration [H+].

In this way the two ions H+(aq) and OU~(aq)
are connected in the chemistry of water. When
[H+] is high, [OH-] must be low. When [OH"]
is high, [H+] must be low.

Let's consider an example. Suppose we dis-
solve 0.10 mole of hydrogen chloride in 1.0 liter
of water. Since HC1 is a strong electrolyte, 0.10
mole of HC1 forms 0.10 mole of H+(aq) ions
and 0.10 mole of C\~(aq) ions in the one liter
volume. The concentrations of H+(aq) and
C\~(aq) due to the HC1 are equal:*

[H+] = [CI"] - 0.10 M (77)

If the concentration [H+] is 0.10 M, expres-
sion (75) allows us to calculate the concentration
[OH-]:

roH-i = -^ = im x 10~" = 10Q x 10~14
L J [H+] o.io " l.o x 10"1

[OH"] = 1.0 X 10-" (18)

We see that adding the 0.10 mole of HC1 to 1.0
liter of water lowered the OH" concentration
from 10-7 M (its value in pure water) to 10-13 M,
a change by a factor of a million!

EXERCISE 11-1

Show that the addition of 0.010 mole of solid
NaOH to 1 .0 liter of water reduces the concen-
tration of H+(aq) to 1.0 X 10"12 M.

EXERCISE 11-2

Suppose that 3.65 grams of HC1 are dissolved in
10.0 liters of water. What is the value of [H+] ?
Use expression (75) to show that [OH-] =
1.00 X l0~nM.

Thus we see that the concentrations of the two

* There is a small additional concentration of H+(aq),
owing to the dissociation of water, but it will be quite
negligible compared with the 0.10 M concentration pro-
vided by the HC1.

ions, H+(aq) and OH~(aq), always remain re-
lated through the equilibrium relation (8). In
pure water they are equal, [H+] = [OH-]. In a
solution of HC1, in which [H+] is high, [OH"]
must be low. In a solution of NaOH, in which
[OH-] is high, [H+] must be low. Furthermore,
as shown in the calculation (18) and Exercise
11-1, with rather low concentrations of HC1
and NaOH, the concentration of H+(aq) [or
OH-(aq)] can be varied from 0.1 M to 10"13 M,
by the immense factor of 1012 (a million million).

This ease with which we can control and vary
the concentrations of H+(aq) and OH~(aq)
would be only a curiosity but for one fact. The
ions H+(aq) and OH~(aq) take part in many im-
portant reactions that occur in aqueous solution.
Thus, if H+(aq) is a reactant or a product in a
reaction, the variation of the concentration of
hydrogen ion by a factor of 1012 can have an
enormous effect. At equilibrium such a change
causes reaction to occur, altering the concentra-
tions of all of the other reactants and products
until the equilibrium law relation again equals
the equilibrium constant. Furthermore, there are
many reactions for which either the hydrogen
ion or the hydroxide ion is a catalyst. An ex-
ample was discussed in Chapter 8, the catalysis
of the decomposition of formic acid by sulfuric
acid. Formic acid is reasonably stable until the
hydrogen ion concentration is raised, then the
rate of the decomposition reaction becomes very
rapid.

In these ways, by affecting equilibrium condi-
tions and by changing reaction rates, the con-
centrations of H+(aq) and OH~(aq) give us
immense leverage in controlling the chemistry
of aqueous solutions.

EXERCISE 11-3

The color of a solution of potassium chromate,
K2Cr04, changes to the color of a solution of
potassium dichromate, K2Cr207, when a few
drops of HC1 solution are added. Write the
balanced equation for the reaction between
CrO;2(aq) and H+(aq) to produce Cr207"2 and
explain the color change on the basis of Le
Chatelier's Principle.

SEC. 11-2 I EXPERIMENTAL INTRODUCTION TO ACIDS AND BASES

183

11-2 EXPERIMENTAL INTRODUCTION TO ACIDS AND BASES

The significant influence that H+(aq) and
OH~(aq) exert on aqueous solution chemistry
was recognized long ago. Consequently chemists
have long identified by the class names "acids"
and "bases" those substances that change the
concentrations of these two ions. We shall see,
later in this chapter, that the meanings of the
terms "acid" and "base" are evolving and have
become more general in modern usage. This is a
natural and desirable development as chemists
attempt to relate the chemistry of aqueous solu-
tions to the chemistry that occurs in other
solvents and in other phases. For the moment,
however, we shall keep our attention focused on
aqueous solutions and consider how a chemist
recognizes acids and bases.

11-2.1 Properties of Aqueous Solutions
off Acids

Consider the following compounds:

HC1 : hydrochloric acid (or, hydrogen

chloride),
HN03 : nitric acid,

CH3COOH : acetic acid,
H2SO4 : sulfuric acid,

H3PO4 : phosphoric acid.

Each of these five compounds is called an
acid. They all share this name because they have
the following important properties in common.

Hydrogen Containing. Each of these com-
pounds contains hydrogen.

Fig. 11-2. Some familiar acids.

Electrical Conductivity. Each of these com-
pounds dissolves in water to form solutions that
conduct electricity. Ions are present in these
aqueous solutions.

Liberation of Hydrogen Gas. The aqueous
solutions of each compound produce hydrogen
gas, H2, if zinc metal is added.

Color of Litmus. Litmus, a dye, is red in color
when placed in these aqueous solutions.

Taste. The dilute, aqueous solutions of each
compound are sour tasting.*

Because the aqueous solutions of these com-
pounds have these properties in common, the
compounds are conveniently classed together
and identified as acids. In fact, these properties
constitute the simplest definition of an acid. They
provide a basis for deciding whether some other
compound should be classified as an acid.

What is the common factor that makes these
different substances behave in the same ways?
In water they all form conducting solutions; we
conclude that they all form ions in water. Each
substance contains hydrogen and each reacts
with zinc metal to produce hydrogen gas. Per-
haps all of these aqueous solutions contain the
same ion and this ion accounts for the formation
of H2(g). It is reasonable to propose that the
common ion is H+(aq). We postulate: a sub-
stance has the properties of an acid if it can re-
lease hydrogen ions.

11-2.2 Properties off Aqueous Solutions

of Bases

Consider the following compounds:

NaOH

: sodium hydroxide,

KOH

: potassium hydroxide,

Mg(OH)2

: magnesium hydroxide,

Na2C03

: sodium carbonate,

NH3

: ammonia.

Citric

Acetic

Tartaric

* Many chemicals, including some acids and some
bases, are poisons. Therefore chemists rarely use taste as
a voluntary means of deciding if an unidentified solution
is acidic or basic.

184

AQUEOUS ACIDS AND BASES | CHAP. 11

Each of these five compounds is called a base.
They all share this name because they have the
following important properties in common.

Electrical Conductivity. Like acids, these com-
pounds dissolve in water to form conducting
solutions. Ions are present in an aqueous solu-
tion of a base.

Reaction with Acids. When one of these com-
pounds is added to an acid solution, it destroys
the identifying properties of the acid solution —
all but electrical conductivity.

Color of Litmus. The dye, litmus, is blue in
color when placed in an aqueous solution of any
of these compounds.

Taste. The dilute aqueous solutions taste bit-
ter.*

Feel. The aqueous solutions feel "slippery."

Again, these properties constitute the simplest
definition of a base. These properties provide a
basis for deciding whether some other compound
should be classified as a base.

11-2.3 An Explanation of the Properties
of Bases

Using the same argument we used for acids, we
can seek a common factor that accounts for the
similarities of bases. Because of the electrical
conductivity, we might seek an ion. Because of
the ability to counteract the properties of acids,
we ought to seek an ion which can remove the
hydrogen ion, H+(aq), since hydrogen ion ac-
counts for the properties of acids.

Sodium hydroxide, NaOH, when dissolved in
water, gives a solution with the properties of a
base. The hydroxides of many elements — those
from the left side of the periodic table — behave
in the same way. Perhaps they dissolve to form
ions of the sort

NaOHfsj q=t Na+faq; + OH~(aq) (19)

KOH(s) +±: K+(aq) + OH-(aq) (20)

Mg(OH)2fsJ =?=*: Mg»(aq) + 20Br(aq) (21)

Ca(OH)2fsj *=± C***(aq) + 20H-(aq) (22)

The hydroxide ion, OH-(aq), could react with

* If you have forgotten the danger of tasting chemicals,
reread the preceding footnote.

hydrogen ion to account for the second property
of bases, the removal of acid properties:

OH'(aq) + H+(aq) +±: H20 (23)

The similarities among the hydroxides are ob-
vious. Let's compare sodium carbonate and am-
monia. Sodium carbonate, Na2C03, dissolves in
water to give a solution with the properties that
identify a base. Quantitative studies of the solu-
bilities of carbonates show that carbonate ion,
C03"2, can react with water. The reactions are

Na2COj(sJ z^=± 2Na+( aq) + CO;2(aq) (24)

CO^(aq) + H20 3=fc

HC03- (aq) + OH-(aq) (25)

Reaction (25) indicates that the presence of car-
bonate ion in water increases the hydroxide ion,
OH~, concentration. This is a constituent that
is present in the solutions of NaOH, KOH,
MgCOH),, and Ca(OH)2.

Reaction (25) also provides a basis for under-
standing the removal of acid properties. If C03~2
readily forms bicarbonate ion,t HCOf, then re-
action (26) is likely:

C03-2f aq) + H+(aq) z<=±: HCO^(aq) (26)

We see that the existence of the stable bicar-
bonate ion, HCO3" (aq) produces the chemical
species, OH~(aq) in common with solutions of
the hydroxides. We can postulate that OH~(aq)
accounts for the slippery "feel" and bitter taste
of the basic solutions. The stability of bicarbo-
nate ion also explains the removal of acid prop-
erties through reaction (26).

The fifth substance listed as a base is ammonia,
NH3. Ammonia readily forms ammonium ion,
NH^. Ammonia can react with water,

NHsf aq) + H20 +± NHt(aq) + OH~( aq) (27)
and with hydrogen ion,

NH,(aqJ + H+(aq)

NHt(aq) (28)

Once again, the formation of NH^ accounts for
ammonia having the properties of a base. Reac-
tion (27) produces hydroxide ion, which, by our
postulate, accounts for the taste and "feel" prop-
erties of solutions of bases. Reaction (28) shows

tBicarbonate ion, HCOf, is also called "hydrogen
carbonate" or "monohydrogen carbonate."

SBC. 11-2 I EXPERIMENTAL INTRODUCTION TO ACIDS AND BASES

185

how ammonia can act to destroy the acid proper-
ties of a solution containing hydrogen ions,
H+(aq).

Investigation of the reactions of other com-
pounds that have the properties of a base shows
that each compound can produce hydroxide ions
in water. The OH~(aq) ions may be produced
directly (as when solid NaOH dissolves in water)
or through reaction with water (as when Na2C03
and NH3 dissolve in water):

NaOHfsJ
COfYagj + H20

NlUfaq) + H20

zt Na+(aq) + OH-(aq) (19)

~HCOa(aq) + OH-(aq) (25)

±NHtJaq) + OH-(aq) (27)

Furthermore, any substance that can produce hy-
droxide ions in water also can combine with
hydrogen ions:

OH-(aq) + H+(aq) z<=h H20 (23)

CO^(aq) + H+(aq)
mitfaq) + H+(aq)

HCO; (aq)

NOt(aq)

(26)
(28)

Since production of OH-(aq) and reaction with
H+(aq) go hand in hand when we are dealing
with aqueous solutions, a base can be described
either as a substance that produces OH~(aq) or
as a substance that can react with H+(aq). In
solvents other than water, the latter description
is more generally useful. Therefore, we postulate:
a substance has the properties of a base if it can
combine with hydrogen ions.

U-2.4 Acids and Bases: Summary

Let us repeat our two definitions and explana-
tions.

Fig. 11-3. Some familiar bases.

DEFINITIONS

An acid is a hydrogen-containing substance that
has the following properties when dissolved in
water:

it is an electrical conductor;
it reacts with Zn to give H2(gJ;
it makes litmus red;
it tastes sour.

A base is a substance that has the following
properties when dissolved in water:

it is an electrical conductor;

it reacts with an acid, removing the acidic

properties;
it makes litmus blue;
it tastes bitter;
it feels slippery.

EXPLANATIONS

A substance is an acid if it can release hydrogen
ions, H+(aq).

A substance is a base if it can react with hy-
drogen ions, H+(aq).

11-2.5 The Nature of H+(aq)

Both of our explanations of the properties of acids and
bases involve the hydrogen ion, H+(aq). This species has
great importance in the chemistry of aqueous solutions,
so we shall consider what is known about it.

Before considering what a chemist means by the sym-
bols H+(aq), we must discuss more generally the inter-
action of ions with water. Lithium chloride provides a
good example. Lithium chloride dissolves in water spon-
taneously at 25°C, forming a conducting solution. At
equilibrium, it has a high solubility:

LiC\(s)

Li+(aq) + a-(aq) t = 25°C (29)

Lithium chloride melts spontaneously above 613°C and
forms a liquid that conducts electricity:

LiClfsj

Li+(7J + Cl-(l) t = 613°C (30)

Sodium carbona-te

Ammonia

Third

Equilibrium in any reaction is determined by a compro-
mise between tendency toward minimum energy ("golf
balls roll downhill") and tendency toward maximum
randomness. Reaction (29) and reaction (30) both involve
increase in randomness since the regular solid lattice dis-
solves or melts to become part of a disordered liquid
state. Both reactions produce ions. But reaction (29)
proceeds readily at 25°C, whereas reaction (30) does not

186

AQUEOUS ACIDS AND BASES | CHAP. 1 1

Li + Caq)
Li+-4If20

Fig. 11-4. Possible tetrahedral * arrangements of

water molecules around Li* and H* ions.

occur until the solid is quite hot, 613°C. The difference
must be in the special stabilities of Li+ and Cl~ ions in
water. The high melting point of lithium chloride shows
that the crystal is very stable. The high solubility of
lithium chloride in water can be explained only by saying
that Li+ (aq) and Or(aq) must also be very stable. This
means that water must interact strongly with these ions.
A similar situation exists for hydrochloric acid, HC1.
This gaseous compound dissolves readily in water at
25°C:

ions as lA+(aq), H+(aq), and C\~(aq). The notation (aq)
reminds us that the ions interact strongly with the solvent.
The symbol is purposely vague because, in most cases,
chemists are not certain of the arrangement of the water
molecules around a given ion. For Li+(aq) and H+(aq),
there may be simple packing of four water molecules
around each ion, as shown in Figure 11-4. This figure
suggests similarity between Li+(aq) and H+(aq).

However, most chemists feel there is quite a difference
between these two ions. After all, H+ is unique — the
proton has no electrons. Many chemists suggest that the
proton attaches itself strongly to one molecule of water,
forming a new molecular species, HjO+ (aq). Figure 11-5

HC\(g) *=fc H+(aq) + C\-(aq) t = 25°C (57) ^ /;_5 A mode, of hydronium ion> h3Q .

The HC1 molecule is a stable one— it must be heated to a
few thousand degrees before the atoms will separate.
Even then, neutral atoms are obtained and still higher
temperatures are needed before gaseous ions are ob-
tained.

HCl(g)

H(g) + Cl(g) t very high (32)

The high temperature required to separate the two atoms
of a molecule of HC1 shows that HC1 is very stable.
Again, we can explain the solubility of HC1 in water by
saying H+(aq) and C\~(aq) must also be very stable.
Water must interact strongly with these ions.
This is why we have been symbolizing these aqueous

* The word tetrahedral means four-sided. If the oxygen
atoms of the four water molecules are connected by lines,
the lines form a four-sided figure.

SEC. 11-2 I EXPERIMENTAL INTRODUCTION TO ACIDS AND BASES

187

shows a model of this proposed ion — called hydronium
ion — surrounded by solvent. Notice that the three hy-
drogen atoms are pictured as equivalent. There is a
pleasing similarity to the formation of the well-established
ammonium ion, NH<+, with its four equivalent hydrogen
atoms:

H,0 + H+ +± HjOVaqj (33)

NH3(aqj + H+ +=± NH<+(aq) (34)

Still other chemists feel there are probably several ar-
rangements of water molecules around the proton. In
addition to HjO+, there may be molecules such as HsO,j+,
H7CV, H9CV, etc.

Unfortunately, the experimental data do not provide
a definite answer to the nature of H+(aq). The hydronium
ion, shown in Figure 11-5, does exist in certain crystal
structures.* Spectroscopic studies f indicate that several
species are present in water. Thermal and electrical con-

* Hydronium ion, HsO+, is a structural unit in solid
perchloric acid hydrate, HC104 • H2Of as shown by nuclear
magnetic resonance studies.

t Spectroscopy refers to the study of the absorption
of light — in this case, by aqueous solutions of acids such
as HC1.

ductivities of aqueous acid solutions have been inter-
preted to indicate the presence of a molecular unit, H90«+.

Figure 11-4 shows the hydrogen ion surrounded by four
oxygen atoms, each part of a water molecule. We could
write the formula for this arrangement as H+-4H20 or
H9(X+. But Figure 11-6 shows how an H30+ ion can serve
as the basis for another structure with the formula
HjO+-3HjO, again giving H9CV. Still, there seems to be
no compelling experimental basis for preference.

So we are faced with at least three plausible structures
for the species H+(aq), as shown in Figures 11-4, 11-5,
and 11-6. Under these circumstances, convenience in dis-
cussing the experimental properties of H+(aq) governs
our choice.

So far in this chapter, we have referred to the aqueous
hydrogen ion as H+(aq). Later in the chapter we will
consider a more general theory of acids and bases. Then
it will be more convenient to designate the aqueous hy-
drogen ion as H30+(aq). The reason will be clear— such
usage aids us in seeing regularity in the behavior of a

Fig. 11-6. A possible model for H»(V based upon

HaO*.

188

AQUEOUS ACIDS AND BASES I CHAP. 1 1

larger class of acids and bases. That is sufficient basis for
the use of any theory.

11-2.6 Acid-Base Titrations

We have noted that the two concentrations [H+]
and [OH"] are "tied together." If 0.100 mole of
HC1 is dissolved in 0.100 liter of water to raise
[H+] to LOOM, then [OH"] goes down to
1.00 X 10-14 M and the product [H+] X [OH"]
remains equal to Kw = 1.00 X 10-14. Perhaps
you are wondering what happens to the hydrox-
ide ions to reduce their concentration from
1.00 X 10-7 M (their concentration in pure wa-
ter) to the new value, 1.00 X 10~14 M. The an-
swer is that they are consumed through reaction
with the added H+(aq):

OR-(aq) do not satisfy the equilibrium expres-
sion. Their product far exceeds 1.00 X 10~14:

OH(aq) + H+(aq)

H20

(35)

This is in accord with Le Chatelier's Principle.
Addition of HC1 to water raises [H+]. By Le
Chatelier's Principle, processes take place that
tend to counteract partially the imposed change.
Reaction with OH~(aq) does tend to counteract
the raised concentration of H+(aq).

In the case we have considered, the actual
amount by wh'ch [H+] is reduced is extremely
small. To reduce [OH"] from 1.00 X 10~7 M to
1.00 X 10~14M (by a factor of 107), reaction
(35) must consume about 10-7 mole of hydrox-
ide ion for every liter of solution. Since one mole
of OH~(aq) reacts with one mole of H+(aq), the
amount of H+(aq) required is also 10-7 mole
for every liter of solution. Subtracting 10-7
mole/liter from a concentration near 1 mole/liter
causes such a small change in [H+] that it need
not be considered in calculations (such as in
Exercises 11-1 and 11-2).

HC1 AND NaOH IN THE SAME

solution: excess HC1

Suppose that to the 0.100 liter of LOOM HC1
solution we add 0.090 mole of solid sodium hy-
droxide. Now we have added both H+(aq) and
OH~(aq) in high concentration to the same solu-
tion. What will happen?

Immediately after the sodium hydroxide dis-
solves, the concentrations of H+(aq) and

initial [H+] = LOOM

0.090 mole

initial [OH"] =

0.100 liter

= 0.90 M

initial product, [H+] X [OH"] = 9.0 X 10"1
(far exceeding 1.00 X 10~14)

Again Le Chatelier's Principle tells us quali-
tatively what will occur and the equilibrium
expression tells us quantitatively. If we add
OH-(aq), a change will take place that tends to
counteract partially the resulting increase in
[OH-]. This occurs through the reaction be-
tween OH-(aq) and H+(aqJ, consuming both
ions and reducing the value of [H+] X [OH-].
Reaction continues until this product reaches the
equilibrium value, Kw = 1.00 X 10-14.

Since Kw is so small, the reaction consumes
almost all of one of the constituents (H+ or
OH~) if the other is present in excess. In our
example, [H+] initially exceeds [OH-] by 0.10
mole/liter:

initial [H+] - initial [OH"] = excess [H+]
LOOM - 0.90 M = 0.10 M

In Section 11-1.3 we calculated that if the
concentration [H+] = 0.100 M, then at equilib-
rium, [OH"] = LOO X 10"13 M. Thus the rather
small excess of hydrogen ion, 0.10 M, is sufficient
to guarantee that the reaction between H+(aq)
and OH~(aq) consumes most of the 0.90 M
OH", reducing [OH"] to 1.0 X 10"13 M.

EXERCISE 11-4

Suppose that 0.099 mole of solid NaOH is
added to 0.100 liter of 1.00 M HC1.

(a) How many more moles of HC1 are present in
the solution than moles of NaOH?

(b) From the excess number of moles and the
volume, calculate the concentration of excess
H+(aqj.

(c) Calculate the excess concentration of H+(aq)
from the difference between the initial con-
centrations of HC1 and NaOH.

SEC. 11-2 I EXPERIMENTAL INTRODUCTION TO ACIDS AND BASBS

189

(d) Calculate the concentration of OH~(aq) at
equilibrium (see your calculations for Exer-
cise 11-2).

HC1 AND NaOH IN THE SAME

solution: excess NaOH

Returning to our original 0.100 liter of 1.00 M
HC1, let us now consider the addition of 0.101
mole of solid NaOH. Again we have added both
H+(aq) and OH~(aq) to the same solution, and
the concentrations immediately after mixing do
not satisfy the equilibrium expression:

initial [H+] - LOOM

•»• i rAu-i 0.101 mole , A1 „,,
initial OH = ...... — = 1.01 M

L J 0.100 liter

initial product, [H+] X [OH"] = 1.01
(far exceeding 1.00 X 10~14)

This solution contains excess hydroxide ion.
Since OH~(aq) is in excess, almost all of the
H+(aq) will be consumed, forming water:

initial [OH"] - initial [H+] = excess [OH"]
1.01 M - LOOM = 0.01 M

In Exercise 11-1 we calculated the equilibrium
concentration [H+] in a solution containing
[OH-] = 0.01 M. The result is [H+] = 1.0 X
10"12 M.

HC1 and NaOH in the samb
solution: no excess of either

In each example used in this section, a number
of moles of NaOH was added to 0.100 liter of
1 .00 M HC1 with one or the other constituent in
excess. Reaction between H+(aq) and OH~(aq)
consumes almost all of the constituent not in
excess. Let us now consider the case in which
there is excess of neither HC1 nor NaOH.

Suppose we add 0. 100 mole of NaOH to 0. 100
liter of LOOM HC1. The initial values of [H+]
and [OH-] are equal and their product far ex-
ceeds 1.00 X 10-14

initial [H+] = LOO M

initial [OH-] = ai,00niole = LOOM

0.100 liter

initial product, [H+] X [OH~]
(far exceeding 1.00 X 10-14)

1.00

Reaction between H+(aq) and OH~(aq) must
occur, forming water:

OH-(aq) + H+(aq) 3=fc H20

Since one mole of OH~(aq) consumes one mole
of H+(aq), the concentrations [H+] and [OH-]
remain equal as reaction (35) proceeds. When
equilibrium is reached, they will still be equal.
This is exactly the situation in pure water. As we
saw in Section 11-1.2,

[H+] = [OH-] - VY„ = LOO X 10"7 M

A solution containing exactly equivalent
amounts of acid and base is neither acidic nor
basic. Such a solution is called a neutral solu-
tion.*

progressive addition of NaOH
to HC1: a titration

Now we have considered the progressive addi-
tion of more and more NaOH to a fixed amount
of HC1 solution. The results are compiled in
Table 11-11.

Table 11-11 shows that the concentration of
hydrogen ions changes drastically as the amount
of NaOH nears the equivalent amount of HC1.
There is a change of [H+] by a factor of 1010 as
the initial concentration of OH~(aq) is changed
from 0.99 to 1.01. Thus the concentration
changes by a huge factor near the point at which
the amounts of acid and base are equivalent.
Because of this, the progressive addition of a base
to an acid, a titration, furnishes a sensitive means
of comparing the concentrations of an acid and
a base solution.

An acid-base titration is carried out by adding
carefully measured amounts of a base solution
to a known volume of the acid solution. The acid
solution contains some substance that provides
visual evidence of the magnitude of [H+], The
dye litmus is such a substance. As mentioned in
Sections 11-2.1 and 11-2.2, litmus is red in solu-
tions containing excess [H+]. Litmus is blue in

* This use of the word "neutral" for a solution with
equal amounts of H+ and OH~ has its disadvantages
because the same word is used in reference to electrica'
neutrality. Aqueous solutions are always electrically neu-
tral whether there is an excess of either H+ or OH" or
an excess of neither.

190

AQUEOUS ACIDS AND BASES I CHAP. 11

Table 11 -II. the concentrations [h ] and coh -] in solutions containing

BOTH HCI AND NaOH

INITIAL
CONC.

H+

INITIAL
CONC.

OH-

EXCESS
CONC.

H+ or OH-

CALC.

[H+]

CALC.
[OH"!

1.00 M

none

1.00MH+

1.00 M

1.0 X 10-» M

1.00

0.90 M

0.10MH+

1.0 X io-1

1.0 X 10"u

1.00

0.99

0.01 MH+

1.0 x io-j

1.0 x io-»

1.00

none

1.0 X io-7

1.00

1.01

0.01 MOH-

1.0 x io-»

1.00 x io-»

solutions in which [H+] is less than [OH-]. A
small amount of litmus in the solution will tint
the solution red until the point in the titration
at which the number of moles of OH~(aq) be-
comes equal to the number of moles of H+(aq).
Then, the slightest addition of more OH~(aq)
causes a drastic reduction in [H+] and the solu-
tion color turns to blue. A dye whose color is
sensitive to the change of [H+] (such as litmus)
is called an acid-base indicator.

11-2.7 pH

For compact expression of H+(aq) concentrations, chem-
ists use a quantity, pH, defined by the equation

pH = -log.o [H+]

Since [H+] = 10~7 M in a neutral solution at 25°C, it
follows that for such a solution

pH = -log.o [IO"7] = -(-7) = +7

This result helps us to understand the symbol pH, first
defined by a Danish chemist, Sorenson. He used the p to
stand for the Danish word poienz (power) and H to stand
for hydrogen. After a change of sign, pH is the power of
ten needed to express the hydrogen ion concentration in
moles per liter. In acidic solutions, p\\ is less than 7
(pH < 7); and in basic solutions, pH is greater than 7
(jjH > 7). Table 1 1 -III expresses the results of Table
11-11 in terms of pH.

Table 11 -III

CONCENTRATIONS OF H aq AND
OH (aq) EXPRESSED IN TERMS OF pH

ACIDITY OR

BASICITY

[H+]

acidic

1.00

acidic

io-»

acidic

io-1

neutral

io-7

basic

io-»s

11-3 STRENGTHS OF ACIDS

Earlier in this chapter, strong and weak electro-
lytes were distinguished in terms of the degree
to which the dissolved material forms ions. As
a particular case, such distinctions can be made
in terms of acids, furnishing a quantitative basis
for defining the strength of an acid.

11-3.1 Weak Acids

We have contrasted the electrical conductivities
of 0.1 M aqueous solutions of hydrochloric acid

and acetic acid (see Figure 11-1). In water, hy-
drochloric acid dissociates completely to ions;
HCI is a strong electrolyte. Because one of the
ions released is H+(aq), HCI is also called a
strong acid. Acetic acid, on the other hand, dis-
sociates to ions only to a slight extent; acetic
acid is a weak electrolyte. Because one of the ions
released is H+(aq), acetic acid is called a weak
acid.

These qualitative ideas can be expressed more
usefully in terms of the principles of equilibrium.

SEC. 11-3 I STRENGTHS OF ACIDS

191

For example, let us contrast the behavior of two
weak acids, acetic acid and hydrofluoric acid:

CH3COOHfaqJ +±

H+(aq) + CH3COO(aq) (36)

HF(aq) 3=fc H+(aq) + ¥~(aq) (37)

Measurements of the electrical conductivities of
0.10 M solutions of these two acids show that
there are more ions present in the HF solution
than in the acetic acid solution. We can conclude
that acetic acid is a weaker acid than HF. This
information is conveyed quantitatively in terms
of the equilibrium constants for reactions (36)
and (37):

y [H+][CH3COO-] , 8 v in_6

*CH>C00H = [CHaCOOH] = L8 X 10

KuF - [H+][F1

KHF - [HF]

6.7 X 10-<

(38)

(39)

Since A"Hf is a larger number than KCHiC00n,
hydrofluoric acid dissociates in water to a larger
extent than does acetic acid. Though HF is a
weak acid (only partially dissociated), it is a
stronger acid than is acetic acid.

We can express these ideas in terms of a
general acid, HB. The acidic nature of HZ? is con-
nected to its ability to release hydrogen ions,

HB(aq) +±: H+(aq) + B(aq) (40)

The equilibrium constant for reaction (40) meas-
ures quantitatively the ease with which Hi? re-
leases H+(aq) ions,

Ka = [H+]ffl-]
A [HB]

(41)

Tables of KA furnish a quantitative measure of
acid strengths with which we can compare dif-
ferent acids and predict their properties. Several
values of KA are given in Table 1 1-IV.

We see in Table 1 1-IV that the equilibrium
view of acid strengths suggests that we regard
water itself as a weak acid. It can release hydro-
gen ions and the extent to which it does so is
indicated in its equilibrium constant, just as for
the other acids. We shall see that this type of
comparison, stimulated by our equilibrium con-
siderations, leads us to a valuable generalization
of the acid-base concept.

EXERCISE 11-5

Which of the following acids is the strongest
acid and which the weakest?

nitrous acid, HN02; Kym0i = 5.1 X 10-4
sulfurous acid, H2S03; Kns0l = 1.7 X 10~2
phosphoric acid, H3P04; /vH,po, = 7.1 X 10~3

Table 11-IV. relative strengths of acids

AT ROOM TEMPERATURE

~ [H±MJ
AA [HB]

ACID

STRENGTH

N AQUEOUS SOLUTION

REACTION

HC1

HNO3

H2S04

HSOr

CH3COOH

H2CO3 (C02 + H20)

H2S

NH<+

HCO,-

H20

very strong

. \. t .. .

strong

We a

weak

very weak

HC\(g)

HNCVsJ

H2S04

HSOc(aq)

HF (aq)

CH3COOH(aqJ

H2COj(aq)

H£(aq)

KHS(aq)

HC03~(aq)

H2Q(aq)

H+(aq)
H+(aq)
H+(aq)

H+(aq)
H+(aq)
H+(aq)
H+(aq)
H+(aq)
H+(aq)
H+(aq)
H+(aq)

+ C\~(aq)
+ N03~(aq)
+ HSO4- (aq)

+ SOrYaqJ

+ r-(aq)

+ CH3COO(aq)
+ HCOr(aq)
+ HS-(aq)
+ NH3(aq)
+ COr'(aq)
+ OH-(aq)

very large
very large
large

1.3 X 10-*

6.7 X 10-<

1.8 X 10-'

4.4 X 10-7

1.0 x 10-7

5.7 X 10-10

4.7 X 10-"

1.8 X 10-'« •

A',,

1.00 X 10"u

*The equilibrium constant, KA, for water equals 7777b; = — — 777 See Section 11-1.1.

[H2UJ jj.j

192

AQUEOUS ACIDS AND BASES I CHAP. 11

11-3.2 Equilibrium Calculations of Acidity

The acidity of a solution has pronounced effects
on many chemical reactions. It is therefore im-
portant to be able to learn and control the hy-
drogen ion concentration. This control is ob-
tained through application of the Equilibrium
Law. Common types of calculation, based on
this law, are those needed to determine KA from
experimental data and those using KA to find
[H+]. We will illustrate both of these types, using
benzoic acid, C6H6COOH, as an example.

DETERMINATION OF KA

To apply the Equilibrium Law to acid solutions,
a chemist must know the numerical value of the
equilibrium constant, KA. Experiments which
provide this information require the measure-
ment of hydrogen ion concentration. Acid-sensi-
tive dyes, such as litmus, offer the easiest esti-
mate of [H+].

A typical example is as follows. Benzoic acid,
C6H5COOH, is a solid substance with only mod-
erate solubility in water. The aqueous solutions
conduct electric current and have the other prop-
erties of an acid listed in Section 1 1-2.1. We can
describe this behavior with reaction (42) leading
to the equilibrium relation (43):

C6H5COOH(aqJ

H+(aq) + C6UbCOO-(aq)
(42)

= [H+][C6H5COO-]
[QH5COOH]

(43)

The following experiment was performed to
determine the equilibrium constant in (43). A
1.22 gram sample of benzoic acid was dissolved
in 1.00 liter of water at 25°C. With dyes whose
color is sensitive to acidity (indicators) the con-
centration of H+(aq) was estimated to be
8 X 10-" M.

To make use of these data, we must first ex-
press all quantities in terms of moles. The mo-
lecular weight of benzoic acid, C6H5COOH, is
122.1 grams/mole. Hence,

1.22 g

1.22 g QH5COOH =

122.1 g mole
= 0.0100 mole C6H5COOH

zoic acid if we assume very little of it has reacted
to form H+(aq) according to reaction (42):

moles 0.0100 mole

[C6H5COOH]
[QH6COOH]

volume 1.00 liter
0.0100 M = 1.00 X 10~2M

and, by measurement,

[H+] = 8 X 10-"M

(Notice that [H+] is less than

(44)

(45)

is less man 10% of
[C6H5COOH], verifying our assumption that
very little of the acid reacted.) Now we know two
of the concentrations in expression (43) and, to
complete the calculation, we must know the con-
centration of benzoate ion, [C6H5COO_]. Since
the benzoic acid was dissolved in pure water, the
only source of C6H5COO~ is reaction (42). This
is also the source of hydrogen ion, H+(aq). Since
these two ions are both produced only by reac-
tion (42), their concentrations must be equal in
this solution. That is,

[H+] = [QH5COO-]

Thus, if [H+] = 8 X 10~4, then, also

[QH5COO-] = 8 X 10-" (46)

Now we can complete the calculation by sub-
stituting (44), (45), and (46) into (43):

= [H+][C6H5COO] = f8 X 10~41 X f8 X 10~4]
A [C6H5COOH] [1.0 X 10-2]

= 64 X 10'8

1.0 x 10-2

KA = 64 X 10-« = 6.4 X 10-B

CALCULATION OF [H+]

Having established experimentally the numerical value of
KA, we can use it in calculations of equilibrium concen-
trations.

As an example, suppose a chemist needs to know the
hydrogen ion concentration in a solution containing both
0.010 M benzoic acid, C6H6COOH, and 0.030 M sodium
benzoate, C6H6COONa. Of course, he could go to the
laboratory and proceed to investigate the colors of in-
dicator dyes placed in the solution. However, it is easier
to calculate the value of [H+], using the accurate value
of Ka listed in Appendix 2.

Sodium benzoate is a strong electrolyte; its aqueous
solutions contain sodium ions, Nd+(aq), and benzoate
ions, QHsCOO-faqJ. Hence the equilibrium involved is
the same as before:

Now we can calculate the concentration of ben- c6VUCOOH(aq) -<-*" H+faqJ + C6HbCOQ-(aq) (42)

SEC. 11-3 I STRENGTHS OF ACIDS

193

At equilibrium, the concentrations must be in accord with
the equilibrium expression. That is,

KA =

rH+irc6Hscoo-]

[C6HsCOOH]

= 6.6 X 10-*

(43)

First, let us assume that very little [H+] is formed through
dissociation of benzoic acid. This assumption implies that
the concentrations of benzoate ion and benzoic acid are
very little affected by reaction (42). Assuming this, we see
that two of the concentrations in (39) are already speci-
fied:

[C6H6COOH] = 0.010 M
[QH6COO-] - 0.030 M

_ rH+irC^COO-] = rH+1(0.030)
A [C6H5COOH] (0.010)

Rearranging (47), we obtain
[H+] = 2.2 X 10-*

(47)

0.030

The calculation cannot be considered complete until
we check the assumption. Was it reasonable to assume
that the concentrations of benzoate ion and benzoic acid
were not changed by reaction (42)1 To decide, we com-
pare the magnitude of [H+], 2.2 X 10~5 M, to the ben-
zoate ion and benzoic acid concentrations. We find that
[QH6COOH] = 0.010 M is about 500 times larger than
the concentration change necessary to form 2.2 X 10-5 M
H+. The same argument applies to [QH6COO-]. The
assumption is valid.

11-3.3 Competition for H Among Weak Acids

We have explained the properties of acids in
terms of their abilities to release hydrogen ions,
H+(aq). Thus acetic acid is a weak acid because
of the slight extent to which reaction (48) re-
leases U+(aq):

CH3COOH(aq)

H+(aq) + CH3COO-(aq) (48)

We have explained the properties of bases in
terms of their abilities to react with hydrogen ion.
Thus ammonia is a base because it can react as
in (49):

NH3(aq) + Y\+(aq) Zf±: NH + (aq) (49)

Now consider the result of mixing aqueous solu-
tions of acetic acid and ammonia. The reaction
that occurs can be compared to a sequence of
reactions,

CH3COOH(aq)
NH3(aq) + H+(aq)

H+(aq) + CH3COO-(aq)

NH4+ (aq)

Net reaction

CH3COOH(aq) + NH3f aq) +±

CH3COO ~(aq) + WHt(aq) (50)

Practically, the result of reactions (48) and (49)
is reaction (50). In reaction (50), we see that
acetic acid acts as an acid in the same sense that
it does in (48). In either case, it releases hydrogen
ions. In (48) acetic acid releases hydrogen ions
and forms H+(aq) and in (50) it releases hydro-
gen ions to NH3 and forms NH^ . In the same
way, ammonia acts as a base in (50) by reacting
with the hydrogen ion released by acetic acid.
So reaction (50) is an acid-base reaction, though
the net reaction does not show H+(aq) explicitly.
Now by taking one more step we can view
acid-base reaction in a broader sense. Suppose
we mix aqueous solutions of ammonium chlo-
ride, NH4CI, and sodium acetate, CH3COONa.
A sniff indicates ammonia has been formed. Re-
action occurs,

NH4+ (aq) + CH3COO~( aq) =e=t

CHaCOOHfaqj + NH3(aq)

(51)

Reaction (57) is just the reverse of reaction
(50). Inspection of this reaction reveals that re-
action (51), too, is an acid-base reaction! Once
again there is an acid that releases H+ — it is NH^
— and a base that accepts H+ — the base is
CH3COO-. Once again the net effect of the re-
action is transfer of a hydrogen ion from one
species to another. We see that the acid-base
reaction between acetic acid and ammonia gave
two products, one an acid, NH^, and one a
base, CH3COO-. A little thought will convince
you that every acid-base reaction does so. The
transfer of a hydrogen ion from an acid to a
base necessarily implies that it might be handed
back. The reaction of handing it back, the re-
verse reaction, is just as much a hydrogen ion
transfer, hence an acid-base reaction, as is the
original transfer.

Notice that we are now referring to reactions
in which a hydrogen ion is transferred from an
acid to a base without specifically involving the
aqueous species H+(aq). A hydrogen ion, H+, is

194

AQUEOUS ACIDS AND BASES I CHAP. 11

nothing more than a proton. Consequently we
can frame a more general view of acid-base
reactions in terms of proton transfer. The main
value of this view is that it is applicable to a
wider range of chemical systems, including non-
aqueous systems.

We generalize our view of the acid-base type
of reaction as follows. In our example, reaction
(50),

CH3COOH + NH3

an acid a base

NH4+ + CH3COO- (50)

an acid a base

The acetic acid reacts as an acid, giving up its
proton, to form acetate, CH3COO~, a substance
that can act as a base. We can write (50) in a
general form:

HB! + B2
Acidi -f Base2

HB2 + Bi
Acid2 + Basex

(52)
(53)

We see that an acid and a base react, through
proton transfer, to form another acid and another
base*

We can use this more general view to discuss
the strengths of acids. In our generalized acid-
base reaction (52), the proton transfer implies
the chemical bond in HBi must be broken and
the chemical bond in HB2 must be formed. If the
HB! bond is easily broken, then HBi will be a
strong acid. Then equilibrium will tend to favor
a proton transfer from HBi to some other base,
B2. If, on the other hand, the HB! bond is ex-
tremely stable, then this substance will be a weak
acid. Equilibrium will tend to favor a proton
transfer from some other acid, HB2, to base Bi,
forming the stable HBi bond.

11-3.4 Hydronium Ion in the Proton Transfer
Theory off Acids

In the proton transfer view of acid-base reac-
tions, an acid and a base react to form another
acid and another base. Let us see how this theory
encompasses the elementary reaction between
H+(aq) and OH~(aq) and the reaction of disso-

* This more general view of acids and bases is named
the Bronsted-Lowry theory after the two scientists who
proposed it, J. N. Brpnsted and T. M. Lowry.

ciation of acetic acid, reactions (54) and (55):

H+(aq) + OH(aq) q=*z H20 (54)

CH3COOHfaqj z^r

H+(aq) + CH3COO-f aqj (55)

It does so by making a specific assumption
about the nature of the species H+(aq). It is
considered to have the molecular formula
H30+(aq). Thus, when HC1 dissolves in water,
the reaction is written

HCl(g) + H20
instead of
HCl(g)

H3Q+(aq) + Cl-(aq) (56)

H+(aq) +C\-(aq) (57)

Whenever H+(aq) might appear in an equation
for a reaction, it is replaced by the hydronium
ion, H30+, and a molecule of water is added to
the other side of the equation. We write (55) in
the form

CH3COOH(aqj + H20 :p±

H3Q+(aq) + CH3COO-(aq)

(58)

Now the dissociation of acetic acid can be re-
garded as an acid-base reaction. The acid
CH3COOH transfers a proton to the base H20
forming the acid H30+ and the base CH3COO_.
The reaction (54) now takes the form

H3Q+(aq) + OH-(aq)

H20 + H20 (59)

In (59) the acid H30+ transfers a proton to the
base OH-, forming an acid, H20, and a base,
H20. We see that within the proton transfer
theory, the molecule H20 must be assigned the
properties of an acid and, as well, those of a base.
This designation of the species H+(aq) in
terms of hydronium ion, H30+, is not necessi-
tated by experimental evidence that proves the
unique existence of this molecule, H30+, in dilute
aqueous solutions (see Section 1 1-2.5). Neverthe-
less, the convenience of this assumption, as an
aid in correlating acid-base behavior, amply
justifies its use.

11-3.5 Contrast off Acid-Base Definitions

In this chapter we first identified acids and bases
in aqueous solution by investigating the proper-

SEC. 11-3 I STRENGTHS OF ACIDS

195

ties possessed by acid solutions and base solu-
tions. By so doing, we defined an acid in terms
of the properties of solutions of acids. Now we
are explaining the behavior of an acid in terms
of the process of proton transfer. Because this
explanation fits a large number of experimental
facts well and conveniently, it has come into
common use. If a chemist is asked what sub-
stances are acids, he is liable to refer to the ex-
planation rather than to the identifying proper-
ties. At this point, he has shifted to a new
definition. Let us compare these two definitions.
Definition 1 . An acid is a substance that has
the properties listed below when dissolved in
water:

it is an electrical conductor;
it reacts with Zn to give H2(g);
it makes litmus red;
it tastes sour.

Definition 2. An acid is a substance that can
release protons.

The first definition is of the type called an
"operational definition." To understand this
term, consider the intended meaning of the word
"definition." According to the dictionary, "defi-
nition" means "a statement of what a thing is."
With a definition to help, it is possible to sort
the universe into two piles — one pile containing
those objects that fit the definition, and another
pile containing those that do not. The "state-
ment" gives criteria by which this sorting can be
carried out. An "operational definition" is, then,
a definition that lists, as criteria, measurements
or observations (that is, "operations") by which
you could decide whether a given object is "in"
or "out."

The second definition is a "conceptual defini-
tion." It defines the class in terms of an expla-
nation of why the class has its properties. To see
the difference — and the relative merit — of the
two kinds of definition, let us consider an anal-
ogy.

Contrast the following pair of definitions.

(1) A "star" is an athlete who regularly scores

an unusually large number of points or who,
in clever defensive acts, repetitiously and ad-
vantageously contributes to the welfare of
his team.
(2) A "star" is an athlete with unusual muscular
coordination.

The first definition is "operational." It ex-
plicitly states criteria for deciding whether a
player is a "star." He is one who "regularly
scores an unusually large number of points" or
who "in clever defensive acts, repetitiously
and. • • -"To decide if a player is a "star," you
count his scoring or, alternatively, number the
outstanding defensive plays he makes.

The second definition might be called a "con-
ceptual definition." It offers an explanation of
why the athlete enjoys special success in his
sport. It is one step removed from things that
show on the scoreboard.

The first definition gives no clue why one
player is a "star" and another player is not. Its
value is that it is practical, useful, and surely
correct. It does not matter how awkward a
basketball player is; if he scores 30 points per
game he is assured of a considerable amount of
public acclaim.

The value of the second (conceptual) definition
is that it contains more information about "star-
dom." If accurate, it has the deeper significance.
It might help the basketball coach more in de-
veloping the optimum characteristics of his
squad. It permits him to predict athletic skill in
advance of the first game.

Returning to our two definitions of an acid,
the first, the operational definition, gives clearcut
instructions on how to decide whether a given
substance is an acid. Dissolve it in water and see
if it has certain properties. The second (concep-
tual) definition, however, has the deeper sig-
nificance since it includes our knowledge of why
an acid has these particular properties. It pro-
vides a basis for finding hidden likenesses be-
tween acid-base reactions in water and other
reactions in other solvents. Each type of defini-
tion has its merit ; neither is the definition.

196

AQUEOUS ACIDS AND BASES I CHAP. 11

QUESTIONS AND PROBLEMS

1 . What is the concentration of H+(aq) in an aque-
ous solution in which [OH~] = 1.0 X 10~3 Ml

2. 100 ml of the HC1 solution described in Exercise
11-2 (p. 182) is diluted with water to 1.00 liter.
What is the concentration of H+(aq)l What is
[OH-] in this solution?

3. Vinegar, lemon juice, and curdled milk, all taste
sour. What other properties would you expect
them to have in common?

4. Give the name and formula of three hydrogen
containing compounds that are not classified as
acids. State for each compound one or more
properties common to acids that it does not
possess.

5. As a solution of barium hydroxide is mixed with
a solution of sulfuric acid, a white precipitate
forms and the electrical conductivity decreases
markedly. Write equations for the reactions that
occur and account for the conductivity change.

6. An eyedropper is calibrated by counting the
number of drops required to deliver 1.0 ml.
Twenty drops are required.

(a) What is the volume of one drop?

(b) Suppose one such drop of 0.20 M HC1 is
added to 100 ml of water. What is [H+] ?

Answer, (c) X 1000.

7. Suppose drops (from the same eyedropper) of
0.10 M NaOH are added, one at a time, to the
100 ml of HC1 in Problem 6b.

(a) What will be [H+] after one drop is added?

(b) What will be [H+] after two drops are
added?

8. Calculate [H+] and [OH-] in a solution made
by mixing 50.0 ml 0.200 M HC1 and 49.0 ml
0.200 M NaOH.

Answer. [OH"] = 5 X 10"12 M.

9. Calculate [H+] and [OH-] in a solution made by
mixing 50.0 ml 0.200 M HC1 and 49.9 ml
0.200 M NaOH.

10. How much more 0.200 M NaOH solution need
be added to the solution in Problem 9 to change
[H+] to 10~7 Ml

11. An acid is a substance HB that can form H+(aq)
in the equilibrium :

HB(aq) *=±: H+(aq) + B~{aq)

(a) Does equilibrium favor reactants or products
for a strong acid?

(b) Does equilibrium favor reactants or prod-
ucts for a very weak acid?

m [H+J[Z?r] K = [H+][B2~]

12. (a) Which of the following acids is the strongest

and which is the weakest ?

ammonium ion, NH/ (in an NH4C1

solution);
bisulfate ion, HS04" (in a KHS04

solution);
hydrogen sulfide, H2S.
(b) If 0.1 M solutions are made of NH»C1,
KHSO4, and H£, in which will [H+] be
highest and in which will it be lowest?

13. (a) Nitric acid is a very strong acid. What is

[H+] in a 0.050 M HNO3 solution?
(b) Hydrogen peroxide, H202, is a very weak
acid. What is [H202] in a 0.050 M H202
solution?

14. From a study of Appendix 2, what generaliza-
tion can you make concerning acids which con-
tain more than one atom of hydrogen in their
molecules or ions?

15. A 0.25 M solution of benzoic acid (symbolize it
HB) is found to have a hydrogen ion concentra-
tion [H+] = 4 X 10-» M.

(a) Assuming the simple reaction HB(aq) +±
H+(aq) + B~(aq), calculate KA for benzoic
acid.

(b) Compare the values of [HB], [H+], [B~],
and KA used in this problem to the corre-
sponding quantities in the benzoic acid cal-
culation presented in the text, Section 1 1-3.2.

QUESTIONS AND PROBLEMS

197

16. If 23 grams of formic acid, HCOOH, are dis-
solved in 10.0 liters of water at 20°C, the [Hi
is found to be 3.0 X 10 3 M. Calculate KA.

17. A chemist dissolved 25 grams of CH3COOH in
enough water to make one liter of solution. What
is the concentration of this acetic acid solution?
What is the concentration of Hi aq>? Assume
a negligible change in [CH3COOH] because of
dissociation to H(aq).

18. When sodium acetate, CH3COONa, is added to
an aqueous solution of hydrogen fluoride, HF,
a reaction occurs in which the weak acid HF
loses H+.

(a) Write the equation for the reaction.

(b) What weak acid is competing with HF for
H+?

Answer, (b) CH3COOH, acetic acid.

19. (a) Write the equation for the reaction that

shows the acid-base reaction between hy-
drogen sulfide, H2S, and carbonate ion,
C03 2.

(b) What are the two acids competing for H+?

(c) From the values of KA for these two acids
(see Table 11 -IV), predict whether the equi-
librium favors reactants or products.

Answer, (c) Products.

20. Write the equations for the reaction between
each of the following acid-base pairs. For each
reaction, predict whether reactants or products
are favored (using the values of KA given in
Appendix 2).

(a) HNO/aqj + NH3f«f,J +±:

(b) NH4+(ac/J + F(aq) zp±

Answer, (a) HNChfaqJ + NH3faqJ

q=± NOt-(aq) + NHt+(aq).
Products, N02 ~(aq) and
NH4+((iqJ favored.

21. Write the equations for the reactions between
each of the following acid base pairs. For each
reaction, predict whether reactants or products
are favored.

(a) H:SO,(aq) + HC03(aq) +=±

(b) HCOztaq) + S03"laq) +±
(C) HS03(aq) + SO3 2(aq) Zf±:

22. If the p\\ of a solution is 5, what is [H+] ? Is the
solution acidic or basic?

23. What is [H+] in a solution of p\\ = 8? Is the
solution acidic or basic? What is [OH-] in the
same solution?

24. Devise an operational and also a conceptual
definition of a gas.

SVANTE AUGUST ARRHENIUS 1859-192)

During a full life this great Swedish chemist met practically
all the important men of science of his day and won their
affection as well as their highest regard. He is said to have
had a genius for friendship. Nevertheless, his early career
was filled with a battle for acceptance.

At 22 Arrhenius had performed many experiments con-
cerned with the passage of electricity through aqueous
solutions and he decided to continue this work in prepara-
tion for his doctorate. For two years he collected voluminous
data on hundreds of solutions and concentrations while
working in the laboratory of the University of Upsala. He
then formulated a carefully considered hypothesis that
aqueous solutions contain charged species, ions. This was a
revolutionary suggestion and his professors found it so
different from their own ideas that they only grudgingly
granted his degree.

Undiscouraged, Arrhenius sent copies of his thesis to
other scientists. Although few took his radical idea seri-
ously, the great German scientist, Ostwald, became so
excited that he traveled to Sweden to meet Arrhenius.
Encouraged by this support, Arrhenius traveled and studied
in Germany and Holland. Finally, in 1889, his paper "On
the Dissociation of Substances in Aqueous Solutions" was
published.

He was invited to come to Leipzig as a professor at the

University but chose to return to Sweden as a lecturer and
teacher at a high school in Stockholm. His theory was still
not generally accepted, and those who did not agree with it
dubbed its proponents "the wild horde of the Ionians."
Even Arrhenius' assignment as professor at Stockholm in
1893 was questioned until a storm of protest came from
scientists in Germany. Within two years of this appoint-
ment he was elected President of the University and was
named a Nobel laureate, the award being only the third
such in chemistry. Arrhenius finally received the acclaim he
had so long deserved. He was offered the coveted position
of professor of chemistry at Berlin but the King of Sweden
founded the Nobel Institute for Physical Chemistry and, in
1905, Arrhenius became its director. He continued as a
tireless experimenter and an extremely versatile scientist
until his death in 1927.

Arrhenius' success in science must be credited not only
to his brilliance as a scientist but also to his conviction in
his views. His understanding of the electrical properties of
aqueous solutions was so far ahead of contemporary thought
that it would have been ignored but for his confidence in
the usefulness of his theory and his refusal to abandon it.
It is fitting tribute that the ionic model of aqueous solutions
has changed permanently the face of inorganic chemistry.

CHAPTER

Oxidation- Reduction
Reactions

Chemical thermodynamics enables one to state what may happen when
two substances react.

WENDELL M. LATIMER, 1953

We have now made use of the principles of equi-
librium in two general types of reactions. In
Chapter 10 we considered reactions involving a
solid and a solution: dissolving and precipita-
tion. In Chapter 1 1 we turned to reactions occur-

ring entirely in solution and involving proton
transfer. Now we shall take a more general view
of equilibrium in aqueous solutions, a view pro-
vided by an investigation of the chemistry of an
electrochemical cell.

12-1 ELECTROCHEMICAL CELLS

Electrochemical cells are familiar — a flashlight
operates on current drawn from electrochemical
cells called dry cells, and automobiles are started
with the aid of a battery, a set of electrochemical
cells in tandem. The last time you changed the
dry cells in a flashlight because the old ones were
"dead," did you wonder what had happened
inside those cells? Why does electric current flow
from a new dry cell but not from one that has
been used many hours? We shall see that this is
an important question in chemistry. By studying
the chemical reactions that occur in an electro-
chemical cell we discover a basis for predicting
whether equilibrium in a chemical reaction fa-

vors reactants or products. The reactions are of
the type called oxidation-reduction reactions,
which is the subject of this chapter.

12-1.1 The Chemistry of an Electrochemical
Cell

Let's begin our investigation of an electrochemi-
cal cell by assembling one. Fill a beaker with a
dilute solution of silver nitrate (about 0.1 M will
do) and another beaker with dilute copper sul-
fate. Put a silver rod in the AgN03 solution and
a copper rod in the CuS04 solution. With a wire,
connect the silver rod to one terminal of an

199

200

OXIDATION-REDUCTION REACTIONS I CHAP. 12

im: Cu.so4

. I JvC CuS04

Fig. 12-1. An electrochemical cell.

SEC. 12-1 I ELECTROCHEMICAL CELLS

201

ammeter to measure the electric current. Con-
nect the other terminal of the ammeter through
a wire resistance, R, to the copper rod. Finally,
connect the two solutions to complete the elec-
tric circuit. Figure 12-1 shows suitable equip-
ment. These two drawings show how a connec-
tion can be made between the two solutions to
complete the electric circuit. A glass tube con-
taining a sodium nitrate solution furnishes an
electrical path. It is called a salt bridge.

As soon as the last connection is made, things
start to happen. The ammeter needle deflects —
electric current is moving through the meter and
the wire resistance, R. The direction of current
flow is that of electrons moving from the copper
rod to the silver rod. The resistance becomes
warm — the cell is doing work as it forces elec-
trons through R. In the beakers, the copper rod
dissolves and the silver rod grows. As time goes
by, the ammeter shows less and less current flow
until, finally, there is none.

Now let's be more quantitative. Let's repeat
the experiment, weighing the metal rods before
and after the test. The weighing shows that dur-
ing the test the copper rod has become 0.635
gram lighter and the silver rod has become 2.16
grams heavier. Chemical reaction has occurred
and, as any good chemist will do, we imme-
diately ask, "How many moles of copper and
silver are involved?"

Moles Cu dissolved =

wt Cu dissolved

atomic wt Cu

= 0.635 g
63.5 g/mole

= 0.0100 mole

Moles Ag deposited = wt Ag deposited
atomic wt Ag

2.16 g
108 g/mole

= 0.0200 mole

We see that there is a simple relationship between
the weight of copper dissolved and the weight of
silver deposited. One mole of copper dissolves in
the right beaker for every two moles of silver
deposited in the left beaker. Copper ions,
Cu+2(aq), are formed in the right beaker from

the neutral copper metal atoms. This means
atoms of copper release electrons into the copper
rod. These electrons move into the wire, through
the resistance, and through the ammeter. They
arrive at the silver rod in the left beaker, where
silver metal is formed from silver ions, Ag+(aq).
Here, the positive silver ions draw electrons from
the silver rod to become neutral silver metal
atoms. Summarizing these processes, we have:

In the right beaker, Cu(s) — >-

Oi^(aq) + 2e~ (7)

In the left beaker, 2Ag+(aqj + 2e~ — >- 2Ag(s) (2)

Overall reaction, Cu(s) + 2Ag+(aq) — ►■

2Ag(s) + Oi-"(aq) (5)

The overall reaction describes what goes on in
the entire electrochemical cell. In half of the cell,
the right beaker, reaction (7) occurs. In the other
half of the cell, the left beaker, reaction (2)
occurs. Hence, reactions (7) and (2) are called
half-cell reactions or half-reactions.

There are several interesting features about
these half-reactions:

(1) The two half-reactions are written separately.
In our electrochemical cell the half-reactions
occur in separate beakers. As the name im-
plies, there must be two such reactions.

(2) Electrons are shown as part of the reaction.
Our ammeter shows that electrons are in-
volved. They flow when the reaction starts,
and do not flow when the reaction stops. The
meter also indicates that the electrons leave
the copper rod, pass through the wire, and
enter the silver rod.

(3) New chemical species are produced in each
half of the cell. The copper rod is converted
to copper ions (the rod loses weight) and the
silver ions are changed to metal (the silver
rod gains weight). The new species can be
explained in terms of gain of electrons (by
silver) and loss of electrons (by copper).

(4) The half-reactions, when combined, express
the overall, or net, reaction.

The net reaction (3) is obtained by combining
(7) and (2) so as to cause the exact balancing of
electrons lost by copper atoms, in (7), and elec-
trons gained by silver ion, in (2). This cancella-

202

OXIDATION-REDUCTION REACTIONS I CHAP. 12

tion is necessary because electrical measurements
show that the electrochemical cell operates with-
out accumulation or consumption of electric
charge. The reaction mixture always remains
electrically neutral. The number of electrons lost
equals the number of electrons gained.

We see that the overall chemical reaction that
occurs in an electrochemical cell is conveniently
described in terms of two types of half-reactions.
In one, electrons are lost; in the other, they are
gained. To distinguish these half-reactions we
need two identifying names.

The half-reaction in which electrons are lost is
called oxidation.

Oxidation Cu(s) — >- Cu^(aq) + 2e~ (/), (4)*

The half-reaction in which electrons are gained
is called reduction.

Reduction 2Ag+(aq) + 2e~

2Ag(s) (2), (5)

The overall reaction is called an oxidation-
reduction reaction.

Oxidation-reduction reaction

Cu(s) + 2Ag+(aq)

Cu+*(aq) + 2Ag(s) (J), (6)

It is often convenient and usually informative
to treat oxidation-reduction in terms of half-
reactions. When it is convenient, oxidation is in-
volved in the half-reaction showing loss of elec-
trons, and reduction is involved in the half-
reaction showing gain of electrons.

12-1.2 Oxidation-Reduction Reactions
in a Beaker

These ideas, developed for an electrochemical
cell, have great importance in chemistry because
they are also applicable to chemical reactions
that occur in a single beaker. Without an elec-
tric circuit or an opportunity for electric cur-
rent to flow, the chemical changes that occur in
a cell can be duplicated in a single solution. It is
reasonable to apply the same explanation.

* In this and later chapters, each equation is assigned
a consecutive number (given on the right). If the equation
has occurred earlier in the chapter, the earlier number is
given as well (on the left).

COPPER OXIDIZED BY Ag+(aq)
IN A BEAKER

We can easily demonstrate that reaction (J) can
occur even when the half-cells are not separated.
You did this in Experiment 7. A copper wire
immersed in AgN03 solution caused copper to
dissolve [the blue color of Cu+2(aq) appeared]
and metallic silver was precipitated. The ratio of
(Cu dissolved) to (Ag formed) was the same as
that in our cell, hence the net reaction was the
same.

EXERCISE 12-1

Compare the mole ratio Ag/Cu derived from
your own data for Experiment 7 to the electro-
chemical data given in Section 12-1.1.

The moles of silver deposited per mole of cop-
per dissolved are the same whether reaction (3)
is carried out in an electrochemical cell or in a
single beaker, as in Experiment 7. If, in the cell,
electrons are transferred from copper metal
(forming Cu+2) to silver ion (forming metallic
silver), then electrons must have been transferred
from copper metal to silver ion in Experiment 7.

Fig. 12-2. Oxidation-reduction reactions can occur
in a beaker.

Cu(s)-+ Cufaf) + 2e
oxt'dairion

2Af(aqh2e--+ 2A9 (s)
reduction.

SEC. 12-1 I ELECTROCHEMICAL CELLS

203

Thus, Experiment 7 involved the same oxidation-
reduction reaction but the electron transfer must
have occurred locally between individual copper
atoms (in the metal) and individual silver ions
(in the solution near the metal surface). This
local transfer replaces the wire "middleman" in
the cell, which carries electrons from one beaker
(where they are released by copper) to the other
(where they are accepted by silver ions).

ZINC OXIDIZED BY U+(aq) IN A BEAKER

Many oxidation-reduction reactions (nicknamed
"redox" reactions) take place in aqueous solu-
tion. One of these was mentioned in Section
11-2.1 when we characterized acids:

Zn(s) + 2H+(aq) — ►- Zn^-(aq) + HJg) (7)

Each zinc atom loses two electrons in changing
to a zinc ion, therefore zinc is oxidized. Each
hydrogen ion gains an electron, changing to a
hydrogen atom, therefore hydrogen is reduced.
(After reduction, two hydrogen atoms combine
to form molecular H2.) As before, reaction (7)
can be separated into two half-reactions:

Zn(s) — ►- Zn**(aq) + 2e~ (8)

2H+(aq) + 2e~ — > HJg)* (9)

Net reaction

Zn(s) + 2H+

Zn+2 + H2(g) (7), (10)

Thus, the reaction by which a metal dissolves in
an acid is conveniently discussed in terms of oxi-
dation and reduction involving electron transfer.
The reaction can be divided into half-reactions
to show the electron gain (by H+ ions) and the
electron loss (by metal atoms).

Not all metals react with aqueous acids.
Among the common metals, magnesium, alumi-
num, iron, and nickel liberate H2 as zinc does.
Other metals, including copper, mercury, silver,
and gold, do not produce measurable amounts
of hydrogen even though we make sure that the
equilibrium state has been attained. With these
metals, hydrogen is not produced and it is surely
not just because of slow reactions. Apparently

* From this point on in this chapter we will consider
only aqueous solutions, hence we will not specify (aq)
for each ion.

Copper
me"tal

1M HCl

Fig. 12-3. Some metals release electrons to H* and
others do not.

some metals release electrons to H+ [as zinc does
in reaction (70)] and others do not.

ZINC OXIDIZED BY C\l+2(aq)
IN A BEAKER

As a third oxidation-reduction example, suppose
a strip of metallic zinc is placed in a solution of
copper nitrate, Cu(N03)2. The strip becomes
coated with reddish metallic copper and the
bluish color of the solution disappears. The
presence of zinc ion, Zn+2, among the products
can be shown when the Cu+2 color is gone. Then
if hydrogen sulfide gas is passed into the mixture,
white zinc sulfide, ZnS, can be seen. The reaction
between metallic zinc and the aqueous copper
nitrate is

Zn(s) + Cu+2

Zn+2 + Cu(s)

Zinc has lost electrons in reaction (77) to form
Zn+2:

Zn(s) — *~ Zn+2 + 2e- (8), (12)

Zinc is oxidized. If zinc is oxidized, releasing
electrons, something must be reduced, accepting
these electrons. Copper ion is reduced:

Cu+2 + 2e-

Cu(s)

U3)

This time, copper ion gains electrons from the
zinc, in contrast to the behavior in Experiment 7,
where copper metal lost electrons to silver.

204

OXIDATION-REDUCTION REACTIONS I CHAP. 12

j .1M Cu.S04

Cu, (s) — — CuT + 2i

.llvf Zn (wo3)2

.lKf CilSO+

Cu + 2e-^-~Cu.(s)

Fig. 12-4. Two electrochemical cells involving copper: with silver, copper is oxidized; with zinc, Cu*2 is reduced.

SEC. 12-1 I ELECTROCHEMICAL CELLS

205

What about the state of equilibrium for the
reaction represented by equation (77)? Let us
place a strip of metallic copper in a zinc sulfate
solution. No visible reaction occurs and attempts
to detect the presence of cupric ion by adding
H2S to produce the black color of cupric sulfide,
CuS, fail. Cupric sulfide has such low solubility
that this is an extremely sensitive test, yet the
amount of Cu+2 formed cannot be detected. Ap-
parently the state of equilibrium for the reaction
(77) greatly favors the products over the reac-
tants.

12-1.3 Competition for Electrons

These reactions can be viewed as a competition
between two kinds of atoms (or molecules) for
electrons. Equilibrium is attained when this com-
petition reaches a balance between opposing re-
actions. In the case of reaction (3), copper metal
reacting with silver nitrate solution, the Cu(s)
releases electrons and Ag+ accepts them so
readily that equilibrium greatly favors the prod-
ucts, Cu+2 and Ag(s). Since randomness tends to
favor neither reactants nor products, the equi-
librium must favor products because the energy
is lowered as the electrons are transferred. If we
regard reaction (3) as a competition between
silver and copper for electrons, stability favors
silver over copper.

The same sort of competition for electrons
is involved in reaction (77), in which Zn(s) re-
leases electrons and Cu+2 accepts them. This time
the competition for electrons is such that equi-
librium favors Zn+2 and Cu(s). By way of con-
trast, compare the reaction of metallic cobalt
placed in a nickel sulfate solution. A reaction
occurs,

Co(s) + Ni+2 +±: Co+2 + Ni(s) (14)

At equilibrium, chemical tests show that both
Ni+2 and Co+2 are present at moderate concen-
trations. In this case, neither reactants (Co and
Ni+2) nor products (Co+2 and Ni) are greatly
favored.

This competition for electrons is reminiscent
of the competition for protons among acids and
bases. The similarity suggests that we might de-
velop a table in which metals and their ions are

listed by tendency to release electrons just as we
did in Table 11-IV (p. 191) in which the acid
strength indicates tendency for an acid to re-
lease H+.

We can already make some comparisons. We
might begin by listing some of the half-reactions
we have encountered in this chapter. We shall
write them to show the release of electrons and
then arrange them in order of their tendency to
do so. First we considered reaction (5) and dis-
covered that copper releases electrons to silver
ion. Therefore, we shall write our first two half-
reactions in the order

Cu(s) — >- Cu+2 + 2e~ (1), (15)

Ag(s) — >- Ag+ + e~ (16)

Listing the Cu-Cu+2 half-reaction first indicates
that it releases electrons more readily than does
the Ag-Ag+ half-reaction.

Now consider reaction (77). Since zinc releases
electrons to copper ion, we know that we must
add it to our list at the top:

Zn(s) — *- Zn+2 + 2e~ (8), (17)

Cu(s) — >- Cu+2 + 2e~ (1), (18)

Ag(s) —*■ Ag+ + e~ (16), (19)

Listing the Zn-Zn+2 half-reaction first tells us
that it releases electrons more readily than does
the Cu-Cu+2 half-reaction. But if this is true,
then the Zn-Zn+2 half-reaction must also release
electrons more readily than does the Ag-Ag+
half-reaction. Our list leads us to expect that zinc
metal will release electrons to silver ion, reacting
to produce zinc ion and silver metal.

We should test this proposal! We dip a piece
of zinc metal in a solution of silver nitrate. The
result confirms our expectation; zinc metal dis-
solves and bright crystals of metallic silver ap-
pear.

Our data allow us to make one more addition
to the list. By reaction (7), zinc reacts with H+
to give Zn+2 and U2(g). The half-reaction
H2-2H+ must be placed below the Zn-Zn+2 half-
reaction. How far below? To answer that, re-
member that copper does not react with H+ to
produce H2. This indicates that the half-reaction

206

OXIDATION-REDUCTION REACTIONS I CHAP. 12

H2-2H+ releases electrons more readily than
does the half-reaction Cu-Cu+2. Now we can
expand our list to that given in Table 12-1.

Table 12-1

SOME HALF-REACTIONS LISTED IN
ORDER OF TENDENCY TO RELEASE
ELECTRONS

Zn(s)

Cu(s)
Agfa,)

Zn+2 + 2e~
2H+ + 2e~
Cu+2 + 2e-
Ag+ + e~

(8), (20)

(21)

(/), (22)

(16), (23)

EXERCISE 12-2

From the statement in the text that nickel metal
reacts with H+ to give H2(g) and the additional
information that zinc metal reacts readily with
nickel sulfate solution, decide where to add the
half-reaction Ni-Ni+2 in our list.

The value of this list is obvious. Any half-
reaction can be combined with the reverse of
another half-reaction (in the proportion for
which electrons gained is equal to electrons lost)
to give a possible chemical reaction. Our list
permits us to predict whether equilibrium favors
reactants or products. We would like to expand
our list and to make it more quantitative. Elec-
trochemical cells help us do this.

12-1.4 Operation off an Electrochemical Cell

Now let's take a more detailed look into the
electrochemical cell. Figure 12-5 shows a cross-
section of a cell that uses the same chemical re-
action as that depicted in Figure 12-1. The only
difference is that the two solutions are connected
differently. In Figure 12-1 a tube containing a
solution of an electrolyte (such as KN03) pro-
vides a conducting path. In Figure 12-5 the silver
nitrate is placed in a porous porcelain cup. Since
the silver nitrate and copper sulfate solutions can
seep through the porous cup, they provide their
own connection to each other.

-Porous cup

Fig. 12-5. The operation of an electrochemical cell.

Before examining the processes in a cell, we
should name the parts of a cell and clear away
some language matters. The electrons enter and
leave the cell through electrical conductors — the
copper rod and the silver rod in Figure 12-5 —
called electrodes. At one electrode, the copper
electrode, electrons are released and oxidation
occurs. The electrode where oxidation occurs is
called the anode. At the other electrode, the
silver electrode, electrons are gained and reduc-
tion occurs. The electrode where reduction oc-
curs is called the cathode.

As electrons leave the cell from the anode
(electrons are released where oxidation occurs),
positively charged Cu+2 ions are produced. Nega-
tive charge is leaving (by means of the electron
movement) and positive charge is produced (the
Cu+2 ions) in this half of the cell. How is elec-
trical neutrality maintained? It must be main-

SEC. 12-2 I ELECTRON TRANSFER AND PREDICTING REACTIONS

207

tained by the movement of ions through the
solution. Negative ions drift toward the anode
and positive ions move away. It is because nega-
tive ions in a cell always drift toward the anode
that negative ions are called anions (pronounced
an'ions). Since positive ions drift away from the
anode and toward the cathode, positive ions are
called cations (pronounced cations).
Here is our electrochemical glossary:

Electrodes: The conductors at which reactions
occur in an electrochemical cell.

Anode: The electrode at which oxidation

occurs.

Cathode: The electrode at which reduction
occurs.

Anion: Negatively charged ion.

Cation: Positively charged ion.

With the verbal matters out of the way, let's
take an electrical tour through the cell shown in
Figure 12-5. We'll start at the surface of the
copper rod and follow the process around the
entire circuit and back to the copper rod. Let us
begin with a particular copper atom that loses
two electrons:

Cu(s) — ►- Cu+2 + 2e~ (/), (24)

The Cu+2 ion drifts away into the solution but
the electrons remain in the copper rod. They
move up through the copper anode, through the
wire, and enter the silver cathode. At the surface
of this rod, the electrons encounter Ag+ ions in
the solution. The electrons react with Ag+ ions
to give neutral silver atoms which remain on the
rod as silver metal:

2Ag4 + 2e-

2kg(s) (2), (25)

Now there is an excess of positive charge in the
solution near the copper anode and a deficiency
of positive charge in the solution near the silver
cathode. Two negative charges have been moved
from the anode half-cell through the wire to the
cathode half-cell. This charge movement causes
all of the negative ions (anions) in the solution
(S04~2 and N03") to start drifting toward the
anode. All of the positive ions (cations) start
drifting toward the cathode. When the movement
of all of these ions amounts to the charge move-
ment of two negative charges from the cathode
porous cup to the anode beaker, our tour of the
cell is completed. Electrical neutrality has been
restored and the net reaction is

Cu(s) + 2Ag+

Cu+2 + 2Ag(sj (3), (26)

12-2 ELECTRON TRANSFER AND PREDICTING REACTIONS

The usefulness of Table 12-1 is clear. Qualitative
predictions of reactions can be made with the
aid of the ordered list of half-reactions. Think
how the value of the list would be magnified if
we had a quantitative measure of electron losing
tendencies. The voltages of electrochemical cells
furnish such a quantitative measure.

12-2.1 Electron Losing Tendency

The circuit shown in Figure 12-1 includes a wire
resistance coil, R. As the current flows through
R, heat is generated. The cell is doing electrical
work in forcing the electron current through R.
Again we apply the Law of Conservation of
Energy. As the electrons leave R, they must have
lower potential energy than they had when they

entered. As they fall to the lower potential en-
ergy, the energy change appears as heat. This
potential energy change is measured by voltage.
Just as lowering a mass from a higher altitude
decreases its potential energy, moving an electric
charge to a lower voltage lowers its potential
energy.

So the voltage of an electrochemical cell meas-
ures its capacity for doing electrical work. Dif-
ferent cells show different voltages. To see the
importance of this voltage, consider the experi-
ment shown in Figure 12-6. In Figure 12-6A we
have a cell based upon reaction (27):

Zn(s) + Ni+2 — ►- Zn+2 + NifaJ (27)

If the concentrations in the cell are 1 M, the
voltage in the cell is 0.5 volt. We will call this

208

OXIDATION-REDUCTION REACTIONS | CHAP. 12

Fie. 12-6. Two cells In opposition. In which direction will current flow?

SEC. 12-2 I ELECTRON TRANSFER AND PREDICTING REACTIONS

209

voltage E° (Zn-Ni+2). As the cell is shown, elec-
trons move clockwise through the meter. On the
right in Figure 12-6B is a cell based upon reac-
tion (28). The voltage of this cell, E° (Zn-Ag+),
is 1.5 volts for 1 M concentrations.

Zn(s) + 2Ag+

Zn+2 + 2Ag(s) (28)

The electrical hookup in this cell causes electrons
to move counterclockwise through the meter.

Then, in Figure 12-6C, the two cells are recon-
nected in opposition to each other. The zinc-
nickel cell pushes the electrons clockwise and
the silver-zinc cell pushes the electrons counter-
clockwise. In each cell, the zinc electrode has a
tendency to dissolve, releasing electrons into the
external circuit. But electrons cannot flow in
both directions. Which cell will win out? Ex-
periment shows that electron current flows coun-
terclockwise from the zinc electrode of the
Zn-Ag+ cell into the zinc electrode of the
Zn-Ni+2 cell. We can explain this intuitively by
saying the stronger (1.5 volts) cell has over-
powered the weaker (0.5 volt) cell. The Zn-Ag+
cell proceeds to react in the sense it would if it
were alone in the circuit. The other cell is forced
to react in the direction opposite to that it would
take spontaneously. The reaction that generates
the higher voltage prevails. The reaction which
generates the higher voltage has the greater tend-
ency to proceed and the observed voltage meas-
ures the tendency.

We can learn one more valuable lesson from
these two cells working against each other. The
measured voltage of our double cell (Figure 12-6)
is 1.0 volt. Since the voltage for each cell is a
measure of the tendency to send out electrons,
we could calculate the net voltage (1.0 volt) by
combining the individual cell voltages. In this
case we must subtract the two because they are
hooked up to oppose each other. So we find

£2et = £°(Zn-Ag+) - £°(Zn-Ni+2)
= 1.5 - 0.5
= 1.0 volt

It is interesting to write an overall reaction for
these two cells when they are operating as in
Figure 12-6:

Zn- Ag+ cell :

Zn(s) — ►- Zn« + 2e~ (8), (29)

2Ag+ + 2e~ -+ 2Ag(s) (2), (30)

Zn-Ni+2cell:

Zn+2 + 2e~ — ►- Zn(s) (8), (31)

NifsJ — ►- Ni+2 + 2e~ (32)

Net reaction in both cells:

(29) + (30) + (31) + (32)

NifsJ + 2Ag+ — >- Nr" + 2AgfsJ (33)

But the net reaction is just the reaction that oc-
curs in a Ni-Ag+ cell! What is the voltage of
such a cell? Experiment shows that such a cell
has a voltage of 1 .0 volt. We find our double cell
(Figure 12-6) has a voltage identical to that of a
Ni-Ag+ cell alone. The tendency of zinc to re-
lease electrons through reaction (29) must have
influenced the voltage of the Zn-Ag+ cell. The
same tendency of zinc to release electrons must
also have influenced the voltage of the Zn-Ni+2
cell. When these cells are put in opposition, the
tendency of zinc to release electrons exactly
cancels.

This shows that the voltage of a given cell may
be thought of as being made up of two parts, one
part characteristic of one of the half-reactions
and one part characteristic of the other half-
reaction. Chemists call these two parts "half-cell
potentials," a term that emphasizes the relation
between voltage and potential energy. The half-
cell potentials are symbolized E°.

Thus we write for the Zn-Ag+ cell

E\ = £°(Zn-Zn+2) - £°(Ag-Ag+) = 1.5 volts (34)

and for the Zn-Ni+2 cell

El = £°(Zn-Zn+2) - £°(Ni-Ni+2) = 0.5 volt (35)

If we place these batteries in opposition, the net
voltage will be

E°3 = E\ - E°2 = [£°(Zn-Zn+2) - £°(Ag-Ag+)]

- [£°(Zn-Zn+2) - £°(Ni-Ni+2)]
El = ^(Ni-Ni*2) - £°(Ag-Ag+) = 1.0 volt (36)

and also

El = E\ - El = (1.5) - (0.5) = 1.0 volt

We see that:

(a) the value £°(Zn-Zn+2) cancels in (36);

(b) (36) is just the expected sum of half-reactions
for a Ni-Ag+ cell;

210

OXIDATION-REDUCTION REACTIONS I CHAP. 12

MEASURING HALF-CELL POTENTIALS

We would like to measure the contribution each
half-reaction makes to the voltage of a cell. Yet
every cell involves two half-reactions and every
cell voltage measures a difference between their
half-cell potentials. We can never isolate one
half-reaction to measure its E°. An easy escape
is to assign an arbitrary value to the potential of
some selected half-reaction. Then we can com-
bine all other half-reactions in turn with this
reference half-reaction and find values for them
relative to our reference. The handiest arbitrary
value to assign is zero and chemists have decided
to give it to the half-reaction

Wg)

2H+ + 2e~

E° = 0.00 • • • volt (27), (37)

We must control concentration during these
measurements of E°, since the voltage of a cell
changes as concentrations change. For example,
in the laboratory we studied a cell based on re-
action (38):

Zn(s) + Cu+2 — >- Zn+2 + Cu(s) (77), (38)

We found that the voltmeter readings were dif-
ferent for different concentrations. Were these
readings in agreement with predictions we would
make on the basis of equilibrium principles?
Increasing the concentration of Cu+2 ion in re-
action (38) should increase the tendency to form
the products (Le Chatelier's Principle). Experi-
mentally, we find an increase in voltage when
more solid copper sulfate is dissolved in the solu-
tion around the copper electrode. Conversely,
decreasing the concentration of the copper ion
should decrease the tendency to form the prod-
ucts. Again in agreement, voltage readings de-
crease when the concentration of Cu+2 ion is
reduced by precipitation of CuS. The voltage
shows the tendency for reaction to occur.

What happens to any cell or battery as it
operates? The voltage decreases until, finally, it
reaches zero. Then we say the cell is dead. Equi-
librium has been attained and the reaction that

has been producing the energy has the same
tendency to proceed as does the reverse reaction.
Again the voltage measures the net tendency for
reaction to occur. At equilibrium there is a bal-
ance between forward and reverse reactions,
hence there is no net tendency for further reac-
tion either way. Therefore, the voltage of a cell
at equilibrium is zero.

Since concentration variations have measura-
ble effects on the cell voltage, a measured voltage
cannot be interpreted unless the cell concentra-
tions are specified. Because of this, chemists
introduce the idea of "standard-state." The
standard state for gases is taken as a pressure of
one atmosphere at 25°C; the standard state for
ions is taken as a concentration of 1 M ; and the
standard state of pure substances is taken as the
pure substances themselves as they exist at 25°C.
The half-cell potential associated with a half-
reaction taking place between substances in their
standard states is called E° (the superscript zero
means standard state). We can rewrite equation
(37) to include the specifications of the standard
states:

H2(g, 1 atm) — >- 2H+(aq, 1 M) + 2er

E° = 0.00 • • • volt (assigned) (39)

Now if we combine a Zn-Zn+2 half-cell in its
standard state with a H2-2H+ half-cell in its
standard state, the voltage (potential) we meas-
ure (0.76 volt) is the value assigned to the half-
reaction:

Zn(s) — >- Zn+2(aq, 1 M) + 2e~

E° = +0.76 volt (40)

Similarly, if we combine a Cu-Cu+2 half-cell in
its standard state with a standard H2-2H+ half-
cell, the voltage (potential) we measure (0.34
volt) is the value assigned to the half-reaction:

Cu(s) — »- Cu+^aq, 1 M) + 2e~

E° = -0.34 volt (41)

Chemists have determined a large number of
these half-cell potentials. The magnitude of the
voltage is a quantitative measure of the tendency
of that half-reaction to release electrons in com-
parison to the H2-2H+ half-reaction. If the sign
is positive, the half-reaction has greater tendency
to release electrons than does the H2-2H+ half-

METALS

\ NON METALS

•

»«

-- ■

•

— *

•

°"

■^

•

' ■ .J

•

OXIDES

/Vate /. (Above) The elements in the periodic table. (Below) Compounds in the periodic table. CHLORIDES

MISCELLANEOUS

SEC. 12-2 I ELECTRON TRANSFER AND PREDICTING REACTIONS

211

reaction. If the sign is negative, the half-reaction Hi(g) — *- 2H+ + 2e
has less tendency.

Cursj — *- Cu+2 + 2e-

AgfsJ — >- Ag+ + e~

E° = 0.000 volt (21), (43)

£? = -0.34 volt (/), (44)

E° = -0.80 volt (16), (45)

TABLE OF HALF-CELL POTENTIALS

Table 12-11 gives the values of the standard oxi-
dation potentials for a number of half-reactions.

A more complete table is given in Appendix 3. The half-reactions, listed in order of decreasing

We have not added the information "1 M" for half-cell potentials, are in the same order as in

each ion since this is implied by the symbol E°. Table 12-1, which was dictated by laboratory ex-

For the same reason, 25°C and 1 atmosphere penence.

pressure of gases are understood.

EXERCISE 12-3

Table 12-11

SELECTED STANDARD OXIDATION
POTENTIALS FOR HALF-REACTIONS*

HALF-REACTION E°

Reduced state Oxidized state (volts)

Zn(a)
Co(a)

Nir«;

H*(g)
Cu(s)
21-

Agf«;

2Br~

Mn+l + 2H,0

2C1-

Mn+l + 4H,0

2e~ + Zn+'
2e~ + Co+*
2e~ + Ni4*
2e~ + 2H+
2e~ + CU+*
2e~ + h(s)
e~ + Ag+
2e~ + Br2(l)
2e~ + MnO, + 4H+
2e~ + Ch(g)
5e~ + MnO*- + 8H+

+0.76
+0.28
+0.25

0.000
-0.34
-0.53
-0.80
-1.06
-1.28
-1.36
-1.52

* A more complete table is given in Appendix 3.

Let's examine this table to see if it agrees with
our laboratory experience. Table 12-1 summa-
rized some of these results in a qualitative way.
Extracting these four half-reactions from Table
12-11, we find

Zn(s)

Zn« + 2e~

E° = +0.76 volt (8), (42)

In Exercise 12-2 you placed the Ni-Ni+2 half-
reaction into Table 12-1. Check your placement
by examining the half-cell potential of this half-
reaction in Table 12-11.

Another way to verify the usefulness of Table
12-11 is to compare its voltage predictions to one
we have measured. For example, we found a
value of approximately 0.5 volt for a cell based
on reaction (46):

Zn(s) + Ni+2 — >- Zn+2 + N\(s) (27), (46)

This cell involves the following two half-
reactions:

Zn(s)

Ni(s)

Zn+2 + 2e-

Ni+2 + 2e~

E° = +0.76 volt (8), (47)

E° = +0.25 volt (32), (48)

By the values of E°, we conclude that the
Zn-Zn+2 half-reaction has the greater tendency
to release electrons. Hence it will tend to transfer
electrons to nickel, forcing half-reaction (48) in
the reverse direction. Our net reaction will be
obtained by subtracting half-reaction (48) from
(47):

minus

ZnfsJ
Kx(s)

Zn+2 + 2e"
Ni« + 2e~

Net reaction Zn(s) + Ni"« — >- Zn^ + Nifsj

£? = +0.76 volt (47), (49)

E\ = +0.25 volt (48), (50)

E° = E\ - El

E° = (+0.76) - (+0.25)

£° = +0.51 volt (51)

212

OXIDATION-REDUCTION REACTIONS j CHAP. 12

Table 12-11 predicts the cell will operate so as to
dissolve metallic zinc and deposit metallic nickel,
and its voltage will be +0.51 volt This is exactly
what occurs in such a cell. Predicting is fun-
let's try it again! Another cell we studied is based
on reaction (52):

Zn(s) + 2Ag+ — >- Zn+2 + 2AgfsJ (28), (52)

The two half-reactions involved are

does not depend upon how many moles we con
sider. Thus:

A%(s) — >- Ag+ + e~

El = -0.80 volt (16), (58)

Zn(s)

Ag(s)

Zn+2 + 2e~

Ag+ + e~

E° = +0.76 volt (8), (53)
E° = -0.80 volt (16), (54)

2Ag(s) — *- 2Ag+ + 2e~

E°2 = -0.80 volt (16), (59)

You might wonder what we would have
learned if we had assumed that either of these
two cells operates with the reverse reaction. Sup-
pose we had proposed a cell based on oxidation
of nickel and reduction of zinc:

minus

Net reaction Nifsj + Zn+2

NifsJ — >- Ni+2 + 2e~ E\ = +0.25 volt (32), (60)

Zn(s) — ►■ Zn+2 + 2e~ E°2 = +0.76 volt (8), (61)

E° = E° - E°2

= (+0.25) - (+0.76)
Ni+2 + Zn(s) E° = -0.51 volt (62)

By the half-cell potentials, we conclude the
Zn-Zn+2 half-reaction has the greater tendency
to release electrons. It will tend to transfer an
electron to silver ion, forcing (54) in the reverse
direction. Hence we obtain the net reaction by
subtracting (54) from (53). But remember that
this subtraction must be in the proportion that
causes no net gain or loss of electrons. If two
electrons are lost per atom of zinc oxidized in
(53), then we must double half-reaction (54) so
that two electrons will be consumed.

Our assumption concerning the chemistry leads
us to the reverse reaction of that we obtained
earlier — reaction (57) — and to an equal voltage
but with opposite sign. The significance of the
negative voltage ( — 0.51 volt) is that equilibrium
in the reaction favors reactants, not products. We
obtain the same prediction we did before — since
the voltage is negative, the reaction will tend to
operate a cell in the reverse direction — to dissolve
zinc metal and to precipitate nickel metal. The
reaction will occur in the direction written (con-

minus

Zn(s)
2Agfs;

Zn+2 + 2e~
2Ag+ + 2e-

El = +0.76 volt
El = -0.80 volt

Net reaction Zn(s) + 2AgH

E° = El- El

= (+0.76) - (-0.80)
Zn+2 + 2Ag(sJ £° = +1.56 volts

(53), (55)
(54), (56)

(28), (57)

Our conclusions are again in agreement with
experiment. The cell operates so as to dissolve
zinc metal and precipitate silver metal. The volt-
age is indeed about 1.5 volts. Finally, experiment
shows that one mole of zinc does react with two
moles of silver ion, as required by the balance of
electrons.

Notice that we did not double El for (54) in
obtaining (57). The voltage of a half-reaction

suming nickel and precipitating zinc) only if the
cell is "overpowered" by an opposing cell of
higher voltage than 0.51 volt (as was done in the
experiment pictured in Figure 12-6).

12-2.2 Predicting Reactions from Table 12-11

The ideas we have developed for reactions occur-
ring in electrochemical cells are also applicable

SEC. 12-2 I ELECTRON TRANSFER AND PREDICTING REACTIONS

213

to reactions that occur in a beaker. Therefore,
chemists use half-cell potentials to predict what
chemical reactions can occur spontaneously.

If a chemist wishes to know whether zinc can
be oxidized if it is placed in contact with a solu-
tion of nickel sulfate, the values of E° help him
decide. The half-cell potential for Zn-Zn+2 is
+0.76 volt, which is greater than that for
Ni-Ni+2 (which is +0.25 volt). The difference,
+0.51 volt, is positive, indicating that zinc has
a greater tendency to lose electrons than does
nickel. Therefore, zinc can transfer electrons to
Ni+2. The chemist predicts: zinc will react with
Ni+2, zinc being oxidized and nickel being re-
duced.

12-11 (something which can be oxidized), and
it must involve a substance from the right
column (something which can be reduced).
2. A substance in the left column of Table 12-11
tends to react spontaneously with any sub-
stance in the right column which is lower in
the Table.

Applying these rules, we would predict: cop-
per metal could be oxidized to Cu+2 by Br2(l) or
Mn02(s) but not by Ni+2 or Zn+2. Of course,
copper metal cannot be oxidized by either zinc
metal or nickel metal because neither zinc
metal nor nickel metal can accept electrons to
be reduced (as far as we know from Table 12-11).

an example: copper and silver

Suppose the question is whether silver will be
oxidized if it is immersed in copper sulfate. The
half-cell potential for Ag-Ag+ is —0.80 volt and
that for Cu-Cu+2 is -0.34 volt. The first value,
— 0.80 volt, is more negative than the second,
—0.34 volt. The difference, then, is still negative:
-0.80 - (-0.34) = -0.46 volt. The negative
answer shows that Ag-Ag+ has less tendency to
lose electrons than does Cu-Cu+2. The reaction
will not tend to proceed spontaneously. Silver
will not be oxidized to an appreciable extent in
copper sulfate.

EXERCISE 12-4

Use the values of E° to predict whether cobalt
metal will tend to dissolve in a 1 M solution of
acid, H+. Now predict whether cobalt metal will
tend to dissolve in a 1 M solution of zinc sulfate
(reacting with Zn+2).

GENERALIZING ON PREDICTIONS WITH E°

We can generalize now on the use of Table 12-11.
A substance on the left in Table 12-11 reacts by
losing electrons. A substance on the right reacts
by gaining electrons. We may draw the following
conclusions:

1. An oxidation-reduction reaction must involve
a substance from the left column of Table

EXERCISE 12-5

Use Table 12-11 to decide which substances in
the following list tend to oxidize bromide ion,
Br~: Ch(g), H+, Ni+2, Mn(V.

EXERCISE 12-6

Use Table 12-11 to decide which substances in
the following list tend to reduce Br2(l): C\~,
H2(g), Ni(s), Mn+2.

PREDICTIONS AND THE EFFECT
OF CONCENTRATIONS

All of our predictions have been based upon the
values of E° that apply to standard conditions.
Yet we often wish to carry out a reaction at
other than standard conditions. Our prediction
then must be adjusted in accord with Le Cha-
telier's Principle as we change conditions from
standard conditions to others of interest to us.
For example, by comparing £°'s, we predicted
that zinc would dissolve in nickel sulfate. These
£°'s show that zinc metal will dissolve if zinc ion
and nickel ion are both present at 1 M concen-
tration:

Zn(s) + NiH

Zn+2 + Ni(s) (27), (63)

In our case, however, no zinc ion is present at
all. How does this affect our prediction? By

214

OXIDATION-REDUCTION REACTIONS I CHAP. 12

Le Chatelier's Principle, the removal of Zn+2
tends to shift equilibrium toward the products.
Therefore, removing Zn+2 increases the tendency
for reaction (63) to occur. Our prediction of re-
action is still valid.

This will not always be the case, however.
Consider the question: "Will silver metal dis-
solve in 1 M H+?" According to Table 12-11,

12-2.3 Reliability of Predictions

There is one more limitation on the reliability of
predictions based upon £°'s. To see it, we shall
consider the three reactions
Cu(s) + 2H+

Fe(s) + 2H+

-*- Cu+2 + H,(g)
E° = -0.34 volt
■*- Fe+2 + U2(g)
E° = +0.44 volt

(65)

(66)

2Ag(s)
2H+(1 M) + 2e~

2Ag+(l M) + 2e~

E° = -0.80 volt
E° = 0.00 volt

Net reaction 2Ag(s) + 2H+(1 M)

2Ag+(l M) + H,(g) E° = -0.80 volt

(64)

The negative voltage shows that the state of
equilibrium favors the reactants more than the
products for the reaction as written. For stand-
ard conditions, the reaction will not tend to
occur spontaneously. However, if we place Ag(s)
in 1 M H+, the Ag+ concentration is not 1 M —
it is zero. By Le Chatelier's Principle, this in-
creases the tendency to form products, in opposi-
tion to our prediction of no reaction. Some silver
will dissolve, though only a minute amount be-
cause silver metal releases electrons so reluctantly
compared with H2. It is such a small amount,
in fact, that no silver chloride precipitate forms,
even though silver chloride has a very low solu-
bility.

That some silver does dissolve to form Ag+
can be verified experimentally by adding a little
KI to the solution. Silver iodide has an even
lower solubility than does silver chloride. The
experiment shows that the amount of silver that
dissolves is sufficient to cause a visible precipi-
tate of Agl but not of AgCl. This places the Ag+
ion concentration below 10-10 M but above
10~17 M. Either of these concentrations is so
small that we can consider our prediction for the
standard state to be applicable here too — silver
metal does not dissolve appreciably in 1 M HC1.
In general, the question of whether a prediction
based upon the standard state will apply to other
conditions depends upon how large is the mag-
nitude of E°. If E° for the overall reaction is only
one- or two-tenths volt (either positive or nega-
tive), then deviations from standard conditions
may invalidate predictions that do not take into
account these deviations.

3Fe(s) + 2N03- + 8H+
+ 2NO(g) + 4H..O

-+- 3Fe+2
E° = +1.40 volts

(67)

The three values of E° are easily calculated from
half-cell potentials. Then, we can predict with
confidence that reaction (65) will not occur to an
appreciable extent if solid copper is immersed in
dilute acid. The negative value of E° (—0.34 volt)
indicates that equilibrium in (65) strongly favors
the reactants, not the products.

