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PROFESSOR: -- that this is not
a quick clicker question, so

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you actually need to maybe
write something down to

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figure out the answer.
 

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So if you haven't clicked in
yet, or if you want to change

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your answer, keep in mind that
you might need to jot down, for

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example, a Lewis structure
before you can answer

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this question.
 

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-- and get in your final answer
to the clicker question.

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For those of you just walking
in now, you might not have a

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chance to get all of the
thought process that you need

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in on this clicker question,
because it is based on a Lewis

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structure, so we
will go over it.

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But for those of you that have
been here for 30 seconds or

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longer, see if you can get
the right answer in here.

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All right, so OK, it looks like
we have a very mixed response

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in terms of the answer to
the clicker question today.

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Raise your hand if you didn't
have time to figure out

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the Lewis structure.
 

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OK, so that accounts
for some of you.

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Let's go over the correct
answer this question.

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So, who got the right
answer -- 31% of us.

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This is not the kind of
percentages we're looking

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for, so let's go over this.
 

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In class on Monday, we did go
over the geometries, and the

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geometries themselves are very
straightforward, once you know

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what the Lewis structure is,
but remember, you can't just

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always look at a molecule
and automatically know

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the Lewis structure.
 

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We actually need to think about
those valence electrons.

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So, here we're dealing with
selenium hydride, so s e h 2.

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I told you that there's 6
valence electrons in selenium,

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so we have 6 there, plus 1
for each of the hydrogen.

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So what we should have is
8 valence electrons in

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our Lewis structure.
 

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How many electrons do we need
to have full valence shells?

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STUDENT: [INAUDIBLE]
 

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PROFESSOR: 12, that's right.
 

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We need 8 plus 4 is
12 for full shells.

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That means that we need
12 minus 8, or 4 bonding

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electrons in our structure.
 

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So I tried to give you one that
you could draw pretty quickly,

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it just has hydrogens in It.
 

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So we have s e h 2.
 

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We can do our 4
bonding electrons.

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How many valence electrons
do we have left?

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STUDENT: [INAUDIBLE]
 

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PROFESSOR: 4, that's right.
 

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So, we need to fill our octet
for selenium, so 1, 2, 3, 4.

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So, this is our Lewis structure
here, hopefully you can

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see why it's not linear.
 

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If it were linear, which 32% of
you seem have thought, that

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would have meant that our Lewis
structure had no lone pairs in

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it, right, and that's
not the case.

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We have two lone pairs, so if
we thought about what the

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bonds were everywhere,
it would be 109 .

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5, but it's bent because we're
only looking at the bonds,

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we're not counting the Lewis
structures in naming our

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geometry, but they do
affect the angles.

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And it turns out that it's
actually less than 109 .

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5, because those lone pairs
are pushing the bonds

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even further away.
 

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So, does this make sense
to everyone if you think

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about it this way?
 

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Pretty much, okay.
 

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So, we'll try another clicker
question like this later.

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Some of these questions that
look very straightforward just

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naming the geometry, you have
to remember to do the first

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step before you jump in and
go ahead with a naming.

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So you'll got plenty of
practice with this on

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your problem-set if
you haven't already.

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All right.
 

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So before we start in with
today's notes, I do want to

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mention that this morning the
Nobel Prize in chemistry

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was announced.
 

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This is an exciting week in
science in general because we

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got to hear another Nobel Prize
every morning pretty much.

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So, let's settle down for a
second and start listening.

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All right.
 

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So, today we're going to be
talking about molecular orbital

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theory, but first I wanted to
just mention, in case some of

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you didn't hear what the Nobel
Prize was this morning, and

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this was in chemistry, it went
to three different chemists.

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Osamu Shimomura, who's a
Japanese chemist, and then

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Martin Chalfi who's at
Columbia, and Robert

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Chen who's at U.
 

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C.
 

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San Diego.
 

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The three of these chemists
split the Nobel Prize this

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morning for their discovery
and/or their application of

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using green fluorescent
protein, which is

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also called GFP.
 

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How many of you have
heard of GFP before?

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Oh, that's so great.
 

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OK.
 

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Many of you have heard of GFP.
 

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For some of you that haven't
I'll just say that it's a

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protein, it's 238 amino acids,
which means that it's about

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1,000, actually more than 1
atoms in size, and this

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protein is fluorescent.
 