Furthermore, we can predict that reactions

(66) and (67) might occur. The positive values of
E° (+0.44 and +1.40 volts) show that equilib-
rium strongly favors products in these reactions.
Again, experiments are warranted. A piece of
iron is immersed in dilute acid — bubbles of hy-
drogen appear. Reaction (66) does occur. A piece
of iron is immersed in a one molar nitric acid
solution — though bubbles of hydrogen may ap-
pear, no nitric oxide, NO, gas appears. Reaction

(67) between iron and nitrate ion does not im-
mediately occur: the reaction rate is extremely
slow. This slow rate could not be predicted from
the £°'s.

So the equilibrium predictions based on E0,s
do not make all experiments unnecessary. They
provide no basis whatsoever for anticipating
whether a reaction will be very slow or very fast.
Experiments must be performed to learn the
reaction rate. The £°'s do, however, provide
definite and reliable guidance concerning the
equilibrium state, thus making many experi-
ments unnecessary; the multitude of reactions
that are foredoomed to failure by equilibrium
considerations need not be performed.

SEC. 12-2 I ELECTRON TRANSFER AND PREDICTING REACTIONS

215

12-2.4 E and the Factors That
Determine Equilibrium

We see that E° furnishes a basis for predicting the equi-
librium state. In Chapter 9 and in subsequent chapters,
equilibrium was treated in terms of two opposing tenden-
cies— toward minimum energy and toward maximum
randomness. What is the connection between E° and
these two tendencies?

Consider two reactions for which E° shows that prod-
ucts are favored, one an exothermic reaction, and the
other an endothermic reaction. For the exothermic reac-
tion, when the reactants are mixed they are driven toward
equilibrium in accord with the tendency toward minimum
energy. Now contrast the endothermic reaction for which
E° shows that equilibrium favors products. When these
reactants are mixed, they approach equilibrium against
the tendency toward minimum energy (since heat is ab-
sorbed). This reaction is driven by the tendency toward
maximum randomness.

Summarizing, E° measures quantitatively the difference
between the tendency to minimum energy and tendency to
maximum randomness under the standard state conditions.

12-2.5 Oxidation Numbers — An Electron
Bookkeeping Device

The reaction between ferric ion, Fe+3, and cu-
prous ion, Cu+, to produce ferrous ion, Fe+2,
and cupric ion, Cu+2, is plainly an oxidation-
reduction reaction:

Fe+3 + Cu+

Fe+2 + CuH

(68)

It is readily separated into two half-reactions
showing electron transfer:

Oxidation (loss of electrons)
Cu+

Reduction (gain of electrons)

Fe+3 + e-

Cu+2 + e- (69)

Fe+2 (70)

Because of the presence of Cu+ ion, ferric ion is
reduced. Chemists say that Cu+ ion acts as a
reducing agent in this reaction — Cu+ ion is the
"agent" that caused the reduction of ferric ion.
At the same time, Cu+ is oxidized because of the
presence of ferric ion. Hence, Fe+3 is called an
oxidizing agent in this reaction.
Another reaction by which ferric ion can be

reduced involves bisulfite* ion, HS03 . The bal-
anced equation is

H20 + HS03- + 2Fe+3 — ►■

2Fe+2 + HS04" + 2H+ (71)

Again half-reaction (70) describes what happens
to ferric ion:

2Fe+3 + 2e-

2FeH

(72)

Since two electrons are gained by the two ferric
ions in half-reaction (72), two electrons must be
released by the remaining constituents in (71).
The other half-reaction can be found by sub-
tracting (72) from (71) to give,

H20 + HSO3- - 2e~ — ►- HSOf + 2H+

H20 + HSOi

HS04- + 2H+ + 2e~ (73)

The combination of H20 and HS03~ acts as a
reducing agent toward Fe+3. Since water solu-
tions of Fe+3 are quite stable, HS03~ is con-
sidered to be the actual reducing agent.

Half-reaction (73) differs from the others we
have looked at. This is not the same simple situa-
tion in which a single atom is oxidized by re-
leasing electrons or is reduced by accepting them.
Here a complicated ion, HS03~, furnishes elec-
trons to Fe+3 in an intricate process in which an
oxygen atom is transferred from H20 to HS03" ,
forming HSCXf and two hydrogen ions. We can-
not assign the electron loss to a particular atom.
In this situation, it is convenient to have a book-
keeping device that at least keeps track of the
number of electrons so that none is forgotten.
The bookkeeping device is called assigning oxi-
dation numbers.

Since we don't know the locations of the elec-
trons held by a molecule such as HS03~, we
assume that the hydrogen atom has a -f 1 charge,
that each oxygen atom has a —2 charge, and
that the sulfur atom has all the rest of the elec-
trons in the molecule. Of course, if the charges
on all the atoms in HS03~ are added together,
they must sum to —1, the molecular charge.
Since there are three oxygen atoms in HS03~ , the
algebra looks like this:

* This ion, HSOf , is also called hydrogen sulfite.

216

OXIDATION-REDUCTION REACTIONS I CHAP. 12

f charge on \ /charge on\ /charge on\ ,^. , x

\ atom /

atom

(charge on
sulfur
atom

charge on
sulfur
atom

(-D
(+4)

This fictitious charge is called the oxidation num-
ber of sulfur:

Oxidation number sulfur = +4 in HS03~ (74)

The same process can be applied to HS04~.
Again assuming hydrogen has a charge of +1
and each of the four oxygen atoms has a charge
of —2, we calculate a fictitious charge on the
sulfur atom of +6:

Oxidation number sulfur = +6 in HS04~ (75)

According to the oxidation number bookkeep-
ing, the two electrons released in the HS03_-
HS04~ half-reaction (73) are associated with the
change in oxidation number of sulfur from +4
to +6.

This arbitrary scheme of assigning oxidation
numbers turns out to be quite useful, provided
we don't forget that the oxidation number is a
fictitious charge. We can see the usefulness by
considering a reaction related to the oxidation
of HSO3-.

Sulfur forms two oxides, S02 (a gas at normal
conditions) and S03 (a liquid that boils at
44.8°C). Under suitable conditions, S02 reacts
with oxygen to form S03:

2SO,fgj + Q2(g)

2SOi(g)

(76)

Is this an oxidation-reduction reaction? His-
torically, it surely is, for the term "oxida-
tion" originally referred specifically to reactions
with oxygen. Yet our electron-transfer view of
oxidation-reduction reactions provides no help
in deciding so. Where in reaction (76) is there
any evidence of electrons being gained or lost?
In such a doubtful case, our oxidation number
scheme provides an answer. Applying the same
assumptions used in treating the HS03~~-HS04~

half-reaction, we can calculate the oxidation
numbers of sulfur in S02 and S03.

In S02:

(charge on \ /charge on
oxygen J + I sulfur
atom / \ atom

/ molecular \
\ charge /

(-2)
assumed

(oxidation \
number 1 =
sulfur /

(0)

Oxidation number sulfur = +4 in S02 (77)

In S03:

charge on

oxygen

atom

(charge on
sulfur
atom

(oxidation\
number 1
sulfur /

Oxidation number sulfur

/ molecular \
\ charge /

(0)
+6 in S03 (75)

Thus, in reaction (76) the sulfur atom changes
oxidation number from +4 to +6, just as it did
in the HS03~-HS04- half-reaction. The oxida-
tion number gives a basis for connecting the
S02-S03 change to the oxidation of HS03" to
HS04~. Both changes are considered to be ex-
amples of oxidation.

Oxidation-reduction reactions occurring in
aqueous solutions are conveniently treated in
terms of half-reactions showing transfer of elec-
trons. Under more general conditions (gaseous
state, other solvents, etc.), it is more convenient
to treat oxidation-reduction reactions in terms
of oxidation numbers, based upon the arbitrary
scheme of assigning charge +1 to a hydrogen
atom bound to an unlike atom and charge —2
to an oxygen atom when it is bound to un-
like atoms. Generally, then, an oxidation'
reduction reaction is one in which oxidation
numbers change.

SEC. 12-3 BALANCING OXIDATION-REDUCTION REACTIONS

217

EXERCISE 12-7

The reactions by which S02 and S03 dissolve in
water are not considered to be oxidation-
reduction reactions:

SO/* J + H,0 — +■ HSCV (aq) + H+(aq) (79)
SOJg) + H,0 — >- HS04" (aq) + U+(aq) (80)

Convince yourself that none of the atoms in
either (79) or (80) changes oxidation number.

EXERCISE 12-8

In reaction (76) the oxidation number of sulfur
changes from +4 to '+6. According to this, two
electrons are released by each sulfur atom oxi-
dized. Show that these electrons are gained by
oxygen if we assume oxygen has oxidation num-
ber equal to zero in 02.

12-3 BALANCING OXIDATION-REDUCTION REACTIONS

When a reaction is properly written, it expresses
certain conservation laws. A chemical reaction
does not destroy or produce atoms. Therefore
there must be the same number and types of
atoms among the reactants as among the prod-
ucts. A chemical reaction also does not destroy
or produce electric charge. Therefore the sum of
the charges appearing on the reactant species
must be the same as the sum of the charges
appearing on the products. The process by which
we write a correct chemical reaction, making
sure these two conservation laws are followed,
is called balancing the chemical reaction.*

Oxidation-reduction reactions must be bal-
anced if correct predictions are to be made. Just
as in selecting a route for a trip from San Fran-
cisco to New York, there are several ways to
reach the desired goal. Which route is best de-
pends to some extent upon the likes and dislikes
of the traveler. We will discuss two ways to
balance oxidation-reduction reactions — first, us-
ing half-reactions and, next, using the oxidation
numbers we have just introduced.

12-3.1 Use of Half-Reactions for Balancing
Oxidation-Reduction Reactions

Suppose we want to describe by an equation
what happens when pure lithium metal is added
to a 1 M HC1 solution. Our first step must be to
decide what products will be obtained. This can
be determined only by experiment. Often you

* Some chemists prefer to call this process balancing
the equation for the reaction.

will know already what the experiment would
give. Sometimes what the products will be is not
obvious but it is easily learned from a reference
book. Sometimes you must do the experiment
yourself or proceed on the basis of an assump-
tion.

For lithium metal in 1 M HC1, the observed
facts are that the metal dissolves spontaneously
and a gas bubbles out of the solution. From
Appendix 3 we select the two half-reactions
(notice that the half-reactions are already "bal-
anced" in both charge and number of atoms):

U(s) — ►- Li+ + e~ (81)

H2(g) -+ 2H+ + 2e~ (21), (82)

Both of these half-reactions show production of
electrons. But we know there must be an electron
used for each produced, so one of the equations
must be reversed. Experiment shows us it is the
second because hydrogen gas is evolved from the
solution. The first equation is correct as written.
Lithium metal dissolves and is converted to ions.
Thus,

Li(s) — ►- Li+ + e~ (81), (83)

2H+ + 2e- — >- H2(g) (9), (84)

These equations now correctly express the ob-
served facts and they must be properly com-
bined. The first step is to make them indicate the
same number of electrons produced and used.
We "balance the electrons." By inspection we
find that we achieve balance by doubling the
first equation.

2U(s) — >- 2Li+ + 2e~ (81), (85)

2H+ + 2e

2LiH
H*f*J

(9), (86)

218

OXIDATION-REDUCTION REACTIONS I CHAP. 12

Now we add the two equations to get the net
reaction,

2H+ + 2Li(s) — *- 2Li+ + U,(g) (87)

Note that the electrons cancel — we took care
that this would happen because we know elec-
trons are neither consumed nor produced in the
net reaction.

As a final check, let us verify the conservation
of charge:

The electrons cancel out. The atoms and charges
are properly balanced, but there is one remaining
disturbance. There are H+ ions included among
the reactants as well as the products. Cancelling
the excess, we obtain the final, balanced reaction:

5H£(g) + 2Mn04" + 6H+ — ►-

5S(s) + 2MH+2 + 8H20 (95)

Before leaving the equation, let us check the
electric charge balance:

5HS(g) + 2Mn04- + 6H+
5(0) + 2(-l) +6(+l)
-2 + +6
+4

2H+ + 2Li(s) — ►- 2Li+ + H2(g)
+2+0 +2+0

+2 = +2

As a more complex case, suppose we want to
write the equation for the reaction that occurs
when hydrogen sulfide gas, H2S, is bubbled into
an acidified potassium permanganate solution,
KMn04. When we do this, we observe that the
purple color of the MnCV ion disappears and
that the resulting mixture is cloudy (sulfur par-
ticles). From Appendix 3 we find the two half-
reactions

H£(g) — ►- S(s) + 2H+ + 2e~ (88)
Mn+2 + 4H20 — ■*- Mn04" + 8H+ + 5e~ (89)

Since we know that sulfur is formed, we will use
equation (88) as it is written. However, the purple
Mn04~ is being changed to almost colorless Mn+2
so we will rewrite equation (89) as the reverse of
what we obtained from Appendix 3:

H2S(g) — ►- S(s) + 2H+ + 2e~ (88), (90)
Mn04- + 8H+ + 5e~ — >- Mn+2 + 4H20 (91)

Balancing the electrons is the next step, but this
time it is a little more difficult. We see that if we
multiply equation (90) by 5 and equation (97)
by 2, there will be 10 electrons in each case.
Arithmetic teachers call this finding the least
common multiple:

5H2Sfs; — >-
2Mn04" + 16H+ + 10e" — ►-

->- 5S(s) + 2Mn+2 + 8H20 (95)

5(0) + 2(+2) + 8(0)
+4
= +4

12-3.2 Balancing Half-Reactions

When potassium chlorate solution, KC103, is added to
hydrochloric acid, chlorine gas is evolved. Although we
can find the half-reaction, 2C1~ = C\2(g) + 2e~, in Ap-
pendix 3, we find no equation with CIO3" ion involved.
We can surmise that CIO3" is accepting electrons and
changing into chlorine. Let us write a partial half-reaction
in which we indicate an unknown number of electrons
and in which we have conserved only chlorine atoms:

CIO3 +xe-

hCUg)

(96)

From experience we note that in acid solution the oxygen
in such oxidizing agents as MnOi" and Cr20f2 ends up
as water and that H+ is consumed. Let us include this
notion by showing 6H+ among the reactants and 3H20
among the products:

CIO3" + 6H+ + xe~

\C\2(g) + 3H20 (97)

Finally, we have to remember that charge is conserved.
Since the products are neutral molecules, x will have to
be 5 in order that the total charge represented among the
reactants is zero. Our desired half-reaction is

CIO3" + 6H+ + 5e~ — ►- $CU(g) + 3H20 (98)

Now we can return to working out the equation for the
reaction we observed. The least common multiple be-
tween 1 and 5 is 5. Writing the half-reactions to involve 5
electrons and adding them, we obtain

5CI- — >- iCh(g) + Se~ (99)
CIO3" + 6H+ + 5e~ — >■ hCh(g) + 3H20 (700)

5C1- + CIO3" + 6H+

3Cl2fgJ + 3H20 (101)

Again demonstrate to yourself that atoms and charge are
conserved.

5S(s) + 10H+ + \0e~
2Mn+2 + 8H20

5HS(g) + 2Mn04" + 16H+ — >- 5S(s) + 10H+ + 2Mn+2 + 8H20

(92)
(93)

(94)

SEC. 12-3 | BALANCING OXIDATION-REDUCTION REACTIONS

219

12-3.3 Use of Oxidation Number in Balancing
Oxidation-Reduction Reactions

We have already introduced oxidation numbers
as a device for assigning a fictitious charge to an
atom in a molecule. According to this scheme,
oxidation-reduction reactions involve changes of
oxidation numbers. Consideration of conserva-
tion of charge reveals that there must be a bal-
ance between changes of oxidation number. Con-
sequently, oxidation numbers provide just as
good a basis for balancing equations as do half-
reactions.

ASSIGNING OXIDATION NUMBERS

For the present, we will limit ourselves to mole-
cules containing hydrogen and/or oxygen along
with the element to which we wish to assign an
oxidation number. The rules we will utilize are
as follows:

(1) The oxidation number of a monatomic ion
is equal to the charge on the ion.

(2) The oxidation number of any substance in
the elementary state is zero.

(3) The oxidation number of hydrogen is taken
to be + 1 (except in H2, which is the elemen-
tary state).

(4) The oxidation number of oxygen is taken to
be —2 (except in 02; ozone; 03; and perox-
ides).

(5) The other oxidation numbers are selected to
make the sum of the oxidation numbers
equal to the charge on the molecule.

(6) Reactions occur such that the net change of
oxidation numbers is zero. (This last rule is
really a result of the conservation of charge.)

Do not worry about the exceptions included
within parentheses in rules 3 and 4. Your atten-
tion will be called to them later when substances
involving them are considered.

EXERCISE 12-9

Show that the oxidation number of nitrogen is
+ 5 in each of the two species N03" and N206.

BALANCING REACTIONS

Just as before, the first step in balancing a reac-
tion must be to decide the products. Again, ex-
periment provides the answer. Let us reconsider
one of the same examples we balanced previously
by the half-reaction method. For these we al-
ready know the products.

In the second example of Section 12-3.1, we
find H2S gas reacts with Mn04~ to give solid
sulfur and Mn+2:

MntV + H£(g) gives S(s) + Mn« (702)

First, we assign oxidation numbers to each ele-
ment, using rules 1-5. We find

Mn04" + HS(g) S(s) + Mn«

+7 -2 0 +2

with changes,

Oxidation
number

for manganese +7
for sulfur

-5

-2 — *- 0

+ 2

If the gain in oxidation number by sulfur is to
equal the loss by manganese, then five atoms of
sulfur must react with two atoms of manganese:

2Mn04" + 5H$(g) gives 5S(s) + 2Mn+2

(not balanced) (103)

2(+7)

2(-5)=-10

+ 2(+2)

5(-2)

5( + 2) = +10

^5(0)

Now we proceed to ensure conservation of oxy-
gen atoms. There are eight oxygen atoms on the
left in (103), hence we must add eight molecules
of H20 to the right. (The reaction occurs in
aqueous solution, so there is plenty of H20.)

2Mn04" + 5H2S(g) gives

5S(s) + 2Mn+2 + 8H20 (not balanced) (104)

Next we must ensure conservation of hydrogen
atoms. On the left, there are 10 hydrogen atoms
(in 5H2S) and on the right 16 (in 8H20). In
aqueous solutions (in neutral or acidic solution)
we assume that these six hydrogen atoms needed
on the left are provided by H+:

2Mn04~ + 5HS(g) + 6H+ — >-

5S(s) + 2Mn+2 + 8H20 (95), (105)

The equation is balanced now but experience
dictates that a check should always be made on
the basis of charge balance:

220

OXIDATION-REDUCTION REACTIONS I CHAP. 12

2(-l) + 5(0) + 6(+l) 5(0) + 2(+2) + 8(0) Of course, the oxidation number method gives

— 2 +6 +4 the same balanced equation as the half-reaction

-|-4 = -|-4 method.

12-4 ELECTROLYSIS

So far in this chapter we have dealt with reac-
tions that proceed spontaneously. But the same
ideas and names are applied to reactions that we
force to take place, against their natural tend-
ency, by supplying energy with an externally
applied electric current. Such a process is termed
electrolysis or "separation by electricity."

We have dealt with electrolysis before — every
time we discussed or measured the electrical con-
ductivity of an electrolyte solution. To see this,
let's consider the processes that occur when we
cause electric charge to pass through an aqueous
solution of hydrogen iodide.

A distinguishing property of ionic solutions is
electrical conductivity, just as it is a distinguish-
ing property for metals, but the current-carry-
ing mechanism differs. Electric charge moves
through a metal wire, we believe, by means of

electron movement. Electrons flow through the
wire without changing the metal chemically.
In contrast, the movement of electric charge
through an aqueous solution of an electrolyte
causes significant chemical changes.

Figure 12-7 shows, on the right, the behavior
of an aqueous hydrogen iodide solution during
conduction. The two carbon rods are connected
by wires to the terminals of a 2 volt battery.
Electrons flow from the battery through the left
carbon rod, entering the solution. An equal num-
ber of electrons leave the solution through the
right carbon rod to return to the battery. The
hydrogen ion, H+, has the ability to accept an

Fig. 12-7. A schematic view of electrolytic
conduction.

QUESTIONS AND PROBLEMS

221

electron at the left electrode, where electrons are
in excess. The H+ ion is changed chemically to a
neutral atom. An iodide ion, I-, has one excess
electron that can be released at the right elec-
trode, where electrons are in deficiency. The I-
ion changes chemically to a neutral atom. The
net result of these two occurrences is the process
we call electrolysis.

Let us sum up the process during the move-
ment of one electron through the entire circuit
shown on the left in Figure 12-7.

(1) An electron on the left carbon rod is gone;
another electron has shown up at the right
carbon rod.

charge. This drift- of the ions through the
solution, positive ions in one direction and
negative ions in the other, explains the con-
duction in aqueous solutions.
(4) The battery performs work in forcing current
to flow through the solution and in causing
chemical changes to occur that would not
proceed spontaneously.

The net reaction is
2H+ + 21"

H,fej + I2

006)

Reaction (106) is just an oxidation-reduction re-
action and it is readily separated into the two
half-reactions

2H+ + 2e~ = H,(g)

21- = \,(g) + 2e~

E° = 0.000 volt
E° = -0.53 volt

Overall reaction 2H+ + 21" = H2(g) + h(g) E° = -0.53 volt

(106), (107)

(2) One W+(aq) ion and one l~(aq) ion are
gone; one H atom and one I atom have been
formed.

(3) As this process continues to take place, the
H+ ions in the solution at the left tend to be
used up and the same occurs for I" ions in
the solution at the right. Since only positive
ions are used up at the left, the remaining
negative ions are electrically repelled from
this region and are attracted toward the
solution at the right where positive ions are
plentiful. Here, negative ions are used up, so
the remaining positive ions are repelled and
they are attracted toward the left where nega-
tive ions are plentiful. Iodide ions, I~ move
from left to right through the solution,
carrying negative charge. At the same time,
hydrogen ions, H+, move from right to
left through the solution, carrying positive

The negative value of E° = —0.53 volt tells us
that the reaction will not occur spontaneously as
written. This voltage tells us further that elec-
trolysis will occur only if a cell with a voltage
exceeding —0.53 volt is placed in the external
circuit so as to oppose the voltage generated by
the cell itself.

EXERCISE 12-10

From Appendix 3, estimate the minimum voltage
required to cause electrolysis of 1 M HC1, form-
ing H-2(g) and 02(g), each at 1 atmosphere pres-
sure. Show that at this voltage electrolysis to
produce H2(g) and C\2(g) will not occur.

QUESTIONS AND PROBLEMS

1. One method of obtaining copper metal is to let 2. (a) If a neutral atom becomes positively charged,

a solution containing Cu"*"2 ions trickle over scrap
iron. Write the equations for the two half-
reactions involved. Assume the iron becomes
Fe~2. Indicate in which half-reaction oxidation
is taking place.

has it been oxidized or reduced? Write a
general equation using M for the neutral
atom.

(b) If an ion X'1 acquires a —2 charge, has it

222

OXIDATION-REDUCTION REACTIONS I CHAP. 12

been oxidized or reduced? Write a general
equation.

3. Aluminum metal reacts with aqueous acidic
solutions to liberate hydrogen gas. Write the
two half-reactions and the net ionic reaction.

4. When copper is placed in concentrated nitric
acid, vigorous bubbling takes place as a brown
gas is evolved. The copper disappears and the
solution changes from colorless to a greenish-
blue. The brown gas is nitrogen dioxide, N02,
and the solution's color is due to the formation
of cupric ion, Cu+2. Using half-reactions from
Appendix 3, write the net ionic equation for this
reaction.

5. Nickel metal reacts with cupric ions, Cu+2, but
not with zinc ions, Zn+2; magnesium metal does
react with Zn+2. In each case of reaction, ions
of +2 charge are formed. Use these data to
expand the table of reactions on p. 206.

6. In acid solution the following are true : H2S will
react with oxygen to give H20 and sulfur. H2S
will not react in the corresponding reaction with
selenium or tellurium. H2Se will react with sulfur
giving H2S and selenium but it will not react
with tellurium. Arrange the hydrides of column
VI, H:0, H2S, H2Se, and H2Te, in order of their
tendency to lose electrons to form the elements,
02, S, Se, and Te.

7. If you wish to replate a silver spoon, would you
make it the anode or cathode in a cell ? Use half-
reactions in your explanation. How many moles
of electrons are needed to plate out 1.0 gram
of Ag?

8. Figure 12-5 shows electrons leaving the Cu(s)
and going to the Ag(s). Experimentally, both
half-cells are found to be electrically neutral
before current flows and to remain so as the cell
operates. Explain this.

9. In the electrolysis of aqueous cupric bromide,
CuBr2, 0.500 gram of copper is deposited at one
electrode. How many grams of bromine are
formed at the other electrode? Write the anode
and cathode half-reactions.

Answer. 1.26 grams of Br2(l)

10. Complete the following equations. Determine
the net potential of such a cell and decide
whether reaction can occur.

(a) Zn + Ag+ — ■*-

(b) Cu + Ag+ — >-

(d) Hg + H+ — ►-

11. For each of the following,

(i) write the half-reactions ;
(ii) determine the net reaction ;
(iii) predict whether the reaction can occur giv-
ing the basis for your prediction :

(a) Mg(s) + Sn+2 — >-

(b) Mn(s) + Cs+ — >-

(d) Zn(s) + Fe+2 — ■+-

(e) Fe(s) + Fe+3 — >-

12. A half-cell consisting of a palladium rod dipping
into a 1 M Pd(N03)2 solution is connected with
a standard hydrogen half-cell. The cell voltage
is 0.99 volt and the platinum electrode in the
hydrogen half-cell is the anode. Determine E°
for the reaction

Pd+2 + 2e~

13. Suppose chemists had chosen to call the 2I~ — »-
I2 + 2e~ half-cell potential zero.

(a) What would be E° for Na — )- Na+ + e~l

(b) How much would the net potential for the
reaction 2Na + I2 — >- 2Na+ + 2I~ change?

14. If a piece of copper metal is dipped into a solu-
tion containing Cr+3 ions, what will happen?
Explain, using E°s.

15. What would happen if an aluminum spoon is
used to stir an Fe(N03)2 solution? What would
happen if an iron spoon is used to stir an A1C13
solution?

16. Can 1 M Fe2(S04)3 solution be stored in a con-
tainer made of nickel metal? Explain your an-
swer.

17. Suppose water is added to each of the beakers
containing copper sulfate in the two electro-
chemical cells shown in Figure 12-4 (p. 204).
What change will occur in the voltage in each
cell? Explain.

18. Determine the oxidation numbers of carbon in
the compounds carbon monoxide, CO, carbon
dioxide, C02, and in diamond.

QUESTIONS AND PROBLEMS

223

19. Determine the oxidation number of uranium in
each of the known compounds: U03) U308)
U2Ofi, U02, UO, K2U04, Mg2U207.

20. By use of half-reactions, give a balanced equa-
tion for each of the following reactions:

(a) H202 + I" + H+

gives H20 + I2

(b) Cr.,CV2 + Fe+2 + H+

gives Cr+3 + Fe+3 + H20

gives Cu+2 + NO + H20

(d) MnOr + Sn+2 + H+

gives Mn+2 + Sn+4 + H20

21. By use of oxidation numbers, give a balanced
equation for each of the following reactions:

(a) HBr + H2S04 gives S02 + Br2 + H20

(b) N03- + CI- + H+

gives NO + Cl2 + H.O

gives Zn+2 + N02 + H.O

(d) BrO" gives Br~ + Br03"

22. Use oxidation numbers to balance the reaction
between ferrous ion, Fe+2, and permanganate
ion, MnO^~, in acid solution to produce ferric
ion, Fe+3, and manganous ion, Mn+2.

23. Show the arbitrariness of oxidation numbers by
balancing the reaction discussed in Problem 22
with the assumption that the oxidation number
of manganese in Mn04~ is +2. Compare with
the result obtained in Problem 22.

24. In order to make NafsJ and C\2(g), an electric
current is passed through NaCl(l). What does
the energy supplied to this reaction do?

CHAPTER

Chemical
Calculations

The sceptical chemist • • • draws conclusions regarding chemical ma-
terials • • • chiefly on the basis of quantitative chemical analysis
which • • • is the touchstone of all chemical hypothesis.

G. T. MORGAN, 1930

Chemistry is a quantitative science. This means
that a chemist wishes to know more than the
qualitative fact that a reaction occurs. He must
answer questions beginning "How much . . .?"
The quantities may be expressed in grams, vol-
umes, concentrations, percentage composition,
or a host of other practical units. Ultimately,
however, the understanding of chemistry re-
quires that amounts be related quantitatively to
balanced chemical reactions. The study of the
quantitative relationships implied by a chemical
reaction is called stoichiometry.

Stoichiometric calculations are based upon
two assumptions. First, we assume that only a
single reaction need be considered to describe the
chemical changes occurring. Second, we assume
that the reaction is complete. For example, con-
sider the question, How much iron is produced
per mole of Fe203 reacted with aluminum in the
following reaction?

2Al(.sJ + Fe203(sj

Al03(s) + 2Fe(s) (/)

We base our calculations upon the two assump-
tions above, whether they are stated or not. First,
we assume this is the only reaction that occurs.
Reaction (2), for example, is assumed to be un-
important.

2AlfsJ + 3Fe203fsj q=^r 6FeO(s) + Al203(sj (2)

Sometimes such an assumption is based upon
experience, sometimes upon hope. Furthermore,
it is assumed that a mole of Fe203 reacts com-
pletely according to reaction (/). None of the
Fe203 remains unreacted at the finish, either be-
cause of equilibrium, because of mechanical
losses of some sort, or because insufficient alumi-
num was added. In practice these conditions are
sometimes difficult to obtain.

If these assumptions are valid, however, stoi-
chiometric calculations provide a reliable basis
for quantitative predictions. It is important to be
able to make these calculations with ease. Fortu-
nately, they all can be made with a single pattern
based upon the mole concept.

224

SEC. 13-1 I A PATTERN FOR STOICHIOMETRIC CALCULATIONS

225

13-1 A PATTERN FOR STOICHIOMETRIC CALCULATIONS

The equation for a chemical reaction speaks in
terms of molecules or of moles. It contains the
basis for stoichiometric calculations. However,
in the laboratory a chemist measures amounts in
such units as grams and milliliters. The first step
in any quantitative calculation, then, is to con-
vert the measured amounts to moles. In mole
units, the balanced reaction connects quantities
of reactants and products. Finally, the result is
expressed in the desired units (which may not
necessarily be the same as the original units).

Let's put this down schematically. Suppose
two substances, A and B, combine according to
a known reaction. We wish to know how much
B will react with (or, be produced from) a meas-
ured quantity of A. The solution to this typical
problem of stoichiometry consists of three steps.

We shall apply this scheme to a series of types
of calculations to show its general applicability.
The calculations are all connected with the man-
ufacture of sulfuric acid, H2S04, one of the most
important commercial chemicals.

Step I.

Amount of A in
measured units

Convert to moles of A
Mol wt of A

Moles of A

(3)

Step II.

Moles of A

Convert to moles of B
Balanced reaction

Moles of B

(4)

Step III.

Moles of B

Convert to amount of B
Mol wt of B

Amount of B in
desired units

(5)

13-2 THE MANUFACTURE OF SULFURIC ACID

Many millions of tons of sulfuric acid, H2S04,
are produced every year. Its uses are so wide that
the amount consumed per year by a country can
be taken as a crude index of the technological
development of that country. Two manufactur-
ing processes have widespread industrial im-
portance and both will be described. These
processes are so highly perfected that the cost
of this useful chemical is only about S22.00 per
ton!

EXERCISE 13-1

If H2S04 is purchased at a price of $22.00 per
ton, how many moles are obtained for a penny?
(Note: 1 pound = 453.6 grams.)

The chemical reactions appear simple. They
begin with pure sulfur (which occurs in natural
deposits in the elemental state). First, sulfur is
burned to give gaseous sulfur dioxide, S02. Next,
the S02 is further oxidized, catalytically, to sulfur
trioxide, SO.t. Finally, addition of water forms
sulfuric acid. The reactions are:

8so,o; + 40/gj

8SO,teJ + 8H,0(IJ

8SO/*J

(<5)

=± 8so,oj

(catalyst needed) (7)

8H>S04(7J

(8)

Overall reaction

Ss(s)+ I202fgj + 8H,.0(/J

8H.S04(7J (9)

Now let's investigate some of the quantitative
questions that are connected with this important
process.

226

CHEMICAL CALCULATIONS I CHAP. 13

13-2.1 Weight-Weight Calculations

A shovelful of sulfur containing 1 .00 kg is placed
in the hopper to be converted to sulfuric acid.
What weight of H2SOA will be formed?

This practical question is of a familiar form.
We wish to calculate the weight of product re-
sulting from a specified weight of reactant. Our
calculational pattern is applicable.

First, a reaction must be assumed. According
to the intent of the process, the overall reaction
is (9):

Srfs) + I202(g) + 8H2Of/j

8H2SO//j (9,70)

Our calculation proceeds on the assumption that
equation (70) is the only reaction that occurs and
that the entire kilogram of sulfur is consumed
in it.

Step I. We must convert grams of sulfur to
moles. The molecular weight of sulfur is
needed.

Mol wt Sg = 8 X (atomic wt of sulfur)

= 8 X (32.1) = 256.8 g mole (77)

Now the number of moles of S8 in 1 .00 kg of
sulfur is

Moles S = (wt sulfur) = (1-00 X 10»g)
8 (mol wt) (256.8 g mole)

= 3.89 moles

(12)

Step II. The next step is to decide how many
moles of H2SO4 can be produced from 3.89
moles of S8. The balanced reaction (70) tells
us that

one mole of S8 forms eight moles H2SO4

hence

3.89 moles of S8 form 8 X (3.89) moles H2S04
3.89 moles of S8 form 37.7 moles H2SOA (13)

Step III. The amount of sulfuric acid formed is
31.1 moles. How much does this weigh?

Wt H2S04 = (moles H2S04Xmol wt of H2S04)
= (31.1 moles)(98.1 g/mole)
= 3051 g

Wt H,S04 = 3.05 kg

Answer. 1.00 kg of sulfur produces 3.05 kg H2SO4
by reaction (10). (14)

13-2.2 Weight-Gas Volume Calculations

The first step in the manufacture of H2S04 is to
burn sulfur to sulfur dioxide. Sulfur burns spon-
taneously in air, liberating heat.

S*(s) + 802(g) z^± SSOJg)

AH = -70.96 kcalAnole S02 (5), (75)

An important economic feature of the modern
processes is the utilization of this heat in another
step in which heat is absorbed.

Of course, a plant designer must anticipate the
weights and volumes of the constituents at each
stage of the process. Hence, he must be able to
answer such a question as, "What weight of sulfur
will burn to produce 100,000 liters of pure SO2 at
500° C and one atmosphere pressure ?"

Again, we must assume reaction (75) can be
carried out exclusively and completely. This time
the calculation begins with a specified amount of
a product and we wish to calculate the corre-
sponding amount of a reactant. Note that we are
immediately confronted with a question of sig-
nificant figures. How many significant figures are
intended in the volume, 100,000 liters? In the
absence of other information, let's let common
sense dictate. Would it be of value to make the
calculation to six significant figures? Undoubt-
edly not. Two, or at most three, significant fig-
ures will probably suffice for the purposes of the
plant designer. Other conditions, such as the
weight of sulfur and the temperature, can't be
easily controlled to more than this accuracy.
Let's carry three significant figures to be sure we
have enough. The volume of S02 is, then,
1.00 X 105 liters.*

Step I. We must convert the specified volume of
S02 into moles. A convenient way to do this
is to calculate the volume this gas would oc-
cupy under conditions at which we know the
volume occupied by one mole of gas. For ex-
ample, we know that one mole of gas occupies
22.4 liters at 0°C and one atmosphere pressure.
Increasing the temperature of a gas at constant

* This volume, 1.00 X 105 liters, is about the volume
of a small room — a practical dimension for a reaction
chamber.

SEC. 13-2 I THE MANUFACTURE OF SULFURIC ACID

227

pressure increases the volume in proportion to
the absolute temperature:

0°C = 273°K
500°C = 273 + 500 = 773°K
1.00 mole of gas occupies 22.4 liters

at 273°K, 1 atm
1.00 mole of gas occupies 22.4 X m liters

at 773°K, 1 atm
1.00 mole of gas occupies 63.4 liters
at 773°K, 1 atm

63.4 liters S02 at 773°K, 1 atm contain
1.00 mole S02

hence
1.00 X 10s liters contain

1'°° * 10& X 1.00 mole S02
63.4

1.00 X 105 liters contain

1.58 X 103 moles S02 at 773°K, 1 atm (16)

Step II. Next we decide how many moles of S8
are needed to produce 1.58 X 103 moles of
S02. The balanced reaction (75) indicates

8 moles S02 are produced from 1 mole S8
1 mole S02 is produced from £ mole S8
1.58 X 103 moles S02 are produced from
Kl-58 X 103) moles S8

7.55 X 10s moles S02 are produced from
198 moles Si (77)

Step III. How much do 198 moles of S8 weigh?

Wt S8 = (moles S8)(mol wt S8)

= (198 moles)(256.8 g/mole)
Wt S8 = 50.8 X 103 g

Answer. 1.00 X 705 liters of SOi are produced
from 50.8 kilograms of sulfur by re-
action (75). (75)

13-2.3 Gas Volume-Gas Volume Calculations

After sulfur dioxide is produced by combustion
of sulfur, further oxidation is needed in the
manufacture of H2S04. The reaction, producing
sulfur trioxide, SO3, is exothermic; heat is re-
leased:

SO,(g) + hO,(g) ^= SCVgj

AH = -23.5 kcal/mole S03 (7), (79)

Yet the reaction is quite slow, even at high tem-
peratures. Evidently the rate is controlled by a
high activation energy. In fact, the practical use
of reaction (79) depends upon the presence of a
catalyst to provide a reaction path with a lower
activation energy. The two important commer-
cial methods for manufacture of H2S04 differ
principally in the choice of catalyst for this step.

The older process is called the lead chamber
process. It uses a mixture of gaseous oxides of
nitrogen — nitric oxide, NO, and nitrogen diox-
ide, N02 — as the catalyst. This process has been
in use and under development for over 200 years.
It is named after the large room-like chambers
lined with lead in which the gaseous reactions
are carried out. The lead walls react with the
acid and become coated with an inert protective
coating of lead sulfate.

The newer process uses a solid catalyst for
reaction (79). Either finely divided platinum or
vanadium pentoxide, V2Os, is effective. Because
catalysis occurs where the gas contacts the sur-
face of the catalyst, this process is called the
contact process.

EXERCISE 13-2

Reaction (79) is carried out at a high temperature
(about 500°C in the contact process). How does
temperature affect equilibrium, according to Le
Chatelier's Principle? In view of your answer,
propose an explanation of why the temperature
is kept high.

Reaction (79) requires the reaction of oxygen
from air and sulfur dioxide. What volume of air,
at 500°C and one atmosphere pressure, is needed
to react with the 1.00 X 70s liters of S02 produced
from 50.8 kilograms of sulfur?

Step I. In considering a chemical reaction be-
tween gases, we can apply Avogadro's Hy-
pothesis: Equal volumes of gases contain
equal numbers of molecules (at the same pres-
sure and temperature). The volume of the S02,
1.00 X 10s liters, is already a measure of the
number of moles of S02.

228

CHEMICAL CALCULATIONS | CHAP. 13

H2S0+ +
catalyst (NV.N02) Steam

Lead- lined chamber-

Air

NO+NOz +
H20

Catalyst
recovery

H1SQ4 prodi

FUNCTION:

Burn sulfur

S6 + 802 — 8502

Catalyst (NO,
N02) added

Fig. 75-7. The "lead chamber" process for H:SO, man-
ufacture.

Step II. By reaction (79), 1 mole of S02 reacts
with \ mole of 02. By Avogadro's Hypothesis,
1 liter of S02 reacts with \ liter of 02 (if they
are at the same temperature and pressure).
Hence 1.00 X 105 liters of S02 react with
1(1.00 X 105) liters of 02 if both S02 and 02
are pure and measured at 500°C and one at-
mosphere.

Answer. We need 0.500 X 10h liters of pure 0> at
500°C and one atmosphere to oxidize
LOO X 105 liters of S02. (20)

Step III. Now we must convert to the desired
units. We need the number of moles of oxygen
that are present in 0.500 X 105 liters (at 500°C,
1 atm), but we wish to use air instead of pure
oxygen. If air contains about 20% oxygen (by
volume), then it takes 5 liters of air (at 500°C,

Ca ta ly tic oxida tion
of S02j reaction yvith H20

S02 + !io2 — -sos

S03 + H20 H2S04

1 atm) to provide the amount of oxygen in

1 liter of pure oxygen (at 500°C, 1 atm). If

0.500 X 105 liters of pure 02 are needed, then

5(0.500 X 105) = 2.50 X 105 liters of air are

needed.

Answer. 2.50 X 10h liters of air (at 500° C, 1 atm)
react with 1.00 X 10h liters of S02 (at
500°C, 1 atm). (21)

13-2.4 Weight-Liquid Volume Calculations

The last step in the preparation of commercial
sulfuric acid is to allow the sulfur trioxide to
react with steam:

S03(g) + H20(g) +± H,SCVZJ (9), (22)

This results in a concentrated sulfuric acid solu-
tion that contains 98% H2S04. It is a viscous,
colorless liquid. When it is mixed with water, so
much heat is liberated that the operation must
be carried out very cautiously. The sulfuric acid

SEC. 13-2 | THE MANUFACTURE OF SULFURIC ACID

229

is slowly poured into the water, not the reverse.
The density of this concentrated sulfuric acid
solution (98%) is 1.84 grams/ml and its con-
centration is 18.3 M.

One of the important uses of sulfuric acid is
that of an oxidizing agent. For example, when
heated, it will even dissolve carbon. The reac-
tion is

C + 2H,.S04 +± CO. + 2H20 + 2S02 (23)

EXERCISE 13-3

Verify that reaction (23) is an oxidation-reduc-
tion reaction and that the oxidation number
change of carbon is balanced by the oxidation
number change of the sulfur.

How many liters of concentrated sulfuric acid
would be consumed in reaction (23) to oxidize
J. 00 kg of carbon ?

Fig. 13-2. The "contact" process for HjSOt

manufacture.

Step I. We wish to oxidize 1.00 kg of carbon.
The number of moles of carbon is

K. , . (wt C) 1.00 X 10>g

Moles carbon = r1 £- = — — — ; — f

(at wt C) 12.01 g/mole

= 83.2 moles

Moles carbon = 83.2 (24)

Step II. Now we must decide how many moles
of sulfuric acid are needed. Reaction (23)
shows that

1 mole C reacts with 2 moles H2S04

83.2 moles C react with 2(83.2) moles H2S04

83.2 moles C react with 166 moles H£0A (25)

Step III. We need 166 moles of H2S04. What
volume of concentrated sulfuric acid (18.3 M)
is needed?

18.3 moles H2S04 are present in 1.00 liter
98 % H2S04

166 moles H2S04 are present in — — liters

98 % H2S04
166 moles H>SOA are present in 9.07 liters

A SQ**

S02 + 02+ impurities Jy 02+ H2

H2S04

S02

Air

FUNCTION : Burn, sulfur. Remove impurities

Ss + BOj — 8S02 that might

"poison." catalyst

H2504 + H20

Dry
S02 - 02
mixture

H2S0^-S05 .
"fuming" sulfuric acid

Catalyticatly Absorb S03(y)
burn. S02 , in H2SO+

S02 +%o2~so3

230

CHEMICAL CALCULATIONS I CHAP. 13

Answer. 1.00 kg of carbon is oxidized by 9.07
liters of concentrated HtSO* in reac-
tion (23). (26)

13-2.5 Liquid Volume-Volume Calculations

A second major use of sulfuric acid of commerce
is in reactions with bases. In laboratory use it is
diluted to a much lower concentration and can
be used as a standard acid. A typical problem
would be the titration of a base solution of
unknown concentration using a sulfuric acid
solution of known concentration. For example,
"What is the concentration of a sodium hydroxide
solution if 25.43 ml of the NaOH solution just
reacts with 18.51 ml of 0.1250 M H^SOi (to pro-
duce a neutral solution) ?"

Step I. We are given the concentration and vol-
ume of H2S04 solution. How many moles of
H2SO4 are present?

Moles H0SO4
present = (volume)(concentration)

= (18.51 X 10-3 liter)(0.1250 NT)
Moles H2S04 = 2.314 X 10-3 mole

Step II. We are interested in the reaction be-
tween H+(aq) and OW~(aq). Sulfuric acid
gives 2 moles of H+(aq) per mole of H2S04
dissolved in water.

Hence

2H+(aqj + SOr*(aq) (27)

Moles H~(aq) = 2(moles H,S04)
= 2(2.314 X 10-')
Moles H+(aq) = 4.628 X 10~3 (28)

Now we are concerned with the acid-base
reaction

H+(aq) + OH-(aq)

By reaction (29),

H20

(29)

one mole of OH~(aq) ion reacts with one
mole of H+(aq) ion. 4.628 X 10~3 mole of
OH~(aq) ion reacts with 4.628 X 10~3 mole oj
H+(aq) ion. (30)

Step III. We can now calculate the concentra-
tion of hydroxide ion. We now know that
4.628 X 10-3 mole of OR-(aq) is contained in
25.43 ml of sodium hydroxide solution:

Concentration
hydroxide ion

(moles OH~ ion)
(volume)

roH_1 = (4.628 X 10~3 mole)
L J (25.43 X 10-3 liter)

[OH-] = 0.1820 M

Answer. The sodium hydroxide solution has a con-
centration of 0.1820 M. (31)

QUESTIONS AND PROBLEMS

1 . In Experiment 7, would the ratio between moles
of copper atoms used and moles of silver atoms
formed change if silver sulfate, Ag2S04, had been
used rather than silver nitrate, AgN03? Explain.

2. Although sodium carbonate is needed in the
manufacture of glass, very little is found in
nature. It is made using two very abundant
chemicals, calcium carbonate (marble) and so-
dium chloride (salt). The process involves many
steps, but the overall reaction is

CaC03 + 2NaCl — -»- Na2C03 + CaCl2

(a) How many grams of sodium chloride react
with 1.00 kg of calcium carbonate?

(b) How many grams of sodium carbonate are
produced?

3. Some catalysts used in gasoline manufacture
consist of finely divided platinum supported on
an inert solid. Suppose that the platinum is
formed by the high temperature reaction between
platinum dioxide, Pt02, and hydrogen gas to
form platinum metal and water.

(a) What is the oxidation number of platinum
in platinum dioxide?

(b) Is hydrogen an oxidizing or reducing agent
in this reaction?

QUESTIONS AND PROBLEMS

231

(d) How many moles of water are produced
along with 1.0 gram of Pt?

(e) How many grams of water are produced
along with 1.0 gram of Pt?

Answer, (e) 0.18 gram of H:0

4. Hydrazine, NjH4, and hydrogen peroxide, H.Oj,
are used together as a rocket fuel. The products
are N2 and H^O. How many grams of hydrogen
peroxide are needed per 1.00 X 103 grams of
hydrazine carried by a rocket ?

Answer. 2.12 X 103 grams of H02

5. Iodine is recovered from iodates in Chile salt-
peter by the reaction

HS03- + IO3- gives I, + SO4"2 + H+ + HO

(a) How many grams of sodium iodate, NaI03,
react with 1.00 mole of KHSO3?

(b) How many grams of iodine, I2, are pro-
duced ?

6. The hourly energy requirements of an astronaut
can be satisfied by the energy released when 34
grams of sucrose are "burned" in his body. How
many grams of oxygen would need to be carried
in a space capsule to meet this requirement?

sucrose + oxygen gives carbon dioxide + water
CnH«Ou+ 02 gives C02 + H.O

7. The chlorine used to purify your drinking water
was possibly made by electrolyzing molten
NaCl to produce liquid sodium and gaseous
chlorine.

(a) How many grams of sodium chloride are
needed to produce 355 grams of chlorine
gas?

(b) What volume would this gas occupy at STP?

8. A reaction involved in the production of iron
from iron ore is

Fe203 + CO gives Fe + C02

AH = -4.3 kcal/mole Fe2Os

(a) How many grams of CO must react to re-
lease 13 kcal?

(b) How many liters of CO(STP) are needed to
produce 1.0 kg of Fe?

9. More C8Hi8, a hydrocarbon that is useful in
gasoline, can be obtained from petroleum if this
reaction takes place:

Ci6H32(g) + mjg) — *- 2C8H18fgJ

(a) How many grams of C8Hi8 can be made
using 224 liters of H2 at STP?

(b) What pressure conditions favor production
ofQH^gJ?

Answer, (a) 1.14 X 103 g CgH18

10. How many liters of oxygen gas, at STP, will be
released by decomposing 14.9 grams of NaOCl
to produce 02(g) + C\~(aq) (as in Experiment
14a)?

11. A compound found in kerosene, a mixture of
hydrocarbons, is decane, Q0H22. A stove might
burn 1 .0 kg of kerosene per hour. Assume kero-
sene is Q0H22 and answer the following:

(a) How many liters (STP) of oxygen are needed
per hour?

(b) How many liters (STP) of carbon dioxide
are produced per hour?

12. How many grams of zinc metal are needed to
react with hydrochloric acid to produce enough
hydrogen gas to fill an 11.2 liter balloon at STP?
What would be the volume of this balloon at
27°C and 680 mm Hg pressure? How many
grams of zinc would be needed if sulfuric acid
were used ?

1 3. How many liters of air (STP) are needed to burn
2.2 liters (STP) of methane, CH4, gas in your
laboratory burner? How much heat is released?
The AH for combustion of CH4 is —210
kcal mole of CH4. Assume air is 20 % oxygen.

14. In the reaction

NHafeJ + 02(g) gives NO(g) + H20(g),
if 4.48 liters of ammonia gas measured at STP
are used, how many liters of oxygen measured
at STP will be needed to react with all the
ammonia?

Answer. 5.60 liters of 02 at STP

15. The following reaction is carried out with all gas
volumes measured at the same pressure and tem-
perature:

CMl0(g) + 02(g) gives C02(g) + H20(g)

(a) How many liters of oxygen are required to
produce 2.0 liters of C02?

(b) If 15 liters of oxygen are used, how many
liters of butane, C4Hi0, will be burned?

Answer, (a) 3.2 liters of 0>

232

CHEMICAL CALCULATIONS I CHAP. 13

16. What volume of Cl2 gas at 37°C and 753 mm
could be obtained from 58.4 liters of HC1, also
measured at 37°C and 753 mm, if the following
reaction could be carried effectively to comple-
tion?

UC\(g) + 02(g) gives H20(g) + C\2(g)

17. Suppose 105 liters of NH3 and 285 liters of 02
are allowed to react until the reaction :

NH3teJ + 02(g) gives H20(g) + NO/gj

is complete. The temperature and pressure are
maintained constant at 200°C and 0.30 atmos-
phere during all volume measurements. What
gas and what volume of it measured at the stated
conditions remains unreacted?

18. A 6 volt lead storage battery contains 700 grams
of pure H2S04( I) dissolved in water.

(a) How many grams of solid sodium carbonate,
Na2C03, would be needed to neutralize this
acid (giving C02 gas and H20) if it were
spilled ?

(b) How many liters of 2.0 M Na2COs solution
would be needed?

19. Nitric acid, HN03, is made by the process

3NQ2(g) + H2OfIJ ++ 2HN03([j + NOfe)

Commercial concentrated acid contains 68 % by
weight HN03 in water. The solution is 15 M.
How many liters of concentrated acid are needed
to react with 0.100 kg of copper metal?

Cu(s) + H+(aq) + NO^(aq) gives

Cxx+Haq) + N02(gj + H20(/;

Answer. 0.42 liter of HN03

20. How many grams of silver metal will react with
2.0 liters of 6.0 M HN03? The reaction is

Ag(s) + H+(aq) + N03_ (aq) gives

Ag+(aq) + NO(g) + H20(JJ

21. A measured volume, 10.00 liters, of the waste
process water from a cotton mill require 23.62
ml of 0.1000 M hydrochloric acid to produce a
neutral solution. What is the hydroxide ion con-
centration in the waste ?

22. What weight of silver chloride may be obtained
from 1 .0 liter of 1 .0 M AgN03, if 1 2 ml of 0. 1 5 M
NaCl are added?

23. How many milliliters of a 0.050 M KMn04 solu-
tion are required to oxidize 2.00 grams of FeS04
in a dilute acid solution?

Answer. 53 ml of KMn04 solution

CHAPTER

Why We Believe
in Atoms

From the time of Dalton • • • the history of the atom has been a march
of triumph. Wherever the concept of the atom was employed for the in-
terpretation of observational measurements, it supplied lucid explanation;
conversely, such success became overwhelming evidence for the existence
of the atom.

HANS REICHENBACH, 1951

In Chapter 2 you were introduced to atoms and
in Chapter 6 they were described in more detail.
You were told that the atom contains charged
particles, that it has a nucleus made up of neu-
trons and protons, and that the nucleus is sur-
rounded by electrons. The atom is incredibly
small but the nucleus is even smaller. But also
you were told that every theory (including the
atomic theory) should be thought about and
criticized — the evidence upon which it is based
should be examined and understood. It is one
thing to ask "Do we believe in atoms?" and
quite another to ask "Why do we believe in
atoms?" In this chapter we shall try to answer
this last, harder question.

Let's begin with an unpretentious example that
shows how we make such decisions in day-to-day
living.

A new tenant is told by his neighbor that the
garbage collector comes every Thursday, early in
the morning. Later, in answer to a question from

his wife about the same matter, the tenant says,
"I have been told there is a garbage collector and
that he comes early Thursday morning. We shall
see if this is true." The tenant, a scientist, ac-
cepts the statement of the neighbor (who has had
opportunity to make observations on the sub-
ject). However, he accepts it tentatively until he
himself knows the evidence for the conclusion.

After a few weeks, the new tenant has made a
number of observations consistent with the ex-
istence of a Thursday garbage collector. Most
important, the garbage does disappear every
Thursday morning. Second, he receives a bill
from the city once a month for municipal serv-
ices. And there are several supplementary ob-
servations that are consistent — often he is
awakened at 5:00 a.m. on Thursdays by a loud
banging and sounds of a truck. Occasionally the
banging is accompanied by gay whistling, some-
times by a dog's bark.

The tenant now has many reasons to believe

233

234

WHY WE BELIEVE IN ATOMS I CHAP. 14

in the existence of the garbage collector. Yet he
has never seen him. Being a curious man and a
scientist, he sets his alarm clock one Wednesday
night to ring at 5:00 a.m. Looking out the win-
dow Thursday morning, his first observation is
that it is surprisingly dark out and things are
difficult to see. Nevertheless, he discerns a
shadowy form pass by, a form that looks like a
man carrying a large object.

Seeing is believing! But which of these pieces
of evidence really constitutes "seeing" the gar-
bage collector? Which piece of evidence is the
basis for "believing" there is a garbage collector?
The answer is, all of the evidence taken together,
constitutes "seeing." And all of the evidence
taken together, furnishes the basis for accepting
the "garbage collector theory of garbage dis-
appearance." The direct vision of a shadowy
form at 5:00 a.m. would not constitute "seeing
a garbage collector" if the garbage didn't dis-
appear at that time. (The form might have been
the paper boy or the milkman.) Neither would
the garbage disappearance alone consist of "see-
ing" the garbage collector. (Perhaps a dog comes
by every Thursday and eats the garbage. Remem-
ber, a dog's bark was heard!) No, the tenant is
convinced there is a garbage collector because
the assumption is consistent with so many ob-
servations, and it is inconsistent with none. Other
possible explanations fit the observations too,
but not as well (the tenant has never heard a dog
whistle gaily). The garbage collector theory

passes the test of a good theory — it is useful in
explaining a large number of experimental ob-
servations. This was true even before the tenant
set eyes on the shadowy form at 5:00 a.m.

Yet we must agree, there are advantages to the
"direct vision" type of experiment. Often more
detailed information can be obtained this way.
Is the garbage collector tall? Does he have a
mustache? Could the garbage collector be a
woman? This type of information is less easily
obtained from other methods of observation. It
is worthwhile setting the alarm clock, even after
we have become convinced there is a garbage
collector.

At the beginning of this course you were a new
tenant. You were told that chemists believe in
atoms and you were asked to accept this proposal
tentatively until you yourself knew the evidence
for it. Since that time, we have used the atomic
theory continuously in our discussions of chemi-
cal phenomena. The atomic theory passes the
test of a good theory: it is useful in explaining a
large number of experimental observations. We
have become convinced there are atoms.

Now we are going to review the types of evi-
dence that form the basis for belief in the atomic
theory. We shall include a number of experi-
ments that are close, in concept, to the "direct
vision" type. These are particularly convincing
and they provide detailed information that is less
readily obtained in other ways.

14-1 CHEMICAL EVIDENCE FOR THE ATOMIC THEORY

Let us begin by looking again at the kinds of
evidence we already have for the existence of
atoms — the evidence from chemistry. We shall
consider, in turn, the definite composition of
compounds, the simple weight relations among
compounds, and the reacting volumes of gases.
Each behavior provides experimental support
for the atomic theory.

14-1.1 The Law of Definite Composition

Compounds are found to have definite composi-
tion, no matter how prepared. For example,

2.016 grams of hydrogen are found combined
with 16.00 grams of oxygen in the compound
water whether the water is prepared by burning
hydrogen in oxygen, by decomposing gaseous
nitrous acid, by heating barium chloride dihy-
drate, or by some other process:

H*(g) + hOJg) — >■ H2OfeJ

2.016 g hydrogen/16.00 g oxygen (/)

2HN02(gJ — >■ NOfgJ + N02(g) + H2OfgJ

2.016 g hydrogen/ 16.00 g oxygen (2)

SEC. 14-1 I CHEMICAL EVIDENCE FOR THE ATOMIC THEORY

235

/ molecule 0?

I mole 02

32 g

+
+

Z molecules H2
Z moles H2

1^" w

2 molecules HzO
2 moles H30

36 g

M,0

g oxygen _ 5£ __fl
g hydr-ogen~ -f 1

1 molecule Oz

1 molecule. H2

1 mole 02

I mole Hz

32 g 02

2 g Hz

1 molecule H202
1 mole H2Oz
34 g H2Oz

g oxygen _ 32 _ 16
g hydrogen- 2 ~ 1

^BaCl2 2H20]fsJ — >- \BaCUJs) + U20(g) Fig. 14-1. Simple multiple proportions of oxygen to

2.016 g hydrogen 16.00 g oxygen (5) hydrogen in H.0 and H4)»

The atomic theory provides a ready explana-
tion for the definite composition of chemical
compounds. It says that compounds are com-
posed of atoms, and every sample of a given
compound must contain the same relative num-
ber of atoms of each of its elements. Since the
atoms of each element have a characteristic
weight, the weight composition of a compound
is always the same.* Thus, the definite composi-
tion of compounds provides experimental sup-
port for the atomic theory.

14-1.2 The Law off Simple Multiple Proportions

In many cases, two elements brought together
under different conditions can form two or more
different compounds. In addition to water, hy-
drogen and oxygen can form a compound called
hydrogen peroxide, H202, in which the weight of

* This statement applies if the naturally occurring dis-
tribution of isotopes is hot disturbed.

hydrogen is ^ the weight of oxygen in the com-
pound. This weight ratio in water is £, exactly
twice as large. What a simple relationship! The
weight ratio is one to sixteen in one compound
of hydrogen and oxygen and one to eight in
another. Such simple numerical relationships are
always found among different compounds of a
set of elements. This is explained very clearly
within the atomic theory. Each molecule of hy-
drogen peroxide contains two atoms of hydrogen
and two atoms of oxygen. The ratio of the num-
ber of hydrogen atoms to oxygen atoms is
2/2 = 1. In contrast, a molecule of water con-
tains two atoms of hydrogen and only one atom
of oxygen. The ratio of hydrogen atoms to
oxygen atoms is 2/1 = 2. Since there are twice
as many hydrogen atoms per oxygen atom in
water as in hydrogen peroxide, of course the
weight ratio of hydrogen to oxygen in water is
twice that in hydrogen peroxide.

236

WHY WE BELIEVE IN ATOMS I CHAP. 14

In general, different compounds of the same
two elements have different atomic ratios. Since
these atomic ratios are always ratios of integers,
1/1, 1/2, 2/1, 2/3, etc., the weight ratios will be
simple multiples of each other. Thus the atomic
theory explains the observation that different
compounds of the same two elements have rela-
tive compositions by weight that are simple
multiples of each other.

This success of the atomic theory is not sur-
prising to a historian of science. The atomic
theory was first deduced from the laws of chemical
composition. In the first decade of the nineteenth
century, an English scientist named John Dalton
wondered why chemical compounds display such
simple weight relations. He proposed that per-
haps each element consists of discrete particles
and perhaps each compound is composed of
molecules that can be formed only by a unique
combination of these particles. Suddenly many
facts of chemistry became understandable in
terms of this proposal. The continued success of
the atomic theory in correlating a multitude of
new observations accounts for its survival. To-
day, many other types of evidence can be cited
to support the atomic postulate, but the laws of
chemical composition still provide the corner-
stone for our belief in this theory of the structure
of matter.

EXERCISE 14-1

Two compounds are known that contain only
nitrogen and fluorine. Careful analysis shows that
23.67 grams of compound I contain 19.00 grams
of fluorine and that 26.00 grams of compound
II contain 19.00 grams of fluorine.

(a) For each compound, calculate the weight of
nitrogen combined with 19.00 grams of fluo-
rine.

(b) What is the ratio of the calculated weight of
nitrogen in compound II to that in I?

(c) Compound I is NF3. This compound has one
atom of nitrogen per three atoms of fluorine.
How many atoms of nitrogen are there per
three atoms of fluorine for each of the mo-
lecular formulas N2F2 and N2F4? Compare

these atom ratios to the weight ratio ob-
tained in part (b) and convince yourself that
compound II could have the formula N2F4
but not N2F2.

14-1.3 The Law of Combining Volumes

Gases are found to react in simple proportions
by volume, and the volume of any gaseous prod-
uct bears a whole-number ratio to that of any
gaseous reactant. Thus, two volumes of hydrogen
react with exactly one volume of oxygen to pro-
duce exactly two volumes of water vapor (all at
the same temperature and pressure). These in-
teger relationships naturally suggest a particle
model of matter and, with Avogadro's Hypothe-
sis, are readily explained on the basis of the
atomic theory.

Once again it is no surprise that the simple
integer volume ratios are readily explained with
the atomic theory. The atomic theory was de-
vised for this purpose, as is indicated in Chap-
ter 2.

To summarize, we find that the weight and
volume relations that are observed in chemical
changes provide an experimental foundation for
the atomic theory. All of contemporary chemical
thought is based upon the atomic model and,
hence, every successful chemical interpretation
strengthens our belief in the usefulness of this
theory.

14-1.4 Chemical Evidence for the Electrical
Nature of Atoms

You have been told that the atomic nucleus bears
a positive charge and is surrounded by a number
of negatively charged particles called electrons.
Also, the nucleus is supposed to contain most of
the mass of the atom and to be made of protons
and neutrons, each of which has nearly two
thousand times the mass of the electron. How
do we know that atoms are built this way? How
do we know that there is such a particle as an
electron? Again, weight relations associated with
chemical reactions provide key evidence.
In Chapter 12 we discussed the operation of

SEC. 14-1 I CHEMICAL EVIDENCE FOR THE ATOMIC THEORY

237

Hs(l)

Ha +z(aq)

Ma CI (l) tn molten CaCl,

T IT-I'I'I1'1!

Al203(l) in molten Na3ALF6

6 . 03 g Hg produced

6.03

— .0300 mole

2 01

1. 38 g Ma produced

1.38
3 Q = 0600 mole

.538 g A.I produced
.538

Z6.9

.0200 mole

.0300

No.

0600

.0200

an electrochemical cell. We successfully inter-
preted the chemical changes brought about by
the movement of electric charge in terms of the
atomic theory. To understand the full impact of
these experiments on the development of the
atomic theory, we must turn back the scientific
clock to the views held in the nineteenth century.
When Michael Faraday first performed his elec-
trolysis experiments (in the early 1830's), the
atomic theory had been proposed but no one had
yet suggested the existence of electrons. There
was no reason to suspect that electricity con-
sisted of individual units. Faraday observed that
the quantity of electricity necessary to deposit a
given weight of an element from solutions of its
different compounds was always equal to a con-
stant, or some simple multiple of this constant.
For example, the amount of electricity that will
deposit 6.03 grams of metallic mercury from a
solution of mercuric perchlorate, Hg(CIO^)2, will
deposit the same number of grams of mercury
from a solution of mercuric nitrate, Hg(N03)2.

Fig. 14-2. Weights of different elements deposited by

a given amount of electricity.

On the other hand, this same amount of elec-
tricity will deposit exactly twice as much mer-
cury, 2 X (6.03) =12.1 grams, from a solution
of mercurous perchlorate, Hg2(004)2. If we re-
state Faraday's experimental finding in terms of
the atomic theory, we see that the number of
atoms of mercury deposited by a certain quan-
tity of electricity is a constant or a simple
multiple of this constant. Apparently this certain
quantity of electricity can "count" atoms. A
simple interpretation is that there are "packages"
of electricity. During electrolysis, these "pack-
ages" are parcelled out, one to an atom, or two
to an atom, or three.

The second of Faraday's observations was that
the weights of different elements that were de-
posited by the same amount of electricity formed
simple whole-number ratios when divided by the
atomic weights of these elements. For example,

238

WHY WE BELIEVE IN ATOMS I CHAP. 14

suppose electric current is passed through the
three electrolysis cells pictured in Figure 14-2.
The two ammeters have the same reading, show-
ing that the current entering the cell at the right
is identical to that leaving the cell at the left.
Thus, the electric circuit guarantees that the same
amount of electricity passes through each of the
three cells.

In the first cell the net reaction is the produc-
tion of metallic mercury and gaseous oxygen
through electrolysis of aqueous mercuric nitrate:

2Hg+*(aq) + 2H20 — *-

2Hg(l) + 02(g) + 4U+(aq) (4)

After current has passed through the cell for a
definite time, the weight of the mercury produced
is found to be 6.03 grams.

In the second cell molten sodium chloride is
electrolyzed.* The net reaction is

NaClflJ —^ Naflj + \&jg) (5)

The same current that produced 6.03 grams of
mercury is found to produce 1.38 grams of
molten sodium.

The third cell represents another industrial
process, the electrolytic process for manufactur-
ing aluminum. Here, A1203 is electrolyzedf and
the net reaction in the cell is

A\203(l)

2Al(l) + %02(g)

(6~)

Here we find that the same current that produced
6.03 grams of mercury produces 0.538 gram of
aluminum.

Thus after the same amount of electricity is
passed through the three cells, the weights of
metals produced are found to be

6.03 g Hg

1.38 gNa
0.538 g Al

How are these weights related? Faraday realized

* In practice, calcium chloride must be added to such
a cell to lower the melting point of the salt mixture and,
even then, the temperature must be high (600°C). This is
the commercial method for manufacturing metallic so-
dium.

t This is the basis for the commercial manufacture of
aluminum. Another salt is added as solvent to lower the
melting point. A mixture of AUOj and NaiAlFe (cryolite)
. can be electrolyzed at 950°C

that simple numbers result in such a case if each
weight is divided by the appropriate atomic
weight:

T^tJ- " 5Sr^T = 00300 mole
at wt Hg 201 g/mole

wt Na 1.38 g nft,m ,

. . VT = rr^ — t-S-t- = 0.0600 mole
at wt Na 23.0 g/mole

wt Al

0.538 g

at wt Al 26.9 g/mole

= 0.0200 mole

These numbers are simply related to each
other as shown by the ratios:

0.0300

0.0600

0.0200

Within the atomic theory, this result means
that a certain amount of electricity will deposit
a fixed number of atoms, or some simple mul-
tiple of this number, whatever, the element. Thus,
in both of Faraday's experiments, we find that
an atom can carry only a fixed quantity of
charge, or some simple multiple of this quantity.
Therefore, electric charge comes in packages. An
atom can carry one package, two packages, pos-
sibly three packages, of charge, but not 1.5872
packages. Whatever the package of charge is, it
is the same for all atoms. The realization that
electric charge comes in packages led to the
proposal that electricity is composed of particles.
Since atoms carry electric charges, atoms must
contain these particles.

EXERCISE 14-2

As current is passed through the cells shown in
Figure 14-2, the oxygen produced in the first cell
is collected and its volume is compared with the
volume of chlorine produced in the center cell
(the volumes being compared at identical tem-
peratures and pressures). The volume of chlorine
is found to be exactly double that of oxygen.
Applying Avogadro's Hypothesis, explain how
this result shows that electricity can "count"
atoms.

sec. 14-2

SEEING PARTS OF ATOMS

239

14-2 "SEEING" PARTS OF ATOMS

Despite the convincing support for the atomic theory
provided by chemical evidence, there is intuitive appeal
to evidence that is closer to the "direct vision" type. From
such experiments comes a much more detailed view of
the atom and its make-up.

14-2.1 "Seeing" Electrons

The Faraday experiments were the original basis for the
suggestion that electricity consists of individual charges
called electrons. Other experiments involving the passage
of electricity through gases provide further evidence that
electrons do exist.

Consider the apparatus shown in Figure 14-3. A glass
tube is fitted with electrodes so that a potential difference
of 10,000 volts can be applied across a space filled with a
desired gas at various pressures. Suppose neon, for ex-
ample, is placed in the tube. With the voltage applied,
the gas will begin to conduct electricity when its pressure
is reduced to about 0.01 atmosphere. The tube then glows
with the familiar color of a "neon sign." If a different
gas is used, the color is different, but otherwise, the be-
havior is about the same. If the pressure is reduced still
further to about 10-6 atmosphere, the glow from the gas

To vacu.
pump

Fig. 14-3. An electric discbarge through a gas.

Fig. 14-4. An electric discharge tube, very low pres-
sure. Electrons travel from the negative
electrode to the positive electrode; some of
them pass through the triangular hole to
produce a triangular spot on the fluores-
cent screen.

voltage can be applied to plates A and P2 when switch S
is closed. When switch Si is open, the fluorescence appears
at position A, just as in Figure 14-4. When Si is closed,
however, the fluorescent spot moves to position B. Ap-
parently the fluorescent spot is caused by particles that
are attracted to the positively charged electrode, P2.
Therefore these particles must be negatively charged.
Light cannot be deflected in this way, hence the fluo-
rescent glow cannot be caused by light. Such experiments
with discharge tubes show that negatively charged par-
ticles exist. These particles are now known to be electrons.
This experiment does have some features of a "direct
vision" observation. First, the glowing spot is directly
visible. Second, it is easy to imagine an invisible stream
of particles hurtling through the triangular hole in the
electrode to crash against the fluorescent screen in a burst

disappears but there remains a fluorescent glow from the
glass walls of the tube. The apparatus shown in Figure
14-4 demonstrates that the fluorescent glow is either
caused by particles or by light rays that travel from the
negative electrode and past the positive electrode (in this
apparatus, through the triangular hole in the center of
the electrode). When the tube operates, a glowing area
appears on the glass wall at position A directly opposite
the hole in the positive electrode and just the same shape
as the hole. Because the glowing area is "shadowed" by
the positive electrode, it must be caused by rays that travel
in straight lines.

Figure 14-5 shows this same apparatus fitted with an
auxiliary pair of electrode plates, Pi and Pt. An electrical

Fig. 14-5. An electric discharge tube with deflection
electrodes.

240

WHY WE BELIEVE IN ATOMS I CHAP. 14

of light. Third, the experiment conveys detailed informa-
tion about these particles, information difficult to obtain
any other way. The electric charge on a particle is clearly
evident from the deflection experiments. Accurate meas-
urements of such deflections even lead to a measure of
the ratio of electron charge to electron mass.

Yet there is a real difference between this experiment
and the "direct vision" of the shadowy form of the gar-
bage collector. We don't see the electron directly; rather
we see a burst of light on the fluorescent screen. The light
isn't considered to be the electron: the burst of light is
caused by molecular damage to the screen— damage re-
sulting from the electron crash. A more apt comparison
from our analogy would be our seeing some footprints
in the garden. We assume the footprints are caused by
the garbage collector. Then from certain properties of the
footprints— size, depth, spacing— we form a detailed
image of his height, weight, stride. You will find that this
is typical of most of the experiments that might be called
"seeing" atoms and their components. We see their "foot-
prints"—bursts of light on a screen, marks on a photo-
graphic plate, discharges in a Geiger counter, etc. These
"footprints" substantiate in every way the atomic theory
and furnish detailed information on the nature of atoms

THE RATIO OF ELECTRON CHARGE
TO ELECTRON MASS, e/m

By the action of a magnetic field electrons may be made
to follow a curved path. Such experiments lead to a deter-
mination of the ratio of the electric charge of an electron
e, to its mass, m.

Consider the apparatus shown in Figure 14-6. The
equipment is similar to that shown in Figure 14-4 except
a fluorescent screen within the tube reveals the trajectory
of the particles that pass through the slot in the positive
electrode. When a magnetic field is added, the electron
trajectory is curved. A mathematical analysis of the
curvature permits an interpretation of this experiment
that leads to a determination of e/m.

Fig. 14-6. The effect of a magnetic field on the elec-
tron beam.

To vacuum
put

Fluorescent- screen

The calculations can be made only if the strength of
the magnetic field is known and if the field is uniform.
Therefore, apparatus more suitable than that shown in
Figure 14-6 is needed. An electric current flowing through
a wire coil generates a magnetic field that is easily meas-
ured and readily made uniform (by making the coil large
compared to the apparatus).

Substituting, then, a large coil for the magnet shown
in Figure 14-6, we can proceed with our measurement.
The beam of electrons passes through the positive elec-
trode and strikes the far end of the tube, producing a
fluorescent spot. When the magnetic field is turned on
(by passing current through the coil), the fluorescent spot
moves. The spot moves because a charged particle moving
in a uniform magnetic field has a path which is an arc of a
perfect circle. From the deflection of the spot and the
length of the apparatus, the radius of this circular path
can be determined. This radius we shall call r.

This radius is useful to us because it is related to the
mass, m, charge, e, and velocity, v, of the electron. It is
determined also by the strength of the magnetic field, B,
as follows:

e B

(7)

This equation shows that the greater the mass or velocity
of the particle, the less curved is its path (a small value
of r describes a highly curved path). On the other hand,
the path of the particle becomes more curved if the mag-
netic field is made stronger.

We can rearrange equation (7) to the form

m r B

(8)

Magnet

Equation (8) shows us how to calculate the charge/mass
ratio for the electron if r is measured and both v and B
are known.

However, the velocity, v, of the electron is still un-
known. We must calculate this quantity from the work
done on the electron as it was accelerated, moving from
the negative electrode to the positive electrode. The work
done on the electron is the product of the charge on the
electron times the voltage difference, V, between the
electrodes:

Work done on electron = e X V (9)

This work is used to accelerate the electron, giving it
kinetic energy. Also, we know that the kinetic energy of
the electron can be expressed in terms of its mass and
velocity :

Kinetic energy of a moving electron = \rn\~1 (10)

We must equate (9) and (10): the work done, eV, equals
the kinetic energy the electron acquires, \rn\-:

eV=$m\* (11)

The voltage, V, we obtain from a voltmeter reading.
Expressions (8) and (//) both relate the ratio e/m to

SEC. 14-2 I "seeing" parts of atoms

241

the electron velocity and the measured quantities, r, B,
and V. We can calculate e/m if v is eliminated from the
two equations. This is an algebraic process that can be
done several ways. Here is one way.
Since (//) involves v2, let us square expression (8):

g2 v2
m2 ~ r-B2

Now let us multiply both sides of (12) by m/e:

e my2

r-B-

Now we can rearrange (//) to the form

e
And, finally, we can substitute (14) into (/J):

m e r-B2 r2B2

(12)

(13)

(14)

2V
r2B2

(15)

We measure V (from the voltmeter reading), r (from
the deflection of the spot), B (from the current through
the magnet coil windings), substitute them into (15), and
calculate

coulombs

- = 1.759 X 10s

m gram

THE CHARGE ON THE ELECTRON

(16)

Experiments like those described in Figures 14-3 to 14-6
establish that the electron is a negatively charged particle
and that it is present in all substances. Further confirma-
tion of the particulate nature of electricity comes from
experiments that were conducted by an American physi-

Fig. 14-7. Millikun's oil-drop apparatus for determin-

ing the electron charge.

Telescope

Adjustable

high

voltage

cist, Robert Millikan, in 1906 to determine the charge on
the electron. The apparatus used for his experiment is
shown schematically in Figure 14-7.

Tiny droplets of oil or some other liquid are sprayed
into the upper part of the apparatus. A few droplets
fall through a small hole into the lower chamber. During
its production, an oil drop is very likely to become
charged by friction. When an oil drop enters the lower
chamber, a voltage is applied to the metal plates. If the
oil drop is charged, its fall can be completely stopped by
adjusting the voltage so that the electrical force on the
charged drop is just equal and opposite to the force of
gravity.

Millikan made thousands of determinations of the
charge on drops of oil, glycerol, and mercury. The charge
on the drop was sometimes positive and sometimes nega-
tive, but in every case its magnitude was some integral
multiple of 1.602 X 10~19 coulomb.* In no case was the
charge any less than this. These experiments give a clear
demonstration that the fundamental unit of electricity
must be a charge of 1.602 X 10~19 coulomb. If the elec-
tron carries this fundamental unit of electricity (as we
believe it does), the value of the charge on the electron
must be 1.602 X 10~19 coulomb:

e = 1.602 X 10-'9 coulomb

(17)

We may use this value of the charge on the electron
to calculate the mass of an electron. To do so, it is neces-
sary to know the ratio of (electron charge/electron
mass) = e/m. This ratio is measured with apparatus
based on principles displayed in Figures 14-4 and 14-6.
Using the result e/m = 1.759 X 10s coulombs/g, the
mass of an electron is found to be

= 1.602 X IP"19 coulomb /electron
1.759 X lO^coulomb/g

m = 9.11 X 10-28 g/electron

(/*)

EXERCISE 14-3

Suppose five measurements of oil-drop charges give the
values listed below:

4.83 X 10"19 coulomb
3.24 X 10-'9
9.62 X 10"19
6.44 X'10-»
4.80 X 1019

* The coulomb is a unit of electric charge. Its magni-
tude can be appraised by its relation to the ampere. One
ampere is an electric current of one coulomb of charge
passing a point in a wire every second. One mole of elec-
trons has, then, 96,500 coulombs of charge. In a wire
carrying 10 amperes, it takes about two and one half
hours for one mole of electrons to pass any point.

242

WHY WE BELIEVE IN ATOMS | CHAP. 14

(a) Divide each charge by the smallest value to investi-
gate the relative magnitudes of these charges.

(b) Assuming each measurement has an uncertainty of
±0.04 X 10-19, decide what electron charge is indi-
cated by these experiments alone.

14-2.2 "Seeing" Positive Ions

Experiments can be conducted in which positive ions are
detected and their properties measured (charge and mass).
These experiments are similar to those we have described
for electrons. A gas discharge tube, such as was shown
in Figure 14-3, can be used because measurements show
that positive ions are present as well as electrons. Whereas
electrons are accelerated toward the positive electrode,
the positively charged ions are accelerated in the opposite
direction, toward the negative electrode. These ions can
be removed from the apparatus as a beam in the same
way that the electron beam is removed in the apparatus
of Figure 14-4. In such fashion, we obtain a beam of posi-
tive ions. By deflecting these beams in electric and mag-
netic fields, the charges and masses of the positive ions
can be measured.

The results of experiments of this type show two very
important differences from measurements on electrons.

(1) The charge/mass ratio for positive ions changes when
the gas in the tube is changed. When the (e/m) meas-

urement is made for electrons, the same value is
obtained no matter what gas is introduced.
(2) The charge/mass ratio for positive ions is very much
smaller than (elm) for electrons. These facts are in-
terpreted to mean that the positive ions are ions
formed from the gas in the tube. The electric charge
is considered to arise from the removal of one or more
electrons from an atom or a molecule. Thus the value
of the ratio (charge/mass) for positive ions depends
upon the gas because each type of atom (or molecule)
has a distinctive mass.

"weighing" positive ions, the mass
spectrograph

A mass spectrograph is an instrument with which the
masses of individual atomic or molecular ions can be
measured. One type of mass spectrograph is shown in
Figure 14-8. Positive ions are accelerated through a
slotted negative electrode and then passed through a
uniform magnetic field. The left view in Figure 14-8 shows
the apparatus supported between the pole faces of a
strong magnet. The right view is an enlargement of the
spectrograph with the magnetic field directed vertically
through the figure. This is the view of the mass spectro-

Fig. 14-8. A mass spectrograph and the mass spec-
trum of neon.

Pole face

To vacuum pump

Vacuu.m charrCber

To vacuum.

Magnet

Electron beam.

Photographic plate
rC__r*" To vacf.um. pump

Neon gas en-try

Plate III. A simple spectrograph and the spectrum of a hot tungsten ribbon.

»s»\

0\ ■

»0

h
>

q
*
n

SEC. 14-2

SEEING PARTS OF ATOMS

243

graph seen by an ant sitting on one pole face of the mag-
net looking toward the other pole face.

The positive ions can be produced with a glow dis-
charge tube like that shown in Figure 14-3. More usually,
however, gaseous atoms or molecules are bombarded
with an electron beam as shown in Figure 14-8. If the
bombarding electrons have enough energy, they cause
positive ion formation when collisions occur with gas
molecules. The figure shows neon gas entering at the
bottom. The gas passes through the electron beam and
some of the atoms collide with electrons to form neon
ions. Both Ne+ and Ne+2 ions are formed and they are
accelerated by the slotted electrode. As the positive ions
enter the magnetic field, they follow a circular path. They
have a large radius of curvature if the mass is high, a low
radius of curvature if the charge is high. Thus each posi-
tive ion follows a circular path fixed by its mass and
charge. After circling through an arc of 180°, the ions are
collected on a photographic plate. The impact of the ions
with the photographic plate causes a reaction that leads
to a darkening of the sensitized surface, just as exposure
to light does. Such a record shows a spot for each ion at
a position fixed by the charge/mass ratio. Measurement
of the position of each spot reveals the masses of the
ions. The record is called a mass spectrum.

When neon gas is put in the spectrograph, the mass
spectrum consists of two widely separated groups of three
spots each. The three spots corresponding to large radii
are caused by neon ions with a single positive charge,
while the three spots corresponding to small radii are
caused by doubly charged ions. For each ionic charge
there are three slightly separated spots which indicate that
neon consists of atoms with three different masses.
This shows that ordinary neon consists of three different
isotopes. The relative abundances of these isotopes can
be determined by measuring the intensity of the spots
caused by each of the ion beams.

EXERCISE 14-4

When chlorine, Cls, is examined in a mass spectrograph,
CI2", Cl+, and Cl+2 ions are formed. Remembering that
there are two isotopes in chlorine, 35 (75%) and 37
(25%), describe qualitatively the appearance of the mass
spectrum. Which ion will produce lines at the largest
radius? Which at the smallest radius? How many lines
will each ion produce?

Figure 14-8 shows a mass spectrograph that uses photo-
graphic detection. Nonphotographic detection is also
possible. The ions, after being sorted according to mass
and charge, can be "counted" by a charge measuring
device. The advantage of such a detector is that tbe result
can be presented continuously on a paper chart, thus
eliminating the cumbersome and slow photographic
process.

THE RATIO OF CHARGE TO POSITIVE

ion mass, e/m

In a mass spectrograph, the factors that determine the
trajectory of the ions are the same as those we discussed
when we considered the measurement of (film) for the
electron. In that discussion we derived equation (79):
e 2V

m = 7* (79)

where e = electron charge,

m = electron mass,

V = accelerating voltage,

B = magnetic field strength.

For positive ions, the charge is e or 2e or, in general,
some integer, n, times e, the electron charge. We might
write a capital M for mass to indicate that the mass of a
positive ion is involved. Then we can solve (19) for the
mass as a function of V, B (experimental conditions we
control), n (which will be one, two, three, or some low
integer), and r, the radius that we measure on the photo-
graphic plate:

neriBi

M =

(20)

With equation (20), we can verify quantitatively our
identifications of the ionic masses and their charges. The
two sets of three spots immediately suggest that one set
is caused by three isotopes, each with the same ionic
charge, and the second set by the same three isotopes with
a different ionic charge. Measurement of the radii shows
that one set has a larger radius by just the square root of
two. This ratio is consistent with the assignment of +1
to one set of ions and +2 to the other set. Calculations
based on the assumption that n = 1 for the outer set,
together with the accurately measured radii of the
paths corresponding to the three spots on the plate
give isotopic masses corresponding to 20, 21, and 22
grams/mole, the three stable isotopes of neon. If the true
value of n for this outer set of spots had been n = 2, the
calculated masses would correspond to isotopic masses of
40, 42, and 44 grams/mole. The choice between the two
assumptions is usually easily made on the basis of chemi-
cal arguments about the possible atoms or molecules
present.

EXERCISE 14-5

Suppose a mass spectrograph is used to measure the
charge/mass ratio for fluorine ions. Fluorine has only one
stable isotope and its atomic weight is 19.0 grams/mole.
From the measured charge/mass ratio, 5.08 X 10s cou-
lombs/gram, and the assumption that the ion has one
electron charge, calculate the mass of one ion. Repeat the
calculation assuming the ion has two electron charges.
Now calculate Avogadro's number from the weight of a
mole of fluorine ions, using each of your two calculations.
Which assumption about ion charge do you prefer?
Could the other be correct as well?

244

WHY WE BELIEVE IN ATOMS I CHAP. 14

Lead block

Fig. 14-9. Rutherford's apparatus for observing the
scattering of alpha particles by a metal foil.
{The entire apparatus is enclosed in a

vacuum chamber.)

14-2.3 "Seeing" the Nucleus: Structure
off the Atom

So far we have described experiments that indicate that
atoms exist and that indicate atoms are composed of
charged particles. We know also that all the positively
charged part of the atom is located in a very small but
dense region which we call the nucleus. The negatively
charged electrons spend most of their time at relatively
great distances from the nucleus. The story of how this
nuclear model of the atom was first proposed gives a
fascinating view of how science progresses.

The first detailed model of the atom, proposed by
J. J. Thomson in 1898, was based upon the expectation
that the atom was a sphere of positive electricity in which
electrons were embedded like plums in a pudding. This
picture of the atom was not particularly satisfying be-
cause it was not useful in predicting or explaining the
chemical properties of the atom. Finally, in 1911, a series
of experiments performed in the McGill University labo-
ratory of Ernest Rutherford showed that Thomson's
picture of the atom had to be abandoned.

The experiment conducted by Rutherford and his co-
workers involved bombarding gold foil with alpha par-
ticles, which are doubly charged helium atoms. The
apparatus used in their experiment is shown in Figure
14-9. The alpha particles are produced by the radioactive
decay of radium, and a narrow beam of these particles
emerges from a deep hole in a block of lead. The beam
of particles is directed at a thin metal foil, approximately
10,000 atoms thick. The alpha particles are detected by
the light they produce when they collide with scintilltaion
screens, which are zinc sulfide-covered plates much like
the front of the picture tube in a television set. The screen

Unde-viat-ed alpha particle beam

is mounted on an arm in such a way that it can be moved
around in a circle whose center is in line with the point
where the alpha particles strike the foil. A telescope is
mounted behind the screen so that the very small flashes
of light produced when individual alpha particles strike
the scintillation screen can be detected and counted. The
apparatus operates in a vacuum chamber in order that
no deflections are caused by the impact of alpha particles
upon gaseous molecules.

The first observation made with this apparatus was that
apparently all the alpha particles passed through the foil
undeflected. Let us see if this result is consistent with the
model of the atom proposed by Thomson. You will recall
that Thomson's picture of the atom assumed that the
positive charge is distributed evenly throughout the entire
volume of the atom with the negative electrons embedded
in it. Since the electrons weigh so little, the positive part
accounts for nearly all of the mass of the atom. Thus the
Thomson model pictures the atom as a body of uniform
density.

Imagine what our thin metal foil would be like if it
were to be made up of Thomson atoms. The physical
properties of a solid suggest that the atoms lie very close
together, so the metal foil would look something like the
diagram shown in Figure 14-10. Of course, the real foil
is 10,000 atoms thick. What would happen to the alpha
particles if they were shot into a solid of such uniform
density? At first we might think that they would be
stopped or deflected back upon colliding with the atoms.
Since it was observed that the alpha particles went straight
through the metal foil, we must reconsider the problem.
When we shoot at a paper target with a high-powered
rifle, the projectile forces its way through the paper. The
alpha particles produced by radium have very high kinetic
energy and are very much like bullets from a high-
powered rifle. Perhaps the very high kinetic energy allows
an alpha particle to force its way right through the atoms
of the metal foil. Since a rifle bullet fired into paper passes
through undeflected, it seems reasonable to conclude that
the alpha particle would also pass through the metal foil
undeflected.

SEC. 14-3 I MEASURING DIMENSIONS OF ATOMS AND MOLECULES

245

Alpha particle

Fig. 14-10. The scattering of alpha particles by a

metal foil made of Thomson atoms.

Atomic nucleus

In summary, in the Thomson model a metal foil is
considered to have essentially uniform density. If this is
true, there is no way for bombarding alpha particles to
be deflected through large angles. At best, the alpha par-
ticles might suffer slight deflections from many collisions
with many atoms. The model predicts the scattering dis-
tribution shown in Figure 14-10.

The first results of Rutherford's experiments seemed
to be quite consistent with the Thomson picture of the
atom. On more careful examination, an astounding dis-
covery was made. By moving the screen around the metal
foil, Rutherford and his co-workers were able to observe
that a very few scintillations occurred at many different
angles; some of these angles were nearly as large as 180°.
It was as if some of the alpha particles had rebounded
from a head-on collision with an immovable object. In
the words of Rutherford, "It is about as incredible as if
you had fired a 15 inch shell at a piece of tissue paper
and it came back and hit you." It was impossible to ex-
plain the simultaneous observation of large-angle and
small-angle deflections by using the Thomson atom.

In order to explain his experimental results, Rutherford
designed a new picture of the atom. He proposed that the
atom occupies a spherical volume approximately 10~8 cm
in radius and at the center of each atom there is a nucleus
whose radius is about 10~12 cm. He further proposed that
this nucleus contains most of the mass of the atom, and
that it also has a positive charge that is some multiple
of the charge on the electron. The region of space outside
the nucleus must be occupied by the electrons. We see
from Figure 14-1 1 that Rutherford's picture requires that
most of the volume of the atom be a region of very low
density.

Using this kind of model of the atom, we can account
for the alpha particles that are deflected through both

Fig. 14-11. The scattering of alpha particles by a foil
made of Rutherford nuclear atoms.

large and small angles. If we allow alpha particles to
impinge upon a metal foil composed of atoms based on
Rutherford's model, only a few of tne particles would be
appreciably deflected by the foil. The heavy, fast moving
alpha particles can brush past the lighter electrons with-
out being deflected. Since most of the volume of the metal
foil is relatively empty space, the greatest number of
alpha particles pass through the metal undeflected. It is
possible, however, foi a few particles to be scattered
through very large angles. Since both the alpha particle
and the nucleus of the atom are positively charged, they
exert a force of repulsion on each other. This force be-
comes large only when the alpha particle comes quite
close to the nucleus. Since the nucleus in question is much
heavier than the alpha particle, it can deflect the alpha
particle considerably, just as a steel post can deflect a
rifle bullet.

Besides providing a qualitative picture of the atom,
Rutherford's experiments provided a way of measuring
the charge of the nucleus. The force that a nucleus exerts
on an alpha particle depends upon the magnitude of the
charge on the nucleus. Rutherford showed how to relate
the number of alpha particles scattered at any angle to
the magnitude of the charge on the nucleus. The first
measurements of the nuclear charge by this method were
not very accurate, but they did show that different ele-
ments have different nuclear charges. By 1920, however,
the alpha particle scattering experiments were so refined
that they could be used to determine nuclear charge
accurately.

14-3 MEASURING DIMENSIONS OF ATOMS AND MOLECULES

There are several ways by which sizes of atoms
in molecules and in solids can be estimated.
These methods are classified as "spectroscopic"

methods because they involve the interaction of
light with matter. The measurements show
atomic size in the sense that they show how

246

WHY WE BELIEVE IN ATOMS I CHAP. 14

closely the atoms pack together. These packing
distances, as measured spectroscopically, have
provided the dimensions for the atomic models
you have seen.

14-3.1 Light and the Frequency Spectrum

Light can be characterized by its frequency or
its wavelength. To understand the meaning of
these terms, consider water waves approaching
and breaking on a beach. Figure 14-12 shows
two measurements we might make, the distance
between crests and the time between waves. The
distance between crests is called wavelength and
it can be expressed in centimeters. The time be-
tween waves, r, indicates how often waves pass
a fixed point. Usually the reciprocal, 1/r, is

Fig. 14-12. Waves can be characterized by wavelength
or by time between waves.

specified. This quantity, with the dimension
waves per second, is called frequency and is sym-
bolized v ("nu"). Light, which is an electromag-
netic disturbance traveling through space, has
properties much like water waves. The electro-
magnetic disturbance varies periodically, as does
the water disturbance, hence it can be charac-
terized by its frequency. Also, light travels
through space with definite distance between the
"wave crests" where the electromagnetic dis-
turbance is greatest. This distance is called the
wavelength.

Figure 14-13 shows a spectrograph — an in-
strument that reveals the frequency composition
of light. The light entering a narrow slit is
focused into a beam by the lens. This beam is
passed through the prism. All of the light is
refracted (bent) by the angular prism, but dif-
ferent frequencies (colors) are bent through
different angles. The result is that the frequency

Time
betrtveen
waves

SEC. 14-3 | MEASURING DIMENSIONS OF ATOMS AND MOLECULES

247

Blue
Violet-
Ultraviolet

Hot

tungsten

ribbon

Fig. 14-13. A simple spectrograph.

composition of the light entering the slit can be
learned from the pattern focused on a photo-
graphic film. The light source is a tungsten rib-
bon heated to a temperature near 1000°C by an
electric current.

This separation of light into its component
frequencies produces a spectrum. This spectrum
is recorded on the photographic film because the
darkening of the film (on development) is deter-
mined by the light intensity. From the spectrum
we learn that different colors correspond to dif-
ferent frequencies. Blue light is found to have a
frequency of about 7.5 X 10u waves per second

Ph o togrraph ic plate

or, in more usual terminology, v = 7.5 X 1014
cycles per second. The red light has a lower
frequency; v is about 4.3 X 10u cycles per sec-
ond.

The experiment shown in Figure 14-13 points
up another extremely important fact. The photo-
graphic film is darkened at larger angles than
that at which blue light appears and at smaller
angles than that at which red light appears. This
implies that the light emitted by the hot ribbon
includes frequencies that are not detected by the
human eye. The frequencies lower than the fre-
quency of red light are called infrared frequen-

Fig. 14-14. The complete light spectrum.

G-arnma rays
X-rays

Ultraviolet
Visible
Infra red

Mic ro wa ve
Radio

Frequency. '22
cycles/second

JO'

10 "

Molecular rotation
b/folecular vibration
Electron excitation

248

WHY WE BELIEVE IN ATOMS i CHAP. 14

cies. The frequencies higher than the frequency of
violet light are called ultraviolet frequencies.

Scientists have now realized that the electro-
magnetic phenomenon called light extends over
an enormous range of frequencies — much wider
than the rather narrow region in which the
human eye is sensitive. Figure 14-14 shows the
range that is commonly studied and the familiar
names given to various spectral regions.

There are three spectral regions or ranges of
light frequencies that are particularly useful to
chemists in learning atomic sizes. We shall dis-
cuss each briefly.

14-3.2 X-Ray Diffraction Patterns

X-Rays are light waves of frequencies near 1018
cycles per second and wavelengths near 10~8 cm.
Such light waves, when reflected from the surface
of a crystal, give patterns on a photographic
film. The pattern is fixed by the spacings of the
atoms of the crystal and their spatial arrange-
ment. The pattern is obtained only with X-rays
because it results from scattering effects that
occur only if the wavelength of the light is close
to the atomic separations within the crystal.
Therefore, a knowledge of the wavelength of the
X-ray light permits an interpretation of the pat-
tern in terms of atomic packing.

Figure 14-15 shows three X-ray diffraction
patterns obtained from small crystalline particles
of metallic copper, aluminum, and sodium. The
qualitative similarity of the patterns given by
copper and aluminum shows that they have the
same crystal packing. Careful measurements of
the spacing of the lines indicate that the atoms

Fig. 14-15. X-Ray diffraction patterns of finely di-
vided metallic copper, aluminum, and
sodium.

CtM.

in copper, though occupying the same relative
positions as the atoms in aluminum, are closer
together. In contrast, the pattern of lines pro-
duced by sodium does not resemble either of the
preceding patterns. Sodium atoms are packed
differently in the metallic sodium crystal.

The X-ray diffraction method is applicable to
solids and provides such detailed views of crystal
geometry as those shown for sodium chloride
solid in Figure 5-10, p. 81.

14-3.3 Microwave Spectroscopy:
Molecular Rotation

Radio waves are light rays of macroscopic wavelengths
(that is, wavelengths of many meters). Using techniques
similar to those used in generating radio waves, "micro-
wave" light can be produced with wavelengths in the
range 1 mm to 10 cm. In this spectral region, gaseous
molecules absorb light because of excitation of rotational
movements of the molecules. The frequencies of these
rotational motions, shown in Figure 7-8 (p. 118), depend
upon the distance of the atoms from the molecular center
of gravity.

Microwave spectroscopy is applicable only to gases but
it is capable of extremely high accuracy. Interatomic dis-
tances and structures have been so measured for many
molecules containing only a few atoms.

14-3.4 Infrared Spectroscopy:
Molecular Vibration

Light found in the spectrum just beyond the red
end of the visible spectrum is called infrared
light. The frequencies are in the range 2 X 1013
to 12 X 1013 cycles per second (with wavelengths
between 1.5 X 10-3 to 2.5 X lCM cm). Mole-
cules absorb light in this spectral region, and
analysis of the frequencies shows that the ab-
sorptions are associated with the excitation of
vibrational motions. These to-and-fro motions of
the atoms occur at natural frequencies just like
the natural vibrational frequencies of a ball-and-
spring model of a molecule. These natural fre-
quencies are fixed by the masses of the atoms,
the molecular shape, and the strengths of the
chemical bonds that link the atoms together.
Again the frequencies absorbed by gaseous mole-
cules provide information about the molecular

SEC. 14-3 I MEASURING DIMENSIONS OF ATOMS AND MOLECULES

249

moments of inertia, the molecular geometry, and
the chemical bonds. In addition, infrared study
can be extended easily to the liquid and solid
state, hence it finds widespread use in chemistry.
Figure 14-16 contrasts the infrared absorption
spectra of hydrogen bromide gas, HBr, and
deuterium bromide gas, DBr. The horizontal
scale shows frequency. For a given frequency,
the vertical scale shows the percentage of light
of that frequency transmitted by the sample.
Thus a reading of 100% means all of the light is
transmitted; hence, no light is absorbed at that
frequency. Plainly, gaseous HBr absorbs in only
one spectral region, that near 7.9 X 1013 cycles
per second. This one absorption corresponds to
the vibrational excitation of the chemical bond
in HBr. There is only one bond, hence only one
absorption. The spectrum of gaseous DBr is
similar but the absorption occurs near 5.9 X 1013
cycles per second. Of course, the chemistries of
DBr and HBr are identical, hence the chemical
bond in DBr is identical to that in HBr; never-

Fig. 14-16. Infrared absorption spectra of gaseous
HBr and DBr.

theless, the vibration frequencies of these two
molecules differ because the atom masses differ.
Since the deuterium atom is heavier than the
hydrogen atom, it vibrates more slowly.

More complicated molecules, with two or
more chemical bonds, have more complicated
absorption spectra. However, each molecule has
such a characteristic spectrum that the spectrum
can be used to detect the presence of that par-
ticular molecular substance. Figure 14-17, for
example, shows the absorptions shown by liquid
carbon tetrachloride, CC14, and by liquid carbon
disulfide, CS2. The bottom spectrum is that dis-
played by liquid CC14 containing a small amount
of CS2. The absorptions of CS2 are evident in the
spectrum of the mixture, so the infrared spec-
trum can be used to detect the impurity and to
measure its concentration.

The value of infrared spectra for identifying
substances, for verifying purity, and for quan-
titative analysis rivals their usefulness in learning
molecular structure. The infrared spectrum is as
important as the melting point for characterizing
a pure substance. Thus infrared spectroscopy
has become an important addition to the many
techniques used by the chemist.

100%

50%
0%

HBr (9)

........ i i i

lOxlo13

8xW13

6xl013

4xl013

2xl013

loo%r

50%- DBr ($)

Wxio13

8xl015

6xW13

4 *1013

2 xjo

Frequency , cycles/second

250

WHY WE BELIEVE IN ATOMS I CHAP. 14

100%
50%

: cs2 (i)

WxlO13

8 xW13

6 xlO13

4xl013

2X10

100%,
50%

: cci+ (i)

lOxlO13

8xl013

GxlO13

4xW13

2 xlO'

ioo%

V \f "N

\/~^\ /

50%

- CCI4 + 5% CS2 (I)

cs2

lOxlO13 8 xlO13

6 xlO13

-4 xlO13

2 xlO13

Frequency , cycles /second

Fig. 14-17. Infrared absorption spectra of liquid carbon tetrachloride, CCU, carbon disulfide, CS«, and a
mixture of the two.

QUESTIONS AND PROBLEMS

1. A compound of carbon and hydrogen is known
that contains 1.0 gram of hydrogen for every 3.0
grams of carbon. What is the atomic ratio of
hydrogen to carbon in this substance?

2. There are two known compounds containing
only tungsten and carbon. One is the very hard
alloy, tungsten carbide, used for the edges of
cutting tools. Analysis of the two compounds
gives, for one, 1.82 grams and, for the other,
3.70 grams of tungsten per 0.12 gram of carbon.
Determine the empirical formula of each.

3. John Dalton thought the formula for water was
HO (half a century passed before the present
formula for water was generally accepted). What
relative weights did he then obtain for the weight
of oxygen and hydrogen atoms?

4. Nitrogen forms five compounds with oxygen in
which 1.00 gram of nitrogen is combined with
0.572, 1.14, 1.73, 2.28, 2.85 grams of oxygen,
respectively. Show that the relative weights of
the elements in these compounds are in the ratio
of small whole numbers. Explain these data us-
ing the atomic theory.

QUESTIONS AND PROBLEMS

251

5. Using Appendix 3, list two metals that could
have given the same number of moles as alumi-
num did in the experiment shown in Figure 14-2.

6. If n coulombs will deposit 0.1 19 gram of tin from
a solution of SnS04, how many coulombs are
needed to deposit 0.1 19 gram of tin from a solu-
tion of Sn(S04)2?

7. Suppose two more cells were attached to the
three in Figure 14-2. In one cell, at one of the
electrodes copper is being plated from CuS04
solution and at one of the electrodes in the other
cell, bromine, Rr-iig), is being converted to bro-
mide ion, Br~. How many grams of Cu and Br~
would be formed during the same operation
discussed in the figure?

8. Carbon monoxide absorbs light at frequen-
cies near 1.2 X 10", near 6.4 X 1013, and near
1.5 X 1015 cycles per second. It does not absorb
at intermediate frequencies.

(a) Name the spectral regions in which it ab-
sorbs (see Figure 14-14).

(b) Explain why carbon monoxide is colorless.

9. The wavelength and frequency of light are re-
lated by the expression X = c/v, where X =
wavelength in centimeters, v = frequency in
cycles per second, and c = velocity of light =
3.0 X 1010 cm/second. Calculate the wavelength
corresponding to each of the three frequencies
absorbed by CO (see Problem 8). Express each
answer first in centimeters, and then in Ang-
stroms (1 A = 10-8 cm).

Answer. 1.5 X 1015 cycles/sec:

2.0#X 10~6 cm/cycle = 2.0 X 10s
A/cycle.

10. The oxygen molecule carries out molecular vi-
bration at a frequency of 2.4 X 1013 cycles/sec-
ond. If the pressure is such that an oxygen
molecule has about 109 collisions per second,
how many times does the molecule vibrate be-
tween collisions?

11. When several oil drops enter the observation
chamber of the Millikan apparatus, the voltage
is turned on and adjusted. One drop may be
made to remain stationary, but some of the
others move up while still others continue to
fall. Explain these observations.

12. Dust particles may be removed from air by pass-
ing the air through an electrical discharge and
then between a pair of oppositely charged metal
plates. Explain how this removes the dust.

1 3. How many electrons would be required to weigh
one gram? What would be the weight of a
"mole" of electrons?

14. About how many molecules would there be in
each cubic centimeter of the tube shown in Fig-
ure 14-3 when the glow appears? When the glow
disappears again because the pressure is too
low?

15. Describe the spectrum produced on a photo-
graphic plate in a mass spectrograph if a mixture
of the isotopes of oxygen (160, "O, and 180) is
analyzed. Consider only the record for + 1 and
+2 ions.

16. Hydroxylamine, NH2OH, is subjected to elec-
tron bombardment. The products are passed
through a mass spectrograph. The two pairs of
lines formed indicate charge/mass ratios of
0.0625, 0.0588 and 0.1250, 0.1176. How can this
be interpreted?

17. Platinum and zinc have the same number of
atoms per cubic centimeter. Would thin sheets
of these elements differ in the way they scatter
alpha particles? Explain.

18. Assume that the nucleus of the fluorine atom is
a sphere with a radius of 5 X 10~13 cm. Calculate
the density of matter in the fluorine nucleus.

19. An average dimension for the radius of a nucleus
is 1 X 10~12 cm and for the radius of an atom is
1 X 10-8 cm. Determine the ratio of atomic vol-
ume to nuclear volume.

CHAPTER

Electrons and the
Periodic Table

It is the behavior and distribution of the electrons around the nucleus that
gives the fundamental character of an atom: it must be the same for mole-
cules.

c. a. coulson, 195 1

We have seen that much is known about the
structure of the atom. A small nucleus containing
protons and neutrons accounts for most of the
mass of the atom. Electrons occupy the space
around the nucleus like bees around a hive.
In the electrically neutral atom, the number of
electrons is equal to the number of protons.

Looking back to Chapter 6, we have discov-
ered marvelous regularity among the elements.
Of the 100 or so elements, six are unique in their
absence of chemical reactivity. Those six ele-
ments, the inert gases, provide the key to the
most important correlation of chemistry, the
periodic table. Not only do these elements fur-
nish the cornerstone for the periodic table but,
also, their electron populations seem to play a
dominant role in the chemistry of the other ele-
ments in the table. An element just preceding an
inert gas in the table (one of the halogens) has a
strong tendency to acquire an extra electron.
The resulting negatively charged ion has, then,
the number of electrons possessed by an atom of
its inert gas neighbor. In striking contrast, an
252

element just following an inert gas (one of the
alkalies) releases electrons quite readily. The re-
sulting positively charged ion has, then, the
number of electrons possessed by an atom of its
inert gas neighbor. In each type of element, the
halogens and the alkalies, the chemistry can be
discussed in terms of the tendency of atoms to
acquire or release electrons so as to reach the
special stability of the inert gases. The impor-
tance of this tendency is revealed in the dramatic
differences that exist between the chemistry of
the halogens and the chemistry of the alkalies.
This special stability associated with the inert
gas electron populations was found to pervade
the chemistry of every element of the third row
of the periodic table (see Section 6-6.2). Each
element forms compounds in which it contrives
to reach an inert gas electron population. Ele-
ments with a few more electrons than an inert
gas are apt to donate one or two electrons to
some other more needy atom. Elements with a
few less electrons than an inert gas are apt to
acquire one or two electrons or to negotiate a

SEC. 15-1 I THE HYDROGEN ATOM

253

communal sharing with other atoms. In all cases,
the number of electrons transferred or shared is
understandable in terms of the inert gas stability.
In this chapter we shall explore our current
understanding of this behavior. We are guided
in this exploration by a regularity presented in
Chapter 6 and reproduced in Table 15-1. The

Table 15-1

REGULARITY AMONG THE ELECTRON
POPULATIONS OF THE INERT GASES

INERT GAS

ELECTRONS

DIFFERENCES

helium

neon

10 - 2 = 8

argon

18 - 10 = 8

krypton

36 - 18 = 18

xenon

54 - 36 = 18

radon

86 - 54 = 32

regularity of the differences 2, 8, 8, 18, 18, 32 is a
clue of magnificent proportions. The electron
populations of the inert gas atoms have startling

regularity. This clue has led scientists to a de-
tailed and quantitative understanding of the
atomic properties that give rise to the periodic
table.

EXERCISE 15-1

To see that these numbers have regularity, con-
sider the series of numbers 2-8-18-32. (We shall
forget, for the time being, that 8 and 18 each
appear twice in the series.)

(a) If you were to consider this series incom-
plete, would you expect the next number
(after 32) to be even or odd?

(b) The numbers 2-8-18-32 were obtained by
subtracting electron populations (that is, by
taking differences). Take differences again
and use them to predict the next number
beyond 32 in the series.

(c) Divide the numbers 2-8-18-32 by two. Use
these numbers as a basis for predicting the
next number beyond 32 in the series by an-
other method than taking differences.

15-1 THE HYDROGEN ATOM

Just as the inert gases form the cornerstone of
the structure we call the periodic table, the sim-
plest atom, hydrogen, provides the key that un-
locks the door of this structure. Atoms of every
other element mimic the hydrogen atom. To see
that this is so, we must examine the interaction
of hydrogen atoms with light. The light emitted
(or absorbed) by hydrogen atoms is called the
atomic hydrogen spectrum. This spectrum ex-
plains the existence of the periodic table.

15-1.1 Light— A Form off Energy

Before we can analyze the spectrum of hydrogen
atoms, we must become more familiar with light.
In Chapter 14 light was characterized by fre-
quency or wavelength. (Reread Section 14-3.1.)
Now we shall consider another property of light

— a property less obvious than the "water wave"
characteristics. Light is a form of energy.

The statement, "Light is a form of energy,"
is consistent with quite a bit of common experi-
ence. Many of you have used a hand lens to
focus light rays on a paper, setting it afire. This
experiment is put to work in the huge solar
furnaces that achieve temperatures of many
thousands of degrees and that melt the most
refractory material. The temperature rise in the
paper or in the refractory material is caused by
the absorption of light. A temperature rise means
energy has been absorbed. This energy must have
been carried by the light.

We do not need a hand lens to "feel" the
energy of light rays. Just remember those lazy
afternoons you spent last summer soaking up
the warmth of the sun. The afternoon pleasure

254

ELECTRONS AND THE PERIODIC TABLE I CHAP. 15

came from the absorption of the energy by the
tissues of your skin. The sunburn pain you suf-
fered that night was caused by the chemical
reactions energized by the treacherous light rays.
Here we have a personal basis for claiming that
light carries energy.

As one other familiar reference, consider pho-
tosynthesis. You have undoubtedly heard many
times that this is the chemical process by which
a plant "stores" the energy of the sun. Much is
known about the chemical reactions of photo-
synthesis and it is indeed true that they result in
formation of chemical compounds with higher
heat content than the starting substances. These
reactions will not occur in the absence of light —
the light supplies the energy required to raise the
reactants to the higher heat content of the
products.

We have all of this familiar experience to build
upon, but it is all qualitative. We need a quan-
titative relationship. How much energy is carried
by light? The answer is simple in form, but not
in concept. Light, too, comes in packages. Each
package, called a photon, contains an amount of
energy determined by the frequency. This state-
ment is contained in the famous equation

E = hv (/)

The quantity h is called Planck's constant. It is

merely a conversion factor that reexpresses fre-
quency, v, in energy units. The experimental
evidence that led to equation (/) furnishes a
fascinating story — a story that you will hear in
your physics class. Our interest is in the applica-
tion of this equation to the interpretation of the
atomic hydrogen spectrum. This spectrum gives
a record of the frequencies that are emitted by a
hydrogen atom. By equation (7), then, the spec-
trum also gives information about the energies a
hydrogen atom can possess.

15-1.2 The Light Emitted by Hydrogen Atoms

Figure 15-1 shows again the spectrograph de-
scribed in Section 14-3.1. This time the light
source is a gas discharge tube such as was shown
in Figure 14-3 on p. 239. Just as before, some of
the light emitted by the source passes through a
narrow slit and is focused to a beam by the lens.
Again this light beam is refracted (bent) by the
angular prism. The spectrum, or frequency com-
position of the light, appears on the photo-
graphic film.
When hydrogen gas is admitted to the dis-

Fig. 15-1. The spectrum of light emitted by a hydro-
gen discharge tube.

Photographic ptaire

SEC. 15-1 I THE HYDROGEN ATOM

255

Hot tungsten ribbon emits
ligrh-t airatl frequencies

Visible region

v^L

Ultra vio le t

region

JJ)

Hydrogen discharge i~u.be emits
lighT eft special frequencies only

Fig. 15-2. Contrast between the continuous spectrum
of a hot tungsten ribbon and the line
spectrum of a hydrogen discharge tube.

charge tube and a high voltage is applied, light
is emitted. To the eye, the light appears magenta
in color but the spectrograph tells a surprising
story! Instead of a fairly continuous darkening
across the photographic film (as obtained from
the hot tungsten ribbon) the film shows a series
of lines! Each line corresponds to a particular
frequency emitted by hydrogen atoms. Each
space between two lines on the film corresponds
to a frequency range in which no light is emitted
by the hydrogen atom.

Fig. 15-3. The spectrum of the hydrogen atom.
Visible region Ultraviolet- region

VO «0 On N.*>«xo Energy ~\ CQ Q^Os

S £ ^ ££'£ of photons, Vj
(k cal/m. ole) <\

vO ^0 Q ^Vrv^>, Frequency, ^
g 3 S S§£ § (cycles/sec) \

* $ ^ ^ £§ Wavelength, ^

o ^

ty ^Ov^o)Q

S<s^S

? s>*3

<i ©\OCOn

Two peculiarities of the hydrogen discharge
spectrum are immediately evident — they were
evident to scientists as long ago as 1840. First,
hydrogen atoms are very particular about the
frequencies they emit. Only special frequencies
are observed in the spectrum. Second, the fre-
quencies corresponding to the lines on the film
are spaced systematically. There are two groups
of lines, one group in the visible part of the light
spectrum and another group in the ultraviolet
part. Within each group there is a regular de-
crease in the spacing between successive lines as
the frequency increases. The measured frequen-
cies are shown in Figure 15-3.

EXERCISE 15-2

Complete the following table for all of the ultra-
violet lines listed in Figure 15-3. Plot the energy
spacing against the arbitrary spacing number
assigned in the last column to convince yourself
that there is regularity in the spacings of these
lines. Make the same sort of a table for the lines
in the visible group.

ENERGY PER

ENERGY

SPACING

GROUP

MOLE OF PHOTONS

SPACING

NUMBER

235.2 kcal

Ultraviolet

278.8

43.6 kcal

lines

294.0

2
3

To see how these results might be explained,
let us translate these spectral lines into energy
terms. The hydrogen atom just before light emis-
sion has some amount of energy — let us call it
E2. Light of frequency ■> is emitted, carrying
away energy hv. The hydrogen atom is left with
less energy — let us call it Ex. As is habitual, we
assume energy is conserved, so the energy lost
by the hydrogen atom must be exactly equal to
that carried away by the light:

E2 - £, = hv (2)

To "explain" why hydrogen atoms emit the
line spectrum, we must seek a model with the

256

ELECTRONS AND THE PERIODIC TABLE I CHAP. 15

No-tch. tt 1

ZSS.Z Ac at

same sort of properties. What sort of a system
has the property that its initial energy, E2, tells
it how much energy it can release so as to make
the difference E2 — Ex just exactly one of the
special energies, hvl Fortunately, we have just
such a system at hand, though it deals in weights,
not energies.

Picture your triple beam balance. The front
beam has a sliding hanger that permits you to
balance any weight placed on the pan up to the
full scale reading on the beam. This front beam
might be compared to the hot tungsten ribbon.
The beam can be used to balance any small
weight change made on the pan: the hot tungsten
ribbon emits light of any energy (that is, of any
frequency).

But now consider the other two beams — they
are quite different. They are notched so that only
particular hanger positions can be selected. Each
position corresponds to a particular weight and
intermediate weights cannot be balanced without
the front beam. The notched beam has properties
in common with the hydrogen atom. The weight
added to the notched beam is fixed by the
notches — intermediate weights cannot be meas-
ured. In the hydrogen atom the energy is some-
how "hung on a notched beam" and the amount
of energy it can absorb or release must corre-
spond to the energy difference between two of
these "notches."

If we pursue this analogy, we can see how

Fig. 15-4. The notches implied by light emission at
235.2 kcal.

scientists have deduced the energy properties of
the hydrogen atom from its line spectrum. We
can use the observed energies, listed in Figure
15-3, to construct a notched beam that would
"deliver" the observed spectrum. This beam
must have a scale calibrated not for weight but
for energy. There must be notches on this scale
to match the observed lines in the spectrum. We
shall find it convenient to begin with the line
observed at wavelength 1216 A or, in energy
units, 235.2 kcal/mole of photons. This energy
represents the difference in energy between two
notches. If we number the notches #1 and #2,
changing an imaginary hanger from notch #1 to
notch #2 delivers 235.2 kcal, as shown in Figure
15-4.

Now consider the other lines in the ultraviolet
group. Exercise 15-2 showed that they are related
to the one at 235.2 kcal in their systematic spac-
ings. Let's seek regularity among these energies
by making a simple assumption. We shall assume
that each line in the ultraviolet group corre-
sponds to a movement of the hanger from notch
#1 to a new notch. This assumption is arbitrary
and will be kept only if it turns out to be helpful.
(Remember the lost child and his rule, "Cylin-
drical objects burn"?)

SEC. 15-1 I THE HYDROGEN ATOM

257

Fig. 15-5. The notches required by the ultraviolet
group of lines if all changes begin in notch

So we can add four more notches to the beam,
as shown in Figure 15-5. Notch #3 is cut at a
scale position 278.8 kcal relative to notch #1.
Notch #4 is at 294.0 kcal, and so on.

The appearance of this notched beam is un-
familiar, to say the least. Can we find any basis

for corroboration or contradiction of this inter-
pretation? To seek such evidence, let's base a
prediction upon the model. One such prediction
concerns the energies that this notched beam
would deliver if we investigated movements of
the hanger from notch #2 to notch #3. Notch #2
is 235.2 kcal from notch #1, and notch #3 is

Fig. 15-6. The energy change implied by a change
from notch #2 to notch #3.

Notch. # /

3 45 6

(278.8 -235.2)
-43.6 Ucat

258

ELECTRONS AND THE PERIODIC TABLE I CHAP. 15

278.8 kcal from notch #1. Therefore, the energy
difference going from notch #2 to #3 is the dif-
ference between these two numbers:

Energy change from notch #2 to #3

= 278.8 - 235.2 = 43.6 (5)

Calculation (3) indicates that our notched beam
model implies light would be emitted with en-
ergy at 43.6 kcal as well as at 235.2 kcal and
278.8 kcal. We refer back to the spectrum in
Figure 15-3 and, indeed, there is light emitted at
43.6 kcal — this is one of the group of lines in the
visible region!

With this encouragement, let's calculate the
other lines implied by changes beginning in
notch #2.

NOTCH

ENERGY CHANGE

CHANGE

CALCULATED

OBSERVED

#2 — >-#3

278.8

- 235.2 = 43.6 kcal

43.6 kcal

#2 —^#4

294.0

- 235.2 = 58.8

58.8

#2 — *-#5

301.1

- 235.2 = 65.9

65.9

#2 -+#6

304.9

- 235.2 = 69.7

69.7

Plainly, the agreement between calculation
and experiment is too good to be accidental.
Our notched beam is a useful basis for interpret-
ing the spectrum of the hydrogen atom.

EXERCISE 15-3

Plot the energy of each line of the ultraviolet
group against notch number, n, using the higher
of the two notch numbers assigned to that line
in Figure 15-5. For example, plot 235.2 kcal on
the vertical axis against 2 on the horizontal axis.
Assign to notch #1 the arbitrary value zero and
draw a smooth curve through all of the points,
including the point for notch #1. Estimate the
energy value that would be observed for a notch
with very high notch number, as suggested by
the curve. (Call this "notch # infinity," n = <x>.)

300

200

Energy based cm
notch** 1=0

lOO

0 -i

301.1
294.0

278.8

235.2

\-12.5
\-19.6

-34.d

73.4

-313.6

-100

JSneryy based on
notch** 00=0

-200

-300

Fig. 15-7. The energy level scheme of the hydrogen
atom.

Historically, the visible emission lines shown in Figure
15-3 were the first atomic hydrogen lines discovered. They
were found in the spectrum of the sun by W. H. Wollaston
in 1802. In 1862, A. J. Angstrom announced that there
must be hydrogen in the solar atmosphere. These lines
were detected first because of the lesser experimental dif-
ficulties in the visible spectral region. They are called the
"Balmer series" because J. J. Balmer was able to formu-
late a simple mathematical relation among the frequencies
(in 1?<?^). The ultraviolet series shown in Figure 15-3 was

SEC. 15-1 I THE HYDROGEN ATOM

259

actually predicted prior to its discovery (in 1915 by
T. Lyman).

Hence our analysis and prediction, selected for logical
clarity, reverse the actual chronology. Our prediction of
the visible spectrum from the ultraviolet spectrum is more
straightforward than was the reverse prediction. Notice
that the ultraviolet spectrum utilizes all of the notches
involved in producing the visible spectrum. In contrast,
none of the visible frequencies involves notch #1, the key
to the ultraviolet spectrum.

15-1.3 The Energy Levels off a Hydrogen Atom

Scientists deduced the notched energy scale of
the hydrogen atom from the spectrum in just the
same way we did.

Of course, more sophisticated language is gen-
erally used. For example, the energy possibilities
are usually shown on a vertical scale and they
are called energy levels rather than notches. Fig-
ure 15-7 shows the energy level scheme of the
hydrogen atom. Each energy level is character-
ized by an integer, n, the lowest level being given
the number 1 .

Two energy scales are shown in Figure 15-7.
On the left is a scale based upon the energy zero
for the n = 1 level. On this scale, the levels
appear at energies corresponding to our notched
beam spacings. The right-hand scale is displaced
upward so that the n = 1 level corresponds to a
negative energy, —313.6 kcal. The zero has been
moved upward by this amount so that the zero
of the energy scale corresponds to the "notch #
infinity" that you estimated in Exercise 15-3.
Obviously, the positioning of the energy levels is
not affected by this arbitrary change of the zero
of energy, so either scale can be used. For rea-
sons that will become evident later in this chap-
ter, the right-hand scale is more convenient and
is the one generally used.

EXERCISE 15-4

Calculate the energy change that occurs between
notch #1 and notch #2, using the right-hand scale
in Figure 15-7. (Remember that the energy
change is the energy of the final level minus the
energy of the initial level. Pay careful attention
to algebraic signs.) Repeat the calculation for the

energy change between notch #2 and #3. Com-
pare your calculations with the numbers shown
in Figure 15-6.

Long after this energy level diagram for the
hydrogen atom had been established, scientists
still pondered its significance. Finally, in the late
1920's, a mathematical scheme was developed
that explained the facts. The mathematical
scheme is called quantum mechanics.

15-1.4 Quantum Mechanics and the
Hydrogen Atom

Prior to the development of quantum mechanics,
the spectrum of the hydrogen atom posed quite a
dilemma. To see the problem, and how it was
resolved, let's go back about fifty or sixty years
and trace the history of this problem. This is a
valuable example because it shows how science
advances.

By the year 1912 it was known that the hydro-
gen atom is composed of a proton and an elec-
tron. These two particles are attracted to each
other by reason of their electric charges. Physi-
cists felt they should be able to calculate the
properties of such a combination. After all, the
laws of motion of macroscopic bodies had been
studied for centuries. The behavior of electrically
charged bodies was also thoroughly understood
on the macroscopic scale. Yet scientists could
not explain why a hydrogen atom exists, let
alone why it would have only particular values of
energy. In fact, the laws that had been deduced
for macroscopic bodies gave the firm (though
incorrect) prediction that the nuclear atom is un-
stable and the electron should collapse into the
nucleus.

At this point a Danish physicist, Niels Bohr,
decided to take a fresh start. In effect, he faced
the fact that an explanation is a search for like-
nesses between a system under study and a well-
understood model system. An explanation is not
good unless the likenesses are strong. Niels Bohr
suggested that the mechanical and electrical be-
havior of macroscopic bodies is not a completely
suitable model for the hydrogen atom. He pro-

260

ELECTRONS AND THE PERIODIC TABLE I CHAP. 15

ceeded to seek a new model that did not contra-
dict the known facts.

He began by supposing that the structure of
an atom (the arrangement of the electrons
around the nucleus) is determined by its energy.
To agree with the facts, Bohr proposed that only
special atomic structures can exist — he called
these special structures "stationary states." Each
such state is characterized by a particular energy,
and since a set of special atomic structures exists,
a corresponding set of energies will be found.
Here Bohr departed from the older atomic
models (those of classical physics) that permitted
structures corresponding to all possible energies.

The most stable state of the atom would be
expected to be the one in which the atom has the
lowest energy. Bohr reasoned that since we ob-
serve that the nuclear atom does exist then it
must be a fundamental fact of nature that an
atom can exist in its most stable state indefinitely.
Even though this fact could not be rationalized
(remember, the earlier laws of physics predicted
the atom should collapse) it had to be accepted
because it was a result of experiments.

Bohr also proposed that although the lowest
energy state of the atom is its most stable state,
the atom can be excited to its higher allowed
energy states (by absorbing light or through a
violent collision with other atoms or electrons).
The excited atom does not remain in this condi-
tion for long; it loses its extra energy by emitting
light. Since only certain levels of energy exist,
only certain energy changes can occur. The en-
ergy change of the atom must be equal to the
energy of the light emitted, in accord with equa-
tion (2), E2 — Ei = hv. Consequently, the fre-
quency of the light emitted by an atom is entirely
determined by the values of the allowed energies
of its electrons.

These ideas were so revolutionary that they
would not have been accepted except for the fact
that Bohr was able to propose a way to calculate
exactly the energy levels for the hydrogen atom.
Within ten years Bohr's calculational methods
were completely replaced by better techniques,
but his postulate that only certain atomic energy
states are possible has been repeatedly shown to
be correct.

In this example there is much food for thought con-
cerning the development of science. The wide applicability
of the laws of motion and of electromagnetics made it
natural for scientists to assume that these same laws,
without change, applied to the atom. True, this repre-
sented an extrapolation, for the laws were deduced on the
macroscopic level. Yet, the same laws that described the
motions of the planets also described the motions of
tennis balls — why not also the motions of electrons?
Many experimental facts said no, but physicists held to
the expectation that a way would be found to explain
these facts within the framework of the known (and
almost sacred) laws. When Bohr finally broke away from
the established laws, he still used them for guidance,
proposing only those changes required by the discordant
facts. Perhaps the most ironic part is that Bohr's principal
weapon in gaining acceptance for his new attack was his
mathematical success in predicting the energy levels of
hydrogen, though his model has since been discarded
completely. The model he used proved to fit only the
hydrogen atom and no other.

Don't be too eager, though, to scoff at this example.
Instead, you may rest assured that some of the theories
you will find in this book are waiting to be swept aside
as we learn more about nature. The rub — no, the excite-
ment— in the game is that we don't know which theories
are fated to go. That remains for some of you to discover!

15-1.5 The Hydrogen Atom and
Quantum Numbers

The modern theory of the behavior of matter,
called quantum mechanics, was developed by
several workers in the years 1925-1927. For our
purposes the most important result of the quant-
um mechanical theory is that the motion of an
electron is described by the quantum numbers
and orbitals. Quantum numbers are integers
that identify the stationary states of an atom;
the word orbital means a spatial description of
the motion of an electron corresponding to a par-
ticular stationary state.

THE PRINCIPAL QUANTUM NUMBER, n

If we return to our notched beam analogy, as
shown in Figure 15-5, we find that we numbered
the notches, 1, 2, 3, •••. These numbers serve
as natural identifying designations. They are the
"quantum numbers" of the balance beam.

In Exercises 15-2 and 15-3 you observed that
the energy levels of hydrogen vary systematically
with the quantum (or notch) number. The

SEC. 15-1 I THE HYDROGEN ATOM

261

smooth curves suggest that the energy could be
conveniently expressed mathematically in terms
of n, the quantum number. Using the right-hand
scale in Figure 15-7, we see that for any value
of n, E is always negative. As n becomes larger,
the energy rises and approaches zero on the
scale. Investigation of the actual energy values
shows that the energy levels of Figure 15-7 are
exactly determined by n according to the relation

£„ = — * kcal mole (4)

En = energy level with quantum number n
n = 1,2,3, ••■ x

The number n is called the principal quantum
number.

The mathematical relationship (4) is the one Bohr was
able to deduce. Current quantum mechanical methods
also deduce this relationship, of course, but with a model
that is in fundamental discord with the one used by Bohr.

ORBITALS AND THE PRINCIPAL
QUANTUM NUMBER, n

Quantum mechanics provides a mathematical
framework that leads to expression (4). In addi-
tion, for the hydrogen atom it tells us a great
deal about how the electron moves about the
nucleus. It does not, however, tell us an exact
path along which the electron moves. All that
can be done is to predict the probability of find-
ing an electron at a given point in space. This
probability, considered over a period of time,
gives an "averaged" picture of how an electron
behaves. This description of the electron motion
is what we have called an orbital.

Thus, an orbital description of the motion of
an electron contains the same information con-
veyed by the holes made by darts in a dartboard.
After the board has been used in many games,
the distribution of holes shows how successful
earlier players had been in their scoring. There
are many holes near the bullseye and, moving
away from it, there is a regular decrease in
the number of holes per square centimeter of
dartboard. At any given distance from the bulls-
eye, the "density" of dart holes (number per
square centimeter) is a measure of the probability
that the next throw will land there.

We see that the holes in the dartboard tell us
only the probability that a given throw will land
a particular distance from the bullseye. It does
not tell us the order in which the holes were made
in the dartboard. In the case of the electron dis-
tribution, the orbital tells us the probability that
an experiment designed to locate the electron
will find it a particular distance from the nucleus.
It does not tell us how the electron moves from
point to point— its trajectory.

Though quantum mechanics does not tell us
the electron trajectory, it does tell us how the
orbital changes as n increases. It also indicates
that for each value of n there are n2 different
orbitals. For the hydrogen atom, the n2 orbitals
for a particular value of n all have the same en-
ergy,

313.6 kcal
n- mole

S ORBITALS

Consider the lowest energy level of a hydrogen
atom, n = 1 . We have just learned that there are
n2 levels with this energy, and since n = 1, there
is but one level. It corresponds to an electron
distribution that is spherically symmetrical
around the nucleus, as shown in Figure 15-8. It
is called the Is orbital.* An electron moving in
an s orbital is called an s electron.

The term "spherically symmetric" and the
picture of an s orbital (Figure 15-8) must be
clearly understood. They indicate that if we were
to look for the electron somewhere on the sur-
face of a sphere of a particular radius, ru which
has the nucleus at its center, the probability of
finding the electron at any one point on the r\
sphere is the same as the probability of finding
it at any other point on the rx sphere. The same
would be true at a different radius, r2, but the
probability of finding the electron somewhere on
the r2 sphere would not be the same as that on
the rx sphere. The chance of finding the electron
does depend upon the radius of the sphere on

* In the symbol \s, the number 1 tells us that n = 1.
The letter s tells us that the orbital is spherically sym-
metric. Since the letter s has been used for an orbital that
is spherically symmetric, we might as well think of it as
an abbreviation : s = spherical.

262

ELECTRONS AND THE PERIODIC TABLE I CHAP. 15

1 S orbita.1

-313.6/4 = -78.4 kcal/mole. One of them is
again spherically symmetric and it is called the
25 orbital. Figure 15-8 shows the atomic 2s
orbital. Here we find the reasonable result that
the higher energy of the 2s electron permits the
electron to spend more time far from the nucleus.
For every value of n, there is one spherically
symmetric orbital. As n increases, the ns orbitals
place the electron, on the average, farther and
x farther from the nucleus.

2 s orbi-tal
Fig. 15-8. Atomic orbitals: Is and 2s orbitals.

which we search. A \s electron can be found any-
where from right at the nucleus to a great dis-
tance away — but it is most likely to be found
approximately 10~8 cm from the nucleus.

The next energy level corresponds to n = 2.
According to our rule, there are n2 = 22 = 4 dif-
ferent spatial arrangements with the same energy,

p ORBITALS

There is only one orbital corresponding to n = 1 ,
the Is orbital. For n = 2, there are four different
spatial arrangements and we have described one
of them, the 2s orbital. The other three are called
2p orbitals. An electron in a p orbital behaves
in such a way that it is most likely found in either
of two regions located on opposite sides of the
nucleus. The motion of a p electron creates an
electron distribution that is shaped somewhat
like a dumbbell. We can place the axis of the
dumbbell along one of the three perpendicular
cartesian coordinate axes. Just as there are three
distinct coordinate axes, there are three distinct
p orbitals, each with its axis perpendicular to the
other two. They are sometimes referred to as the
pz, py, and pz orbitals to emphasize their direc-
tional character. The p£ orbital is concentrated
in the x direction — a px electron is more apt to
be found near the x axis than anywhere else. The
pu orbital, on the other hand, is concentrated
along the y axis. These directional characteristics
are useful in explaining the geometrical proper-
ties of molecules. Figure 1 5-9 shows the electron
distribution of the 2p orbitals.*

Every energy level with n above 1 has three p
orbitals. As n increases, the np orbitals place the
electron, on the average, farther and farther from
the nucleus, but always with the axial directional
properties shown in Figure 1 5-9.

d AND / ORBITALS

At this point we might recast the hydrogen atom
energy level diagram to express what we know

* The three p orbitals can be considered to extend along
the three perpendicular axes, x, y, and r. Hence, p might
be thought of as an abbreviation: p = /perpendicular.

SEC. 15-1 I THE HYDROGEN ATOM

263

2p x
orbital

Fig. 15-9. Atomic orbitals: the 2p orbitals.

about orbitals. Figure 15-10 shows each orbital
as a pigeon hole. For a given value of n, there
are n2 total orbitals. For n = 3, there are 32 = 9
orbitals, five more than accounted for by one 3s
orbital and the three 3p orbitals. These five or-
bitals are called d orbitals and they have more
complicated spatial distribution than do the p
orbitals.

EXERCISE 15-5

From the information that the numbers of s, p,
and d orbitals are 1, 3, and 5, how many of the
next higher (/) orbitals would you expect?
Verify your answer by calculating n2 for n = 4
and comparing to your sum of the numbers of
s, p, d, and /orbitals.

15-1.6 The Hydrogen Atom and
the Periodic Table

At last we are ready to return to the periodic
table. At last we are able to begin answering
those who are "wondering why" about the spe-
cial properties of the electron populations in
Table 15-1. Let us reproduce Table 15-1 together
with the numbers of orbitals of the hydrogen
atom. The suggestion of a connection is irresist-
ible, as seen in Table 15-11.

The hydrogen atom orbitals give us the num-
bers 2, 8, 18, and 32 — the numbers we find
separating the specially stable electron popula-
tions of the inert gases. It was necessary to
multiply n2 by two — an important factor that
could not have been anticipated. Furthermore,
it will be necessary to find an explanation for the
occurrence of eight-electron differences both at
neon and at argon and eighteen-electron differ-
ences both at krypton and at xenon.

Nevertheless, we seem to have a significant
start toward explaining the periodic table. We

Table 15-II. stable electron populations and the hydrogen atom

THE INERT GASES THE HYDROGEN ATOM

number

number of

element

electrons

differences

orbitals

2X«'

helium

n = 1

n2 = 1

2X1=2

neon

10 - 2 = 8

n = 2

W2 - 4

2X4=8

argon

18 - 10 = 8

n = 3

„2 = 9

2 X 9=18

krypton

36 - 18 = 18

n = 4

rt2 = 16

2 X 16 = 32

xenon

54 - 36 = 18

radon

86 - 54 = 32

264

ELECTRONS AND THE PERIODIC TABLE I CHAP. 15

-100

Energy,
heal/mole

©• •©©© ■•@®@®@

-t

-200

-300

Us)

Fig. 15-10. The energy level scheme of the hydrogen
atom.

can understand the chemical trends within the
first ten elements in terms of the hydrogen atom
orbitals with the following two assumptions:

(1) that atoms of these elements have orbitals
and energy levels that are qualitatively like
those of the hydrogen atom;

(2) that a single orbital (of any element) can
accommodate, at most, two electrons.

The first assumption springs from the similarities
noticed in Table 15-11 and is well substantiated
by a wealth of spectral study of these atoms. The
second assumption is stimulated by the factor of
two needed in the last column of Table 15-11.
With these two assumptions, we can propose
the electronic arrangement of lowest energy for
each atom. We do so by mentally placing elec-
trons successively in the empty orbitals of lowest
energy. The electron orbital of lowest energy is
the Is orbital. The single electron of the hydro-
gen atom can occupy this orbital. In the helium

SEC. 15-2 | MANY-ELECTRON ATOMS

265

atom, there are two electrons and the nuclear
charge is two. Since each orbital can accommo-
date two electrons, both of the electrons can go
into the Is orbital. The resulting electronic ar-
rangement is described by the notation Is2, which
means that there are two electrons in the Is
orbital. The notation Is2 is called the electron
configuration.

Now let us consider what will happen when
there are three electrons near a triply charged
nucleus, as in the lithium atom. The first two of
the three electrons go into the lowest energy
orbital, the Is orbital. When this orbital has two
electrons in it, it is completely filled, according
to our second assumption. Any additional elec-
trons must be placed in orbitals of higher energy.
Therefore, the third electron in lithium goes into
the 2s orbital, and we write the electron configu-
ration as ls22s'. Despite the nuclear charge of
three in the lithium atom, this last electron is
rather weakly bound because the 2s electron in
lithium spends most of its time farther away from
the nucleus than do the Is electrons. This elec-
tron should be easily removed to give Li+, which
is, indeed, the characteristic behavior of an
alkali element.

The beryllium atom has one more electron
than does the lithium atom. The fourth electron
that enters the beryllium atom can occupy the 2s
orbital to give a configuration of ls22s2. The two
2s electrons will be most easily removed, tending

to form the Be+- ion.

There is no more room in the 2s orbital for a
fifth electron, which appears when we move on
to the boron atom. However, another orbital
with principal quantum number 2 is available.
A 2p orbital accepts the fifth electron, giving the
configuration \s22s22pl. Continuing this process,
we obtain the following configurations:

carbon atom

Is2

2s22p\2p\

nitrogen atom

Is2

2s22p\2p\2p\

oxygen atom

Is2

2s22p\2p\2p\

fluorine atom

Is2

2s22p]2pl2p\

neon atom

Is2

2s22plx2p\2p\

If we proceed to the next element, sodium atom,
we are again forced to use an orbital with the
next higher quantum number:

sodium atom Is2 2s22p\2p\l2p\ 3s1

Again there is one electron which spends most
of its time farther away from the nucleus than
any of the others. This one electron could be
easily removed to give Na+, and we return to the
type of chemistry shown by lithium.

The hydrogen atom energy levels, together
with our two assumptions, have provided a good
explanation of some of the properties of the first
eleven elements. We shall see that they explain
the entire periodic table.

15-2 MANY-ELECTRON ATOMS

All atoms display line spectra. In general these
spectra are much more complicated than the
atomic hydrogen spectrum shown in Figure 15-3.
Nevertheless, these spectra can be interpreted in
terms of the concepts we have developed for the
hydrogen atom.

15-2.1 Energy Levels of Many-Electron Atoms

Analysis of the spectra of many-electron atoms
shows the following similarities to the hydrogen
atom case.

(1) All atoms have "stationary states" and can
hold only particular values of energy.

(2) The atomic spectra can be understood in
terms of transitions between energy levels
corresponding to these particular values of
energy.

(3) The energy level diagrams resemble the hy-
drogen atom level diagram except that the n2
levels with the same value of n no longer all
have the same energy.

Figure 15-11 shows a schematic energy level
diagram of a many-electron atom. Blue patterns

Arbitrary

energy

Scale

65 ooo

Fig. 15-11. A schematic energy level diagram of a many-electron atom.

SEC. 15-3 | IONIZATION ENERGY AND THE PERIODIC TABLE

267

indicate levels with the same hydrogen atom
principal quantum number. The effect of placing
many electrons into the energy levels of the hy-
drogen atom is a "tilting" of the diagram. The/?
orbitals are slightly higher in energy than are the
s orbitals of the same value of n. The d orbitals
and / orbitals of this same value of n are suc-
cessively even higher. The result is that the 3d
orbitals are raised approximately to the energy
of the 45 and 4p orbitals. The 35 and 3p orbitals
are left more or less isolated in energy. They are
much higher in energy than are the 25 and 2p
orbitals but much lower than is the cluster of
45, 4/7, and 3d orbitals.

15-2.2 The Periodic Table

Now we can see the development of the entire
periodic table. The special stabilities of the inert
gases are fixed by the large energy gaps in the
energy level diagram, Figure 15-11. The number
of orbitals in a cluster, multiplied by two because
of our double occupancy assumption, fixes the
number of electrons needed to reach the inert
gas electron population. The numbers at the

right of Figure 15-11 are exactly the numbers we
found as differences between inert gas electron
populations (see Table 15-1).

It is important to recognize what we have
done. We have shown how the energy level dia-
gram of hydrogen was deduced from the atomic
hydrogen spectra. We have seen that the energy
level diagram of a many-electron atom can be
regarded as a tilted hydrogen atom diagram.
Such a diagram contains clusters of energy levels
with large energy gaps between. These clusters of
energy levels provide a basis for explaining the
electron populations of the inert gases, provided
we assume two electrons can occupy each orbital.

We have not explained why two, but no more
than two, electrons can occupy each orbital. This
is not known and is accepted because the facts
of nature require it. This assumption is called
the Pauli Principle.

We have given much explanation of why the
clusters and energy gaps of the energy level dia-
gram should give rise to a periodicity of chemical
properties as we move across the periodic table.
We shall seek such an explanation in terms of
the energy necessary to remove an electron from
an atom, the ionization energy.

15-3 IONIZATION ENERGY AND THE PERIODIC TABLE

The amount of energy required to remove the most
loosely bound electron from a gaseous atom is
called the ionization energy. We can repre-
sent this process by the equation

gaseous atom + energy

gaseous ion

+ gaseous electron

In terms of our energy level diagram it is the
energy necessary to lift an electron from the
highest occupied orbital the rest of the way up to
the limit corresponding ton = <». Figure 15-12
shows the ionization process for lithium atom in
terms of an energy level diagram. Each orbital
is shown as a pigeon hole, designated O. If this
orbital is occupied by one electron, it can be
indicated by a diagonal line across the pigeon
hole: 0. If two electrons occupy the orbital,
crossed diagonal lines are shown, one for each

electron: <g>. Of course we assume the Pauli
Principle that only two electrons can occupy a
given orbital. Hence, an orbital shown as <g> is

Fig. 15-12
o

-500

^ -1,000

,* -isoo

The ionization of a lithium atom

0tooo /p

J24 kcal

OOO 2P
2s

Li ($) + energy — »■ Li*(g) ■/■ e~(g)

268

ELECTRONS AND THE PERIODIC TABLE I CHAP. 15

filled; an orbital shown as Q or 0 is half-
occupied ; an orbital shown as O is empty.

The term ionization energy is also applied to
the removal of the most loosely bound electron
from an atom that has already lost one or more
electrons (that is, an ion). Hence the "ionization
energy" of, say, Mg+ is the energy of the process

Mg+ (g) -f energy -* Mg+2 (g) + e~(g)

Since the process removes the second electron
from a magnesium atom, the ionization energy
of Nig"1" is called the "second ionization energy"
of magnesium.

15-3.1 Measurement of Ionization Energy

The ionization energy provides a basis for under-
standing the periodicity of the chemistry of the
elements. Owing to the stimulation of Bohr's
ideas, many systematic determinations of ioniza-
tion energies were carried out between 1914 and
1920. The first determinations were done by
bombarding an atomic vapor with electrons
whose kinetic energy was known accurately.
When the kinetic energy of the bombarding elec-
trons is increased to a certain critical value,
singly charged positive ions can be detected elec-
trically. These ions result from collisions between
atoms and the bombarding electrons that have
been given just enough kinetic energy to cause
the most weakly bound electron to be ejected
from the atom. This critical value was found to
be characteristic of the substance being investi-
gated. Table 15-111 shows some of the measured
ionization energies for some of the lighter ele-
ments.*

Table 75-///

IONIZATION ENERGIES
OF THE ELEMENTS

ATOMIC
NUMBER

ELEMENT

IONIZATION ENERGY

(kcal/mole)

313.6

566.7

124.3

214.9

191.2

259.5

335

313.8

401.5

497.0

118.4

175.2

137.9

187.9

241.7

238.8

300

363.2

100.0

following it. This is followed by a slow rise in the
ionization energy (another regularity) as we
proceed across a row of the periodic table. The
best way to see the trend is in a graphical pre-
sentation of the data of Table 15-111, as shown
in Figure 15-13. We see that the ionization en-

Fig. 15-13. Ionization energy as a function of the
atomic number.

15-3.2 Trends in Ionization Energies

Examine the figures in the last column of Table
15-111 and search for regularities. The most ob-
vious one is the dramatic change in ionization
energy between each inert gas and the element

* The ionization energy is given in this book in units
of kilocalories per mole, the energy that would be re-
quired to remove an electron from each one of a mole of
atoms. These units allow an easy comparison between
ionization energies and the energy changes that occur in
ordinary chemical reactions.

600

4-00

ZOO -

2 4

6 8 10 1Z 14 16 16 20 ZZ
Atomic number

SEC. 15-3 | IONIZATION ENERGY AND THE PERIODIC TABLE

269

ergy increases more or less regularly across a
row of the periodic table, reaching a maximum
at the inert gas. As soon as we encounter an
alkali metal, we notice that the ionization energy
is quite small, and in subsequent elements the
general upward trend repeats itself. There is a
startling similarity between the regularities we
have found in the ionization energies and the
periodicity of chemical properties. We shall see
that this is not an accident: the trend in chemical
behavior as we move across the periodic table
can be explained in terms of the trends in the
ionization energies.

Let us begin by contrasting the ionization
properties of sodium and chlorine:

Nate; — *• Na+teJ + e~(g)

AH = +118.4 kcal/mole
Cite; —*■ C\+(g) + e-(g)

AH = + 300 kcal/mole

Since sodium has a low ionization energy, 1 18.4
kcal/mole, a relatively small amount of energy
is required to remove an electron. This is con-
sistent with the chemical evidence that sodium
tends to form compounds involving the ion, Na+.
The ease of forming Na+ ions can be explained
in terms of the low ionization energy of the
sodium atom. In contrast, it requires a large
amount of energy, 300 kcal/mole, to remove an
electron from a chlorine atom. It is not surpris-
ing, then, that this element shows little tendency
to lose electrons in chemical reactions. Instead,
chlorine commonly acquires electrons to form
negative ions, CI-.

Between sodium and chlorine, there is a slow
rise in ionization energy. For magnesium and
aluminum the ionization energy is still rather
low. Hence electrons are readily lost and positive
ions can be expected to be important in the

chemistry of these elements. As the ionization
energy rises, the chemistries of silicon, phos-
phorus, and sulfur show a trend toward electron
sharing. For these elements, an inert gas electron
population cannot be reached by losing electrons
because the ionization energy required is too
high. They seek the inert gas stability by sharing
electrons or, for sulfur and chlorine, by acquiring
electrons to form negative ions.

These correlations between ionization energy
and chemical properties confirm the idea that
the electronic structure of an element determines
its chemical behavior. In particular, the most
weakly bound electrons are of greatest im-
portance in this respect. We shall call the
electrons that are most loosely bound, the valence
electrons.

15-3.3 Ionization Energies and
Valence Electrons

It is possible to remove two or more electrons
from a many-electron atom. Of course it is
always harder to remove the second electron
than the first because the second electron to come
off leaves an ion with a double positive charge
instead of a single positive charge. This gives an
additional electrical attraction. Even so, the
values of successive ionization energies have
great interest to the chemist.

Consider the three elements, sodium, magne-
sium, and aluminum. For each of these elements
we know several ionization energies, correspond-
ing to processes such as the following:

NateJ
Na+teJ

Na+teJ + e-(g)

Ei = first ionization energy

(5)

Na+Ygj + e-(g)

E2 = second ionization energy (6)

Table 15-IV. successive ionization energies of Na, ms. and ai

(KILOCALORIES PER MOLE)

ELEMENT

ELECTRON CONFIGURATION,
NEUTRAL ATOM

sodium

Is*

2s>2?

3sl

118

1091

1653

—

magnesium

Is1

2y*2/7«

3j»

175

345

1838

2526

aluminum

Is*

2$lp

3^3/7'

138

434

656

2767

270

ELECTRONS AND THE PERIODIC TABLE I CHAP. 15

Na+3(gJ

Na+3fg; + e-(g)

E3 = third ionization energy

(7)

Na+"(gj + e-(g)

Ei = fourth ionization energy 8)

The experimental values of these energies are
shown in Table 15-IV. Let us begin by comparing
sodium and magnesium. For each, the first
ionization process removes a 3s electron, the
most weakly bound. Nevertheless, the ionization
energies are somewhat different:

Na(g J +H8 kcal — + Na+(g) + er(g) (9)
Mg(g) + 175 kcal — *- Mg+(g) + e~(g) (10)

The difference is caused by the higher nuclear
charge of magnesium. Magnesium is element 12,
hence it has twelve protons in the nucleus, com-

pared to eleven protons in the nucleus of the
sodium atom. Of course the valence electrons are
more strongly attracted to the +12 nucleus of
Mg than the +11 nucleus of Na.

The second ionization energy, however, re-
verses the situation:

Na+fgj + 1091 kcal
Mg+(g) + 345 kcal

Na+Ygj + e~(g)
Mg-W + e-(g)

(12)

For sodium, it takes three times as much energy
to remove the second electron as it does for
magnesium. We can understand the energies in

Fig. 15-14. Energy level diagram representation of

the ionization of magnesium and sodium
atoms.

-1,000

-2,000±r

t OOO 3*

175 kcal

2j»

OOO

345 kcal

OOO

1838 kcal

o OOO

T®

T® /* T®

Mg(9) + 175 kcal— Mg+(9) + e~(g) MS+2(g) + 1838 kcal~Mo+3(9) + e~(S)

Aig+(g) + 345 kcal 2*fg+2(9) + e~(9)

lOOO 33p

118 kcal
m ®®® ±Zf

OOO

1091 kcal

®®0

^ -1,000

-2,000}?

J® Is T<

Na(g) + 118 kcal—Na+(g) + e~ (g)

Na+ (g) + 1091 kcal -Na.+Z(g) + e~ (g)

OOO

SEC. 15-3 | IONIZATION ENERGY AND THE PERIODIC TABLE

271

(77) and (72) in terms of two principles. First,
for magnesium the second ionization energy (345
kcal) exceeds the first (175 kcal) because reaction
(72) takes a 3s electron away from the positively
charged ion, Mg+, whereas reaction (70) removes
a 3s electron from an initially neutral atom. This
same factor is operative in sodium, of course,
but in addition, the second ionization energy of
sodium must remove an electron from a 2p
orbital instead of the 35 orbital. This 2p orbital
is one of a cluster much lower in energy than
the 35-3/j cluster. Figure 15-14 shows that this
behavior is readily understood with the aid of
the energy level diagram. We conclude that it is
difficult to remove two electrons from sodium
but it is relatively easy to remove two electrons
from magnesium. This is why we say sodium has
one valence electron and magnesium has two.
Removing more than one electron from sodium
or more than two from magnesium is very diffi-
cult because orbitals much lower in the energy
diagram are involved.

Continuing to aluminum, we see that its first
ionization energy is below the first ionization
energy of magnesium — despite the fact that alu-
minum has the higher nuclear charge. We find
the explanation, however, in the second column
of Table 15-1V. For aluminum we remove a 3p
electron to produce Al+ whereas a 35 electron
is removed from Mg to form Mg+. A glance at
the energy level diagram, Figure 15-11, shows
that the 3>p level is somewhat above the 35 level,
hence it should be easier to remove the 3p
electron.

If we continue to remove electrons from alu-
minum, we discover a very large increase in
ionization energy when the fourth electron is re-
moved. Again this is because the fourth electron
must be withdrawn from a 2p orbital, an orbital
much lower on the energy level diagram. We
conclude that three electrons, the two 35 and the
one 3p, are more easily removed than the others.
Since aluminum has three easily removed elec-
trons, aluminum is said to have three valence
electrons.

Remembering how we placed electrons in the
lowest empty orbitals, two per orbital, we can
now generalize concerning the number of valence

electrons a given atom possesses. We count the
electrons placed in the orbitals that form the
highest partially filled cluster of energy levels.
These electrons are most easily removed and
they determine the chemistry of the atom.

EXERCISE 15-6

Explain why chemists say that boron has three
valence electrons and that chlorine has seven.
How many valence electrons has fluorine? Oxy-
gen? Nitrogen?

15-3.4 The Fourth Row of the Periodic Table

Turn back to Figure 15-1 1, the energy level dia-
gram of a many-electron atom, and consider the
occupied orbitals of the element potassium. With
19 electrons placed, two at a time, in the orbitals
of lowest energy, the electron configuration is

potassium atom Is2 2s*2p6 3s*3p6 4s1

Potassium has one valence electron. It is the first
member of the fourth row, the row based on the
cluster of orbitals with about the same energy as
the 45 orbital. There are nine such orbitals, the
45 orbital, the three Ap orbitals, and the five 3d
orbitals. Hence the fourth row of the periodic
table will differ from the second and third rows.
The fourth row, as seen in the periodic table,
consists of eighteen elements.

Calcium is the second element of the fourth
row. It has two electrons more than the argon
inert gas population and these two electrons both
occupy the 45 orbital:

calcium atom Is2 2522p6 3523p6 4s2

When we add the next electron to form the
element scandium, the orbital of lowest energy
that is available is one of the 3d orbitals (since
the 3d orbitals are slightly lower in energy than
the 4p orbitals). As succeeding electrons are
added to form other elements, they enter the 3d
orbitals until the ten available spaces in these
orbitals are filled.

The elements that are formed when the 3d elec-
trons are added are called the transition metals

272

ELECTRONS AND THE PERIODIC TABLE I CHAP. 15

or the transition elements. Since their chemi-
cal properties and electronic structures are unlike
those of any of the lighter elements we have
discussed, it is reasonable that these transition
elements should head a new set of ten columns
of the periodic table. The next orbitals in line
for occupancy are the 4/7, and it is not surprising
that the chemical properties of the element gal-
lium, which has one 4/? electron, resemble those
of aluminum, which has one 3p electron. This
row of the periodic table is completed when the
4/7 orbitals are entirely filled. We see that the
reason the fourth row of the periodic table con-
tains eighteen elements is that the five 3d orbitals
have energies that are approximately the same
as those of the 45 and 4/7 orbitals. The ten extra
spaces for electrons provided by the 3d orbitals
increase the length of the period from eight to
eighteen.

Just as the long fourth row of the periodic table arises
from filling the 4s, 3d, and 4p orbitals, the fifth row, which
also consists of eighteen elements, comes from filling the
5s, 4d, and 5p orbitals. In the sixth row, something new
happens. After the 6s and the first one of the 5d electrons
have entered, subsequent electrons go into the 4) orbitals.
The fact that there are seven 4/ orbitals means that
fourteen electrons can be accommodated in this manner.
Filling the 4/orbitals gives rise to a series of elements with
almost identical chemical properties called the rare earth

elements. Once the 4/orbitals are full, electrons enter the
5dand 6p orbitals until the sixth period is completed with
the inert gas radon.

We see that the rows of the periodic table arise
from filling orbitals of approximately the same
energy. When all orbitals of similar energy are
full (two electrons per orbital), the next electron
must be placed in an s orbital that has a higher
principal quantum number, and a new period of
the table starts. We can summarize the relation
between the number of elements in each row of
the periodic table and the available orbitals of
approximately equal energy in Table 15-V.

Table 15-V

THE NUMBER OF ELEMENTS IN EACH
ROW OF THE PERIODIC TABLE

ROW OF

NO. OF

LOWEST ENERGY ORBITALS

TABLE

ELEMENTS

AVAILABLE TO BE FILLED

2s, 2p

3s, 3p

4s, 3d, 4p

5s, 4d, 5/7

6s, 4/, 5d, 6/7

Is, 5f, 6d, Ip

QUESTIONS AND PROBLEMS

1. Which of the following statements concerning
light is FALSE?

(a) It is a form of energy.

(b) All photons possess the same amount of
energy.

(d) It includes the part of the spectrum called
X-rays.

2. Use the energy level diacram in Figure 15-7 to
calculate the energy required to raise the electron
in a hydrogen atom from level #1 to level #2;
from level #1 to level #3; from level #1 to level
#4. Compare these energies with the spectral
lines shown in Figure 15-3, p. 255.

Your plot in Exercise 15-2 suggested that the
energy levels given in Figure 15-7 are systemati-
cally related. To explore this relationship further,
divide the energy of each level by that of the
first level (using the right-hand scale). How are
the fractions so obtained related to the numbers
of the energy levels?

Calculate, using frequency units (cycles per sec-
ond) and Figure 15-3, the lines predicted by the
notched beam due to changes beginning in notch
§7>. Use the complete light spectrum shown in
Figure 14-14 (p. 247) to decide in what spectral
region these additional lines were found. (It was
this sort of prediction that actually led to the
discovery of this set of lines.)

QUESTIONS AND PROBLEMS

273

5. According to the quantum mechanical descrip-
tion of the \s orbital of the hydrogen atom, what
relation exists between the surface of a sphere
centered about the nucleus and the location of
an electron ?

6. What must be done to a 25 electron to make it a
3s electron? What happens when a 35 electron
becomes a 25 electron?

7. If the energy difference between two electronic
states is 46.12 kcal/mole, what will be the fre-
quency of light emitted when the electron drops
from the higher to the lower state?

Planck's constant

= 9.52 X 10~14 (kcal sec)/mole

8. Determine the value of En for n — 1, 2, 3,
4, for a hydrogen atom using the relation
En = — 313.6/h2. For each En, indicate how
many orbitals have this energy.

9. The quantum mechanical description of the Is
orbital is similar in many respects to a descrip-
tion of the holes in a much used dartboard. For
example, the "density" of dart holes is constant
anywhere on a circle centered about the bullseye,
and the "density" of dartholes reaches zero only
at a very long distance from the bullseye (effec-
tively, at infinity). What are the corresponding
properties of a 15 orbital?

In the light of your answer, point out errone-
ous features of the following models of a hydro-
gen atom (both of which were used before
quantum mechanics demonstrated their inade-
quacies).

(a) A ball of uniform density.

(b) A "solar system" atom with the electron
circling the nucleus at a fixed distance.

10. Name the elements that correspond to each of
the following electron configurations

l52

l52 251

l52 2522p1

I52 2s22p3

\s2 2522p6 3523p6 4a1

1 1 . Make a table listing the principal quantum num-
bers (through three), the types of orbitals, and
the number of orbitals of each type.

12. The electron configuration for lithium is l5225'
and for beryllium it is l52252. Estimate the ap-
proximate ionization energies to remove first
one, then a second, electron. Explain your esti-
mates.

13. What trend is observed in the first ionization
energy as you move from lithium down the col-
umn I metals? On this basis, can you suggest a
reason why potassium or cesium might be used
in preference to sodium or lithium in photo-
electric cells?

14. Consider these two electron populations for
neutral atoms:

A. I52 2522/?6

B. I52 2522p6

351;
6sl.

Which of the following is FALSE?

(a) Energy is required to change A to B.

(b) A represents a sodium atom.

(d) Less energy is required to remove one elec-
tron from B than from A.

15. How many valence electrons has carbon? Sili-
con? Phosphorus? Hydrogen? Write the elec-
tron configurations for neutral atoms of each
element.

16. The first four ionization energies of boron atoms
are as follows:

Ex = 191 kcal/mole
£2 = 578
£3 = 872
£4 - 5962

Explain the magnitudes in terms of the electron
configurations of boron and deduce the number
of valence electrons of boron.

CHAPTER

Molecules in the
Gas Phase

For the nature of the chemical bond is the problem at the heart of all
chemistry.

BRYCE CRAWFORD, JR., 1953

A molecule is a cluster of atoms that persists
long enough to have characteristic properties
which identify it. The questions we would like
to answer are "Why does the cluster of atoms
persist?" and "Why does the clustering result in
characteristic properties?" In this chapter we

will restrict our attention to molecules as they
exist in the gas phase. Then, in Chapter 17, we
shall consider what additional ideas we need in
order to understand the forces which cause the
formation of liquids and solids.

16-1 THE COVALENT BOND

When two atoms become fixed together they are
said to form a molecule. In "explaining" the
properties of a molecule we use models such as
two styrofoam spheres glued together, or two
wooden spheres held together by a stick, or
perhaps by a spring. In each model it is necessary
to provide a connection — glue, stick, or spring.
It is natural to assume that there is a connection
between the atoms in a molecule. This connec-
tion is called the chemical bond.

16-1.1 The Hydrogen Molecule

Under normal conditions of temperature and
pressure, hydrogen is a gas. By weighing a meas-
274

ured volume of hydrogen gas and applying
Avogadro's Hypothesis, we discover that a mole-
cule of hydrogen contains two hydrogen atoms.
Only if the temperature is raised to several thou-
sand degrees are the collisions with other mole-
cules sufficiently energetic to knock a hydrogen
molecule apart:

H2(g) +±: U(g) + U(g) AH = 103.4 kcal (/)

Since energy is absorbed in reaction (7), the
molecule H2 is more stable (has a lower energy)
than two separated atoms. This chemical bond
(and every chemical bond) forms because the
energy is lower when the atoms are near each
other.

SEC. 16-1 I THE COVALENT BOND

275

'lA

Fig. 16-1. The formation of a molecule of hydrogen,
Hs.

THE ORIGIN OF THE STABILITY
OF THE CHEMICAL BOND

To see why the energy is lower when the atoms
are near each other, we must examine interac-
tions among the electric charges of the atoms.
Figure 16-1 shows the reverse of reaction (7) in
a schematic way. Quantum mechanics tells us
that the Is orbital of each hydrogen atom has
spherical symmetry before reaction. This is sug-
gested by the shading in Figure 16-1. Yet, at any
instant, we picture the electron at some particu-
lar point, as shown by the negative charge of
electron 1 located a distance riA from nucleus A.
The energy of hydrogen atom A can be explained

1A\

in terms of the average attraction between elec-
tron 1 and nucleus A. This is fixed by the average
of the distance between the two, riA. The same
is true of hydrogen atom B — electron 2 and
nucleus B attract each other. Now consider the
new electrical interactions present after the two
atoms have moved close together. Now electron
1 feels the attraction of both protons. Electron 2,
as well, feels the attraction of both protons. This
is the "glue" that holds the two atoms together.
The chemical bond in H2 forms because each oj
the two electrons is attracted to two protons
simultaneously. This arrangement is energetically
more stable than the separated atoms in which

Fig. 16-2. Attractive and repulsive forces in the hy-
drogen molecule.

Electron- proton, distances:
attractive forces

Electron.- electron and

proton -proton distances:

repulsive forces

■■J

■^

-o-

New

®-

276

MOLECULES IN THE GAS PHASE I CHAP. 16

each electron is attracted to only one proton.
Figure 16-2 A shows a possible set of the
electron-proton distances as they might be seen
if it were possible to make an instantaneous
photograph. Such distances fix the attractions
that cause the chemical bond. But it is well to
remember there are also repulsions caused by
the approach of the two atoms, as shown in
Figure 16-2B. The two electrons repel each other
and the two protons do the same. These repul-
sions tend to push the two atoms apart. Which
are more important, the two new attraction

terms or the two new repulsion terms illustrated
in Figure 16-2? Experiment shows that the new
attraction terms dominate — a stable chemical
bond is formed. That is not to say the repulsions
are not felt. In fact, the proton-proton repulsions
prevent the two hydrogen atoms from approach-
ing even closer. The stable bond length in the
hydrogen molecule is fixed by a balance between

Fig. 16-3. Schematic representation of the interaction
between two atoms.

y\.. A simplified representation of a Is orbi-ta.1:

wilt be shown as

with the addition

of shading to
indicate occupancy

B. A simplified representation of orbifral occu.pa.ncy.

Is orbital
emp<ty

Is orbital

containing

one electron ■

"half filled "

Is orbital
con -ta in ing
■two ele c tro n s ■
"filled "

C. Overlap and bonding of the hydrogen molecule :

D. Absence of overlap for two helium atoms .

<=]

2He

SEC. 16-1 I THE COVALENT BOND

277

Electron. -pro-ton distances •'
attractive forces

rV"'S ^4*

the forces of attraction (Figure 16-2 A) and the
forces of repulsion (Figure 16-2B).

We see in Figure 16-2 that bringing two hydrogen
atoms together produces two new repulsions and two new
attractions. Experiment shows that a chemical bond is
formed ; the energy of attraction predominates over the
energy of repulsion. Why is this so? An explanation is
found in the mobility of the electrons. The electrons do
not occupy fixed positions but move about through the
molecule. They take advantage of this mobility to remain
as far from each other as possible. They preferentially
occupy positions like those shown in Figure 16-1 in which
each electron is closer to both nuclei than to the other
electron. The two electrons preferentially move away
from positions in which they would be near each other.
Thus they are said to "correlate" their motion so as to
remain apart, reducing the electron-electron repulsion.

OVERLAP AND THE CHEMICAL BOND

We can simplify our discussion of the chemical
bonding in the hydrogen molecule with the aid
of Figure 16-3. First, in Figure 16-3 A, we picture
the electron distribution in cross-section. The
electron distribution extends outward far from
the nucleus and uniformly in all directions. But
the distribution is concentrated near the nucleus
so we ought to focus our attention on the center
region of the Is orbital. We do so by representing
the Is orbital by a circle with radius large enough
to contain most of the electron distribution.

An orbital can accommodate either one or two
electrons but no more. Figure 16-3B shows a way
of differentiating an empty \s orbital, a Is or-
bital containing one electron, and a Is orbital
containing two electrons.

Now, in Figure 16-3C, consider the interaction

Electron.- electron and
proton-proton distances ■'
repulsive forces

4r.

,-■4-

New /

! ^ '' '

jV«w

Fig. 16-4. Attractive and repulsive forces when helium
atoms approach.

of two hydrogen atoms. Each atom has a single
electron in a b orbital. As the two hydrogen
atoms approach each other, the circles overlap
each other. In this region of overlap the two
electrons are shared by the two protons (as
shown by the crosshatched and shaded area).
This sharing, which permits the two electrons to
be near both protons a good part of the time,
causes the chemical bond. When a bond arises
from equal sharing, it is called a covalent bond.*

16-1.2 Interaction Between Helium Atoms

A measurement of the density of helium gas
shows that it is a monatomic gas. Molecules of
He2 do not form. What difference between hy-
drogen atoms and helium atoms accounts for the
absence of bonding for helium? The answer to
this question also must lie in the attractive and
repulsive electrical interactions between two he-
lium atoms when they approach each other.
Figure 16-4 A shows the attractive forces in one
of our hypothetical instantaneous snapshots.
There are, of course, four electrons and each is
attracted to each nucleus. In Figure 16-4B we
see the repulsive forces. Taking score, we find
in Figure 16-4A eight attractive interactions, four

* The prefix "co" in "covalent" conveys the notion of
"sharing" as it does in the words "coworker," "co-
author," etc. The stem of the word, "-valent" refers to
"combination."

278

MOLECULES IN THE GAS PHASE I CHAP. 16

more than in the separated atoms. In addition,
we count seven repulsive interactions, five more
than in the separated atoms. Again we appeal to
experiment and we learn that the four new at-
tractive terms are not sufficient to outweigh the
five new repulsions. A chemical bond does not
form.

Thus we find that an explanation of the bond-
ing in H2 and the absence of bonding for He2 lies
in the relative magnitudes of attractive and re-
pulsive terms. Quantum mechanics can be put to
work with the aid of advanced and difficult
mathematics to calculate these quantities, to tell
us which is more important. Unfortunately,
solving the mathematics presents such an ob-
stacle that only a handful of the very simplest
molecules have been treated with high accuracy.
Nevertheless, for some time now chemists have
been able to decide whether chemical bonds can
form without appealing to a digital computer.

Figure 16-3D shows the simplified representa-
tion of the interaction of two helium atoms. This
time each helium atom is crosshatched before the
two atoms approach. This is to indicate there are
already two electrons in the Is orbital. Our rule
of orbital occupancy tells us that the Is orbital
can contain only two electrons. Consequently,
when the second helium atom approaches, its
valence orbitals cannot overlap significantly. The
helium atom valence electrons fill its valence
orbitals, preventing it from approaching a sec-
ond atom close enough to share electrons. The
helium atom forms no chemical bonds.*

16-1.3 Representations of Chemical Bonding

We propose, then, that chemical bonds can form
if valence electrons can be shared by two atoms
using partially filled orbitals. We need a short-
hand notation which aids in the use of this rule.
Such a shorthand notation is called a represen-
tation of the bonding.

* Each helium atom does have, of course, vacant 2s
and 2p orbitals which extend farther out than the filled
Is orbital. The electrons of the second helium atom can
"overlap" with these vacant orbitals. Since this overlap
is at great distance, the resulting attractions are extremely
small. This type of interaction presumably accounts for
the attractions that cause helium to condense at very low
temperatures.

ORBITAL REPRESENTATION
OF CHEMICAL BONDING

Our rule about covalent bond formation can be
applied quite simply through an orbital repre-
sentation:

o ooo

"■ \a O OOO

Bond can form. (2)

o OOO
O OOO

Bond c an.no f form.

In this representation there is no need to con-
sider the next higher energy level cluster — the
2s, 2/7 orbitals. For hydrogen and helium these
are much higher in energy and can give rise only
to extremely weak attractions.

ELECTRON DOT REPRESENTATION
OF CHEMICAL BONDING

The sharing of electrons can be shown by rep-
resenting valence electrons as dots placed be-
tween the atoms:

H +H

H:H

(i)

We shall use both orbital and electron dot rep-
resentations to show chemical bonding.

16-1.4 The Bonding of Fluorine

Our explanation of chemical bonding is of value
only if it has wide applicability. Let us examine
its usefulness in considering the compounds of
the second-row elements, beginning with fluo-
rine.

Under normal conditions of temperature and
pressure, fluorine is a gas. From gas density
experiments we discover that a molecule of fluo-
rine contains two atoms. There is a chemical
bond between the two fluorine atoms. Let us see
if our expectations agree with these experimental
facts.

SEC. 16-1 I THE COVALENT BOND

279

ORBITAL REPRESENTATION OF THB
BONDING OF FLUORINB

A fluorine atom has the orbital occupancy shown
below:

pacity and a stable compound is formed, HF.

In each of these cases, F2 and HF, we find that

the fluorine atom forms one bond — in F2 it is to

a second fluorine atom, in HF it is to a hydrogen

* ® ® <g®0 o ooo

We see that the neutral fluorine atom has seven
valence electrons; that is, seven electrons occupy
the outermost partially filled cluster of energy
levels. This cluster of energy levels, the valence
orbitals, contains one electron less than its ca-
pacity permits. Fluorine, then, has the capacity
for sharing one electron with some other atom
which has similar capacity. If, for example, an-
other fluorine atom approaches, they might share
a pair of electrons and form a covalent bond:

Is Zs 2p

atom. We describe this single bonding capacity
by saying, fluorine is univalent.

ELECTRON DOT REPRESENTATION
OF THE BONDING OF FLUORINE

In the electron dot method of showing chemical
bonds it is necessary to show only the valence
electrons. In fluorine there are seven — the pair
of electrons in the Is orbital is so tightly bound

* <g> <g> <8®0 O OOO

* <8> ® ®®Q O OOO

(5)

After the second atom approaches, sharing its
electron, each fluorine atom now has "filled" all
of its valence orbitals. No additional bonding
capacity remains. Hence F2 does not add a third
or fourth atom to form F3, F4, etc.

Now consider the possibility of the bonding
that might occur if a fluorine atom encounters
a hydrogen atom. Again fluorine has an oppor-
tunity to share electrons with a second atom
having a partially filled valence orbital:

Is 2s 2p

that it plays little role in the chemistry of fluo-
rine. In this representation, then, we show the
reaction between two fluorine atoms as follows:

:F- + -F:

:F:F:

(7)

From (7) we conclude that a covalent bond can
form between two fluorine atoms. Furthermore,
a census of the number of electrons owned or

* ® <g> ®80

O OOO
O OOO

«5)

Fluorine now has, by sharing a pair of electrons
with hydrogen, "filled" its valence orbitals. It
has no further bonding capacity. The same is
true of the hydrogen atom, though for this atom
there is but one valence orbital, the Is orbital.
Since no partially filled valence orbitals remain
for either atom, there is no more bonding ca-

shared by either of the fluorine atoms shows that
the valence orbitals are filled. For example, the
fluorine atom on the left feels the electrical at-
tractions of eight electrons near at hand,* as

* Remember, we are omitting from the discussion the
tightly bound h electrons of each fluorine atom.

280

MOLECULES IN THE GAS PHASE I CHAP. 16

shown in Figure 16-5. Since eight electrons is just
the capacity of the 2s, 2p valence orbitals, each
fluorine atom has reached the energetically stable
arrangement of an inert gas.

The fluorine atom The fluorine atom.

on the left on the right

feels eight- electrons feels eight electrons

: F : -)F :

: F I: F :

v. •• /

Fig. 16-5. The electrons near each fluorine atom in
F:.

Hydrogen fluoride can be represented by the
electron dot picture

:F + H

F:H

(5)

Again, a census of the number of electrons near
each of the atoms shows that this is a stable
arrangement. True, the hydrogen atom has, close
at hand, only two electrons whereas fluorine has
eight. This is energetically desirable, however,
because hydrogen has only one valence orbital,
the Is orbital. Two electrons just fill this orbital.

The fluorine atom The hydrogen atom

feels eight electrons feels two electrons

[: F Ah

F \\ H )

Fiq. 16-6. The electrons near each atom in HF.

We see that the bonding of a fluorine atom to
another fluorine atom or to a hydrogen atom can
be explained in terms of sharing electrons so as
to fill the partially filled valence orbitals. This

sharing makes the molecule F2 (or HF) energeti-
cally more stable than the separated atoms would
be. The energy stability results from the shared
electrons being attracted simultaneously to both
positive nuclei. A chemical bond results.

ELECTRON AFFINITY OF THE FLUORINE ATOM

Experiment shows that a gaseous fluorine atom can
acquire an electron to form a stable ion, F~(g). We can
discuss the energy of formation of this ion in the same
way that we treated ionization energies. The first ioniza-
tion energy of fluorine atom is the energy required to
remove an electron from a neutral atom in the gas phase.
We shall call this energy E\. Then the heat of reaction
can be written in terms of Ex :

*(g) — *■ F+(gj + trig) AH = E, (9)

The second ionization energy, E2, refers to reaction (10):

F+(g) — >- F**(e) + e-(g) AH = E2 (10)

Now we can add a new process with an energy which
might logically be called E0:

?-(g)

F(g) + e-ig) AH = Eo

Comparing equations (9), (10), and (//), we see that £o
is just the ionization energy of F~(g). By usual practice,
however, the reverse of reaction (11) is usually considered.
Of course the heat of reaction (12) is just the negative of
that of reaction (//):

F(g) + e'ig)

F-(g)

AH = -En

(12)

The energy change of reaction (12) is called the electron
affinity of the fluorine atom. It is symbolized by E and,
as defined here, is a negative quantity if heat is released
when the ion is formed:

E= -Eo

U3)

Electron affinities are difficult to measure and are
known reliably for only a small fraction of the hundred
or so elements. The electron affinity of fluorine is one

that is known:

E = — 83 kcal, mole

(14)

Eo = +83 kcal mole

(15)

The experimental quantities shown in (14) and (15) indi-
cate that the F~ ion is more stable than a fluorine atom
and an electron. Energetically, a fluorine atom "wants"
another electron. It is profitable to express reaction (12)
in terms of orbital occupancy:

eg) (g) ®g0 O * •"<*"» —

0<75>

SEC. 16-2 I BONDING CAPACITY OF THE SECOND-ROW ELEMENTS

281

The neutral fluorine atom has seven valence electrons;
that is, seven electrons occupy the highest partially filled
cluster of energy levels. This cluster of energy levels thus
contains one fewer electron than its capacity permits. The
electron affinity of fluorine shows that the addition of this
last electron is energetically favored. This is in accord
with much other experience which shows that there is a
special stability to the inert gas electron population.

In view of the electron affinity of a fluorine atom, we
can speculate on what would be the result of a collision
between two fluorine atoms. Will a reaction occur? The
energy is one of the factors which determines the answer.
First let us consider a reaction that does not occur spon-
taneously.

F(g) + F(g)

F+(g) + F-(g)

(17)

Reaction (17) can be rewritten in two steps:

F(g)
F(g) + e-(g)

fluorine atom number 1 shares one valence electron of
fluorine atom number 2. Thus a part of the electron affin-
ity of fluorine atom number 1 is "satisfied" though it was
not necessary to take the electron away from atom num-
ber 2. Meanwhile, fluorine atom number 2 is getting the
same sort of energy benefit from the valence electron of
atom number I. Each fluorine atom has acquired another
electron at least a part of the time. We have gained a part
of the stability of reaction (19) without paying for reac-
tion (18). The most energy that could be released by such
an electron sharing would be double the electron affinity
of fluorine, 2 X 83 = 166 kcal. This takes no account of
the work done in bringing the two positive nuclei near
each other. Nor can we expect to gain the whole electron
affinity under conditions of electron sharing. Yet the
energy released when two fluorine atoms form a bond is
36.6 kcal/mole, a reasonable fraction of the maximum
possible.

F+(g) + e-(g)
F~(g)

&H = Ex = 401.5 kcal/mole
AH - -Eo = -83

Net F(g) + F(g)

F+(g) + F~(g)

AH =

318.5

(18)
U9)

(17)

We see that reaction (77) is energetically unfavorable.
The stability of F~ is more than outweighed by the diffi-
culty of removing an electron from another fluorine atom.
There is another possible consequence of a collision
between two fluorine atoms. The two atoms can remain
together to form a molecule. Each atom has a valence
electron in a "half-filled" orbital. We can imagine these
two atoms orienting so that these "half-filled" orbitals
overlap in space. Then the "half-filled" valence orbital of

Now we can say why the chemical bond forms between
two fluorine atoms. First, the electron affinity of a fluorine
atom makes it energetically favorable to acquire one more
electron. Two fluorine atoms can realize a part of this
energy stability by sharing electrons. All chemical bonds
form because one or more electrons are placed so as to feel
electrostatic attraction to two or more positive nuclei
simultaneously.

16-2 BONDING CAPACITY OF THE SECOND-ROW ELEMENTS

In Chapter 6 we saw that the chemical com-
pounds of the third-row elements display a
remarkable regularity. Return to Chapter 6 and
reread Section 6-6.2. The same simple trend in
chemical formulas is discovered in the second
row of the periodic table. Now we have a basis
for explaining why these trends are found.

16-2.1 The Bonding Capacity of Oxygen Atoms

The neutral oxygen atom has eight electrons. Six
of these occupy the 2s, 2p orbitals and are much
more easily removed than the two in the Is
orbital. Therefore oxygen has six valence elec-
trons. The 25, 2p orbitals are the valence orbitals.
They can accommodate the valence electrons in
two ways, as follows:

®so o ooo

(20)

p ® <g> &O0 O COO

(21)

282

MOLECULES IN THE GAS PHASE I CHAP. 16

Remember the spatial arrangement of the p or-
bitals? Each one protrudes along one of the
three cartesian axes (as shown in Figure 15-9).
If the electrons have the orbital occupancy of
(20), then two electrons occupy the p orbital

atom has partially filled valence orbitals. Elec-
tron sharing can occur, placing electrons close
to two nuclei simultaneously. Hence a stable
bond can occur. This is shown in representations
(22) and (23).

3s 3P

O OOO

o ooo

2s 2p

(22)

:0.

:0'.K

(23)

protruding along the x axis (px) and two elec-
trons occupy the/? orbital protruding along the>>
axis (pv). In the occupancy of (21), two electrons
occupy px, only one electron occupies pu, and the
last is in />,. Thus (21) differs from (20) by the
movement of one of the electrons from pv to a
different region of space, pz. Since electrons repel
each other, we can expect that the configuration
which keeps the electrons farther apart, (21), is
the lower in energy. Experiment shows that it is
and we shall base our discussion of the bonding
of oxygen on orbital occupancy (21). However,
the occupancy represented by (20) also contrib-
utes to the chemistry of oxygen atoms.

Suppose a hydrogen atom approaches an oxy-
gen atom in its most stable state, (27). Each

In either representation, (22) or (23), we see
that there is residual bonding capacity remaining
in the species OH. In (22) the third 2p orbital has
a single electron but a capacity for two. This
means more bonding can occur. In (23) a census
of the electrons near the oxygen atom indicates
there are only seven. The oxygen atom would be
more stable if it could add one more electron.
With either representation, we conclude that OH
should be able to react with another hydrogen
atom. See representations (24), (25).

Now we have the compound H20. By either
representation, the bonding capacity of oxygen
is expended when two bonds are formed. Oxygen
is said to be divalent, and the compound H20 is
extremely stable. Each of the atoms in H20 has
filled its valence orbitals by electron sharing.

O (XT)
O CCO

o ooo

(24)

OiH

•O'.K
H

(25)

SEC. 16-2 I BONDING CAPACITY OF THE SECOND-ROW ELEMENTS

283

reaction between two OH molecules The electron dot representation is

Though OH is reactive, it is a cluster of atoms
with sufficient stability to be identified as a mole-
cule. It is present in a number of high tempera-
ture flames, for example. Its chemistry might be
expected to be like that of fluorine atoms. Com-
pare the electron dot formulas

0:H

Since two fluorine atoms react, forming a cova-
lent bond, we can expect two OH molecules to
do the same sort of thing

H H

6: — *- :6 6:

o.+

(26)

Reaction (26) yields the compound H202. This
is the formula of the well-known substance,
hydrogen peroxide. By these considerations of
chemical bonding, we see that the structure of
HjOj must involve an oxygen-oxygen bond:

I
O— O

I
H

(27)

EXERCISE 16-1

Predict the structure of the compound S2CI2 from
the electron dot representation of the atoms.
After you have predicted it, turn back to Figure
6-12, p. 103, and check your expectation.

:OF:

:F:'"

Again we find that oxygen is divalent.

(29)

EXERCISE 16-2

Draw orbital and electron dot representations of
each of the following molecules: OF, F202, HOF,
HF02. Which of these is apt to be the most
reactive?

16-2.2 The Bonding Capacity
of Nitrogen Atoms

For the same reason we discussed for oxygen
atoms, the nitrogen atom is most stable when it
has the maximum number of partially filled
valence orbitals. This keeps the electrons as far
apart as possible. The most stable state of the
nitrogen atom is as follows:

rJV

0 000 v»

It is now straightforward to predict that nitrogen
will form a stable hydrogen compound with for-
mula NH3. Nitrogen is trivalent. A similar com-
pound, NF3, will be formed with fluorine. The
electron dot formulas are

OXYGEN-FLUORINE COMPOUNDS

It is a simple matter to predict that oxygen will
form a stable compound with two fluorine
atoms, FsO. The orbital representation is*

Is 2s 2x>

0
0
0

0 00Q
0 0&0

(28)

* Henceforth we will omit empty orbitals much higher
in energy than the valence orbitals.

:F:

(31)

:N:H

and : N : F :

ammonia

:F:

nitrogen trifluoride

(32)

EXERCISE 16-3

The molecule NH2 has residual, unused bonding
capacity and is extremely reactive. The molecule
N2Hi (hydrazine) is much more stable. Draw an
electron dot representation of the bonding of
hydrazine. Draw its structural formula (show
which atoms are bonded to each other).

284

MOLECULES IN THE GAS PHASE I CHAP. 16

16-2.3 The Bonding Capacity off Carbon Atoms

There are a number of orbital occupancies that
we might consider for the carbon atom:

First let us compare (33) and (34). By our con-
ventional argument that electrons repel each
other, configuration (34) should be more stable
than (33). The second, (34), places one electron
in each of the px and pv orbitals whereas (33)
places two electrons in the same orbital, px. It is
an experimental fact that (34) is more stable
than (33).

Now we can predict the chemistry of the car-
bon atom in this state. It should be divalent,
forming compounds CH2 and CF2. Let us con-
sider one of these, say CH2.

C'.H
H

(36)

Here is a situation we haven't met before.
After using the two available partially filled
orbitals to form covalent bonds with hydrogen
atoms, there remains a vacant valence orbital.
In the electron dot formulation (36) we see that
the carbon atom finds itself near only six elec-
trons in CH2. The valence orbitals will accom-
modate eight electrons. Because one valence or-

bital is completely vacant, we can expect C7/2 to
be reactive.

In fact, both CH2 and CF2 are considered to
be stable* but extremely reactive molecules.
Though there is reaction mechanism evidence
verifying the existence of each species, it is not
possible to prepare either substance pure. This
great reactivity shows that energy considerations
favor the use of all four of the valence orbitals if
possible. This argument leads us to consider a
third orbital occupancy:

0 020 <">

With orbital occupancy (37), a carbon atom
has four half-filled valence orbitals. True, (37) is
somewhat less stable than (34) because an elec-
tron was raised from the 25 energy level to the
slightly higher 2p energy level. This process is
called "promoting" the electron. On the other
hand, the promotional energy is not very large,
and in return for it the carbon atom acquires the
capacity to form four covalent bonds. Each co-
valent bond increases stability, more than com-
pensating for the energy investment in promoting
one of the 2s electrons. With orbital occupancy
(37), carbon can share pairs of electrons with,
for example, four hydrogen atoms or four fluo-
rine atoms. Hence, carbon is tetravalent:

H:C:H

methane

:F:

: F : C : F :

":F:"

carbon tetrafluoride

(38)

(39)

* The molecule CH2 is stable in the sense that it does
not spontaneously break into smaller fragments. It is
reactive because other molecules formed from this group
of atoms have much lower energy.

SEC. 16-2 | BONDING CAPACITY OF THE SECOND-ROW ELEMENTS

285

EXERCISE 16-4

Draw electron dot formulas for the molecules
CH3, CF3, CHF3, CH2F2, CH3F. Which will be
extremely reactive?

EXERCISE 16-5

Draw an electron dot and a structural formula
for the molecule C2H6 (ethane) which forms if
two CH3 molecules are brought together. Explain
why C2H6 is much less reactive than CH3.

is
B <g)

2 s 2P

0"0(20

(44)

16-2.4 The Bonding Capacity of Boron Atoms

The boron atom presents the same sort of option
in orbital occupancy as does carbon:

Is 2s 2p

H H H

sB (g)

** ® 0 00O w

The electron configuration (41) is somewhat
higher in energy than (40). It is necessary to
promote a 2s electron to the 2p state to obtain
(41). In return, however, the boron atom gains
bonding capacity. Whereas a boron atom can
form only one covalent bond in configuration
(40), it can form three in configuration (41).
Since each bond lowers the energy, the chemistry
of boron is fixed by the electron configuration
(41).

Now we can expect that boron will be tri-
valent. We predict that there should be molecules
such as BH3 and BF3

B:H
H

B:F:

We cannot help noticing, however, that there
remains a completely vacant valence orbital. For
example, for BH3 the orbital representation
would be as shown in (44).

The last vacant 2p orbital is reminiscent of the
configuration found for CH2 [see (35)]. Since
CHj is very reactive, presumably BH3 will be the

same. Such is the case. There is only indirect
evidence establishing the existence of BH3. In-
stead boron forms a series of unusual compounds
with hydrogen, the simplest of which is called
0OO W diborane, B2H6.

You might wish to predict the structure of diborane
(which is now known) but do not be discouraged if you
are not able to. Its structure, once elucidated, came as
quite a surprise to even the most sophisticated chemists.
The explanation of the structure is, even today, composed
of a large proportion of words and a small proportion of
understanding.

Boron is an obliging element. On the one hand
it conforms to our expectation that the unused
valence orbital will affect the bonding capacity
of boron, as shown by the reactivity of BH3. On
the other hand boron conforms to our expecta-
tion that electron configuration (41) will make
boron trivalent. The compound BF3 is a stable,
gaseous compound and, in contrast to BH3, is
readily prepared in a pure form. The explanation
of how the fluorine atoms are able to satisfy in
part the bonding capacity of the vacant 2p or-
bital (though hydrogen atoms cannot) must wait
until a later chemistry course. For our interest
here, BF3 is the most stable boron-fluorine com-
pound and it demonstrates the trivalent bonding
capacity of boron.

16-2.5 The Bonding Capacity
of Beryllium Atoms

The beryllium atom, like boron and carbon, can
promote an electron in order to form more
chemical bonds:

(42)

(43)

,Be

0 000 <«>

286

MOLECULES IN THE GAS PHASE I CHAP. 16

Therefore we should expect in the gaseous state
to find molecules such as BeH2 and BeF2. These
molecules have been detected. On the other
hand, beryllium has the trouble boron has, only
in a double dose. It has two vacant valence or-
bitals. As a result, BeH2 and BeF2 molecules, as
such, are obtained only at extremely high tem-
peratures (say, above 1000°K). At lower temper-
atures these vacant valence orbitals cause a
condensation to a solid in which these orbitals
can participate in bonding. We shall discuss these
solids in the next chapter.

16-2.6 The Bonding Capacity of Lithium Atoms

There is little new to be said about the bonding
capacity of a lithium atom. With just one valence
electron, it should form gaseous molecules LiH
and LiF. Because of the vacant valence orbitals,
these substances will be expected only at ex-
tremely high temperatures. These expectations
are in accord with the facts, as shown in Table
16-1, which summarizes the formulas and the
melting and boiling points of the stable fluorides
of the second-row elements. In each case, the
formula given in the table is the actual molecular
formula of the species found in the gas phase.

16-2.7 Valence

Very often the word valence is used in discussins
the nature of chemical bonding. Unfortunately
this word has been used as a noun to mean a
number of different things. Sometimes valence
has been used to mean the charge on an ion,
sometimes it has meant the total number of
atoms to which a particular atom will bond, and
at other times, the word valence has been used
to mean oxidation number. Perhaps the most
widely accepted definition of the word is that it
is the number of hydrogen atoms with which an
atom can combine, or release, in a chemical re-
action. It is clear that a word with so many mean-
ings might confuse a discussion of chemical
bonding. For this reason we have avoided and
will continue to avoid using the word as a noun
in this book.

We have, however, made a careful definition
of the term "valence electrons" ("the electrons
that are most loosely bound"; see p. 269).
We have also used carefully the term "valence
orbitals" to mean the entire cluster of orbitals
of about the same energy as those which are
occupied by the valence electrons. In both of
these uses, the word valence is used as an
adjective.

Table 16-1. the fluorides of the elements in the second row

OF THE PERIODIC TABLE

Li Be B C N

Formula

LiF

BeFj

BF,

CF4

NF,

F,0

Melting point (°K)

1143

1073

146

Boiling point (°K)

1949

—

172

145

153

128

16-3 TREND IN BOND TYPE AMONG THE SECOND-ROW FLUORIDES

We have termed the chemical bond in the hydro-
gen molecule, H2, a covalent bond. This indicates
that electrons are shared so that they are simul-
taneously and, on the average, equally near two
nuclei. This makes the system more stable and a
chemical bond results.