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The protein was first
discovered and first isolated

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from this jellyfish here.
 

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This was done by Shimomura,
and he did this in

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the 60's and 70's.
 

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He actually recognized and, of
course, many people recognized

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that this jellyfish was
fluorescent, and he isolated

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the actual protein, and
determined and proved that this

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protein was sufficient to
cause this fluorescence.

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So, in terms of thinking about
applications of why it's so

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exciting that you have this
fluorescent protein, other than

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it's always really fun to look
at things that are fluorescent,

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we can think about in terms of
biology why it's so exciting,

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you can actually tag this
protein to any other protein

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that you're studying and now
you have a visual handle

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on what's going on.
 

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So, for example, if you were
interested in some protein

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involved in cancer, you
could tag it with GFP.

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You could watch where it
localizes in the cell, there

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are fluorescent assays you
could use to determine what

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other proteins it
interacts with.

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You could see, for example,
when it's expressed in a cancer

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cell, and where it's expressed.
 

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There are all sorts of things
you can do once you can

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visualize something
with fluorescence.

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The reason it is so exciting
that it's a protein, and it's a

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protein, this is the structure
here, it's a ribbon structure

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so you can kind of see what it
looks like, it's made up of

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all natural amino acids.
 

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So this means we can code for
it in DNA, you don't have

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to worry how am I going
to get into the cell.

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All you have to do is mutate
the DNA, which is very

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straightforward to do in
molecular biology, and now you

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can tag absolutely any protein
that you're interested in.

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So, as I said, this was first
discovered and isolated

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from the jellyfish.
 

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This was done by Shimomura --
that was kind of the first step

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in this process of having it
become such a useful tool.

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And then, many years later, not
until 1994 did Martin Chalfi,

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at Columbia, show that yes, I
can, in fact, take the DNA and

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put it into a different
organism, and he put it

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into e coli, a bacteria.
 

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And what he could show was that
it could be expressed, this is

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a picture from his 1994 science
paper, in that e coli, and it

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is going to fluoresce green.
 

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Now, the first application
tends to not be quite as

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exciting as, for example, all
the other organisms people have

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put it in since then -- you can
have flies, you see transgenic

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mice that are glowing
green with this GFP.

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Of course, that's not the
useful application for it, it's

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more of a proof of principle,
but it does show you that you

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can put it in for studies
in any organisms.

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And the field was really pushed
forward by the discoveries of

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Robert Chen at UC San Diego,
and what he did was he actually

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figured out how it was that
this protein fluoresced, what

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caused the actual fluorescence.
 

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And once he did that, both he
and many other scientists,

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could then, once they
understood what caused the

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fluorescence, make little
changes to the actual protein,

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and tune what the properties of
that fluorescent protein were.

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So now, for example, there are
a whole range, just a whole

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rainbow of fluorescent
proteins that can be used.

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And I'm sure you can imagine
that if you want to label onw

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protein green and one red and
one yellow, now you can start

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looking at really complex
biological processes.

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So, it's pretty rare that
chemistry makes the

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every day news.
 

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So hopefully you'll all look in
the normal papers today, not

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just the scientific journals,
and get to read something

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about chemistry.
 

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It's always fun to see how its
described in The New York

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Times or in The Boston Globe.
 

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The other thing I wanted to
mention and I'm not sure if

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the exhibit's still there.
 

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But there was an exhibit of
jellyfish, I know at least

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until last year, at the Boston
Museum of Science, and all of

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you guys can get in there free.
 

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And it's neat to see the
glowing jellyfish and think

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about the fluorescent
protein that's in them.

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So, I encourage you to do that
the next time you have some

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free time on your hands, maybe
at IAP or some time like that.

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All right, so let's move
into today's notes.

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Today we're talking about
molecular orbital theory.

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This is a shift, this is a new
topic that we're starting.

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So far we've exclusively been
using Lewis structures any

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time we've tried to describe
bonding within molecules.

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Lewis structures are really
useful, we use them all

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the time in chemistry.
 

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And they're useful because,
first of all, they're easy to

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depict, they're easy to draw
-- relatively easy once we

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get all the rules down.
 

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And also they're accurate
over 90% of the time.

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But they're not accurate all
the time in predicting bonding

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within molecules, and the
reason for this is because

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Lewis structures are
not, in fact, based

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on quantum mechanics.
 