All chemical bonds occur because electrons
can be placed simultaneously near two nuclei.
Yet it is often true that the electron-sharing
which permits this is not exactly equal sharing.
Sometimes the electrons, though close to both
nuclei, tend to distribute nearer to one nucleus

SEC. 16-3 I TREND IN BOND TYPE AMONG THE FIRST-ROW FLUORIDES

287

than to the other. We can see why by contrasting
the chemical bonding in gaseous fluorine, F2, and
in gaseous lithium fluoride, LiF.

16-3.1 The Bonding in Gaseous
Lithium Fluoride

We have already treated the bonding in an F2
molecule. Since neither fluorine atom can pull an
electron entirely away from the other, they com-
promise by sharing a pair of electrons equally.
How does the chemical bonding in the lithium
fluoride molecule compare?

As we have mentioned earlier, lithium has one
valence electron, hence can share a pair of elec-
trons with one fluorine atom:

<8>

When the bonding electrons move closer to one
of the two atoms, the bond is said to have ionic
character.

In the most extreme situation, the bonding
electrons move so close to one of the atoms that
this atom has virtually the electron distribution
of the negative ion. This is the case in gaseous
LiF. In an electron dot representation, we might
show

Li :F:

Li+ F-

(50)

(51)

When a formula like (50) or (51) provides a use-
ful basis for discussing the properties of a mole-

Is
2s 2P

<s ooo

(46)

or Ll'.F'.

(47)

Thus we can expect a stable molecular species,
LiF. The term "stable" again means that energy
is required to disrupt the molecule. The chemical
bond lowers the energy because the bonding
electron pair feels simultaneously both the lith-
ium nucleus and the fluorine nucleus. That is not
to say, however, that the electrons are shared
equally. After all, the lithium and fluorine atoms
attract the electrons differently. This is shown by
the ionization energies of these two atoms:

F(g) —+ F+(g) + e-(g)

A// = 401.5 kcal/mole (48)

Li(g) — >- Li+fgj + e-(g)

AH = 124.3 kcal/mole (49)

Clearly the fluorine atom holds electrons much
more strongly than does the lithium atom. As a
result, the electron pair in the lithium fluoride
bond is more strongly attracted to the fluorine
atom than to the lithium. The energy is lower if
the electrons spill toward the fluorine atom.

cule, the bond in that molecule is said to be an
ionic bond.

CONTRAST OF COVALENT AND
IONIC BONDS

The fluorine molecule is held together by the
energy gain resulting from placing a bonding
pair of electrons near both fluorine nuclei simul-
taneously. The electrons move about in the
molecule in such a manner that, on the average,
they are distributed symmetrically between and
around the identical fluorine nuclei. This sym-
metrical distribution is reasonable, since the two
fluorine nuclei attract the bonding electrons
equally. The lithium fluoride molecule also is
held together by the energy gain resulting from
placing a bonding pair of electrons near the
lithium and fluorine atoms simultaneously. In
this case, however, the electrons move in such a
way as to remain closer to the fluorine than the

288

MOLECULES IN THE GAS PHASE ! CHAP. 16

Covalent bond

F-F

Covalent bond
yvith partial
ionic character

F-Ct

Ionic bond

F-Li

<5k

.jss>

<=*-

Fig.

76-7. Electron

distributions m various bond

lithium atom. Fluorine attracts the bonding elec-
trons more strongly than does lithium.

We see again that there is but one principle
which causes a chemical bond between two
atoms: all chemical bonds form because electrons
are placed simultaneously near two positive nuclei.
The term covalent bond indicates that the most
stable distribution of the electrons (as far as
energy is concerned) is symmetrical between the
two atoms. When the bonding electrons are
somewhat closer to one of the atoms than the
other, the bond is said to have ionic character.
The term ionic bond indicates the electrons are
displaced so much toward one atom that it is a
good approximation to represent the bonded

atoms as a pair of ions near each other. Figure
16-7 shows schematically how the electron dis-
tributions are pictured in covalent, partially
ionic, and ionic bonds. The figure also shows
how the electrons might look in an instantaneous
snapshot. In each type of bond, the electron-
nucleus attractions account for the energy sta-
bility of the molecule.

THE ELECTRIC DIPOLE
OF THE IONIC BOND

The spilling of negative electric charge toward
one of the atoms in an ionic bond causes a
charge separation. This can be represented
crudely as in the last drawing in Figure 16-8.
The molecule is electrically positive at the lithium
end and electrically negative at the fluorine end.
It is said to possess an electric dipole. The mole-
cule is then called a polar molecule. The forces
between molecules possessing electric dipoles are
much stronger than those between nonpolar
molecules. We shall see in Chapter 17 that these
forces, too, involve the same electrical interac-
tions we have discussed here.

The last representation of Figure 16-8 is com-
monly used and it is the simplest way of showing
a bond dipole. The arrow means that the nega-
tive charge is mainly at one end of the bond. The
directional property of the arrow implies that
the force this molecule exerts on another mole-
cule depends upon the direction of approach of
the second molecule.

16-3.2 Ionic Character in Bonds to Fluorine

We can expect the effects just discussed to be at
work in the bonds fluorine forms with other ele-

Fig.

16-8. Representations of the electric dipole of

gaseous lithium fluoride.

SEC. 16-3 I TREND IN BOND TYPE AMONG THE FIRST-ROW FLUORIDES

289

ments. The ionization energies of the elements
give us a rough clue to the electron-nuclear
attractions. Table 16-11 compares the ionization
energies of each element of the second row with
that of fluorine. The last column describes the
type of chemical bond.

The trend in bond type shown in Table 16-11
has important influence on the trend in proper-
ties of the fluorine compounds. The trend arises
because of the increasing difference between
ionization energies of the two bonded atoms.

gen cannot be predicted from its measured ioniza-
tion energy.

Examination of the properties of a number of
compounds involving hydrogen indicates that
the ionic character of bonds to hydrogen are
roughly like those of an element with ionization
energy near 200 kcal/mole. Thus the hydrogen
fluorine bond in HF is ionic and chemists believe
that the electrons are spilled toward the fluorine
atom, leaving the hydrogen atom with a partial
positive charge. Hydrogen acts like an element

Table 16-11. bond types in some fluorine compounds

IONIZATION ENERGIES (kcal/mole)

COMPOUND

BOND

ELEMENT BONDED TO F

FLUORINE

BOND TYPE

F— F

401.5

Covalent
slightly

OF2

O— F

313.8

401.5

ionic

NF3

N— F

335

401.5

increasing
ionic

CF,

C— F

259.5

401.5

character increasing

covalent

character

BF3

B— F

191.2

401.5

slightly

BeF2

Be— F

214.9

401.5

covalent

LiF

Li— F

124.3

401.5

' Ionic

16-3.3 Ionic Character in Bonds to Hydrogen

In Chapter 6 the element hydrogen was charac-
terized as a family by itself. Often its chemistry
distinguishes it from the rest of the periodic
table. We find this is the case when we attempt
to predict the ionic character of bonds to hy-
drogen.

The ionization energy of the hydrogen atom,
313.6 kcal/mole, is quite close to that of fluorine,
so a covalent bond between these two atoms in
HF is expected. Actually the properties of HF
show that the molecule has a significant electric
dipole, indicating ionic character in the bond.
The same is true in the O — H bonds of water
and, to a lesser extent, in the N — H bonds of
ammonia. The ionic character of bonds to hydro-

with lower ionization energy than fluorine. The
same is true but in decreasing amount for hydro-
gen when it is bonded to oxygen and to nitrogen.
The carbon-hydrogen bond has only a slight
ionic character. At the other end of the periodic
table, gaseous lithium hydride is known to have
a significant electric dipole but now with the
electric dipole turned around. In LiH the elec-
trons are spilled toward the hydrogen atom,
leaving the lithium atom with a partial positive
charge. This is in accord with the low ionization
energy of lithium, 124.3 kcal/mole, well below
the value of 200 kcal/mole that we have assigned
to hydrogen. For our purposes, it suffices to dis-
cuss the bonding of hydrogen in terms of an
apparent ionization energy near 200 kcal/mole.

290

MOLECULES IN THE GAS PHASE I CHAP. 16

16-3.4 Bond Energies and Electric Dipoles

It is found experimentally that a bond between two
atoms with very different ionization energies tends to be
stronger than a bond between atoms with similar ioniza-
tion energies. Since electric dipoles are caused by differ-
ences in the ionization energies of bonded atoms, we can
conclude that strong bonds are expected in molecules
with electric dipoles.

For example, contrast the bond energies of the gaseous
molecules Na2, Clj, and NaCl :

Na2fs;

CUfg)

NaClfgj

2NafgJ
2C\(g)

Nate; + cite;

Aff(Na2) = 17 kcal
Atf(Oi) = 57 kcal
A//(NaCl) = ?

A rough estimate of the bond energy of NaCl could be
based upon the bond energies of Na2 and Cl2, 17 kcal
and 57 kcal. Since the 17 kcal bond energy of Na2 is
derived from the sharing of an electron pair between two
sodium atoms, the one sodium atom in NaCl might be
expected to contribute one-half this amount to the bond
energy of NaCl, V = 8.5 kcal. In a similar way, the
single chlorine atom in NaCl might contribute one-half
the bond energy of Cl2, Af = 28.5 kcal. Thus we arrive
at an estimate of A#(NaCl):

^(Nad) (estimated) = W* + "*X>
= 8.5 + 28.5 = 37.0

Experimentally we discover that A#(NaCl) is much
larger, 98.0 kcal— a discrepancy of 98 - 37 = 61 kcal.
This discrepancy is explained in terms of the large dif-
ference in ionization energy of sodium and chlorine
atoms:

Cite; — *• Cl+te; + e~ Ei = 300 kcal
Nate; — *■ Na+te; + er E, = 118 kcal

The large difference implies that the energy is lowered
even more than 37 kcal because the electron pair need not
remain equally shared between the two atoms in NaCl.
Instead they can concentrate nearer the atom that holds
electrons more tightly (the chlorine atom) if a net lower-
ing of the energy results.

Table 16-1 1 1 collects some data. The existence of a
correlation between the ionization energy difference,
Ei(X) — E\(Y), and the bond energy discrepancy,
AHxy — i(A/fxi + A//y,), is obvious.

Needless to say, if ionic character affects the energy
stability of a chemical bond it also affects the chemistry
of that bond. The tendency toward minimum energy is
one of the factors that determine what chemical changes
will occur. As a bond becomes stronger, more energy is
required to break that bond to form another compound.
Hence we see that ionic bonds are favored over covalent
bonds and that ionic character in a bond affects its
chemistry.

EXERCISE 16-6

From the following bond energy data and the ionization
energies given in Table 15-111, calculate the entries in the
last two columns of Table 16-111 for the compounds LiF
and LiBr. The ionization energy, Eu for bromine atom is
273 kcal/mole.

Li2te;

F2fg;

Br2te;

LiFte;

LiBrte;

2Lite;
2Fte;

2Brte;

Lite; + Fte;

Lite; + Brte;

AH = 25 kcal
AH = 36
AH = 45.5
AH = 137
AH = 101

Table 16-111. ionization energy differences and bond

ENERGY DISCREPANCIES

MOLECULE

X Y XY Ei(X)-Ei(Y)

AHxy - \(AHx* + AHy,)

NaKte;

18 kcal/mole

0 kcal/mole

ClBrte;

LiClte;

176

NaClte;

182

Kcite;

200

66.5

LiFte;

277

106

16-4 MOLECULAR ARCHITECTURE

The properties of a molecule are primarily deter-
mined by the bond types which hold it together

and by the molecular "architecture." By archi-
tecture we mean the structure of the molecule—

SBC. 16-4 I MOLBCULAR ARCHITBCTURB

291

the shape of the molecule. We shall investigate
what is known about the molecular structures of
the second-row hydrides and fluorides.

16-4.1 The Shapes of H,0 and FtO

The orbital representations of the bonding in
H20 and in F20 suggest that two p orbitals of
oxygen are involved in the bonding [see repre-
sentation (24)]. Figure 16-9 shows the spatial
arrangement we assign to the p orbitals (assum-
ing they are like hydrogen atom orbitals). If the

to the ball-and-stick model of the molecule. A
line is drawn between the oxygen atom and each
hydrogen atom to indicate that a chemical bond
holds these two atoms together. No line is drawn
between the two hydrogen atoms since we feel
they are not directly bonded to each other. Now
we can apply our discussion of the role of elec-
trons in bonding to add to the meaning of this
line representation. A line is drawn between two
atoms to indicate that a pair of electrons is
shared between these two atoms, resulting in a
chemical bond.

Fig. 16-9. The expected shape of the HJO molecule:
p* bonding.

spatial arrangement persists after the bonds
form, the molecular shape would be fixed, as
shown. The molecule would be bent, with an
angle near 90°. The same would be true for F20.
The measured bond angles are as follows:

F ^^ F

ZH-0-H = 1O4.S0
(52a)

ZF-O-F m 202 •

(52b)

It is generally true that a divalent atom with
twop orbitals as valence orbitals forms an angular
molecule. Since this prediction is reliable, the
bonding is usually characterized by identifying
the valence orbitals. Oxygen is said to use p2
(read, "p two") bonding in water and FjO.

Notice that the structural formulas (52a) and
(52b) use another representation of the bonding.
It is quite familiar, of course, since it corresponds

16-4.2 The Shapes of NH, and NF,

In NH3 and NF3, three p orbitals are involved in
the bonding [see representation (30)]. Figure
16-10 shows the spatial arrangement implied by
assuming persistence of the hydrogen atom or-
bitals after bonding. We expect, then, that am-

Fig. 16-10. The expected shape of the NH, molecule:
p* bonding.

monia has a pyramidal shape (a pyramid with
a three-sided base). The bond angles should be
near 90°. Both NH3 and NF3 do have pyramidal
shapes. The measured bond angles are as fol-
lows:

Hz:—f-.._7^a ZH-K-H- 107°

(53a)

F-""

ZF-N-F = 102*
(53b)

292

MOLECULES IN THE GAS PHASE I CHAP. 16

Again, experiments show that it is generally
true that a trivalent atom with three p orbitals as
valence orbitals forms a pyramidal molecule. The
bonding is called pz bonding (read, "/? three").

16-4.3 The Shapes of CH< and CF4

The bonding of methane, CH4, and that of car-
bon tetrafluoride, CF4, involve four valence or-
bitals, the 2s orbital and the three 2p orbitals.
Four bonds are formed by carbon and, as before,
we characterize the bonding by naming the va-
lence orbitals: sp% (read, "sp three"). This time,
however, the assumption of the persistence of
the spatial distributions of the hydrogen atom
orbitals does not indicate directly what bond
angles to expect. Experiment shows, however,
that spz bonding always gives bond angles which
are exactly or very close to tetrahedral angles.
This means that the angle between any two
carbon-hydrogen bonds is 109°28'. The structure
is called tetrahedral because the four hydrogen
atoms occupy the positions of the corners of a
regular tetrahedron (a four-sided figure with
equal edges). The structure is shown in Figure
16-11.

16-4.4 The Shape of BF,

The boron atom in BF3 uses the 2s and two 2p
orbitals in bonding. Therefore the bonding is

Fig. 16-11. The tetrahedral bonding of carbon: sp*
bonding.

Fig. 16-12. The structure of BF3: sp- bonding.

called sp"1. Again we must let experiment tell us
the bond angles which are found with sp2 bond-
ing. The structure of BF3 is that of an equilateral
triangle. The structure, shown in Figure 16-12,
is planar and each of the three fluorine atoms is
the same distance from the boron atom as are
the others.

16-4.5 The Shape of BeF;

Beryllium atom in gaseous BeF2 uses the 2s and
only one 2p orbital in bonding. The bonding is
called sp. Experiment shows that the molecule is
linear and symmetrical, as shown in Figure
16-13. The structure of gaseous BeH2 is un-
doubtedly linear and symmetric as well, by
analogy to BeF2.

Fig. 16-13. The structure of BeF2: sp bonding.

-h£>-

SEC. 16-4 I MOLECULAR ARCHITECTURE

293

16-4.6 Summary of Bonding Orbitals
and Molecular Shape

From the data presented here, the orbitals in-
volved in bonding correlate with the molecular
architecture. The relationships are summarized
in Table 16-IV.

THE MOLECULAR D1POLE OF LlF

The lithium fluoride bond is highly ionic in char-
acter because of the large difference in ionization
energies of lithium and fluorine. Consequently,
gaseous lithium fluoride has an unusually high
electric dipole.

Table 16-IV. bonding orbitals, bonding capacity, and molecular shape

ELEMENT

BONDING
ORBITALS

BONDING
CAPACITY

MOLECULAR SHAPE
OF THE FLUORIDE

EXAMPLE

none

monatomic

linear, diatomic molecule

LiF

linear

BeF,

sp2

planar, triangular

BF,

sp3

tetrahedral

CF«

pyramidal

NF,

bent

OF,

linear, diatomic molecule

none

monatomic

16-4.7 Molecular Shape and Electric Dipoles

Consider the fluorides of the second-row ele-
ments. There is a continuous change in ionic
character of the bonds fluorine forms with the
elements F, O, N, C, B, Be, and Li. The ionic
character increases as the difference in ionization
energies increases (see Table 16-11). This ionic
character results in an electric dipole in each
bond. The molecular dipole will be determined
by the sum of the dipoles of all of the bonds,
taking into account the geometry of the mole-
cule. Since the properties of the molecule are
strongly influenced by the molecular dipole, we
shall investigate how it is determined by the
molecular architecture and the ionic character
of the individual bonds. For this study we shall
begin at the left side of the periodic table.

THE MOLECULAR DIPOLE OF BeF2

The beryllium-fluorine bond is also highly ionic
in character. However, there are two such Be-F
bonds and the electrical properties of the entire
molecule depend upon how these two bonds are
oriented relative to each other. We must find
the "geometrical sum" of these two bond dipoles.
The geometrical sum of two arrows can be
understood simply with the aid of Figure 16-14.
Figure 16-14A shows how two arrows pointing
in the same direction combine to give a longer
arrow. Figure 16-14B shows how two arrows
oppositely directed combine to give a shorter

Fig.

16-14. The geometrical sum of dipoles: both

length and direction are important.

combine to fftve

combine -to give

294

MOLECULES IN THE GAS PHASE I CHAP. 16

Fig. 16-15. The absence of a molecular dipole in
BeFt.

arrow. Figure 16-14C shows how two arrows
that are not parallel add to give an arrow in a
new direction.

Now we can apply the process of combination
shown in Figure 16-14 to BeF2. In the linear,
symmetric BeF2 molecule, the two bond dipoles
point in opposite directions. Since the two bonds
are equivalent, there is a complete cancellation,
as shown in Figure 16-15. Hence the molecule
has no net dipole; the molecular dipole is zero.

THE MOLECULAR DIPOLES
OF BF3 AND CF4

Both of these molecules are thought to have

= O

zero. Careful consideration of the geometry
shows that there is a complete cancellation by
the bond dipoles in each molecule. This cancel-
lation is shown in Figure 16-16 for BF3. The
molecular dipole is zero.

THE MOLECULAR DIPOLE IN F20

Since F20, with p- bonding, is a bent molecule,
the two bond dipoles do not cancel each other
as they do in BeF2. On the other hand, the ioni-
zation energies of oxygen and fluorine are not
very different, so the electric dipole of each bond
is small in magnitude. These add together, ac-
cording to the geometry, to give a polar mole-
cule, as shown in Figure 16-17.

moderate amounts of ionic character in each Fig. 16-16. The absence of a molecular dipole in

bond. Yet the molecular dipoles are each exactly fiF,.

^^ J^

+ /*<

— bd = o

SEC. 16-5 I DOUBLE BONDS

295

•V

•s = ss>

16-5 DOUBLE BONDS

In deciding the bonding capacity of a given atom
from the second row, we have counted the num-
ber of hydrogen atoms or fluorine atoms with
which it would combine. Thus oxygen combines
with two hydrogen atoms to form water, H20.
Oxygen is said to be divalent. Oxygen shares two
pairs of electrons, one pair with each hydrogen
atom. Each of these shared pairs forms a single
bond.

16-5.1 Bonding in the Oxygen Molecule

Now let us investigate the oxygen molecule,
which experiment tells us has the molecular for-
mula 02. We might begin by considering forma-
tion of a single bond between two oxygen atoms,
as represented by the orbital representation

<8> <8ES>

* <8> ®

(54)

We see that each oxygen atom has residual bond-
ing capacity. Each atom could, for example,
react with a hydrogen atom to form hydrogen
peroxide, as shown in electron dot representa-
tion (26). Each oxygen atom could react with a
fluorine atom to form F202. In short, each oxy-
gen atom is in need of another atom with an
electron in a half-filled valence orbital so that it
can act as a divalent atom.

Fig. 16-17. The molecular dipole of FaO.

But suppose oxygen can find no hydrosen
atoms or fluorine atoms. Then, it does the lazy
thing: the two atoms, already bound by one
bond, form a second bond with each other. The
result might be shown in the orbital representa-
tion (55).

** 2s 2P

0 <8> (8) ®SQ
0 <8> <8> <800

(55)

There is strong evidence supporting this pro-
posal. The bond in the oxygen molecule is
stronger than the oxygen-oxygen bond in hy-
drogen peroxide (more energy is required to
break it). The vibrational frequency of the oxy-
gen molecule is higher than that of a normal
single bond, showing that there is additional
bonding (see Section 14-3.4). The bond length
in the 02 molecule is 1.21 A. In the gaseous
hydrogen peroxide molecule, the oxygen-oxygen
distance is 1.48A. The short bond length in the
02 molecule shows that the two oxygen atoms
are drawn together more effectively than in
HOOH, suggesting there are extra bonding elec-
trons in 02.

Because all of the evidence we have examined
is consistent with the orbital representation (55),

296

MOLECULES IN THE GAS PHASE I CHAP. 16

the bond in 02 is called a double bond. An elec-
tron dot representation can be written as follows

:o: :o:

(56)

The representation (56) shows two pairs of elec-
trons shared. Each oxygen atom finds itself near
eight electrons. There is, on the one hand, a
stable molecule, because all of the bonding
capacity of each oxygen atom is in use. On the
other hand, this special aspect of the bonding of
oxygen undoubtedly contributes to the reactivity
of oxygen.

16-5.2 Ethylene: A Carbon-Carbon
Double Bond

Ethylene is a simple compound of carbon and
hydrogen with the formula C2H4. Thus it has two
less hydrogen atoms than does ethane, C2H6.
This means that to write a structure of ethylene
we must take account of two electrons that are
not used in C — H bond formation. Suppose we
write an electron dot representation involving
only single bonds

H H
H:C:C:H (57)

This formula has two unpaired electrons, repre-
senting unused bonding capacity. This objection-
able situation can easily be rectified by allowing
the two unpaired electrons to pair, and thus form
an additional two-electron bond. Now the car-
bon atoms are joined by a double bond, just as
the oxygen atoms in 02 are double bonded to
each other

H . . H

:c: :c:

H H

(58)

CHEMICAL REACTIVITY OF ETHYLENE

In ethane, C2H6, all of the bonds are normal
single bonds. Experiment shows that ethane is a
fairly unreactive substance. It reacts only when
treated with quite reactive species (such as free
chlorine atoms), or when it is raised to excited
energy states by heat (as in combustion).
Ethylene, on the other hand, reacts readily

with many chemical reagents. Having four elec-
trons forming the carbon-carbon bond, the elec-
trons of the double bond seem to be accessible
to attack. We find that the typical reactions of
ethylene are those with reagents that seek
electrons. For example, oxidizing agents are
electron-seeking, and we would expect the dou-
ble bond to be readily oxidized. This is indeed
the case. Ethylene will reduce (that is, be oxidized
by) such oxidizing agents as potassium perman-
ganate or potassium dichromate at ordinary
temperatures. Under these same mild conditions
ethane is completely unreactive to the same re-
agents.

GEOMETRICAL FEATURES OF ETHYLENE

The shape of the ethylene molecule has been learned by a
variety of types of experiments. Ethylene is a planar
molecule — the four hydrogen and the two carbon atoms
all lie in one plane. The implication of this experimental
fact is that there is a rigidity of the double bond which
prevents a twisting movement of one of the CH: groups
relative to the other. Rotation of one CH2 group relative
to the other — with the C — C bond as an axis — must be
energetically restricted or the molecule would not retain
this flat form.

CIS-TRANS ISOMERISM OF ETHYLENE
DERIVATIVES

It is possible to replace hydrogen atoms of ethylene by
halogen atoms. For example, one such compound has the
formula C2H2C12. In preparing such a compound, chem-
ists discovered long ago that they could obtain three dif-
ferent pure substances with this same formula, QH2Q2.
Different compounds with the same molecular formulas are
called isomers. The existence of three separate QH2CI2
isomers is readily explained in terms of the molecular
geometry. The three structures possible with this formula
are shown in Figure 16-18. They are all called dichloro-
ethylene.

Two types of isomerism are involved. Formula (59)
differs from (60) and (61). In formula (59) both chlorine
atoms are attached to the same carbon atom. In both

(60) and (61) there is one chlorine atom attached to each
carbon atom. The difference between (59) and the other
pair, (60) and (61), is indicated by calling these molecules
structural isomers.

The pair of isomers (60) and (61) differ in another way.
Though each has one chlorine atom attached to each
carbon atom, in (60) they are on the "same side" of the
double bond. This relationship is called the cis form. In
(67) the chlorine atoms are across from each other. This
relationship is called the trans form. Formulas (60) and

(61) identify cis and trans isomers of dichioroethylene.

QUBSTIONS AND PROBLEMS

297

Fig. 16-18. The isomers of dichloroethylene.

Experiment shows that it is exceedingly difficult to
convert (59) into (60) or (61). To make such a conversion,
bonds must be broken and reformed. Such reactions are
almost always quite slow because the activation energies

must be almost as large as the energies of the bonds being
broken. In contrast, the conversion of (60) into (61) (or
the reverse) can be accomplished merely by heating the
substance. No bonds must be broken completely — only
a rotation around the carbon-carbon double bond is
necessary. This process has a much lower activation
energy and the reaction occurs at moderate temperatures.

QUESTIONS AND PROBLEMS

1. Which one of the following statements is FALSE
as applied to this equation ?

Wg)

H(g) + H(g) AH = 103.4 kcal

(a) The positive AH means the reaction is endo-
thermic.

(b) Two grams of H(g) contain more energy
than 2 grams of H2(g).

(d) The spectrum of H2(g) is the same as the
spectrum of H(g).

2. What are the molecular species present in gase-
ous neon, argon, krypton, and xenon? Explain.

3. Determine the number of attractive forces and
the number of repulsive forces in LiH.

4. What energy condition must exist if a chemical
bond is to form between two approaching
atoms ?

298

MOLECULES IN THE GAS PHASE I CHAP. 16

5. What valence orbital and valence electron condi-
tions must exist if a chemical bond is to form
between two approaching atoms ?

6. Give the orbital and also the electron dot repre-
sentations for the bonding in these molecules:
Cl2, HG1, C120.

7. Using the electron dot representation, show a
neutral, a negatively charged, and a positively
charged OH group.

8. Draw the orbital representation of the molecule
N2H4, hydrazine.

9. Knowing the orbitals carbon uses for bonding,
use the periodic table to predict the formula of
the chloride of silicon. What orbitals does silicon
use for bonding?

10. Draw the orbital representations of

(a) sodium fluoride,

(b) beryllium fluoride, BeF2.

11. In general, what conditions cause two atoms to
combine to form:

(a) a bond that is mainly covalent;

(b) a bond that is mainly ionic;

12. What type of bonding would you expect to find
in MgO? Explain.

13. Considering comparable oxygen compounds,
predict the shape of H2S and H2S2 molecules.
What bonding orbitals are used?

14. Predict the formula and molecular shape of a
hydride of phosphorus.

15. Draw an electron dot representation for the
NH4" ion. What shape do you predict this ion
will have?

16. Predict the type of bonding and the shape of the
ion BF^~ .

17. Consider the two compounds CH3CH3 (ethane)
and CH3NH2 (methylamine). Why does CH3NH2
have an electric dipole while CH3CH3 does not ?

18. Consider the following series: CH4, CH3C1,
CH2C12, CHCla, CCL. In which case(s) will the
molecules have electric dipoles? Support your
answer by considering the bonding orbitals of
carbon, the molecular shape of the molecules,
and the resulting symmetry.

19. Predict the structure of the compound N2Fj from
the electron dot representation of the atoms and
the molecule.

20. Which of the isomers of dichloroethylene shown
in Figure 16-18 will be polar molecules?

21. Draw structural formulas for all the isomers of
ethylene (C2H4) in which two of the hydrogen
atoms have been replaced by deuterium atoms.
Label the cis and the trans isomers.

LINUS C. PAULING, 19 0 1

No other living chemist has contributed more to our under-
standing of chemical bonding than Linus C. Pauling. His
ideas pervade every aspect of chemistry. These ideas have
won him some seventeen medals and high awards, including
the 1954 Nobel Prize in Chemistry. His international re-
nown is bespoken by his election to honorary membership
in sixteen scientific societies in ten different countries.

Linus Pauling was born in Portland, Oregon, and his
hobbies as a boy were largely scientific. At 11 he began an
insect collection, which led to reading books on entomology.
At 13 Linus found a chemistry book in the family library
and founded a laboratory in the family basement. By the
time he entered Oregon State College he was determined
to become a chemical engineer. In 1922 Pauling received
the B.S. degree, and this was followed by the Ph.D. at
California Institute of Technology. By now his interest had
turned to the fundamental aspects of chemistry and after a
year of post-doctoral s.udy in Europe he returned to the
faculty at California Institute of Technology. There he
established and pursued his illustrious career.

Linus Pauling's prodigious scientific productivity has
broadly influenced the face of chemistry. His interest fo-
cused on the chemical bond and he was one of the earliest
of chemists to recognize the importance of the quantum
mechanical point of view. He gave quantitative meaning to
the electronegativity concept. He discussed the mixing of
the ionic and covalent character of chemical bonds and
introduced the term "resonance" — a concept that manv

chemists criticize but that most chemists use regularly.
Pauling gave detailed consideration to the effective sizes
of atoms in molecules and crystals. He became an authority
on hydrogen bonding and he proposed a theory of metallic
bonding. He advocated and, with his colleagues, established
the existence of helical structures in proteins. And, while
publishing over 300 papers, he authored several books that
won wide acceptance: Introduction to Quantum Me-
chanics (with E. B. Wilson, Jr.), The Nature of the
Chemical Bond, General Chemistry, and College Chem-
istry.

Pauling has a deep sensitivity to the welfare of mankind
and he has worked energetically to awaken the conscience
of society to its new responsibilities in a nuclear age. At-
tempting to inform the public of the pressing need for
lasting peace, he has written a book entitled No More War.
He has engaged in many public forums and debates, even
though these activities sometimes attracted ridicule in times
when fear made his cause an unpopular one. In recognition
of these activities he was awarded the 1962 Nobel Peace
Prize, and thus became the second person in history to
receive the Nobel Prize twice.

There is no chemistry course given today that is not in-
fluenced by the ideas of Linus C. Pauling. He is a man of
broad imagination, dramatic personality, and boundless
inspiration. Mankind will long benefit because he chose to
explore the frontiers of science.

CHAPTER

The Bonding in
Solids and Liquids

// is • ' ' possible to discuss the structure of any substance • • • by
describing the types • • • of its bonds and in this way to account for its
characteristic properties.

LINUS PAULING, 1939

Any pure gas, when cooled sufficiently, will con-
dense to a liquid and then, at a lower tempera-
ture, will form a solid. There is great variance
in the temperature at which this condensation
occurs. Apparently there is a corresponding vari-
ance of the forces in liquids and solids. For
example, lithium fluoride gas at one atmosphere
pressure condenses when cooled below 1949°K.
When the temperature is lowered to 1 143°K, the
liquid forms a clear crystal. In contrast, lithium
gas at this pressure must be cooled to 1599°K
before it forms a liquid and this liquid does not
solidify until the temperature reaches 453°K. The
solid is a white, soft metal, not resembling
crystalline lithium fluoride at all. Fluorine gas is
equally distinctive. At one atmosphere pressure
it must be cooled far below room temperature
before condensation occurs, at 85°K. Then the
liquid solidifies to a crystal at 50°K. Why do
these three materials behave so differently? Can
we understand this great variation? Let us begin
by finding a common point of departure.

Two or more atoms remain near each other
in a particular arrangement because the energy
300

favors that arrangement. This is true whether
the cluster of atoms is strongly or weakly bound,
whether it contains a few atoms or 1023 atoms,
whether the arrangement is regular (as in a
crystal) or irregular (as in a liquid). The cluster
of atoms is stable if and only if the energy is
lower when the atoms are together than when
they are apart.

Furthermore, there is but one reason that two
or more atoms have lower energy when they are
in proximity. In this way electrons can be close
to two or more positive nuclei simultaneously.
However, the magnitude of the attractive forces
varies greatly, depending on how close the elec-
trons are able to approach these positive nuclei.
This approach distance is fixed by the electron
occupancy of the valence orbitals.

Thus the occupancy of the valence orbitals is
the clue we shall follow in our attempt to predict
when to expect a substance to be a high-melting,
salt-like crystal, when to expect a metal, when
to expect a low-melting, molecular crystal. This
is an ambitious program. Let's see how far we
can go, beginning with the pure elements.

SEC. 17-1 I THE ELEMENTS

301

17-1 THE ELEMENTS

The examples just mentioned include two ele-
ments, fluorine and lithium. Fluorine forms a
weakly bound molecular solid. Lithium forms a
metallic solid. Let us see how we can account for
this extreme difference, applying the principles
of bonding treated in Chapter 16.

17-1.1 van der Waals Forces

The diatomic molecule of fluorine does not form
higher compounds (such as F3, F4, • • •) because
each fluorine atom has only one partially filled
valence orbital. Each nucleus in F2 is close to a
number of electrons sufficient to fill the valence
orbitals. Under these circumstances, the diatomic
molecule behaves like an inert gas atom toward
other such molecules. The forces that cause
molecular fluorine to condense at 85°K are,
then, the same as those that cause the inert gases
to condense. These forces are named van der
Waals forces, after the Dutch scientist who
studied them.

When the outer orbitals of all of the atoms in
the molecules are filled — giving inert gas con-
figurations— then the electrons of another mole-
cule cannot approach the nuclei closely. When
molecules of this sort approach each other, the
energy is lowered only a few tenths of a kilo-
calorie per mole. This weak interaction is typical
of van der Waals forces.

We have, now, a simple rule for predicting

when a weakly bound molecular liquid and a
low-melting crystal will be formed by a given
element. If the element forms a molecule that
gives each atom the orbital occupancy of an inert
gas, then only van der Waals interactions among
such molecules remain.

We shall take up later in this chapter the fac-
tors that determine the magnitude of van der
Waals forces. For the moment, we will merely
observe that the elements forming van der Waals
liquids and solids are concentrated in the upper
right-hand corner of the periodic table (see Fig-
ure 17-1). These are the elements able to form
stable molecules that satisfy completely the
bonding capacity of each atom.

EXERCISE 17-1

Gaseous phosphorus is made up of P4 molecules
with four phosphorus atoms arranged at the
corners of a regular tetrahedron. In such a
geometry, each phosphorus atom is bound to
three other phosphorus atoms. Would you ex-
pect this gas to condense to a solid with a low
or high melting point? After making a prediction
on the basis of the valence orbital occupancy,
check the melting point of phosphorus in Table
6-V1II, p. 101.

Fig. 17-1. Elements that form molecular crystals

bound by van der Waals forces.

I«

Clz

Brz

Air,

1 1 I 1

1 1 1 I

. 1

302

THE BONDING IN SOLIDS AND LIQUIDS I CHAP. 17

Fig. 17-2. Carbon forms network solids: diamond
and graphite.

17-1.2 Covalent Bonds and Network Solids

Fluorine, F2, oxygen, 02, and nitrogen, N2, all
form molecular crystals but the next member of
this row of the periodic table, carbon, presents
another situation. There does not seem to be a
small molecule of pure carbon that consumes
completely the bonding capacity of each atom.
As a result, it is bound in its crystal by a network
of interlocking chemical bonds.

With one 2s and three 2p orbitals available for
the bonding of carbon, we can expect it to form
a lattice in which each atom forms four bonds.

Furthermore, sp* bonding is connected with
tetrahedral bond angles (as in Figure 16-11).
These expectations are consistent with the ex-
perimentally determined structure of diamond,
shown in Figure 17-2.

Diamond is a naturally occurring form of
pure, crystalline carbon. Each carbon atom is
surrounded by four others arranged tetrahe-
drally. The result is a compact structural network
bound by normal chemical bonds. This descrip-
tion offers a ready explanation for the extreme
hardness and the great stability of carbon in this
form.

Fig. 17-3. Elements that form solids involving cova-
lent bonding.

(re

r<>

CI,

Br,

Atz

SEC. 17-1 I THE ELEMENTS

303

Graphite is another solid form of carbon. In
contrast to the three-dimensional lattice struc-
ture of diamond, graphite has a layered structure.
Each layer is strongly bound together but only
weak forces exist between adjacent layers. These
weak forces make the graphite crystal easy to
cleave, and explain its softness and lubricating
qualities.

The elements that form network solids lie on
the right side of the periodic table, bordering
the elements that form molecular crystals on
one side and those that form metals on the other.
Thus they are intermediate between the metals
and the nonmetals. In this borderline region
classifications are sometimes difficult. Whereas
one property may suggest one classification, an-
other property may lead to a different conclu-
sion. Figure 17-3 shows some elements that form
solids that are neither wholly metallic nor wholly
molecular crystals.

17-1.3 Metallic Bonding

We have considered solid forms of the elements
fluorine, oxygen, nitrogen, and carbon. In each
case, a solid is formed in which the bonding
capacity is completely satisfied. The remaining
elements of the second row. that is, beryllium,

and lithium, are metallic. These elements do not
have enough electrons to permit the complete
use of the valence orbitals in covalent bonding.
Furthermore, the ionization energies of these
elements are quite low. We find there are two
conditions necessary for metallic bonding:
vacant valence orbitals and low ionization en-
ergies.

CHARACTERISTIC PROPERTIES
OF METALS

Perhaps the most obvious metallic property is
reflectivity or luster. With few exceptions (gold,
copper, bismuth, manganese) all metals have a
silvery white color which results from reflecting
all frequencies of light. We have said previously
that the electron configuration of a substance
determines the way in which it interacts with
light. Apparently the characteristic reflectivity of
metals indicates that all metals have a special
type of electron configuration in common.

A second characteristic property of metals is
high electrical conductivity. The conductivity is
so much higher than that of aqueous electrolyte
solutions that the charge movement cannot in-
volve the same mechanism. Again we find a

Fig. 17-4. The metallic elements.

AT,

CI,

Co.

/Si

Go.

Br2

La-
Lu

«f

At,

Ac-

La.

Er-

Ant

Ctn

L*v

304

THE BONDING IN SOLIDS AND LIQUIDS I CHAP. 17

metallic behavior that suggests there is a special
electron configuration.

Metals also possess unusually high thermal
conductivity, as anyone who has drunk hot
coffee from a tin cup can testify. It is noteworthy
that among metals the best electrical conductors
are also the best thermal conductors. This is a
clue that these two properties are somehow
related and, again, the electron configuration
proves to be responsible.

Though the mechanical properties of the vari-
ous metals differ, all metals can be drawn into
wires and hammered into sheets without shatter-
ing. Here we find a fourth characteristic property
of metals: they are malleable or workable.

LOCATION OF METALS IN THE
PERIODIC TABLE

The location of the metals in the periodic table
is shown in Figure 17-4. We see that the metals
are located on the left side of the table, while the
nonmetals are exclusively in the upper right
corner. Furthermore, the elements on the left
side of the table have relatively low ionization
energies. We shall see that the low ionization
energies of the metallic elements aid in explain-
ing many of the features of metallic behavior.

ELECTRON BEHAVIOR IN METALS

What is the nature of the metallic bond? This
bond, like all others, forms because the electrons
can move in such a way that they are simultane-
ously near two or more positive nuclei. Our
problem is to obtain some insight into the special
way in which electrons in metals do this.

Consider a crystal of metallic lithium. In its
crystal lattice, each lithium atom finds around
itself eight nearest neighbors. Yet this atom has
only one valence electron, so it isn't possible for
it to form ordinary electron pair bonds to all of
these nearby atoms. However, it does have four
valence orbitals available so its electron and the
valence electrons of its neighbors can approach
quite close to its nucleus. Thus each lithium
atom has an abundance of valence orbitals but
a shortage of bonding electrons.

Consider the dilemma of the valence electron
of a particular lithium atom. It finds eight neigh-

bor nuclei nearby and complete freedom of
movement in the empty valence orbitals around
its parent nucleus. Everywhere the electron
moves it finds itself between two positive nuclei.
All of the space around a central atom is a
region of almost uniformly low potential energy.
Under these circumstances, it is not surprising
that an electron can move easily from place to
place. Each valence electron is virtually free to
make its way throughout the crystal.

This type of argument leads us to picture a
metal as an array of positive ions located at the
crystal lattice sites, immersed in a "sea" of
mobile electrons. The idea of a more or less
uniform electron "sea" emphasizes an important
difference between metallic bonding and ordi-
nary covalent bonding. In molecular covalent
bonds the electrons are localized in a way that
fixes the positions of the atoms quite rigidly. We
say that the bonds have directional character —
the electrons tend to remain concentrated in
certain regions of space. In contrast, the valence
electrons in a metal are spread almost uniformly
throughout the crystal, so the metallic bond does
not exert the directional influence of the ordinary
covalent bond.

We can obtain some idea of the effectiveness
of this electron "sea" in binding the atoms to-
gether if we compare the energy necessary to
vaporize one mole of a metal to the free atoms
with the energy required to break one mole of
ordinary covalent bonds. We find that the energy
necessary to vaporize a mole of one of the alkali
metals is only one-fourth to one-third of the
energy needed to break a mole of ordinary co-
valent bonds. This is not too surprising. The
ionization energy of a free alkali metal atom is
small ; this means that the valence electron in the
free atom does not experience a strong attraction
to the nucleus. Since the electron is not strongly
attracted by one alkali metal atom, it is not
strongly attracted by two or three such atoms
in the metallic crystal. Thus, the binding energy
between electrons and nuclei in the alkali metal
crystals is rather small, and the resulting metallic
bonds are rather weak. We might expect, how-
ever, that the metallic bond would become
stronger in those elements which have a greater

SEC. 17-1 I THE ELEMENTS

305

number of valence electrons and a greater nu-
clear charge. In these cases there are more elec-
trons in the "sea," and each electron is more
strongly bound, owing to the increased nuclear
charge. This argument is in accord with the ex-
perimental heats of vaporization shown in Table
17-1.

Table 17-1

HEATS OF VAPORIZATION OF METALS
(kcal/mole)

Second Row

Third row

Fourth Row

Fifth Row

Sixth Row

32.2

Be
53.5

129

Na
23.1

Mg
31.5

Al
67.9

K
18.9

36.6

Rb
18.1

Sr
33.6

16.3

35.7

To pick a specific case, let us compare the
heats of vaporization of magnesium and alumi-
num. The higher value for aluminum shows that
the metallic bond is indeed stronger when the
number of valence electrons and the charge on
the nucleus increase. Thus the strength of the
metallic bond tends to increase as we go from
left to right along a row in the periodic table.
The transition metal elements are harder and
melt and boil at higher temperatures than the
alkali or alkaline earth metals.

EXPLANATION OF THE PROPERTIES
OF METALS

The nonlocalized or mobile electrons account for
the many unique features of metals. Since metal-
lic bonds do not have strong directional charac-
ter, it is not surprising that many metals can be
easily deformed without shattering their crystal
structure. Under the influence of a stress, one

plane of atoms may slip by another, but as they
do so, the electrons are able to maintain some
degree of bonding between the two planes.
Metals can be hardened by alloying them with
elements which do have the property of forming
directed covalent bonds. Often just a trace of
carbon, phosphorus, or sulfur will turn a rela-
tively soft and workable metal into a very brittle
solid.

*^W£%

Fig. 17-5. Slippage of planes of metal atoms.

Metals conduct electricity because some va-
lence electrons are free to move throughout the
solid. At the same time, these mobile electrons
are effective in holding the crystal together be-
cause wherever they move, they are simultane-
ously close to two or more nuclei. In covalently
bonded solids the electrons are strongly localized
in the space between a particular pair of atoms.
In order for these substances to conduct elec-
tricity, a great deal of energy must be supplied
to remove the electrons from this region between
the atoms. This energy is not available in normal
electric fields, so covalent substances do not
normally conduct electricity.

The excellent heat conductivity of metals is
also due to the mobile electrons. Electrons which

306

THE BONDING IN SOLIDS AND LIQUIDS | CHAP. 17

are in regions of high temperature can acquire
large amounts of kinetic energy. These electrons
move through the metal very rapidly and give
up their kinetic energy to heat the crystal lattice
in the cooler regions. In substances where the
electrons are highly localized, heat is conducted
as small amounts of energy are transferred from
one atom to its immediate neighbor; this is a
slower process than electron energy conduction.

To complete our discussion of metallic bond-
ing we must explain why metallic properties
eventually disappear as we proceed from left to
right along a row in the periodic table.

We have seen that the reasons for the mobility
of electrons in metals are that they are readily
removed from the atom (the ionization energy is

low) and that they can be close to two or more
positive nuclei just about anywhere in the crystal
(there are numerous vacant valence orbitals). As
the nuclear charge on atoms increases and the
vacant orbitals become filled, the regions imme-
diately between two nuclei become relatively
more attractive to the electron, compared with
all other regions. Electrons tend to be more and
more localized in these regions, and normal
covalent bonds with their directional character
appear.

In summary we can say that the metallic bond
is a sort of nondirectional covalent bond. It
occurs when atoms have few valence electrons
compared with vacant valence orbitals and when
these valence electrons are not held strongly.

17-2 COMPOUNDS

We have seen that the pure elements may solidify
in the form of molecular solids, network solids,
or metals. Compounds also may condense to
molecular solids, network solids, or metallic
solids. In addition, there is a new effect that does
not occur with the pure elements. In a pure ele-
ment the ionization energies of all atoms are
identical and electrons are shared equally. In
compounds, where the most stable electron dis-
tribution need not involve equal sharing, electric
dipoles may result. Since two bonded atoms may
have different ionization energies, the electrons
may spend more time near one of the positive
nuclei than near the other. This charge separa-
tion may give rise to strong intermolecular forces
of a type not found in the pure elements.

17-2.1 van der Waals Forces and
Molecular Substances

Though charge separations are possible in com-
pounds, there are many molecules that do not
have appreciable electric dipoles. On cooling,
these molecules behave much like the molecules
of pure elements. If the bonding capacity of
each atom is completely satisfied, then only the
weak van der Waals forces remain between mole-

cules. These weak interactions give low melting
solids and low boiling liquids that retain many
of the properties of the gaseous molecules.

There are three factors that seem to be par-
ticularly important in determining the magni-
tudes of van der Waals forces: the number of
electrons, the molecular size, and the molecular
shape. These factors are effective both for ele-
ments and compounds, though greater variety is
found for compounds.

VAN DER WAALS FORCES AND
NUMBER OF ELECTRONS

We have already observed in Chapter 6 that the
melting and boiling points of the inert gases in-
crease as the number of electrons increases (see
Figure 6-3). Elements and compounds with co-
valent bonding behave in the same way. Figure
17-6 shows this in a graphical presentation.
Figure 17-6 A shows the melting and boiling
point trends among the inert gases and among
the halogens. The horizontal axis shows the row
number, which furnishes an index of the total
number of electrons of the respective elements.
Figure 17-6B refers to compounds with formulas
CX4. Again the horizontal axis shows the row
number but now of the outermost atoms in the

SEC. 17-2 I COMPOUNDS

307

-400

300

T, °K

200

100

400

300

200

100

* Sotting point
• Melting point

• Boiling point

CH+ CFj •- — — • Melting point

12 3 4 5 6

Row number of JC in. CX

Fig. 17-6. The melting and boiling points of some
molecular compounds and the halogens.
A. The inert gases and the halogens. B.
The carbon compounds of formula CXt.

molecule since these are the atoms which "rub
shoulders" with neighboring molecules. As far
as van der Waals forces are concerned, it is quite
important that CBr4 has atoms from the fourth
row of the periodic table on the "surface" of the
molecule and somewhat less important that the
central atom, carbon, is from the second row.
The outermost atoms are most influential in fix-
ing intermolecular forces.

VAN DER WAALS FORCES AND
MOLECULAR SIZE

If comparisons are made among similar mole-
cules, then the larger the molecule, the higher
is its melting point. For example, if we compare
methane, CH4, and ethane, C2H6, the exterior
atoms are the same — hydrogen atoms. Still, the
boiling point of ethane, 185°K, is higher than
that of methane, 112°K. This difference is at-

tributed to the fact that there must be greater
contact surface between two ethane molecules
than between two methane molecules. The same
effect is found for C2F6 (boiling point, 195°K)
and CF4 (boiling point, 145°K); for C2Br6 (this
substance decomposes at 483°K before it reaches
its boiling point) and CBr4 (boiling point,
463°K).

Notice that the two factors just mentioned, number of
electrons and molecular size, might lead to another gen-
eralization— that the boiling point goes up in proportion
to molecular weight. The molecular weight, the molecular
size, and the number of electrons all tend to increase
together. This molecular weight-boiling point correlation
has some usefulness among molecules of similar com-
position and general shape but chemists do not feel that
there is a direct causative relation between molecular
weight and boiling point.

VAN DER WAALS FORCES AND
MOLECULAR SHAPE

Substances whose structures have a high degree
of symmetry generally have higher melting points
than closely related compounds that lack this
symmetry. There are striking examples of this

308

THE BONDING IN SOLIDS AND LIQUIDS | CHAP. 17

Tabic 17-11. THE EFFECT OF MOLECULAR SHAPE ON MELTING POINT: cis-
A N D trans- ISOMERS

m.p. cis-

m.p. trans-

1,2-dichloroethylene CICH=CHCI -80°C

butenoic acid CH3CH=CHCOOH 15°

fumaric, maleic acids HOOCCH=CHCOOH 130°

-50°C (nans- 30° higher m.p.)

72° (trans- 57° higher m.p.)

290° (trans- 160° higher m.p.)

among the double-bonded compounds. The cis-
and trans- isomers of many such compounds
have melting point and boiling point differences
that can be traced to the differences in molecular
shape. For example, the higher melting point of
trans-\,2 dichloroethylene than that of the iso-
meric cis-\,2 dichloroethylene may be partly ex-
plained by arguing that the long and symmetrical
trans- form can pack into an orderly crystal
lattice in a neater and more compact fashion
than the "one-sided" cis- form molecules. This
and two other examples of this are shown in
Table 17-11. (The dichloroethylene structures are
shown in Figure 16-18.)

Another example of the influence of molecular
symmetry on physical properties is found in two
structural isomers of the formula C6Hi2. These

are called normal pentane and neopentane and
their molecular shapes differ drastically as
shown in Figure 17-7.

The extended molecule, H-pentane, has a zig-
zag shape. We see that van der Waals forces act
between the external envelope of hydrogen atoms
of one molecule and those of adjacent molecules.
This large surface contact gives a relatively high
boiling point. On the other hand, this flexible,
snake-like molecule does not pack readily in a
regular lattice, so its crystal has a low melting
point.

Contrast the highly compact, symmetrical neo-

Fig. 17-7. Molecular shape, a factor that influences
melting and boiling points.

CH*

H,C

Normal pen-tane
b.p. 36°C

TTL.p.

-130°C

Neopenrtane

b.p. 9°C

m.p. -200C

SEC. 17-2 I COMPOUNDS

309

pentane. This ball-like molecule readily packs
in an orderly crystal lattice which, because of its
stability, has a rather high melting point. Once
melted, however, neopentane forms a liquid that
boils at a temperature below the boiling point of
«-pentane. Neopentane has less surface contact
with its neighbors and hence is more volatile.
It is well to add that most of the compounds
of carbon condense to molecular liquids and
solids. Their melting points are generally low
(below about 300°C) and many carbon com-
pounds boil below 100°C. The similar chemistry
of the liquid and solid phases shows the reten-
tion of the molecular identities.

17-2.2 Covalent Bonds and Network
Solid Compounds

Compounds can form network solids and, since
two or more different atoms are involved, there
is much greater variety among the network solid
compounds than among the network solid ele-
ments. Silica, with empirical formula Si02, is a
network solid. Silica and other silicon-oxygen
compounds make up about 87 % of the earth's
crust. Almost all common minerals contain sub-
stantial amounts of silicates, the general term for
silicon-oxygen solids. These are network solids
but with interesting and important variations.
Figures 17-8, 17-9, and 17-10 show in a schematic
way the three types of network solids formed by
silicon. The silicon is always tetravalent but in
some of its compounds it forms infinite silicon-
oxygen-silicon chains; in some it forms infinite
interlinked sheets; and, in some, it forms an
infinite three-dimensional network solid.

Many properties of silicates can be understood
in terms of the type of network lattice formed.
In the "one-dimensional" networks, shown in
Figure 17-8, the atoms within a given chain are
strongly linked by covalent bonds but the chains
interact with each other through much weaker
forces. This is consistent with the thread-like
properties of many of these silicates. The asbes-
tos minerals are of this type.

In a similar way, the sheets of the "two-
dimensional" network silicates, shown in Figure

17-9, are held together weakly. Hence these
minerals cleave readily into thin but strong
sheets. The micas have this type of structure.
Clays also have this structure, and their slippery
"feel" when wet can be explained in terms of the
hydration of the planes on the outside of the
crystals. The three-dimensional network shown
in Figure 17-10 is silica (quartz). Like diamond,
it is hard and it has a high melting point.
The various minerals that make up granite are
of this type.

17-2.3 Metallic Alloys

We have already learned that metals may be deformed
easily and we have explained this in terms of the absence
of directional character in metallic bonding. In view of
this principle, it is not surprising that two-element or
three-element metallic crystals exist. In some of these,
regular arrangements of two or more types of atoms are
found. The composition then is expressed in simple in-
teger ratios, so these are called metallic compounds. In
other cases, a fraction of the atoms of the major con-
stituent have been replaced by atoms of one or more
other elements. Such a substance is called a solid solution.
These metals containing two or more types of atoms are
called alloys.

ELECTRICAL CONDUCTIVITY

Electrical conductivity in metals apparently depends upon
the smooth and uninterrupted movement of electrons
through the lattice. This is suggested by the fact that small
amounts of impurities reduce the conductivity very much.
We shall see, in Chapter 22, that copper is purified com-
mercially to 99.999% and the reason is directly connected
to the consequent gain in electrical conductivity.
Table 17-111 shows some conductivities of copper with

Table 17-I/I

CONDUCTIVITY OF COPPER ALLOYS
(ALL AT 20 C UNLESS NOTED)

PERCENT
PERCENT ALLOYING ALLOYING CONDUCTIVITY

COPPER ELEMENT ELEMENT (ohm-CmV"1

100.00

—

5.9 X 10s

Mn (0°C)

0.98

2.1 X 10s

95.8

4.2

0.56 X 10s

Al (0°C)

1.2 X 10s

Al (0°C)

0.79 X 10s

0.56 X 10s

Fig. 17-8. One-dimensional network silicates:

the asbestos minerals.

Fig. 17-9. Two-dimensional network silicates.

the mica and clay minerals.

**Qk,

• Silicon
O Oxygen
X Other atoms

Fig. 17-10. Three-dimensional network silicates:

granitic minerals.

SEC. 17-2 | COMPOUNDS

311

.Vj

12 3 4

Percervfr Jrfti, by weight

Fig. 17-11. Electrical resistivity of copper containing
manganese.

various impurities.* Figure 17-11 shows the data for
copper-manganese alloys graphically. The figure shows
resistivity, the reciprocal of conductivity, plotted against
percent manganese (by weight). The importance of puri-
fication of copper for electrical wire is evident in this
figure if we remember that the power lost in a conductor
is proportional to resistance (for a given current). In a
conductor hundreds of miles long, a factor of two reduc-
tion in resistivity is a lucrative gain to a company selling
electrical power.

EXERCISE 17-2

Use Figure 17-11 to estimate the resistivities of two metal
samples, one made of pure copper and the other of a
copper-manganese alloy containing one atom of manga-
nese for every one hundred copper atoms. Calculate the
ratio of the cost due to power loss from wire of the impure
material to the cost due to the power loss from wire of
the pure material.

HARDNESS AND STRENGTH

Alloys are harder and stronger than pure metals as usu-
ally prepared. The most familiar example is steel and pure
iron. The tensile strength of pure iron can be increased

* The conductivities are given in units (ohm-cm)-1. The
reciprocal of this number is the resistance one would find
(in ohms) for a wire 1 cm long and with a cross-section
of 1 cm*.

ten-fold by the addition of only a percent of carbon and
smaller amounts of nickel or manganese. The tensile
strength of brass (65-70% Cu, 35-30% Zn) is more than
twice that of copper and four times that of zinc.

The hardness and strength of alloys can be explained
in terms of bonding. The impurity atoms added may form
localized and rigid bonds. These tend to prevent the
slippage of atoms past each other, which results in a loss
of malleability and an increase in hardness.

17-2.4 Ionic Solids

Thus far we have not considered the effects that
arise from charge separations. The most extreme
case is represented by the formation of ionic
solids. Usually, these can be looked on as arrays
of positive and negative ions, neatly stacked so
that each positive ion has only negative ion
neighbors and each negative ion has only posi-
tive ion neighbors. Figure 5-10 (p. 81) shows
such a crystal arrangement, that of sodium
chloride. Why does such a solid form and what
are its properties? These are the questions we
shall try to answer here.

THE STABILITY OF IONIC CRYSTALS

In discussing the bonding in the gaseous LiF
molecule, the electric dipole of the molecule is
explained in terms of the different ionization
energies of Li and F atoms. Though the molecule
holds together because the bonding electrons are
near both nuclei, the energy favors an electron
distribution concentrated toward the fluorine.
A stable and polar molecule is formed. Stable,
perhaps, but in the gaseous state, reactive! The
valence orbitals of the lithium atom are almost
vacant. According to our experience (for exam-
ple, with CH2, BH3, carbon atoms, metal atoms)
the presence of empty valence orbitals implies
that additional electron sharing can occur. Lith-
ium fluoride molecules are, then, more stable
when they condense so as to place each lithium
atom simultaneously near several fluorine atoms.
Just as in metals, an atom with vacant orbitals
is more stable with several neighbors. Then the
electrons held by the neighbor atoms can be near
two or more nuclei at once. There is, however,
a significant difference from metals — in solid
lithium fluoride, half of the atoms have high

312

THE BONDING IN SOLIDS AND LIQUIDS I CHAP. 17

Fig. 17-12. Sodium chloride crystals.

ionization energies. Fluorine atoms hold their
electrons tightly. Therefore the characteristic
electron mobility of metals is not present in the
ionic solids. The absence of mobile electrons
implies that none of the metallic properties is
expected. Let us see what properties such a solid
does have.

PROPERTIES OF IONIC CRYSTALS

Ionic solids, such as lithium fluoride and sodium
chloride, form regularly shaped crystals with well
defined crystal faces. Pure samples of these solids
are usually transparent and colorless but color
may be caused by quite small impurity contents
or crystal defects. Most ionic crystals have high
melting points.

Molten lithium fluoride and sodium chloride
have easily measured electrical conductivities.
Nevertheless, these conductivities are lower than
metallic conductivities by several factors of ten.
Molten sodium chloride at 750°C has a conduc-
tivity about 10-5 times that of copper metal at
room temperature. It is unlikely that the electric
charge moves by the same mechanism in molten
NaCl as in metallic copper. Experiments show
that the charge is carried in molten NaCl by
Na+ and Cl~ ions. This electrical conductivity
of the liquid is one of the most characteristic

properties of substances with ionic bonds. In
contrast, molecular crystals generally melt to
form molecular liquids that do not conduct elec-
tricity appreciably.

17-2.5 Effects Due to Charge Separation

We have considered the weak van der Waals
forces that cause the condensation of covalent
molecules. The formation of an ionic lattice
results from the stronger interactions among
molecules with highly ionic bonds. But most
molecules fall between these two extremes.
Most molecules are held together by bonds that
are largely covalent, but with enough charge
separation to affect the properties of the mole-
cules. These are the molecules we have called
polar molecules.

Chloroform, CHC13, is an example of a polar
molecule. It has the same bond angles as meth-
ane, CH4, and carbon tetrachloride, CCL». Car-
bon, with spz bonding, forms four tetrahedrally
oriented bonds (as in Figure 16-11). However,
the cancellation of the electric dipoles of the four
C — CI bonds in CCI4 does not occur when one
of the chlorine atoms is replaced by a hydrogen
atom. There is, then, a molecular dipole remain-
ing. The effects of such electric dipoles are
important to chemists because they affect chemi-
cal properties. We shall examine one of these,
solvent action.

SEC. 17-2 I COMPOUNDS

313

SOLVENT PROPERTIES AND
MOLECULAR DIPOLES

The forces between molecules are strongly af-
fected by the presence of molecular dipoles. Two
molecules that possess molecular dipoles tend to
attract each other more strongly than do mole-
cules without dipoles. One of the most important
results of this is found in solvent properties.
Table 17-IV shows some solubility data of

Table 17-IV

SOLUBILITIES IN CARBON TETRA-
CHLORIDE, CCI„ AND IN ACETONE,

CHCOCH (25°C, moles liter)

SOLVENT

SOLUTE

CCh

CH3COCH3

SOLUTE

POLARITY

{nonpolar)

(polar)

CH«, methane

nonpolar

0.029

0.025

CjH6, ethane

nonpolar

0.22

0.13

CH3C1,

chloromethane

polar

1.7

2.8

CH3OCH3,

methyl ether

polar

1.9

2.2

various solutes in the two solvents, carbon tetra-
chloride, CCLj and acetone, CH3COCH3. These
two solvents differ in their polar properties. In
CCU the central carbon atom is surrounded by
four bonds that form a regular tetrahedron like
that pictured in Figure 16-11. With this molecu-
lar shape, CCLj has a zero molecular dipole. In
contrast, acetone has a bent structure and the
oxygen atom gives it a significant electric dipole.

Contrast the solubilities in Table 17-IV. The
first two substances, CH4 and C2H6, have zero
molecular dipoles. In each case, the solubility in
CCU exceeds the solubility in CH3COCH3. The
next two substances, CH3C1 and CH3OCH3, have
nonzero molecular dipoles. In each of these
cases, the solubility in acetone is the larger.

There is a reasonable explanation of the data
in Table 17-IV. When a solute dissolves, the
solute molecules must be separated from each
other and then surrounded by solvent molecules.
Furthermore, the solvent molecules must be
pushed apart to make room for the solute mole-

cules. Since dipoles interact strongly with each
other, a polar mblecule such as CH3CI is en-
ergetically more stable when surrounded by
solvent molecules that are also polar. Hence,
CH3CI has the higher solubility in the polar
solvent acetone than in carbon tetrachloride. On
the other hand, a nonpolar molecule such as
CH4 will find it difficult to wedge in between the
strongly interacting molecules of a polar solvent
— more difficulty than it will encounter in dis-
solving in a nonpolar solvent. Hence CH4 has
higher solubility in the nonpolar solvent, CCI4.

SOLUBILITY OF ELECTROLYTES
IN WATER

The dissolving of electrolytes in water is one of
the most extreme and most important solvent
effects that can be attributed to electric dipoles.
Crystalline sodium chloride is quite stable, as
shown by its high melting point, yet it dissolves
readily in water. To break up the stable crystal
arrangement, there must be a strong interaction
between water molecules and the ions that are
formed in the solution. This interaction can be
explained in terms of the dipolar properties of
water.

When an electric dipole is brought near an ion,
the energy is lower if the dipole is oriented to
place unlike charges in proximity. Hence water
molecules tend to orient preferentially around
ions, the positive end of the water dipole pointing
inward if the ion carries negative charge and the
negative end pointing inward if the ion carries
positive charge. Figure 17-13 shows this process
schematically: it is called hydration.

There are two effects of the orientation of
water dipoles around the ions. First, the energy
is lowered because the orientation serves to bring
unlike charges near each other. This tends to
encourage the ions to leave the sodium chloride
crystal and enter the solution. Also, however,
there is an effect on randomness whose magni-
tude is difficult to predict. The orientation of the
water molecules around the ion, fixing them rela-
tive to the ion, constitutes an orderly arrange-
ment. Since all systems tend toward maximum
randomness, the orientation effect works against
molecules leaving the crystal to enter the solu-

314

THE BONDING IN SOLIDS AND LIQUIDS I CHAP. 17

Fig. 17-13. Hydration of ions: orientation of water
dipoles around ions in aqueous solutions.

tion. These two effects of ion-hydration, lower-
ing the energy of an electrolyte solute while
decreasing the change in randomness as it dis-
solves, give water distinctive properties as an
electrolyte solvent. It helps explain, for example,
why some salts absorb heat as they dissolve in
water (for example, NH4C1) while some release
heat as they dissolve (for example, NaOH). For
most solvents the crystal has lower energy than
the solution, and heat is absorbed as solid dis-
solves. In water, however, the hydration effects
can cause the solution to have the lower energy,
so heat can be evolved during the dissolving
process.

17-2.6 Hydrogen Bonds

In Figure 17-6 A we saw that the boiling points of
symmetrical molecules increase regularly as we
drop down in the periodic table. Figure 17-14
shows the corresponding plot for some molecules
possessing electric dipoles.

Consider first the boiling points of HI, HBr,
HC1, and HF. The last, hydrogen fluoride, is far
out of line, boiling at 19.9°C instead of below
— 95°C as would be predicted by extrapolation
from the other three. There is an even larger
discordancy between the boiling point of H20
and the value we would predict from the trend
suggested by H2Te, H2Se, and H2S.

Could the extremely high boiling points of HF
and H20 be due to the fact that these are the
smallest molecules of their respective series? No,

SEC. 17-2 I COMPOUNDS

315

this does not appear to be the explanation, for
corresponding discrepancies are not present in
the data plotted in Figure 17-6A. There must be
some other explanation for these exceptional
boiling points. There must be forces of some new
kind between the molecules of H20 and of HF
that tend to keep them in the liquid phase.

These same forces are recognized in solid com-
pounds. The most familiar example is solid H20,
or ice. Ice has a crystal structure in which the
oxygen and hydrogen atoms are distributed in a
regular hexagonal crystalline lattice that some-
what resembles the diamond lattice (see Figure
17-2). Each oxygen atom is surrounded by four
other oxygen atoms in a tetrahedral arrange-
ment. The hydrogen atoms are found on the lines
extending between the oxygen atoms.

/
O— H----0

/ \

The attractive force between — OH and O must
be the bond that joins the water molecules to-
gether into the crystal lattice of ice. This bond
is a hydrogen bond.

ENERGY OF HYDROGEN BONDS

The hydrogen bond is usually represented by

O — H O in which the solid line represents

the original O — H bond in the parent compound
(as in water, HOH, or methyl alcohol, CH3OH).

Fig. 17-14. The boiling points of some hydrides.
ICO "'°

Bo i tiny poin-t,

SO -

-100

The dotted line shows the second bond formed
by hydrogen, the bond called the hydrogen bond.
It is usually dotted to indicate that it is much
weaker than a normal covalent bond. Considera-
tion of the boiling points in Figure 17-14, on the
other hand, shows that the interaction must be
much stronger than van der Waals forces. Ex-
periments show that most hydrogen bonds re-
lease between 3 kcal/mole and 10 kcal/mole
upon formation:

AH = -3 to -7 kcal/mole

(/)

The energy of this bond places it between van
der Waals and covalent bonds. Roughly speak-
ing, the energies are in the ratio

van der Waals

attractions

hydrogen bonds : covalent bonds
10 : 100

WHERE HYDROGEN BONDS ARE FOUND

Hydrogen bonds are found between only a few
atoms of the periodic table. The commonest are
those in which H connects two atoms from the
group F, O, and N, and less commonly CI.

The hydrogen bond to fluorine is clearly evi-
dent in most of the properties of hydrogen fluo-
ride. The high boiling point of HF, compared
with those of the other hydrogen halides, is one
of several pieces of data that show that HF does
not exist in the liquid compound as separate HF
molecules. Instead there are aggregates of mole-
cules, which we describe in general terms as
(HF)*. Gaseous hydrogen fluoride contains the
molecular species H2F2, H3F3, and so on up to
H6F6 as well as some single HF molecules.

These species can be represented in a descrip-
tive formula such as the following:

H— F • • H— F H— F H— F (2)

An extreme example of the fluorine-hydrogen
bond is found in the hydrogen difluoride ion,
HF2" . This ion exists in acidic solutions of fluo-
rides,

H+(aq) + ¥~(aq)
HF(aq) + F-(aq)

HF(aq)
HF2" (aq)

(i)
00

Row number of X
in HX and H2X

and in the ionic crystal lattice of salts such as
KHF2. The HF2~ ion may be regarded as con-

316

THE BONDING IN SOLIDS AND LIQUIDS | CHAP. 17

sisting of two negatively charged fluoride ions
held together by a proton:

F © F

It is not typical, however, for few hydrogen
bonds form with the proton equidistant from
the two atoms to which it bonds.

this difference is that intramolecular hydrogen bonds can
exist between the two — COOH groups for maleic acid
but not for fumaric acid:

(5) maleic acid 0=C

OH -O

C— O

(8)

H C H

\ / \ /

fumaric acid C C O

/ \

0=C H

(9)

INTER- AND INTRAMOLECULAR
HYDROGEN BONDS

One of the factors connected with the formation of strong
hydrogen bonds is the acidic character of the hydrogen
atom involved. Thus the hydrogen bond formed by hy-
drogen fluoride is one of the strongest known. Acetic acid,
CH3COOH, is a representative of an important class of
acidic hydrogen bonding compounds. All of the members
of this class possess the structural unit called the car-
boxylic acid group:

(6)

O— H

For this type of compound, the formation of hydrogen
bonds can lead to the coupling of the molecules in pairs,
to form a cyclic structure:

O...H— O

2CH3COOH^rCH3— C C— CH3

\ S

O— H O

AH = -14kcal (7)

Here the favorable geometrical arrangement with two
hydrogen bonds contributes 14 kcal to the stability of the
hydrogen bonded product, (7). These are called inter-
molecular hydrogen bonds (inter means between).

Hydrogen bonds can also exist when the O — H group
and the other bonding atom are close together in the
same molecule in such positions that a ring can be formed
without disturbing the normal bond angles. These are
called w/ramolecular hydrogen bonds (intra means
within).

An example of intramolecular hydrogen bonding
is provided by the cis- and trans- forms of the acid
HOOC— CH=CH— COOH. The trans- form, fumaric
acid, has a higher melting point than the cis- form, maleic
acid. In addition to the general effect of molecular shape
(mentioned earlier in this chapter), another reason for

This intramolecular bonding in maleic acid, (8), halves
its ability to form intermodular bonds. In fumaric acid,
on the other hand, all of the hydrogen bonds form between
molecules (intermolecular bonds) to give a stronger, in-
terlinked crystal structure.

THE NATURE OF THE HYDROGEN BOND

In the hydrogen bond we find the hydrogen atom
attached to two other atoms. Yet our bonding
rules tell us that the hydrogen atom, with only
the \s orbital for bond formation, cannot form
two covalent bonds. We must seek an explana-
tion of this second bond.

The simplest explanation for the hydrogen
bond is based upon the polar nature of F — H,
O — H, and N — H bonds. In a molecule such as
H20, the electron pair in the O — H bond is dis-
placed toward the oxygen nucleus and away from
the hydrogen nucleus. This partial ionic charac-
ter of the O — H bond lends to the hydrogen
atom some positive character, permitting elec-
trons from another atom to approach closely to
the proton even though the proton is already
bonded. A second, weaker link is formed.

THE SIGNIFICANCE OF THE
HYDROGEN BOND

Hydrogen bonds play an important part in de-
termining such properties as solubility, melting
points, and boiling points, and in affecting the
form and stability of crystal structures. They
play a crucial role in biological systems. For ex-

QUESTIONS AND PROBLEMS

317

ample, water is so common in living matter that
it must influence the chemical behavior of many
biological molecules, most of which can also
form hydrogen bonds. Water can attach itself by
hydrogen bonding, either by providing the pro-
ton, as in

O— H

o=c

(10)

or by accepting the proton, as in

H
\ /

N— H

<")

Furthermore, intramolecular hydrogen bonding
is one of the chief factors in determining the
structure of such important biological sub-
stances as proteins, as discussed in Chapter 24.

QUESTIONS AND PROBLEMS

1. Make a table that contrasts the melting points
and boiling points of LiF, Li, and F2, expressing
the temperatures on the Centigrade scale.

2. Without looking in your textbook, do the fol-
lowing.

(a) Draw an outline of the periodic table, indi-
cating the rows but not the individual ele-
ments.

(b) Place a number at the left of each row in-
dicating the number of elements in that row.

(d) Draw two diagonal lines across the table to
separate it into three regions. Write in each
region one of the words "metals," "non-
metals," "covalent solids."

(e) Now compare your diagram to Figure 17-4.

Sulfur exists in a number of forms, depending upon the temperature and, sometimes, upon the past
history of the sample. Three of the forms are described below. A is the room temperature form and
it changes to B above the melting point of A, 113°C. B changes to C on heating above 160°C.

Crystalline solid

Yellow color, no metallic

luster
m.p. = 113°C
Dissolves in CSj, not in

water
Electrical insulator

113°C

Liquid

Clear, straw color
Viscosity (fluidity) about
the same as water

Electrical insulator

^200°C

Liquid

Dark color

Very viscous (syrupy)

Electrical insulator

Which of the following structures would be most likely to account for the observed properties of each
of the three forms described above?

(a) a metallic crystal of sulfur atoms;

(b) a network solid of sulfur atoms;

(d) a molecular crystal of Ss molecules;

(e) a metallic liquid like mercury;

(0 a molecular liquid of Ss molecules;

(g) a molecular liquid of Sn chains, with n = a

very large number;
(h) an ionic liquid of S+ and S" ions.

318

THE BONDING IN SOLIDS AND LIQUIDS I CHAP. 17

4. Contrast the bonds between atoms in metals, in
van der Waals solids, and in network solids in
regard to:

(a) bond strength;

(b) orientation in space;

5. Aluminum, silicon, and sulfur are close together
in the same row of the periodic table, yet their
electrical conductivities are widely different.
Aluminum is a metal; silicon has much lower
conductivity and is called a semiconductor; sul-
fur has such low conductivity it is called an
insulator. Explain these differences in terms of
valence orbital occupancy.

6. Sulfur is made up of Sg molecules; each molecule
has a cyclic (crown) structure. Phosphorus con-
tains P4 molecules; each molecule has a tetra-
hedral structure. On the basis of molecular size
and shape, which would you expect to have the
higher melting point?

7. Discuss the conduction of heat by copper (a
metal) and by glass (a network solid) in terms
of the valence orbital occupancy and electron
mobility.

8. The elements carbon and silicon form oxides
with similar empirical formulas : C02 and Si02.
The former sublimes at — 78.5°C and the latter
melts at about 1700°C and boils at about 2200°C.
From this large difference, propose the types of
solids involved. Draw an electron dot or orbital
representation of the bonding in C02 that is con-
sistent with your answer.

9. How do you account for the following properties
in terms of the structures of the solids?

(a) Graphite and diamond both contain carbon.
Both are high melting yet the diamond is
very hard while graphite is a soft, greasy
solid.

(b) When sodium chloride crystals are shattered,
plane surfaces are produced on the frag-
ments.

10. If you were given a sample of a white solid,
describe some simple experiments that you
would perform to help you decide whether or
not the bonding involved primarily covalent
bonds, ionic bonds, or van der Waals forces.

11.

12.

13.

14.

If elements A, D, E, and J have atomic numbers,
respectively, of 6, 9, 10, and 11, write the formula
for a substance you would expect to form be-
tween the following:

(a) D and J;

(b) A and D;

(d) E and E;

(e) J and J.

In each case describe the forces involved between
the building blocks in the solid state.

Consider each of the following in the solid state :
sodium, germanium, methane, neon, potassium
chloride, water. Which would be an example of

(a) a solid held together by van der Waals forces
that melts far below room temperature;

(b) a solid with a high degree of electrical con-
ductivity that melts near 200°C;

(d) a nonconducting solid which becomes a good
conductor upon melting;

(e) a substance in which hydrogen bonding is
pronounced ?

Predict the order of increasing melting point of
these substances containing chlorine: HC1, Cl2,
NaCl, CC14. Explain the basis of your prediction.

Identify all the types of bonds you would expect
to find in each of the following crystals:

(a) argon,

(b) water,

(d) carbon monoxide,

(e) Si,

(0 Al,
(g) CaCl2,
(h) KCIO3,
(i) NaCl,
(j) HCN.

15. Each of three bottles on the chemical shelf contains a colorless liquid. The labels have fallen off the
bottles. They read as follows.

Label No. 1

Label No. 2

//-butanol
CH3CH2CH2CH2OH

mol wt = 74.12

w-pentane

CH3CH2CH2CH2CH|
mol wt = 72.15

Label No. 3

diethyl ether
CH3CH2OCH2CH3

mol wt = 74.12

QUESTIONS AND PROBLEMS 319

The three bottles are marked A, B, and C, and a series of measurements were made on the three liquids
to permit identification, as follows.

solubility
m.p. b.p. density AH vap'n in water

Liquid^ -131. 5°C 36.2°C 0.63 g/cc 85 cal/g 0.036 g/ 100 ml

Liquid B -116 34.6 0.71 89.3 7.5

Liquid C -89.2 117.7 0.81 141 7.9

Which liquid should be given Label No. 1, Label No. 2, Label No. 3? Explain how each type of meas-
urement influenced your choices.

16. Maleic and fumaric acids are cis- and trans- Maleic acid gives up its first proton more readily

isomers having two carboxyl groups, than does fumaric acid. However, the opposite

is the case for the second proton. Account for
HOOC— CH=CH— COOH this in terms of structure.

During his brilliant scientific career, Peter Debye has
added richly to our knowledge of the structure of physical
chemistry. His research contributions have won for him
awards and honorary degrees from many countries and he
has earned unbounded respect wherever men seek a deeper
understanding of nature.

Born in Maastrecht, the Netherlands, he graduated in
electrical engineering, did his early research in theoretical
physics, and received his doctorate at the University of
Munich in 1908. Three years later, at the age of 27, he
accepted a full professorship at the University of Zurich
where his immediate predecessor was Albert Einstein. Dur-
ing this year he developed two of his most lasting and
fundamental studies, establishing still accepted theories of
the specific heat of solids and of the interactions among
polar molecules. Shortly thereafter, he returned to the
Netherlands as professor of theoretical physics at Utrecht.
As his scientific contributions multiplied, he occupied pro-
fessorships successively at the Universities of Goettingen
{Germany), Zurich {Switzerland), Leipzig {Germany), and
Berlin {Germany). In Berlin he was appointed director of
the Max Planck Institute. During these fruitful years, his
research ranged through X-ray scattering, interatomic dis-
tances, the theory of electrolytes, magnetic cooling, and
dipole theory; this work won for him the Nobel Prize in
Chemistry for 1936.

With the onset of World War II, politics began to inter-
fere with his research. Debye was actually forbidden to
enter the Max Planck Institute which he directed because
he refused to accept German citizenship. Despite obstruc-
tion by the German government, he left Germany by way
of Italy and came to the United States. In 1940 he was
appointed professor of chemistry and head of the depart-
ment of chemistry at Cornell University. Six years later,
he became an American citizen. During the war years his
research turned toward the structure and particle size of
high polymers.

Attacking this new field with his usual deep insight and
characteristic originality, Debye made fundamental and
important contributions in the study of macromolecules.
Now professor emeritus, Debye is in great demand as a
consultant and lecturer. He has rare ability in presenting
the most complicated subjects in a fashion that gets to the
heart of the problem with penetrating clarity. Whenever he
speaks at a scientific meeting, the auditorium is filled to
capacity with an audience confident they will hear new and
interesting ideas. Inevitably they leave inspired and stimu-
lated by their contact with this great scientist — a man who
can delve into the most profound aspects of nature and
bring to them light and understanding.

CHAPTER

The Chemistry of
Carbon Compounds

The synthesis of brazilin would have no industrial value; its biological
importance is problematical, but it is worthwhile to attempt it for the
sufficient reason that we have no idea how to accomplish the task.

ROBERT ROBINSON, 1947

The compounds of carbon furnish one of the
most intriguing aspects of all of chemistry. One
reason they interest us is that they play a domi-
nant role in the chemistry of living things, both
plant and animal. Another reason is that there
are innumerable carbon compounds useful to
man — dyes, drugs, detergents, plastics, perfumes,
fibers, fabrics, flavors, fuels — many of them
tailored to suit particular needs. Manufacture of

these compounds has given rise to a huge chemi-
cal industry requiring millions of tons of raw
materials every year.

Where do we find the enormous quantities of
carbon and carbon compounds needed to feed
this giant industry? Let's begin our study of
carbon chemistry by taking a look at the chief
sources of carbon and carbon compounds.

18-1 SOURCES OF CARBON COMPOUNDS

18-1.1 Coal

Coal, a black mineral of vegetable origin, is
believed to have come from the accumulation of
decaying plant material in swamps during pre-
historic eras when warm, wet climatic conditions
permitted rapid growth of plants. The cycles of
decay, new growth, and decay, caused successive
layers of plant material to form and gradually
build up into vast deposits. The accumulation of
top layers of this material and of sedimentary

rocks excluded air from the lower material and
subjected it to enormous pressures. In time the
layers were compressed into hard beds composed
chiefly of the carbon that was present in the origi-
nal plants, and containing appreciable amounts
of oxygen, hydrogen, nitrogen, and some sulfur.
Thus, coal is not pure carbon. The "hardest"
coal, anthracite, may contain from 85 to 95%
carbon; the "softest," peat, is not really coal at
all but one of the early stages in the geological

321

322

THE CHEMISTRY OF CARBON COMPOUNDS I CHAP. 18

history of coal. Peat still contains unchanged
plant remains and may contain no more than 50
to 60% carbon.

When coal is heated to a high temperature in
the absence of air, it undergoes decomposition:
volatile products (coal gas and coal tar) distill
away and a residue called coke remains. Coke is
a valuable industrial material which finds its
chief use in the reduction of iron ore (iron oxide)
to iron for the manufacture of steel. Coke is
essentially carbon that still contains the mineral
substances that are present in all coals (and form
the ash that results when coal or coke is burned).

About eight gallons of coal tar are obtained
from a ton of coal. Coal tars are very complex
mixtures ; over 200 different carbon compounds
have been isolated from them. While the great
value of coal to mankind has been as a fuel, a
source of energy, the many substances in coal
gas and coal tar make coal also an important
source of chemical raw materials.

18-1.2 Petroleum

Petroleum is a complex mixture which may
range from a light, volatile liquid to a heavy,
tarry substance. Petroleum also has its origin in
living matter that has undergone chemical
changes over the course of geological time. It is
found in porous rock formations called oil pools,
between impervious rock formations that seal
off the pools. When a pool is tapped, the oil
flows through the porous structure (driven by
subterranean gas or water pressure) and so is
brought to the surface.

18-1.3 Natural Gas

Natural gas is a mixture of low molecular weight
compounds of hydrogen and carbon (hydrocar-
bons) found in underground "fields" of sand-
stone or other porous rock. This gas escapes to
the surface of the earth when the field is tapped
by drilling.

18-1.4 Certain Plant and Animal Products

Plants and animals are themselves highly effec-
tive chemical factories and they synthesize many
carbon compounds useful to man. These include
sugars, starches, plant oils and waxes, fats, gela-
tin, dyes, drugs, and fibers.

Because all of these sources of carbon com-
pounds ultimately find their origin in living
matter, plant or animal, the chemistry of carbon
is called organic chemistry. Compounds contain-
ing carbon are called organic compounds.
This term includes all compounds of carbon
except C02, CO, and a handful of ionic sub-
stances (for example, sodium carbonate, Na2C03,
and sodium cyanide, NaCN). You may wonder
how many organic substances are known. The
number is actually so large it is difficult to pro-
vide a reliable estimate. A great many more
compounds of carbon have been studied than of
any other element except hydrogen (hydrogen is
present in most carbon compounds). There are
undoubtedly over one million different carbon
compounds known. The number of new organic
compounds synthesized in one year (about
100,000 per year) exceeds the total number of
compounds known that contain no carbon!

18-2 MOLECULAR STRUCTURES OF CARBON COMPOUNDS

How can there be so many compounds contain-
ing this one element? The answer lies in the
molecular structures. We shall find that carbon
atoms have an exceptional tendency to form
covalent bonds to other carbon atoms, forming
long chains, branched chains, and rings of atoms.
Each different atomic arrangement gives a mole-

cule with distinctive properties. To understand
why a particular substance has its characteristic
properties, its structure must be known. Thus
the determination of the molecular structure of
carbon compounds is one of the central prob-
lems of organic chemistry. Let's see how it is
done.

SEC. 18-2 I MOLECULAR STRUCTURES OF CARBON COMPOUNDS

323

18-2.1 The Composition and Structure
of Carbon Compounds

Ethane and ethanol* are two common carbon
compounds. Ethane is a gas that usually makes
up about 10% of the household gas used for
heating and cooking. Its useful chemistry is al-
most wholly restricted to the combustion reac-
tion. Ethanol is a liquid that takes part in a
variety of useful chemical reactions. It has great
value in the manufacture of chemicals and it
bears little chemical resemblance to ethane. Yet,
the similarity of the two names, ethane and
ethanol, suggests that these compounds are re-
lated. This is so. To understand how they are
related and why their chemistries are so different,
we must learn their molecular structures. We
must find out what kinds of atoms are present in
each substance, how many atoms there are per
molecule, and their bonding arrangement. Usu-
ally many experiments must be performed before
the molecular structure of a compound is known
with certainty. This fascinating problem of car-
bon chemistry involves three basic experimental
steps: to determine the empirical formula, then
the molecular formula, and finally the structural
formula. First we shall review the information
conveyed by each of these formulas, using eth-
ane as an example. Then, in Section 18-2.2 we
will consider what experiments are used in the
determination of each type of formula, using
ethanol as an example.

EMPIRICAL FORMULA

The empirical formula tells only the relative
number of atoms of each element in a molecule.
For example, consider ethane. Analysis shows
that this is a compound of carbon and hydrogen
and that there are three hydrogen atoms for

* Ethanol is another name for the substance ethyl
alcohol.

H C

/
H

C—H

CH3 CH3

Fig. 18-1. Structural formulas of ethane, CV/e.

every carbon atom. Its empirical formula, there-
fore, is CH3.

MOLECULAR FORMULA

The molecular formula tells the total number of
atoms of each element in a molecule. Ethane is
found to have a molecular weight of 30. This
molecular weight together with the empirical
formula tells us the molecular formula. It cannot
be CH3 — this compound would have a molecular
weight of 15. The molecular formula C2H6 also
has three hydrogen atoms per carbon atom.
Since it has a molecular weight of 30, it has both
the correct empirical formula and the correct
molecular weight. Ethane has the molecular for-
mula C2H6.

EXERCISE 18-1

Write the molecular formula for the carbon-
hydrogen compound containing two carbon
atoms and having empirical formula CH2. What
is its molecular weight?

STRUCTURAL FORMULA

The structural formula tells which atoms are
connected in the molecule. In ethane, the two
carbon atoms are linked and three hydrogen
atoms are attached to each carbon atom. Various
ways of representing its structural formula are
shown in Figure 18-1.

The formulas in Figure 18-1 all represent the
same structure; the choice of which formula to
use depends upon what feature of the structure
is to be emphasized. The first and second draw-
ings emphasize the three-dimensional nature of

324

THB CHEMISTRY OF CARBON COMPOUNDS I CHAP. 18

ethane; the third is a simpler way of doing the
same thing; and the last formula merely shows
that three hydrogens are attached to each carbon
atom. It is not at all difficult to decide that
CH3CH3 must be the structural formula for eth-
ane. By the bonding rules developed in Chapter
16, we know that carbon is always surrounded
by four electron pair bonds and that a hydrogen
atom forms only one covalent bond. There is no
structure other than the one shown in Figure
18-1 in which two carbon atoms and six hydro-
gen atoms can be bound together and satisfy all
the bonding rules.

18-2.2 Experimental Determination
of Molecular Structure

We have seen three steps in fixing molecular
structure. What experiments are involved in each
of these steps? Let's investigate the nature of
these experiments using ethanol as a second ex-
ample.

DETERMINING THE EMPIRICAL FORMULA

In order to determine the empirical formula of a
compound, you must first find out just what ele-
ments are present in it. Sometimes this is done
simply by burning some of the compound in
pure oxygen. If the compound contains only
carbon and hydrogen, only carbon dioxide and
water will be produced. If the compound con-
tains some nitrogen as well, nitrogen gas or one
of the nitrogen oxides will be produced in the
combustion and can be identified. Another way
of finding out which elements are in a compound
is to allow the compound to react with hot,
liquid sodium metal. If the compound contains
nitrogen, sodium cyanide, NaCN, will be
formed; if it contains sulfur, sodium sulfide,
Na2S, will be a product. Once such reactions
show which elements are in the compound, rela-
tive numbers of atoms of each element (the
empirical formula) can be determined.

The usual method for finding the empirical
formula is simply illustrated with ethanol, a
compound containing only carbon, hydrogen,
and oxygen. A weighed amount of the pure com-
pound is completely burned in oxygen to give
carbon dioxide (from the carbon) and water

(from the hydrogen). The weight of the carbon
dioxide reveals how much carbon was in the
weighed amount of sample. The weight of water
reveals how much hydrogen was in the sample.
The remainder of the sample must have been
oxygen. (There is no need to discuss here the
procedures for compounds containing halogens,
nitrogen, or sulfur, for they are quite similar.)
Suppose we burn 46 grams of ethanol. Collec-
tion of the products yields 88 grams of carbon
dioxide and 54 grams of water. We wish to learn
the relative numbers of carbon, hydrogen, and
oxygen atoms in the compound, and we can do
this by calculating the number of moles of car-
bon dioxide and water produced by the combus-
tion of the 46 gram sample. Therefore, we
calculate:

Number of = 88 g = 88 g
moles C02 mol wt C02 44 g/mole

= 2.0 moles C02

54 g

Number of _
moles H20 mol wt H20 18 g/mole

= 3.0 moles H20

Now we can make the following statements
about the compound ethanol:

46 grams of ethanol yield two moles of C02 and
three moles of H20

46 grams of ethanol contain two moles of carbon
atoms and six moles of hydrogen atoms

46 grams of ethanol contain 24 grams of carbon
atoms and 6 grams of hydrogen atoms.

Thus we have accounted for (24 + 6) = 30 g
of the 46 g we started with. The remainder of the
sample, (46 — 30) = 16 g, must have been oxy-
gen. This is just

16 g 16 g

at. wt oxygen 16 g/mole

= 1.0 mole oxygen atoms

Summarizing, we know that 46 grams of the
compound ethanol contain

two moles of carbon atoms,
six moles of hydrogen atoms,
one mole of oxygen atoms.

SEC. 18-2 I MOLECULAR STRUCTURES OF CARBON COMPOUNDS

325

Since the relative number of atoms of each ele-
ment in the compound is the same as the relative
number of moles of atoms of each element in the
sample, we can say that the empirical formula
of ethanol is

QHeO, or C2HeO

This example has been much simplified by our
selection of 46 grams of sample. In practice, less
than a gram of sample is used and whole num-
bers of moles are not obtained. A typical set of
experimentally obtained data is given in Exer-
cise 18-2.

EXERCISE 18-2

Automobile antifreeze often contains a com-
pound called ethylene glycol. Analysis of pure
ethylene glycol shows that it contains only car-
bon, hydrogen, and oxygen. A sample of ethylene
glycol weighing 15.5 mg is burned and the
weights of C02 and H20 resulting are as follows:

weight of sample burned = 15.5 mg,
weight of C02 formed = 22.0 mg,
weight of H20 formed = 13.5 mg.

What is the empirical formula of ethylene glycol?

DETERMINING THE MOLECULAR
FORMULA

Now we know that the relative numbers of atoms
in ethanol are two carbon to six hydrogen to one
oxygen. We do not know yet whether the molec-
ular formula is C2HeO, C,Hi202, C6Hi,03, or
some other multiple of the empirical formula,
C2HeO.

This returns us to the problem of the deter-
mination of molecular weight. Avogadro's Hy-
pothesis provides one of the convenient ways of
measuring molecular weight if the substance can
be vaporized.

This was exactly the measurement you made
in Experiment 6 — it is called the vapor-density
method for molecular weight determination. To

apply the method to a liquid, such as ethanol,
a temperature above the boiling point is needed.
A weighed amount of liquid is placed in a gas
collecting device held at an easily regulated
temperature. For example, a steam condenser
around the device provides a convenient way of
holding the temperature at 100°C. When the
substance has vaporized completely, its pressure
and volume are measured (perhaps using equip-
ment like that shown in Figure 9-1 of the Labo-
ratory Manual). This provides a measurement of
the weight per unit volume of gaseous ethanol
at a known temperature and pressure. Again,
this weight is compared with the weight of the
same volume of a reference gas (usually 02) at
the same temperature and pressure.

Suppose such a vapor-density measurement
shows that a given volume of ethanol at 100°C
and one atmosphere weighs 1.5 times as much
as the same volume of oxygen gas at 100°C and
one atmosphere. Since equal volumes contain
equal numbers of molecules at the same tem-
perature and pressure (Avogadro's Hypothesis),
one molecule of the unknown gas must weigh
1.5 times the weight of a molecule of 02. There-
fore,

mol wt of the unknown gas = (1.5) X (molwt02)

= 1.5 X 32 = 48g/mole

Even though this number is not very accurate,
it will suffice for the purpose of deciding that the
molecular formula is C2HeO (with molecular
weight 46.07 g/mole), not (C2H60)2 (with mo-
lecular weight 92.14 g mole), or (C2H60)3 (with
molecular weight 138.21 g mole), or any higher
multiple of empirical formula units.

There are two other common methods for learning the
molecular weight of an unknown compound. They are
important in the study of compounds which are not
readily vaporized (as is required in the vapor-density
method). These methods are called the boiling point eleva-
tion and freezing point lowering methods. We have already
mentioned in Section 5-2.1 that a solution of salt water
has a higher boiling point than that of pure water. The
boiling point elevation for a given solute concentration
(expressed in moles) is almost independent of the choice
of solute. Hence this temperature measurement can be
readily interpreted in terms of a molar concentration.
From the weight of sample used in preparation of the
solution, the molecular weight can be calculated.

326

THE CHEMISTRY OF CARBON COMPOUNDS I CHAP. 18

Exactly the same type of behavior is found for the
freezing point of a solution except that the freezing
point is lower than that of the pure solvent. Thus we have
two methods for molecular weight determination which
are applicable to compounds with such low vapor pres-
sure or which decompose so readily that the vapor density
method cannot be used.

EXERCISE 18-3

Ethylene glycol, the example treated in Exercise
18-2, has an empirical formula of CHsO. (Is this
what you obtained?) A sample weighing 0.49
gram is vaporized completely at 200°C and at
one atmosphere pressure. The volume measured
under these conditions is 291 ml. This same vol-
ume, 291 ml, of oxygen gas at 200°C and one
atmosphere weighs 0.240 gram. What is the
molecular formula for ethylene glycol, CH^O,
C2H602, C3H9O3, C4Hl2Ou or some higher mul-
tiple of CH30?

ethane is a saturated* compound. Neither can
the oxygen atom just be attached somehow to a
hydrogen atom. Each hydrogen atom in ethane
has its bonding capacity already satisfied.

Rather than trying to find the structural for-
mula of ethanol by tacking an oxygen atom to
ethane, let us start with the oxygen atom and see
how we might build a molecule around it, using
the two carbon atoms and six hydrogen atoms
which are at our disposal. We already know that
the oxygen atom is commonly divalent, and that,
as in water, it makes bonds to hydrogen atoms.
Let us start our molecular construction with a
bond between one hydrogen atom and the oxy-
gen atom:

O— H

The other bond which the oxygen atom can
make must be to a carbon atom, since if it were
to another hydrogen atom we would simply have
a water molecule. Therefore we write

determining the structural
formula: ethanol

The determination of how the atoms of a mole-
cule are connected is the most important prob-
lem in identifying an unknown compound. It can
be as exciting as a detective story, with the
chemical and physical properties furnishing the
clues. With the right collection of clues, the
chemist can ascertain the identity of the mole-
cule.

What, then, is the structure of ethanol? First
we must learn its empirical formula. Analysis
shows that the empirical formula is C2H60. The
molecular formula is fixed by a vapor density
measurement. The molecular weight is found to
be 46, showing that the molecular formula is the
same as the empirical formula, C2H60. It remains
to discover the arrangement and connections of
the atoms. We might begin by eliminating some
structures which we can be sure are incorrect.
Ethanol is not simply ethane with an oxygen
atom somehow attached to a carbon atom, be-
cause in ethane all of the four bonds of each
carbon are satisfied, and so there is no way in
which an additional bond can form. We say that

O— H

The carbon atom we have added must form
three additional bonds to satisfy its tetravalent
bonding capacity. If all of these bonds were to
hydrogen atoms, we would have the completed
molecule CH3OH, and would also have two hy-
drogen atoms and a carbon atom left over.
Therefore, one of the bonds our first carbon
atom forms must be to the other carbon atom,
and the two other bonds must be to hydrogen
atoms. We have then

/
C— C— H

\
O— H

We can easily complete the structure by adding
three bonds from our last carbon atom to the

* This usage of the word "saturated" shows that chem-
ists, like other people, sometimes use the same word with
two entirely different meanings. On p. 164 this word was
used to describe a solution which contains the equilib-
rium concentration of a dissolved substance. As used
here, in reference to organic compounds, it means that
all bonds to carbon are single bonds and they are all
formed with hydrogen or other carbon atoms.

SEC. 18-2 I MOLECULAR STRUCTURES OF CARBON COMPOUNDS

327

three hydrogen atoms we have left. The result is

H H

\ /

H— C— C— H

H O— H

We have now used all six hydrogen atoms, the
two carbon atoms and the oxygen atom which
the molecular formula of ethanol requires. All
the bonding rules are satisfied, so the structure
we have written must be taken as a possible
structural formula for ethanol. However, we
must also decide whether this is the only possible
structural formula for a molecule with molecular
formula C2H60. A little reflection shows it is not.
Instead of having the oxygen atom form one
bond to carbon and one to hydrogen, why not
start with two oxygen-carbon bonds?

O
/ \

c c

We have six hydrogen atoms at our disposal, and

each of the carbon atoms must form three more

bonds. Therefore we complete the structure by

writing

H /

\ /
C

/ \
H H

Satisfy yourself that this structure violates no
bonding rules, and conforms to the empirical
and molecular formula of ethanol.

We have now found all possible structural
formulas for the ethanol molecule. The oxygen
atom is either directly bonded to one carbon
atom or to two carbon atoms. Once a choice
between these two possibilities is made, the
structure of the rest of the molecule can be deter-
mined from the molecular formula and the bond-
ing rules. The two possible structures are shown
in Figure 18-2. Such compounds with the same
molecular formula but different structural formu-
las are called structural isomers. The existence
of the two compounds 1 and 2 was known long
before their structures were clarified. Hence the
existence of these isomers perplexed chemists for
decades. Now we recognize the crucial impor-

CH3 CH2 OH

(1)

CH30CH3 (2)

Fig. 18-2. The structures of the C~HaO isomers.

tance of learning the structural formulas (as well
as molecular formulas) to identify the isomers.

So our problem is to decide whether ethanol
has structure 1 or structure 2. How can we tell
which is correct? Let us see what preliminary
ideas we can get from an examination of the
structural formulas.

In structure 2, all of the hydrogen atoms are
the same — each hydrogen atom is bonded to a
carbon which is, in turn, bonded to the oxygen
atom. In structure 1, one of the hydrogen atoms
is quite different from any of the others: it is
bonded to oxygen and not to carbon. Of the
remaining five, two are similarly placed, on the
carbon bonded tc oxygen, and three are on the
other carbon. Structures 1 and 2 should have

328

THE CHEMISTRY OF CARBON COMPOUNDS | CHAP. 18

quite different chemistries. Which one should
correspond to the chemistry of ethanol?

We can offer several kinds of evidence. Some
comes from the behavior of ethanol in chemical
reactions and some from the determination of
certain physical properties. Let's consider the
reactions first.

Clean sodium metal reacts vigorously with
ethanol, giving hydrogen gas and an ionic com-
pound, sodium ethoxide, with empirical formula
C2H5ONa. The reaction is quite similar to the
behavior of sodium and water which yields hy-
drogen and the ionic compound sodium hy-
droxide, NaOH. This suggests, but certainly does
not prove, that ethanol shows some structural
similarity to water. In water we have two hydro-
gen atoms bonded to oxygen atoms, and in struc-
ture 1 we have one hydrogen atom bonded to
oxygen. This bit of chemical evidence suggests
ethanol has structure 1 .

More quantitative evidence can be obtained
by carrying out the reaction between an excess
of sodium and a weighed amount of ethanol and
measuring the amount of hydrogen gas evolved.
When this is done it is found that 46 grams of
ethanol (one mole) will produce only § mole of
hydrogen gas. We can therefore write a balanced
chemical equation for the reaction of sodium
with ethanol:

NafsJ + C2H60(J J — ►- \Wg) + C2H5ONa(sj (7)

This equation expresses the fact that one mole
of ethanol produces \ mole of hydrogen gas.
Hence, one mole of ethanol must contain one
mole of hydrogen atoms that are uniquely ca-
pable of undergoing reaction with sodium. Ap-
parently one molecule of ethanol contains one
hydrogen atom that is capable of reacting with
sodium and five that are not. Let us now consider
structures 1 and 2 in the light of this information.
In structure 2 all six of the hydrogen atoms are
structurally equivalent, whereas in structure 1
there is one hydrogen atom in the molecule
which is structurally unique — that is, the one
bonded to the oxygen atom. Structure 1 is there-
fore consistent with the experimental fact that
only one hydrogen atom per molecule of ethanol
will react with sodium and structure 2 is not.

We can find further evidence that structure 1,
CH3CH2OH, is the correct structural formula
for the substance known as ethanol. It is known
that compounds which contain only carbon and
hydrogen (such as ethane, C2H6) do not react at
all readily with metallic sodium to produce hy-
drogen gas. In these compounds the hydrogen
atoms are all bonded to carbon atoms (see Fig-
ure 18-1), so we can make the deduction that, in
general, hydrogen atoms which are bonded to
carbon atoms do not react with sodium to
produce hydrogen gas. In structure 2, CH3OCH3,
all the hydrogen atoms are bonded to carbon
atoms, so we do not expect a compound with
this structure to react with sodium. Ethanol re-
acts with sodium, so it is unlikely that ethanol
has structure 2.

Let us consider one other reaction of ethanol.
If ethanol is heated with aqueous HBr, we find
that a volatile compound is formed. This com-
pound is only slightly soluble in water and it
contains bromine: its molecular formula is found
by analysis and molecular weight determination
to be C2H5Br (ethyl bromide, or bromoethane).
With the aid of the bonding rules, we can see
that there is only one possible structure for this
compound. This result is verified by the fact that
only one isomer of C2H5Br has ever been dis-
covered.

Now we can ask how this chemical reaction
furnishes a clue to the structure of ethanol.
Structure 1 could give structure 3 in Figure 18-3
merely by breaking the carbon-oxygen bond.

Fig. 18-3. The structural formula of ethyl bromide
(bromoethane).

CHjCH2Bf (3)

SEC. 18-2 I MOLECULAR STRUCTURES OF CARBON COMPOUNDS

329

Convince yourself of this fact by writing an
equation using the structural formulas 1 and 3.
In contrast, bromoethane can be obtained from
structure 2 only through a complicated rear-
rangement. Two carbon-oxygen and one carbon-
hydrogen bond would have to be broken. Expe-
rience shows that such complicated reshufflings
of atoms rarely occur. Therefore, the reaction
between ethanol and hydrobromic acid, HBr, to
form bromoethane provides more evidence that
ethanol has structure 1.

The evidence cited so far has been associated
with the chemistry of ethanol. Its boiling point
provides a different sort of information also
leading to structure 1. Ethanol is a liquid with a
boiling point of 78°C. This can be compared
with the boiling point of ethane, C2HC, which is
-172°C, and to that of water, 100°C. Plainly,
the substance ethanol is more like water than
like ethane, as far as boiling point is concerned.
Once again this can be understood better in
terms of structure 1. Structure 1 has, in common
with H20, oxygen linked to hydrogen. The high
boiling point of water is explained in terms of an
abnormally large intermolecular attraction of
such an O — H group to surrounding water mole-
cules. The interaction is called hydrogen bonding
(see Section 17-2.6). If ethanol also has the O — H
group (as in structure 1) then it too can exert the
same abnormally large attraction to neighboring
ethanol molecules. Thus structure 1 provides an
explanation of the fact that the boiling point of
ethanol is so high.

This possibility of forming hydrogen bonds
should cause a strong attraction between water
and a compound of structure 1. If there is strong
attraction, then ethanol should have high solu-
bility in water. Experiment shows that they are
miscible — they dissolve in all proportions. Again
the evidence tends to strengthen belief in struc-
ture 1.

Notice that our attempt to determine the
structural formula of ethanol has involved the
consideration of a variety of types of evidence.
Others could be listed as well — for example, the
infrared spectrum of the liquid and the X-ray
diffraction pattern of the solid add strong sup-
port for structure 1 . No one fact by itself gives

absolute proof of the structure, but all the facts
considered together show that 1 is unquestion-
ably the correct structure for ethanol. A com-
parable set of experiments shows that another
compound with the formula C2H60 has proper-
ties consistent with structure 2. This compound
is called dimethyl ether.

EXERCISE 18-4

Ethylene glycol has empirical formula CH30 and
molecular formula C2H602. Using the usual
bonding rules (carbon is tetravalent; oxygen is
divalent; hydrogen is monovalent), draw some
of the structural formulas possible for this com-
pound.

EXERCISE 18-5

Decide which of your structures in Exercise 18-4
best fits the following list of properties observed
for pure ethylene glycol.

(a) It is a viscous (syrupy) liquid boiling at
197°C.

(b) It is miscible with water, that is, it dissolves,
forming solutions, in all proportions.

(d) It reacts with sodium metal, producing hy-
drogen gas.

(e) A 6.2 gram sample of ethylene glycol reacts
with an excess of sodium metal to produce
2.4 liters of hydrogen gas at one atmosphere
pressure and 25°C.

18-2.3 The Ethyl Group

All of the reactions and the physical properties
of ethanol have been explained on the basis of
the behavior of the OH group in structure 1,
CH3CH2OH. This is true of most of the reactions
of ethanol — the reaction centers at the OH group
(which is called the hydroxyl group), and the
remainder of the molecule, C2H5 — , remains in-
tact. The reactions lead to the suggestion that
there are two parts in the molecule of ethanol, the
H H

I I
H — C — C — group, which is unchanged during

H H

330

THE CHEMISTRY OF CARBON COMPOUNDS I CHAP. 18

reactions, and the —OH group, which can
change. This concept of the structural integrity
of the hydrocarbon group is an important one
in organic chemistry. It focuses attention on the
groups that do change, the so-called functional
groups. If the chemistry of a particular functional
group is understood for one compound, it is a
good assumption that this same chemistry will be
found for another compound containing this
same functional group. Thus, compounds with
the OH group are given a family name, alcohols.
The rest of the molecule, the carbon skeleton,
has relatively little effect and it remains intact
through the reactions of the functional group.
We have mentioned earlier that when ethanol
reacts with hydrogen bromide, ethyl bromide is
formed. Similar treatment of ethanol with hy-
drogen chloride or hydrogen iodide gives us the
corresponding ethyl halides:

CH3CH>OH + HBr — >- CH3CH2Br + H20 (2)

CH3CH2OH + HC1
CH3CH2OH + HI

CH3GH2CI + H20 (3)
CH3CH2I + H20 (4)

We say that the hydroxyl group has been dis-
placed, and the halogen atom substituted for it.
You can see that the group CH3CH2— has re-
mained intact in all of these reactions. Indeed,
this group has appeared in most of our discussion
so far, sometimes attached to oxygen (as in
ethanol and sodium ethoxide), sometimes at-
tached to other atoms (as in the ethyl halides).
You will recall that earlier we became acquainted

with ethane, C2H6. Looking at the structural
formula of ethane, you see that it is simply the
CH3CH2 — group attached to hydrogen:

H H

H— C— C— H or CH3CH2

I I
H H

This group, CH3CH2 — (also written C2H5 — ), is
called the ethyl group.

Because ethyl bromide, ethyl alcohol (etha-
nol), etc., can be thought of as being derived
from ethane by the substitution of one of its
hydrogens by — Br, — OH, etc., we speak of these
as derivatives of ethane, and we say that ethane
is the parent hydrocarbon for a series of related
compounds.

The name ethyl is derived from the name of
the parent hydrocarbon, ethane. In the same
way the name of the methyl group (CH3 — ) is
derived from that of methane, CH4 ; the name of
the propyl group (CH3CH2CH2 — ) is derived
from propane, CH3CH2CH3; etc.

It is important to recognize that these groups
are not substances that can be isolated and
bottled. They are simply parts of molecules that
remain intact in composition and structure dur-
ing reactions. We find this way of classifying
organic groups a useful and convenient one, but
we must keep in mind that in the reactions we
have described, the ethyl group is not actually
formed as a distinct substance. Table 18-1 gives
more examples of group names (see p. 338).

18-3 SOME CHEMISTRY OF ORGANIC COMPOUNDS

18-3.1 Chemical Behavior off Ethyl and
Methyl Bromide

We can use these bromine compounds to illus-
trate one kind of organic reaction. Ethyl bromide
is not particularly reactive but it does react with
bases such as NaOH or NH3. If we mix ethyl
bromide and aqueous sodium hydroxide solu-
tion and heat the mixture for an hour or so, we
find that sodium bromide and ethanol are
formed.

C2H5Br + OH(aq) — >■ C2H5OH + Br-(aq) (5)
This reaction may seem similar to the reaction
between aqueous HBr and NaOH but there are
two important differences. The ethyl bromide
reaction is very slow (about one hour is needed
for the reaction) and it occurs between a covalent
molecule (C2H5Br) and an ion (OH"). In con-
trast, the reaction between HBr and NaOH in
water occurs in a fraction of a second and it
involves ions only, as shown in reaction (<5).

SEC. 18-3 I SOME CHEMISTRY OF ORGANIC COMPOUNDS

331

H+(aq) + OH(aq) — ►- H20 (6)

Let's describe the course of reaction (5) in
terms of a model. We will use methyl bromide
to make the description simpler but the reaction
of ethyl bromide is of the same type.

CH3Br + OH" — -*- CH3OH + Br" (7)

First of all, let us recount a few of the experi-
mental facts.

(1) Methyl bromide is a compound in which the
chemical bonds are predominantly covalent.
An aqueous solution of methyl bromide does
not conduct electricity, hence it does not
form ions (such as CH3+ and Br~ ions) in
aqueous solutions.

(2) The reaction takes a measurable time for
completion.

(3) Experiments show that the rate of the reac-
tion is increased by increasing the concentra-
tion of OH- and also by raising the tem-
perature.

OH-(<tq)

CH5Br-

A.. Re octants approaching

These observations remind us of Chapter 8,
in which we considered the factors that deter-
mine the rate of a chemical reaction. Of course,
the same ideas apply here. We can draw qualita-
tive information about the mechanism of the
reaction by applying the collision theory. With
quantitative study of the effects of temperature
and concentration on the rate, we should be
able to construct potential energy diagrams like
those shown in Figure 8-6 (p. 134).

Figure 18-4 illustrates the mechanism chemists
have deduced for this reaction. This picture
shows: (A) the approach of the hydroxide ion,
(B) the atomic arrangement thought to be the
activated complex, and (Q the final products.
In the activated complex the O — C bond is be-
ginning to form and the C— Br bond is beginning

Fig. 18-4. The mechanism and potential energy dia-
grams for the reaction
CHaBr + OH(aq) — »» CH3OH + Br-(aq)

Reaction coordinate

Reaction, coordinate
S, A possible form, of the activated complex (Notice the unstable positions of the hydrogen atoms)

HOCH3

Sr-(a<J)

Reaction coordinate

C. Products separating (Notice the new, stable positions of the hydrogen atoms.)

332

THE CHEMISTRY OF CARBON COMPOUNDS I CHAP. 18

to break. The potential energy curves for the
reaction are shown alongside the molecular
models. The slow rate shows that activation
energy is needed. One of the reasons why activa-
tion energy must be supplied is that in the ac-
tivated complex the bond angles have been
distorted from their favorable (stable) configura-
tions and forced into an unstable condition.

18-3.2 Oxidation of Organic Compounds

By far the majority of the million or so known
compounds of carbon also contain hydrogen and
oxygen. There are several important types of
oxygen-containing organic compounds and they
can be studied as an oxidation series. For in-
stance, the compound methanol, CH3OH, is very
closely related to methane, as their structural
formulas show. Methanol can be regarded as the
first step in the complete oxidation of methane
to carbon dioxide and water.

Fig. 18-5. Structural formulas of methane and
methanol.

H C

formaldehyde, HCHO

H-

metha-nol, CH3OH

/
c

acetaldehyde, CH3CH0

Fig. 18-6. Structural formulas of formaldehyde and
acetaldehyde.

ALDEHYDES

Methanol (and other alcohols) react with com-
mon inorganic oxidizing agents such as potas-
sium dichromate, K2Cr207. When an acidic,
aqueous solution of potassium dichromate reacts
with methanol, the solution turns from bright
orange to muddy green, owing to the production
of the green chromic ion, Cr+3. The solution then
has a strong odor easily identified as that of
formaldehyde, CH20. This formula represents
the structure at the top in Figure 18-6. Notice
that the bond between carbon and oxygen is a
double bond (see Section 16.5), and that all the
atoms lie in the same plane.

SEC. 18-3 I SOME CHEMISTRY OF ORGANIC COMPOUNDS

333

The balanced net reaction for the formation of
formaldehyde is

3CH3OH + Cr2Of 2(aq) + 8H +(aq) — >-

3CH.O + 2Cr "(aq) + 7H,0 (5)

Since the dichromate ion on the left side of the
equation has been reduced to chromic ion, Cr+3,
on the right side, the conversion of methanol to
formaldehyde must involve oxidation. To show
more clearly that methanol has been oxidized,
let us balance this reaction by the method of
half-reactions. We have encountered the half-
reaction involving dichromate and chromic ions
before (Problem 20b in Chapter 12). It is

Cr207-2(aq) + \4H+(aq) + 6e~ — ►-

2Cr+3(aq) + 7H20 (9)

To balance the methanol-formaldehyde half-
reaction we write, as a start,

CH3OH gives CH20 (10a)

This statement does not yet show the fact that
hydrogen atoms are conserved in the reaction,
since there is a deficiency of two hydrogen atoms
on the right. This can be remedied by adding
two hydrogen ions,

CH3OH gives CH20 + 2U+(aq) (10b)

Now the equation is chemically balanced, but
not electrically balanced. The addition of two
electrons to the right-hand side completes the
balancing procedure, and the completed half-
reaction is

CH3OH — >- CH20 + 2H+( aq) + 2e~ (10c)

This equation shows that the methanol molecule
has lost electrons and thus has been oxidized.
Formaldehyde is the second member in the oxi-
dation series of methane.

In a similar manner, ethanol can be oxidized
by the dichromate ion to form a compound
called acetaldehyde, CH3CHO. The molecular
structure of acetaldehyde, which is similar to
that of formaldehyde, is shown at the bottom in
Figure 18-6. We see that the molecule is struc-
turally similar to formaldehyde. The methyl
group, — CH3, replaces one of the hydrogens of
formaldehyde. The balanced equation for the
formation of acetaldehyde from ethanol is

3CH3CH2OH + Cn07-2(aq) + m+(aq) —+

3CH3CHO + 2Cr+3(aq) + 7H20

CARBOXYLIC ACIDS

(11)

Another oxidation product can be obtained from
the reaction of an acidic aqueous solution of
potassium permanganate with methanol. The
product has the formula HCOOH, and is called
formic acid. The structural formula of formic
acid is shown in Figure 18-7. The structure of
formic acid is also related to the structure of
formaldehyde. If one of the hydrogen atoms
of formaldehyde is replaced by an OH group, the

Fig. 18-7. Structural formulas of formic acid and
acetic acid.

H C

formic acid, HCOOH

H—

acetic acid, CHjCOOH

334

THE CHEMISTRY OF CARBON COMPOUNDS I CHAP. 18

resulting molecule is formic acid. The balanced
equation for its formation from methanol is

5CH3OH + 4Mn04" (aq) + \2H+(aq) —*-

5HCOOH + 4Mn+*(aq) + 11H20 (12)

The half-reaction involving methanol and formic
acid can be obtained by using the three steps
outlined in our previous example:

CH3OH gives HCOOH (13a)

Chemical balance:
CH3OH + H20 gives HCOOH + 4H+( aq) (13b)

Charge balance: CH3OH + H20 — >-

HCOOH + 4H+(aq) + 4e~ (13c)

From this completed half-reaction we see that
the conversion of methanol to formic acid in-
volves the loss of four electrons. Since the oxida-
tion of methanol to formaldehyde was only a
two-electron change, it is clear that formic acid
is a more highly oxidized compound of carbon
than formaldehyde or methanol.

EXERCISE 18-6

Balance the half-reaction for the conversion of
formaldehyde, HCHO, to formic acid, HCOOH.

Just as methanol can be oxidized to formic
acid [reaction (72)], ethanol can be oxidized to
an acid, CH3COOH, called acetic acid. The
molecular structure of acetic acid is shown in
Figure 18-7. The atomic grouping — COOH is
called the carboxyl group and acids containing
this group are called carboxylic acids.

The balanced equation for production of
acetic acid from ethanol is

5CH3CH2OH + 4Mn04" (aq) + UH+(aq) — ■
5CH3COOH + AMn^aq) + 11H20

(14)

Acetic acid can also be obtained by the oxidation
of acetaldehyde, CH3CHO:

5CH3CHO + 2Mn04-faqJ + 6H+(aq) — ►-

5CH3COOH + 2Mn+i(aq) + 3H20 (75)

The oxidation of acetic acid is difficult to accom-
plish. It does not react in solutions of K2Cr207
or KMn04. Vigorous treatment, such as burning,

causes its complete oxidation to carbon dioxide
and water. Formic acid also can be oxidized to
carbon dioxide and water by combustion with
oxygen.

EXERCISE 18-7

There is a compound called propanol with struc-
tural formula CH3CH2CH2OH. If it is oxidized
carefully, an aldehyde called propionaldehyde is
obtained. Vigorous oxidation gives an acid called
propionic acid. Draw structural formulas like
those shown in Figures 18-6 and 18-7 for pro-
pionaldehyde and propionic acid.

EXERCISE 18-8

Balance the half-reaction involved in the oxida-
tion of ethanol to acetic acid. Compare the
number of electrons released per mole of ethanol
with the number per mole of methanol in the
equivalent reaction (75c). How many electrons
would be released per mole of propanol in the
oxidation to propionic acid?

KETONES

The bonding rules permit us to draw two accept-
able structural formulas for an alcohol contain-
ing three carbon atoms, CH3CH2CH20H and
CH3CHOHCH3. In the first of these isomers
(the one considered in Exercises 18-7 and 18-8),
the OH group is attached to the end carbon
atom. In the second, the OH group is attached
to the second carbon atom. They are both called
propanol because they are both derived from
CH3CH2CH3, propane. They are distinguished
by numbering the carbon atom to which the
functional group, the OH, is attached. Thus,
CH3CH2CH2OH is called 1 -propanol because
the OH is on the first (the end) carbon atom in
the chain. The other alcohol, CH3CHOHCH3,
is called 2-propanol because the OH is on the
second carbon atom. The structures of these two
alcohols are shown in Figure 18-8.

We have already considered the oxidation of
1 -propanol in Exercise 18-7. The second isomer,

SEC. 18-3 I SOME CHEMISTRY OF ORGANIC COMPOUNDS

335

l-propanol, CH3CH2CH2OH

Z-propanol. CH3CHOHCH3

Fig. 18-8. The molecular structures of l-propanol
and 2-propanol.

2-propanol, can also be oxidized and the product
is called acetone:

3CH3CHOHCH3 + Cr JO f2(aq) + SH+(aq) — >-

3CH3COCH3 + 2Ct+3(aq) + 7H20 {16)

Acetone has the structure shown in Figure 18-9.
Acetone is the simplest member of a class of
compounds called ketones. They are quite similar
in structure to the aldehydes, since each contains
a carbon atom doubly bonded to an oxygen
atom.* They differ in that the aldehyde has a

* The group C=0 is called the carbonyl group.

hydrogen atom attached to this same carbon
atom whereas the ketone does not. (Compare
Figures 18-6 and 18-9.) Since this hydrogen atom
is not present, a ketone cannot be oxidized
further to an acid.

Figure 18-10 summarizes the successive oxida-
tion products that can be obtained from alcohols.
When the hydroxyl group, OH, is attached on
an end carbon atom, an aldehyde and a car-
boxylic acid can be obtained through oxidation.
When the hydroxyl group is on a carbon atom
attached to two other carbon atoms, oxidation
gives a ketone. Huge amounts of aldehydes and
ketones are used industrially in a variety of
chemical processes. Furthermore, these func-
tional groups are important in chemical syn-
theses of medicines, dyes, plastics, and fabrics.

18-3.3 The Functional Group

The reactive groups we have encountered thus
far, such as —Br, —OH, — CHO, — COOH, are
called functional groups. They are the parts from
which the molecules get their characteristic
chemical behavior.

For example, it is a general behavior of alco-
hols to undergo a reaction in which the — OH
group is displaced by a halogen atom, such as
— Br [as in reaction (77)].

Fig. 18-9. The molecular structure of acetone, the

simplest ketone.

acetone, CH3C0CH3

336

THE CHEMISTRY OF CARBON COMPOUNDS I CHAP. 18

1- alcohol ». aldehyde ^. car-boxylic acid — »~ carbon dioxide -h tva-ter-

-2e-

c —

2 ~ alcohol

II
O

ketone

o=c=o + H\

carbon, dioxide ~h ->va~ter

It is a general characteristic of aldehydes to be
oxidizable to acids, as in reaction (79).

These common types of behavior are shown
by using the general symbol, R — , to stand for
the part of the molecule which does not change,
and writing a reaction in such a way as to focus
attention on the functional group. For example:

RCH2OH + HBr — ■»- RCH2Br + H20 (77)

RCH2Br + OU(aq) — *■

RCH2OH + Br-(aq) (75)

3RCHO + Cr,07-2(aq) + SH+(aq) — >-

3RCOOH + 2Cr+*(aq) + 4H20 (79)

The symbol R — represents any alkyl group
(such as CH3 — , C2H6 — ) in these formulas.

18-3.4 Amines

Alcohols can be related to water by imagining
that an alkyl group (such as — CH3) has been
substituted for one of the two hydrogen atoms

Fig. 18-10. Summary: The successive steps in the oxi-
dation of an alcohol.

of water. In the same way, amines are related to
ammonia:

H H H

N— H

N— R

an amine

N— CH2CH3

ethylamine

Amines can be prepared by direct reaction of
ammonia with an alkyl halide, such as CH3Br or
CH3CH2I. Iodides react fastest and an excess of
ammonia is often used to help control formation
of undesired alternate products:

CH3CH2NH2 + NHJ (20)

ethylamine

RNH2 + NHJ (27)

CH3CH2I + 2NH3

ethyl iodide

R— I + 2NH3

Equations (20) and (27) represent a net change
that occurs when an excess of ammonia reacts

SEC. 18-3 I SOME CHEMISTRY OF ORGANIC COMPOUNDS

337

with an alkyl iodide. The actual reaction goes by
two successive steps. The first step is analogous
to the attack of the hydroxide ion on an alkyl
halide (see Figure 18-4):

NH3 + RI — »- RNH3+ (aq) + l(aq) (22)

The second step is a proton transfer reaction (see
Section 11-3.3):
NH3 + RNH3+ (aq) — ►- RNH2 + NH + (aqj (23)

18-3.5 Acid Derivatives: Esters and Amides

We see from reactions (75) and (79) that oxida-
tion of an aldehyde gives an organic acid. All
of these acids contain the functional group
— COOH, the carboxyl group. The bonding in
this group is as follows:

— C

•

O— H

The carboxyl group readily releases a proton, so
it is an acid. For example, acetic acid dissolves
in water and the solution is conducting, it turns
blue litmus red, it is sour, and it shows all the
other properties of an acid. The reaction

CH3COOH + H20 — >-

CH3COO(aq) + H30+(aq) (24)

has an equilibrium constant of 1.8 X 10~5.

In addition to this acidic behavior, an im-
portant characteristic of carboxylic acids is that
the entire OH group can be replaced by other
groups. The resulting compounds are called acid
derivatives. We will consider only two types of
acid derivatives, esters and amides.

ESTERS

Compounds in which the — OH of an acid is
transformed into —OR (such as — OCH3) are
called esters. They can be prepared by the direct
reaction between an alcohol and the acid. For
example,

CH,OH + CH3C

methyl , acetic
alcohol acid

The method of naming is indicated by the bold
face parts of the names of the reactants.

EXERCISE 18-9

Write equations for the reaction of (a) ethanol
and formic acid; (b) propanol and propionic
acid; (c) methanol and formic acid. Name the
esters produced.

When equilibrium is reached in reaction (25),
appreciable concentrations of all of the reactants
may be present. If methyl acetate (the product
on the right) alone is dissolved in water, it will
react with water slowly to give acetic acid and
methanol until equilibrium is attained:

CH3C

+ H20

O— CH3

CH3OH + CH3C

(26)

Of course, the usual equilibrium considera-
tions apply. For example, if we add the substance
methanol, equilibrium conditions will shift, con-
suming the added reagent (methanol) and acetic
acid to produce more methyl acetate and water,
in accord with Le Chatelier's Principle. Thus a
large excess of methanol causes most of the
acetic acid to be converted to methyl acetate.

EXERCISE 18-10

Write the equilibrium expression relating the
concentrations of reactants and products in re-
action (26). Notice that the concentration of
water must be included because it is not neces-
sarily large enough to be considered constant.

CH3C

+ H,0

(25)

O— CH3
methyl
acetate

+ water

338

THE CHEMISTRY OF CARBON COMPOUNDS f CHAP. 18

Reaction (25) between methanol and acetic acid
is slow, but it can be speeded up greatly if a
catalyst is added. For example, addition of a
strong acid such as hydrochloric acid or sulfuric
acid will speed up the reaction by catalysis. As
mentioned in Section 9-1.4, the catalyst does not
alter the equilibrium state (that is, the concentra-
tions of the reactants at equilibrium), but only
permits equilibrium to be attained more rapidly.

EXERCISE 18-11

A strong acid such as hydrochloric acid or sul-
furic acid will catalyze reaction (25). Explain
why this implies that these acids will catalyze
reaction (26) as well. (Consult Section 8-2.3.)

Esters are important substances. The esters of
the low molecular weight acids and alcohols have
fragrant, fruit-like odors and are used in per-
fumes and artificial flavorings. Esters are useful
solvents; this is the reason they are commonly
found in "model airplane dope" and fingernail
polish remover.

AMIDES

A compound in which the — OH group of an
acid is replaced by — NH2 is called an amide.
When the —OH is replaced by — NHR, the
product is called a nitrogen-substituted amide or,
abbreviated, an N-substituted amide. One way
to make amides is to react ammonia (or an
amine) with an ester:

CH3C

+ NH3

O— CH3

methyl

acetate

CH3C + CHjOH (27)

ace t amide NHj

CH3C

+ H2N— CH2CH3 — >-

ethylatnine

O— ( H

methyl

acetate

CH3C + CH3OH (28)

N— CH2CHS

JV-ethyl 1
acetamide n.

Table 18-1. regularities in names of alkanes, alcohols, and amines

NUMBER OF
CARBON ATOMS

ALCOHOLS

AMINES

CH<

CH,OH

CH,NH2

methane

methanol
methyl alcohol

methylamine

CH.CH,

CH,CH2OH

CH,CH2NH,

ethane

ethanol
ethyl alcohol

ethylamine

CHjCHjCHi

CH,CH2CH2OH

CH,CH2CH2NH,

propane

1-propanol
propyl alcohol

1 -propylamine

CH3CHSCH2CH1

CH,CH2CH2CH2OH

CH3CH2CH2CH2NH2

butane

1-butanol
butyl alcohol

1-butylamine

chkch^ch,

CH,(CH2),CH2OH

CH,(CH2)6CH2NHf

octane

1-octanol
octvl alcohol

1-octylamine

SEC. 18-4 I NOMENCLATURE

339

Note the similarity of the two reactions. Amides
are of special importance because the amide

grouping

— C

NH—

is the basic structural element in the long-chain
molecules that make up proteins and enzymes
in living matter. Hydrogen bonding between two
amide groups helps determine the protein struc-
ture, a topic that will be dealt with later, in
Chapter 24.

18-4 NOMENCLATURE

The names of organic compounds have some
system. Each functional group defines a family
(for example, alcohols, amines) and a specific
modifier is added to identify a particular example
(for example, ethyl alcohol, ethyl amine). As an
alternate naming system, the family may be
named by a general identifying ending (for ex-
ample, alcohol names end in -ol) and a particular
example is indicated by an appropriate stem
(ethyl alcohol would be ethanol). These naming
systems are illustrated in Tables 18-1 and 18-11.
Scrutiny of these tables reveals that the key

to the system is the name of the alkane which is
modified in a systematic way to get the names
that carry over into the acid derivatives. Starting
at pentane (C6Hi2) and hexane (C6Hu) the alkane
names are themselves quite regular. They are
derived from Greek words for the number of
carbon atoms.

The compounds with more complicated shapes
and more than one functional group are de-
scribed by a straightforward numbering system
that you will learn in later chemistry courses.
Other functional groups will be studied then too.

Table 18-11. regularities in names of acids, amides, and esters

NUMBER OF

ESTERS (ACID

CARBON ATOMS ACIDS

AMIDES

WITH METHANOL)

HCOOH

formic acid

HCONH2

formamide

HCOOCH,

methyl formate

CH.COOH

acetic acid

CH,CONH,

acetamide

CH,COOCH,

methyl acetate

CH,CH2COOH

propionic acid

CH,CH2CONH2

propionamide

CH,CH2COOCH,

methyl propionate

CHjCH^CHjCOOH

butyric acid

CH,CH,CH2CONH2

butyramide

CH,CH2CH2COOCH,
methyl butyrate

CH,(CH,),COOH

octanoic acid
caprylic acid

CH^CH^CONH,

octanamide
caprylamide

CH,(CH2),COOCH,

methyl octanoate
methyl caprylate

340

THE CHEMISTRY OF CARBON COMPOUNDS I CHAP. 18

18-5 HYDROCARBONS

Compounds that contain only hydrogen and carbon
are called hydrocarbons. The hydrocarbons
that have only single bonds all have similar
chemistry and they are called, as a family, the
saturated hydrocarbons. If there are carbon-
carbon double bonds, the reactivity is much
enhanced. Hence hydrocarbons containing one
or more double bonds are named as a distinct
family, unsaturated hydrocarbons. Both saturated
and unsaturated hydrocarbons can occur in
chain-like structures or in cyclic structures. Each
of these families will be considered.

18-5.1 Saturated Hydrocarbons

We have already remarked that ethane is a mem-
ber of a family of compounds called the saturated
hydrocarbons. This term identifies compounds
that contain only carbon and hydrogen in which
all bonds to carbon are single bonds formed with
hydrogen or other carbon atoms. They occur in
chains, branched chains, and cyclic structures.

Fig. 18-11. Structural formulas for some five-carbon
saturated hydrocarbons.

n -pentane

H H

\/
C

/\
H—C C

r/\ A

H H

/ \

C—K

H H

w \/

c c-

-H

/ \ /

H-

-c c

HX\ /V

H ^c

H A

a k

ieo -pe.rtta.viG

H H

\ I

9—h

cycLope.yita.YLe

SEC. 18-5 I HYDROCARBONS

341

The chain and branched chain saturated hydro-
carbons make up a family called the alkanes.
Some saturated hydrocarbons with five carbon
atoms are shown in Figure 18-11. The first ex-
ample, containing no branches, is called normal-
pentane or, briefly, /i-pentane. The second ex-
ample has a single branch at the end of the chain.
Such a structural type is commonly identified by
the prefix "iso-". Hence this isomer is called
/50-pentane. The third example in Figure 18-11
also contains five carbon atoms but it contains
the distinctive feature of a cyclic carbon struc-
ture. Such a compound is identified by the prefix
"cyclo" in its name — in the case shown, cyclo-
pentane.

normal conditions — their boiling points are
shown in Table 18-111. The composition of gaso-
line is mainly highly branched alkanes with from
six to ten carbon atoms. Paraffin waxes are usu-
ally alkanes with from twenty to thirty-five
carbon atoms.

The saturated hydrocarbons are relatively in-
ert except at high temperatures. For example,
sodium metal is usually stored immersed in an
alkane such as kerosene (8 to 14 carbon atoms)
to protect it from reaction with water or oxygen.
Combustion is almost the only important chemi-
cal reaction of the alkanes. That reaction, how-
ever, makes the hydrocarbons one of the most
important energy sources of our modern tech-
nology.

EXERCISE 18-12

What are the empirical formulas of the three
compounds shown in Figure 18-11? The molecu-
lar formulas? Which are structural isomers?

EXERCISE 18-13

There is one more alkane with molecular formula
C6Hi2, called neopentane. Draw its structural
formula.

EXERCISE 18-14

Using the data given in the last column of Table
18-111, plot the heat released per carbon atom
against the number of carbon atoms for the
normal alkanes. Consider the significance of this
plot in terms of the molecular structures of these
compounds.

The alkanes are the principal compounds pres-
ent in natural gas and in petroleum. The low
molecular weight compounds are gases under

In a sense, the absence of reactivity of satu-
rated hydrocarbons, whether cyclic or not, is a
crucial aspect of their chemistry. This inertness
accounts for the fact that the chemistry of

Table 18-111. some properties of saturated hydrocarbons

SATURATED
HYDROCARBON

MOLECULAR
FORMULA

MELTING
POINT

BOILING
POINT

HEAT OF
COMBUSTION OF

gas (kcal/mole)

methane

CH<

-182.5°C

-161.5°C

-212.8

ethane

CHjCH,

-183.3°C

-88.6°C

-372.8

propane

CH3CH2CH}

-187.7°C

-42. rc

-530.6

n-butane

CH3CH2CH2CH,

-138.4°C

-0.5°C

-687.7

isobutane

CH,— CH— CH3
CH,

-159.6°C

-11.7°C

-685.7

n-hexane

CtHu

-95.3°C

68.7°C

-1002.6

cyclohexane

C«Hi2

+6.6°C

80.7°C

-944.8

H-octane

CsHu

-56.8°C

125.7°C

-1317.5

w-cctadecane

CisHj8

+28.2°C

316. rc

-2891.9

342

THB CHBMISTRY OF CARBON COMPOUNDS I CHAP. 18

organic compounds is mainly concerned with the
functional groups. The functional groups are
usually so much more reactive than the carbon
"skeleton" that it can be assumed that the skele-
ton will remain intact and unchanged through
the reaction.

18-5.2 Unsaturated Hydrocarbons

Unsaturated compounds are those organic com-
pounds in which less than four other atoms are
attached to one or more of the carbon atoms.
Ethylene, C2H4, is an unsaturated compound.
Because ethylene involves only carbon and hy-
drogen, it is an unsaturated hydrocarbon. Pro-
pylene, the next more complicated unsaturated
hydrocarbon, has the molecular formula C3H6.

The structural formulas of ethylene and propyl-
ene are shown in Figure 18-12. Cyclic hydrocar-
bons also can involve double bonds. The
structural formula of a cyclic unsaturated hydro-
carbon is shown also in Figure 18-12.

Unsaturated hydrocarbons are quite reactive
— in contrast to the relatively inert saturated
hydrocarbons. This reactivity is associated with
the double bond. In the most characteristic re-
action, called "addition," one of the bonds of
the double bond opens and a new atom becomes
bonded to each of the carbon atoms. Some of
the reagents that will add to the double bond are

Fig. 18-12. Structural formulas of some unsaturated
hydrocarbons.

H H

\ /

C = C

H C

\
H

propylene

\ /

SEC. 18-5 I HYDROCARBONS

343

H2, Br2, HC1, and H20. Examples are shown
below for ethylene.

H H

\ /

C=C

/ \

H H

\ /

C=C

/ \
H H

+ H2

+ Br2

H
\

/
H

\ /

C=C + HC1

/ \

\ /

/ \

H
H

/
\

+ H20

\ /
H— C— C— H

/
H

\
H

H— C-

/
-C— H

/
Br

\
H

H— C-

-C— H

/
H

\
CI

\
H— C-

-C— H

/
H

\
OH

(29a)

(29b)

(29c)

(29d)

Oxidizing agents also attack the double bond.
When a reaction between an unsaturated com-
pound and the permanganate ion occurs, the
violet color of permanganate fades. This reac-
tion, as well as reaction (29b) in which a color
change also occurs, is used as a qualitative test
for the presence of double bonds in compounds.

18-5.3 Benzene and Its Derivatives

There is another important class of cyclic com-
pounds that is still different from the two classes
just described. The simplest example is the com-
pound benzene, a cyclic compound with six car-
bon atoms in the ring and formula C6H6. The
benzene ring is found to be planar with 120°
angles between each pair of bonds formed by a
given carbon atom. Thus, experiment tells us the
molecule is a regular hexagon with the following
atomic arrangement:

I
C

/ \
H— C C— H

I
H— C C— H

\ /
C

I
H

(

S**

A difficulty arises when we attempt to represent
the bonding in benzene and its derivatives. We
might write

(30)

Both of these structures satisfy the formal va-
lence rules for carbon, but each has a serious
fault. Each structure shows three of the carbon-
carbon bonds as double bonds, and three are
shown as single bonds. There is a wealth of
experimental evidence to indicate that this is not
true. Any one of the six carbon-carbon bonds in
benzene is the same as any other. Apparently the
fourth bond of each carbon atom is shared
equally with each adjacent carbon. This makes
it difficult to represent the bonding in benzene
by our usual line drawings. Benzene seems to be
best represented as the "superposition" or "aver-
age" of the two structures. For simplicity, chem-
ists use either one of the structures shown in
(30) usually expressed in a shorthand form (31)
omitting the hydrogen atoms:

(3D

Still another shorthand symbol sometimes used
is

(32)

Whichever symbol is used, (30), (31), or (32), the
chemist always remembers that the carbon-
carbon bonds are actually all the same and that
they have properties unlike either simple double
bonds or simple single bonds.

THE SUBSTITUTION REACTION
OF BENZENE

Benzene shows neither the typical reactivity nor
the usual addition reaction of ethylene. Benzene
does react with bromine, Br2, but in a different
type of reaction:

344

THE CHEMISTRY OF CARBON COMPOUNDS I CHAP. 18

+ Br2

H H

+ HBr (33)

'\
H
H

bromobenzene

In this reaction, called bromination, one of the
hydrogen atoms has been replaced by a bromine
atom. Notice that the double bond structure is
not affected— this is not an addition reaction.
Nitric acid causes a similar reaction, called nitra-
tion:

+ HON02

NO,

+ H20 (34)

Reactions of the type shown in (33) and (34) are
called substitution reactions. The substitution re-
action is the characteristic reaction of benzene
and its derivatives and is the way in which a
multitude of compounds are prepared by the
organic chemist. By this means he is able to
introduce functional groups, which can then be

modified in various ways. Benzene and its deriva-
tives are commonly called aromatic com-
pounds.

MODIFICATION OF FUNCTIONAL
GROUPS ON THE BENZENE RING

One of the most important derivatives of benzene is
nitrobenzene. The nitro group is — NOj. Nitrobenzene is
important chiefly because it is readily converted into an
aromatic amine, aniline, by reduction. One preparative
procedure uses zinc as the reducing agent:

NOj

3ZnfsJ + f' J + 6H+(aq) —*-

nitrobenzene

NH,

3Za»(aq) + 2H,0 + J

aniline

Aniline and other aromatic amines are valuable indus-
trial raw materials. They form an important starting point
from which many of our dyestuffs, medicinals, and other
valuable products are prepared. For example, you have
used the indicator, methyl orange, in your laboratory
experiments. Methyl orange is an example of an aniline-
derived dye, although it is used more as an acid-base
indicator than for dyeing fabrics. The structure of methyl
orange is as follows:

CH,

\ /=

CH,

y\^

N=N

_i i

methyl orange

SO,H

The portions of the methyl orange molecule set off by
the dotted lines come from aromatic amines like aniline.
Aniline is indeed the starting material from which methyl
orange and related dyes ("azo dyes") are made.

Another useful aniline derivative is acetanilide, which
is simply the amide formed from aniline and acetic acid :

CH3C

O H

+ N-

OH H

acetic acid + aniline

CH,C

/
H

acetanilide

+ H,0

+ water

(36)

SEC. 18-5 I HYDROCARBONS

345

Acetanilide has been used medicinally as a pain-killing zene, which is chlorinated as a first step. Reaction of
remedy. chlorobenzene with base gives phenol:

PHENOL AND ITS USES

Another important constituent of coal tar is hydroxy-
benzene, or phenol:

phenol

Most of our phenol is now made industrially from ben-

Cl
CI — ►- | i +HC1 (37)

chlorobenzene

. _.. , . heat

+ OH(aq) ►-

pressure

+ a~(aq) (38)

Table 18-IV. structures and uses of some benzene derivatives

STRUCTURE NAME USE

OCH,

/ \
H O

CH3CH,0

vanillin

flavoring material

phenacetin

I
N O

/ \ •

H C

\
CH,

pain-reliever (in headache
remedies)

hydroquinone photographic developer

NH2

"Novocaine"

local anaesthetic

(procaine)

C QH6

• \ /

O OCH2CH2N

CH=CH2

styrene

monomer for preparation

\s>

of polystyrene plastics

346

THE CHEMISTRY OF CARBON COMPOUNDS I CHAP. 18

Phenol is a germicide and disinfectant, and was first
used by Lister in 1867 as an antiseptic in medicine. More
effective and less toxic antiseptics have since been dis-
covered.

Perhaps the most widely known compound prepared
from phenol is aspirin. If phenol, sodium hydroxide, and
carbon dioxide are heated together under pressure, sali-
cylic acid is formed (as the sodium salt):

+ CO, + NaOH

OH
— COO~Na+

{39)

0%™+™

r- OH
I— COOH

salicylic acid

+ Na+fooJ (40)

Salicylic acid is quite useful. Its methyl ester has a sharp,
characteristic odor and is called "oil of wintergreen."

The acid itself (or the sodium salt) is a valuable drug in
the treatment of arthritis. But the most widely known
derivative of salicylic acid is aspirin, which has the fol-
lowing structure:

o— c

y \

CH,

COOH

aspirin

You will see, by examining this structure, that aspirin is
an ester of acetic acid. Aspirin is mankind's most widely
used drug. Somewhat over 20 million pounds of aspirin
are manufactured each year in the United States alone!
This amounts to something like 150 five-grain tablets for
every person in the country !

Table 18-IV shows the structures of a few simple ben-
zene derivatives that are important commercial products.
Study these structures so that you can see their relation-
ship with the simple compounds from which they are
derived.

18-6 POLYMERS

Table 18-111, p. 341, shows that the melting
points of the normal alkanes tend to increase as
the number of carbon atoms in the chain is
increased. Ethane, C2H6, is a gas under normal
conditions; octane, C8Hi8, is a liquid; octadec-
ane, Ci8H38, is a solid. We see that desired physi-
cal properties can be obtained by controlling the
length of the chain. Functional groups attached
to the chain provide additional variability, in-
cluding chemical reactivity. In fact, by adjusting
the chain length and composition of high mo-
lecular weight compounds, chemists have pro-
duced a multitude of organic solid substances
called plastics. These have been tailored to meet
the needs of a wide variety of uses, giving rise
to an enormous chemical industry.

The key to this chemical treasure chest is the
process by which extended chains of atoms are
formed. Inevitably it is necessary to begin with
relatively small chemical molecules — with car-
bon chains involving only a few atoms. These
small units, called monomers, must be bonded
together, time after time, until the desired mo-
lecular weight range is reached. Often the desired

properties are obtained only with giant mole-
cules, each containing hundreds or even thou-
sands of monomers. These giant molecules are
called polymers and the process by which they
are formed is called polymerization.

18-6.1 Types of Polymerization

Polymerization involves the chemical combina-
tion of a number of identical or similar molecules
to form a complex molecule of high molecular
weight. The small units may be combined by
addition polymerization or condensation poly-
merization.

Addition polymers are formed by the reaction
of the monomeric units without the elimination
of atoms. The monomer is usually an unsatu-
rated organic compound such as ethylene,
H2C=CH2, which in the presence of a suitable
catalyst will undergo an addition reaction to
form a long chain molecule such as polyethylene.
A general equation for the first stage of such a
process is

SEC. 18-6 POLYMERS

347

c=c

H H

o=c

I
H— C-

H H

C— C=C

H H

The same addition process continues and the
final product is the polymer, polyethylene:

I
H— C-

I— C=C

H \H/ „

in which n is a very large number.

"Lucite," and "Plexiglas" result. It is thus pos-
sible to create molecules with custom-built prop-
erties for various uses as plastics or fibers.

Condensation polymers are produced by reac-
tions during which some simple molecule (such
as water) is eliminated between functional groups
(such as alcoholic OH or acidic COOH groups).
In order to form long chain molecules, two or
more functional groups must be present in each
of the reacting units. For example, when ethylene
glycol, HOCH2CH2OH, reacts with />araphthalic

acid HOOC

COOH a polyester of

high molecular weight called "Dacron" is pro-
duced. The equation below shows the first
stages of this process:

H-

H H r

-O— C— C— O-f-H +

H H

HO-f-C— / V-C— '

!l \=/ll'
' O O l.

, OH + H+>

H H

When one or more of the hydrogens are re-
placed by groups such as fluorine, F; chlorine,
CI; methyl, CH3 or methyl ester, COOCH3;
synthetic polymers such as "Teflon," "Saran,"

Fig. 18-13. Molecular structures of a-amino acids.

group : a -ammo acid group

an tx - amino acid,
(general formula)

glycine

C CHZ CH2

glutawiic acid

18-6.2 "Nylon," a Polymeric Amide

"Nylon," the material widely used in plastics
and fabrics, is a condensation polymer. It con-
sists of molecules of extremely high molecular
weight and it is made up of small units joined
one to another in a long chain of atoms. The
reaction by which the units become bonded to-
gether is a conventional amide formation. There
is, however, the additional requirement that the
reaction must take place time after time to form
an extended, repeating chain. To accomplish
this, we select reactants with two functional
groups. Thus polyamides can be made from one
compound with two acid groups,

HOOC— CH2— CH2— CHr- CH*— COOH

adipic acid

and another with two amine groups,
H2N— CH2— CH2— CH2— CHz— CH2— CH2— NH2

1 , 6-diami noh e xane

348

THE CHEMISTRY OF CARBON COMPOUNDS I CHAP. 1

These molecules can react repeatedly, each time
removing water, and forming amide linkages at
both ends:

O O

C-(CH2)4-C

H— O

N-(CH2)6-N

N— (CH2)6— N

I I

H H

O O

\ S

C— (CH2)4— C H

/ \ /

N— (CH2)6— N

Other polyamides can be made from different
acids and other amines, giving a variety of prop-
erties suited to a variety of uses.

18-6.3 Protein, Another Polymeric Amide

A most important class of polyamides is that of
the proteins, the essential structures of all living
matter. In addition, they are a necessary part in
the diet of man because they are the source of
the "monomeric" units, the amino acids, from
which living protein materials are made.

Proteins are polyamides formed by the poly-
merization, through amide linkages, of a-amino
acids. Three of the 25-30 important natural
a-amino acids are shown in Figure 18-13. Each
acid has an amine group, — NH2, attached to
the a-carbon, the carbon atom immediately ad-
jacent to the carboxylic acid group.

The protein molecule may involve hundreds
of such amino acid molecules connected through
the amide linkages. A portion of this chain might
be represented as shown in Figure 18-14.

Fig. 18-14. The structure of protein showing the am-
ide chain.

An amide is decomposed by aqueous acids to
an acid and an amine (produced as the ammo-
nium salt):

CH3C

+ H30+(aq)

NHCH3

A'-methyl

acetamide

CH3C

•

+ CH3NH3+(aqJ (41)

acetic
acid

methyl

ammonium

ioD

In this same type of reaction, a protein can be
broken down into its constituent amino acids.

It is in this way that most of what we call the
"natural" amino acids have been discovered.
Proteins from many sources — egg yolk, milk,
animal tissues, plant seeds, gelatin, etc. — have
been studied to learn what amino acids compose
them. In this way about thirty of the "natural"
amino acids have been identified.

When you consider how many different ways
thirty (or even fewer) different amino acids can
be combined in long chains of a hundred or
more, you can see why there are so many pro-
teins known, and why different living species of

glycine

QUESTIONS AND PROBLEMS

349

plants and animals can have in their tissues a
great many different proteins. Enzymes, the bio-
logical catalysts, are also proteins. Each enzyme,
of course, has its own particular structure, de-
termined by the order and spatial arrangement
of the amino acids from which it is formed.
Perhaps the most marvelous part of the chem-
istry of living organisms is their ability to syn-
thesize just the right protein structures from the
myriad of structures possible.

EXERCISE 18-15

Take the letters A, B, C, and see how many dif-
ferent three unit combinations you can make;
for example, ABC, BAC, AAC, CBC, etc. This
will convince you that a chain made of hundreds
of groups with up to thirty different kinds of
units in each group can have an almost unlimited
number of combinations.

QUESTIONS AND PROBLEMS

1. What information is revealed by the empirical
formula? The molecular formula? The struc-
tural formula? Demonstrate, using ethane, C2H6.

2. Write the balanced equation for the complete
burning of methane.

3. Draw the structural formulas for all the C2H3C13
compounds.

4. Draw the structures of two isomeric compounds
corresponding to the empirical formula C3HgO.

5. Draw the structural formulas of the isomers of
butyl chloride.

6. What angle would you expect to be formed by
the C, O, H nuclei in an alcohol molecule?
Explain.

7. When 0.601 gram of a sample having an em-
pirical formula CH20 was vaporized at 200°C,
and one atmosphere pressure, the volume occu-
pied was 388 ml. This same volume was occupied
by 0.301 gram of ethane under the same condi-
tions. What is the molecular formula of CH20?

One mole of the sample, when reacted with
zinc metal, liberated (rather slowly) \ mole of
hydrogen gas. Write the structural formula.

Answer. The molecular formula is C2H402.

8. A 100 mg sample of a compound containing only
C, H, and O was found by analysis to give 149
mg C02 and 45.5 mg H20 when burned com-
pletely. Calculate the empirical formula.

9. How much ethanol can be made from 50 grams
of ethyl bromide? What assumptions do you
make in this calculation?

10. Write the balanced equation for the production
of pentanone from pentanol, using dichromate
ion as the oxidizing agent.

11. One mole of an organic compound is found to
react with \ mole of oxygen to produce an acid.
To what class of compounds does this starting
material belong?

12. Using the information given in Table 7-II, deter-
mine the reaction heat per mole of C-Mt(g) for
the complete combustion of ethane.

13. An aqueous solution containing 0.10 mole/liter
of chloroacetic acid, ClH2CCOOH, is tested wiih
indicators and the concentration of H+(aq) is
found to be 1.2 X 10~2 M. Calculate the value
of KA (if necessary, refer back to Section 11-3.2).
Compare this value with KA for acetic acid — the
change is caused by the substitution of a halogen
atom near a carboxylic acid group.

14. Give simple structural formulas of

(a) an alcohol,

(b) an aldehyde, and

each derived from methane; from ethane; from
butane; from octane.

15. Write the equations for the preparation of
methylamine from methyl iodide.

16. Write equations to show the formation of the
esters, methyl butyrate and butyl propionate.

17. Given the structural formula

H O H H H

I / I I I
H— C— C— O— C— C— C— H

I III

H H H H

for an ester, write the formula of the acid and
the alcohol from which it might be made.

350

THE CHEMISTRY OF CARBON COMPOUNDS I CHAP. 18

18. How much acetamide can be made from 3.1
grams of methyl acetate? See equation (27),
p. 338. Assume the ester is completely converted.

Answer. 2.5 grams acetamide.

19. An ester is formed by the reaction between an
acid, RCOOH, and an alcohol, R'OH, to form
an ester RCOOR' and water. The reaction is
carried out in an inert solvent.

(a) Write the equilibrium relation among the
concentrations, including the concentration
of the product water.

(b) Calculate the equilibrium concentration of
the ester if K = 10 and the concentrations
at equilibrium of the other constituents are:

[RCOOH] = 0.1 M;
[R'OH] = 0.1 M;
[H20] = 1.0 M. '

[RCOOH] = 0.3 M;
[R'OH] = 0.3 M\
[H20] = 1.0 M.

20. Give the empirical formula, the molecular for-
mula, and draw the structural formulas of the
isomers of butene.

21. There are three isomers of dichlorobenzene (em-
pirical formula C3H.C1). Draw the structural
formulas of the isomers.

/OH

22. Consider the compound phenol,

(a) Predict the angle formed by the nuclei C, O,
H. Explain your choice in terms of the
orbitals used by oxygen in its bonds.

(b) Predict qualitatively the boiling point of
phenol. (The boiling point of benzene is
80°C.) Explain your answer.

(d) In a LOW aqueous solution of phenol,
[H+] - 1.1 X 10-5. Calculate K.

SIR ROBERT ROBINSON, !•••-

Sir Robert Robinson will always be recognized as one of
the outstanding British scientists. With admiration and
pride, his country knighted him in 1937. His international
recognition in organic chemistry was signaled in 1947, when
he received the Nobel Prize in Chemistry.

The son of a surgical dressing manufacturer, Robinson
was born in Chesterfield, Derbyshire, England. He received
the Ph.D. at the University of Manchester, and his early
scientific promise led to a Professorial appointment at the
University of Sydney when he was only 26. Three years
later, he returned to England as Professor at the University
of Liverpool. As his reputation grew, he moved from uni-
versity to university, until, in 1930, he accepted one of the
most coveted academic positions in England— Professor of
Chemistry at Oxford.

During a prolific career that has produced an astounding
bibliography of about 600 publications, Robinson has found
leisure time to scale many of the mountain peaks in Switzer-
land, Norway, and New Zealand. This zeal for moun-
taineering couples well with his interest in photography.
He is a formidable chess player and a lover of music. His
wife, also a chemist of note, aided him in his career while
raising his family, a son and a daughter.

Sir Robert is a master of organic synthesis. His research
entered on the synthesis and the determination of struc-

res of biochemically important substances. These include

tremely complicated molecules with as many as five

sed rings and a variety of functional groups whose posi-
tioning and spatial relationships are critically important
in fixing biological activity. His name is linked with our
knowledge of certain alkaloids— a family of nitrogen con-
taining compounds that includes some powerful drugs, such
as strychnine and morphine. His crowning achievements
may have been in the total synthesis of such important
compounds as cholesterol, cortisone, and other related
structures, called steroids. During the second world war,
Professor Robinson led the Oxford scientific team that
studied the chemistry and structure of the antibiotic,
penicillin.

These are but a few pinnacles of success in a brilliant
scientific career. Like the Alpine peaks that Sir Robert
Robinson loves to climb, these accomplishments tower high,
to challenge and inspire this and future scientific genera-
tions.

CHAPTER

The Halogens

5
B

6
C

6
0

11
Ha

15
P

18
Ar

23
V

24
Cr

28
Hi

29
Cu

30
Zn

32
Ge

33
As

34-
Se

35
Br-

S3
I

36
Kr

30
Sr

40
Zr

■41

4-2

43
Tc

44
Fu

•45

4-b

4-7

48
Ctt

Te.

55
Cs

57-71

72
Hf

77
If

70
Pt

79
Au

82
Pb

64
Po

66
Fn

68
Fa

89-

i > '

> i •

■ ■ j j

The halogens are a family of elements appearing
on the right side of the periodic table, in the
column just before the inert gases. The elements
in this group — fluorine, chlorine, bromine, io-
dine, and astatine — show some remarkable simi-
larities and some interesting trends in chemical
behavior. The similarities are expected since the

electron populations of the outer levels are
analogous. Each element has one electron less
than an inert gas arrangement. The trends, too,
are understandable in terms of the increases in
nuclear charge, number of electrons, and atomic
size, going from top to bottom of this column
of the periodic table.

19-1 PROPERTIES OF THE HALOGENS

You have already studied some properties of the
halogens. They are very reactive elements that
exist under normal conditions as diatomic mole-
cules with covalent bonds. These molecules all
are colored. Gaseous fluorine is pale yellow;
gaseous chlorine is yellow-green; gaseous bro-
mine is orange-red (remember the film equilib-
rium); gaseous iodine is violet. The halogens are
all toxic and dangerous substances. Fluorine, F2,
is the most hazardous; the danger decreases as
the atomic number of the halogen becomes
352

larger. Even iodine, I2, should be handled with
caution.*

19-1.1 Electron Configurations off the Halogens

Table 19-1 shows the electron configurations of
the halogens, using the symbols introduced in

* These elements produce nasty "burns" that are slow
to heal. The mucous membranes are attacked especially,
and chlorine "poisoning" is really a lung inflammation.
Under no circumstances should inexperienced people
handle these substances without close guidance.

SEC. 19-1 ! PROPERTIES OF THE HALOGENS

353

Br2

all inner
orbital s
filled

Fig. 19-1. Orbital representations of the bonding in
Fa and Br2.

Section 15-1.6. The superscripts give the number
of electrons in a particular type of orbital; the
letters s, p, and d indicate the shape of the
orbital; the number before the letter indicates
the principal quantum number of the orbital.
The important point to note is that each of these
halogens has one less electron than the number
required to fill the outermost cluster of energy
levels. In each case the shortage is in the outer-
most p orbitals in which six electrons can be
accommodated. Therefore, one electron could be
added to each of the halogen atoms without re-
quiring the population of additional, higher en-
ergy orbitals. Sharing an electron that spends
some time in a valence orbital of another atom
produces a covalent bond to the halogen atom.
Organic compounds (such as ethyl bromide) and
the halogen molecules (F2, Cl>, Br2, I2) contain
covalently bonded halogen atoms. The similar
orbital representations of the bonding in F2 and
Br2 are contrasted in Figure 19-1.
Electron dot representations of the electron

sharing in Br2 and I2 raise the question of how
many valence electrons need be shown. Each of
these two elements appears in a row of the
periodic table in which the inert gas element has
18 electrons in the orbitals that fix the chemistry
of the row. Nevertheless, it is a convenience to
emphasize the similarity among the halogens by
showing only the valence electrons in the outer-
most s and p orbitals. Thus the 3d (valence)
electrons of bromine are usually omitted so that
the electron dot representations of F2 and Br2
will appear alike, as shown in Figure 19-2.
Table 19-1 also shows the ionization energies

: F : F :

: Br ' Br :

Fig. 19-2. Electron dot representations of the bond-
ing in Fs and Br2. Note the omission, for
convenience, of the 3d valence electrons of
bromine.

Table 19-1. electron configurations and ionization energies of the

H A LO G E N S

ELECTRON CONFIGURATION

NUCLEAR
CHARGE

inner
electrons

valence
electrons

IONIZATION
ENERGY, E\

(kcal/mole)

fluorine

Is2

2s-2p>

401.5

chlorine

+ 17

\s22s-2fr

3^3^

300

bromine

+ 35

■ ■ ■ 3^3/j6

3d*4s*4pr

273

iodine

+53

■■■4s-Aff

ldl05s-5fr

241

354

THE HALOGENS I CHAP. 19

of the gaseous halogen atoms. They decrease
significantly as we move downward in the pe-
riodic table. Nevertheless, all of these ionization
energies are very large compared with those of
the alkali metals (compare sodium, whose ioni-
zation energy is 118.4 kcal). Hence when any of
the halogens reacts with an alkali metal, an
ionic solid is formed. These ionic solids, or salts,
contain halide ions, F-, Cl~, Br-, or I~, each
with the appropriate inert gas electron popula-
tion.

19-1.2 The Sizes of Halogen Atoms and Ions

The "size" assigned to an atom or ion requires
a decision about where an atom "stops." From
quantum mechanics we learn that an atom has
no sharp boundaries or surfaces. Nevertheless,
chemists find it convenient to assign sizes to
atoms according to the observed distances
between atoms. Thus, atomic size is defined op-
erationally— it is determined by measuring the
distance between atoms.

For example, Figure 19-3 contrasts the dimen-
sions assigned to the halogens in the elementary
state. One-half the measured internuclear distance
is called the covalent radius. This distance
indicates how close a halogen atom can approach

Fig. 19-3. Covalent radii and van der Waals radii

(in parentheses) of the halogens (in Ang-
stroms).

another atom to which it is bonded. To atoms to
which it is not bonded, a halogen atom seems
to be larger. We can take as a measure of this
size one-half the distance between neighboring
molecules in the solid state. This defines an
effective radius, the van der Waals radius, and
it is shown by the black lines in Figure 19-3.
It is an effective radius, not a real radius, be-
cause the electron distribution actually extends
far out from the atom.

These distances aid us in explaining and pre-
dicting bond lengths in other covalent halogen
compounds. For example, when a chlorine atom
is bonded to a carbon atom (as in carbon tetra-
chloride, CCLi), the bond length can be expected
to be about the sum of the covalent radius of the
carbon atom plus the covalent radius of the
chlorine atom. The covalent radius of carbon is
taken as 0.77 A (from diamond), so the C — CI
bond length might be near (0.77 + 0.99) =
1.76 A. Experiment shows that each bond length
in COWs 1.77 A.

EXERCISE 19-1

Using the carbon atom covalent radius 0.77 A
and the covalent radii given in Figure 19-3, pre-
dict the C — X bond length in each of the follow-
ing molecules: CF4, CBr4, CI4. Compare your
calculated bond lengths with the experimental
values: C— F in CF4 = 1.32 A, C— Br in CBr4 =
1.94 A, C— IinCI4 = 2.15 A.

ct,

Br,

SEC. 19-1 I PROPERTIES OF THE HALOGENS

355

F~ cr

Fig. 19-4. Ionic radii of the halide ions (in Ang-
stroms).

Figure 19-4 contrasts the effective sizes of the
halide ions. Each of these dimensions is obtained
from the examination of crystal structures of
many salts involving the particular halide ion.
The effective size found for a given halide ion is
called its ionic radius. These radii are larger than
the covalent radii but close to the van der Waals
radii of neutral atoms.

The covalent, van der Waals, and ionic radii
are collected in Table 19-11 together with some
physical and chemical properties of the halogens.
We see some interesting trends. For each type
of radius we find a progressive increase from the
top of the column to the bottom. This increase
in size reflects the fact that as atomic number
rises, higher energy levels are used to accommo-
date the electrons. In addition, Table 19-11 shows
a trend of increasing melting and boiling points
as we move downward in the periodic table.
This trend is appropriate for a series of molecular
solids in which van der Waals forces are the
principal ones holding the molecules in proxim-

ity. This type of force is higher for more complex
molecules with more electrons.

The last column of Table 19-11 shows the dis-
sociation constants for the reactions of the type

Xt(g) +=£ 2X(g) U)

with equilibrium constant

where

[X] = partial pressure of X atoms,

[Xz] = partial pressure of X2 molecules.

Again the available data show a trend; K in-
creases in the series CU, Br2, I2. The equilibrium
conditions are fixed by two factors: tendency to
minimum energy and tendency to maximum
randomness. The randomness factor is about the
same for the three halogens, so the trend in equi-
librium constants is largely determined by the
energy effects. In each dissociation reaction (of
CI2, Br2, and I2) energy is absorbed. However,
more energy is absorbed to break the Br2 bond
than to break the I2 bond, and still more to
break the Cl2 bond. The energy absorbed ac-
counts for the trend in equilibrium constants.

Table 19-11. sizes of the halogen atoms, melting points, boiling points,

AND DISSOCIATION PROPERTIES OF THE HALOGEN MOLECULES

HALOGEN, X

COVALENT
RADIUS
IN Xi

VAN DER WAALS
RADIUS IN Xt

IONIC
RADIUS

(-1 ion)

M.P. OF

B.P. OF

BOND

DISSOCIATION

ENERGY

CONSTANT OF

OF Xt

Xt AT 1000°C

fluorine, F

0.72 A

1.35 A

1.36 A

55°K

85°K

36 kcal

—

chlorine, CI

0.99

1.80

1.81

172°K

238.9°K

57.1

io-»

bromine, Br

1.14

1.95

265.7°K

331.8°K

45.5

8 X 10-»

iodine, I

1.33

2.15

2.16

387°K

457°K

35.6

10-'

356

THE HALOGENS I CHAP. 19

Why should the bond energy be greater for Cl2
than for Br2, and greater for Br2 than for I2?
Presumably the size effect is a factor. Two halo-
gen atoms remain together because a pair of
electrons is simultaneously near both nuclei.
However, the larger the halogen atoms, the more
distant are those bonding electrons from the
nuclei. Since the electrical forces decrease with
distance, the bond energy lessens.

The dissociation constant of F2 at 1000°C is
not known and the bond energy of F2 was only
learned recently. Almost all chemists were sur-
prised when experiment showed that the energy

necessary to break the bond in F2 is much lower
than that in Cl2. This is still not well explained
so, rather than abandon familiar arguments con-
cerning halogen properties based on the trend
in size, chemists treat fluorine as a special case.

EXERCISE 19-2

On the basis of the trend in atomic size, what
trend is expected in the ionization energy Ex of
the halogen atoms? Compare your prediction
with the actual trend in Eu given in Table 19-1.

19-2 HALOGEN REACTIONS AND COMPOUNDS

Most of the reactions of the halogens are of the
oxidation-reduction type. The halogens are so
reactive that they do not occur uncombined in
nature and they must be made from halide com-
pounds (salts). We shall consider briefly the
preparation of the elements and then explore
some of the very interesting chemistry of this
family.

made by electrolysis of molten sodium chloride
(dissolved in CaCl2 to lower the melting point).
Figure 19-5 shows the components of the elec-
trolysis cell. At one electrode molten sodium is
produced, and at the other, chlorine gas is
collected.

19-2.1 Preparation of the Halogens

Oxidation through electrolysis is used to make
fluorine and chlorine. Chlorine, for example, is

Fig. 19-5. Preparation of chlorine and sodium by-
electrolysis of molten NaCl.

Na(t)

EXERCISE 19-3

At the left electrode in Figure 19-5 the half-
reaction occurring is CI" — *- \C\i(g) + e~, and
at the right electrode the half-reaction is
Na+ + e~ — >- Na(l). Which electrode is the
anode and which is the cathode? With these
half-reactions, balance the net reaction occurring
in the electrolysis cell.

Gaseous fluorine is also prepared by electroly-
sis of molten fluoride salts but simpler methods
are available for the preparation of bromine and
iodine. Chemical oxidation, usually with chlo-
rine as the oxidizing agent, provides Br2 and I2
economically because chlorine is a relatively in-
expensive chemical. The reactions are

I/a CI (t) in molten CaCl2

2l-(aq) + Ch(g)
2Br(aq) + Cl2(g)

h(s) + 2Cl-(aq) (2)

Bu(g) + 2C\-(aq) (3)

SEC. 19-2 I HALOGEN REACTIONS AND COMPOUNDS

357

EXERCISE 19-4

From the £°'s for the half-reactions of the type
2X~ — ►■ X2 + 2e~, show that Cl2 can be used
to produce Br. from Br~ and I2 from I~ but not
to produce F2 from F~.

19-2.2 Reduction of the Halogens

A substance that is readily reduced is a good
oxidizing agent. The oxidizing abilities of the
halogens vary in a regular manner, fluorine being
the strongest and iodine the weakest. On the
other hand, the iodide ion sometimes acts as a
reducing agent, while fluoride ion never does.
These statements are reflected in the £° values
for the oxidation half-reactions:

2F"
2C1-
2Br"

21"

F, + 2e~
Cl, + 2e~
Br . + 2e-
h + 2e~

E°

E°
E°
E°

-2.87 volts (4)

-1.36 (5)

-1.07 (6)

-0.53 (7)

The trend in oxidizing ability is quantitatively
expressed in the trend in £°'s. The E° for reac-
tion (4), —2.87 volts, indicates that fluoride ion,
F~, has little tendency to release electrons. Con-
versely, F2 has a high tendency to acquire elec-
trons. Hence, F2 is readily reduced — F2 is a
powerful oxidizing agent. At the bottom of the
list, E° for reaction (7) is —0.53 volt. Iodide
ion, I-, has a moderate tendency to release
electrons, oxidizing I- to I2. Thus iodide ion has
a moderate tendency to be oxidized, acting as a
reducing agent. Conversely, I2 has only a moder-
ate tendency to acquire electrons. Iodine has a
moderate tendency to be reduced — 12 can act as
an oxidizing agent.

How can this trend in half-cell potentials and oxidizing
abilities be explained? Let us imagine that reaction (7),
as an example, is carried out in a hypothetical series of
simpler steps:

2l-(aq)

2l~(g)

21(g)

h(g)

2\~(g)

2\(g) + 2e~(g)

Ug)

U(s)

Dehydration (5)

Electron removal (9)

Molecule formation (70)

Condensation (77)

2\~(aq)

h(s) + 2e~(g) Net reaction

U2)

We can consider the energy effect that accompanies the
net reaction (72) in terms of the energy effects of the
hypothetical steps. Since the tendency toward minimum
energy is one of the factors that fixes equilibrium, the
energy change is one of the factors influencing E°. Step
(8) is a dehydration step in which the I~ ions are pulled
out of the water into the gas phase. In this step, heat is
absorbed; since the system gains energy, AH is positive
for step (8). In step (9), electrons are pulled off the gaseous
I~ ions to give neutral gaseous I atoms; this step also
requires energy, and again AH is positive. In step (70),
two atoms of iodine come together to form a diatomic
molecule. This is the opposite of breaking the bond in I»;
energy is liberated and the system loses energy, so AH is
negative for this step. In step (77), the gaseous U mole-
cules form solid iodine crystals; this is the reverse of
vaporizing the solid and also corresponds to liberation of
energy. The question as to which halide ion is the best
reducing agent depends primarily upon the net energy
change considering all the steps involved in the half-
reaction. The halide ion that requires the least energy to
convert it to the halogen should be the best reducing
agent. Table 19-111 shows the experimentally measured
requirement for each of the above steps for the four
halogens (using ^ as a general symbol for a halogen
atom).

Notice that in Table 19-111 dehydration and electron
removal are the steps that involve the largest energy
changes. The amount of energy required for each of these
processes diminishes as atomic weight increases for the
three large halogens. For fluorine, dehydration accounts
for more than half the total energy, but electron removal
is still a major part of the positive energy change. The

Table 19-111. the energy required for some halogen reactions

ELECTRON REMOVAL MOLECULE CONDENSATION
HALO- DEHYDRATION 2X~(g) ►■ FORMATION X%(g) >- X2 OVERALL REACTION

gen 2X~(aq) — »- 2X~(g) 2X(g) + 2e~(g) ?X(g) — >■ X2(g) (normal state) 2X~(aq) — ►- Xt + 2e~

+246 kcal

+ 162 kcal

-37 kcal

-1.6 kcal

+ 369 kcal

+ 178kcal

+ 171 kcal

-58 kcal

— 4.4 kcal

+287 kcal

+ 162 kcal

+ 161 kcal

-46 kcal

-7 kcal

+270 kcal

+ 144 kcal

+ 146 kcal

-36 kcal

-10 kcal

+244 kcal

358

THE HALOGENS I CHAP. 19

values in the last column show that the energy change
for the overall reaction forms a regular series, roughly
comparable to the variation of E°. The half-reactions
having the most negative E° are those that require the
most energy, and that show the least tendency to proceed
as written from left to right. Therefore, the energies listed
in the last column do help to explain why the iodide ion
is a better reducing agent than the fluoride ion. This also
explains why F2 oxidizes compounds better than do the
other halogens. In aqueous solution the affinity of fluorine
for electrons, plus the rather strong attraction between
water molecules and F~, make F2 a good oxidizing agent.

The halogens are reactive even without water.
All the halogens react quite vigorously with most
of the metals to produce simple halide salts.
Copper and nickel, however, appear to be quite
inert to F2. This apparent inertness is attributed
to the fact that a thin layer of the fluoride salt
forms on the surface of each of those metals and
protects it from further attack by fluorine.

19-2.3 lodimetry

The I~-I2 half-reaction has many applications in
aqueous solution chemistry. The use of I- as a
reducing agent and I2 as an oxidizing agent,
particularly for quantitative purposes, is called
iodimetry.

This half-reaction possesses an E° of —0.53
volt; neither is I~ a particularly powerful re-
ducing agent nor is I2 a particularly powerful
oxidizing agent:

I2 + 2e~ E° = -0.53 volt (13)

There are many half-reactions below this one in
Appendix 3, so there are quite a few substances
that will oxidize I~. For example, iodide ion can
be quantitatively oxidized to I2 by Fe+3, Br2,
Mn02, Cr2Of2, Cl2, and Mn04", On the other
hand, there are many half-reactions above
E° = —0.53 volt in Appendix 3. For example,
I2 can be quantitatively reduced to I- by Sn+2,
H2S03, and Cr+2. The usefulness of the /~-/2 re-
action derives from the fact that all of the sub-
stances mentioned react rapidly and without side
reactions.

To top off this versatility, iodine possesses an
unusually sensitive and specific indicator. Iodine
reacts with starch to give a blue-colored complex.
This complex is so intensely colored that I2 can
be detected at a concentration as low as 10-6 M

and it furnishes the basis for the qualitative test
for iodine known as the "starch-iodine" test.
More important, the complex serves as a sensi-
tive indicator in oxidation-reduction titrations
based upon the I~-I2 half-reaction.

EXERCISE 19-5

Balance the reaction that occurs when I- is
oxidized to I2 by Mn04~ in acid solution, produc-
ing Mn+2.

19-2.4 Positive Oxidation States

of the Halogens: The Oxyacids

The halogens, except fluorine, can be oxidized to
positive oxidation states. Most commonly you
will encounter these positive oxidation states in a
set of compounds called "halogen oxyacids" and
their ions.

Compounds of the type HC103, HC104, etc.,
are examples of the halogen oxyacids. Chlorine
and iodine form a series of these acids in which
the halogen oxidation number can be +1, +3,
+5, or +7. For chlorine the series is made up
of HCIO (hypochlorous acid), HC102 (chlorous
acid), HCIO3 (chloric acid), and HC104 (per-
chloric acid). Although it is not easy to handle
these unstable substances, aqueous solutions of
these acids have been examined to find out how
strong they are as proton donors. HCIO is a weak
proton donor, HC102 is somewhat stronger,
HCIO3 is quite strong, and HC104 is the strongest
of all. (Perchloric acid, HC104, is, in fact, one
of the strongest acids known.)

Now we might wonder how to account for the
observed trend in terms of structure and bond-
ing. Figure 19-6 shows the presumed positions
of the atoms in these molecules. It is evident that
in each case we need to break a hydrogen-
oxygen bond to split off the proton. A regular
decrease in the strength of the hydrogen-oxygen
bond as we proceed from chlorous to perchloric
acid would explain the trend in acidity. How do
we account for the fact that the strength of this
bond varies as we go through the sequence?
Formally, we say that the oxidation number of
chlorine ranges from +1 to +3 to +5 to +7

SEC. 19-2 ' HALOGEN REACTIONS AND COMPOUNDS

359

NAME

FORMULA BALL-AMD- STICK MODEL SPACEFILLING MODEL

Hypochlorous acid HO CI

Chlorous acid HO CIO

(HClOz)

Chloric acid

HOCIOZ
(HClOs)

Perchloric acid HOClOj

(HCIO+)

Fig. 19-6. Presumed structures of the chlorine oxy acids.

360

THE HALOGENS I CHAP. 19

across the set, but actually there are no Cl+1,
Cl+3, Cl+5, or Cl+7 ions in these acids. The oxida-
tion number is only an artificial way of keeping
count of electric charges, as we learned in Sec-
tion 12-3.3. What is more to the point (and this
is really why the oxidation number cnanges in
the first place) is that there is an increasing num-
ber of oxygen atoms bonded to the central
chlorine. Each time an additional oxygen bonds
to the chlorine atom, some electron charge is
drawn off the chlorine and hence away from the
original O — CI bond. This, in turn, draws elec-
trons from the adjacent H — O bond and thereby
weakens it.

This increase in acid strength with oxidation
number is a general phenomenon. For example,
nitric acid (HNO3, in which the oxidation num-
ber of N equals +5) is stronger than nitrous
(HN02, oxidation number +3); sulfuric acid
(H2SO4, in which the oxidation number of S
equals +6) is stronger than sulfurous (H2S03,
oxidation number +4). A consistent, hence use-
ful, explanation is found. When an oxygen atom
is added to the central atom, there is a reduction
of the strength of O — H bonds in attached OH
groups.

The halogen oxyacids and their anions are
quite easily reduced — they are good oxidizing
agents. When one of these acts as an oxidizing
agent, the halogen is reduced to a lower oxida-
tion number. Just what oxidation number is
attained depends upon a variety of factors, in-
cluding acidity of the solution, strength of the
reducing agent, amount of reducing agent, and
temperature.

For example, the common use of sodium hy-
pochlorite solution, NaOCl, as a bleaching solu-
tion depends upon the oxidizing action of hypo-
chlorite, OCl~. Iodate ion, I03", also furnishes
a strong oxidizing power, as shown by E° for the
half-reaction 12-10^

I2 + 6H20 — >- 2IO3- + 12H+ + 10e-

E° = -1.2 volts (14)

The chlorine and bromine counterparts, chlo-
rate, CIO3", and bromate, Br03", have E0,s that
are even more negative. Hence these ions are
even stronger oxidizing agents than iodate ion.
It is not unusual in halogen chemistry to find
striking differences in the chemistry of acidic and

basic solutions. For example, iodine in an acidic
solution is quite stable, but in a basic solution
it is spontaneously oxidized to oxidation number
+ 5 in the IO:i~ ion. The reason for this can be
seen by considering the half-reactions (15) and
(16):

21- — >- h + 2e~

E° = -0.53 volt (15)

h + 6H-.0 — ►■ 2IO3- + 12H- + 10«r

E° = -1.2 volts (16)

The E° for reaction (75) is more negative than
that of reaction (75), hence iodate ion can react
with I~. In 1 M acid solution, the E° for the net
reaction

6I2 + 6H20

2IO3- 4- 101- -h 12H+ (17)

is (-1.2 volts) - (-0.53 volt) or -0.7 volt.
Since E° of the net reaction is so negative, we
expect that the reaction will not proceed spon-
taneously, and it doesn't. Let us now see what
happens when the H+ concentration becomes so
low that the OH- concentration is 1 M. When a
reaction occurs in basic solution it is conven-
tional to show OH- (rather than H+) in the
balanced reaction. Therefore, for the reaction of
iodine in basic solution, half-reaction (16) be-
comes

I2 + 120H-

2IO3- + 6H20 + 10<?- (18)

Le Chatelier's Principle aids us in predicting how
the tendency for I2 to release electrons in (18)
will be affected if we raise the hydroxide ion
concentration. Raising the concentration of a
reactant (such as OH-) tends to favor products.
Hence E° for (18) (1 M OH~) will be more posi-
tive than E° for (76) (1 M H+). In 1 M OH~,
E° = —0.23 volt. Since this E° is now more posi-
tive than E° for reaction (75) (E° = -0.53 volt),
the net reaction (79) can occur:

6I2 + 120H-

21O3- + 101- + 6H20 (79)

This reaction now has an E° of ( — 0.23 volt) —
(-0.53 volt) or +0.30 volt. With this positive
value of E°, we can expect that in basic solution
the reaction will proceed spontaneously (if the
rate is rapid), and it does.

The industrial preparation of bromine takes
advantage of this effect of hydrogen ion concen-
tration on the direction of the spontaneous re-

SEC. 19-2 I HALOGEN REACTIONS AND COMPOUNDS

361

action. A dilute solution of bromine is produced
by chlorination of salt well brines:

EXERCISE 19-6

Cl2 + 2Br-

Br2 + 2C1-

(20)

The liberated bromine is carried by a stream of
air into an alkaline solution of sodium carbonate
where it dissolves as a mixture of bromide and
bromate [the analogy of reaction (19)]. This last
step serves to concentrate the product, and free
bromine is obtained by subsequent acidification
of the solution, through the reaction

6H+ + 5Br~ + BrOi

3Br2 + 3H20 (27)

An alternative process for preparing bromine
from sea water begins again with reaction (20).
The liberated bromine is produced at a very low
partial pressure and it is necessary to concentrate
it. This is accomplished through the reaction
between S02 and Br2

Br2(gj + SO,(g) + 2H20(g) — »-

2HBrfgJ + 2HSO<(g)

(22)

The resulting acid vapors have a great affinity
for water (do you remember how rapidly HC1
dissolved in water in the film, gases and how
they combine?). Hence the HBr rapidly dis-
solves in water and concentrations as high as
0.5 M can be reached, a thousandfold more
concentrated than original sea water. With this
concentration, chlorine is again introduced to
produce Br2 by reaction (20).

19-2.5 Self -Oxidation-Reduction:
Disproportionate

In reaction (79) the iodine shown on the left has
an oxidation number of zero. After the reaction,
some of the iodine atoms have oxidation number
+5 and some —1. In other words, the iodine
oxidation number has gone both up and down
in the reaction. This is an example of self-
oxidation-reduction, sometimes called dispropor-
tionation. It is a reaction quite typical of, but not
at all restricted to, the halogens.

When chlorine gas is bubbled into a solution
of NaOH, self-oxidation-reduction occurs to give
hypochlorite ion, C10~, by the reaction

Cl2 + 20H- — +■ a- + CIO" + H20 (23)

Show that when an aqueous solution of NaCl
is being electrolyzed, vigorous stirring in the cell
might permit reaction (23) to occur.

Going one step further, if a basic solution
containing hypochlorite ion is heated, the CIO-
can again disproportionate,

3C10- — ►- 2C1" + C103- (24)

this time to produce the chlorate ion, C103".

19-2.6 Special Remarks on Fluorine

Because of the small size of the atom, fluorine is
rather special in the halogen group. We have
already seen that it is a strong oxidizing agent
in aqueous solution, and that a large part of this
arises because of the large hydration energy
associated with the fluoride ion. Another way in
which fluorine reveals its special character is in
the properties of hydrogen fluoride compared
with the other hydrogen halides. These are ex-
plained in terms of a special attraction of fluorine
for protons — an attraction called hydrogen
bonding (reread Section 17-2.6).

The strong attraction of fluorine for protons
shows up in another way. In aqueous solution,
HF is a weak acid whereas HC1, HBr, and HI
are strong acids. The dissociation constant of
HF is 6.7 X 10-4, so hydrofluoric acid is less
than 10% dissociated in a 0.1 M HF solution.

Another contrast between HF and the other
hydrogen halides, HC1, HBr, and HI, is found
in the reactivity with glass. Hydrofluoric acid
cannot be stored in glass bottles because it etches
the silica, Si02, in glass. On the other hand, even
the most concentrated hydrochloric acid solu-
tions can be stored indefinitely in glass without
any evidence of a comparable reaction. To store
HF solutions, we must use either polyethylene or
wax containers (rather than glass) because of
this reactivity with silica. Silicon bonds more
strongly to fluorine than to oxygen and hence
silica dissolves in a solution of HF by the reac-
tion

Si02 + 6HF — >- SiF6"3 + 2H,0+ (25)

362

THE HALOGENS | CHAP. 19

Hydrofluoric acid is a polar material, as water
is, and it behaves as an ionizing solvent when it
is scrupulously free of water. Salts that dissolve
readily in liquid HF include LiF, NaF, KF, AgF,
NaN03, KN03, AgN03, Na2S04, K2S04, and
Ag2S04. Liquid HF also dissolves organic com-
pounds and is used as a solvent for a variety of
reactions.

A very stable bond involving fluorine is the
carbon-fluorine bond. The strength of this C — F
bond is comparable to the C — H bond, and has
led to the existence of a series of compounds
known as the fluorocarbons. These are analogous
to the hydrocarbons and can be imagined as
being derived from them by substituting F atoms
for H atoms. For example,

F— C— C— F

/ \

F F

is the fiuorocarbon analogue of ethane. It is
called perfluoroethane. Many of the fluorocar-
bons are quite inert and their uses exploit this
property. CC12F2, Freon, is a volatile, nonpoison-
ous, noncorrosive material used in refrigerators
and as the propellant in some aerosol cans.
Increasingly important also are the polymeric
fluorocarbons, such as Teflon, which are derived
from perfluoroethylene, CF2=CF2, by polymeri-
zation. They can be used, for example, as
gaskets, valves, and fittings for handling ex-
tremely corrosive chemicals.

QUESTIONS AND PROBLEMS

1. Give the electron configuration for each of the
trio F-, Ne, Na+. How do the trios CI", Ar, K+,
and Br-, Kr, Rb+ differ from the above?

2. Table 19-11 contains values for the covalent
radii and the ionic radii of the halogens. Plot
both radii versus row number. What systematic
changes are evident in the two curves ?

3. Using the data from Table 19-11, plot on one set
of axes the melting and boiling points of the
halogens versus row number.

4. For astatine, use your graphs from Problems 2
and 3 as a basis for a prediction of its covalent
radius, ionic radius of the —1 ion, melting point,
and boiling point.

5. Predict the molecular structures and bond
lengths for SiF4, SiCLj, SiBr4, and Sil4, assuming
the covalent radius of silicon is 1.16 A.

6. Explain in terms of nuclear charge why the K+
ion is smaller than the Cl~ ion, though they are
isoelectronic (they have the same number of
electrons).

7. Can aqueous bromine, Br2, be used to oxidize
ferrous ion, Fe+*(aq), to ferric ion, Fe+3(aq)
(use Appendix 3)? Aqueous iodine, I2?

8. What will happen if F2 is bubbled into 1 M NaBr
solution? Justify your answer using E° values.

9. Using E° values, predict what will happen if, in
turn, each halogen beginning at chlorine is added
to a 1 M solution of ions of the next lower halo-
gen: Cl2 to Br~, Br2 to I". Which halogen is
oxidized and which is reduced in each case?

10. Write a balanced equation for the reaction of
dichromate and iodide ions in acid solution.
Determine E° for the reaction

Ct2Oj 2(aq) + l-(aq) + H+(aq) gives

Cr+3(aq) + I2 + H20

Answer. E° = +0.80 volt.

11. Balance the equation for the reaction of iodine
with thiosulfate ion:

k-2/

>-2/

I2 + SA" (aq) gives S4Oe («<?J + ^~(»<l)

thiosulfate tetrathionate

ion ion

What is the oxidation number of sulfur in the
tetrathionate ion?

12. How many grams of iodine can be formed from
20.0 grams of KI by oxidizing it with ferric
chloride (FeCl3)? Determine E°.

Answer. 15.3 grams of I2.

13. Balance the equation for the reaction between
S02 and I2 to produce S04~2 and I" in acid solu-
tion. Calculate E°. From Le Chatelier's Prin-
ciple, predict the effect on the E° in this reaction
if H+ = 10~7 M is used instead of H+ - 1 M.

QUESTIONS AND PROBLEMS

363

14. What is the oxidation number of the halogen
in each of the following: HF, HBr02, HI03,
CIO3" , F2) CIO," ?

15. Comparable half-reactions for iodine and chlo-
rine are shown below.

|I, + 3H.O — >- ICV + 6H+ + Se~

E° = -1.195 volts
2C12 + 3H20 — *- CIO3- + 6H+ + 5e~

E° = -1.47 volts

(a) Which is the stronger oxidizing agent, iodate,
IO3- , or chlorate, C103_ ?

(b) Balance the equation for the reaction be-
tween chlorate ion and I" to produce I2 and
Cl2.

16. Two half-reactions involving chlorine are

2C1- — ►- Cl2 + 2tr

E° = -1.36 volts
Cl2 + 2H20 — ►- 2HOC1 + 2H+ + 2e~

E° = -1.63 volts

(a) Balance the reaction in which self-oxidation-
reduction of Cl2 occurs to produce chloride
ion and hypochlorous acid, HOC1.

(b) What is the oxidation number of chlorine in
each species containing chlorine?

(d) Explain, using Le Chatelier's Principle, why
the self-oxidation-reduction reaction occurs
in 1 M OH~ solution instead of 1 M H+.

17. How many grams of Si02 would react with
5.00 X 102 ml of 1.00 M HF to produce SiF4?

18. A water solution that contains 0.10 M HF is 8%
dissociated. What is the value of its KA1

Answer. 6.9 X 10~4.

19. From each of the following sets, select the sub-
stance which best fits the requirement specified.

(a) Strongest acid

(b) Biggest atom

(d) Best reducing agent

(e) Weakest acid

HOC1, HOCIO,

HOCIO,
F, CI, Br, I

F, CI, Br, I
F", CI", Br", I-
HF, HC1, HBr, HI

(0 Best hydrogen bonding HF, HC1, HBr, HI

20. Describe two properties that the halogens have
in common and give an explanation of why they
have these properties in common.

CHAPTER

The Third Row of
the Periodic Table

/
H

z
He

4
**

s «

B C

6
O

10
Ne

IS
At

14
Si

15 1 16

P \ s

23
V

29
M*

27
Co

SO
In

34
St

Z5
Br-

36
Kf

SS
Sr

4t
Ub

42
Mo

4$
Cd

to
Sn

Si
Sb

S3
I

ST-fi

76
09

77
I*

«4

«*•

* 1 * »

In earlier chapters we recognized that strong
chemical similarities are displayed by elements
which are in the same vertical column of the
periodic table. The properties which chlorine
holds in common with the other halogens reflect
the similarity of the electronic structures of these
elements. On the other hand, there is an enor-
mous difference between the behavior of ele-
ments on the left side of the periodic table and
those on the right. Furthermore, the discussions
in Chapter 15 revealed systematic modification

in certain atomic properties, such as ionization
energy, as we proceed from left to right along a
row of the periodic table. Our purpose in this
chapter is to examine the chemical behavior of
the elements in the third row of the periodic
table to look for trends in chemical properties.
Specifically, we will consider the physical prop-
erties of the elements themselves, their per-
formance as oxidizing or reducing agents, and
the acid-base behavior of their hydroxides.

20-1 PHYSICAL PROPERTIES OF THE ELEMENTS

All the elements in Row 3 are commercially
available or can be easily prepared in the labo-
ratory. Try to examine as many of these elements
as possible in the laboratory as you study this
chapter. If all the elements are available to you,
364

arrange them in order of atomic number and
compare them. You can hardly imagine a more
varied set of appearances. At one extreme we
have the metals sodium, magnesium, and alumi-
num. When they are freshly cut these solids show

SEC. 20-1 | PHYSICAL PROPERTIES OF THE ELEMENTS

365

the bright luster or reflectivity typical of metals.
They are soft: sodium is so soft it can be cut
with a knife; magnesium and aluminum bend in
your fingers and can be easily scratched by a
sharp object. Silicon also shows the metallic
luster, but is much harder than magnesium or
aluminum. Phosphorus in the form known as
white phosphorus is a yellowish, waxy solid with
a distinctly nonmetallic appearance. Black phos-
phorus, obtained by subjecting white phosphorus
to high pressure, is a dark gray solid which does
have some of the luster which characterizes
metals. By the time we get to sulfur, however,
it is very clear we have a nonmetal. The two
gases chlorine and argon complete the trend
away from metallic appearance.