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So, molecular orbital theory,
on the other hand, is based

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on quantum mechanics.
 

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And specifically, MO theory
is the quantum mechanical

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description of wave
functions within molecules.

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So, saying wave functions
within molecules might sound a

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little confusing, but remember
we spent a lot of time talking

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about wave functions within
atoms, and we know how to

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describe that, we know that a
wave function just means

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an atomic orbital.
 

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It's the same thing with
molecules -- a molecular

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wave function just means
a molecular orbital.

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So, we'll start today talking
about the two kinds of

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molecular orbitals, we can
talk about bonding or

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anti-bonding orbitals.
 

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Then we're going to actually
use MO theory to describe

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bonding within these molecules,
and we'll start with

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homonuclear diatomic molecules.
 

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Diatomic mean it's di atomic,
it's made up of two atoms, and

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homonuclear means that those
two are the same atoms.

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Then at the end, we'll look at
an example with a heteronuclear

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diatomic molecules.
 

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So, again, the same thing,
but now two different atoms.

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So, I will point out, in terms
of MO theory, because it

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rigorously does take into
account quantum mechanics, it

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starts to become complicated
once we go beyond

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diatomic molecules.
 

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So we're going to limit in
our discussion in 511-1 for

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molecular orbital theory
to diatomic molecules.

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However, on Friday we will use
a different approach so we can

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talk about bonding within atoms
that have more than two atoms,

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molecules with more
than two atoms.

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All right, so one thing that I
first want to point out about

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MO theory that is a big
difference from Lewis

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structures, is that in MO
theory valence electrons are

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de-localized over the
entire molecule.

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So, when we talked about Lewis
structures, we actually

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assigned electrons to
individual atoms or

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to individual bonds.
 

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Whereas in molecular orbital
theory, what I'm telling you is

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instead we understand that the
electrons are spread all over

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the molecule, they're not just
associated with a single

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atom or a single bond.
 

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So specifically, what we do
associate them instead is

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within molecular orbitals, and
what we say is that they can be

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either in bonding or
anti-bonding orbitals.

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And again, I want to point out
that a molecular orbital, we

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can also call that a wave
function, they're

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the same thing.
 

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And these orbitals arise
from the combination of

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individual atomic orbital.
 

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So, if we have two atomic
orbitals coming together from

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two different atoms and they
combine, what we end up forming

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is a molecular orbital.
 

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The reason that we can talk
about this is remember that

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00:12:06 --> 00:12:08
we're talking about wave
functions, we're talking

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about waves, so we can have
constructive interference in

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which two different orbitals
can constructively

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interfere, we can also have
destructive interference.

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00:12:18 --> 00:12:20
So, we'll start by taking a
look at constructive

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interference, and another way
to explain this is just to say

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00:12:24 --> 00:12:27
again, molecular orbitals are
a linear combination

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of atomic orbitals.
 

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So, let's start our discussion
of a bonding orbital.

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Our simplest case that we can
look at would be if we had two

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1 s orbitals coming together.
 

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So let's say, for example,
in a hydrogen atom.

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So in hydrogen atom a, I'll
depict that here where the

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nucleus is this dot, and then
the circle is what I'm

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00:12:46 --> 00:12:49
depicting as the wave function.
 

272
00:12:49 --> 00:12:52
It makes sense to draw the wave
function as a circle, because

273
00:12:52 --> 00:12:55
we do know that 1 s orbitals
are spherically symmetric.

274
00:12:55 --> 00:12:57
So, we can say that a circle
is a good approximation

275
00:12:57 --> 00:12:59
for a 1 s wave function.
 

276
00:12:59 --> 00:13:02
Similarly, with the second
hydrogen atom, we've got the

277
00:13:02 --> 00:13:06
nucleus in the middle, and the
1 s b wave function around it.

278
00:13:06 --> 00:13:08
So these are atomic orbitals.
 

279
00:13:08 --> 00:13:11
What we're going to do in
forming a molecule is just

280
00:13:11 --> 00:13:15
bring these two orbitals close
together such that now we have

281
00:13:15 --> 00:13:18
their nucleus, the two nuclei,
at a distance apart that's

282
00:13:18 --> 00:13:21
equal to the bond length.
 