20-1.1 Sodium, Magnesium, and Aluminum:
Metallic Solids

Can we explain the wide variation in appearance
and physical properties of these elements? We
have already said in Chapter 17 that metals are
found at the left of the periodic table. The low
ionization energy and vacant valence orbitals of
one of these elements lead to a sea of highly
mobile valence electrons. The mobile electrons
hold the atoms together in the metallic crystal
and, at the same time, are responsible for the
ease of conduction of heat and electricity. We
also remarked that the metallic bond becomes
stronger as the number of valence electrons per
atom and their ionization energy increase. The
trend in the physical properties of the third-row
metals seems to be well explained in terms of the
increasing number and increasing ionization en-
ergy of the valence electrons.

Table 20-1

HEATS OF VAPORIZATION AND
BOILING POINTS OF METALS

A//xap (kcal/mole)
b.p. (°C)

23.1
889

31.5
1120

67.9
2327

20-1.2 Silicon: A Network Solid

Even though silicon is metallic in appearance, it
is not generally classified as a metal. The elec-
trical conductivity of silicon is so much less than
that of ordinary metals it is called a semicon-
ductor. Silicon is an example of a network solid
(see Figure 20-1) — it has the same atomic ar-
rangement that occurs in diamond. Each silicon
atom is surrounded by, and covalently bonded
to, four other silicon atoms. Thus, the silicon
crystal can be regarded as one giant molecule.
Almost all of the valence electrons in the
silicon crystal are localized in the covalent bonds
and are not free to conduct heat or electricity by
moving throughout the solid. On the other hand,
in the solid there are always a few valence elec-
trons which have acquired enough energy to be
nonlocalized and these few electrons account for
the small, but noticeable, electrical conductivity
of silicon. Again, we can rationalize the behavior
of silicon in terms of its atomic structure and
ionization energy. The fact that the silicon atom
has four electrons (3s23p2) in its valence orbitals
accounts for its tendency to form four covalent
bonds. The increase in ionization energy and

Fig. 20-1 . The crystal structures of silicon and
diamond.

EXERCISE 20-1

Write out the electron configuration of sodium,
magnesium, and aluminum and find the ioniza-
tion energies for all their valence electrons (Table
20-IV, p. 374). Account for the trend in the heats
of vaporization and boiling points (Table 20-1)
of these elements. Compare your discussion with
that given in Section 17-1.3.

diamond
C-C distance =1.54 A

silicon
Si -Si dirtance = 2.35 A.

366

THE THIRD ROW OF THE PERIODIC TABLE I CHAP. 20

absence of vacant valence orbitals as we proceed
along the row accounts for the increasing locali-
zation of the valence electrons into covalent
bonds and the almost complete disappearance of
electrical conductivity.

20-1.3 Phosphorus, Sulfur, and Chlorine:
Molecular Solids

Since the ionization energy of the phosphorus
atom is still higher than that of the silicon atom,
it is not surprising that the common forms of
phosphorus are nonmetallic molecular solids.
White phosphorus consists of discrete P4 mole-
cules (see Figure 20-2) and weak van der Waals
forces between the separate molecules are re-
sponsible for the stability of the solid. The
electronic structure of the phosphorus atom
provides an explanation of the formula and
structure of the P4 molecule. Phosphorus has the
electron configuration ls22s22/?63.s23/?3, and we
can suggest that since the 3p orbitals are half-
filled, phosphorus should be able to form three
covalent bonds. The geometry should be like
that in ammonia, NH3, in which the three N — H
bonds form a pyramid with a three-sided base
(see page 291). As shown in Figure 20-2, in the
P4 molecule each phosphorus atom does make

Fig. 20-2. The structure of a P« molecule.

Fig. 20-3. The structure of an S. molecule.

three bonds, and each atom is at one apex of a
pyramid.

The electron configuration in the valence or-
bitals of the sulfur atom (3s23/?4) suggests that it
will form two covalent bonds by making use of
two half-filled 7>p orbitals. This is, in fact, ob-
served in the molecule S8, which is present in the
common forms of solid sulfur. The S8 molecules
assume the form of a puckered ring, as shown in
Figure 20-3. As with the phosphorus, the sta-
bility of this crystalline form of sulfur is due to
van der Waals forces between discrete molecules.

The electronic structure of the chlorine atom
(3s23/?5) provides a satisfactory explanation of the
elemental form of this substance also. The single
half-filled 2>p orbital can be used to form one
covalent bond, and therefore chlorine exists as
a diatomic molecule. Finally, in the argon atom
all valence orbitals of low energy are occupied
by electrons, and the possibility for chemical
bonding between the atoms is lost.

EXERCISE 20-2

Using the principles discussed in Chapter 17,
attempt to arrange the third-row elements from
silicon through argon in order of increasing boil-
ing point, starting with the element you think
has the lowest boiling point. Be prepared to
defend in a class discussion your choice of the
position you assign each element in this se-
quence.

SEC. 20-2 I THE ELEMENTS AS OXIDIZING AND REDUCING AGENTS

367

20-2 THE ELEMENTS AS OXIDIZING AND REDUCING AGENTS

In the last section your attention was called to
the extreme variation in the physical appearance
and properties of the third-row elements. We
might start this section with a similar statement
about the behavior of the elements as oxidizing
and reducing agents. The outstanding chemical
characteristic common to metallic sodium, mag-
nesium, and aluminum is strong reducing power.
Their tendency to lose electrons and to react with
other elements is so great that they are found in
nature only in compounds, never as free ele-
ments. All three of these metals will react with
water to give hydrogen. In the case of sodium
this reaction is fast and all of the sodium is con-
sumed if sufficient water is present. For mag-
nesium and aluminum, the reaction produces a
thin layer of oxide on the metallic surface. This
oxide layer has low solubility in water and the
oxide adheres strongly to the metal. Hence it
forms a protective layer that prevents further
contact between water (or air) and the metal.
This protection accounts for the notable resist-
ance of aluminum to weathering, upon which
most of the structural uses of aluminum depend.
If either magnesium or aluminum comes in con-
tact with mercury, the protective layer is re-
moved and rapid reaction takes place.

As we saw in Chapter 19, chlorine represents
the other extreme in chemical reactivity. Its most
obvious chemical characteristic is its ability to
acquire electrons to form negative chloride ions,
and, in the process, to oxidize some other sub-
stance. Since the tendency to lose or gain elec-
trons is a result of the details of the electronic
structure of the atom, let us try to explain the
chemistry of the third-row elements on this basis.

20-2.1 Sodium, Magnesium, and Aluminum:
Strong Reducing Agents

A glance at Appendix 3, the table of £°'s for
half-reactions, should convince you that sodium,
magnesium, and aluminum are among the
strongest reducing agents available. Their £°'s
are also listed in Table 20-11. Part of this strong

Table 20-11

THE HALF-CELL POTENTIALS
OF THIRD-ROW METALS

NafsJ — >- Na+(aq) + tr

E° = 2.71 volts

Mg(s) — >- Mg^(aq) + 2e~

E° = 2.37 volts

A\(s) — >- A\+3(aq) + 3e~

E° = 1.66 volts

tendency for metals to lose electrons and become
positive ions in aqueous solutions is a result of
the fact that the valence electrons of their atoms
are not very strongly bound, as is shown by their
low ionization energies.

However, this is not the complete explanation for their
reducing properties. Let us analyze the energy require-
ments of the process

Nafsj

Na+(aq) + e~

in much the same way as we treated the half-reactions of
the halogens in Section 19-2.2. Reaction (/) can be dis-
cussed in terms of a series of hypothetical steps:

NafsJ

Natej

Na+f|?J

Nafgj

vaporization (2)

Na+fgJ + e~(g) ionization

Na+(aqJ

hydration

(i)
(4)

Step (2) is just the vaporization of solid sodium to a gas.
This step requires energy ; AH is positive for this reaction.
In step (3) an electron is removed from a gaseous atom
to form a gaseous ion; AH is positive since this step re-
quires energy also. Step (4) is the hydration of a positive
ion; energy is evolved in this process, so AH is negative.
Sodium metal is a good reducing agent, primarily because
the energy required to carry out reaction (/) is small. In
other words, the energy we put in to cause steps (2) and
(5) is small and is somewhat compensated by the energy
we get out in step (4). We can use this example to set up
general criteria for a good metallic reducing agent: (a) the
metallic crystal must not be too stable [otherwise the
energy required for step (2) will be large] ; (b) the ioniza-
tion energy of the gaseous atom should be small; (c) the
hydration energy, the energy evolved in step (4), should
be large.

We have already mentioned that the stability of the
metallic crystal and the ionization energies of the atom
tend to increase in the series sodium, magnesium, and
aluminum. In spite of this, aluminum is still an excellent
reducing agent because the hydration energy of the Al+I
ion is very large (Table 20-111).

368

THE THIRD ROW OF THE PERIODIC TABLE | CHAP. 20

Table 20-111

HYDRATION ENERGIES OF SOME
THIRD-ROW IONS (kcal/mole)

NaH
97

460

Al+3
1121

electrolysis

MgCW — ► Mg(s) + a,(g) (5)

The magnesium metal is thus recovered for re-
peated use in reaction (7). Chlorine produced in
reaction (8) is also put to use in the manufacture
of TiCU, the other reactant in reaction (7).

The third-row metals also show their strong
reducing properties in reactions which do not
take place in aqueous solution. For instance,
magnesium metal ignited in air will react with
carbon dioxide, reducing it to elemental carbon:
2Mg(s) + C02(g) —>- 2MgO(sj + C(s) (5)

If aluminum metal is mixed with a metal oxide
such as ferric oxide, Fe203, and ignited, the oxide
is reduced and large amounts of heat are evolved:

2 Alfs; + Fe203(s) — »- 2Fefsj + A\203(s)

kcal

AH = -203

mole Fe203

(6)

These reactions are possible because of the great
stability, or low energy, of the oxides of mag-
nesium and aluminum. The oxides are ionic com-
pounds whose great stability can be attributed
to strong electrostatic forces between small, posi-
tively charged Mg+2 ions (or A1+3 ions) and
negatively charged oxide ions, O-2. The fact that
the Mg+2 and Al+3 ions are very small allows
these positive ions to approach closely the nega-
tive oxide ions, resulting in strong attractive
forces.

The ease of oxidation of magnesium is im-
portant in the commercial manufacture of tita-
nium metal. Titanium, when quite pure, shows
great promise as a structural metal, but the
economics of production have thus far inhibited
its use. One of the processes currently used, the
Kroll process, involves the reduction of liquid
titanium tetrachloride with molten metallic
magnesium:

TxCUl) + 2Mg(l) — >- Ti(s) + 2MgCl2r/J (7)

Titanium has a very high melting point (1812°C),
so the magnesium chloride can be vaporized and
distilled away from the solid titanium. The gase-
ous magnesium chloride is condensed and then
electrolyzed to regenerate magnesium and chlo-
rine:

20-2.2 Silicon, Phosphorus, and Sulfur:
Oxidizing and Reducing Agents
of Intermediate Strengths

Silicon also can act as a reducing agent, as we
might expect from the properties of sodium,
magnesium, and aluminum. It reacts with mo-
lecular oxygen to form silicon dioxide, Si02. This
network solid is held together by very strong
bonds. However, because of the rather high
ionization energy of the silicon atom, and the
great stability of the silicon crystal, its reducing
properties are considerably less than those of the
typical metals.

Phosphorus continues the trend away from
strong reducing properties. Elemental phospho-
rus will react with strong oxidizing agents like
oxygen and the halogens,

?<(s) + 502(g)
Pt(s) + 6C\2(g)

P«O10(sj
4PC13(!J

(9)
(10)

but will also react with strong reducing agents
such as magnesium to form phosphides:

Pt(s) + 6Mg(s) — *- 2Mg,P2(s; (//)

Therefore, elemental phosphorus shows both re-
ducing and oxidizing properties. This interme-
diate behavior can be explained in terms of the
electron occupancy and ionization energy of the
phosphorus atom. The fact that the ionization
energy of phosphorus is greater than the previous
elements in Row 3 suggests it will be a poorer
reducing agent than, for example, aluminum.
The combination of a noticeable affinity for elec-
trons (as evidenced by the ionization energy) and
three half-filled 3p orbitals provides an explana-
tion for the appearance of a weak oxidizing tend-
ency in phosphorus. Since phosphorus neither
loses nor gains electrons readily, it is neither a
strong reducing agent nor a strong oxidizing
agent.

SEC. 20-2 I THE ELEMENTS AS OXIDIZING AND REDUCING AGENTS

369

From this discussion you should not conclude
that phosphorus is an inert element. Many of the
reactions of elemental phosphorus take place
very rapidly. For instance, when white phospho-
rus is exposed to the air it reacts rapidly accord-
ing to reaction (9). Evidently this reaction has a
low activation energy, accounting for the fact
that though phosphorus is not a strong reducing
agent, yet, with oxygen, it reacts rapidly.

The ionization energy of the sulfur atom shows
that it is even more reluctant than phosphorus
to lose electrons. The common compounds of
sulfur are the sulfides, which may be formed by
reactions of elemental sulfur with a large number
of metals. Typical reactions are

SZn(s) + SgfsJ — *- SZnS(s) (12)

16Ag(sJ + Sa(s) — »- 8Ag2S(s) (13)

These reactions show sulfur in the role of an
oxidizing agent. The properties of compounds
such as ZnS suggest they contain the sulfide ion,
S~2. The formation of this ion again can be ex-
pected on the basis of the fact that the neutral
sulfur atom has two electrons less than enough
to fill the valence orbitals. Acquisition of two
electrons completely fills the low energy valence
orbitals and solid ionic compounds can be
formed.

Sulfur reacts with molecular oxygen to form
compounds in which sulfur is assigned positive
oxidation numbers, +4 and +6. The reactions
are those used in the manufacture of sulfuric
acid (see Chapter 13):

\S6(s) + 02(g)
S02(g) + iO/gj

SCVgJ
SQ3(g)

(14)
(15)

Of these two oxides, S02 (sulfur dioxide) is gase-
ous at ordinary temperatures and pressures,
while S03 (sulfur trioxide) is a solid with a rather
high vapor pressure. Gaseous sulfur trioxide con-
sists of discrete S03 molecules. Even though
sulfur forms these and other compounds in which
it is assigned a positive oxidation number, it does
so only by reacting with the strongest of oxidiz-
ing agents. Therefore we see that sulfur is
somewhat like phosphorus — its strength as a
reducing agent or as an oxidizing agent is in-
termediate.

EXERCISE 20-3

Water saturated with S02 gas is a relatively mild
but quite useful reducing agent. Which of the
following aqueous ions might be reduced by it?

(a) Fe+3 to Fe+2

(b) Cu+2 to Cu+

(d) Hg+2 to Hg(l)

20-2.3 Chlorine: A Strong Oxidizing Agent

The formation of several oxidation states is
typical of the elements on the right side of the
periodic table. We have already discussed in
Chapter 19 the fact that chlorine can exist in the
+ 1, +3, +5, and +7 oxidation states as well as
in the — 1 state. In its compounds, chlorine is
most often found in the — 1 state. This prepon-
derance of — 1 compounds shows that elemental
chlorine behaves as an oxidizing agent in most
of its reactions.

When we review the oxidation-reduction prop-
erties of the Row 3 elements we do see a rather
smooth trend in behavior from the strong reduc-
ing agent sodium, through the elements like
phosphorus and sulfur which are neither strong
reducing agents nor strong oxidizing agents,
finally ending with the strong oxidizing agent
chlorine, where the reducing tendency is very
low. This trend is quite consistent with the
ionization energies and orbital occupancy of the
atoms. This rationalization of chemical behavior
in terms of electronic structure aids in remember-
ing the chemistry of these elements, and in
making predictions of the chemistry of other
elements in the table.

EXERCISE 20-4

Water containing dissolved Cl2 is a useful oxidiz-
ing agent. Which of the following aqueous ions
might be oxidized by it?

(a) Fe+2 to Fe+3

(b) Cu+ to Cu+2

(d) Mn+2 to Mn02(s)

(e) Mn+2 to Mn04_

370

THE THIRD ROW OF THE PERIODIC TABLE I CHAP. 20

20-3 THE ACIDIC AND BA6IC PROPERTIES OF THE HYDROXIDES

In the last section we saw that the variation in
chemical reactivity of elements in the third row
of the periodic table can be understood in terms
of electronic structure of the atoms. Those ele-
ments which have both high ionization energies
and vacant valence orbitals tend to gain electrons
and act as oxidizing agents, while those with
lower ionization energies tend to lose electrons
and act as reducing agents. We wish now to ex-
plain the acid-base behavior of the hydroxides
of the third-row elements, again using the sim-
plest possible ideas of atomic structure. We
already encountered a similar problem in Chap-
ter 19 when we explained the acidity of the
oxyacids of chlorine. In this chapter we will
again deal with a series of compounds which
contain the group

M— O— H

Here M might be any of the third-row elements.
Compounds of this structure may, of course, act
as bases by releasing hydroxide ions, breaking
the M — OH bond. The hydroxide ion, then, can
accept a proton from an acid HB

M— O— H

OH" (aq) + KB

M+(aq) + OH-(aq) (16)
H20 + B-(aq) (17)

Chemists recognize another way in which
M — O — H compounds can act as bases. The
base M — O — H can react directly with VLB:

M— O— H + HS^±

M(OH2)+(aq) + B-(aq) (18)

In reaction (18), one of the unshared pairs of
electrons on the oxygen atom of the MOH group
has accepted a proton, and so MOH can act as
a base without actually releasing hydroxide ion.
In addition, MOH compounds can act as
acids, breaking the MO — H bond:

M— O— H +±: M-O-(aq) + H+(aq) (19)
or,

M— O— H(s) + H20 +±:

M-O-(aq) + HaO+(aq) (19a)

We can attempt to predict whether a com-
pound containing the MOH group will behave
as an acid or base by considering the strength
with which the element M binds electrons. If the
atom M has a high ionization energy, it holds
electrons strongly, so we would not expect the
M — O bond to break and form M+ and OH-
ions. Also, if M holds electrons strongly, we
expect it to draw electrons away from the oxygen
atom in the MOH group, with the result that
the oxygen atom will be less able to acquire and
bind a proton as indicated in reaction (18).
Therefore, as the ionization energy of M in-
creases, we expect MOH to be less able to act
as a base, either by reaction (76) or (18).

An increase in strength with which M binds
electrons has another consequence, as we men-
tioned in Chapter 19. As M draws electrons
toward itself, the O — H bond may become
weakened and the compound would be expected
to display acidic properties.

20-3.1 Sodium and Magnesium Hydroxides:
Strong Bases

Let us apply these ideas to the third-row ele-
ments. On the left side of the table we have the
metallic reducing agents sodium and magnesium,
which we already know have small affinity for
electrons, since they have low ionization energies
and are readily oxidized. It is not surprising,
then, that the hydroxides of these elements,
NaOH and Mg(OH)2, are solid ionic compounds
made up of hydroxide ions and metal ions.
Sodium hydroxide is very soluble in water and
its solutions are alkaline due to the presence of
the OH- ion. Sodium hydroxide is a strong base.
Magnesium hydroxide, Mg(OH>2, is not very
soluble in water, but it does dissolve in acid
solutions because of the reaction

Mg(OH)2fS; + 2H+(aq) — >-

Mg+2f aq) + 2H20 (20)

Magnesium hydroxide is a strong base.

SEC. 20-3 | THE ACIDIC AND BASIC PROPERTIES OF THE HYDROXIDES

371

20-3.2 Aluminum Hydroxide: Both an Acid
and a Base

We can tell from the ionization energy of alumi-
num that this atom holds its second and third
valence electrons rather firmly. With this fact in
mind, we can see why aluminum hydroxide,
Al(OH)3, would not be as strongly basic as are
the hydroxides, NaOH and Mg(OH)2. Alumi-
num hydroxide has extremely low solubility in
neutral aqueous solutions but does react with
strong acids according to the reaction

Al(OH)3(s) + 3H+(aq) — ►- Al+3(aq) + 3H20 (27)

Al(OH)3f sj + 3H30+f aq)

A\(OH^3(aq) (22)

These reactions say the same thing, but reaction
(22) emphasizes the fact that the A1+3 ion is
hydrated in aqueous solutions. In any case, the
fact that Al(OH)3 reacts with acids shows that
it has the properties of a base.

Aluminum hydroxide will also react with hy-
droxide ion to dissolve according to the equation

Al(OH)3(sJ + OH-(aq) — »- Al(OH)4-(aqJ (23)

Reaction (23) shows that Al(OH)3 has the prop-
erties of an acid, since it reacts with the base
OH~. A substance that acts as an acid under some
conditions and as a base under other conditions
is said to be amphoteric. The electronic situa-
tion in Al(OH)3 is such that it can either accept
a proton (act as a base) or react with OH~ (act
as an acid). We will see in Chapter 22 that
several other hydroxides also show amphoteric
behavior.

20-3.3 Silicon, Phosphorus, Sulfur,
and Chlorine Oxyacids

Both our original prediction about the effect of
ionization energy on acid-base behavior and the
trend which we have observed in the first three
elements lead us to expect that the hydroxide or
oxide of silicon should not be basic, but perhaps
should be weakly acidic. This is in fact observed.
Silicon dioxide, Si02, can exist as a hydrated
solid containing variable amounts of water,

Si02 xH20. This hydrated oxide does react with
hydroxide ion to form soluble silicate ions:

Si02xH,0(solid hydrate) + 20H (aq) — >-

SiO,->qJ+H,0 (24)

This reaction shows that the hydrated oxide
Si02 xH20 is acidic, since it reacts with a base.

As we mentioned earlier, phosphorus can be
found in four different oxidation states. The
"hydroxides'' of the +1, +3, and +5 states of
phosphorus are hypophosphorous acid, H3P02,
phosphorous acid, H3P03, and phosphoric acid,
H3P04. Their structures are shown in Figure
20-4. As suggested by their names, these com-
pounds are distinctly acidic, and are of moderate
strength. The equilibrium constant for the first
ionization of each acid is approximately 10-2:

hypophosphorous acid:

H3P02 — >- H+(aq) + H2P02~ (aq)

K = 1 X 10"2 (25)
phosphorous acid:

H3POs -+ H+(aq) + H2P03" (aq)

K = 1.6 X 10-2 (26)
phosphoric acid:

H3P04 — ->- H+(aq) + H2P04- (aq)

AT = 0.71 X 10-2 (27)

Notice that this series differs from the oxyacids
of chlorine, shown in Figure 19-6. In the chlorine
oxyacids, oxygen atoms add successively and the
first ionization constants become larger. In the
phosphorus oxyacids, hydrogen atoms are suc-
cessively replaced by O — H groups but the first
ionization constants change very little.

The most common oxyacid of sulfur is H2S04,
sulfuric acid. In dilute aqueous solutions this
substance is almost completely dissociated into
ions according to the equation

H2S04 — -»- H+(aq) + HS04- (aq) (28)

so sulfuric acid is classified as a strong acid. The
bisulfate ion, HS04~ (also called the hydrogen
sulfate ion), is also an acid, since the equilibrium
constant for the reaction

HS04"

H+(aq) + SO^(aq) (29)

is approximately 10~2. From these equilibrium
constants it is clear that sulfur in the +6 oxida-

372

THE THIRD ROW OF THE PERIODIC TABLE I CHAP. 20

NAME

FORMULA. BALL-AND -STICK MODEL SPACE- FILLING- MODEL

Hypophosphorous acid H3P02

Phosphorous acid

H3P0s

Phosphoric acid

H3PO+

Fig. 20-4. Presumed structures of the phosphorus
oxyacids.

tion state is even more acidic than silicon and
phosphorus.

Sulfur in the +4 oxidation state also forms
an oxyacid, sulfurous acid (H2S03). This com-
pound is not as strong an acid as H2S04. The
equilibrium constant for the reaction

H2SO3

H+(aq) + HSOz(aq) (30)

is approximately 10~2. The species HS03~ is
called the bisulfite ion or the hydrogen sulfite
ion. It too is a weak acid and dissociates in water
to form the sulfite ion, S03-2

HSO3- — *■ H+(aq) + SO;*(aq) (31)

Therefore we see that the oxyacids of sulfur
continue the trend of increasing acidity which

we have observed in the successive third-row
elements.

The oxyacids (or, hydroxides) of chlorine were
discussed in Section 19-2.4 and an explanation
of their acid strengths was given there. The ar-
guments used to explain the increase in acid
strength of the oxyacids of chlorine with in-
creasing oxidation number connect with those
used in this chapter to explain the increase of
acidity which occurs as we successively consider
the oxygen compounds of the third-row ele-
ments. As a halogen atom adds oxygen atoms,
electron charge is drawn away from the halogen.
This, in turn draws electrons from the O — H
bond and weakens it. Acid strength increases.
As we move to the right in the periodic table, the
ionization energy increases and, again, attrac-
tion of electrons by the central atom increases.
This, too, draws electrons from the O — H bond
and weakens it. Acid strength increases.

SEC. 20-4 I OCCURRENCE AND PREPARATION OF THE THIRD-ROW ELEMENTS

373

20-4 OCCURRENCE AND PREPARATION OF THE THIRD-ROW ELEMENTS

Except for argon, the third-row elements make
up an important fraction (about 30%) of the
earth's crust. Silicon and aluminum are the sec-
ond and third most abundant elements (oxygen
is the most abundant). Both the occurrence and
the mode of preparation of each element can be
understood in terms of trends in chemistry dis-
cussed earlier in this chapter.

20-4.1 Occurrence in Nature

Sodium (fifth most abundant element) is found
principally as Na+ ion in water soluble salt
deposits, such as NaCl, and in salt waters. The
element reacts rapidly with water and with at-
mospheric oxygen, hence is not found in an
uncombined state in nature.

Magnesium (eighth most abundant element) is
found principally as Mg+2 ion in salt deposits,
particularly as the slightly soluble carbonate,
MgC03, and also in sea water. The element is
oxidized by atmospheric oxygen and is not found
in an uncombined state in nature.

Aluminum (third most abundant element) is
found as the A1+3 ion in oxides and as the com-
plex ion A1F6~3. Important minerals are bauxite,
which is best described as a hydrated aluminum
oxide, A1203 xH20, and cryolite, Na3AlF6. The
element is readily oxidized and is not found in
an uncombined state in nature.

Silicon is the second most abundant element
in the earth's crust. It occurs in sand as the
dioxide Si02 and as complex silicate derivatives
arising from combinations of the acidic oxide
Si02 with various basic oxides such as CaO,
MgO, and K20. The clays, micas, and granite,
which make up most soils and rocks, are silicates.
All have low solubility in water and they are
difficult to dissolve, even in strong acids. Silicon
is not found in the elemental state in nature.

Phosphorus (eleventh most abundant element)
occurs mostly as the phosphate anion, PCX,-3, in
such minerals as "phosphate rock," which is a

complex mixture of Ca3(PO,)2 and CaF2. Most
phosphates have low solubility in water. The
element is not found uncombined in nature.

Sulfur (fourteenth most abundant element) oc-
curs in minerals either in an oxidized state as
sulfate anion, S04~2, or in a reduced state as
sulfide anion, S~2. Gypsum, CaS04-2H20, with
low solubility in water, and Epsom salt,
MgS04-7H20 with high solubility in water, are
two common sulfate minerals. Galena, PbS, iron
pyrites, FeS2, and zinc blende, ZnS, are impor-
tant sulfide minerals. Sulfur occurs as the free
element in large underground deposits.

Chlorine (sixteenth most abundant element) is
found as Cl~ in water soluble salt deposits, such
as NaCl, and in salt waters. The element, Cl2, is
not found in the atmosphere.

Argon is found only in the elemental state.
Air contains about 1 % argon.

20-4.2 Mode of Preparation of the Element

Sodium is prepared by electrolysis of molten
NaCl (giving chlorine as a by-product) or of
molten NaOH.

Magnesium is an important structural metal.
It can be prepared through the sequence of
steps: precipitation of Mg+2 from sea water by
OH~ to form Mg(OH)2; conversion of Mg(OH)2
to MgCl2; electrolysis of molten MgCl2.

Aluminum, though the third most abundant
element, was quite expensive until about 1886,
when a practical commercial electrolysis process
was developed by a young American chemist,
C. M. Hall. Bauxite, A1203jcH20, is dissolved
at about 1000°C in molten cryolite, Na3AlF6, and
electrolyzed.

Silicon in the elemental state has important
electronic applications as a semiconductor that
were developed only during the last decade. The
discovery of these uses was possible only after
methods were developed for preparing silicon of
extremely high purity. Reduction of Si02 with

374

THE THIRD ROW OF THE PERIODIC TABLE ] CHAP. 20

carbon in an electric furnace is one process for
manufacture of silicon. Very pure silicon is made
by decomposing SiCl4. Still further purification
of the element is based upon the "zone-melting"
technique in which a rod of silicon is heated to
melting in a thin zone. This molten zone is
gradually moved along the length of the rod.
The impurities dissolve in the liquid and move
along with the zone, leaving metal of ultra-high
purity.

Phosphorus is prepared by heating a mixture
of Ca3(P04)2, sand, and carbon (coke). White
phosphorus, P4, distills out and can be cooled
and collected under water.

Sulfur is pumped out of natural underground
deposits in the molten state after it is melted
with water heated under pressure to about
170°C.

Chlorine is prepared by the electrolysis of
molten NaCl or of aqueous NaCl.

Argon is obtained through fractional distilla-
tion of liquefied air.

20-4.3 Some Properties of the Second-
and Third-Row Elements

The first ionization energies of elements 1 to 19
are shown in Table 15-1 II. The energies to re-
move successive electrons from gaseous Na, Mg,
and Al atoms are shown in Table 20-IV.

Trends in the properties of AH of vaporization
and boiling point for the second- and third-row
elements are compared in Table 20-V.

Table 20-IV

SUCCESSIVE IONIZATION ENERGIES
OF SODIUM, MAGNESIUM,
AND ALUMINUM (kcal/mole)

ELEMENT

(1st e~)

E2
(2nd e~)

Ez
(3rd e~)

(4th e~)

118

1091

1653

—

175

345

1838

2526

138

434

656

2767

Table 20-V.

TRENDS IN PROPERTIES OF SECOND

A//Vap B.P.

element (kcal/mole) (°C)

QUESTIONS AND PROBLEMS

AND THIRD-ROW ELEMENTS

A//v.p B.P.

element (kcal/mole) (°C)

32.2

1326

23.1

889

53.5

2970

31.5

1120

128.8

~3900

67.9

2327

170

~4000

(105)

2355

0.67

-196

3.0

280

0.81

-183

2.5

445

0.78

-188

4.9

-34.1

0.42

-246

1.6

-186

Make a graph with an energy scale extending
on the ordinate from zero to 3000 kcal/mole and
with the abscissa marked at equal intervals with
the labels Na, Mg, and Al. Now plot and con-
nect with a solid line the first ionization energies,
£w of these three elements (see Table 20-IV).
Plot E2 and connect with a dashed line, E3 with
a dotted line, and EA with a solid line. Draw a

circle around each ionization energy that identi-
fies a valence electron.

2. Plot the ionization energy of the first electron
removed from the atoms of both the second- and
third-row elements against their atomic number
(abscissa). What regularity do you observe?

3. Silicon melts at 1410°C and phosphorus (white)

QUESTIONS AND PROBLEMS

375

at 44°C. Explain this very great difference in
terms of the structures of the solids.

4. Recalling the chemistry of nitrogen, write for-
mulas for phosphorus compounds correspond-
ing to

(a) ammonia,

(b) hydrazine,

5. Write the formula for the fluoride you expect to
be most stable for each of the third-row ele-
ments.

6. The heat of reaction for the formation of
MgO(s) from the elements is —144 kcal/mole
of MgO(s). How much heat is liberated when
magnesium reduces the carbon in C02 to free
carbon? See Table 7-II.

Answer. AH = — 97 kcal/mole MgO.

7. Magnesium oxide is an ionic solid that crystal-
lizes in the sodium chloride type lattice.

(a) Explain why MgO is an ionic substance.

(b) How many calories would be required to
decompose 8.06 grams of MgO? (Use the
data in problem 6.)

8. Aluminum oxide (A1203) is thought to dissociate
at high temperature (1950°C) according to the
equation: 2Al203(sJ — >■ 4A10(gj + 02(g). The
total vapor pressure at 1950°C is about 1 X 10~6
atm.

(a) Which element is oxidized and which is re-
duced in this reaction?

(b) Write the equation for the equilibrium con-
stant.

9. Explain the observation that phosphorus acts
both as a weak reducing agent and as a weak
oxidizing agent.

10. (a) What are the oxidation numbers of phos-
phorus in the two compounds phosphorous
acid, H3PO3, and phosphoric acid, H3P04?
(b) From the £° values in Appendix 3, decide
which of the following substances might be
reduced by phosphorous acid: Fe+2; Sn+4;
I2; Cr".

H20 + H3P03 — *■ H3PO4 + 2H+ + 2e~

E° = 0.276 volt

11. Answer the following in terms of electron con-
figuration and ionization energy:

(a) Which elements in the second and third rows
are strong

(i) oxidizing agents?
(ii) reducing agents?

(b) What properties do strong oxidizing agents
have?

12. Of the elements Na, Mg, Al, which one would
you expect to be most likely to

(a) form a molecular solid with chlorine?

(b) form an ionic solid with chlorine?

13. One kilogram of sea water contains 0.052 mole
of magnesium ion. What is the minimum num-
ber of kilograms of sea water that would have
to be processed in order to obtain 1 kg of
Mg(OH)2?

Answer. 3.3 X 102 kg of sea water.

14. Why is aluminum hydroxide classed as an am-
photeric compound?

15. Some of the following common compounds of
the third-row elements are named as hydroxides
and some as acids :

NaOH sodium hydroxide

Mg(OH)2 magnesium hydroxide

Al(OH)3 aluminum hydroxide

Si(OH)4 silicic acid (usually written H4Si04)

P(OH)3 phosphorous acid (usually written

H3P03)
S(OH)2 not known
Cl(OH) hypochlorous acid (usually written

HOC1)

(a) Explain why these compounds vary system-
atically in their acid-base behavior.

(b) Write equations that show the reactions of
each of these substances either as acids, as
bases, or both.

16. A solution containing 0.20 M H3P03, phospho-
rous acid, is tested with indicators and the
H+(aq) concentration is found to be 5.0 X
10-2 M. Calculate the dissociation constant of
H3P03, assuming that a second proton cannot
be removed.

376

THE THIRD ROW OF THE PERIODIC TABLE I CHAP. 20

17. Elemental phosphorus is prepared by the reduc-
tion of calcium phosphate, Ca3(P04)j, with coke
in the presence of sand, Si02. The products are
phosphorus, calcium silicate, CaSi03, and car-
bon monoxide.

(a) Write the equation for the reaction.

(b) Using 75.0 kg of the ore, calcium phosphate,
calculate how many grams of P4 can be ob-
tained and how many grams of coke (as-
sumed to be pure carbon) will be used.

CHAPTER

The Second Column
of the Periodic Table:
The Alkaline Earths

z
He

u
Ha

4
Bt

s
B

13
At

15
P

16
S

18
Ar

21
Sc

.22

23
V

24
Cr

25
Mn

27
Co

29
Cu

30
Zn

31
Ga

32
Ge

33
As

34*
Se

35
Br

36
Kr

43
Tc

44
Ru

45-

48
Cd

49
In

so
Sn

S3
I

ss
Cs

17-71

76
■09

79
Au

62
Fb

34
Po

86
Rn

66
Ra

69-

iii'

i i i »

In the preceding chapter we looked at the ele-
ments of the third row in the periodic table to
see what systematic changes occur in properties
when electrons are added to the outer orbitals
of the atom. We saw that there was a decided
trend from metallic behavior to nonmetallic,
from base-forming to acid-forming, from simple
ionic compounds to simple molecular com-
pounds. These trends are conveniently discussed

in terms of the ionization energies and orbital
occupancies.

There are similar, but smaller, trends in the
properties of elements in a column (a family) of
the periodic table. Though the elements in a
family display similar chemistry, there are im-
portant and interesting differences as well. Many
of these differences are explainable in terms of
atomic size.

21-1 ELECTRON CONFIGURATION OF THE ALKALINE EARTH ELEMENTS

The elements of the second column and their
electron configurations are given in Table 21-1.
For each element, the neutral atom has two more

electrons than an inert gas. We can expect these
two electrons to be easily removed, to give the
stability of the inert gas electron configuration.

377

378 THE SECOND COLUMN OF THE PERIODIC TABLE: THE ALKALINE EARTHS I CHAP. 21

Table 21-1. electron configurations of the alkaline earth elements

ELECTRON ARRANGEMENT

ELEMENT

SYMBOL

NUCLEAR
CHARGE

INNER LEVELS

OUTER LEVELS

beryllium

1j»

magnesium

\s*

2s*2pt

calcium

•••2jj2/7<

3s23p«

45*

strontium

•••3s*3/>» 3rf10

4s*4p>

5s2

barium

•••4^4/j« 4dl°

5s*5(fi

6s1

radium

■ ■ ■tpWSpfi 5d10

6526/7«

7s*

EXERCISE 21-1

On the basis of the electron configurations and
positions in the periodic table, answer the follow-
ing questions.

(a) Is calcium likely to be a metal or nonmetal ?

(b) Is calcium likely to resemble magnesium or
potassium in its chemistry?

EXERCISE 21-2

Predict the chemical formula and physical state
at room temperature of the most stable com-
pound formed by each alkaline earth element
with (a) chlorine; (b) oxygen; (c) sulfur.

Exercises 21-1 and 21-2 pose some of the
simplest questions we can ask about the alkaline
earths. The periodic table arranges in a column
elements having similar electron configurations.
We can expect elements on the left side of the
periodic table to be metals (as magnesium is).
Furthermore, we can expect that the elements in
a given column will be more like each other than
they will be like elements in adjacent columns.
Thus, when we find that the chemistry of mag-
nesium is almost wholly connected with the
behavior of the dipositive magnesium ion, Mg+2,
we can expect a similar situation for calcium,
and for strontium, and for each of the other
alkaline earth elements. This proves to be so.

Remembering, then, that the alkaline earths
are classed as a family because of general simi-
larity, we shall investigate the detailed differences
among them.

21-2 TRENDS IN PHYSICAL PROPERTIES

21-2.1 Atomic Radii in Solids

The size of an atom is defined in terms of the
interatomic distances that are found in solids
and in gaseous molecules containing that atom.
For an atom on the left side of the periodic table,
gaseous molecules are obtained only at very high
temperatures. At normal temperatures, solids
are found and there are two important types to
consider, metallic solids and ionic solids. Table
21-11 shows the nearest neighbor distances in the

pure metals, in the gaseous oxide molecules, and
in the solid oxides (which, except for BeO, have
the sodium chloride crystal structure pictured in
Figure 5-10, p. 81). This table also shows the
corresponding radii assigned to each alkaline
earth atom, first, in the metallic state, second,
in the gaseous molecule (assuming that oxygen
has the same size as in 02) and third, in the state
of a -f-2 ion (assuming that the oxide ion, O-2,
should be assigned a radius of 1.32 A).

SEC. 21-2 I TRENDS IN PHYSICAL PROPERTIES

379

Table 21-11. trends in interatomic distances

NEAREST NEIGHBOR DISTANCE

(Angstroms)

ALKALINE EARTH ATOMIC SIZE

IN THE PURE

IN THE GASEOUS

IN THE SOLID

METALLIC

DOUBLE-BOND

IONIC

ELEMENT

METAL

OXIDE

RADIUS

2.23

1.33

1.64

1.11

0.73

0.32

3.20

1.75

2.10

1.60

1.15

0.78

3.95

1.82

2.40

1.97

1.22

1.08

4.30

1.92

2.57

2.15

1.32

1.25

4.35

1.94

2.76

2.17

1.34

1.44

We see that, no matter what type of bonding
situation is considered, there is a trend in size
moving downward in the periodic table. The
alkaline earth atoms become larger in the
sequence Be < Mg < Ca < Sr < Ba. These
atomic sizes provide a basis for explaining trends
in many properties of the alkaline earth elements
and their compounds.

21-2.2 Ionization Energies

Table 21 -III shows the first three ionization en-
ergies of the alkaline earths.

Table 21-111

ionization energies of the
alkaline earth elements

ionization energy (kcal/mole)
element Et (1st e~) £2 (2nd e~) E3 (3rd e~)

214

420

3533

175

345

1838

140

274

1173

132

253

986

120

230

811

EXERCISE 21-4

If the ionization energy £\ is regarded as a
measure of the distance between the electron
and the nuclear charge, what do the ionization
energies of Be and Ba indicate about the relative
sizes of the two atoms?

From Exercise 2 1 -4 we see that the decreasing
ionization energies observed for the alkaline
earth atoms are readily explained in terms of
their increasing size moving down in the periodic
table. Notice that the ionization energy trend
going down in the periodic table is the same as
the trend going to the left in the periodic table.

EXERCISE 21-5

From the ionization energies, predict which
solid substance involves bonds having the most
ionic character: BeCl2, MgCl2, CaCl2, SrCl2,
BaCl2. Which substance is expected to have most
covalent character in its bonds?

EXERCISE 21-3

For each of the alkaline earths, calculate the
ratio Ei/E\. Account for the results in terms of
the charges on the ions formed in the two ioniza-
tion steps.

21-2.3 Metallic Properties

Table 21 -IV shows some properties of the metals
and their crystal forms. Since different crystal
forms are involved in the series, trends in the
properties are obscured. Figure 21-2 shows scale
representations of the crystal structures of metal-
lic beryllium, calcium, and barium.

380 THE SECOND COLUMN OF THE PERIODIC TABLE: THE ALKALINE EARTHS | CHAP. 21

Metal, M

Gaseous oxide, Jvf-O

Metallic radius

- - , -

Double bond radius

1.08

Ionic radius

Fig. 21-1. Sizes of the alkaline earth atoms with various bond types (in Angstroms).

SEC. 21-3 ; TRENDS IN CHEMICAL PROPERTIES

381

Be
Hexagonal

Ca
Face-- centered cubic

Ba
Body-cen-t-ered cubic

Fig. 21-2. Scale representations of the crystal struc-
tures of Be, Ca, and Ba.

Table 21 -IV. properties of the alkaline earths in the metallic state

CRYSTAL
STRUCTURE

DENSITY

(g ml)

MELTING

HEAT OF

ELECTRICAL

POINT

VAPORIZATION

CONDUCTIVITY

(°C)

(kcal/mole)

(ohm-cm)-1

hexagonal

1.85

1283

1.69 X 10s

hexagonal

1.75

650

2.24 X 10s

face-centered cubic

1.55

850

2.92 X 10*

face-centered cubic

2.6

770

0.43 X 10"

body-centered cubic

3.5

710

0.16 X 10s

EXERCISE 21-6

the metallic state. Compare the trend in these
From the density of each element, calculate the molar volumes with the trend in the metallic
volume occupied by one mole of its atoms in radii shown in Table 21-11.

21-3 TRENDS IN CHEMICAL PROPERTIES

We have already observed (in Exercise 21-2) that
the alkaline earths have similar chemistry. As
shown in Table 21-1, they have similar electron
configurations. Table 21 -III shows that each ele-
ment has two valence electrons. With these basic
likenesses in mind we shall explore the chemical
trends among these elements.

21-3.1 Oxidation and Reduction

All of the alkaline earths are strong reducing

agents, since they readily release electrons. The
values of E° are collected in Table 21-V.

Table 21-V

THE HALF-CELL POTENTIALS
FOR THE ALKALINE EARTHS

Mg
Ca
Sr
Ba

Be+2 + 2e-
Mg+2 + 2e-
Ca+2 + 2e-
Sr+2 + 2e-
Bh+2 + 2c-

E° = +1.85 volts

E° - +2.37

E° = +2.87

E° = +2.89

E> = +2.90

382 THE SECOND COLUMN OF THE PERIODIC TABLE: THE ALKALINE EARTHS I CHAP. 21

EXERCISE 21-7

The ease of removal of an electron from a gase-
ous atom, the ionization energy, is one of the
factors that is important in fixing E°. Refer back
to Table 21-111 and predict the trend in E° that
this factor would tend to cause.

21-3.2 Acid and Base Properties

We have explained the trends in acid-base char-
acter across the periodic table by considering
the increasing ionization energy of the metal
atom. As the atom M in a structure M — O — H
attracts electrons more and more strongly, there
is increasing tendency toward acidic properties.
As the ionization energy of M goes down, there
is increasing tendency toward basic properties.

Moving down in a column is equivalent in
many respects to moving to the left in the peri-
odic table. Since we find basic properties pre-
dominant at the left of the periodic table in a
row, we can expect to find basic properties
increasing toward the bottom of a column. Thus
the base strength of the alkaline earth hydroxides
is expected to be largest for barium and stron-
tium. The greatest acid strength is expected for
beryllium hydroxide.

Experimentally we find that strontium and
barium hydroxides are indeed strong bases. All
of the alkaline earth hydroxides dissolve readily
in acidic solutions, showing that they are all
bases to some extent:

CaO(s) + H20(l)

Ca(OH)2f8J

A7/ = -15.6 kcal (7)

Be(OH)2(sJ +2H+(aq)
Mg(OH)2(sJ + 2H+(aq)
Ca(OH)2fsJ + 2H+(aq)
Sr(OH)2(sJ +2H+(aq)
Ba(OH)2fsJ +2H+(aq)

Be« +2H20 (7)

Mg+2 + 2H20 (2)

Ca4* +2H20 (3)

Sr^ +2H20 (4)

Ba+2 +2H2Q (5)

Only beryllium hydroxide dissolves appreciably
in strong base solutions,

Be(OH)2(s) + 20H-

BeO-2 + 2H20 (5)

These hydroxides are formed from the cor-
responding oxides. For example, calcium oxide,
or lime, reacts with water as in reaction (7).

The process is called "slaking" the lime and it is
used by plasterers in preparing mortar, which
requires Ca(OH)2. As water is added to lime
there is a considerable evolution of heat, as
evidenced by wisps of steam that rise from the
sample.

Because all of the alkaline earth oxides react
with water to form basic hydroxides, they are
called basic oxides. The reactions and their
heats are as follows:

BeO(s) + H20(l)
MgO(s) + H20(Zj
CaOfsJ + H20(l)
StO(s) + H20(%)
BaO(s) + HiOd)

Be(OH)2(s)

AH = -2.5 kcal (8)

Mg(OH)/sJ

AH = -8.9 kcal (9)

Ca(OH)2(sJ

AH = -15.6 kcal (70)

Sr(OH)2(sJ

AH = -19.9 kcal (77)

Ba(OH)./sJ

AH = -24.5 kcal (72)

Notice the progressively increasing exothermic
reaction heat, moving downward in the series.

EXERCISE 21-8

How much heat is evolved if one pound (454
grams) of lime is slaked according to reaction
(70)? How many grams of water can be evapo-
rated with this heat? (The heat of vaporization
of water is about 10 kcal/mole.)

21-3.3 Solubilities of Alkaline Earth
Compounds in Water

We encountered the solubilities of alkaline earth
salts in Chapter 10 and discovered some interest-
ing trends. Before looking back to Figures 10-5
and 10-6, see how much you can recall about
these solubilities.

SEC. 21-3 I TRENDS IN CHEMICAL PROPERTIES

383

EXERCISE 21-9

In your notebook indicate one of the four
answers

(i) none of the alkaline earth ions;
(ii) all alkaline earth ions;
'iii) Be+2, Mg+2, and Ca+2, but not Sr+2, Ba^2,

or Ra+2;
(iv) Sr+2, Ba+2, and Ra+2, but not Be+2, Mg+2,
or Ca+2

for each of the following:

(a) || form compounds of low solubilities
with C1-, Br~, and I".

(b) IHfl torni compounds of low solubilities
with sulfate, S04~2.

(d) || form compounds of low solubilities
with hydroxide, OH~.

(e) || form compounds of low solubilities
with carbonate, C03-2.

Now compare your answer with Figures 10-5
and 10-6.

Table 21 -VI

THE SOLUBILITY PRODUCTS OF THE
ALKALINE EARTH HYDROXIDES

COMPOUND

Ktp

Mg(OH)2

8.9 X 10-"

Ca(OH)j

1.3 X 10-6

Sr(OH)2

3.2 X 10-*

Ba(OH)2

5.0 X 10~«

few ions in solution; the larger values correspond
to higher concentration in a saturated solution —
that is, higher solubility.

EXERCISE 21-10

Suppose you have a solution in which the con-
centration of hydroxide ion is 1 M. How many
moles per liter of the different alkaline earth ions
listed in Table 21 -VI could you have (at equilib-
rium) in this solution? If the concentration of
hydroxide ions were 0.5 M, how would your
answers change?

THE HYDROXIDES

When a hydroxide such as calcium hydroxide is
added to water in sufficient amount, we get a
saturated solution containing Ca+2 and OH~ in
equilibrium with excess undissolved solid. The
equilibrium

Ca(OH)/S/) ^± Ca+Yaqj + 20H~( aq) (13)

can also be established by mixing Ca+2 ions (for
example, from a solution of CaCl2) with OH~
ions (for example, from a solution of NaOH)
until a precipitate of Ca(OH)2 forms. In either
case, at equilibrium the concentrations of Ca+2
and OH- ions are such that the equilibrium
expression is satisfied:

[Ca«][OH-p = K„

(14)

In Table 21 -VI the numerical values of
[M+2][OH~]2 are listed for some of the alkaline
earth hydroxides. Small values indicate relatively

Exercise 21-10 demonstrates that there is a
regular trend in the solubilities of the alkaline
earth hydroxides.

THE CARBONATES AND SULFATES

Although the hydroxides of the alkaline earth
elements become more soluble in water as we go
down the column, the opposite trend is observed
in the solubilities of the sulfates and carbonates.
For example, Table 21 -VI I shows the solubility
products of the alkaline earth sulfates.

Table 21 -VI I

THE SOLUBILITY PRODUCTS OF THE
ALKALINE EARTH SULFATES

COMPOUND

MgSO«

soluble (K.p » 10"1)

CaSCX

2.4 X 10-6

SrSO«

7.6 X 10-7

BaSO*

1.5 X 10-»

384 THE SECOND COLUMN OF THE PERIODIC TABLE: THE ALKALINE EARTHS | CHAP. 21

The solubility of calcium carbonate is such
that in a saturated solution the product of
ion concentrations [Ca+2] [C03_ 2] is 5 X lO"9.
Though this may seem quite small, it is large
enough to be important to man, especially if he
lives in a region of the earth where there are
extensive limestone deposits. Calcium carbonate
can be dissolved in water, especially if it contains
much dissolved C02. This is objectionable be-
cause soap added to water which contains even
traces of Ca+2 forms a precipitate of calcium
stearate. This is the ring that is so difficult to
remove from the bathtub.

The dissolving of limestone by ground water
is another example of chemical equilibrium. The
behavior of this system depends upon the chemi-
cal equilibrium between CaC03 and its dissolved
ions and the equilibrium between carbonate ion
and dissolved C02 in the water. When CaC03
dissolves in water it establishes the equilibrium

CaC03(s) +± Ca+2(aq) + C03" 2(aq) (15)

The carbonate ion, a base, can accept a proton
from water, an acid,

CO3- 2(aq) + H.O(l) +±

HC03 (aq) + OH( aq) (16)

Thus, solutions of carbonates are found to be
basic. Aqueous solutions of carbon dioxide are,
on the other hand, acidic. The reactions in this
equilibrium are

C02(g) + H20(l) =?=*: H«C03(aq) (17)

H«C03(aq) z^± HCO3- (aq) + H+(aq) (18)

The combination of reaction (18) and (16) shows
how carbon dioxide enhances the solubility of
calcium carbonate by removing carbonate ion to
form bicarbonate ion,

H,C03(aq) + CO3- 2(aq) z<±t 2HC03 (aq) (19)
or, in the net reaction,

CaCCVsJ + CO,(g) + H2Of/J q=b

Ol+2(aq) + 2HC03- (aq) (20)

The result is that we get an appreciable concen-
tration of Ca+2 in the water, giving so-called
hard water — hard on the soap and hard on the
people who use it. Caves in limestone regions are
formed essentially by the combination of the two
equilibria above. In contrast, the weird icicle-like
projections (stalactites) found hanging from the
roofs of such caves are formed by the reverse of
these reactions. On standing, a droplet of a satu-
rated solution containing Ca+2 and HC03~ may
lose some C02 and H20 by evaporation. Loss of
C02 and H20 from the equilibrium (20) en-
hances the reverse-directed change, resulting in
the deposit of a fleck of CaC03. The same change
occurs when hard water of this kind is boiled in
a pot or heated in a boiler. The white scum you
may see forming on the surface of boiling water
is often due to these equilibria.

Fig. 21-3. Stalactites: solubility equilibria at work.

Limestone '■ CaCOj
CaC03(s) * C02 + H20 -* Ca+2 + 2HC03~

; HZ0(f)

CaZ * ZHCO;^ Co. C03 (s) + C02 +H20

21-4 OCCURRENCE AND PREPARATION OF THE ALKALINE
EARTH ELEMENTS

As we did in the preceding chapter, we conclude
by summarizing some information about the
occurrence in nature and the modes of preparing
the alkaline earth elements.

21-4.1 Occurrence in Nature

All of the alkaline earth elements exist in nature
as the M+2 cations.
Beryllium (forty-fourth most abundant ele-

SEC. 21-4 I OCCURRENCE AND PREPARATION OF THE ALKALINE EARTH ELEMENTS 385

ment) is rather rare and occurs mostly as
an aluminum beryllium silicate called beryl,
Be3Al2Si6Oi8. Beryl containing traces of chro-
mium has a beautiful green color and is called
emerald.

Magnesium (eighth most abundant element) is
found principally as Mg+2 ion in salt deposits,
particularly as the slightly soluble carbonate,
MgC03, and also in sea water. The natural de-
posits of MgC03 with CaC03 are called dolo-
mite. Magnesium is present as a cation in the
asbestos silicates.

Calcium (sixth most abundant element) is
found in limestone, CaC03, and gypsum,
CaS04-2H20. Bones are made up of calcium
phosphate, Ca3(P04)2.

Strontium (thirty-eighth most abundant cle-
ment) is rather rare and is found principally as
the mineral strontianite, SrC03.

Barium (eighteenth most abundant element)
is also rather rare; it occurs as the mineral
barite, BaS04.

Radium is radioactive and extremely rare. It
occurs in trace amounts (one part in 1012) in
uranium ores such as pitchblende (mainly U308).

EXERCISE 21-11

What property held in common by the following
compounds accounts for their presence in natu-
ral mineral deposits: MgC03, CaC03, SrC03,
BaS04, and (in bones) Ca3(P04)2?

EXERCISE 21-12

What property held in common by the alkaline
earth elements accounts for the fact that the free
elements are not found in nature?

21-4.2 Mode of Preparation of the Element

Only magnesium is produced in any substantial
quantity in the elemental form. The reaction
sequence used is given in Section 20-4.2.

The general method for preparing the alkaline
earth elements is to convert the mineral to a
chloride or a fluoride by treatment with HC1 or
HF. Then the molten salt is electrolyzed or, as
in the case of BeF2, reduced with a chemical
reducing agent such as Mg.

Alfred E. Stock was one of the greatest inorganic chemists
of the twentieth century. His research investigations, which
resulted in over 250 publications, were characterized by
brilliant experimental technique and convincing thorough-
ness. Both of these rich qualities were needed for his
hazardous studies of the hydrides oj boron, an overlooked
area of chemistry in which he was recognized as the single
world authority for a period of at least a decade. It is fitting
that his name is perpetuated in the "Stock system" of in-
organic chemical nomenclature (in which Roman numerals
indicate oxidation numbers).

Stock was born in Danzig, Poland, and his aptitude for
science was displayed early in his boyhood collections of
salamanders, butterflies, and plants. He studied at the
University of Berlin where the chemistry facilities of the
day were so limited that this brilliant experimentalist-to-be
had to wait till his third semester to approach a laboratory
bench. He received the Ph.D. at the University of Berlin
in 1899, graduating magna cum laude.

Shortly after 1900, young Alfred Stock began his life-
time work: study of the chemistry of boron. He reasoned
that this neighbor of the versatile carbon atom could not
possibly have the dull and limited chemistry popularly as-
sumed at the time. He entered this study stimulated by his
own desire to know — despite advice by the laboratory di-

rector to select another area because the chemistry of
boron was already thoroughly investigated. His persistence
was rewarded by discoveries of a succession of hydrides
of boron, such as diborane, B^Hs, tetraborane, BaH\0,
pentaborane, BhH*, and decaborane, BwHu. The structures
and even the very existence of these compounds baffled
chemists for many years. Even to the date oj Stock's death,
theoreticians had no convincing explanation of the absence
of the prototype molecule, BH3, and their discussions of the
nature of the bonding in diborane were based upon an
assumed structure that was later shown to be incorrect.
Stock's amazing exploratory study went far beyond the
expectations and predictions of other inorganic chemists
of his day. This work, that culminated in his book, Hydrides
of Boron and Silicon, presaged the rapidly opening field
of"unusuar inorganic chemistry now so actively pursued.
Alfred E. Stock was always zealous in recognizing the
contributions and help of his coworkers and subordinates
during a period in which this was an uncommon virtue. He
was not only an outstanding scientist, but also a considerate
and thought Jul human being as revealed by his comment:
"The most important problem for the scientific mind to
solve will be how to free mankind from political, social,
and economic limitations and how to give it a purer,
broader-minded understanding of humanity. . . ."

CHAPTER

The Fourth-Row
Transition Elements

/
H

4
Be

5
B

8
0

10
Ne

15
At

15
P

18
Ar

Transition. elements ^

22
Ti

23
V

24
Cr

25
Mn

27
Co

28\ 29
Ni 1 C u

30
Zn

31
Go.

32
Ge

33
As

35
Br

36
Kr

38
Sr

39
Y

40
Zr

4b
Pd

47
A9

49
In

so
Sn

53
I

55
Cs

56
Ba

57-71

76
Ft

Ify

82
Pb

33
Bt

04
P0

56
Rn

87
Fr

88
Ra

89-

■ >

■ • •

In the preceding chapters we have studied the
chemistry of the elements across the top of the
periodic table and down the two sides. Now we
shall consider the elements in the middle. These
are usually referred to as the transition elements
because chemists once believed that some ele-
ments behaved in a way intermediate between
the extremes represented by the left and right

sides of the periodic table. Today, the term
"transition element" remains mostly a useful
way of designating elements in this particular
region of the periodic table, even though we can-
not pinpoint a specific set of properties and say
that all the transition elements have all these
properties.

22-1 DEFINITION OF TRANSITION ELEMENTS

There is some disagreement among chemists as
to just which elements should be called transi-
tion elements. For our purposes, it will be con-
venient to include all the elements in the columns
of the periodic table headed by scandium
through zinc.

Across the top, as the first row of the transi-

tion region, we have the elements scandium (Sc),
titanium (Ti), vanadium (V), chromium (Cr),
manganese (Mn), iron (Fe), cobalt (Co), nickel
(Ni), copper (Cu), and zinc (Zn). On the left, we
have the scandium column which includes, be-
sides Sc, yttrium (Y, 39), lanthanum (La, 57),
and actinium (Ac, 89). For reasons that we shall

387

Arbitrary

erterqy

scale.

ooo oqS©g oogS

ooo

4 t>

ooo

Z *

t'ii;. 22-1 . The fourth-row transition elements in the enersv level diagram.

SEC. 22-1 I DEFINITION OF TRANSITION ELEMENTS

389

take up in the next chapter, we group with
lanthanum the fourteen elements that follow La
(Z = 58 through Z = 71); these we call the
lanthanide elements. On the right, the transition
elements end with the zinc column. Besides zinc,
this includes cadmium (Cd, 48) and mercury
(Hg, 80). It is strongly advisable during the dis-
cussion that follows to look back at the periodic

Co.

Is
Is2

09 ®g® <g>

2s'

Js2

table frequently to see where each particular
element is placed.

22-1.1 Electron Configuration

There are two immediate questions we ask about
the transition elements once we know where they
are in the periodic table: (1) Why do we consider
these elements together? (2) What is special
about their properties? These questions are
closely related because they both depend upon
the electron configurations of the atoms. What,
then, is the electron configuration we might ex-
pect for these elements?

To answer this question, we need to review
some basic ideas on the electronic buildup of
atoms. We saw in Chapter 15 that as we progres-
sively add electrons to build up an atom, each
added electron goes into the lowest energy level

that is not already fully occupied. With this prin-
ciple as a guide, let us consider the electron
configurations as we build up the first row of
transition elements from scandium through zinc.
Looking at the periodic table, we see that calcium
comes just before scandium. The twenty elec-
trons in a calcium atom are distributed as shown
in the following arrangement:

J> 4s

+sz

OOOOOOCO o

In element number 21, we must accommodate
one more electron. At first sight we might predict
that the 21st electron goes into the 4p orbital, as
the next higher energy level after 45. The 4p
orbital is of higher energy than the 45 but, more
important, there is a set of five 3d orbitals in
between. The 21st electron goes into a 3d orbital
as the level of next higher energy. This is shown
in Figure 22-1 (which is just Figure 15-11 re-
produced here for convenient reference).

EXERCISE 22-1

Draw on one line a set of orbitals from Is
through Ad. Under this give the orbital occu-
pancy for Al, Sc, and Y. Account for the fact
that yttrium is much more like scandium than is
aluminum.

Table 22-1.

THE ELECTRON CONF

GURAT

ONS OF '

ELEM ENTS

ATOMIC

ELEMENT

SYMBOL

NUMBER, Z

scandium

titanium

vanadium

chromium

manganese

iron

cobalt

nickel

copper

zinc

ROW TRANSITION

ELECTRON CONFIGURATION

l*2 2s2 2pf> 3s1 3pfi

3d' 4s1

3d* 4s2

3d3 4s*

3d* 4s1

Each fourth-row transi-

3d* 4s2

tion element has these

3d* 4s*

levels filled.

3d1 4s1

3</Mj2

3dv>4s*

3du>4s*

390

THE FOURTH-ROW TRANSITION ELEMENTS I CHAP. 22

There are five 3d orbitals available, all more
or less of the same energy. Putting a pair of elec-
trons in each of these five orbitals means that a
total of ten electrons can be accommodated
before we need to go to a higher energy level.
Not only scandium but the nine following ele-
ments can be built up by adding electrons into
3d orbitals. Not until we get to gallium (element
number 31) do we go up to another set of
orbitals.

EXERCISE 22-2

Again using Figure 22-1, decide which orbital
would next be used after the five 3d orbitals have
been filled. What orbital would next be used
after the Ad set has been filled? What element
does this correspond to in the periodic table?

At this stage, with the help of Figure 22-1, or
with an atomic orbital chart, you should be able
to work out the electronic configuration of most
of the transition elements. You will not be able
to deduce them all correctly because there are
some exceptions resulting from special stabilities
when a set of orbitals is filled or half-filled. The
fourth-row transition elements have the set of
electron configurations shown in Table 22-1.
Notice that chromium (Z = 24) and copper
(Z = 29) provide interruptions to the continuous
buildup. In the case of chromium the whole atom
has lower energy if one of the 4s electrons moves
into the 3d set to give a half-filled set of 3d
orbitals and a half-filled 4s orbital; in the case
of copper, the atom has lower energy if the 3d
set is completely populated by ten electrons and
the 4s orbital is half-filled, instead of having nine
3d electrons and two in the 4* orbital.

EXERCISE 22-3

Make an electron configuration table like Table
22-1 for the fifth-row transition elements —
yttrium (Z = 39) through cadmium (Z = 48).
In elements 41 through 45, one of the 5s electrons
moves over to a 4d orbital. In element 46, two
electrons do this.

In the sixth-row transition elements (lan-
thanum through mercury) there is an additional
complication. There are seven 4/ orbitals which
are very close in energy to the 5d orbitals. Put-
ting electrons into these 4/ orbitals means there
will be fourteen additional elements in this row.
These fourteen elements are almost identical in
many chemical properties. We will discuss them
in the next chapter.

22-1.2 General Properties

What properties do we actually find for the
transition elements? What kinds of compounds
do they form? How can the properties be inter-
preted in terms of the electron populations of
the atoms?

Looking at a sample of each transition element
in the fourth row, we see that they are all
metallic. When clean, they are shiny and lus-
trous. They are good conductors of electricity
and also of heat : some of them (copper, silver,
gold) are quite outstanding in these respects. One
of them (mercury) is ordinarily a liquid; all
others are solids at room temperature.

So far as chemical reactivity is concerned, we
find a tremendous range. Some of the transition
elements are extremely unreactive. For example,
gold and platinum can be exposed to air or water
for ages without any change. Others, such as
iron, can be polished so they are brightly metallic
for a while, but on exposure to air and water they
slowly corrode. Still others are vigorously re-
active and, when exposed to air, produce a
shower of sparks. Lanthanum and cerium, for
instance, especially when finely divided, oxidize
immediately when exposed to air. (Some ciga-
rette lighters have flints containing these metals.)

It is hard to generalize about the chemical
reactivities of a group of elements since reactivi-
ties depend upon two factors: (A) the relative
stability of the specific compounds formed com-
pared with the reactants used up, and, (B) the
rate at which the reaction occurs. In special
cases there are other complications. For exam-
ple, chromium metal (familiar in the form of
chrome plate) is highly reactive toward oxygen.
Still, a highly polished piece of chromium holds

SEC. 22-1 I DEFINITION OF TRANSITION ELEMENTS

391

its luster almost indefinitely when exposed to air.
The explanation is that a very thin, invisible coat
of oxide quickly forms on the surface and pro-
tects the underlying metal from contact with the
oxygen in the air. In other words, bulk chromium
is unstable with respect to oxidation by air, but
the protective layer of oxide cuts the rate of
conversion so much that no reaction is observed.
What about compounds of the transition ele-
ments? Suppose we go into the chemical stock-
room and see what kinds of compounds are on
the shelf for a particular element, say chromium.
First we might find a bottle of green powder
labeled Cr203, chromic oxide, or chromium(III)
oxide. Next to it there would probably be a
bottle of a reddish powder, Cr03, chromium(VI)
oxide. On an amply stocked chemical shelf, we
might also find some black powder marked CrO,
chromous oxide, or chromium(II) oxide. There
would probably also be some other simple com-
pounds such as CrCl3, chromic chloride, or
chromium(III) chloride, a flaky, reddish-violet
solid, and maybe some green CrF2, chromous
fluoride or chromium(II) fluoride. Elsewhere in
the stockroom we would run across K2Cr04, a
bright yellow powder (potassium chromate),
probably next to a bottle of orange potassium
dichromate, K2Cr207. Soon we would get the
idea that the compounds of chromium, at least
the common ones, correspond to oxidation num-
ber of +2 (CrO and CrF2), +3 (Cr2Os and
CrCl3), and +6 (Cr03, K2Cr04, K2Cr207).

EXERCISE 22-4

What is the oxidation number of chromium in
each of the following compounds: Cr207-2,
Cr042, Cr(OH)3, Cr02Cl2?

Along with these simple compounds, we might
also find some rather more complex substances.
For example, we might find next to CrCl3
vials of several brightly colored solids labeled
CrCl3-6NH3, CrCl3 5NH3, CrCl3 4NH3, and
CrCl3-3NH3. Recalling that the dot in these for-
mulas simply indicates that a certain number of
moles of NH3 are bound to one mole of CrCl3,
we would conclude that here also the oxidation
number of chromium is +3. Looking further,
we might find other complex compounds such
as K3CrF6, Na3Cr(CN)6, KCr(S04)2 12H20. In
all these the chromium has a +3 oxidation num-
ber. As a result of our stockroom search, we
would form three conclusions: (1) chromium
forms simple and complex compounds; (2) chro-
mium forms a number of stable solids, most of
them colored; (3) chromium may have different
oxidation numbers, including +2, +3, and +6.
Similar conclusions would have resulted for most
of the other transition elements.

Is there any regularity to the kind of com-
pounds the fourth-row transition elements form?
Table 22-11 shows what chemists have found.

Table 22-11.

SYMBOL

TYPICAL OXIDATION NUMBERS FOUND FOR FOURTH-ROW
TRANSITION ELEMENTS

REPRESENTATIVE COMPOUNDS

COMMON OXIDATION NUMBERS

(most common in bold type)

NUMBER OF VALENCE
3d, 4s ELECTRONS

Sc,03

TiO,

Ti203.

TiO*

+ 2, +3, +4

vo,

v2o3,

vo2, v2o=

+ 2, +3, +4,

CrO,

Cr503,

CrO,

+ 2, +3

MnO

Mn:0.,

Mn02, KjMnCX, KMnQ4

+2, +3, +4

+6, +7

FeO.

Fe>03

+ 2. +3

CoO,

Co,03

+2, +3

NiO.

Ni.Q3

+2, +3

CU;0. CllO

+1, 42

ZnO

392

THE FOURTH-ROW TRANSITION ELEMENTS I CHAP. 22

EXERCISE 22-6

Look through a handbook of chemistry and find
one other compound of each oxidation state
given for the elements in Table 22-11.

Several points should be noted from this table:

(1) For most of the transition elements, several
oxidation numbers are possible.

(2) When several oxidation numbers are found
for the same element, they often differ from
each other by jumps of one unit. For exam-
ple, in the case of vanadium the common
oxidation numbers form a continuous series
from +2 to +3 to +4 to +5. Compare this
with the halogens (Chapter 19). In the case
of chlorine, for example, the common states
are —1, +1, +3, +5, and +7 (jumps of
two units instead of one unit).

(3) The maximum oxidation state observed for
the elements first increases and then de-
creases as we go across the transition row.
Thus we have +3 for scandium, +4 for
titanium, -f-5 for vanadium, +6 for chro-
mium, and +7 for manganese. The +7 rep-
resents the highest value observed for this
transition row. After manganese, the maxi-
mum value diminishes as we continue toward
the end of the transition row.

What explanation can we give for these ob-
servations? Why does the combining capacity
vary from one transition element to another in
such a way that the above pattern of oxidation
numbers develops? The combining capacity of
an atom depends upon how many electrons the
atom uses for bonding to other atoms. The
unique feature of the transition elements is that
they have several electrons in the outermost d
and s orbitals, and the ionization energies of all
of these electrons are relatively low. Therefore
it is possible for an element like vanadium to
form a series of compounds in which from two
to five of its electrons are either lost to, or shared
with, other elements. Consider, for example, the
oxides VO and V203, which contain the V+2 and
the V+3 ions, respectively. While more energy is
needed to form V+3 than to form V+2, the V+3
has, because of its higher charge, a greater attrac-
tion for the O-2 ion than does V+2. This extra
attraction in V203 compensates for the energy
needed to form the V+3 ion, and both oxides (as
well as V02 and V205) are stable compounds.
Notice, moreover, that the maximum oxidation
number of the transition elements never exceeds
the total number of s and d valence electrons.
The higher oxidation states become increasingly
more difficult to form as we proceed along a row,
because the ionization energy of the d and 5 elec-
trons increases with the atomic number.

22-2 COMPLEX IONS

The remaining general point to be made about
the transition elements is that they form a great
variety of complex ions in which other molecules
or ions are bonded to the central transition ele-
ment ion to form more complex units. These are
called complex ions. Take the series already men-
tioned: CrCl3-6NH3, CrCl3-5NH3, CrCl3-4NH3,
and CrCl3-3NH3. How can we account for the
existence of such a series? The answer can be
seen if we consider some of the observed facts
about these complex compounds. For example,
if we dissolve one mole of each in water and add

a solution of silver nitrate in an attempt to
precipitate the chloride as AgCl,

Ag+ + CI" zf±: AgClfsJ (2)

we find that sometimes much of the chloride
cannot be precipitated. The observed results are:

Moles of CI

Moles of Cl~

Compound

precipitated
3 of 3

not precipitated

CrCl3-6NH3

CrCl3-5NH3

2 of 3

1 of 3

CrCl3-4NH3

1 of 3

2 of 3

CrCl3-3NH3

3 of 3

SEC. 22-2 I COMPLEX IONS

393

Evidently, there are two ways in which chlorine
is bound in these compounds, one way which
allows the Cl~ to be precipitated by Ag+ and
another way which does not. In CrCl3-6NH3,
all of the chloride can be precipitated; in
CrCl3 3NH3, none of it can be. Other data also
indicate different types of bonding. For instance,
the freezing point lowering of an aqueous solu-
tion of CrCl3 6NH3 indicates there are four
moles of particles per mole of CrCl3-6NH3; the
solution is highly conducting.* On the other
hand,. the freezing point lowering of CrCl3-3NH3
solution indicates there is one mole of particles
per mole of CrCl3-3NH3; furthermore, the solu-
tion does not conduct at all. The explanation of
this behavior was provided in the early 1900's
by Alfred Werner, who noted that complex com-
pounds of chromium +3 could be accounted
for by assuming each chromium is bonded to
six near neighbors. In CrCl3-6NH3, the cation
consists of a central Cr+3 surrounded by 6 NH3
molecules at the corners of an octahedron;
the three chlorine atoms exist as anions, CI-.
In CrCl3-5NH3, the cation consists of the
central chromium surrounded by the five NH3
and one of the CI atoms; the other two CI
atoms are anions. In CrCl3-4NH3, the chro-
mium Is bound to four NH3 and two CI leaving
one chloride anion. In CrCl3 3NH3, all three
CI atoms and all three NH3 molecules are
tied to the central chromium. The formulas
can be written

[Cr(NH3)6]Cl3
[Cr(NH3)5Cl]Cl2
[Cr(NH3)4Cl2]Cl
[Cr(NH3)3Cl3]

22-2.1 Geometry off Complex Ions

The way that atoms or molecules are arranged
in space around a central atom has a great in-
fluence on whether a given complex aggregate
is stable enough to be observed. What kinds of
arrangements are found in complex ions? What

* From work on simple salts such as NaCl we expect
that the "particles" are ions, and the conductivity con-
firms this.

shapes do these complex ions show? Can we find
any regularity in the transition elements that will
enable us to predict what complex ions will
form?

First, let us introduce a concept useful in
giving spatial descriptions; the coordination
number is the number of near neighbors that
an atom has. For example, in the complex ion
A1F6""3 (which is the anion present in the solid
mineral cryolite), each Al atom is surrounded by
six fluorine atoms at the corners of an octa-
hedron, as shown in Figure 22-2. We say that
aluminum has a coordination number of 6 with
fluorine. In the complex ion AlBr4~, which seems
to be an important intermediate when aluminum
bromide acts as a catalyst for many organic re-
actions, the bromine atoms are arranged around
a central Al at the corners of a regular tetra-
hedron. Figure 22-3 shows the arrangement. The
coordination number of aluminum is 4 with
bromine.

If more than simple atoms are bound to a
central atom, then the coordination number still
refers to the number of near neighbors. For
example, in solid potassium chrome alum,
KCr(S04)2- 12H20, and also in its fresh aqueous
solutions, the chromium-containing cation is

Fig. 22-2. An octahedral complex: aluminum with
coordination number 6.

Ion with
3 charge.

394

THE FOURTH-ROW TRANSITION ELEMENTS I CHAP. 22

Jon -with.
-1 charge

Fig. 22-3. A tetrahedral complex: aluminum with co-
ordination number 4.

Cr(H20)^3. It consists of a central chromium
joined to six H20 molecules, exactly as the fluo-
rines are arranged around aluminum in Figure
22-2. The oxygen portion of each H20 molecule
is turned toward the central chromium and the
H portions point away from the center. The
corners of the cage that holds the central Cr
atom are occupied by six oxygen atoms, each of

Ion with +1 charge

which also holds two H atoms. The shape of
this complex ion is octahedral and we say that
in Cr(H20)t+3 chromium shows a coordination
number of 6 to oxygen.

Notice that in an octahedral complex ion such
as [Cr(NH3).iCl2]+ there is a possibility of ob-
serving isomers. The two chlorine atoms may
occupy octahedral positions which are next to
each other on the same side of the metal atom,
or positions located on opposite sides of the
metal atom (see Figure 22-4). The isomer in
which the two similar groups are located on the
same side of the metal atom is called the cis-
isomer, and the other is called the /ra/is-isomer.

The complex ion Fe(C204)3~3 is formed when
rust stains are bleached out with oxalic acid
solution. It also has a transition element showing
coordination number of 6, even though there are
only three groups (QAf2 groups) around each
iron ion. Figure 22-5 shows the arrangement.
Each C204"2, the oxalate group, uses two of its
oxygen atoms to bind onto the central iron atom.
The number of near neighbors, as viewed from
the iron atom, is six oxygen atoms at the corners
of an octahedron. Picturesquely, a group such as

Fig. 22-4. The isomers of [CV(A///S)4C72]\

Ion -with +1 charge

Cis. isomer

Tr,

ans- isomer

SEC. 22-2 I COMPLEX IONS

395

Fig. 22-5. The structure of Fe(C£>t)»-

oxalate, which can furnish simultaneously two
atoms for coordination, is said to be bidentate,
which literally means double-toothed.

In addition to the tetrahedral and octahedral
complexes mentioned above, there are two other
types commonly found— the square planar and
the linear. In the square planar complexes, the
central atom has four near neighbors at the
corners of a square. The coordination number
is 4, the same number as in the tetrahedral com-
plexes. An example of a square planar complex
is the complex nickel cyanide anion, Ni(CN)^2.

In a linear complex, the coordination number
is 2, corresponding to one group on each side of
the central atom. The silver-ammonia complex,
which generally forms when a very slightly solu-
ble silver salt such as silver chloride dissolves in
aqueous ammonia, is an example, as shown in
Figure 22-6. Another example of a linear com-

Fig. 22-6. A linear complex, Ag{NH3)a\

Ion yvt'th +1 charge

plex is Ag(CN)2", which is formed during the
leaching of silver ores with NaCN solution.

22-2.2 Bonding in Complex Ions

What holds the atoms of a complex ion together?
There are two possibilities. In some complexes,
as in A1F6"3, the major contribution to the bond-
ing comes from the attraction between a positive
ion (Al+3) and a negative ion (F~). The bonding
is ionic. In other complexes, as in Fe(CN)6~3,
there is thought to be substantial sharing of elec-
trons between the central atom and the attached
groups. The bonding is mainly covalent. When
there is such sharing, an electron or an electron
pair from the attached group spends part of its
time in an orbital furnished by the central atom.
In either type, as emphasized in Chapter 16, the
electron is attracted to both atoms in the bond.

For transition elements there are usually
empty d orbitals ready to accommodate elec-
trons from attached groups. This is by no means
always necessary, as witness the case of Zn+2, a
good complex-former even though all its 3d
orbitals are already occupied. Any vacant or-
bital, low enough in energy to be populated, will
serve as a means whereby complex formation
can be accomplished.

The geometry of a complex ion often can be
explained quite reasonably in terms of the or-
bitals of the central atom populated by electrons
from the attached groups. If only one s and one
p orbital is used, the bonding is called sp bond-
ing. We have already seen in Section 16-4.5 that
this bonding situation gives rise to a linear ar-
rangement. Therefore this might explain why
some complexes are linear, as is Ag(NH3)2+. If
one 5 and three p orbitals are used, the complex
uses sp3 bonding. Then a tetrahedral complex
can be expected, as observed for Zn(NH3)4+2.
When d orbitals are involved, other geometries
can be explained (for example, square planar,
dsp2; octahedral, cPsp3).

22-2.3 Significance of Complex Ions

Besides their occurrence in solid compounds,
complex ions such as we have mentioned are

396

THE FOURTH-ROW TRANSITION ELEMENTS I CHAP. 22

important for two other reasons: (1) they may
decide what species are present in aqueous solu-
tions; and (2) some of them are exceedingly im-
portant in biological processes.

As an example of the problem of species in
solution, consider the case of a solution made
by dissolving some potassium chrome alum,
KCr(SO,)r 12H20, in water. On testing, the solu-
tion is distinctly acidic. A currently accepted
explanation of the observed acidity is based upon
the assumption that, in water solution, chromic
ion is associated with six H20 molecules in the
complex ion, Cr(H20)c+3. This complex ion can
act as a weak acid, dissociating to give a proton
(or hydronium ion). Schematically, the dissocia-
tion can be represented as the transfer of a pro-
ton from one water molecule in the Cr(H20)^3
complex to a neighboring H20 to form a hy-
dronium ion, H30+. Note that removal of a
proton from an H20 bound to a Cr+3 leaves an
OH- group at that position. The reaction is
reversible and comes to equilibrium:

Cr(H,0)+3 + HO +±: Cr(HJ0)5OH+- + H30+ (3)

We see that Cr(H2O)0+3 acts as a proton-donor,
that is, an acid.

22-2.4 Amphoteric Complexes

Another reason chemists find the above complex
ion picture of aqueous solutions useful is that
it is easily extended to explain amphoteric be-
havior. Take the case of chromium hydroxide,
Cr(OH)3, a good example of an amphoteric hy-
droxide. It dissolves very little in water, but is
quite soluble both in acid and in base. Presum-
ably it can react with either. How can this
behavior be explained in terms of the complex
ion picture?

First, consider the equilibrium represented by
equation (3) when NaOH is added to solution.
Added OH- combines with the H30+ to form
H20. This removes one of the species shown on
the right side of the equation, so formation of
Cr(H20)5OH+2 is favored. In other words, as
OH- is added to Cr(H2O)0+3 the reaction is
favored which corresponds to pulling a proton
off Cr(H20)+3.

What happens when enough NaOH has
been added to remove three protons from
each Cr(H20)G+3? Removal of three protons
leaves the neutral species Cr(H20)3(OH)3, or
Cr(OH)3-3H20. This neutral species has no
charges to repel other molecules of its own kind
so it precipitates. However, as more NaOH is
added to this solid phase, one more proton can
be removed to produce Cr(H20)2(OH)4~ ; and the
Cr(OH)3-3H20 dissolves. [In principle, more
protons could be removed, perhaps eventually to
form Cr(OH)6"3, but there is as yet no evidence
for this.]

The following equations summarize the steps
believed to occur when NaOH is slowly added
to a solution of chromic ion. Step (4c) corre-
sponds to formation of solid hydrated chromium
hydroxide; step (4cl) corresponds to its dissolving
in excess NaOH.

Cr(H20)G+3 + OH- +±:

Cr(H20)5OH+2 + H20 (4a)
Cr(H20)5OH+2 + OH" z^±:

Cr(H20),(OH)2+ + H.O (4b)
Cr(H20)J(OH)2+ + OH- ^=t

Cr(H20)3(OH)3(sJ + HO (4c)
Cr(H20)3(OH)3(sJ + OH- +±l

Cr(H ,0)2(OH)4- + H20 (4d)

When acid is added to a solution such as in
equation (4cf), the above set of reactions is
progressively reversed, first causing precipitation
of chromium hydroxide by the reverse of reac-
tion (4d) and then its subsequent dissolving by
the reverse of reaction (4c).

22-2.5 Complexes Found in Nature

Complex ions have important roles in certain
physiological processes of plant and animal
growth. Two such complexes are hemin, a part
of hemoglobin, the red pigment in the red cor-
puscles of the blood, and chlorophyll, the green
coloring material in plants. The first of these,
hemoglobin, contains iron and properly fits in a
discussion of complex compounds of the transi-
tion elements; the second, chlorophyll, is a com-
plex compound of magnesium. Magnesium is

SEC. 22-2 I COMPLEX IONS

397

not a transition element, but chlorophyll is dis-
cussed here because it has some features in
common with hemoglobin and because it avoids
the impression that only transition elements form
complexes.

Chlorophyll, as extracted from plants, is actu-
ally made up of two closely related compounds,
chlorophyll A and chlorophyll B. These differ
slightly in molecular structure and can be sepa-
rated because they have different tendencies to
be adsorbed on a finely divided solid (such as
powdered sugar).

five-membered ring. Consider also the vast
amount of knowledge and experimentation that
are summed up in the statement that the struc-
ture of this complicated molecule is known.

EXERCISE 22-7

If a typical plant leaf yields 49.0 mg of chloro-
phyll A, how many milligrams of this will be
magnesium? The molecular weight of chloro-
phyll A is 893.

EXERCISE 22-6

If you wish to prepare some chlorophyll, grind
up some fresh leaves and extract with alcohol.
The alcohol dissolves the chlorophyll, as shown
by the solution color.

To show the complexity of this biologically
important material, the structural formula of
chlorophyll A is shown on the left in Figure 22-7.
You need not memorize it. The most obvious
thing to note is that it is a large organic molecule
with a magnesium atom in the center. Right
around the magnesium atom there are four near-
neighbor N atoms, each of which is part of a

Hemin is shown on the right in Figure 22-7.
It is shown beside the model of chlorophyll A to
emphasize the astonishing similarity. The por-
tions within dotted lines identify the diiferences.
Except for the central metal atom, the differences
are all on the periphery of these cumbersome
molecules. We cannot help wondering how na-
ture managed to standardize on this molecular
skeleton for molecules with such different func-
tions. We cannot avoid a feeling of impatience
as we await the clarification of the possible rela-
tionship, a clarification that will surely be pro-
vided by scientists of the next generation.

Fig. 22-7. The structures of chlorophyll A and hemin.

Chlorophyll A

Hemin

398

THE FOURTH-ROW TRANSITION ELEMENTS ' CHAP. 22

The most important function of hemoglobin
in the blood is that of carrying oxygen from the
lungs to the tissue cells. This is done through a
complex between the iron atom of the hemin
part and an oxygen molecule. Just how the 02
is bound to the hemin is not yet clear, but it must
be a rather loose combination since the 02 is
readily released to the cells. The complex is
bright red, the characteristic color of arterial
blood. When the 02 is stripped off the hemin
group, the color changes to a purplish red, the
color of blood in the veins.

Not only 02 molecules but also other groups
can be bound to the iron atom of hemoglobin.
Specifically, carbon monoxide molecules can be
so attached and, in fact, CO is more firmly
bound to hemoglobin than is 02. This is one
detail of the carbon monoxide poisoning mecha-
nism. If we breathe a mixture of CO and 02
molecules, the CO molecules are preferentially
picked up by the red blood cells. Since the sites

normally used to carry 02 molecules are thus
filled by the CO molecules, the tissue cells starve
for lack of oxygen. If caught in time, carbon
monoxide poisoning can be treated by raising
the ratio of 02 to CO in the lungs (in other words,
administering fresh air or oxygen). The two re-
actions,

02(g) + hemoglobin -<->- complex! (5a)
CO(g) + hemoglobin -<->- complex2 (5b)

have sufficiently similar tendencies to go to the
right that the first reaction can be made to exceed
the second if the concentration of 02 sufficiently
exceeds that of CO. Another remedial measure
is to inject methylene blue directly into the blood
stream. Carbon monoxide bonds more strongly
to methylene blue than to hemin. Equilibrium
conditions then favor the transfer of CO to the
methylene blue, thus freeing hemoglobin for its
normal oxygen transport function.

22-3 SPECIFIC PROPERTIES OF FOURTH-ROW TRANSITION ELEMENTS

The preceding discussion of the transition ele-
ments has been quite general, with the implica-
tion of rather wide applicability. Now we turn
to a consideration of the transition elements and
their compounds as specific individuals.

Table 22-111 collects some of the data ordi-
narily found useful for the transition elements
of the fourth row of the periodic table. The
following are some notes on regularities ob-
served.

The atomic weight increases regularly across
the row except for the inversion at cobalt and
nickel. We would expect the atomic weight of
Ni to be higher than that of Co because there
are more protons (28) in the Ni nucleus than in
the Co nucleus (27). The reason for the inversion
lies in the distribution of naturally occurring
isotopes. Natural cobalt consists entirely of the
isotope 27C0; natural nickel consists primarily of
the isotopes ^Ni and asNi, the 58-isotope being
about three times as abundant as the 60-isotope.

Abundance in the earth's crust. With the excep-

tion of iron, which is very abundant, and tita-
nium, which is moderately abundant, all the
other elements of the first transition row are
relatively scarce. However, some of them, such
as copper, are quite familiar. Copper is one of
the few metallic elements found free in nature.
The existence of deposits of metallic copper
undoubtedly accounts for the fact that man
evolved through the Bronze Age before the Iron
Age. Copper, the essential ingredient of bronze,
did not require the difficult smelting process
needed for iron.

Melting point. Except for zinc at the end of
the row, the melting points are quite high. This
is appropriate, since these elements have a large
number of valence electrons and also a large
number of vacant valence orbitals. Toward the
end of the row, in zinc, the 3d orbitals become
filled and the melting point drops.

Density. There is a steady increase in density
through this row, with some leveling off toward
the right. This trend is closely tied to the almost

SBC. 22-3 I SPECIFIC PROPERTIES OF FOURTH-ROW TRANSITION ELEMENTS

399

constant size of the atoms so the main effect
producing density change is the increasing nu-
clear mass.

Ionization energy. As ionization energies go,
the values found for the transition elements are
neither very high nor very low. They are all
rather similar in magnitude. The sequential in-
crease in nuclear charge, which would tend to
increase the ionization energy, seems to be al-
most offset by the extra screening of the nucleus
provided by the added electrons.

Ionic radius. Ionic radii do not change much
in going across a transition row. The reason for
this is essentially a balance of two effects: (1) As
nuclear charge increases across the row, the
electrons would be pulled in, so the ions ought

to shrink. (2) As more 3d electrons are added,
these electrons repel each other and the ions
ought to swell. These effects just about cancel.
As expected, the size of the +3 ion is smaller
than the size of the +2 ion of that same element.
Keeping nuclear charge constant, removal of one
additional 3d electron would reduce the repul-
sion between the 3d electrons remaining, thus
allowing them all to be pulled closer to the
nucleus.

Color. Many solid compounds of the transi-
tion metals and their aqueous solutions are
colored. This color indicates light is absorbed in

Fig. 22-8. Atomic sizes of the transition elements.

2.5

z.o

1.5

* 1.0

0.5 -

Ck---

Melral

-o o^

19K 20Ca 2lSc 22Ti 23V 24Cr 25Mn 26Fe 27Co 23™ 29Cu 30Zn 31&a

Etlemenrts

400

THE FOURTH-ROW TRANSITION ELEMENTS ! CHAP. 22

the visible part of the spectral region. The energy
levels that account for this absorption are rela-
tively close together and involve unoccupied d
orbitals. The environment of the ion changes the
spacing of these levels, thereby influencing the
color. A familiar example is the Cu+2(aq) ion,
which changes from a light blue to a deep blue
when NH3 is added. The formation of the am-
monia complex alters the energy level spacing
of the central Cu+2 ion to produce the color
change.

E°. The last row of the table gives the values
of the oxidation tendencies for these metals.
Except for scandium (which goes to a +3 state),
the values quoted correspond to the reaction

M(s)

U+Haq) + 2e~

(6)

As can be seen from the table, all the elements
except copper have positive values, which means
these metals are more easily oxidized than is
hydrogen gas, for which E° is zero. Thus, man-
ganese metal should dissolve in acid to liberate
hydrogen gas. The E° for the overall reaction,

Mn(s) + 2H+(aq) +± Un^(aq) + H2(g) (7)

is +1-18 volts, so the reaction should proceed
spontaneously to the right. (Note that this is an
equilibrium consideration and it tells nothing

about the rate. The rate may be slow.) For cop-
per the reaction

Cu(s) + 2H+(aq) ^=fc Cu^(aq) + H2(g) (8)

has a negative E° ( — 0.34 volt), so reaction to
the right is not expected.

Zinc, at the end of the row, has E° (+0.76),
which is intermediate between the values at the
beginning of the row and those toward the end.
We predict, with this value of E°, that Zn should
reduce Fe+2, Co+2, Ni+2, and Cu+2 to the corre-
sponding metals but should not be able to reduce
Sc+3, Ti+2, V+2, Cr+2, or Mn+2 to the metals.

22-3.1 Scandium

Scandium has not yet been available in large
enough amounts to have it develop interesting or
important uses. Neither has it been available for
much experimental work, so there remains much
to be learned about this element.

22-3.2 Titanium

There is intense interest in titanium. This interest
stems from an unusual combination of desirable
properties in one metal. It is strong; it has low
density; and it is remarkably resistant to corro-

Table 22-11 I. some properties of fourth-row transition elements

element Sc Ti V Cr Mn Fe Co Ni Cu Zn

Atomic number

Atomic weight

45.0

47.9

51.0

52.0

54.9

55.9

58.9

58.7

63.5

65.4

Abundance*

(%bywt.)

0.005

0.44

0.015

0.020

0.10

5.0

0.0023

0.008

0.0007

0.01

Melting point (°C)

1400

1812

1730

1900

1244

1535

1493

1455

1083

419

Boiling point (°C)

3900

3130f

3530f

2480|

2087

2800

3520

2800

2582

907

Density (g cm3)

2.4

4.5

6.0

7.1

7.2

7.9

8.9

7.1

First ioniz. energy

(kcal/mole)

154

157

155

171

180

175

176

216

+2 ion radius (A)

—

0.90

0.88

0.84

0.80

0.76

0.74

0.72

0.74

+ 3 ion radius (A)

0.81

0.76

0.74

0.69

066

0.64

0.63

0.62

—

E° (volt)

M — >- M+J + 2e~

2.1**

1.6

1.2

0.90

1.18

0.44

0.28

0.25

-0.34

0.76

* In the earth's crust.

t Estimated

** M — >- M+3 + 3e~

SEC. 22-3 | SPECIFIC PROPERTIES OF FOURTH-ROW TRANSITION ELEMENTS

401

sion. The difficulty has been to find an economi-
cal way of getting it out of its natural minerals:
rutile, Ti02, and ilmenite, FeTi03. This was
solved in part by heating Ti02 in chlorine gas to
convert it to TiCU and then reducing the TiCU
with magnesium metal. Two problems still stand
in the way of large-scale use of this rather abun-
dant element. One is the great sensitivity of its
properties to the presence of trace impurities
(especially H, O, C, and N); the other is the
difficulty of forming it into useful shapes.

22-3.3 Vanadium

This element is important mainly because of its
use as an additive to iron in the manufacture of
steel. A few percent of vanadium stabilizes a
high-temperature crystal structure of iron so that
it persists at room temperature. This form is
tougher, stronger, and more resistant to corro-
sion than ordinary iron. Automobile springs, for
example, are often made of vanadium steel.

Also important is V205, divanadium pentox-
ide, an orange powder which is used as a catalyst
for many reactions of commercial significance.
For example, in the manufacture of sulfuric acid,
V205 catalyzes the step in which S02 is oxidized
to S03. How it works is still in dispute, but the
general belief is that the catalytic action is de-
pendent upon the ability of vanadium to show
various oxidation states. One suggested mecha-
nism is that the solid V20B absorbs an S02 mole-
cule on the surface, gives it an oxygen atom to
convert it to S03, and is itself reduced to V204,
divanadium tetroxide. The V204 in turn is re-
stored to V206 by reaction with oxygen. Catalytic
reactions, especially those involving solid-gas
interfaces, are not very well understood at the
present time.

some reagents — chlorine, for instance. In air,
however, it is inactive, probably because of for-
mation of an impervious oxide coat. Other
metals, such as the alkali and alkaline earth
metals, also form oxide coats but they are not
very effective in protecting the underlying metal
from atmospheric oxidation. The main difference
is that when chromium is converted to oxide
there is a swelling in this oxide layer that arises
from the increase in volume per chromium atom.
This gives a nonporous surface coat of oxide.
On the other hand, when a metal such as calcium
is oxidized, the oxide layer has a smaller volume
per Ca atom than the metal itself. The result is
that the surface layer shrinks, tending to crack
and open up fissures through which oxygen (and
water vapor) can reach the underlying metal.
Many of the transition elements show the kind
of self-protective action found in chromium.

Most of the chromium we see is only a thin
coating on iron or other metals. Such a coating
called chrome plate, is put on in an electrolysis
cell in which the object to be plated is the
cathode of the cell. The essential ingredients of
the plating bath are Cr03, chrorruum(VI) oxide,
and either H2S04 or Cr2(S04)3, chromium(III)
sulfate, but there are various additives, including
such unlikely substances as glue or milk, which
are supposed to give better coatings. Pure bulk
chromium metal is fairly difficult to make. It
can be done by using a reaction called the
Goldschmidt reaction in which aluminum metal
is used as a reducing agent. So much heat is re-
leased when A1203 is formed from the elements
that stable oxides, as for example Cr203, can be
reduced by Al. A mixture of aluminum powder
and Cr203, when ignited, gives a vigorous reac-
tion to produce A1203 and chromium.

22-3.4 Chromium

Interest in this metal comes from its remarkable
inertness to atmospheric corrosion. Also, it is
very hard and thus it forms an ideal protective
coating. On the basis of its E° (1.18 volts higher
than hydrogen) we expect chromium to be quite
reactive; in fact, it is vigorously reactive with

EXERCISE 22-8

Write the equation for the reduction of Cr203
by Al. If it takes 399 kcal/mole to decompose
A1203 into the elements and 270 kcal/mole to
decompose Cr203, what will be the net heat
liberated in the reaction you have just written?

402

THE FOURTH-ROW TRANSITION ELEMENTS I CHAP. 22

Metallic chromium is an ingredient of several
important alloys. Some forms of stainless steel,
for example, contain about 12% Cr. Nichrome,
which is commonly used for heating coils, has
about 15% Cr in addition to 60% Ni and 25%
iron. Both these alloys are quite resistant to
chemical oxidation.

In compounds, the important oxidation num-
bers of Cr are +2, +3, and +6. In all of these
states the chromium ions are colored and, in
fact, the element got its name from this property
(chroma is the Greek word for color). The +2
state is not frequently encountered but it can be
made quite easily as the beautiful blue chromous
ion in solution by dripping a solution containing
Cr* over metallic zinc. Air has to be excluded
since 02 rapidly converts Cr1"2 back into Cr*.

The +3 state of chromium is best represented
by chromium(III) oxide, Cr203, which is a green,
inert solid used as a green pigment. It can be
made in rather spectacular fashion by heating
ammonium dichromate. Once started, the re-
action

(mU)2CT2Ch(s) ++

N2teJ + 4H2Ofg; + Cr203(s) (9)

keeps itself going. The nitrogen and water are
formed as hot gases which blow the light, fluffy
Cr203 about. Another way of getting Cr203 is
by dehydrating "chromium trihydroxide" '
heat:

Ion yvith
-2 charge

with

2Cr(OH)3fsJ *=£ Cr£>3(s) + m£>(g) (70)

There is much argument about how to write an
appropriate formula for "chromium trihydrox-
ide." A gelatinous, green precipitate does form
when base is added to a solution containing
chromic ion (Cr1"3), but it includes a great deal
of excess H20, so pure Cr(OH)3 is never ob-
tained. This is a common problem with the
transition elements. Their hydroxides are not
well-characterized, mainly because of the con-
siderable difficulty of distinguishing between a
water molecule of hydration and an OH group.
For example, many chemists maintain that the
formula is really Cr203nH20, and that n usually
is found to equal 3. This agrees with the empirical
formula Cr(OH),.

EXERCISE 22-9

Calculate the percent chromium in Cr(OH)3 and
inCr203-3H20.

Whatever that green precipitate has for its
chemical formula, it is observed to be ampho-
teric. It dissolves both in excess acid and in
excess base, as explained earlier.

Fig. 22-9. The structures of chromate and dichromate
ions.

chrornaire tori

dichromate ion

;ec. 22-3 I SPECIFIC PROPERTIES of fourth-row transition elements

403

Probably the most common compound of
+ 3 chromium is potassium chrome alum,
KCr(S0,)212H20. We know that the twelve
water molecules are distributed equally, six
around Cr+3 and six around K+. Potassium
chrome alum is just one example of the general
class of solids called alums which have a +1
ion, a +3 ion, two sulfates, and twelve molecules
of water. In the dyeing industry chrome alum
is used for fixing dyes to fabrics.

The +6 state of chromium is represented by
the chromates and dichromates. The chromate
ion is a tetrahedral ion with Cr at the center;
dichromate ion may be visualized as two such
tetrahedra having one oxygen corner in common.
Figure 22-9 shows the arrangements. Chromates
can easily be converted to dichromates by addi-
tion of acid,

2&0;2(aq) + 2H+(aq)

Cr207-2(aq) + H20 (//)

The change can be followed by noting the color
change from yellow (characteristic of chromate)
to orange (characteristic of dichromate). The
reverse conversion from dichromate to chromate
occurs on addition of base.

Both chromates and dichromates are strong
oxidizing agents. One example of their use is in
"cleaning solution," a mixture of K2Cr207 and
concentrated sulfuric acid. Laboratory glassware
can be thoroughly cleaned of grease films by
immersion in cleaning solution.

22-3.5 Manganese

The major use of manganese is in the production
of steel, during which manganese reacts with
oxygen and keeps the gas from forming bubbles
when the iron solidifies. This prevents the forma-
tion of structure-weakening holes in finished
steel. Manganese usually occurs in nature as a
mixture of oxides along with oxides of iron. The
ore, without separation, can be reduced with
carbon in a high temperature furnace to give
alloys of iron and manganese. This so-called
"ferromanganese" is added to the crude molten
iron on its way to becoming steel.
Probably the most commonly encountered

compound of Mn is manganese dioxide, Mn02,
used in the ordinary flashlight cell. What goes on
in a dry cell when it generates electricity is very
much in dispute. The reactions are complicated
and apparently change character depending upon
the amount of electric current being drawn
from the cell. When small currents are involved,
which is what dry cells are designed for, the
reactions are believed to be as follows:

AT THE ANODE

(the terminal in-
dicated as the
negative pole on
commercial cells)

Zn(s)
AT THE CATHODE

the zinc container is oxidized
from the metallic state to the
+2 state, probably as some
complex zinc ion, but writ-
ten for simplicity as Zn+2

Zn+2 + 2e-

U2)

the Mn02 picks up, in a
complicated reaction, an
electron which reduces the
manganese from the +4 to
+3 state in the presence of
NH.Cl

2Mn02 + 2NH4+ + 2e~ — >■

Mn203 + 2NH3 + H20

(13)

The role of the other components in the cell is
not completely understood. Some of these com-
ponents (such as NH4C1, ammonium chloride,
and ZnCl2, zinc chloride, in the center paste) are
involved in other reactions that come into play
at larger current drain.

One other important compound of manganese
is potassium permanganate, KMn04. This is an
intensely violet-colored material much used as
an oxidizing agent in the laboratory. (It is too
expensive to use on a large scale; in industry,
chlorine is more likely to be used.)

22-3.6 Iron

Iron is the workhorse of the metals. It is quite
abundant (ranking fourth of all elements and sec-
ond of the metals, by weight) and easy to make
inexpensively on a large scale; and it has useful
mechanical properties, especially when alloyed
with other elements. Steel, one of our most use-
ful construction materials, is essentially iron

404

THE FOURTH-ROW TRANSITION ELEMENTS I CHAP. 22

containing a small percentage of carbon and
often small amount of other elements.

NATURAL OCCURRENCE OF IRON

Most of the accessible iron is combined either
with oxygen or sulfur. The oxygen compounds
are the common minerals hematite, Fe203, and
magnetite, Fe304, both of which are useful raw
materials for producing iron. Another mineral,
FeS2, is called iron pyrites or "fool's gold," but
this is not a common source since removing all
the sulfur is difficult. (Sulfur impurity in steel
makes it brittle. The sulfur compounds of iron
are low melting and on cooling they stay liquid
longer than does the mass itself and keep the
iron from compacting efficiently.) Iron exists in
the elemental form. in some meteorites. Since
meteorites are believed to come from the
break-up of a planet, the existence of iron me-
teorites is taken as support for the theory that
the core of the earth is largely iron.

MANUFACTURE OF IRON

The production of iron is an excellent example
of chemical reduction on a massive scale. The
process is carried out in a huge vertical reactor,
called a blast furnace, which may be several
stories tall. Raw materials — iron ore, limestone,
and coke — are fed into the top of the furnace
and oxygen is blown in at the bottom. The pur-
pose of the iron ore (let us assume Fe203) is to
provide the iron. The limestone reacts with sand,
Si02, in the iron ore removing it as molten
calcium silicate, called slag. The coke supplies
the reducing agent, carbon, and as it reacts the
heat released maintains the required high tem-
perature.

Here is a simplified version of what goes on in
a blast furnace. As the mixture of ore, limestone,
and coke falls through the furance, it meets the
updraft of oxygen. Carbon monoxide is formed,

2C(s) + 02(g) — >- 2CO(g) + 52.8 kcal (14)

and as this carbon monoxide rises through the
furnace it progressively reduces the Fe203 — first
to Fe304, then to FeO, and eventually to Fe. The
successive reactions take place progressively as
the solid descends:

CO( g) + 3Fe..(VsJ
COte) + Fe304(sj

CO(g) + FeOfsJ

2Fe30/sj + CO-,(g) (15)
3FeO(s) + C02(g) (16)
Fe(l) + CQ2(g) (17)

Since the reactions (15), (16), and (17) require
successively higher temperatures, the blast fur-
nace temperature is kept highest near the bottom
of the furnace. Near the bottom, the temperature
is sufficiently high that the impure iron — satu-
rated with carbon— collects there as a molten
liquid. The slag, which is mainly calcium silicate,
CaSi03, removes any sand in the ore through
reaction with limestone, CaC03.

CaC03(sJ — ►- CaOfsJ + C02(g) (18)

CaO(s) + SiO/sJ — ►■ CaSi03(l) (19)

Molten CaSi03 is less dense than molten iron
and floats on top of it. An average furnace that
produces about 750 tons of iron per day, will
also yield 410 tons of slag. The slag is sometimes
useful in the manufacture of cement and, when
it contains sufficient phosphorus, in the manu-
facture of fertilizer.

When this impure iron cools, the resulting
solid is called pig iron or cast iron. It is quite
brittle and is not useful if high strength is needed.
The impure iron is made into steel by burning
out most of the carbon, sulfur, and phosphorus.
Today there are three common furnace types for
making steel — the open-hearth furnace (85 % of
U.S. production), the electric arc furnace (10%),
and the Bessemer converter (5 %). These furnaces
differ in construction but the chemistry is basi-
cally similar.

The process of burning out the impurities is
slowest in the open-hearth furnace. This implies
there is plenty of time to analyze the melt and
add whatever is needed to obtain the desired
chemical composition. Manganese, vanadium,
and chromium are frequent additives. The prop-
erties of the finished steel depend upon the
amount of carbon left in and upon the identity
and the quantity of other added elements. Soft
steel, for example, contains 0.08-0.18 weight per-
cent carbon; structural steel, 0.15-0.25%; hard
steel or tool steel, 1-1.2%.

The electric arc furnace is used for special
purpose steels. Because the environment can be

SEC. 22-3 ! SPECIFIC PROPERTIES OF FOURTH-ROW TRANSITION llfMENTS

405

controlled, electrically heated crucibles avoid
contamination problems caused by chemical
fuels.

The Bessemer converter is the oldest of the
three methods and the fastest (about 15 minutes
per charge). However, the speed is a mixed bless-
ing because there is not sufficient time to make
analyses and fine adjustments in the amounts
of the alloying elements.

RUSTING

One well-known property of iron is the way in
which it tends to go back to oxide from which
it was derived. In fact, one out of every four men
in the steel industry is concerned essentially with
replacing iron lost by rusting! This shows how
important corrosion is. What is the chemical
nature of rusting, and how can it be controlled?
First, rusting is a special case of corrosion in
which the metal being corroded is iron and the
corroding agent is oxygen. The observed facts
are that H20 and 02 are necessary; H+(aq)
speeds up the reaction; some metals such as Zn
hinder the corrosion, other metals such as Cu
speed it up; and strains (as are produced when
a nail is bent) usually accelerate the reaction.

How can these observations be interpreted?
The most promising mechanism suggested is a
many-step process in which the following se-
quence of events occurs: (1) the iron acts as an
anode to give up two electrons and form Fe+2
(ferrous) ion; (2) the electrons are picked up by
H+(aq) ions to form transient neutral H atoms;
(3) the H atoms are immediately oxidized by 02
to form H20; (4) the Fe+2 is oxidized by 02 in
the presence of H20 to form rust. Rust, in-
cidentally, is not a simple compound but seems
to be an indefinite hydrate of Fe203, so it is fre-
quently given the formula Fe203«H20.

Acids, for example those in fruit juices, cata-
lyze rust formation because they furnish H+(aq)
to accept electrons from the iron, causing it to
dissolve faster. Oxygen gas is necessary to oxidize
Fe+2 to Fe203. The presence of water facilitates
the migration of Fe+2 from the reaction site. The
resulting reduction in Fe+2 concentration allows
more to be formed. Support for these ideas
comes from the frequent observation that when

the 0_> supply is restricted (as under a rivet head)
the iron is eaten away at one spot (shank of the
rivet) but the rust deposits where the 02 is
plentiful (where the rivet head overlaps a plate).
One can surmise that the rivet shank is dissolved
by the half-reaction

Fe(s)

Fe 2 + 2e-

(20)

in some acidic solution, perhaps rain water con-
taining C02. Then the Fe+2 could have been
washed out to the surface where oxidation con-
verted Fe+2 to Fe203/?H20. A similar explana-
tion would hold for the observation that iron
pipes buried in the soil near cinders usually rust
rapidly. Cinders generally contain acid-forming
oxides, which could help speed up the dissolving
of iron.

PREVENTION OF CORROSION

The observed effect of metals on the rate of
rusting also supports the above theory and sug-
gests a way to stop the corrosion. When zinc is
in close contact with iron, the iron does not
corrode but the zinc tends to oxidize away. The
belief here is that the zinc, with a more positive
£° than Fe, gives electrons up to the iron,
effectively preventing Fe from dissolving. This
kind of protection is called cathodic protection
and has a variety of applications. For example,
ship hulls, particularly of tankers, are so pro-
tected in sea water. Magnesium is used rather
than zinc but the principle is the same. Easily
replaceable blocks of magnesium are bolted to
the steel hulls and the magnesium oxidizes in-
stead of the hull. Zinc coated iron ("galvanized
iron") furnishes a second example. The zinc
fortunately does not oxidize very much because
when it reacts with oxygen and water in the
presence of C02 it forms a self-protective coat of
basic zinc carbonate. Thus the zinc is self-
protective and at the same time gives cathodic
protection to the underlying iron.

Some metals such as copper or tin, when in
contact with iron, actually speed up the rate of
rusting. The reason for this is that on these
metals, reaction of electrons with W+(aq) is
more rapid than on iron itself. Thus the effect is
to draw the electrons away from the iron, speed-

406

THE FOURTH-ROW TRANSITION ELEMENTS I CHAP. 22

ing up the rate at which Fe goes to Fe+2. Tin
itself is inert to the atmosphere so a piece of iron
completely covered with tin is safe from rusting.
However, once the protective tin coat is punc-
tured the rusting of the iron will be faster than
if the tin were not there at all. This accounts for
the observation that "tin cans," which are tin-
covered steel, rust very quickly once they start.
One of the easiest ways to prevent rusting of
iron objects is to shut out the supply of 02 and
H20. This can be done by painting the object or
by smearing it with grease. The only caution here
is to do the job thoroughly, since an exclusion
of 02 and H20 that is only partial can do as
much harm as good. Witness the rivet that would
have been saved if there were a good tight
coating of paint to seal the lip against entrance
of the dissolving solution.

22-3.7 Cobalt

This element does not appear in the headlines
very often but it is of practical importance. Prob-
ably its greatest single use is in alloys, including
stainless steels. Pure cobalt is almost as magnetic
as iron and, when alloyed with aluminum, nickel,
copper, and iron, the resulting Alnico alloy has
a permanent magnetization far exceeding that of
iron.

Ni2 Os
cathode

In most of its simple compounds, cobalt has a
+2 oxidation number. This includes the well
known cobaltous chloride. In dilute aqueous
solution this salt is an almost invisible pink; on
dehydration it changes to deep blue. The color
change, which is ascribed to a replacement of
some of the water molecules surrounding Co+2
by Cl~ ions (to form a complex ion), is exploited
in "invisible inks," the writing of which appears
on heat application. Another use is the simple-
weather forecasters in which a swatch of blotting
paper turns pink when the humidity rises, sug-
gesting that rain is supposed to be coming.

In its complex compounds, of which there are
many thousands, Co almost invariably has a +3
oxidation number. Apparently, Co+3 ion accom-
panied by six coordinating groups is particularly
stable. Cobalt complexes are important in bio-
chemistry. Some enzyme reactions go through a
cobalt-complexing mechanism. Although only
small traces are needed, cobalt is essential to the
diet.

22-3.8 Nickel

The five-cent coin, ordinarily called the "nickel,"
is actually 25% nickel (the other 75% is copper).

Fig. 22-10. Cells from Edison and lead storage bat-
teries {schematic) .

Edison Celt

<fr

Lead Cell

SEC. 22-3 | SPECIFIC PROPERTIES OF FOURTH-ROW TRANSITION ELEMENTS

407

This familiar metallic object furnishes an im-
portant example of a nickel alloy. Other impor-
tant nickel alloys include the nickel steels, which
are tough and rust-resistant, Monel metal (60%
Ni, 40% Cu), which is acid-proof, and Nichrome
(60% Ni, 25% Fe, 15% Cr), mentioned in Sec-
tion 22-3.4. Finely divided nickel is used as a
catalyst for hydrogenation, the addition of hy-
drogen to double-bonded carbon atoms. This
process is important in the foods industry to
convert edible vegetable oils, such as cottonseed
oil, into solid fats. The carbon chains of the
vegetable oils have double bonds which have a
high tendency to become oxidized and tend to
develop an unpleasant flavor. When H2 is added
to the double bond, the carbon chain becomes
saturated and the material becomes more attrac-
tive to the cook. Oleomargarine is an example of
such a catalytically hydrogenated compound.

In most of its compounds nickel has a +2
oxidation number, but it is possible to get a
higher state by heating Ni(OH)2 with hypochlo-
rite ion in basic solution. Hypochlorite ion,
CIO-, is one of the stronger oxidizing agents at
our disposal in basic solution. There is consider-
able argument about the formula of the black
solid that is formed, but we shall label it as Ni203
and write the equation

2Ni(OH)2(sj + C\0-(aq) +±:

Ni2CVs,) + 2H20 + C\-(aq) (21)

Nickel(III) oxide is important as the oxidizing
agent in the Edison storage cell, shown in Figure
22-10. Table 22-IV compares the Edison battery
and the more common "lead storage battery."
In both batteries the electrode products are
solids and they adhere to the electrodes; the
batteries can be regenerated by reversing the
flow of electricity with some external device such
as a direct current generator. "Charging a bat-
tery" simply means reversing the half-reaction
at each electrode. It should be noted that on
discharge, in the Edison cell there is no net
consumption of the electrolyte, potassium hy-
droxide, so its concentration stays constant. In
contrast, in the lead cell the sulfuric acid elec-
trolyte is consumed during use of the cell and is
regenerated during charging. This variability of

H2S04 concentration during use of a lead cell
provides the basis for the convenient hydrometer
test of the state of discharge of an automobile
battery. The hydrometer measures the density
of the electrolyte solution, thus indicating how
much of the H2S04 has been consumed. Obvi-
ously, this method cannot be used to check an
Edison cell since the electrolyte concentration is
constant.

Table 22-IV

COMPARISON OF EDISON AND
LEAD STORAGE BATTERIES

EDISON LEAD

Oxidizing agent Ni203 Pb02

Reducing agent Fe Pb

Electrolyte KOH H2SO<

Voltage (one cell) 1.35 volts 2.0 volts
Features light weight heavy

constant variable voltage dur-

voltage ing discharge
expensive inexpensive
rugged voltage and H2SO<
density indicate
when recharge
needed

EDISON CELL (DURING DISCHARGE)

Anode reaction

Ft(s) + 20H-(aq) — >■ Ft(OHh(s) + 2e~ (22a)
Cathode reaction
Ni203(sj + 3H20 -I- 2e~ — >-

2Ni(OH)2rs>l + 20H-(aq) (22b)
Net reaction
Fe(s) + Ni2CVsj + 3H20 — >-

Fe(OH)jfsJ + 2Ni(OH)2rsJ (22c)

LEAD CELL (DURING DISCHARGE)

Anode reaction

PbfsJ + HSOr (aq) ►-

PbSCVs; + H+(aq) + 2e~ (23a)
Cathode reaction
PbCVsJ + HSOr(aq) + 3H+(aq) + 2e~ — )-

PbSCVsJ + 2HsO (23h)
Net reaction

Pb(s) + PbCVsJ + 2H +(aq) + 2HSCV — >-
2PbSO«fsj + 2HjO (23c)

408

THE FOURTH-ROW TRANSITION ELEMENTS I CHAP. 22

Edison batteries cost more than lead storage
batteries, but they have (he advantage of being
lighter, so the amount of electrical energy avail-
able per unit weight is greater. Also they are
more rugged in standing up to mechanical shock.
The difficulty of determining when recharging
is needed and the expense are disadvantages.

22-3.9 Copper

This element occurs in nature in the uncombined
state as native copper and in the combined state
as various oxides, sulfides, and carbonates. The
chief mineral is chalcopyrite, CuFeS2, from
which the element is extracted by roasting (heat-
ing in air) followed by reduction. The roasting
reaction can be written

4CuFeS2(s) + 902(g) — >-

2Cu2SfsJ + 2Fe2Q3(s) + 6SQ2(g)

(24)

It shows the formation of the important by-
product S02, which may be converted to H2S04.
It also shows the formation of Fe203, which is
subsequently removed by adding sand and heat-
ing in a furnace. The sand furnishes Si02 which
combines with the iron oxide to form a low
melting slag of iron silicate. After the slag is run
off, the Cu2S is heated in a current of air to give
consecutively

2Cu2S(sJ + 302(g) — >- 2Cu2OfsJ + 2S02(g) (25)

Cu2S(s) + 2Cu2OfsJ — >- 6Cu(sJ + S02(g) (26)

The copper obtained from this process is about
99% pure, yet this is not pure enough for most
uses, especially those involving electrical con-
ductivity. To refine the copper further, it is made
the anode of an electrolytic cell containing cop-
per sulfate solution. With careful control of the
voltage to regulate the half-reactions that can
occur, the copper is transferred from the anode
(where it is about 99% Cu) to the cathode where
it can be deposited as 99.999% Cu. At the anode
there is oxidation of copper,

Cu(s) — *■ Cu+2(aq) + 2e~ (27)

along with oxidation of any other metal (such as
Fe) which is more readily oxidized than copper.
The elements less readily oxidized than copper

(such as silver and gold) simply crumble off into
a heap under the anode. This so-called "anode
sludge" can then be worked to recover and iso-
late these very valuable by-products. At the
cathode there is reduction but, with well regu-
lated voltage, only of copper.

Cu+*(aq) + 2e~

Cu(s)

(28)

Of course, as time goes on the copper sulfate
solution in the cell has to be replaced because
it collects undesirable ions such as Fe+2, which
have not been reduced because the voltage used
is favorable to the reduction of Cu+2 only.

In compounds, copper usually has a +2 oxi-
dation number, cupric, or copper(II), and oc-
casionally + 1, cuprous, or copper(I). The
most common compound of cupric copper is
CuS04-5H20. The blue color of this solid is due
to the Cu+2 ion hydrated by four of the five
H20 molecules in a surrounding square planar
arrangement. This ion is present in aqueous solu-
tions as well, but with two more distant H20
molecules along the axis perpendicular to this
square. When cupric solutions are treated with
an excess of ammonia, they turn a deep blue.
This is attributed to formation of a tetra-
ammine copper(Il) ion complex, usually written
C^NHs)^2 but probably also containing two
additional H20 molecules coordinated to the
copper. Cupric ion is toxic to lower organisms,
so it is used to suppress the growth of algae in
ponds and fungi and molds on vines. Bordeaux
mixture used to spray grapes and potatoes is
made of copper sulfate and lime.

The + 1 state of copper is found only in com-
plex compounds or slightly soluble compounds.
The reason for this is that in aqueous solution
cuprous ion is unstable with respect to dispro-
portionation to copper metal and cupric ion.
This comes about because cuprous going to
cupric is a stronger reducing agent than copper
going to cuprous. The following exercise in the
use of E° puts this on a more quantitative basis:

Cu(s) — >- Cu+(aqj + e~

E° = -0.52 volt (29)

Cu+(aq) — >- Cu+Yaqj + e~

E° = -0.15 volt (30)

QUESTIONS AND PROBLEMS

409

Since reaction (30) has a more positive E° than
reaction (29), it can force reaction (29) to reverse,
thus in effect transferring an electron from one
Cu+ ion to another Cu+ ion. The net reaction,
obtained by subtracting reaction (29) from re-
action (30) is

2Cu+(aq) +=± Cu(s) + Cu+*(aq) (31)

which has an E° of —0.15 minus —0.52 volt,
or +0.37 volt. Positive £0,s for net reactions
mean the reaction should take place spontane-
ously from left to right.

22-3.10 Zinc

We have already encountered zinc as the irregu-
lar member at the end of the fourth transition
row. We have also mentioned, in Section 17-2.3,
its use as a constituent of the important class of
alloys called brasses and its use in "galvanizing"
iron to protect iron from rusting. Galvanized
iron is made by dipping iron into molten zinc so
as to give a thin adhering layer of Zn over the Fe.
On prolonged exposure to air containing C02 the
zinc forms a thin protective skin of basic zinc

carbonate (zinc hydroxycarbonate). When a hole
forms, penetrating into the iron, the iron does
not rust as would be the case with tin-coated
iron. On the contrary, the fresh Zn surface ex-
posed reacts with C02, 02, and H20 of the air
to form a plug of zinc hydroxycarbonate which
seals the hole.

One other interesting and important com-
pound of zinc is the sulfide. ZnS. It is the mineral
zinc blende, one of the major sources of zinc
and, also, it is the luminescent material on the
face of many television picture tubes. Zinc sulfide
is a semiconductor and, when a beam of elec-
trons strikes the screen, electrons in the solid are
energized so they can wander through the ZnS
much like the electrons in a metal. When these
electrons find an attractive site, usually in the
vicinity of a purposely added impurity atom,
they can be trapped and give off energy as visible
light. This phenomenon, called fluorescence,
makes possible the conversion of one frequency
of light energy to another. The observed color
of the fluorescence depends upon the mode of
preparing the ZnS and on the nature of the
impurity in the ZnS structure.

QUESTIONS AND PROBLEMS

1. Why are the elements with atomic numbers 21 to
30 placed in a group and considered together in
this chapter?

2. Write the orbital representation for

(a) chromium,

(b) molybdenum,

3. What properties of the transition elements are
consistent with their being classified as metals?

4. Ferrous ion, iron(II), forms a complex with six
cyanide ions, CN~; the octahedral complex is
called ferrocyanide. Ferric ion, iron(III), forms
a complex with six cyanide ions; the octahedral
complex is called ferricyanide. Write the struc-
tural formulas for the ferrocyanide and the ferri-
cyanide complex ions.

5. Draw the different structures for an octahedral
cobalt complex containing four NH3 and two
NO: groups.

6. Draw the structures of the compounds

Cr(NH3)6(SCN)3
Cr(NH3)3(SCN)3

(SCN~ is the thiocyanate ion). Consider the oxi-
dation number of chromium to be +3 and the
coordination number to be 6 in both compounds.
Estimate

(a) the solubility of these compounds in water;

(b) their relative melting points;

7. Why does NH3 readily form complexes, but
NH4+ does not?

8. Place a piece of paper over Figure 15-13 and
trace it. Extend the abscissa and add the ioniza-

410

THE FOURTH-ROW TRANSITION ELEMENTS I CHAP. 22

tion energies of the transition elements. Com-
plete the row with the following ionization
energies: Ga, 138; Ge, 187; As, 242; Se, 225;
Br, 273; Kr, 322; Rb, 96 kcal/mole.

9. The volume per mole of atoms of some fourth-
row elements (in the solid state) are as follows:
K, 45.3; Ca, 25.9; Sc, 18.0; Br, 23.5; and Kr,
32.2 ml/mole of atoms. Calculate the atomic
volumes (volume per mole of atoms) for each of
the fourth-row transition metals. Plot these
atomic volumes and those of the elements given
above against atomic numbers.

10. Chromic oxide, Cr203, is used as a green pigment
and is often made by the reaction between
Na2Cr2CVsJ and NH4CIM to give Cr203(sj,
NaClfsJ, N/gJ, and HzO(g). Write a balanced
equation and calculate how much pigment can
be made from 1.0 X 102 kg of sodium dichro-
mate.

11. Chromic hydroxide, Cr(OH)3, is a compound
with low solubility in water. It is usually hy-
drated and does not have the definite composi-
tion represented by the formula. It is quite
soluble either in strong acid or strong base.

(a) Write an equation showing the ions pro-
duced by the small amount of Cr(OH)3 that
dissolves.

(b) Explain, using Le Chatelier's Principle, why
Cr(OH)3 is more soluble in strong acid than
in water.

12. What is the oxidation number of manganese in
each of the following: MnCV (aq); Mn+2( aq);
Mn5CVsJ; MnCVsj; Mn(OH)2fsJ; MnCl2(sj;
MnF/sj?

13. Manganese(III), Mn+3( aq), spontaneously dis-
proportionates to Mn+i(aq) and Mn02(s). Bal-
ance the equation for the reaction.

14. Use the E° values in Table 22-IH to predict what
might happen if a piece of iron is placed in a
1 M solution of Mn+2 and if a piece of manga-
nese is placed in a 1 M solution of Fe+2. Balance
the equation for any reaction that you feel would
occur to an appreciable extent.

15. Iron exists in one cubic crystalline form at 20°C
(body centered cubic, with cube edge length
2.86 A) and in another form at 1100°C (face
centered cubic, with cube edge length 3.63 A).

(a) Draw a picture of each unit cell, showing the
nine atoms involved in a body centered cubic
cell and the fourteen atoms involved in a
face centered cubic cell. (See Figure 21-2.)

(b) Decide the number of unit cells with which
each atom is involved (in each structure).

(c) How many atoms are in each unit cell if we
take into account that some atoms are shared
by two or more adjoining unit cells?

(d) Calculate the volume of the unit cell and,
with your answer to part (c), the volume per
atom (for each structure).

(e) What conclusion can be drawn about the
"effective size" of an iron atom?

16. One of the important cobalt ores is

Co3(As04)2-8H20

How much of this ore is needed to make 1.0 kg
of Co?

17. Nickel carbonyl, Ni(CO)4, boils at 43°C, and
uses the sp3 orbitals of Ni for bonding. Give
reasons to justify the following:

(a) it forms a molecular solid;

(b) the molecule is tetrahedral;

(d) the liquid is a nonconductor of electricity;

(e) it is not soluble in water.

18. Write balanced equations to show the dissolving
of Cu(OH)2(sJ on the addition of NH3(aqJ, and
also the reprecipitation caused by the addition
of an acid.

19. Cupric sulfide, copper(II) sulfide, reacts with hot
nitric acid to produce nitric oxide gas, NO, and
elemental sulfur. Only the oxidation numbers of
S and N change. Write the balanced equation for
the reaction.

20. The solubility of copper(II) iodide, Cul2, is 0.004
g/liter. Determine the value of the solubility
product.

CHAPTER

Some Sixth- and

Seventh-Row

Elements

La\Ce\ Pr \Xd\pm \ Sm \Eu \ Gd\ Tb\Vy\ Ho \Er \Tm\Yb\Lu\
Ac\Th\Pa\l/ \Vp \Pu \Am\On\Bk \Cf\Es \Fm \M<il \

103

£w\

23-1 THE SIXTH ROW OF THE PERIODIC TABLE

The fifth-row transition elements have general
similarity to the fourth-row transition elements.
The electron structure is essentially the same
except that the Ad orbitals are filling instead of
the 3d orbitals. Near the beginning of the sixth-
row transition elements there is a change: the/
orbitals begin to fill to form fourteen elements
before the d orbitals can be occupied to give the
lypical transition elements. This chapter will dis-

cuss these fourteen elements and some of the
seventh-row elements.

23-1.1 The Lanthanides, or Rare Earths

The lanthanides, or rare earths, are lanthanum
and the fourteen chemical elements following.
These fourteen elements, all of which are very
similar to La in chemical behavior, include

411

412

SOME SIXTH- AND SEVENTH-ROW ELEMENTS \ CHAP. 23

cerium, praseodymium, neodymium, prome-
thium, samarium, europium, gadolinium, ter-
bium, dysprosium, holmium, erbium, thulium,
ytterbium, and lutetium (atomic numbers 58 to
71). These elements are called the rare earth
elements because they were extracted from ox-
ides, for which the ancient name was "earth,"
and because those oxides were rather rare.
During the 1940's, techniques for separating
elements were developed to such a degree that
the rare earth elements are no longer so rare.
The most striking property of these elements is
that their chemical properties are almost identical.
For example, they are all reactive metals (about
like calcium). They react with water to give
a vigorous evolution of hydrogen. They all form
basic trihydroxides which are only slightly solu-
ble in water but readily soluble in acid.
For example,

2La(s) + 6H20 — ►■

2La(OH)3(sj + m2(g) (/)

La(OH)3(sj + m+(aq) z<=t La+3(aq) + 3H20 (2)
And, in similar reactions,

2Ce(sj + 6H.O — >-

2Ce(OH)3(s) + 3H2(g) (3)

Ce(OH)3(sj + 3H+(aq) z<=±: Ce+3(aq) + 3H20 (4)

2Pr(s) + 6H.O — ►-

2Pr(OH)/s>) + mjg) (5)
Pr(OHh(s) + 2H+(aq) +±: Pr+3f aq) + 3H.O (6)

The reason usually cited for the great simi-
larity in the properties of the lanthanides is that
they have similar electronic configurations in the
outermost 65 and 5d orbitals. This occurs be-
cause, at this point in the periodic table, the
added electrons begin to enter Af orbitals which
are fairly deep inside the atom. These orbitals
are screened quite well from the outside by outer
electrons, so changing the number of 4/ electrons
has almost no effect on the chemical properties
of the atom. The added electrons do not become
valence electrons in a chemical sense — neither
are they readily shared nor are they readily re-
moved.

The very slight differences that do exist among
these elements are due to small changes in size
brought about by increase of nuclear charge. The
separation of the lanthanide elements from each
other is based upon clever exploitation of these
slight differences in properties. Table 23-1 shows
a comparison of some of the properties of the
various lanthanide elements. As can be seen,
+3 is the common oxidation number and is most
characteristic of the chemistry of these elements.
Another thing to note is the steady decrease in

Table 23-1. some properties of lanthanum and the lanthanide elements

OUTER ELECTRON

OXIDATION

E°

+ 3 ION

ELEMENT

CONFIGURATION

STATES

M — >- M+3 + 3e~

RADIUS

54'

6s2

2.52 volts

1.15 A

Af 5d

6s2

+3,

2.48

1.11

4/s

6s2

+ 3,

2.47

1.09

6s2

2.44

1.08

6s2

2.42

1.06

Af*

6s2

+2,

2.41

1.04

6s2

2.41

1.03

Af 5d

6s2

2.40

1.02

6s2

+3,

2.39

1.00

Af0

6s2

2.35

0.99

Afn

6s2

2.32

0.97

Af1

6s2

2.30

0.96

Af3

6s2

2.28

0.95

Af*

6s2

+2,

2.27

0.94

Af*5d

'6i2

2.25

0.93

SEC. 23-2 THE SEVENTH ROW OF THE PERIODIC TABLE

413

the M+3 ionic size, shown in the last column.
This decrease is called the "lanthanide contrac-
tion" and is due to the fact that the nuclear
charge increases through the series (and, con-
sequently, so is the attraction for the electrons
increased), while the added electrons are not
entering outer orbitals where they would tend to
increase atomic size.

EXERCISE 23-1

(a) Balance the equation for the reaction be-
tween neodymium metal and chlorine gas.

(b) At 0°C and one atmosphere pressure, how
many liters of chlorine gas would react with
14.4 grams of the metal?

EXERCISE 23-2

To a solution containing "tracer" amounts of
radioactive gadolinium, Gd~3(aq) (concentra-

tion less than 10~12 M) is added 0.01 M lantha-
num chloride, LaCl3, and 0.1 M hydrogen fluo-
ride, HF. A precipitate of lanthanum fluoride
forms and it contains most of the radioactive
gadolinium. Explain in terms of the similarity of
lanthanum and gadolinium ions.

23-1.2 Occurrence and Preparation

The most important minerals of the lanthanide
elements are monazite (phosphates of La, Ce,
Pr, Nd and Sm, as well as thorium oxide) plus
cerite and gadolinite (silicates of these elements).
Separation is difficult because of the chemical
similarity of the lanthanides. Fractional crystal-
lization, complex formation, and selective ad-
sorption and elution using an ion exchange resin
(chromatography) are the most successful meth-
ods.

23-2 THE SEVENTH-ROW OF THE PERIODIC TABLE

The last row of the periodic table is unique in
that all of these elements have radioactive nuclei.
However, this property has no direct effect on
the chemistry of these elements. The nuclear
charge and nuclear mass affect the chemistry of
an atom but the fact that the nucleus might ex-
plode at any moment (that is, undergo some sort
of radioactive decay) isn't known by the elec-
trons surrounding the nucleus. We might com-
pare the situation to the packing of eggs in egg
crates. How many eggs can be placed in an egg
crate is fixed entirely by their shape and size. It
is immaterial that eggs have the additional prop-
erty that they can hatch if fertilized. The packing
of fertile eggs is also fixed entirely by their shape
and size, so a crate of fertile eggs is indistinguish-
able from a crate of infertile eggs (unless one of
the fertile eggs happens to hatch while you are
looking at them). In exactly the same way, the
chemistry of an atom has nothing to do with the
stability or instability of its nucleus.

Nevertheless, there is special interest in the
chemistry of the seventh-row elements, a special

interest directly tied to the nuclear instability. As
we have mentioned in Chapter 7, the energy
contents of nuclei are many orders of magnitude
higher than chemical heat contents. If this nu-
clear energy content can be tapped and put to
work, enormous quantities of energy become
available. Thus far. the radioactive nuclei of
some of these seventh-row elements have been
most useful in our attempts to harness nuclear
energy.

Nuclear fuels, like chemical fuels, must be
purified to be most effective. The purification of
the seventh-row elements has presented some
fascinating and difficult problems of chemistry —
so difficult, in fact, that chemists have played as
big a role in the development of nuclear energy
as have physicists.

23-2.1 The Occurrence of the
Seventh-Row Elements

Only five of the seventh-row elements are found
in nature: radium, actinium, thorium, protac-

414

SOME SIXTH- AND SEVENTH-ROW ELEMENTS I CHAP. 23

tinium, and uranium. It is one of the significant
advances of science in the first half of this cen-
tury that ten additional elements of this row have
been synthesized: francium, neptunium, pluto-
nium, americium, curium, berkelium, califor-
nium, einsteinium, fermium, and mendelevium.
The methods for raising the nuclear charge of a
nucleus require ingenious applications of phys-
ics. However, the synthesis of these ten elements
could never have been demonstrated if it had not
been for the solution of some difficult problems
of chemistry, worked on by some of the most
highly skilled chemists in the world. At this time,
elements 102 and 103 have been prepared and
the preparation of elements beyond is under
study.

23-2.2 The Elements Following Actinium

The most interesting elements of the seventh row
are those following actinium. For some of these
elements a large amount of chemistry is known.
The first four, actinium, thorium, protactinium,
and uranium, used to be shown in the periodic

table under lanthanum, hafnium, tantalum, and
tungsten, since they resemble these elements in
many chemical reactions. With the discovery that
one could make elements beyond uranium (those
of higher Z than 92), it was suggested that there
might be a 5/ series of elements analogous to the
4/ series of rare earths (the lanthanides). As a
result, it has become customary to place the ele-
ments following actinium under the rare earths
(see inside front cover) and to call these elements
the "actinides." Table 23-11 collects information
concerning the known oxidation states of the
elements following lanthanum, those following
lutetium (that is, beginning with hafnium), and
those following actinium.

The most striking feature of the contrasts
shown in Table 23-11 is that the seventh-row
elements display the multiplicity of oxidation
states characteristic of transition elements rather
than the drab chemistry of the +3 rare earth
ions. Whereas Ce+3(aq) can be oxidized to
Ce+i(aq) only with an extremely strong oxidizing
agent, Th+YagJ is the stable ion found in tho-
rium salts and 7h+3(aq) is unknown. In a similar

Table 23-11. oxidation numbers found for some sixth-row and

SEVENTH-ROW ELEMENTS

RARE EARTH ELEMENTS

SIXTH-ROW ELEMENTS

SEVENTH-ROW ELEMENTS

FOLLOWING

LANTHANUM

FOLLOWING LUTETIUM

FOLLOWING ACTINIUM

Oxidation

Name

Numbers*

Name

Numbers

Name

Numbers

lanthanum

+ 3

lutetium

-3

actinium

cerium

+3. +4

hafnium

-4

thorium

praseodymiurr

+ 3. +4

tantalum

protactinium

+4, +5

neodymium

tungsten

+2, +3, +4, +5,

uranium

+3, +4, +5, +6

promethium

rhenium

+ 3, ~4. +5,

+6, +7

neptunium

+ 3, +4, +5. +6

samarium

+2,

osmium

+2, +3, +4,

+6 +8

Plutonium

+3, +4. +5, +6

europium

+2,

iridium

-3. +4,

americium

+3, +4, +5, +6

gadolinium

+ 3

platinum

-2. +4

curium

terbium

+ 3. +4

gold

+ 1,

-3

berkelium

+ 3. +4

dysprosium

mercury

+ 1.

californium

holmium

thallium

+1,

einsteinium

+ 3

erbium

lead

+2, +4

fermium

+ 3

thulium

bismuth

+ 3. +5

mendelevium

ytterbium

+2,

polonium

+2, +4,

102

lutetium

astatine

(-1)

103

•The most common oxidation numbers are in blue type.

SEC. 23-2 I THE SEVENTH ROW OF THE PERIODIC TABLE

415

contrast, the most important oxidation state of
protactinium is +5 whereas no compound in-
volving the +5 oxidation state of praseodymium
is known.

Yet it is generally accepted today that the
elements following actinium involve 5/ energy
levels, identifying them as the seventh-row equiv-
alents of the lanthanides.

Spectroscopic investigations of the gaseous
atoms and of ions in suitable crystals furnish
one of the bulwarks of this view. Table 23-111
contrasts the actinide electronic configurations
as they are known today with the configurations
of the corresponding rare earths. Although the
corresponding configurations are not identical
they are quite similar. They offer no hint of
explanation for the varied actinide oxidation
states and the monotonous similarity of the
lanthanides. It may be that one of the most
significant conclusions to be drawn from the
investigations of the actinides is that elements of
an / orbital transition series do not necessarily
resemble each other strongly in chemical be-
havior, contrary to what had been inferred from
studies of the rare earths.

Table 23-III

THE ELECTRON CONFIGURATIONS
OF GASEOUS ACTINIDE AND
LANTHANIDE ATOMS

LANTHANIDES ACTINIDES

5dl

6s2

to1

7s2

4/1

Sd>

6s2

6cP

7s2

4/3

6s2

6dl

7s2

6s2

to1

7s2

4/6

6s2

5f*

&/>

7s2

4/8

6s2

7s2

6s2

7s2

5dl

6s2

&/>

7s2

EXERCISE 23-3

Two half-reactions involving neptunium are
Np+3 — >- Np+4 + e~

E° = -0.147 volt (7)

Np+< + 2H.O — >- Np02+ + 4H+ + e~

E° = -0.75 volt (8)

Find an oxidizing agent in Appendix 3 that could
oxidize Np+3(aq) to Np+i(aq) but not to
Np02+ (aq).,

EXERCISE 23-4

Balance the equation for the reaction between
permanganate, MnO^aq), and plutonium(III),
Pu+3(aq), to form manganous, Mn+2(aq), and
plutonyl ion, Pu0^2faqj, in acid solution.

EXERCISE 23-5

Plutonium(IV), Pu+YagJ, forms a complex ion
with fluoride ion, PuF+3:

PuF^

pU+4 _|_ f-

K = 1.6 X 10"7 (9)

If the F~ concentration is adjusted to 0.10 M in
a solution containing 1.0 X 10-3 M Pu+4, calcu-
late the ratio of the concentration of Pu"1"4 to
PuF+3. What is the equilibrium concentration of
Pu+4?

Chemical investigation of the actinides is made
difficult by the extreme instability of the nuclei —
an instability that increases as we go to higher
atomic numbers. This leads to intense radioac-
tivity (which makes shielding precautions neces-
sary) and small amounts of material to work
with. The heavier actinides are usually made by
bombarding the lighter actinides with neutrons,
helium nuclei (alpha particles), or even carbon
nuclei. With very unstable nuclei, it is hard to
accumulate enough atoms to permit chemical
studies on a macroscopic scale since the element
is disintegrating at the same time it is being
produced. The consequence is that much of the
chemical investigation is done by tracer tech-
niques wherein chemical reactions are performed
on a bulk scale with some other, more available,
element. Clues to the chemistry of the trace ele-
ment are furnished by the behavior of the
radioactivity — whether it follows along with the
bulk carrier or is lost in the chemical operations.

There is reason to believe that not many more
new elements will be created. The instability of
the nuclei of the last few elements suggests that

416

SOME SIXTH- AND SEVENTH-ROW ELEMENTS I CHAP. 23

most or all of the still higher elements have
nuclei so unstable that they will not survive long
enough to be detected.

23-2.3 Nuclear Stability and Radioactivity

The outstanding characteristic of the actinide
elements is that their nuclei decay at a measur-
able rate into simpler fragments. Let us examine
the general problem of nuclear stability. In
Chapter 6 we mentioned that nuclei are made up
of protons and neutrons, and that each type of
nucleus can be described by two numbers: its
atomic number (the number of protons), and its
mass number (the sum of the number of neu-
trons and protons). A certain type of nucleus is
represented by the chemical symbol of the ele-
ment, with the atomic number written at its
lower left and the mass number written at its
upper left. Thus the symbol

represents the nucleus of the plutonium isotope
which contains 94 protons and 239 — 94 = 145
neutrons. Since the forces that exist in the nu-
cleus depend upon the number of protons and
neutrons of which it is composed, we can get a
rough idea of whether a nucleus is stable just
by examining these two numbers.

First of all, let us define what we mean by
stability. Consider an initially pure sample of
292U. Regardless of the physical or chemical
state in which we find the uranium atoms, some
of them will decay each instant to become
thorium atoms by the spontaneous reaction:

*1 J

42He + 2^Th

UO)

Notice that both the electric charge and the
total number of nuclear particles (nucleons) are
conserved in the nuclear decomposition. Careful
study of the rate of this nuclear decay shows that
in a given period of time a constant fraction of
the nuclei present will undergo decomposition.
This observation allows us to characterize or
describe the rate of nuclear decay in a very simple
manner. We simply specify the length of time it
takes for a fixed fraction of the nuclei initially
present to decay. Normally we pick the time for

half the nuclei to decay; this length of time is
known as the half-life of the nucleus. For ex-
ample, measurements show that after 4.5 X 10°
years, half the atoms in any sample of 2H\J will
decay to 2goTh. A nucleus is considered to be
stable if its half-life is much longer than the age
of the earth, which is about 5 X 109 years.
Nuclei that are very unstable are characterized
by half-lives which are quite short — in some
cases only a small fraction of a second.

Now let us turn to the problem of how the composition
of a nucleus affects its stability. The forces that exist
between the particles in the nucleus are very large. The
most familiar of the intranuclear forces is the coulomb
force of repulsion which the protons must exert on one
another. In order to appreciate the magnitude of this
repulsive force, let us compare the force between two
protons when they are separated by 10-8 cm, as they are
in the hydrogen molecule, with the force between two
protons separated by 10-13 cm, as they are in a helium
nucleus. In the first case we have

force

(charge on proton 1 Xcharge on proton 2)

(distance between protons)2

(+gX+f) _jl
(10~s)2 io-

1016e2

where e is the electric charge on one proton. In the
second case we have

force =

(+e)(+g) = JL_
(10"13)2 10-20

= 10-6<?2

U2)

By comparing these two answers we can see that the
repulsive force between two protons in the nucleus is
about ten billion times as great as the repulsive force
between two protons bound together in a hydrogen mole-
cule. In order to overcome these enormous intranuclear
coulomb repulsions and hold the nucleus together there
must exist some very strong attractive forces between the
nucleons. The nature of these forces is not understood
and remains a very important problem in physics.

It is relatively easy to summarize how nuclear
stability (and hence the attractive nuclear forces)
depends upon the numbers of protons and neu-
trons in the nucleus. For atoms with atomic
number less than 20, the most stable nuclei are
those in which there are equal numbers of pro-
tons and neutrons. For atoms with atomic num-
bers between 20 and 83, the most stable nuclei
have more neutrons than protons. For atoms of
atomic number greater than 83, no nucleus can
be considered stable by our definition. These

SEC. 23-2 I THE SEVENTH ROW OF THE PERIODIC TABLE

417

loo

Number of protons , P

Fig. 23-1. The relation between number of neutrons
and protons in stable nuclei. Each dot
identifies a stable nucleus.

statements are demonstrated in Figure 23-1,
where the number of neutrons is plotted against
the number of protons for all stable nuclei. We
see that these stable nuclei form a belt which
deviates increasingly from a neutron to proton
ratio of unity (the dashed line, TV = P) as the
charge on the nucleus increases. Nuclei whose
neutron-proton ratio is such that they lie outside
of this belt of stability are radioactive.

23-2.4 Types of Radioactivity

There are three common ways by which nuclei
can approach the region of stability: (1) loss of
alpha particles (a-decay); (2) loss of beta parti-
cles (j3-decay); (3) capture of an orbital electron.
We have already encountered the first type of
radioactivity, a-decay, in equation (10). Emis-
sion of a helium nucleus, or alpha particle, is a
common form of radioactivity among nuclei
with charge greater than 82, since it provides a
mechanism by which these nuclei ean be con-
verted to new nuclei of lower charge and mass
which lie in the belt of stability. The actinides, in
particular, are very likely to decay in this way.

418

SOME SIXTH- AND SEVENTH-ROW ELEMENTS I CHAP. 23

For example, the most stable isotope of element
100, fermium, has a half-life of only 4.5 days:

?$Fm — >- 2HCf + *He (13)

Nuclei that have a neutron-proton ratio which
is so high that they lie outside the belt of stable
nuclei often decay by emission of a negative
electron (a beta particle) from the nucleus. This
effectively changes a neutron to a proton within
the nucleus. Two examples are

281Cf
?JNa

23*Es + _?<?
2^Mg + _°e

(14)
(15)

This type of radioactivity, /3-decay, is found in
both the light and heavy elements.

Nuclei that have too many protons relative to
their number of neutrons correct this situation
in either of two ways. They either capture one
of their Is electrons or they emit a positron
(a positively charged particle with the same mass
as an electron). Either process effectively changes
a proton to a neutron within the nucleus.

23-2.5 Nuclear Energy

In the previous section we saw that the stability
of a nucleus is affected by its neutron/proton
ratio. Even among those nuclei that we consider
stable, however, there is a variation in the forces
which hold the nucleus together. In order to
study this variation in nuclear binding energy,
let us consider the process of building a nucleus
from protons and neutrons. For an example, let
us look at the hypothetical reaction

2 }H + 2 In — ■*■ |He (16)

First we will compare the masses of the reactants
with those of the product:

30 160

fvfass number

24-0

Fig. 23-2. The binding energy per particle in the
nucleus.

amount of energy must be released (see Section
7-4.1). By the Einstein mass-energy relationship,
E = mc2, we can calculate that in the formation
of 1 mole of helium nuclei there would be

E = mc1 = (o.03 -^-\ X (3 X 1010 — Y
\ mole/ \ sec/

= 2.7 X 1019 *cm\
sec2 mole

E = I 2.7 X 101

sec2 mole,

= 6.4 X 10+11

X 2.4 X 10-8

calories

calories '
g cmVsec2,

mole

(17)

or 640 billion calories of energy released. This
amount of energy can be considered as the bind-
ing energy of a mole of helium nuclei since this
much energy must be supplied to dissociate a
mole of ^He into two moles each of protons and
neutrons.

Similar calculations can be made for other
nuclei. A significant comparison between nu-
clear binding energies can be made if we divide
the total binding energy of each nucleus by the

Mass of 2 protons = 2 X 1.00759 - 2.01518 g mole
Mass of 2 neutrons = 2 X 1.00897 = 2.01794 g mole

Total mass of reactants
Mass of ^He

Mass difference between
products and reactants

Since there is a decrease of 0.03035 gram/mole
of helium formed in this reaction, an equivalent

4.03312 g /mole
4.00277 g/mole

0.03035 g/mole

number of nucleons in the nucleus. This calcula-
tion provides us with the binding energy per

SEC. 23-2 I THE SEVENTH ROW OF THE PERIODIC TABLE

419

particle in the nucleus. The binding energy per
particle varies in a systematic way as the mass
number of the nucleus increases. This variation
is shown in Figure 23-2.

The nuclei that have a mass number of ap-
proximately 60 have the highest binding energy
per nuclear particle, and are therefore the most
stable nuclei. This graph helps us to understand
the existence of the processes of nuclear fission
and nuclear fusion. If the nuclei of the heavier
elements such as uranium and plutonium are
split into two smaller fragments, the binding
energy per nucleon is greater in the lighter nuclei.
As in every other reaction in which the products
are more stable than the reactants, energy is
evolved by this process of nuclear fission. Gener-
ally this fission reaction is induced by the bom-
bardment of a particular isotope of uranium or
plutonium with neutrons,

In + 239|U — »- >&Ba + 39|Kr + 3 J/i (18)

The two nuclei on the right side are just two of
the many possible products of the fission process.
Since more than one neutron is released in each
process, the fission reaction is a self-propagating,
or chain reaction. Neutrons released by one
fission event may induce other fissions. When
fission reactions are run under controlled condi-
tions in a nuclear reactor, the energy released by

the fission process eventually appears as heat.
The energy released by the fission of one pound
of *yu is equivalent to that obtained from more
than 1000 tons of coal.

Figure 23-2 shows that when very light nuclei
such as J H or ,H are brought together to form
heavier elements, the binding energy per nucleon
again increases and energy is released. The graph
also shows that the energy released per nucleon
(therefore per gram of reactant) is considerably
greater in the fusion process than in the fission
reaction. By use of a set of reactions in which
four protons are converted into a helium nucleus
and two electrons, one pound of hydrogen could
produce energy equivalent to that obtained from
10,000 tons of coal. For this reason, and because
of the great abundance of hydrogen, fusion re-
actions are potentially sources of enormous
amounts of energy. Unfortunately fusion reac-
tions take place rapidly only at temperatures
that are greater than a million degrees. These
temperatures have been attained briefly by use
of nuclear fission explosions. At present, at-
tempts are being made to attain the temperatures
required for nuclear fusion by less destructive
means, so that these reactions can be used as an
energy source. The most promising processes are
based upon the fusion of deuterium, ,H, or
tritium, jH, nuclei.

One of the major advances of science in the first half of
this century was the synthesis of ten elements beyond ura-
nium. Glenn T. Seaborg participated in the discovery oj
most of these, a sufficient tribute to his outstanding ability
as a scientist. For the first such discoveries, those of nep-
tunium and plutonium, he shared with Professor Edwin M.
McMillan the Nobel Prize in Chemistry for 1951.

Seaborg rose from humble surroundings. He was born
in a small mining town, Ishpeming, Michigan. During his
boyhood, his Swedish- American parents moved to southern
California. There, two important trails began to become
apparent — his high intellect and his unlimited willingness
to work. Attending the University of California at Los
Angeles, he worked his way through, first as a stevedore,
then picking apricots, then as a linotype apprentice. By his
sophomore year Seaborg had distinguished himselj enough
to be appointed a laboratory assistant. This employment
continued through his undergraduate career and it providea
him his first opportunity to engage in research.

For graduate study, Seaborg moved to the Berkeley
campus of the University of California. He received the
Ph.D. in 1937 and, after two years of work with G. N.
Lewis, began work on the attempted preparation of elements
beyond uranium. This research, following earlier work oj
E. M. McMillan, culminated rather rapidly in the discovery
of plutonium. During the war years, Seaborg had a key
role in working out the chemical processes used to extract
and purify plutonium. Then, after the war, he relumed with
full force to the discovery of new elements. There grew
under his direction a large portion of the now-famous
University of California Radiation Laboratory. With Sea-
borgs stimulation and guidance, highly skilled research
teams prepared, many for the first time, americium,
curium, berkelium, californium, einsteinium, fermium. and
mendelevium. More elements are to come from this excit-
ing laboratory.

In 1958, Glenn T. Seaborg was asked to become Chan-
cellor of the Berkeley campus of the University of Cali-
fornia, a world center of learning. Two years later his
talents were demanded for national service: Seaborg was
appointed Chairman of the U.S. Atomic Energy Commis-
sion. Thus he brings his deep knowledge of science and of
scientists to the service of society in an age when society has
desperate need for guidance by scholars oj this preeminent
stature.

243; [244)

COMPANY

CHAPTER

Some Aspects
of Biochemistry:
An Application
of Chemistry

• • • the elucidation of the structure of biochemicals • • • vv/7/ neces-
sarily lead to a deeper understanding of their junction, and thus finally
to the understanding of the mechanism of life itself.

ALBERTE PULLMAN AND BERNARD PULLMAN, 1962

Living organisms — bacteria, fungi, mosses, al-
gae, plants, animals — are highly organized sys-
tems of chemical compounds. All organisms
derive the energy for their activities, and produce
the substances of which they are built, by means
of chemical reactions.

A century and a half ago men regarded the
chemistry of living organisms as something quite
distinct from the chemistry of rocks, minerals,
and other nonliving things. Indeed, there was in
their minds at that time the inclination to believe
that living things were imbued with some mys-
terious "vital force" that was beyond the power
of men to define and understand.

As time went on, it became apparent that the

mystery in the chemistry of living things was due
to ignorance of the details of what went on, and
with an increased understanding of chemical
principles, the mystery gradually began to dis-
appear. Compounds that were earlier known
only as the products o! plants and animals were
produced in the laboratory from ordinary in-
organic substances. By the middle of the nine-
teenth century the superstitious belief in a chemi-
cal "vital force" had disappeared, and now there
are few chemists who believe that the chemistry
of living organisms is beyond the power of men
to understand.

We still, however, mark off a large area of
chemical study by the term "biochemistry." This

421

422

SOME ASPECTS OF BIOCHEMISTRY: AN APPLICATION OF CHEMISTRY I CHAP. 24

is not because biochemistry is fundamentally
different from chemistry in general. It is because
in order for a chemist to use his talents effectively
to solve certain kinds of problems he must devote
special (but not exclusive) attention to what is
known about a particular field of knowledge.
Biochemists are chiefly concerned with the
chemical processes that go on in living organ-

isms. These scientists must use information from
all branches of chemistry to answer the questions
they ask, but their questions are usually some-
thing like, "What kind of molecules make up
living systems?" "How does a living system
produce the energy it needs?" or "What struc-
tures do biochemicals have?"

24-1 MOLECULAR COMPOSITION OF LIVING SYSTEMS

The chemical system of even the smallest plant
or animal is one of extreme complexity. It has a
multitude of compounds, many of polymeric
nature, existing in hundreds of interlocking equi-
librium reactions whose rates are influenced by
a number of specific catalysts. We will not try
to study such a system. Instead we will show
some parts of it, some examples that have been
well studied and which illustrate the applicability
of chemical principles. All of our knowledge of
biochemistry has come through use of the same
basic ideas and the same experimental method you
have learned in this course.

We shall consider four classes of compounds
that have great importance in biochemistry.
Sugars, fats, and proteins occur in most animals
and plants, while cellulose is more common in
plants. These are all discussed in the following
sections.

24-1.1 Sugars

SIMPLE SUGARS

The word "sugar" brings to mind the sweet,
white, crystalline grains found on any dinner
table. The chemist calls this substance sucrose
and knows it as just one of many "sugars"
which are classed together because they have
related composition and similar chemical re-
actions. Sugars are part of the larger family of
carbohydrates, a name given because many such
compounds have the empirical formula CH20.

EXERCISE 24-1

Glucose, a sugar simpler than sucrose, has a
molecular weight of 180 and empirical formula
CH20. What is its molecular formula?

The structure of the glucose molecule was
deduced by a series of steps somewhat like those
described in Chapter 18 for ethanol. Glucose was

/ /

found to contain one aldehyde group — C

I \.

and five hydroxyl groups ( — OH). These func-
tional groups show their typical chemistry. The
aldehyde part can be oxidized to an acid group.
This reaction is like equation (18-19) (p. 336).
If a mild oxidizing agent (such as the hypobro-
mite ion in bromine water) is used, the aldehyde
group can be oxidized without oxidizing the
hydroxyl groups.

EXERCISE 24-2

Write

the
O

equation for the oxidation of

R — C by hypobromite ion, BrO-, to pro-

\
H

duce Br-. What are the oxidation numbers of

carbon and bromine before and after reaction?

Which element is oxidized? Which is reduced?

SEC. 24-1 I MOLECULAR COMPOSITION OF LIVING SYSTEMS

423

If all the oxygen containing groups are re-
duced, n-hexane results. This test helps establish
that the glucose molecule has a chain structure.
One representation of the structural formula of
glucose, C6Hi206, is

H O

CHOH

I
CHOH

Using a delicate reduction method, the alde-
hyde group can be converted to a sixth hydroxyl
group, giving the substance called sorbitol. This
compound shows the typical behavior of an
alcohol. For example, it forms esters with acids:

CH2OH

I
CHOH

+ 6H3CCOOH

CHOH acetic

acid

CHOH

I
CHoOH

H O

I II

H— C— O— C— CH3

II
H— C— O— C— CH3

And so on for all six carbons.
This is a hexa-acetate.

Another naturally occurring sugar is fructose,
also C6Hi206. It is an isomer of glucose but the

There is another aspect of the structure of
glucose and fructose. They, like other simple
sugars, can exist as a straight chain but this form
is in equilibrium with a cyclic structure. In solu-
tions the latter form prevails. Reaction (2) shows
both forms of glucose.

H .0

CHOH
1
CHOH

H H-?-°H 0 H
\ /• \ /

CHOH
1

/ \?H 9/ \

' C c x

CHOH

HO 1 1 OH
H OH

(2)

CH,OH

The ring form can be written in a simpler way,
showing the hydrogen atoms attached to carbon
atoms by lines only and omitting the symbols for
the ring carbons:

— C— OH

carbon of the C=0 group is at the second

EXERCISE 24-4

At equilibrium in a 0.1 M solution of glucose in
position in the'carbon chain instead of at the water, only 1 % of the glucose is in the straight
end. This makes fructose a ketone (see Section chain form. What is K for reaction (2)?
18-3.2). .

EXERCISE 24-3

Draw a structural formula for the fructose mole-
cule (remember that fructose is an isomer of
glucose). Explain why fructose cannot be oxi-
dized to a six-carbon acid.

DISACCHARIDES

The two sugars we have discussed are mono-
saccharides—they have a single, simple sugar
unit in each molecule. The sugar on your table
is a disaccharide— it has two units. One molecule
of sucrose contains one molecule of glucose
and one of fructose hooked together (losing a

424

SOME ASPECTS OF BIOCHEMISTRY: AN APPLICATION OF CHEMISTRY CHAP. 24

molecule of water in the joining reaction). Fruc- what with the sugar). High solubility in water

tose has a slightly different ring structure because is readily explained because sugars have many

\ functional groups that can form hydrogen bonds.

the C=0 group is not on the end carbon. From your previous study of hydrogen bonding

' (Section 17-2.6) you might recall that about

The formation of sucrose is shown in equa- 5 kca, ar£ rdeased per mde of hydrogen bonds

tion (5).

C — OH

OH +

glucose.

C — OH

H20

C — OH

sitcrose.

PROPERTIES

Sugars occur in many plants. Major commercial
sources are sugar cane (a large, specialized grass
which stores sucrose in the stem) and sugar beet
(as much as 15% of the root is sucrose). In
addition, fruits, some vegetables, and honey con-
tain sugars. On the average every American eats
almost 100 pounds of sugar per year. The nation
requires about 2 X 1010 pounds per year of which
about one-fourth is grown in the United States.
Sugar, at about 10 cents a pound retail, is one
of the cheapest pure chemicals produced.

Sugars are fairly soluble in water, about 5
moles dissolving per liter (solubility varies some-

formed. This energy can then make up part of
that needed to disrupt the structure of the
crystal.

Sugars are easily oxidized. One oxidation re-
action that shows this involves cupric hydroxide,
O

R— C

+ 2Cu(OH);

R— C

•

+ Cu2Q(s) + 2H20 (4)

The reaction is more complicated than shown.
The Cu(OH)2 is not very soluble in the basic

SEC. 24-1 I MOLECULAR COMPOSITION OF LIVING SYSTEMS

425

solution used, so tartaric acid is added to form a
complex ion. The Cu20(s) is a red solid which
does not form such a complex and thus precipi-
tates from solution. The reaction is characteristic
and is used as a qualitative test for simple alde-
hydes and for sugars.

An important metabolic reaction of disaccha-
rides is the reverse of (3). Water, in the presence
of H+(aq), reacts with sucrose to give glucose
and fructose. This process is called hydrolysis,
meaning "reaction with water."

24-1.2 Cellulose and Starch

Cellulose is an important part of woody plants,
occurring in cell walls and making up part of the
structural material of stems and trunks. Cotton
and flax are almost pure cellulose. Chemically,
cellulose is a polysaccharide — a polymer made
by successive reaction of many glucose molecules
giving a high molecular weight (molecular weight
~600,000). This polymer is not basically dif-
ferent from the polymers that were discussed in
Section 18-6:

24-1.3 Fats

Fats, as well as animal and plant oils, are esters.
Actually they are triple esters of glycerol (1,2,3-
propanetriol):

I
H C-

I
O

I
C

I
— C-

V \

H H

H or C3H803

When carboxylic acids, similar to those you
studied in Section 18-3.2, react with glycerol OH
groups, a fat is formed. In natural fats the acids
usually have twelve to twenty carbon atoms,
Ci6 or Ci8 acids being most common.

EXERCISE 24-6

Write the formula for glycerol tributyrate, and
then write the formula of the fat made from
glycerol and one molecule each of stearic
(C17H35COOH), palmitic (C15H3iCOOH), and
myristic (C13H27COOH) acids. How many iso-

— C— OH

I
— C — OH

(5)

— C— OH

Starch is a mixture of glucose polymers, some
of which are water-soluble. This soluble portion
consists of comparatively short chains (molecular
weight ~4000). The portion of low solubility
involves much longer chains and the polymer
chain is branched.

EXERCISE 24-5

The monomer unit in starch and that in cellulose
each has the empirical formula C6Hi0O5. These
units are about 5.0 A long. Approximately how
many units occur and how long are the molecules
of cellulose and of the soluble starch?

mers are possible for the last fat? How many
would be possible if all possible combinations of
the three acids were used? Compare your answer
with that for Exercise 18-15, p. 349.

Common fats (butter, tallow) and oils (olive,
palm, and peanut) are mixed esters; each mole-
cule has most often three, sometimes two, or,
rarely, one kind of acid combined with a single
glycerol. There are so many such combinations
in a given sample that fats and oils do not have
sharp melting or boiling points. Ranges are
found instead.

426 SOME ASPECTS OF BIOCHEMISTRY: AN APPLICATION OF CHEMISTRY | CHAP. 24

An important reaction of fats is the reverse of
ester formation. They hydrolyze, or react with
water, just as disaccharides do. Usually hydroly-
sis is carried out in aqueous Ca(OH)2, NaOH, or
KOH solution. Because of long use in the prepa-
ration of soap from fats, the alkaline hydrolysis
reaction (6) is called saponification.

The metal salts of natural carboxylic acids, like
sodium stearate, are called soaps.

Fats make up as much as half the diet of many
people. Fats are a good source of energy because
when they are completely "burned" in the body
they supply twice as much energy per gram as
do proteins or carbohydrates.

o
I n

— C — O — C — C,,H

nll3S

1 II

— C — O — C — Cl7H35 + 3Na+(a,) + 30H~(aq)

I °

C — O — C— C„H

13 "27

fat

— C — OH

— — C — OH

— C— OH

glycerol

0
2 Na_0-C-C17H35
sodium, stea.ra.ire.

II
Na- O-C— Ci3H27

sodium, myristate.

24-2 ENERGY SOURCES IN NATURE

24-2.1 Some Fundamental Biochemical
Processes

An animal (such as man) expends energy con-
tinuously, to maintain body temperature and to
perform such activities as breathing, circulating
blood, and moving about. What chemical proc-
esses supply this energy?

The chief source of such energy is the combus-
tion of carbon compounds to C02. You know
that man exhales more carbon dioxide than he
inhales in the air he breathes. This extra carbon
dioxide is one of the products of the oxidation
processes by which food is oxidized and energy
is liberated.

One of the important foods of animal organ-
isms is sugar. Man eats sugar in various forms:
as sucrose, as glucose, and as starch, the form
in which sugar is stored in many plant tissues,
such as the potato. Cellulose, although a glucose
polymer, is not a good food for humans because
their digestive chemistry cannot hydrolyze it to
sugar rapidly enough. Termites, however, can
hydrolyze cellulose and they find wood products
quite palatable.

We can regard sucrose and starch as sources
of glucose, for these react with water to form
glucose in the body:

C12H220„ + H20

C6H1206 + C6H1206 (7a)

glucose fructose

HO(C6H10O5)„H + (n - 1)H20

starch

/jC6H1206 (7b)

glucose

Glucose is one of the most important sources
of energy for living creatures of all kinds. We
can illustrate this with the fermentation of sugar.
Yeast, a plant, uses glucose in a chemical reac-
tion

QH1206 — »- 2C2H5OH(*J + 2C02(gj (8)

In this transformation of glucose into alcohol
and carbon dioxide, energy is liberated, and this
energy is used by the yeast plants. Thus glucose
is used as a fuel by the growing organism to
furnish the energy needed for growth.

The process by which yeast breaks down glu-
cose has been carefully studied by biochemists
and the way in which this transformation occurs
is now known in considerable detail. One of the
reasons this process is so interesting is that a
nearly identical process takes place in human
muscle, in this case to furnish energy needed for
muscular activity.

Both yeast and muscle break glucose down
into an acid called pyruvic acid,

SEC. 24-2 I ENERGY SOURCES IN NATURE

427

CH3— C O

O— H

The process requires eight separate steps, all of which are carried out with the aid of biological
catalysts called enzymes. We can picture the series of reactions in the following, much sim-
plified manner:

CH.OH

CHOH

c=o

CHOH

and

converted

broken

CHOH

down
into

CHOH

CH2OH

glucose
(C.HiiOi)

fructose
(CiHuO.)

CH:OH
CHOH

which reacts

with water

to give

glyceraldehyde
2(CiH.0i)

CH2OH"

I
CHOH

I
OH

an intermediate

(9)

CH2OH"

I
CHOH

I
OH

reacts with an enzyme
(E) to produce

CH2OH

I
CHOH

/ \
HO O

glyceric acid

+ reduced enzyme (£H2) (10)

CH2OH

CH2

then

CHOH loses a

C— OH

water

C molecule

\ to give

/ V

+ H20

(11)

CH2

C— OH

rearranges
into

U2)

CH3

c=o

/ \

HO O

pyruvic acid

To this stage, yeast and muscle reactions are the same. Now, however, they proceed differently:

In Yeast

CH3

CH3
C=0 is |

decomposed C + C02

C to / \

/ \ HO

HO O acetaldehyde

U3)

CH3

I
C

/ \
H O

reduced

enzyme

•om (10
above,

" from (W), (EH2)

CH3

+ enzyme (E)
CH.OH

etbanol

(14)

428

SOME ASPECTS OF BIOCHEMISTRY: AN APPLICATION OF CHEMISTRY I CHAP. 24

You will notice that in equation (9) each mole-
cule of six-carbon sugar gave two molecules of
the three-carbon compound, glyceraldehyde.
Thus, each molecule of glucose gives two C02
and two ethanol molecules.

In Muscle
CH3

c=o

+ reduced enzyme (EH2)

-f enzyme (E) (75)

CH3

CHOH

I
C

/ \
HO O

lactic acid

Lactic acid is commonly produced when sugar
is broken down by living cells. Lactic acid is so
named because it is produced when milk sours.

What happens to the energy from the oxida-
tion of glucose? A study of the breakdown of
glucose in the absence of oxygen shows that about
20 kcal are liberated per mole of glucose con-
sumed :

C6H1206fsj — >- 2C2HbOH(l) + 2C02(g)

AH = -20 kcal (76)

This energy is used by the organism to synthesize
other very reactive chemical compounds which
are not shown in our simplified scheme. These
reactive molecules then take part in other proc-
esses (such as muscle action) in which the energy
is released. Part of the energy of glucose is stored
as heat content (or "chemical energy") in the
reactive compounds. This "storage" of energy
in compounds enables living organisms to make
efficient use of the energy of oxidation.

QHuOefsJ + 602fgj — ►■

6C02(gj + 6H20(l) + 673 kcal (77)

Since 673 kcal/mole could be released by com-
plete oxidation, we might wonder why the yeast
cells (and muscle) extract only 20 kcal/mole and
leave so much of the potentially available energy
untouched. This extra energy is there in ethanol
and lactic acid and could be released if these
compounds were oxidized further to C02.

Fermentation in the absence of oxygen, equa-
tion (76), with the liberation of only a small frac-
tion of the total available energy of glucose, is
not the usual chemistry employed by living or-
ganisms. It is a "reserve" mechanism, useful in
tiding over yeast or muscle during times of oxy-
gen shortage. In the case of muscle, it provides a
temporary source of energy when excessive de-
mands require the performance of work at a
faster rate than oxygen can be supplied by the
circulation of blood to the tissue.

Ordinarily, oxygen is used and glucose is oxi-
dized all the way to C02 and H20. Most living
creatures exist in contact with a supply of oxy-
gen, either in the air or dissolved in water. Hence,
most of the metabolic activities of the living
world occur in the presence of oxygen. Under
these conditions, the breakdown of the glucose
molecule proceeds without the use of oxygen
only to the point at which pyruvic acid is formed.
Then the pyruvic acid, instead of being reduced
to lactic acid or ethanol and C02, is oxidized by
oxygen to C02 and water. It is during this oxida-
tion that most of the energy, originally available
in glucose, is liberated.

EXERCISE 24-7

Write a balanced equation for the oxidation of
pyruvic acid to C02 and H20.

24-2.2 Oxidative Metabolism

If a mole of glucose is completely burned in a
calorimeter, a great deal of energy is liberated:

Little of this energy is directly given off as
heat, as it would be if we burned pyruvic acid
with a match. Most of the energy is stored in new

SEC. 24-2 I ENERGY SOURCES IN NATURE

429

chemical compounds that can undergo reactions
leading to the synthesis of fats and proteins and
the other substances of which living matter is
made.

Let us look at a very simple part of the overall
process. The first thing that happens is the break-
down of pyruvic acid into acetic acid and C02
(the "oxidation" process has started!):

CH3C O

oxidation <y

— > CH3C

+ CQ2(g)

U8)

The acetic acid then enters a cycle of reactions

in which it is the fuel that keeps the cycle run-
ning, and C02, water and energy are taken off
along the course of the cycle. This cycle is shown
in Figure 24-1 and its steps are represented by
equations (19a) to (19f).

Each "turn" around this cycle uses up one
molecule of acetic acid and produces two of C02
and two of water. The stages at which oxidation
occurs are not shown in detail, for this is a rather
complex subject and is beyond our present ability
to consider in detail. However, it can be seen
that the oxidation does occur (because acetic
acid and oxygen are fed in and C02 and H20
are discharged), and the energy of "burning"
sugar, the source of the acetic acid, is being
released.

Fig. 24-1. The cycle by which acetic acid is burned as an energy source.

CH,COOH

(l9a)

Wergy + K2 0 -«

H20

/Fumaric) CHCOOH
j acid.

CHCOOH

' C-COOH
I
CH2COOH

M9f)

CH,COOH
I

HO-C-COOH (C£%)

+ 1 °2 ♦*

HO-CHCOOH
I
CH2COOH

k(l9e)

I
CH,COOH

\ /

(19b)

-► COz -h H20 +

eneray

Oxygen
supply

C-COOH

I
CH,

I
CH2COOH

Vz 02 in

CHjCOOH

(19 d) CH,COOH

+ <T°2 iM-

O, ■*■ enerc

(19 c J

430 SOME ASPECTS OF BIOCHEMISTRY: AN APPLICATION OF CHEMISTRY j CHAP. 24

C-COOH

CH,COOH

+ CH*COOH ■► HO-C— COOH

I
CH2C00H

(19 a)

CH2C00K

HO — C — COOH + -j Oz

CHZCOOH

C—COOH
I
CH2 + COz + }izO + energy (l9 b)

CHzCOOH

C—COOH

I
CH2

I
CH,COOH

CH,COOH

CH2COOH

+ COz + energy (l9c)

CH,COOH

I
CH2C00H

+ fo2

CHCOOH

+ HzO + energy (*9d)

CHCOOH

+ H,0

HO — CHCOOH

I
CH2COOH

(l9e)

HO — CHCOOH
I
CH2COOH

+ z02

'**

C-COOH

CH2 COOH

+ HzO r energy (l9f)

C-COOH

CH,COOH

+ CH3COOH *- cycle begins again

(19 a)

24-2.3 Photosynthesis Green plants reverse the process of sugar

What is the source of the vast amount of energy breakdown by synthesizing sugars from C02 and

consumed by our mechanized society? The water:

largest of all of our energy sources is the sun, f£X^g) + 6HjQ(l) + m kcal _^.

C6HI206 + 60/gJ (20)

and energy from the sun is stored in our fuels
(wood, coal, petroleum) as a result of the photo-
synthesis process.

We cannot specify exactly how this energy is

SEC. 24-3 I MOLECULAR STRUCTURES IN BIOCHEMISTRY

431

used by the plant. We know that the reaction
requires a special compound, chlorophyll, which
is the green compound that gives green plants
their color. Chlorophyll does not appear as a
reactant in equation (20) because it is a catalyst.
We also know that the process of incorporat-
ing C02 into complex molecules (that result
finally in the synthesis of sugar) is very similar
to the reverse of the process of sugar breakdown.
These reactions are complicated, however, and
will not be discussed here.

EXERCISE 24-8

The sun provides about 0.50 calorie on each
square centimeter of the earth's surface every
minute. How long would it take ten leaves to
make 1 .8 grams of glucose if the area of each
leaf is 10 cm2 and if only 10% of the energy is
used in the reaction?

EXERCISE 24-9

Normally about 0.03 % of the molecules in air
are carbon dioxide molecules. How many liters
of air (at STP) are needed to provide enough
C02 to form 1.8 grams of glucose in reaction
(20)1

EXERCISE 24-10

Suppose red light of wavelength 6700 A is ab-
sorbed by chlorophyll.

(a) Show that the frequency of this light is
4.5 X 1014 cycles per second.

(b) How much energy is absorbed per mole of
photons absorbed? (See Section 15-1.1;
h = 9.5 X 10"" kcal sec/mole.)

(c) How many moles of photons would be
needed to provide enough energy to produce
one mole of glucose by reaction (20) if all
of the energy were provided by red light?

24-3 MOLECULAR STRUCTURES IN BIOCHEMISTRY

Some of the most exciting recent advances in
biochemistry have come from recognition of the
importance of the structural arrangement of
molecular parts. You saw in Chapter 18 that
the chemistry of a C2H60 compound depends
upon structure. Thus an ether, CH3 — O — CH3,
behaves quite differently from an isomeric alco-
hol, CH3CH2OH. You also learned how inter-
actions between molecules can influence the
properties of water (Sections 17-2.5, 17-2.6) and
arrange the molecules in preferred positions
around an ion (Figure 17-13). This kind of struc-
ture also influences the observed properties.
Both the covalent bond arrangement and inter-
molecular interactions are involved in fixing the
structure of biochemical substances. A few ex-
amples are discussed below.

24-3.1 The Structure of Starch and Cellulose

A striking example of the effect of structure is
shown by cellulose and water-soluble starch.
Both contain the same monomer since hydroly-

sis gives only glucose in each case. But the
glucose ring differs slightly in the arrangement
of the OH groups. This results in two different
polymers. Let us represent the ring structure of
equation (2) by this simplified symbol:

cc -form
(21a)

There is another isomer, identical in all parts,
except for the placement of the right-hand OH
group. Its symbol is

/? -form

(21b)

If we connect a string of the a-form and allow
for the normal 105° angle of bonds to oxygen we
get the polymer called starch:

432

SOME ASPECTS OF BIOCHEMISTRY: AN APPLICATION OF CHEMISTRY CHAP. 24

—in

starch

(22a)

On the other hand, a chain of the /3-form of
glucose gives the polymer called cellulose:

cetitilose

the coiled form of the proteins it contains. A few
moments of thought concerning the profound

v0II

n
(22b)

The very different geometry of the ether link-
ages in starch and cellulose causes these two
polymers to have different chemical properties.

24-3.2 Proteins

In Section 18-6.3 the composition of proteins
was given. They are large, amide-linked poly-
mers of amino acids. However, the long chain
formula (Figure 18-14, p. 348) does not represent
all that is known about the structure of proteins.
It shows the covalent structure properly but does
not indicate the relative positions of the atoms
in space.

The use of X-ray diffraction (Section 14-3.2)
and the principles that describe hydrogen bond-
ing have led to the recognition of a coiled form of
the chain in natural proteins. This model is con-
sistent with other tests also and has received
general acceptance. It is shown in Figure 24-2.
This form has a great deal of regularity — it is not
at all a random shape. Order is not achieved
without some energy factor to maintain it, and
this order results in large part from hydrogen
bonds. These are shown in the figure as dotted
lines, just as they were in Section 17-2.6. When
the hydrogen bonds are broken (by heating or
putting the protein in alcohol), the order dis-
appears and the coiled form loses its shape. Often
this damage cannot be repaired and the coil is
permanently deformed. Cooking an egg destroys

differences between the physical form and be-
tween the chemical potentialities of an egg before
and after it is cooked will suggest the very great
importance of molecular structure in biochem-
istry.

24-3.3 Enzymes

All of the biochemical reactions we have just
been discussing proceed at ordinary tempera-
tures and pressures. Most biochemical reactions
(especially those in the human body) take place
at about 37°C (98°F), and proceed at a rate ade-
quate for the role they play, which is to make
life, growth, and reproduction possible. Most of
them would not proceed at a measurable rate at
this temperature outside of living organisms.
Glucose, pyruvic acid, acetic acid, etc., are very
stable compounds and can remain in contact
with oxygen without apparent change. This is so,
even though their oxidation to carbon dioxide
and water releases large amounts of energy. To
make these reactions proceed, nature uses cata-
lysts to provide new paths with lower activation
energy hills over which the systems can pass so
that measurable reaction rates are achieved.

Biological catalysts are called enzymes. Nearly
every step of the breakdown of a complex mole-
cule to a series of smaller ones, within living
cells, is catalyzed by specific enzymes. For in-
stance, when acetaldehyde is reduced in yeast

SEC. 24-3 I MOLECULAR STRUCTURES IN BIOCHEMISTRY

433

cells to ethanol, in a process we can represent as
follows,

O OH

/ /

CH3— C + 2[H] — *■ CH3— CH2

(23)

the reaction takes place in the presence of a
specific enzyme called "alcohol dehydrogenase."
You can see that the hydrogenaiion of acetalde-
hyde is the reverse of the dehyclrogenation of
ethanol. The enzyme is named for the latter

Fig. 24-2. The coiled or helix form of a protein
molecule.

reaction, but of course it catalyzes the reaction
in either direction (see Section 9-1.4). Con-
ditions at equilibrium are not affected by the
enzyme, but the rate at which the reacting sub-
stances reach the equilibrium state is affected by
the enzyme (as with any catalyst).

Enzymes are protein molecules. While all en-
zymes are proteins, we do not imply that all
proteins can act as enzymes. The protein mole-
cules of enzymes are very large, with molecular
weights of the order of 100,000.* In contrast, the
substance upon which the enzyme acts (called a
substrate) is very small in comparison with the
enzyme. This creates a picture of the reaction in
which the small substrate molecule becomes at-
tached to the surface of the large protein mole-
cule, at which point the reaction occurs. The
products of the reaction then dissociate from the
enzyme surface and a new substrate molecule
attaches to the enzyme and the reaction is re-
peated. We can write the following sequence:

enzyme + substrate — >-

enzyme-substrate complex (24)

enzyme-substrate complex — >-

enzyme + reaction products

(25)

Adding these equations and cancelling gives

substrate — >■ reaction products (26)

Despite the large size of an enzyme molecule,
there is reason to believe that there are only one
or a few spots on its surface at which reaction
can occur. These are usually referred to as
"active centers." The evidence for this view of
enzyme reactions comes from many kinds of
observations. One of these is that we can often
stop or slow down enzyme reactions by adding
only a small amount of a "false" substrate. A
false substrate is a molecule that is so similar to
the real substrate that it can attach itself to the
active center, but sufficiently different that no
reaction and consequently no release occurs.
Thus, the active center is "blocked" by the false
substrate.

* Since so many enzymes are known, this number is
given only to offer a rough idea of size, because actual
molecular weights may range from considerably less than
100,000 to considerably more.

434

SOME ASPECTS OF BIOCHEMISTRY: AN APPLICATION OF CHEMISTRY I CHAP. 24

SPECIFICITY OF ENZYMES

Most enzymes are quite specific for a given sub-
strate. For example, the enzyme "urease" that
catalyzes the reaction

NH2

o=c

+ H20 +=£ C02 + 2NH, (27)

NH2

is specific for urea. If we try to use urease to
catalyze the analogous reaction of a very similar
molecule, W-methylurea, no catalysis is ob-
served :

NHCHj

o=c

\
NH2

AT-methylurea

+ H20 ++ C02 + NH3 + CH3NH2

(28)

This suggests that on the surface of the enzyme
there is a special arrangement of atoms (belong-
ing to the amino acids of which the protein is
constructed) that is just right for attachment of
the urea molecule but upon which the methyl
urea will not "fit."

Specificity is not always perfect. Sometimes an
enzyme will work with any member of a class of
compounds. For example, some esterases (en-
zymes that catalyze the reaction of esters with
water) will work with numerous esters of similar,
but different, structures. Usually, in cases of this
kind, one of the members of the substrate class
will react faster than the others, so the rates will
vary from one substrate to another.

A PRACTICAL APPLICATION OF ENZYME
INHIBITION BY A "FALSE" SUBSTRATE

It is now believed that many of our useful drugs
exert their beneficial action by the inhibition of
enzyme activity in bacteria. Some bacteria, such
as staphylococcus, require for their growth the
simple organic compound paraaminobenzoic

acid and can grow and multiply in the human
body because sufficient amounts of this com-
pound occur in blood and the tissues. The con-
trol of many diseases caused by these (and other)
bacteria was one of the first triumphs of chemo-
therapy,* and the first compound found to be
an effective drug of this type was sulfanilamide:

NH2 NH2

(29)

0=C— OH

0=S— NH

II
O

paraaminobenzoic acid

sulfanilamide

It seems reasonable that an enzyme which
used /xzraaminobenzoic acid as a substrate
might be deceived by sulfanilamide. The two
compounds are very similar in size and shape
and in many chemical properties. To explain the
success of sulfanilamide, it is proposed that the
amide can form an enzyme-substrate complex
that uses up the active centers normally occupied
by the natural substrate.

Usually fairly high concentrations of such a
drug are needed for effective control of an in-
fection because the inhibitor (the false substrate)
should occupy as many active centers as possible,
and also because the natural substrate will prob-
ably have a greater affinity for the enzyme. Thus
the equilibrium must be influenced and, by using
a high concentration of the false substrate, the
false substrate-enzyme complex can be made to
predominate. The bacteria, deprived of a normal
metabolic process, cannot grow and multiply.
Now the body's defense mechanisms can take
over and destroy them.

* Chemotherapy is the control and treatment of disease
by synthetic drugs. Most of these are organic compounds,
often of remarkably simple structure. Sulfanilamide is one
example of an organic compound synthesized by chemists
for the treatment of bacterial infections.

Robert Woodward is surely one of the outstanding Ameri-
can synthetic organic chemists of all time. His astonishing
record of successful syntheses of biologically important
substances has won him ten honorary doctorate degrees
and at least as many major awards here and abroad.

Woodward was born in Quincy, Massachusetts. His in-
terest in chemistry developed at an early age and it seemed
to grow without need for stimulation nor urging. By the
time he entered Massachusetts Institute of Technology at
the age of sixteen, he knew as much organic chemistry as
the average graduating senior. M.I.T. recognized his capa-
bilities and opened the laboratory to him. He passed course
examinations at a rate and with a performance that brought
him the Bachelor s Degree in three years and, in only one
additional year, the Ph.D. Professor J. F. Norris, then
director of the M.I.T. laboratory announced, "We saw we
had a person who possessed a very unusual mind. ■ ■ ■ We
think lie will make a name for himself in the scientific
world."

How richly this prophecy has come true is read in the
scientific literature of organic chemistry. His first major
success was the synthesis of quinine, a problem he began
worrying as a high school student. This compound was
to be typical of the difficult molecules he has successfully

synthesized. Quinine has the formula CxHuOtN*, and it
includes two benzene rings, a tricyclic structure, a double
bond, an OH group, and a methyl ether linkage, all com-
bined in a very particular geometrical configuration. A host
of other, comparably difficult syntheses have been achieved
by Woodward and his large group of students, including
some natural products whose importance has made their
names familiar household words: cholesterol, cortisone,
and chlorophyll. His contributions to the structure deter-
minations of antibiotics reads like a doctor's chemical
shelf, including penicillin, terramycin, and aureomycin. He
has added to our knowledge of the polymerization processes
by which amino acids link into proteins, and some of his
synthetic protein-like polymers have physical properties
quite comparable to those of silk and wool fibers.

All of these accomplishments bespeak a dedication to
chemistry and a capacity for work that are commensurate
with his intellectual capability. Woodward can be found
in his office or in the laboratory after midnight many days
a week. However, the rewards for such intensive effort are
proportionate. To Robert Woodward, as to most chemists,
chemistry is an exciting adventure for which there just
isn't enough time.

CHAPTER

The Chemistry
of Earth,
the Planets,
and the Stars

There may, of course, be types of life with a wholly different chemical
basis to our own, for example, a low temperature life on the outer planets
which is based on reactions in liquid ammonia.

J. B. S. HALDANE, 1960

The advent of space exploration quickens the
pulse of every scientist including, as much as any,
that of the chemist. Chemists are playing many
crucial roles: preparing new fuels, new metals,
and new plastics to cope with a new range of
performance needs; anticipating the environ-
mental chemistry that will face the first space
explorers, devising ways to permit survival under
conditions so extreme that they are not even
present on our planet, collecting and interpreting
data that will reflect onto and illuminate age-old
questions about the origin of the earth, the solar
system, and life itself.

This age is heavily laden with terrestrial prob-
lems for the chemist, as well. Our planet seems
to be shrinking under man's fantastic success in
curbing nature's whim and in bending it to his
436

will. The magnitude of this success is measured
in an awesome and exponential population
growth. This growth portends problems of food
production, fuel consumption, and even of living
space that dwarf those of the past. Suddenly
immense urban areas are threatened by air pol-
lution problems that were completely unknown
thirty years ago. Power consumption is expand-
ing so rapidly that some scientists anticipate the
depletion of fossil fuel supplies and they urge
haste in developing nuclear fuels and in harness-
ing more completely the vast energy of the sun.
It is an exciting age — challenging, yes, but
exciting. It is an age in which we must understand
our planet, Earth, and if we do, we can begin to
venture toward our neighbor planets and beyond
them toward the stars.

SEC. 25-1 ! THE CHEMISTRY OF OUR PLANET, EARTH

437

25-1 THE CHEMISTRY OF OUR PLANET, EARTH

The earth is the source of all the substances
directly available to us. Energy comes to us from
outside the earth— from our sun and to a small
degree from the other stars. The earth, in turn,
radiates energy into outer space. If the amount
of energy radiated is more than the amount
received, the earth will grow colder— if it is less,
the earth will warm. Some of the solar energy is
stored as chemical heat content when new sub-
stances, particularly organic compounds, are
formed. For short periods of time (as measured
in terms of geologic ages) we can use the energy
stored in fossil fuels such as coal and oil or we
can draw on nuclear fuels. In the long run, we
shall have to depend upon the sun as our pri-
mary source of energy.

The substances we can use come primarily
from the earth. In its movement around the sun,
the earth sweeps through space and collects
material from meteors and some cosmic dust,
but the amount of gathered material is small
compared with the amount present in the earth.*
We shall consider the material of the earth and
see how it is put to use by mankind.

25-1.1 The Parts of the Earth

A discussion of the chemistry of the earth is
conveniently broken into three parts, each of
which corresponds to one of the phases solid,
liquid, or gas.

The lithosphere is the solid portion of the
earth. We shall use the term to include the
central core, though there remains controversy
as to whether the core is solid or liquid. The
lithosphere is a sphere of solid material about
4000 miles in radius. We have direct access to
only a minute fraction of this immense ball. The
deepest mine penetrates only two or three miles.
The deepest oil wells are about five miles deep.

* It has been estimated that about five tons of material
are gathered per day as the earth sweeps through space.
Yet the amount collected in a billion years would form
a dust layer only a few millimeters thick if spread evenly
over I he earth's surface.

This relatively thin shell that we can study
directly is called the earth's crust. In view of
seismic observations, we consider the thickness
of this crust to be about 20 miles. The remainder
we shall call the inner lithosphere, which includes
the central part called the core.

About 80% of the earth's surface is covered
with aqueous solution. This liquid layer, the
oceans, is called the hydrosphere. The average
depth of the hydrosphere is about three miles
but at ocean "deeps" or "trenches," it changes
precipitously to depths over twice that.

Surrounding the earth is the third phase, a
gas. The gas mixture surrounding the earth is
called the atmosphere. Over 98% of this gas
(air) is less than 40 miles above the earth's
surface.

25-1.2 Composition and Properties
of the Atmosphere

The composition of the earth's atmosphere dif-
fers from day to day, from altitude to altitude,
and from place to place. The largest variation is
in the concentration of water vapor. Water
evaporates continually from the hydrosphere,
from the soil, from leaves, from clothes drying,
etc. At intervals, parts of the atmosphere become
chilled until the dew point or frost point is
reached and then any vapor in excess of the
saturation amount is precipitated as rain or
snow.

Since the concentration of the water vapor
varies so much, geochemists usually report the
composition of "dry air"— that is, air from
which all the water vapor has been removed.
The composition of a sample of dry air is shown
in Table 25-1. Notice the low concentration of
hydrogen and helium in the air. The earth is a
rather small object in the universe and exerts a
relatively low gravitational attraction on the
gases above it. Hence, most of the hydrogen and
helium originally associated with the matter of
the earth could be lost rather easily. Notice also
that nitrogen is more abundant than oxygen in

438

THE CHEMISTRY OF EARTH, THE PLANETS, AND THE STARS | CHAP. 25

Earth '& c ru.S-t-

20 miles

Fig. 25-1. The parts of the earth.

the air, even though oxygen is much more abun-
dant in the hydrosphere and lithosphere. Notice
also that, except for water and carbon dioxide,
the major components of air are elements.