283
00:13:21 --> 00:13:24
And what we end up forming is a
molecular orbital, because as

284
00:13:24 --> 00:13:27
we bring these two atomic
orbitals close together, the

285
00:13:27 --> 00:13:30
part between them, that wave
function, constructively

286
00:13:30 --> 00:13:33
interferes such that in our
molecular orbital, we actually

287
00:13:33 --> 00:13:39
have a lot of wave function
in between the two nuclei.

288
00:13:39 --> 00:13:42
So we can go ahead and name our
molecular orbital, just like we

289
00:13:42 --> 00:13:45
know how to name our
atomic orbitals.

290
00:13:45 --> 00:13:47
And I'm going to name
this sigma 1 s.

291
00:13:47 --> 00:13:50
The 1 s just comes from the
fact that the molecular orbital

292
00:13:50 --> 00:13:53
is a combination of two
1 s atomic orbitals.

293
00:13:53 --> 00:13:56
And the sigma tells us
something about the symmetry

294
00:13:56 --> 00:13:59
of this molecular orbital,
specifically that it's

295
00:13:59 --> 00:14:01
cylindrically symmetric
about the bond axis.

296
00:14:01 --> 00:14:04
So that is the bond axis
-- it's just the axis

297
00:14:04 --> 00:14:06
between the two nuclei.
 

298
00:14:06 --> 00:14:09
Sometimes it's also called
the internuclear axis.

299
00:14:09 --> 00:14:12
So any time you have two atoms
bonding, the bond axis is just

300
00:14:12 --> 00:14:15
the axis that they're
bonding along.

301
00:14:15 --> 00:14:18
Another thing I want to point
out about every sigma orbital

302
00:14:18 --> 00:14:21
that you see, and it will make
more sense when we contrast

303
00:14:21 --> 00:14:23
it with pi orbitals later.
 

304
00:14:23 --> 00:14:26
But in sigma orbitals, you have
no nodal planes along the bond

305
00:14:26 --> 00:14:29
axis, so if we had a nodal
plane here, we'd see an area

306
00:14:29 --> 00:14:31
where the wave function
was equal to zero.

307
00:14:31 --> 00:14:32
We don't see that.
 

308
00:14:32 --> 00:14:34
It will make more sense when
we can show you one where

309
00:14:34 --> 00:14:36
it does have that area.
 

310
00:14:36 --> 00:14:39
But keep in mind sigma
orbitals have no nodal

311
00:14:39 --> 00:14:41
planes along the bond axis.
 

312
00:14:41 --> 00:14:43
All right.
 

313
00:14:43 --> 00:14:45
So, let's look at this in
another way, sometimes it's

314
00:14:45 --> 00:14:47
hard to picture these
waves combining.

315
00:14:47 --> 00:14:50
So let's think of them a little
bit more by graphing the

316
00:14:50 --> 00:14:53
amplitude of the wave, and
seeing how we can have this

317
00:14:53 --> 00:14:54
constructive interference.
 

318
00:14:54 --> 00:14:57
So again, if we think of a
graph of the wave function, we

319
00:14:57 --> 00:15:01
had the wave function is at its
highest amplitude when it's

320
00:15:01 --> 00:15:04
lined up with the nucleus, and
then as we got further away

321
00:15:04 --> 00:15:08
from the nucleus, the amplitude
of the wave function ends up

322
00:15:08 --> 00:15:12
tapering off until -- it never
hits zero exactly, but

323
00:15:12 --> 00:15:13
it goes down very low.
 

324
00:15:13 --> 00:15:18
So we can draw that for 1 s a,
we can also draw it for 1 s b,

325
00:15:18 --> 00:15:20
and what I'm saying for the
molecular wave function is that

326
00:15:20 --> 00:15:22
we have the interference
between the two, and we have a

327
00:15:22 --> 00:15:25
constructive interference, so
we end up adding these two

328
00:15:25 --> 00:15:27
wave functions together.
 

329
00:15:27 --> 00:15:30
So, we're talking about wave
functions and we know that

330
00:15:30 --> 00:15:35
means orbitals, but this is --
probably the better way to

331
00:15:35 --> 00:15:37
think about is the physical
interpretation of

332
00:15:37 --> 00:15:38
the wave function.
 

333
00:15:38 --> 00:15:41
So what is the wave function
squared going to be equal to?

334
00:15:41 --> 00:15:42
STUDENT: [INAUDIBLE]
 

335
00:15:42 --> 00:15:46
PROFESSOR: Probability
density, yes.