Table 25-1

COMPOSITION OF A CERTAIN
SAMPLE OF DRY AIR*

SUBSTANCE

PERCENT OF

Name

Formula

MOLECULES

nitrogen

N,.

78.09

oxygen

20.95

argon

0.93

carbon dioxide

CO,

0.03

neon

0.0018

helium

0.00052

krypton

0.0001

hydrogen

0.00005

xenon

0.000008

♦Traces of other compounds (less than
0.0002% of the molecules) are also known to be
present.

With the aid of the gas laws we can calculate
the relative concentrations of the components in
moist air. Suppose that on a certain day the

Atm. osph ere
200 miles

Hydro sph ere
~ 3 miles

atmospheric pressure is 750 mm and that after
drying a sample of this air we find the pressure
to be 738 mm. Then the partial pressure of the
water vapor was 750 mm - 738 mm = 12 mm.
Since the partial pressure varies directly with the
number of molecules, we find that the fraction
of the air molecules that are water molecules in
the moist gas is 12 mm/750 mm = 0.016. In
this rather moist gas, 1.6% of the molecules are
water molecules.

The gravitational force on a heavy molecule
exceeds that on a light molecule. Consequently,
there is a tendency for "sedimentation" of the
molecules of high molecular weight relative to
the gas molecules of low molecular weight. This
is opposed by the tendency toward maximum
randomness, which tends to keep the atmosphere
thoroughly mixed. The net result is a slight
change of composition with altitude. Dry air at
sea level contains about 78% nitrogen molecules
and 21% oxygen molecules, but at 60,000 feet
a sample of dry air has about 80% nitrogen
molecules and only 19% oxygen molecules.

Apart from this gravitational effect, there are
composition changes due to chemical reactions
induced by light. These are caused by absorption
of ultraviolet light in the upper atmosphere. For
example, oxygen absorbs ultraviolet light and
the energy taken up by the molecule exceeds the

sec. 25'

THE CHEMISTRY OF OUR PLANET, EARTH

439

bond energy. The bond breaks, to give two oxy-
gen atoms:

o,(g) + hv — ►■ 20(g) (/)

Of course the oxygen atoms so produced are
very reactive. One fate of these atoms is to com-
bine with another oxygen molecule, 02, to form
ozone, 03:

O(g) + 0,(g) — *- 03(g) (2)

Ozone is a highly reactive form of the element
oxygen, though by no means as reactive as oxy-
gen atoms. It is produced in the atmosphere only
at high altitudes because the ultraviolet light of
the frequency required in reaction (7) is so com-
pletely absorbed that it does not reach lower
altitudes. Balloon flights have shown that the
ozone concentration is negligible at sea level but
that it rises to a maximum at a height of about
15 miles.

This small amount of ozone 15 miles above
the earth's surface absorbs most of the ultra-
violet light not absorbed by 02. Thus, 02 and 03
together make the atmosphere opaque in most
of the ultraviolet spectral region. Presumably the
chemistry of life on this planet would have
evolved quite differently if this ultraviolet light
reached the earth's surface. As a single example,
reflect that photosynthesis would have photons
of much higher energy with which to operate if
the atmosphere were transparent in the ultra-
violet region.

EXERCISE 25-1

Suppose that the photosynthesis reaction (20b)
in Chapter 24 (p. 430) could be based upon light
of wavelength 2400 A (this light is absorbed
heavily by ozone). How many moles of these
photons would provide the 673 kcal of energy
needed to produce one mole of glucose? (Re-
member, E = hv, and h = 9.5 X 10~14 kcal
sec/mole). Compare your answer with that of
Exercise 24-10.

At the opposite end of the spectrum, the infra-
red, again the atmosphere becomes virtually
opaque. This is due mainly to absorption by

gaseous water and carbon dioxide. Thus we find
that the air, normally regarded as transparent,
actually serves to filter the sun's rays striking
the earth. The very high energy photons (in the
ultraviolet) and the very low energy photons (in
the infrared) are removed and the spectral region
between is transmitted.

EXERCISE 2S-2

On this planet, of what value would be an eye
that is sensitive only to light in the ultraviolet
spectral region? Discuss the evolutionary sig-
nificance of the facts that the human eye and the
photosynthesis process are both dependent upon
light in the part of the spectrum called the
"visible."

25-1.3 Composition of the Hydrosphere

Waters exposed to air dissolve some of it. In
water, oxygen is twice as soluble as nitrogen but
since nitrogen is four times more abundant than
oxygen in air, more dissolved nitrogen is present
than dissolved oxygen. It is the dissolved ele-
mentary oxygen that is used by living organisms
for their oxidative processes. The concentration
of dissolved carbon dioxide is low because its
concentration in the air is low. But dissolved
carbon dioxide is necessary for photosynthesis
in marine plants. Dissolved carbon dioxide is
responsible for part of the pleasant taste of
water. Boiled water has lost almost all of the
dissolved gases. It tastes "flat."

Ocean water also contains dissolved molecules
from the gases of the air. These can be removed
by boiling, but other solutes remain. When a
kilogram of average ocean water is distilled, 967
grams of water can be collected and 33 grams of
solids (primarily salts) remain behind. Thus, we
may say that 3.3% of the weight of the ocean
water is due to dissolved salts. Actually, more
than forty of the elements have been identified
as being present in ocean water but half of these
are present in very small concentrations "less
than 1 gram per billion, 109, grams of water."

440

THE CHEMISTRY OF EARTH, THE PLANETS, AND THE STARS I CHAP. 25

Table 25-11. an average composition of ocean water (disregarding

DISSOLVED GASES)

ELEM

:NT

PREDOMINANT

NUMBER OF MOLES

Name

Symbol

SPECIES

PER

KILOGRAM

hydrogen
oxygen

H20

53.7

chlorine

C\-(aq)

0.535

sodium

Na+(aq)

0.460

magnesium

Mg+2(aq)

0.052

sulfur

SOT2(aq)

0.028

calcium

Ca+2(aq)

0.010

potassium

K+(aq)

0.010

bromine

Kr-(aq)

0.008

In Table 25-11 are shown the concentrations (in
number of moles per 1000 grams of ocean water)
of water and of the most abundant ions.

Several facts become apparent. There are
fewer Na+ ions than Cl~ ions; other positively
charged ions — Mg+2, Ca+2, and K+ — are also
present. Sulfate ions, S04~2, and bromide ions,
Br~, are other negatively charged ions present in
the water. Thus, ocean water is more than a
solution of sodium chloride. Another fact is that
K+ ions are much less plentiful than Na+ ions
(Na+/K+ is about 46) even though K+ ions are
rather plentiful in the earth (Na+/K+ is about 2).

25-1.4 Composition and Properties
off the Lithosphere

We know very much about the outermost por-
tion of the lithosphere because it is available for
direct study. In contrast, we know almost noth-
ing about the inner lithosphere, though it con-
stitutes over 99.5% of the mass of the earth.

THE INNER LITHOSPHERE

Seismic observations furnish our only probe of
the inner lithosphere. The shock waves initiated
by an earthquake travel through the interior of
the earth in paths that are bent in accordance
with the elastic properties and density of the
medium they penetrate. From these paths, seis-
mologists have been able to determine the ex-
istence of zones within the lithosphere. The

outermost portion, or mantle, is approximately
2000 miles in depth and it is generally thought
to be solid. This solid has a density of about 3
grams per milliliter near the crust and it in-
creases to about 5 grams per milliliter at the
bottom of the mantle. This higher density is
caused by increasing pressure in the depths of
the earth. For a comparison, the pressure at this
depth is thought to be over a million atmos-
pheres, two or three times greater than the
highest pressures recorded in static laboratory
experiments.

The inside of the lithosphere is called the core.
Still higher pressures are expected and the den-
sity may rise as high as 18 grams per milliliter
at the center of the earth. Some of the core may
be liquid but the evidence is not decisive.

The composition of the mantle is probably
rock-like, meaning it is made up of various
silicates. These minerals have densities, com-
pressibilities, and rigidities that match those
indicated by the seismic studies. The core has
long been thought to be largely iron, a view sug-
gested by the composition of meteorites. These
are solid objects that plummet through the at-
mosphere from space and they may be pieces of
exploded planets resembling earth. Hence their
composition furnishes a possible clue to the
composition of the inner lithosphere. Current
speculation ranges from iron to a high density
rock but new evidence is needed.

The temperature of the center of the earth is

SEC. 25-1 I THE CHEMISTRY OF OUR PLANET, EARTH

441

thought to be a few thousand degrees. Though
this would melt rock at the earth's surface, solids
can remain stable at the exceedingly high pres-
sures thought to exist in the core.

Needless to say, very much remains to be
learned about the chemistry of the inner litho-
sphere. It is a high temperature and high pressure
laboratory whose door has not yet been opened.

THE EARTH'S CRUST

Oxygen and silicon are the most abundant ele-
ments in the earth's crust. Table 25-1 II shows
that 60% of the atoms are oxygen atoms and
20% are silicon atoms. If our sample included
the oceans, hydrogen would move into the third
place ahead of aluminum (remember that water
contains two hydrogen atoms for every oxygen
atom). If the sample included the central core

Table 25-111

ABUNDANCE OF ELEMENTS

IN THE EARTH'S CRUST*

NO. OF ATOMS

ATOMIC

PER

RANK ELEMENT NUMBER

10,000 ATOMS

oxygen

6050

silicon

2045

aluminum

625

hydrogen

270

sodium

258

calcium

189

iron

187

magnesium

179

potassium

138

titanium

phosphorus

8.6

carbon

5.5

manganese

3.8

sulfur

3.4

fluorine

3.3

chlorine

2.8

chromium

1.5

barium

0.75

* From calculations of I. Asimov (/. Chem. Ed.) 31,
70 (1954) on the data of B. Gutenberg (Editor) "Internal
Constitution of the Earth" 2nd Ed. Dover Publications,
New York, 1951, p. 87. Reprinted with permission of the
publisher.

of the earth, iron would probably move into
second place ahead of silicon, and magnesium
would be fourth. Thus, the exact order is changed
by the sample (part of earth) chosen. In any of
the lists of elements, the most abundant atoms are
those of elements having low atomic numbers, 26
or less. All of the elements beyond iron (element
number 26) account for less than 0.2% of the
weight of the earth's crust.

25-1.5 Availability off Elements

In our daily life most of us are more concerned
with the availability of the elements than with
their general abundance in earth. The air is all
around us and equally accessible to all. Water is
somewhat more restricted. Some regions have
a surplus of water, whereas other regions are
deficient. Even in regions of abundant rainfall,
the use of water may be so great that the reserves
become depleted gradually. Thus, as the earth's
population increases and more and more water
is being used, water is becoming a natural re-
source to be treated with respect.

Many of the metals used by ancient man —
copper (cuprum, Cu), silver (argentum, Ag),
gold (aurum, Au), tin (stannum, Sn), and lead
(plumbum, Pb) — are in relatively short supply.
Ancient man found deposits of the first three
occurring as the elementary metals. These three
may also be separated from their ores by rela-
tively simple chemical processes. On the other
hand, aluminum and titanium, though abundant,
are much more difficult to prepare from their
ores. Fluorine is more abundant in the earth than
chlorine but chlorine and its compounds are
much more common — they are easier to prepare
and easier to handle. However, as the best
sources of the elements now common to us be-
come depleted, we will have to turn to the ele-
ments that are now little used.

During geological time, a number of separat-
ing and sorting processes — melting, crystalliza-
tion, solution, precipitation — have concentrated
various elements in local deposits. In these, the
elements tend to be grouped together in rather
stable compounds. These are called minerals.
Many of the minerals have compositions similar

442

THE CHEMISTRY OF EARTH, THE PLANETS, AND THE STARS ■ CHAP. 25

to compounds we can make in the laboratory
but most are not so pure. There are large de-
posits of sodium chloride, for example, appar-
ently formed when ancient seas evaporated in
locales where the solid deposits were later pro-
tected from the dissolving action of water. The
ocean itself is an enormous source of sodium
chloride. In contrast, potassium salts have not
been concentrated in a similar way. Many of the
metallic elements are concentrated as sulfide
minerals (for example, lead, PbS; molybdenum,
MoS2; zinc, ZnS). Other elements occur in rather
concentrated oxide deposits (for example, iron,
Fe203; manganese, Mn02). Large deposits of
carbonates (for example, zinc, ZnC03; calcium,
CaC03) and sulfates (for example, barium,
BaS04) of some metallic elements are known.
The minerals sufficiently concentrated to act as
commercial sources of desired elements are called
ores.

25-1.6 The Air as a Source of Elements

We are so used to free air that we do not think
of oxygen as an important chemical. For ex-
ample, we buy natural gas (mostly methane) as
a fuel and burn it in air to furnish heat to us.
If methane were free in the air and oxygen were
scarce so that we had to purchase it we would
consider oxygen to be the "fuel." In either case,
the amount of heat released would be that for
the reaction represented by the following equa-
tion.

CH<(g) + 2Q2(g) — >

C02(g) + 2H20(l) -|- 213 kcal (5)

Despite the general availability of unlimited
quantities of oxygen in the air, tremendous quan-
tities of the pure gas are prepared annually for
industrial and medical use. Billions of cubic feet
of oxygen gas are manufactured every year, by
liquefaction of air followed by fractional distilla-
tion to separate it from nitrogen.

Ores of nitrogen are relatively rare. The best
mineral is sodium nitrate, NaN03, found in large
deposits in Chile. We now prepare the nitrogen
compounds we desire from the nitrogen of the

air. Thus, the air is our best source of oxygen
and nitrogen — two very important elements.

EXERCISE 25-3

Explain in terms of energy the significance of the
fact that a piece of wood is stable in air at room
temperature but if the temperature is raised, it
burns and releases heat.

25-1.7 The Age of the Earth

Part of looking ahead is understanding the past.
One of the most interesting questions man has
asked about the past is, "How old is the earth?"
Of course we are not sure there is an answer. We
shall see, however, that ingenious methods have
been developed that date the earth's crust.
Scientists generally accept the proposal that the
earth's crust, as we now find it, has a finite age.

The most reliable methods for establishing the
age of a long-lasting object (such as a mountain)
depend upon the presence of natural radioac-
tivity. The decay of the radioactive elements can
be likened to a clock that is partially unwound.
By studying the extent to which the clock has
unwound, we cannot tell the age of the clock but
we can measure how long ago it was wound.

For example, consider the chemical composi-
tion of a very old crystal of pitchblende, U308.
We may presume that this crystal was formed
at a time when chemical conditions for its for-
mation were favorable. For example, it may have
precipitated from molten rock during cooling.
The resulting crystals tend to exclude impurities.
Yet, careful analysis shows that every deposit of
pitchblende contains a small amount of lead.
This lead has accumulated in the crystal, be-
ginning at the moment the pure crystal was
formed, due to the radioactive decay of the
uranium.

The sequences of radioactive decays that lead
to lead are well-known and the rates of decay
have been carefully measured. We shall consider
the sequence based upon the relatively slow de-
composition of the most abundant uranium
isotope, mass 238 (natural abundance, 99%):

SEC. 25-1 | THE CHEMISTRY OF OUR PLANET, EARTH

443

"|U — ►■ 2^Th + JHe a-decay, half-life,

/1/2 = 4.6 X 109 years (4)

The products are an a-particle (a helium nu-
cleus), and a thorium isotope that is unstable and
that rapidly decays by emitting successively two
electrons:

-'&Th — >- 23,?Pa + _?«?

/3-decay tm = 24.1 days (5)

23JPa — > 23,<U + _oe

/3-decay ty2 =1.14 minutes (6)

Thus we have returned to an isotope of uranium,
234U, but one of half-life very much shorter than
that of 238U. This isotope begins a succession of
a-decays, each moving the product upward in
the periodic table:

2S$U — *■ 2$Th + £He

a-decay tm = 2.7 X 105 years (7)

a + *2He

a-decay tm = 8.3 X 104 years (8)

(9)

(70)

22JRa

>Rn

L8Po

a-decay til2 = 1.6 X 103 years
2!?Po + 4He

a-decay tu2 = 3.8 days
2^Pb + 4He

a-decay txt2 = 3.1 minutes

(//)

We have at last reached lead — but 2l4Pb is itself
radioactive! This isotope decays in a succession
of /3-decays :

;pb

|Bi +

/3-decay tm = 27 minutes (72)

2^Bi — >- 2^Po + _°e

/3-decay /i/2 = 20 minutes (13)

Again a-decay occurs, returning to the element
lead, but to another isotope that decays by
emitting successively two /3-particles:

2MPo _^ 2,npb + 4He

a-decay /,/2 = 1.5 X 10~4 seconds (14)

2|§Pb — +- 2L°Bi + _?e

/3-decay tm = 22 years (75)

£Bi — >- 2i°Po + _°e

/3-decay tm = 5 days

(16)

This isotope of polonium, 210Po, again decays by
a-decay, but this time giving an isotope that does
not decay further:

2.oPo _^ 2oepb + 4He

a-decay tU2 = 140 days (77)

2^Pb — >- not radioactive tV2 infinite (18)

The products in this long sequence of reactions
accumulate in the stable isotope of lead, 206Pb.
The amount of 206Pb present depends upon how
long the deposit of uranium has decayed since the
crystal (/308 was formed.

There is, fortunately, a rather simple verifica-
tion of the presumption that all of the lead in
the U:iOs came from this tedious sequence of
nuclear reactions, (4) to (77). Lead ores that do
not contain uranium include several isotopes —
206Pb makes up about 26 % of the total and the
rest is 204Pb ( 1 .4 %), 207Pb (2 1 %), and 208Pb (52 %).
Of these, two other isotopes can be formed
through radioactive decay of some other ura-
nium or thorium isotope by a sequence like that
shown for 238U. Of the four stable lead isotopes,
only one is not derived from radioactive decay,
204Pb. Hence, the ratio of the amount of this
isotope to that of 206Pb measures the amount of
206Pb present in excess of the natural abundance.
This excess must have come from decay of 238U.
If there is no 204Pb present, then all of the 206Pb
came from 238U.

Thus, analysis of uranium minerals with the
aid of the mass spectrograph gives information
on the age of the mineral. Though many different
half-lives are involved in forming the lead, only
the longest half-life (the rate-determining step) is
of importance. Combining the lead content with
the mU half-life provides estimates of mineral
ages in the range of five billion years.

What have we learned in this estimate? Surely
we can say the age of the earth cannot be shorter
than 5 X 109 years. That was when the uranium
mineral clock was wound — but the clock could
be much older. To evaluate this number further,
we must look for other types of data.

Fortunately, there are other radioactive ele-
ments in nature that give similar bases for esti-
mates. As a second example, potassium in nature

444

THE CHEMISTRY OF EARTH, THE PLANETS, AND THE STARS | CHAP. 25

includes one radioactive isotope, ™K, which de-
cays by capturing an electron into its nucleus.
The product is ^Ar, a stable isotope:

fsK. + -?e — ►• JsAr electron capture

tm = 1.5 X 1010 years (79)

Once again, the ratio of the abundance in a
crystal of the two isotopes, ^Ar/^K, provides a
clue to the age of the crystal. Mica is a mineral
that has been much studied in this type of min-

eral age estimation. Such estimates also tend to
date the minerals as a few billion years old.
Similar age figures are obtained from the natural
radioactivity of rubidium.

In conclusion, the agreement of all of these
methods based upon radioactive decay furnishes
a strong clue that the earth's crust as we know
it today was formed about five billion years ago.
What preceded is a subject of intense interest and
monumental disagreement.

25-2 THE CHEMISTRY OF THE PLANETS

Our solar system consists of the Sun, the planets
and their moon satellites, asteroids (small plan-
ets), comets, and meteorites. The planets are
generally divided into two categories: Earth-like
(terrestrial) planets — Mercury, Venus, Earth,
and Mars; and Giant planets — Jupiter, Saturn,
Uranus, and Neptune. Little is known about
Pluto, the most remote planet from Earth.

As space exploration begins, we can look for-
ward to a vast multiplication of our present
knowledge of the planets. Conceivably we shall
be analyzing samples of the moon within this
decade. The distances to the other planets are
such that voyages of the order of a few months
suffice to reach them. Again information will
accumulate rapidly.

Contrast our position ten years ago. There lay
the planets — ours to see but not to touch. And
not to see well, either. No, we must view the
planets through the gauze curtain of the at-
mosphere. Small wonder we know rather little
about our nearest neighbors, despite our eager-
ness to pry. Yet we do know enough about the
other planets to say that their surface environ-
ments are totally unlike that on earth. Though
much of what is today thought to be true will be
known to be false tomorrow, it provides stimu-
lating reading.

Table 25-IV begins this survey with a com-
parison of mass, radius, and density of the
planets and the sun. These data are probably the
most reliable facts known about the planets since

they are deduced from the orbital movements in
the solar system.

Table 25-IV

DATA ON THE SOLAR SYSTEM

RELATIVE
MASS

RADIUS

(kilometers)

DENSITY

(g/ml)

Sun

3.32 X 106

695

X 103

1.41

Mercury

0.05

2.5

X 10*

5.1

Venus

0.81

6.2

X 10s

5.0

Earth

(1.00)

6.371

X 103

5.52

Mars

0.11

3.4

X 10*

3.9

Jupiter

3.18 X 102

X 103

1.33

Saturn

X 10*

0.71

Uranus

14.6

25.8

X io»

1.27

Neptune

17.3

22.3

X 103

2.22

Pluto

0.03?

2.9

X 103

25-2.1 Meteorites

We do have some direct evidence concerning the
composition of solid matter outside the earth's
atmosphere. Occasionally a piece of solid, a
meteorite, falls through the atmosphere to give
us a real sample for analysis. Such a piece of
solid may become hot because of its rapid move-
ment through the air and therefore glow brightly.
Many meteors burn up or evaporate but some
are large enough to survive and reach the earth's
surface. These we can examine and analyze.

SEC. 25-2 I THE CHEMISTRY OF THE PLANETS

445

Meteorites are of two kinds: stony meteorites
that are rock-like in character, and metallic
meteorites that consist of metallic elements. The
kinds of substances in the stony meteorites are
very much like the substances in the crust of the
earth, if we allow for the fact that the meteors
could not bring gases or liquids with them. We
feel that the other type, the metallic meteors, give
valuable clues about the nature of the earth's
central core. Experts have long believed that
these meteorites are fragments from exploded
planets that, perhaps, resembled the earth.

Whether this is true or not, meteorites give us
one definite piece of information. Isotopic analy-
sis shows that each element in a meteorite has
the same isotopes in the same percentages that
this same element has on earth. The accepted
explanation for this fact is that meteorites and
the earth share a common origin and that they
became separated after the elements were cre-
ated.

these are the substances that can be detected and
other gases are undoubtedly present as well.

There appears to be a correlation between the
mass of the planets and the mass and composi-
tion of their atmospheres. Generally, only those
planets of high mass were able to retain much
of their atmospheres. Nitrogen, hydrogen, and
helium are probably abundant, though not yet
detected, on the heavier planets. Table 25-V also
reveals a considerable range in the surface tem-
peratures of the planets. The higher tempera-
tures on the terrestrial planets also contributed
to the loss of their atmospheres.

EXERCISE 25-4

Nitrogen is considered to be a likely constituent
of the atmosphere of Jupiter, though it is un-
detected as yet. As a chemist, would you expect
oxygen also to be an important constituent of
Jupiter's CH4-NH3 atmosphere?

25-2.2 The Planetary Atmospheres

Through spectroscopic observations and some-
times tenuous deductions there has accumulated
a significant picture of the makeup of the plane-
tary atmospheres. Doubt pervades much of this
picture, yet it represents our starting place in
knowledge as we venture outside our own atmos-
phere for the first time. Table 25-V summarizes
a part of this information — the maximum surface
temperatures and the chemical compositions.
Naturally, these compositions are incomplete:

Table 25-V

THE PLANETARY ATMOSPHERES

MAX. SURFACE

SOME OF THE

PLANET

TEMPERATURE

(°C)

GASES PRESENT

Venus

430

co2

Mercury

350

none

Earth

02, N2, H20, etc.

Mars

N2, C02, HsO

Jupiter

-138

CHt, NH3

Saturn

-153

CH4, NH3

Uranus

-184

CH4

Neptune

-200

CH4

The composition of the planetary atmospheres
is fairly constant. This is indeed surprising in
view of the fact that molecules such as methane,
ammonia, and carbon dioxide are easily decom-
posed by the ultraviolet radiation from the sun.
Presumably other reactions regenerate those sub-
stances that are light sensitive.

The atmosphere of Venus is chiefly carbon
dioxide in a concentration much higher than
that found on Earth. Surprisingly, no evidence
has been found for carbon monoxide, though
ultraviolet light decomposes C02 to form CO.
The atmosphere of Mars is thought to be largely
nitrogen (around °8 %) and some carbon dioxide.

Recent space-probe and earth-based spectro-
scopic studies of the planet Venus suggest how
much remains to be learned about the other
planets. Earlier estimates of the surface tempera-
ture of Venus placed it near 60°C. The more
detailed studies show, however, that two charac-
teristic temperatures can be identified, — 40°C
and 430°C. The lower temperature is attributed
to light emitted from high altitude cloud tops.
The higher temperature is likely to be the average
surface temperature.

446

THE CHEMISTRY OF EARTH, THE PLANETS, AND THE STARS | CHAP. 25

CHjf , NH3 , H^, He atmosphere

Liquid CH4

Ice and solid N'Hj
Rock and mef-al

v % ■

'--■

Liquid (?)

Solid hydrogen.

Ice, high pressure form
Solid hydrogen, high

pressure form

"•:■ :•

These latest findings provide an estimate that
the atmospheric pressure at the ground surface
is 10 ± 2.5 atmospheres. This is the pressure felt
by a deep-sea diver at a depth of 300 feet, near
the limit of human endurance. The atmospheric
temperature at the surface averages around
430°C, rising in some locales possibly as high as
550°C (near the softening temperature of ordi-
nary glass) whereas in other regions, cool breezes
may blow at temperatures below 350°C (near
the boiling point of mercury and the melting
point of lead). Compare these extremes to the
narrow range in which humans can survive com-
fortably. If you have ever experienced a desert
temperature as high as 43°C (1 10°F) or a winter
as cold as -35°C (-30°F), you will realize that
Venus will not be plagued by too many tourists
from Earth.

The giant planets possess low surface tempera-
tures and have atmospheres that extend several
thousand miles. The markings on Jupiter, the
largest planet, consist of cloud formations com-
posed of methane containing a small amount of
ammonia. The atmosphere of Jupiter absorbs the
extreme red and infrared portions of the spec-
trum. These absorptions correspond to the ab-
sorption spectra of ammonia and methane,
suggesting the presence of these gases in Jupiter's

Fig. 25-2. Two proposed structures of Jupiter.

atmosphere. Free hydrogen and helium are also
thought to be present but this is difficult to
verify. Estimates of the average molecular weight
of the gases in Jupiter's atmosphere are around
3. The atmosphere of Jupiter includes belts or
bands that seem to be indicative of climate
variations similar to our own equatorial, tem-
perate, and polar climates.

Jupiter has a single permanent marking called
the Great Red Spot. This spot is oval in shape,
about 30,000 miles long and about 7,000 miles
wide. The coloring is thought to result from light
reflected from the different layers in the planet's
atmosphere. Theories concerning the origin of
this spot and of Jupiter's brightly colored at-
mospheric belts are imaginative, numerous, and
in general disagreement.

25-2.3 The Planetary Lithospheres

Needless to say, the extreme difficulty we experi-
ence in probing the composition of the earth
beneath us suggests that little is known about
the inner composition of the planets. The evi-
dence available is indirect (average density, sur-

SEC. 25-3 I THE STARS

447

.-■'\bH

¥#

£ NH CaH

; OH YO

• AlO ZrO
TiO

CH
Cx

NH
CN

r,'J

.-'

^$H

AlO

SiH

CaH

Tio

Zro

face composition, etc.) and the interpretations
are conflicting. Nevertheless, we present in Fig-
ure 25-2 two proposals that have been offered as
possible structures for the planet Jupiter. A va-
riety of responses are evoked by these startling
proposals: dismay— that the available data are so
inconclusive; discouragement— that our knowl-
edge is so incomplete; anticipation — for the near
future when some of the uncertainties will be
removed ; sympathy— for the poor astronaut who
is to step out of his space vehicle to plant his
flag in an unfriendly sea of, alas, liquid methane.

25-2.4 The Sun

The surface temperature of the sun is about
5500°C. Moving inward from the surface, the
temperature rises, probably above one million
degrees. At these large temperatures the competi-
tion between opposing tendencies toward lowest
energy (favoring molecules) and toward highest
randomness (favoring atoms) is dominated by
the randomness factor. As a result, only the
simplest molecules are to be expected there.

Fig. 25-3. Some molecules detected in the solar at-
mosphere and the elements they contain.

The solar spectrum is, of course, as well
studied as our planetary atmosphere will permit.
More information will be forthcoming as spectra
from man-made satellites are recorded above the
atmosphere. At this time, the spectra of many
diatomic molecules have been detected. These
are not the familiar, chemically stable molecules
we find on the stockroom shelf. These are the
molecules that are stable on a solar stockroom
shelf. Figure 25-3 shows some of these and the
location in the periodic table of the elements
represented.

Inside the sun, thermal energies are sufficient
to destroy all molecules and to ionize the atoms.
These ions emit their characteristic line spectra
and tens of thousands of lines are observed. The
lines that have been analyzed show the existence
of atoms ionized as far as 0+5, Mn+12, and Fe+13.
At this time, over sixty of the elements have been
detected in the sun through their spectral emis-
sions and absorptions.

25-3 THE STARS

Our knowledge of the stars and of space is en-
tirely obtained through spectroscopy and will be
for the forseeable future. That is not at all to say
we know and shall know little of the other gal-
axies. It is to say that our information will be
incomplete. But man is opportunistic and clever
— small pieces of a spectroscopic jawbone may

permit us to build quite a reasonable likeness of
the Universe.

As evidence that this is so, consider that the
element helium was detected in the sun before
it was found on earth! Though oxygen contains
0.2% of the oxygen- 18 isotope on earth, it, too,
was first detected in a solar spectrum. Two

148

THE CHEMISTRY OF EARTH, THE PLANETS, AND THE STARS I CHAP. 25

chemical species whose spectra were first pro-
vided through photographs of comets are CO+
and C3. Let us look briefly then, and with respect,
at the astronomers' knowledge of stellar chem-
istry.

25-3.1 Stellar Atmospheres

Our sun is, of course, a star. It is a relatively cool
star and, as such, contains a number of diatomic
molecules (see Figure 25-3). There are many
stars, however, with still lower surface tempera-
tures and these contain chemical species whose
presence can be understood in terms of the
temperatures and the usual chemical equilibrium
principles. For example, as the star temperature
drops, the spectral lines attributed to CN and
CH become more prominent. At lower tempera-
tures, TiO becomes an important species along
with the hydrides MgH, SiH, and A1H, and
oxides ZrO, ScO, YO, CrO, AlO, and BO.

Detailed consideration of the chemical equi-
libria among these species provides evidence of
the presence of other molecules that cannot be
observed directly. The chemical properties of the
molecules mentioned provide a firm basis for
predicting the presence and concentrations of
such important molecules as H2, CO, 02, N2,
and NO. Thus the faint light by which we view
the distant stars is rich with information. We
need but learn how to read it.

25-3.2 Interstellar Space

In addition to the stars, with their characteristic

light emissions, the space between them is part
of the astronomical spectroscopy laboratory.
Light from a distant star must traverse fantastic
distances to reach our telescopes and the absorp-
tion of this light by the most minute concentra-
tions of atoms and molecules in space becomes
important and detectable. Absorption spectra
have shown that diatomic molecules such as CH,
CN, and CH+ are present at an average concen-
tration of about one molecule per 1000 liters.
These molecules are probably concentrated in
"clouds" with one molecule per 100 liters.

EXERCISE 2S-S

Calculate the volume in liters of a sphere of
radius 6400 kilometers (the radius of the earth).
How many grams of oxygen would be needed
to fill this volume to a concentration of one
molecule per 1000 liters?

In addition to these molecules, atoms are
present, as shown by absorptions of Ca, Na, K,
Fe, and other atoms. There are some absorptions
that have not been identified but these may be
due to small solid particles. How these particular
molecules and atoms happen to be present in
these almost nonexistent "clouds" and what
other molecules and atoms are there, yet to be
detected, is a matter for wondering. But wonder-
ing is at once the pleasure and the driving force
of science.

APPENDIX

A DESCRIPTION OF A
BURNING CANDLE

A drawing of a burning candle is shown1 in Figure
Al-1. The candle is cylindrical in shape2 and has a
diameter3 of about f inch. The length of the candle
was initially about eight inches4 and it changed
slowly5 during observation, decreasing about half
an inch in one hour6. The candle is made of a
translucent7, white8 solid9 which has a slight odor10
and no taste11. It is soft enough to be scratched with
the fingernail12. There is a wick13 which extends from
top to bottom14 of the candle along its central axis15
and protrudes about half an inch above the top of
the candle16. The wick is made of three strands of
string braided together17.

A candle is lit by holding a source of flame close
to the wick for a few seconds. Thereafter the source
of flame can be removed and the flame sustains itself
at the wick18. The burning candle makes no sound19.
While burning, the body of the candle remains cool
to the touch20 except near the top. Within about
half an inch from the top the candle is warm21 (but
not hot) and sufficiently soft to mold easily22. The
flame flickers in response to air currents23 and tends
to become quite smoky while flickering24. In the
absence of air currents, the flame is of the form
shown in Figure Al-1, though it retains some move-
ment at all times23. The flame begins about £ inch
above the top of the candle26 and at its base the flame
has a blue tint27. Immediately around the wick in a
region about \ inch wide and extending about \ inch
above the top of the wick28 the flame is dark29. This
dark region is roughly conical in shape30. Around
this zone and extending about half an inch above
the dark zone is a region which emits yellow light31,
bright but not blinding32. The flame has rather

JDasiA

Ccr&r^e&uL, Mf^ou^—f^^

T3supfut yt&trus-

&Ma>.

CUnnot /£ ^vcJi

'S;M^m

Fig. Al-1. A burning candle.

sharply defined sides33, but a ragged top34. The wick
is white where it emerges from the candle35, but from
the base of the flame to the end of the wick36 it is
black, appearing burnt, except for the last r\ inch
where it glows red37. The wick curls over about J

449

450

A DESCRIPTION OF A BURNING CANDLE I APP. 1

inch from its end38. As the candle becomes shorter,
the wick shortens too, so as to extend roughly a
constant length above the top of the candle39. Heat is
emitted by the flame40, enough so that it becomes
uncomfortable in ten or twenty seconds if one holds
his finger \ inch to the side of the quiet flame41 or
three or four inches above the flame42.

The top of a quietly burning candle becomes wet
with a colorless liquid43 and becomes bowl shaped44.
If the flame is blown, one side of this bowl-shaped
top may become liquid, and the liquid trapped in the
bowl may drain down the candle's side45. As it
courses down, the colorless liquid cools46, becomes
translucent47, and gradually solidifies from the out-
side48, attaching itself to the side of the candle49. In
the absence of a draft, the candle can burn for hours
without such dripping60. Under these conditions, a
stable pool of clear liquid remains in the bowl-shaped
top of the candle61. The liquid rises slightly around
the wick62, wetting the base of the wick as high as
the base of the flame63.

Several aspects of this description deserve specific
mention. Compare your own description in each of
the following characteristics.

(1) The description is comprehensive in qualitative
terms. Did you include mention of appearance?

smell? taste? feel? sound? (Note: A chemist
quickly becomes reluctant to taste or smell an
unknown chemical. A chemical should be con-
sidered to be poisonous unless it is known not
to be!)

(2) Wherever possible, the description is stated
quantitatively. This means the question "How
much?" is answered (the quantity is specified).
The remark that the flame emits yellow light is
made more meaningful by the "how much" ex-
pression, "bright but not blinding." The state-
ment that heat is emitted might lead a cautious
investigator who is lighting a candle for the first
time to stand in a concrete blockhouse one hun-
dred yards away. The few words telling him
"how much" heat would save him this overpre-
caution.

(3) The description does not presume the importance
of an observation. Thus the observation that a
burning candle does not emit sound deserves to
be mentioned just as much as the observation
that it does emit light.

(4) The description does not confuse observations
with interpretations. It is an observation that the
top of the burning candle is wet with a colorless
liquid. It would be an interpretation to state the
presumed composition of this liquid.

APPENDIX

RELATIVE STRENGTHS OF
ACIDS

IN AQUEOUS SOLUTION AT ROOM TEMPERATURE

All ions are aquated

ACID

STRENGTH

REACTION

perchloric acid

hydriodic acid

hydrobromic acid

hydrochloric acid

nitric acid

sulfuric acid

oxalic acid

sulfurous acid (S02 + H20)

hydrogen sulfate ion

phosphoric acid

ferric ion

hydrogen telluride

hydrofluoric acid

nitrous acid

hydrogen selenide

chromic ion

benzoic acid

hydrogen oxalate ion

acetic acid

aluminum ion

carbonic acid (C02 + H20)

hydrogen sulfide

dihydrogen phosphate ion

hydrogen sulfite ion

ammonium ion

hydrogen carbonate ion

hydrogen telluride ion

hydrogen peroxide

monohydrogen phosphate ion

hydrogen sulfide ion

water

hydroxide ion

ammonia

very strong

::yery strong

strong

I
J

weak

I
»

very weak

HCIO4

HBr

HC1

HNO3

H2SO4

HOOCCOOH

H2SO3

HSO4-

H3PO4

Fe(H20)6+3

H2Te

HN02

H2Se

Cr(H20)6+3

C«H5COOH

HOOCCOO-

CH3COOH

A1(H20)6+3

H2C03

H2S

H2P04-

HSO3-

NH4+

HCO3-

HTe"

H202

HPO4-2

HS"

H20

OH-

NH3

H+ +
H+ +
H+ +
H+ +

H+ +
H+ +

H*
FT
H^
H-
H"»
W
H-
H-

H+ +

H"
H-
H^
H^
H^

H^
HJ
HJ
H*
1-T
H*
H'

CIO4-

I-

Br-

ci-

NO3-

HSO4-

HOOCCOO-

HSO3-

SO4-2

H2P04"

Fe(H20)5(OH)+2

HTe"

NOr

HSe-

Cr(H20)5(OH)+2

QH6COO-

OOCCOO-2

CH3COO-

Al(H20)6(OH)+J

HCO,-

HS-

HPO4-1

SO,"2

NH,

CO3-2

Te"2

HOr

po4-»

s_2

OH-
O2

NH2

[H+][OH"] =

very large
very large
very large
very large
very large
large

5.4 X 10-2
1.7 X 10~2
1.3 X 10-
7.1 X 10-

6.0 X 10-

2.3 X 10-
6.7 X 10-

5.1 X 10"<

1.7 X 10"«

1.5 X 10-

6.6 X 10"

5.4 X 10-

1.8 X 10"s
1.4 X 10-
4.4 X 10-'

1.0 x 10-7

6.3 X 10"8

6.2 X lO-8

5.7 X lO"10
4.7 X 10-11
1.0 X 10"11

2.4 X lO"12
4.4 X 10-"

1.3 X 10"13

1.0 X 10-"
<10"36
very small

451

APPENDIX

STANDARD OXIDATION
POTENTIALS FOR
HALF-REACTIONS

IONIC CONCENTRATIONS, 1 M IN WATER AT 25 °C

All ions are aquated

HALF-REACTION

E° (volts)

Very strong

reducing

agents

Li -

Rb -
K-
Cs-
Ba-

Sr ■
Ca-
Na-
Mg-
Al-
Mn
H,fo) + 20H- ■
Zn-
Cr
H2Te
2Ag + S~2
Fe
H*{g) ■
Cr-2
H2Se
Co
Ni
Sn
Pb

H2S(tf)

Sn4*

Cu+

S02(<?) + 2H20

21-

H,02

e~ + Lr
e- + Rb+
er +K+
e- +Cs+
2e- + Ba41
2*- + Sr«
2e- -fCa4*
e- + Na+
2e~ + Mg42
3e- + Al4*
2e- + Mn41
2e~ + 2H,0
2e- + Zn-5
3e- + Cr-3
2e- + Te + 2H+

• 2e- + Ag:S

■ 2*" + Fe+2

2e~ + 2H+ (10-7 Af)

• e~ + Cr+3

- 2e~ + Se + 2H+

■ 2e~ + Co~-

■ 2e~ + Ni^

- 2e- + Sn-5

- 2<r + Pb*2

• 2e- + 2H+

- 2e~ + S + 2H+

- 2e- + Sn4*

- er + Cu+2

- 2e~ + SOr* + 4H+

- 2e~ + Cu'2

- er + Cu+

- 2e- + I,

- 2e~ + 0,fa) + 2H+

3.00

Very weak

2.92

oxidizing

2.92

agents

2.92

2.90

2.89

2.87

2.71

2.37

1.66

1.18

0.83

0.76
0.74

0.72
0.69

C.44
0.414

0.41
0.40

(jo

0.28

0.25

0.14

0.13

0.00

-0.14

-0.15

-0.17

-0.34

-0.52

-0.53

-0.68

452

APP. 3 I STANDARD OXIDATION POTENTIALS FOR HALF-REACTIONS

453

APPENDIX 3— (Continued)

HALF-REACTION

E° (VOLTS)

-0.77

-0.78

-0.79

-0.80

-0.815

-0.96

-1.00
-1.06

3
OS

-1.23

3
O

-1.28

-1.33

-1.36
-1.50

-1.52

Very strong

-1.77

oxidizing

-2.87

agents

Very weak

reducing

agents

Fe--

N02((?) + H20

Hg(0

HsO

NOfo) + 2H20

Au + 4C1-

2Br-

H20

Mn+2 + 2H20

2Cr+> + 7H20

2C1-

Mn+2 + 4H20

2H20

2F-

e~ + Fe*3

e~ + NCV + 2H+

2e~ + Hg+2

e~ + *Hg2«

e~ + Ag+

2e~ + iO^g) + 2H+ (10-7 M)

3e- + N03" + 4H+

3e" + AuCU-

2e~ + Br2(0

2e~ + hOtig) + 2H+

2e~ + MnO, + 4H+

6e~ + Cr2CV2 + 14H+

2e~ + a,(g)

3e~ + Au+3

5er + MnCV + 8H+

2e~ + H202 + 2H+

2e~ + Ft(g)

APPENDIX

NAMES, FORMULAS,
AND CHARGES OF SOME
COMMON IONS

POSITIVE IONS (CATIONS)

aluminum

ammonium

barium

calcium

chromium (II), chromous

chromium (III), chromic

cobalt (II), cobaltous

copper (I),* cuprous

copper (II), cupric

hydrogen, hydronium

iron (II),* ferrous

iron (III), ferric

lead

lithium

magnesium

manganese (II), manganous

mercury (I),* mercurous

mercury (II), mercuric

potassium

silver

sodium

strontium

tin (II),* stannous

tin (IV), stannic

zinc

Al+»

NH4+

Ba+2

Ca+2

Cr*2

Cr+J

Co"2

Cu"

Cu+2

H+, H3CT

Fe+2

Fe+3

Pb+2

Li+

Mg«

Mn*2

Hgo"2

Hg"2

K+

Ag+

Na+

Sr2

Sn"2

Sir"

Zn"2

NEGATIVE IONS (ANIONS)

acetate CH3COO-

bromide Br-

carbonate CO.T2

hydrogen carbonate ion, bicarbonate HCO3-

chlorate CIO3-

chloride Cl-

chlorite CKV"

chromate Cr04-2

dichromate Cr^Or-2

fluoride F~

hydroxide OH-

hypochlorite CIO-

iodide I~

nitrate NO3-

nitrite NOr

oxalate CjOi-2

hydrogen oxalate ion, binoxalate HGO,-

perchlorate ClOr

permanganate Mn04-

phosphate POj-3

monohydrogen phosphate HPO<-2

dihydrogen phosphate H2POr

sulfate SOi-2

hydrogen sulfate ion, bisulfate HSO,"

sulfide S-2

hydrogen sulfide ion, bisulfide HS"

sulfite S03-a

hydrogen sulfite ion, bisulfite HSO3-

* Aqueous solutions are readily oxidized by air.

Note: In ionic compounds the relative number of positive and negative ions is such that the sum of their electric
charges is zero.
454

Index

Absolute temperature, 57

Kelvin scale, 58
Absolute zero, 58

Abundance of elements in earth's
crust, see Elements, abundance
in earth's crust
Acetaldehyde structure, 332
Acetamide, 338
Acetanilide, 344
Acetic acid

in biochemistry, 428

structure, 333
Acetone

solubility of ethane in, 313

structure, 335
Acetylene, 43
Acid-base

Br0nsted-Lowry theory, 194

contrast definitions, 194

indicators, 190

reactions, 188

titrations, 188
Acids, 183

aqueous, 179

carboxylic, 334

derivatives of organic, 337

equilibrium calculations, 192

experimental introduction, 183

names of common, 183

naming of organic, 339

properties of, 183

relative strengths, 192, 451

strength of, 190

summary, 185

weak, 190, 193
Actinides, 414
Actinium

electron configuration, 415

oxidation number, 414
Activated complex, 134
Activation energy, 132, 134, 369
Activities of Science, /
Addition polymerization, 346
Additivity of reaction heats,

law of, 111
Adipic acid, 347
Adsorption, 138
Age of earth, 442
Air

as source of elements, 442

composition, 438

oxygen content, 228
Alanine, 347
Alcohols, 330

naming. 338

oxidation of, 336
Aldehydes, 332
Alkali metals, 93

atom models, 98

atomic volume, 95

chemistry, 95

ionic bonds, 95

properties, table, 94

reactions with chlorine, 95

reaction with water, 96
Alkaline earth elements

atomic radii, 378

chemical properties, 381

electron configuration, 378

ionization energies, 379

occurrence, 384

preparation, 384

properties of, 377, 381

solubilities, 382
Alkaline earth hydroxides

heat of reaction to form, 382

K.p, 383
Alkaline earth oxides, heat of reac-
tion with water, 382
Alkaline earth sulfates, K,p, 382
Alkanes, 341

naming, 338
Alkyl group, 336
Alloys, 309

Alnico, 406

copper, 71, 309

covalent bonds, and, 305

gold, 71

hardness and strength, 311

nickel, 407
Alnico, 406

Alpha carbon atoms, 348
Alpha decay, 417, 443
Alpha particle, 417

scattering, 245
Aluminum

boiling point, 365

compounds, 102

heat of vaporization, 365

hydration energy, 368

hydroxide, 371

ionization energies, 269, 374

metallic solid, 365

occurrence, 373

properties, 101

preparation, 238, 373

reducing agent, 367
Alums, 403
Americium

electron configuration, 415

oxidation numbers, 414
Amides, 338

naming, 339
Amines, 336

naming, 338
Amino acids, 347, 348, 432

Ammonia

a base, 184

boiling point, 64

complexes, 392, 395, 408

complex with Ag+, 154

Haber process for, 150

and hydrogen chloride, 24

model of, 21

molar volume, 60, 64

production, 150

P-V behavior of, 19, 51,60

solubility, 20
Ampere, 241
Amphoteric, 371

complexes, 396
Analogy

billiard ball, 6, 18

car collision, 126, 129

dartboard, 261

garbage collector, 233

golf-ball, 155

lost child, 3

mountain pass, 132

notched beam, 256
. station-wagon, 155
Angstrom, A. J., 258
Aniline, 344
Anions, 170, 207
Anode, 207
Anode sludge, 408
Anthracite, 321
"Antifreeze," 72
Antimony, 31
Aq notation, 79
Aqueous, 79
Aqueous solutions

of acids, 179

of bases, 179

electrical conductivity of, 78

of electrolytes, 78, 313

precipitation reactions in, 80
Argon, 91

boiling point, 374

heat of vaporization, 105, 374

ionization energy, 268

melting point, 307

occurrence, 373

preparation, 374

properties, 101

use 105
Aromatic compounds, 344
Arrhenius, Svante, 198
Arsenic, ionization energy, 410
Asbestos, 310
Aspirin, 346
Astatine

oxidation number, 414

properties, 97

Boldface numbers refer to definitions. Italic numbers refer to sections.

455

456

INDEX

Astronaut, 231
Atmosphere, 437

composition, 437

one, 54

properties, 437
Atmospheric pressure, 53
Atom

chemical evidence for existence,
234

defined, 21

mass and charge of parts, 87

nuclear model, 86

number per molecule, 26

sizes, 88

structure, 86
Atomic hydrogen spectrum, 253
Atomic number. 88

and periodic table, 89

table, inside back cover
Atomic orbitals, 262, 263
Atomic pile, 120
Atomic theory, 17, 22, 28, 234

as a model, 17

chemical evidence for, 234

of John Dalton, 236

review. 34
Atomic velocity distribution, 130, 131
Atomic volume, 94, 98

alkali metals, 94

halogens. 97

inert gases. 91

third-row elements, 101
Atomic weight. 33

table, inside back cover
Atoms, 21

conservation of, 40

electrical nature of. 236

measuring dimensions of, 245

AVOGADRO, AMADEO

hypothesis, 25, 52
hypothesis and kinetic theory, 58
law. 25
number, 33
Azo dyes, 344

Bacteria, 434
Balancing reactions, 42

by half-reactions, 218

by oxidation number, 219

oxidation-reduction reactions, 217,
219
Balmer, J. J., 258
Balmer series, 258
Barite, 385
Barium

atomic size, 379

chemistry, 382

electron configuration, 378

heat of vaporization, 305

hydroxide, Ksp, 383

ionization energies, 379

occurrence, 385

properties, 381
Barometer, 53

standard, 54
Barrier, potential energy, 134
Bases, 185

aqueous. 179

common, 185

experimental introduction, 183

properties of, 183, 184
Basic oxides, 382
Battery, storage

Edison. 406

lead, 406
Bauxite. 373

Benzene

derivatives, 343

modification of functional group,
344

representation of. 343

substitution reactions, 343
Benzoic acid, 192
Berkelium, oxidation numbers for,

414
Beryl, 385
Beryllium

atomic size, 379

boiling point, 374

bonding capacity, 285

chemistry of, 382

electron configuration, 378

heat of vaporization, 374

ionization energies, 379

occurrence, 384

preparation, 385

properties, 381

structure, 381
Beryllium difluoride, dipole in, 293
Berzelius, Jons, 30
Bessemer converter, 404
Beta decay, 417
Beta particle, 417
Bicarbonate ion, 184
Bidentate, 395
Billiard ball analogy, 6, 18

and kinetic energy, 114
Billiard ball collision, conservation

of energy in, 1 14
Binding energy, 121, 418
Biochemistry, 421
Bismuth, oxidation numbers, 414
Blast furnace, 404
Bohr, Niels, 259
Boiling point, 67

elevation, 325

normal, 68

pure substances, table, 67
Bond

breaking and rate, 125

chemical, 31, 35, 274

covalent, see Covalent bond

covalent and ionic contrasted, 287

covalent and network solids, 302,
309

double, 295

energies and electric dipoles, 290

energy, 119

energy of halogens, 355

hydrogen, see Hydrogen bond

infrared identification, 250

ionic, see Ionic bond

length, 276, 295

origin of stability, 215

overlap, 277

reason for forming, 274

stability, 275

stability and acid strength, 187,
188

type, 286

types in fluorine compounds, 289
Bonding

capacity and molecular shape, 293

dsp\ 395

</V\ 395

electron dot representations, 278

in complex ions, 395

in fluorine, 278

in gaseous lithium fluoride, 287

in oxygen molecule, 295

in solids and liquids, 300

metallic, 303

orbital representation, 278

p>, 291

P3, 291

representations, 278

sp. 292

sp\ 292

sp3, 292

tetrahedral, 292
Bonding capacity of

beryllium, 285

boron, 285

carbon, 284

lithium, 286

nitrogen, 283

second-row elements, 281
Bordeaux mixture, 408
Boron

boiling point, 374

bonding capacity, 285

heat of vaporization, 374

ionization energies, 273
Boron trifluoride, dipole in, 294
Boyle's Law, 17

Brackets, concentration notation, 151
Brass, 311
Bromate ion, 360
Bromine

atomic volume, 410

boiling point, 307

color, 352

covalent radius, 354

electron configuration, 353

electron dot representation, 353

ionic radius, 355

ionization energy, 410

melting point, 307

orbital representation, 353

preparation. 356, 361

properties, 97, 355

van der Waals radius, 354
Bronsted-Lowry theory, 194
Burning

candle, 449

methane, 41

paraffin, 44
Butane, 338

properties, 341
/.ro-butane properties, 341
Butanoic acid, properties, 308
1-Butanol, 338
Butyl alcohol, 338
1-Butylamine, 338
Butyramide, 339
Butyric acid, 339

Calcium

atomic size, 379
atomic volume, 410
chemistry, 382

electron configuration, 271, 378
heat of vaporization, 305
ionization energies, 379
occurrence, 385
properties. 381

Calcium carbonate, 384

Calcium hydroxide, 62, 383

Calculations

based on chemical equations, 44
of H+ concentration, 192
gas volume-gas volume, 227
liquid volume-volume, 230
weight-gas volume. 226
weight-liquid volume, 228
weight-weight, 226

Californium
electron configuration, 415
oxidation number, 414

Boldface numbers refer to definitions. Italic numbers refer to sections.

INDEX

457

Calorie, 10

Calorimeter. ///

Calorimetry. Ill

Candle, description of burning, 449

Caprslamide. 339

Capr\lic acid, 339

Carbohydrates, 422

Carbon

alpha (a), 348

boiling point, 374

bonding in. 284

compounds, sources of, 321

compounds, composition and
structure. 323

co\alent radius, 365

electron configuration. 265

heat of vaporization, 374

ionization energy, 268
Carbon dioxide, 28

molar volume. 51. 60
Carbon disulfide, infrared spectrum,

249
Carbon monoxide

absorption by. 251

boiling point. 64

molar volume. 51. 60, 64

poisoning, 398
Carbon tetrabromide

boiling point. 307

bond length. 354

melting point, 307
Carbon tetrachloride

boiling point, 307

bond length, 354

infrared spectrum 249

melting point, 307

poison. 82

solubility of ethane in, 313

use. 82
Carbon tetrafluoride

boiling point, 307

bond length. 354

dipole in. 294

melting point, 307
Carbon tetraiodide

boiling point, 307

bond length. 354

melting point, 307
Carbonyl group. 335
Carboxyl group. 334, 337
Carboxylic acids, 333, 334, 425
Car collision analogy, 126, 129
Cast iron, 404
Catalysis, 135
Catalysts

action of, 135

enzymes, 138

and equilibrium, 148

examples of, 137

in formic acid decomposition, 137

in manufacture of H2S04, 227

and rusting, 405
Cathode, 207
Cathodic protection, 405
Cations, 170, 207
Cell

dry, 403

electrochemical, 199, 206
Cellulose, 425

structure, 431
Centigrade temperature, 57, 58

relation to °K, 58
Cerite, 413
Cerium

properties, 412

source, 413
Certainty, absence of, //

Cesium, 23

chemistry. 95

heat of vaporization, 305

properties. 94
Chain reaction, 419
Chalcocite, 46
Chalcopyrite, 408
Charge

conservation of, 80

electric, 75

electron, 241

positive, 242, 243

separation, 312
Charge mass ratio of electrons,

240
Charles, Jacques, 57
Charles' law, 58
Chemical bonding, see Bonding
Chemical bonds, see Bond
Chemical change, 38
Chemical energy, 1 19
Chemical equations, see Equations
Chemical equilibrium, law of. 152
Chemical formulas, see Formula
Chemical kinetics. 124
Chemical reactions, see Reactions
Chemical stability. 30
Chemical symbols. 30

not from common names, 31

see inside back cover
Chemotherapy. 434
Chlorate ion, 360
Chloric acid, 359
Chlorides

chemistry of, 99

of alkali metals. 93, 103

of third-row elements, 103
Chlorine

boiling point, 374

color, 352

compounds, 102

covalent radius, 354

electron configuration, 353

heat of fusion, 69

heat of reaction to form atoms, 290

heat of vaporization, 347

ionic radius, 355

ionization energy, 268

melting point, 307

molar volume, 60, 64

occurrence, 373

oxidation numbers, 359

oxidizing agent, 369

oxyacids, 359, 373

preparation, 356, 374

properties, 97, 101, 355

structure, 366

use, 106

van der Waals radius, 354
Chloroacetic acid, 349
Chlorobenzene, 345
Chloromethane

solubility in CC14, 313

solubility in acetone, 313
Chlorophyll, 431

structure, 397
Chlorous acid, 359
Chromate ion, structure, 402
Chromatography, 413
Chromium

electron configuration, 389

oxidation number, 391

properties, 400, 401

radius, 399
Chromium(III) oxide, 402
Chromium trihydroxide, 396, 402
Cis-trans isomerism, 296, 394

Citric acid. 428
Cleaning solution, 403
Coal. 321
Coal gas, 322
Coal tar, 322
Cobalt

atomic radius, 399

electron configuration, 389

ore, 410

oxidation numbers, 391

properties, 400, 406
Coke, 322, 404
Collision theory, 126

concentration effect, 126

and organic mechanism, 331

temperature effect, 129
Color, 399

Colorimetric analysis, 151
Communicating scientific informa-
tion, 12
Combining volumes, law of, 26, 236
Combustion, heat of hydrogen, 40
Complex ions, 392

amphoteric, 396

bonding in, 395

formation. 413

geometry of, 393

in nature, 396

isomers, 394

linear, 395

octahedral, 393

significance of, 395

square planar, 395

tetrahedral, 394

weak acids, 396
Compound, 28

bonding in, 306
Concentration

and equilibrium, 148

and E zero's, 213

and Le Chatelier's Principle, 149

effect on reaction rate, 126, 128

molar, 72

of solids in equilibria expressions,
153

of water in equilibria expressions,
154

pH. 190
Conceptual definition, 195
Condensed phases, 27, 68, 78

electrical properties, 78
Conductivity, electrical

in metals, 81

in water solutions, 78

of solids, 80
Condensation polymerization, 346
Conservation

of atoms, 40

of charge, 80, 218

of mass, 40
Conservation of energy

in a billiard ball collision, 114

in a chemical reaction, 115

in a stretched rubber band, 114

law of, 113, 117, 207
Constant heat summation law, 1 1 1
Contact process, H5SO4, 227
Coordination number, 393
Copper

alloys, conductivity of, 309, 311

atomic radius, 399

electron configuration, 389

heat of fusion, 69

heat of melting, 69

native, 408

occurrence, 408

oxidation numbers, 391

Boldface numbers refer to definitions. Italic numbers refer to sections.

458

I N D L; X

Copper (continued)

properties, 400. 408
Core, of earth, 440
Corrosion, 405
Coulomb, 241
Coulombic forces, 416
Coulson, C. A., 252
Covalent bonds, 274, 277, 288

elements that form solids using,
302

in halogens, 97

vs. ionic, 287

and network solids, 302, 309
Covalent radius, 354

of halogens, 354
Crawford, Bryce, Jr., 274
Cryolite, 393
Crystal

ionic, 81, 311

metal, 304

molecular, 81, 302
Crystallization, 70, 71

and equilibrium, 144

fractional, 413
Curium, oxidation number, 414
Cyclohexane, properties, 341
Cyclopentane, 340
Cyclopentene, 342

d orbitals, 262

dsp1 bonding, 395

d2sp* bonding, 395

"Dacron," 347

Dalton, John, 12, 38, 236

atomic theory, 236
Dartboard analogy, 261
Da Vinci, Leonardo, 1
Debye, Peter J., 320
Decomposition

of formic acid, 137

of water, 40, 115
Definite composition of compounds,

Law of, 234
Definitions, see under word of in-
terest

conceptual, 195

experiment, 2

operational, 195
Delta H (AH), 110
Delta H cross (AHt), 135
Density, 28
Derived quantity, 10
Deuterium, 90, 123, 419
Deuterium bromide, infrared spec-
trum, 248
Diamond, structure, 302, 365
Diatomic, 31
Diborane, 285
Dichloroethylene, isomers of, 297

properties, 308
Dichromate ion, 402
Diffraction

of X-rays, 248

patterns, 248
Dipoles, 288

BF3, 294

BeF2, 293

and bond energy, 290

CF4, 294

F20, 294

geometrical sum, 293

LiF, 293

molecular, 293

and molecular shape, 293

of ionic bond, 288

and solvent properties, 313

Disaccharides, 424
Discharge tube, 239, 255
Disproportionation, 361
Dissociation, 179

constant of HF, 361
Dissolving rate, 164
Distillation, 70
Disulfur dichloride, 103
Divalent, 282
Divanadium pentoxide, 401

DbBEREINER, J. W., 104

Dolomite, 385

Double arrows, use of, 146

Double bonds, 295

Dry cells, 199, 403

Dyes, azo, 344

Dynamic nature of equilibrium, 144,

165
Dysprosium, properties, 412

E = mc\ 121, 418
E°, 209

and equilibrium, 215

table, 452
Earth

age of, 442

chemistry of, 437

core of, 440

crust of, 441

data, 444

parts of, 438
Edison storage battery, 406
Einstein, Albert, 121
Einsteinium, oxidation number, 414
Elastic collision, 6
Electrical nature of atoms, 236
Electrical phenomena, 74
Electrical properties of condensed

phases, 78
Electric arc furnace, 404
Electric charge, 75

detection, 74

effect of distance, 76

in matter, 77

interactions, 75

negative, 77

positive, 77

production, 76

types, 76
Electric conductivity, see Conduc-
tivity
Electric current, 78
Electric dipoles, see Dipoles
Electric discharge, 239
Electric force, 76, 77
Electricity, fundamental unit, 241
Electrochemical cell

chemistry of, 199

and Le Chatelier's Principle. 214

operation, 206

standard half cell, 21C
Electrodes, 207
Electrolysis, 220, 221

apparatus, 40

cells, 238

of water, 40, 115
Electrolytes, 169, 179

strong, 180

weak, 180
Electrolytic conduction, 220, see also

Conductivity, electrical
Electrometer, 75
Electron, 77

affinity, 280

affinity of fluorine atom, 280

behavior in metals, 304

capture, 417

charge, 241

charge/mass ratio, 240

competition for, 205

deflection, 239

evidence for existence, 236

losing tendency, 207

mass of, 87, 241

"sea" of, 304

"seeing," 239

transfer, 201, 202

valence, 269

van der Waals forces and, 306
Electron configuration, 265

of alkaline earths, 378

of halogens, 352

of lanthanides, 415

of transition elements, 389
Electron dot representation of chemi-
cal bonding, 278
Electron populations of inert gases,

92, 263
Elements, 29

abundance in earth's crust, 441

availability, 441

covalent solids formed by, 302

discovery, 30

in ocean, 440

known to ancients, 30, 441

metals, 303

molecular crystals formed by, 301

naming, 30, 31

on Sun, 447

symbols, see inside back cover
e/m, 240, 243
Emerald, 385

Empirical formula, 81, 323, 324
Endothermic reaction, 40, 135
Energy

activation, 132, 134, 369

binding, 121,418

changes on warming, 119

chemical, 119

effect on equilibrium, 167

kinetic, see Kinetic energy

law of conservation of, 113, 117,
207

molecular, 118

nuclear, 119, 418

of hydrogen bond, 315

of light, 253

of motion, see Kinetic energy

of position, see Potential energy

potential, see Potential energy

stored in living organisms, 428

stored in a molecule, 117

stored in the nucleus, 110
Energy levels, 259

notched beam analogy, 256

of H atom, 258, 259, 264

of many-electron atoms, 265, 266
Enzymes, 138, 427,452
Equations

balancing, 42

calculations based upon, 44, 226

chemical, 41

symbols used in, 41, 146

writing, 42
Equilibrium, 67, 144, 145, 158

altering, 147

and chemical reactions, 145

and E°, 215

and Le Chatelier's Principle, 149

and rate, 155

and solubility, 144, 163

attainment of, 148

Boldface numbers refer to definitions. Italic numbers refer to sections.

INDEX

459

Equilibrium (continued)

calculations, 192

constant, 151, table, 154

crystallization and, 144

dynamic nature of, 144, 165

effect of catalyst, 148

effect of concentration, 148; of en-
ergy, 167; of randomness, 166;
of temperature, 67. 148, 167

factors determining, 155, 158

law of chemical, 152, 173

liquid-gas, 66

qualitative aspects of, 142

quantitative aspects of, 151

recognizing, 143

state of, 142, 147

sugars, 423

thermal, 56

use of double arrows, 146
Equilibrium constant, 151

for water, 181

table, 154

table (acids), 191, 451
Equilibrium Law relation, table, 154
Erbium, properties, 412
Esters, 337
Ethane, 323

properties, 341

solubility in CC14; in acetone, 313
Ethanol, 323

boiling point, 329

determining structural formula,
326
Ethylamine, 338
N-Ethyl acetamide, 338
Ethyl bromide, 328

reactions of, 330
Ethyl iodide, 336
Ethylene, 346

chemical reactivity, 296

double bond in, 296
Ethylene glycol, 325
Ethyl group, 329
Europium, properties, 412
Exothermic reaction, 40, 135
Experiment, 2

Experimental errors, see Uncertainty
Eyring, Henry, 124, 141
E zero, 209, table, 452

/orbitals, 262

Fable, Lost Child in Woods, 3

Faraday, Michael, 237

Fats, 425

Fermentation, 426

Fermium, oxidation number, 414

Ferromanganese, 403

Fission, nuclear, 120, 419

Flea, 88

Fluorescence, 239, 409

Fluorides

bonding capacity, 293

bond types in second row, 286

of second-row elements, melting
and boiling points, 286

orbitals, 293

shape, 293
Fluorine

boiling point, 374

bonding, 278

color, 352

complex, 393

covalent radius, 354

electron affinity, 280

electron configuration, 265

electron dot representation, 353

heat of reaction to form atoms, 290

heat of vaporization, 374

ionic character of bonds, 288

ionic radius, 355

ionization energy, 268

melting point, 307

orbital representation, 353

preparation, 356

properties, 355

special remarks, 361

van der Waals radius, 354
Fluorine compounds, bond type, 289
Fluorine oxide

dipole in, 294

molecular shape, 291
Fluorocarbons, 362
Fool's gold, 404
Force, electric, 76, 77
Formaldehyde, 332
Formamide, 339
Formation, heat of, 113
Formic acid, 197, 333

catalytic decomposition, 137
Formula

chemical, 30

empirical, 81, 323, 324

molecular, 31, 323, 325

structural, 31, 323, 326
Fourth row of periodic table, 271,

387
Fractional crystallization, 413
Freezing point lowering, 325, 393
"Freon," 362
Frequency of light, 246

relation to wave length, 251
Fructose, 423
Fumaric acid, 428

properties, 308

structure, 316
Functional groups, 330, 335
Fundamental

property, 78

unit of electricity, 241
Furnace, electric arc, 404
Fusion, heat of, 68

pure substances, table, 69
Fusion, nuclear, 121, 419

Gadolinite, 413
Gadolinium, properties, 412
Gallium

atomic radius, 399

ionization energy, 410
Galena, 373

Galvanized iron, 405, 409
Garbage collector analogy, 233
Gas

elements found as, 65

ideal, 60

inert, 90

and kinetic theory, 49, 53

liquid-gas equilibrium, 66

liquid-gas phase change, 66

measuring pressure, 53

model of, 18, 23

molar volume, 49, 50, 51, 60

molecules in, 274

molecular weight, 34, 51

natural. 322

perfect, 59

pressure, 53

pressure, cause of, 54

properties of, 20

review, 61

solubility, 20, 167

volume, calculation of, 227

volume, change with temperature,
57

volume measurement, 24
Gasoline, 46, 63, 126, 341
Gastric juice, 1 38
Gaviota Pass, 136
Gay-Lussac, Joseph L., 49
Geiger counter, 240
Generalization, 4, 153

melting of solids, 4

reliability of, 59
Geometry

molecular, 290-297

of complex ions, 393
Germanium, ionization energy, 410
Glucose, 422
Glutamic acid, 347
Glyceraldehyde, 427
Glyceric acid, 427
Glycerol, 425
Glycine, 347
Gold

alloy, 71

oxidation numbers, 414
Goldschmidt reaction, 401
Golf ball analogy, 155
Graphite

burning, 46

structure, 302
Gypsum, 385

h (Planck's constant), 254

Hand AH, 110

[H+], calculation of, 192, see also

Hydrogen ion
Haber, Fritz, 151
Haber process, 140, 150
Hafnium, oxidation number, 414
Haldane, J. B. S., 436
Half-cell potentials

effect of concentration, 213

measuring, 210

standard, 210

table of, 211,452
Half-cell reactions, 201
Half-life, 416
Half-reaction, 201

balancing, 218

potentials, 452
Halides

chemistry of, 99

ions, 98
Hall, C. M., 96. 373
Halogens

atom models, 98

bond energies, 355

chemistry, 98

color, 352

covalent bonds, 97

covalent radius, 355

electron configuration, 352

ionization energies, 353

oxyacids, 358

positive oxidation states, 358

preparation. 356

properties, 96; table, 97. 352. 355

reactions, energy required for. 357

reactions of compounds, 356

reduction of, 357

sizes of atoms and ions, 354

toxicity. 352
Hardness of metals and alloys, 31 1
Hard water, 384
Heat and chemical reactions, 108

Boldface numbers refer to definitions. Italic numbers refer to sections.

460

INDEX

Heat content, 110, 116

change during a reaction, 110

of a substance, 109
Heat of combustion of

diamond, 122

graphite, 122

hydrazine, 47

hydrogen, 40

methane, 123
Heat of formation. 113
Heat of reaction, 135

between elements, table, 112

oxidation of HC1, 160

oxidation of sulfur dioxide, 161

predicting, 112
Heat of reaction to form

ammonia, 1 12

Br atoms, 290

carbon dioxide, 112

carbon monoxide, 112

CI atoms, 290

CO + H2, 110

ethane, 112

F atoms, 290

H atoms, 274

hydrogen chloride, 160

hydrogen iodide, 112

iron(lll) oxide, 162

Li atoms, 290

Li + Br, 290

Li + F, 290

Na + CI, 290

NH3 products, 114

Na atoms, 290

NO, 112

N02, 112

N*3 products, 114

P4Oio, 140

propane, 112

sulfur dioxide, 112

sulfuric acid, 112

water, 109, 112
Heat of solution, 166

Cl2 in H20. 167

h in benzene, 167

I2 in ethyl alcohol, 166

I2 in CCU, 166

02 in HoO, 167

N20 in H20, 167
Heat of vaporization, 66; see also

Vaporization
Helium, 91

boiling point, 63

heat of vaporization, 105

interaction between atoms, 277

ionization energy, 268

molar volume, 60

on Sun, 447

source, 91
Hematite, 404
Hemin, structure of, 397
Hess's Law, 111
Heterogeneous, 70

systems and reaction rate, 126
n-Hexane properties, 341
Hibernation, 2

HlLDEBRAND, JOEL H., 163

Holmium, properties, 412
Homogeneous, 70

systems and reaction rate, 126
Hydration, 313
Hydrazine. 46, 47, 231
Hydrides of third-row elements, 102

boiling point of, 315
Hydrocarbons, 340

unsaturated, 342

Hydrochloric acid, 42
Hydrofluoric acid, storage of, 361
Hydrogen, 99

apparent ionization energy, 289

boiling point, 63

bonding in, 274

burning, 25, 41

chemistry, 100

energy level scheme, 258, 264

freezing point, 63

heat of combustion, 40

helium nuclear reaction. 419

ion, and weak acids, 193

ion (aqueous), 187

ion as catalyst, 137

ion, competition for in weak acid,
194

ion concentration, calculation of,
192

ion, hydrated form, 187

ion, nature of, 185

ionic character of bonds, 289

ionization energy, 268

isotopes, 90, 123, 419

molar heat of combustion, 40

molar volume, 60

production, 62

properties, table, 100

solubility, 20

spectrum of, 253, 255
Hydrogenation, 407
Hydrogen atom, 253

energy level scheme, 258, 264

energy levels of, 259

heat of reaction to form, 274

light emitted by, 254

orbitals, 263

periodic table and, 263

quantum mechanics and, 259

quantum numbers and, 260

spectrum of, 253, 255
Hydrogen bonds, 314, 329, 361

energy of, 315

inter- and intramolecular, 316

nature of, 316

and proteins, 432

representation of, 315

significance of, 316

sugars, 424
Hydrogen bromide

boiling point, 315

infrared spectrum, 248
Hydrogen chloride

ammonia mixture, 24

boiling point, 64

heat of oxidation, 160

heat of reaction to form, 160

model of, 21

molar volume, 60, 64

P-V relation, 19
Hydrogen fluoride

boiling point, 315

bonding, 280

melting point, 99
Hydrogen halides, 315
Hydrogen iodide, boiling point, 315
Hydrogen peroxide bonding, 283, 295
Hydrogen selenide, boiling point, 315
Hydrogen sulfide, boiling point, 315
Hydrogen telluride, boiling point,

315
Hydrolysis

of fats, 426

of starch, 426

of sugars, 426
Hydrometer, 407

Hydronium ion, 187

concentration calculation, 192

concentration and /?H, 190

model, 186
Hydroquinone, 345
Hydrosphere, 437

composition, 439
Hydroxide ion, 106, 180
Hydroxides of third row, 371
Hydroxylamine, 251
Hydroxyl group, 329
Hypobromite ion, 422
Hypochlorite ion, 361
Hypochlorous acid, structure, 359
Hypophosphorous acid, 372
Hypothesis, Avogadro's, 25, 52

Ice

bonds in, 315

melting of. 69

structure, 69, 315
Ideal gas, 60

see also Perfect gas
Ilmenite, 401
Indicators

acid-base, 190

litmus, 21, 189
Inductive reasoning, 3
Inert gases. 90

atom models, 98

compounds, 91

electron population, 92, 263

properties, table, 91

stability of structure, 93
Information

accumulating, 75

organizing, 75
Infrared, 247
Infrared spectra, 248

of HBr, DBr, 248

of CCU, CS2, 249
Infrared spectroscopy, 249
Invisible ink, 406
Iodate ion, 360
Iodimetry, 358, 358
Iodine

boiling point, 307

color, 352

covalent radius, 354

electron configuration, 353

ionic radius, 355

melting point, 307

preparation, 231, 356

properties of, 97, 355

solubility in CC14, 166

solubility in ethanol, 163

van der Waals radius, 354
Ion, 78

exchange resin, 413

formation, 86

hydrogen (aqueous), 185

names, formulas, charges of com-
mon, 454
Ionic bond, 287, 288

dipole of, 288

in alkali metal halides, 95

vs. covalent, 287
Ionic character, 287
Ionic crystal, 81. J77
Ionic radius, 355
Ionic solids, 79, 57, i77

electrical conductivity, 80

properties of, 572

solubility in water, 79

stability of, J77

Boldface numbers refer to definitions. Italic numbers refer to sections.

INDEX

461

Ionization

lithium, 267

magnesium, 270

sodium, 270
Ionization energy, 267

alkaline earths. 379

and atomic number, 268

and the periodic table, 267

and valence electrons, 269

halogens, 353

measurement of, 268

successive. 269

table of, 268

trends, 268
Iridium, oxidation numbers, 414
Iron

atomic radius, 399

cast, 404

complex with oxalate, 395

electron configuration, 389

galvanized, 405, 409

manufacture of, 404

occurrence, 404

oxidation numbers, 391

properties. 400. 403

rusting of. 45, 85. .405
Iron(III) oxide, heat of reaction to

form. 162
Iron pyrites, 46, 404
Isomers

cis- and trans-, 296, 394

of complex ions, 394

of C;H60, 327

structural, 327
/.yo-prefix, 341
Isotopes, 90

and mass number, 89

and mass spectrograph, 243

vital statistics, table, 89

Jupiter, data on, 444

magnitude, 154

table, 154
Ka

determination, 192

values, 191
K.p, 174

table, 174
Ku, 181

change with temperature, 181

values at various temperatures, 181
Kcal, 40

Kelvin temperature scale, 58
Ketones, 334
Kerosene, 231, 341
Kilo, 40
Kilocalorie, 40
Kinetic energy, 53, 114

billiard ball analogy, 6, 114

distribution, 130, 131

formula for, 59

of a moving particle, 59

relation to temperature, 56, 131
Kinetic theory, 49, 52, 53

and Avogadro's Hypothesis, 58

review. 61
Kinetics, chemical, 124
Knudsen cell, 63
Kroll process, 368
Krypton, 91

atomic volume, 410

boiling point, 307

heat of vaporization, 105

Boldface numbers refer to definitions. Italic numbers refer to sections.

ionization energy, 410
melting point, 307

Laboratory, 2

Lanthanide elements, 411, 389

contraction, 413

electron configurations, 415

occurrence and preparation, 413

oxidation numbers, 414

properties, 412
Lanthanum

heat of vaporization, 305

properties, 412

source, 413
Latimer, Wendell M., 199
Lattice, 81
Laws, 14, 117

additivity of reaction heats, 111

Avogadro's, 25, 52

Boyle's, 17

Charles', 58

chemical equilibrium, 152, 173

chemical equilibrium derived, 155

combining volumes, 236

conservation of charge, 80

conservation of energy, 113, 117

conservation of mass, 40

constant heat summation, 111

definite composition, 234

Hess's, 111

octaves, 104

of nature, 117

simple multiple proportions, 235
Lead, oxidation numbers, 414
Lead chamber process, H2S04, 227
Lead storage battery, 406
Le Chatelier, Henry L., 149
Le Chatelier's Principle, 149, 181,
188, 337, 360

and concentration, 149

and electrochemical cells, 214

and equilibrium, 149

and pressure, 149

and temperature, 150
Levels, energy, 259
Lewis, G. N., 48, 142
Light, 246

a form of energy, 253

emitted by hot tungsten, 255

emitted by hydrogen atoms, 254

spectrum, 247
Lime, manufacture, 143
Limestone, 385, 404
Linear complex, 395
Line spectra, 255
Liquid, 27, 65

gas equilibrium, 66

gas phase change, 66

solid phase change, 5, 68

volume calculations, 230
Lister, 346
Lithium

bo ling point, 374

bonding capacity, 286

bonding in gaseous, 287

heat of reaction to form atoms, 290

heat of vaporization, 374

ionization energy, 268

ionization, 267

properties, 94
Lithium fluoride

bonding in gaseous, 287

dipole in, 293

heat of reaction to form atoms, 290
Lithosphere, 437

composition, 440

properties, 440
Litmus, 21, 189

color in acids, 183

color in bases, 184
Los Angeles to San Francisco anal-
ogy, 132
Lost Child fable, 3
"Lucite," 347
Lutetium, properties, 412
Lyman, T., 258

M, (symbol for molar concentration),

72
Macroscopic

changes and equilibrium, 143, 147

properties, 118
Magnesium

atomic size, 379

boiling point, 365

chemistry, 382

complex formation, 396

compounds, 102

electron configuration, 378

heat of vaporization, 365

hydration energy, 368

ionization energies, 269, 374

ionization, 270

metallic solid, 365

occurrence, 373, 385

oxide, heat of reaction to form, 375

preparation, 373

properties, 101, 381

reducing agent, 367
Magnesium hydroxide, 370

K,p, 383
Magnesium sulfate, Kap, 383
Magnetic field, 240
Magnetite, 404
Maleic a id

properties, 308

structure, 316
Manganese

atomic radius, 399

electron configuration, 389

oxidation numbers, 391

properties, 400, 403
Manganese dioxide, 403
Manometer, 53
Mantle, 440
Manufacture of

aluminum, 373

ammonia, 140, 150

helium, 91

hydrogen, 62

iodine, 231

lime, 143

magnesium, 373

methanol, 160

nitric acid, 45, 232

nitrogen compounds, 35, 150

phosphorus, 376

sodium, 238

sodium carbonate, 230

sulfuric acid, 225, 369

titanium, 368

uranium, 35
Many-electron atoms

energy level diagram, 266

energy levels of, 265
Mars, data on, 444
Mass

conservation of, 40

of atom and its parts, 87

of electron, 87, 241
Mass defects, 121

462

INDEX

Mass-energy relationship, 121
Mass number, 90, 120
Mass spectrograph, 242, 443
Mass spectrum of neon, 242
Matter

electrical nature, 74, 11

fundamental property, 78
McMillan, Edwin M., 420
Measurement of gas pressure. 53
Mechanism of reaction, 127, 128

CH3Br + OH-, 331

decomposition of formic acid, 138,
139

oxidation of HBr, 128
Melting

of ice, 69

of solids, generalization, 4
Melting, heat of, see Fusion, heat of
Melting point

alkali metals, 94

alkaline earths, 381

halogens, 355 •

inert gases, 91

paradichlorobenzene, 9

sodium chloride, 69

third-row elements, 101

transition elements, 400

water, 69
Mendeleev, Dimitri, 104, 107
Mendelevium, oxidation number, 414
Mercuric perchlorate, 237
Mercurous perchlorate, 237
Mercury, oxidation numbers, 414
Mercury (planet), data on, 444
Metabolism, oxidative, 429
Metallic

alloys, 309

bond, 303

elements, 303

radius, 380

substances, 81
Metals

alkali, 94

characteristic properties, 81, 303

conductivity, 81

electron behavior in, 304

hardness and strength, 311

heats of vaporization, 305

location in periodic table, 304

properties explained, 305
Meteorites, 404, 444
Methane

burning, 41

molar volume, 60

properties, 341

solubility in CCl*; in acetone, 313

structural formula, 332
Methanol, structural formula, 332
Methyl

acetate, 338

alcohol, 338

amine, 338

ammonium ion, 348

bromide, 330

butyrate, 339

caprylate, 339

ether, boiling point, 313

formate, 339

group, 330

octanoate. 339

orange, 344

propionate. 339

salicylate, 346
yV-Methyl acetamide, 348
Meyer, Lothar, 104
Mica, 310

Microscopic processes, and equilib-
rium, 147
Microwave spectroscopy, 249
Millikan, Robert, 241

oil drop experiment, 241
Minerals, 373, 385
Miscible, 176
Model

atomic theory, 34

electron-proton, 76

gases, 23

kinetic theory of gases, 53

nuclear atom, 86

particle, 18

pressure of gases, 18

scientific, 18

system, 7
Models

alkali atoms, 98

halogen atoms, 98

inert gas atoms, 98

molecular, various representations,
32

water, 31
Molar concentration, 72
Molar heat

of combustion, see Heat of com-
bustion

of fusion, see Fusion, heat of

of melting, see Fusion, heat of

of vaporization, see Vaporization,
heat of
Molar volume, 50

alkali metals, 94

gases, comparison, 50, 51, 60

halogens, 97

inert gases, 91

table of, 60

third-row elements, 101
Mole, 32, 32, 33

in calculations, 44, 225

volume of, see Molar volume
Molecular architecture, 290
Molecular crystal, 81
Molecular formula, 31

determination, 325
Molecular rotation, 249
Molecular shape

BF3, 292

BeF2, 292

CF4, 292

CH<, 292

F20, 291

H20, 291

NF3, 291

NH3, 291

and melting point, 308

and van der Waals forces, 307
Molecular size, and van der Waals

forces, 307
Molecular solids, 102, 301, 306
Molecular substances and van der

Waals forces, 306
Molecular structure, experimental

determination of, 324
Molecular velocities, distribution of,

131
Molecular vibrations, 249
Molecular weight, 33, 33

boiling point correlation, 307

calculation, 33

determination, 325
Molecules, 21, 274

energy of, 118

measuring dimensions, 245

models of, 21

number of atoms in, 26

polar, 288

relative weights, 25

representations of, 32

weights of, 22
Momentum, 59
Monazite, 413
Monel metal, 407
Monomer, 346
Monosaccharides, 424
Morgan, G. T., 224
Motion, types of, 118
Mountain pass analogy, 132
Multiple proportions, Law of, 235
Myristic acid, 425

n (principal quantum number), 259,

261
Names

of common ions, 454

organic, 339
Natural gas, 46, 322
Negative charge, 77, see also Electron
Negative ion, 87, 207
Negligible solubility, 73
Neodymium

properties, 412

source, 413
Neon, 91

boiling point, 374

electron configuration, 265

heat of fusion, 69

heat of vaporization, 105, 374

ionization energy, 268

mass spectrum of, 242

melting point, 307

sign, 239
Neopentane, 341

properties, 308
Neptune, data on, 444
Neptunium

electron configuration, 415

oxidation numbers, 414
Net reaction, 201
Network solid, 102

and covalent bonds, 302, 309

diamond and graphite, 302

one-dimensional, 309

three-dimensional, 309

two-dimensional, 309
Neutral, 77, 189
Neutron

properties, 87

ratio to proton, 417

relation to mass number, 90
Newlands, J. A. R., 85, 104
Newton, Isaac, 17
Nichrome, 402
Nickel

atomic radius, 399

alloys of, 407

electron configuration, 389

oxidation number, 391

properties, 400, 406
Nickel carbonyl, 410
Nitration, 344

Nitric acid, manufacture, 45, 232
Nitric oxide

as catalyst, 227

reaction with oxygen, 26

solubility, 20
Nitrobenzene, 344
Nitrogen

boiling point, 63, 374

bonding capacity, 283

Boldface numbers refer to definitions. Italic numbers refer to sections.

INDBX

463

Nitrogen (continued)

compounds, manufacture of, 35,
150

density, 50

electron configuration, 265

freezing point, 63

heat of vaporization, 374

ionization energy, 268

molar volume, 49, 51, 60

reaction with hydrogen, 150
Nitrogen dioxide, as catalyst, 227
Nitrous acid, KA, 191
Noble gases, see Inert gases
Nomenclature

common ions, 454

organic compounds, 339
Normal boiling point, 68
Notched beam analogy to H spec-
trum, 256
Novocaine, 345
M-pentane, 308, 340
nu, 246
Nuclear

atom model, 86

changes, 121

charge, 86

energy, 119,4/8

fission, 120, 419

forces, 87

fuels, 35, 413

fusion, 121, 419

power plant, 82

reactions, 120, 443

stability, 416
Nucleon, 120, 416
Nucleus

energy stored in, 120

radius, 88, 251

"seeing," 244
"Nylon," 347

Observation, 2, 15
Ocean, composition of, 440
Ocean water, composition, 439
w-Octadecane, properties, 341
Octahedral complex, 393
Octane, 46, 338

properties, 341
Octanamide, 339
Octanoic acid, 339
1-Octanol, 338
Octaves, Law of, 104
Octyl alcohol, 338
1-Octylamine, 338

OH molecules, reaction between, 282
Oil-drop experiment, 241
Oil of wintergreen, 346
Oleomargarine, 407
Open hearth furnace, 404
Operational definition, 195
Orbital representation of chemical

bonding, 278
Orbitals

atomic, 262, 263

d and /, 262

molecular shape and, 293

overlap, 277

p, 262

principal quantum number and,
261

s,261
Ores, 442

Organic compounds, 322
Osmium, oxidation numbers, 414
Ostwald, W., 65

Boldface numbers refer to definitions. Italic numbers refer to section*.

Overall reaction, 201
Overlap and bonding, 277
Oxidation, 202

of alcohols, 336

of ammonia, 113

of food, 426

of hydrogen chloride, 160

of organic compounds, 332

of sulfur, 225, 369

of sulfur dioxide, 225, 369
Oxidation numbers, 215, 216

balancing redox reactions using,
219

rules for assigning, 219
Oxidation potentials, 211, 452
Oxidation-reduction, 202, 216

self, 361
Oxidation-reduction reactions, 202

balancing with half-reactions, 217

balancing with oxidation numbers,
219
Oxides

basic, 382

third-row elements, 102
Oxidizing agent, 215
Oxyacids

of chlorine, 359, 373

of phosphorus, 371, 372

of sulfur, 371
Oxygen

abundance in earth's crust, 441

boiling point, 63, 374

bonding capacity, 281

bonding in molecules, 295

bond length, 295

double bond in, 295

electron configuration, 265

fluorine compounds, 283

freezing point, 63

heat of vaporization, 374

ionization energy, 268

isotopes, 90

mass of atom, 33

molar volume, 51, 60

molecule, 27

preparation from KC103, 46

pressure-volume behavior, 18

reaction with nitric oxide, 26

solubility, 20
Ozone, 439

p electron, 262

p orbitals, 262

p2 bonding, 291

p3 bonding, 291

Packing in crystals, 81

Palmitic acid, 425

/raraaminobenzoic acid, 434

/raradichlorobenzene, 8

Paraffin, 43

paraphthallic acid, 347

Partial pressure, 55, 67, 438

Particles, fundamental, table, 78, 87

Pauli Principle, 267

Pauling, Linus C, 299, 300

Peat, 321

«-Pentane, 340

properties, 308
i.vo- Pen tane, 340
Pepsin, 138
Perchloric acid, 359
Perfect gas, 59

molar volume, 60
Pcrfluoroethane, 362
Perfluoroethylene, 362
Periodic table, 85, 267, see inside

front cover

and atomic number, 89

elements in each row of, 272

fourth row, 271, 387

historical development, 103

and hydrogen atom, 263

and ionization energy, 267

second column, 377

seventh column, 352

seventh row, 413

sixth row, 41 1
Permanganate ion

oxidation of Fe+2, 125

oxidation of H2S, 218

oxidation of oxalate ion, 125
Petroleum, 322
pH, 190
Phase, 70
Phase change, 5

liquid-gas, 66

solid-liquid, 5, 68
Phases, 5

condensed, 27, 68, 78
Phenacetin, 345
Phenol, 345
Phosphides, 368
Phosphoric acid, 372

KA, 191
Phosphorous acid, 372
Phosphorus

black, 365

boiling point, 374

chemistry of, 368

compounds, 102

heat of vaporization, 374

ionization energy, 268

melting point, 374

occurrence, 373

oxyacids, 371, 372

preparation, 374, 376

properties, 101

structure, 366

white, 120, 365, 366, 369
Photon, 254

Photosynthesis, 254, 430
Pig iron, 404
Pitchblende, 385, 442
Planck's constant, 254
Planets

atmosphere of, 445

chemistry of, 444

lithosphere of, 446
Plastics, 3<6
Platinum

catalyst, action of, 227

dioxide, 230

oxidation numbers, 414
"Plexiglas," 347
Pluto, data on, 444
Plutonium

electron configuration, 415

oxidation numbers, 414
Polar molecule, 288
Polonium, oxidation numbers, 414
Polyethylene, 347
Polymerization, 346

types of, 346
Polymers, 346, 431, 432
Polysaccharides, 425
Polystyrene, 345
Positive charge, 77, see also Proton

formation of, 242
Positive ions. 207

charge to mass ratio, 243

"seeing," 242
Positron, 218

464

INDEX

Potassium, 93

atomic radius, 399

atomic volume, 410

chemistry, 95

electron configuration, 271

heat of vaporization, 305

ionization energy, 268

properties, 94
Potassium chrome alum, 393
Potassium permanganate, 218
Potential energy, 115

barrier, 134

diagrams, 134, 138, 139, 331
Potentials, half-cell

effect of concentration, 213

measuring, 210

table, 211,452
Praseodymium

properties, 412

source, 413
Precipitation

and equilibrium, 144

for separation, 176

from aqueous solution, 80

rate, 164
Predictions

based on E°'s, 212

of equilibrium conditions, 149

of formation of precipitate, 175

of heat of reaction, 112
Predominant reacting species, 80
Pressure, 55

and Le Chatelier's Principle, 149

and reaction rate, 126

atmospheric, 53

cause, 54

measurement, 53

of gases, 53

partial, 55, 67, 145

standard, 54

vapor, 66, 67, 145
Pressure-volume relationships, 13, 18

of ammonia, table, 19, 51, 60

of hydrogen chloride, table, 19

of other gases, 19

of oxygen, table, 14, 18
Principle

Le Chatelier's, 149

Pauli, 267
Principal quantum number, 260
Procaine, 345
Product, 39

Promethium, properties, 412
Promoting electrons, 284
Propane, 338

properties, 341
1-Propanol, 335
2-Propanol, 335
Protactinium, 415

electron configuration, 415

oxidation numbers, 414
Propionamide, 339
Propionic acid, 338
Propyl alcohol, 338
1-Propylamine, 338
Propylene, 342
Propyl group, 330
Protein, 348, 432

helical structure, 433

hydrogen bonds in, 432
Proton, 77

and atomic number, 89

competition for, 193

donors, 396

mass of, 87, 121

relation to atomic number, 88

transfer, 194

transfer theory of acids, 194
Ptyalin, 138

Pullman, Alberte, 421
Pullman, Bernard, 421
Purification, 70
Pyruvic acid, 426

Qualitative aspects of equilibrium,

142
Qualitative presentation of data, 14
Quantitative aspects of equilibrium,

151
Quantitative analysis, infrared, 250
Quantitative presentation of data, 14
Quantum mechanics, 259, 260

and the hydrogen atom, 259
Quantum number, 260

and hydrogen atom, 260

and orbitals, 261

principal, 260

Radioactive nuclei, 413
Radioactivity, 416, 442

types of, 417
Radium

electron configuration, 378

occurrence, 385
Radius

covalent, 354

double bond, alkaline earths, 380

double bond, oxygen, 295

ionic, 355, 380

metallic, 380

van der Waals, 354, 380
Radon, 91

boiling point, 307

heat of vaporization, 105

melting point, 307
Randall, M., 142
Randomness, effect on equilibrium,
166, 447

in dissolving, 313
Rare earths, 272, see also Lantha-

nides
Rate

determining step, 128

effect of concentration on, 126, 128

of dissolving, 164

of opposing reactions, 148, 155

of reactions, 124
Reactant, 40

Reacting species, predominant, 80
Reaction coordinate, 133
Reaction heat, 135

additivity of, 111

measurement of, 111
Reaction rates, 124

factors affecting, 125
Reactions, 38, 129

acid- base, 188

balancing, 42, 217, 219

calcium carbonate decomposition,
143

chemical, principles of, 38

conservation of energy in, 115

effect of heat on, 108

endothermic, 40, 135

equations for chemical, 41

equilibrium in, 145

exothermic, 40, 135

half-cell, 201

mechanism of, 127, 128

nuclear, 120, 121, 419

oxidation-reduction, 202, 216

predicting from E°, 212

redox, 203

substitution, of benzene, 344
Reactive, 284
Reasoning, inductive, 3
Redox, 203
Reducing agent, 215
Reduction, 202
Regularities, search for, 3
Reichenbach, Hans, 233
Relativity, theory of, 121
Repulsive forces in H2, 275
Resin, ion exchange, 413
Rhenium, oxidation number, 414
Roasting, 46

Robinson, Sir Robert, 321, 351
Rotational motion, 118

and microwave spectroscopy, 249
Rubber band, conservation of energy

in, 114
Rubidium

heat of vaporization, 305

ionization energy, 410

properties of, 94
Rule, 14

Rusting, 45, 85, 405
Rutherford, Ernest, 244
Rutherford

nuclear atom, 244

scattering experiment, 244
Rutile, 401

s electron, 261
s orbitals, 261
sp bonding, 292
sp2 bonding, 292
sp3 bonding, 292
Salicylic acid, 345
Salt bridge, 201
Samarium, properties, 412

source, 413
San Francisco to Los Angeles anal-
ogy, 132
Saponification, 426
"Saran," 347
Saturated

hydrocarbons, 326, 340

solutions, 72

vapor, 145
Saturn, data on, 444
Scandium

atomic radius, 399

atomic volume, 410

electron configuration, 389

heat of vaporization, 305

oxidation number, 391

properties, 400
Scattering of alpha particles, 245
Science, 2, 15

activities of, 1

communicating information, 12

review, 15
"Sea" of electrons, 304
Seaborg, Glenn T., 420
Second column of periodic table, 371
Second-row elements, bonding ca-
pacity. 281
Selenium, ionization energy, 410
Self oxidation-reduction, 361
Separation of charge, 312
Separations

by crystallization, 413

by distillation, 70

by precipitation, 176
Seventh column of periodic table, 352

Boldface numbers refer to definitions. Italic numbers refer to sections.

INDEX

465

Sc\cnih row of periodic tabic. 411
occurrence, 413

Significant figures. 9. 12

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