336
00:15:46 --> 00:15:49
Probability density of finding
an electron within that

337
00:15:49 --> 00:15:52
molecule in some given volume.
 

338
00:15:52 --> 00:15:54
So let's think about that
instead, let's think about

339
00:15:54 --> 00:15:55
probability density.
 

340
00:15:55 --> 00:15:58
So if we're talking about
probability density that's

341
00:15:58 --> 00:16:00
the wave function squared.
 

342
00:16:00 --> 00:16:03
But now we're talking not about
an atomic wave function, we're

343
00:16:03 --> 00:16:05
talking about a molecular
wave function.

344
00:16:05 --> 00:16:08
So to talk about it's squared,
we're going to say it's

345
00:16:08 --> 00:16:11
sigma 1 s squared.
 

346
00:16:11 --> 00:16:13
Oh, and actually before you
skip your page in the notes, I

347
00:16:13 --> 00:16:16
realized I should write out for
you what the addition

348
00:16:16 --> 00:16:18
is to start with.
 

349
00:16:18 --> 00:16:23
So, when we're combining two
waves, what we have is 1 s a

350
00:16:23 --> 00:16:29
that we're adding together with
1 s for b, the second atom.

351
00:16:29 --> 00:16:32
And what we end up for
our molecular wave

352
00:16:32 --> 00:16:34
function is sigma 1 s.
 

353
00:16:34 --> 00:16:38
So this is what we call
our molecular orbital.

354
00:16:38 --> 00:16:41
All right, so that will now
allow you to turn the page,

355
00:16:41 --> 00:16:44
I think, and we can take a
look at the probability.

356
00:16:44 --> 00:16:48
So the probability again,
that's just the orbital

357
00:16:48 --> 00:16:50
squared, the wave
function squared.

358
00:16:50 --> 00:16:54
So when we write that out, we
just write sigma 1 s squared,

359
00:16:54 --> 00:16:57
or we can break it up into its
individual parts, there's

360
00:16:57 --> 00:16:59
no reason we can't
do that as well.

361
00:16:59 --> 00:17:04
So just to say that it's 1
s squared plus 1 s b, all

362
00:17:04 --> 00:17:07
of that together squared.
 

363
00:17:07 --> 00:17:09
So if we write out every term
individually, what we end up

364
00:17:09 --> 00:17:12
with is essentially just the
probability density for the

365
00:17:12 --> 00:17:15
first atom, then the
probability density for the

366
00:17:15 --> 00:17:19
second atom, and then we have
this last term here, and this

367
00:17:19 --> 00:17:23
is what ends up being
the interference term.

368
00:17:23 --> 00:17:25
So in this case where we're
adding it together, this is

369
00:17:25 --> 00:17:27
going to be constructive
interference.

370
00:17:27 --> 00:17:31
So in this case the cross term
represents constructive

371
00:17:31 --> 00:17:37
interference between the two 1
s atomic wave functions.

372
00:17:37 --> 00:17:38
And this again is what
we're going to call

373
00:17:38 --> 00:17:41
a bonding orbital.
 

374
00:17:41 --> 00:17:44
So it can often make a lot
more sense if we think about

375
00:17:44 --> 00:17:45
things in terms of energy.
 

376
00:17:45 --> 00:17:49
We've been discussing energy
diagrams a lot in this class,

377
00:17:49 --> 00:17:51
it's a very good way to
visualize exactly

378
00:17:51 --> 00:17:52
what's going on.
 

379
00:17:52 --> 00:17:55
So, let's think of the energy
of interaction when we're

380
00:17:55 --> 00:17:58
comparing atomic orbitals to
molecular bonding orbitals.

381
00:17:58 --> 00:18:01
And what you find is when you
have a bonding orbital, the

382
00:18:01 --> 00:18:05
energy decreases compared
to the atomic orbitals.

383
00:18:05 --> 00:18:08
So you can see here in this
slide we have the atomic

384
00:18:08 --> 00:18:12
orbitals for the two hydrogen
atoms, each of them have one

385
00:18:12 --> 00:18:16
electron in them, hydrogen has
one electron in a 1 s orbital.

386
00:18:16 --> 00:18:19
So, if we look at the molecular
orbital, that's actually going

387
00:18:19 --> 00:18:22
to be lower in energy than
either of the two

388
00:18:22 --> 00:18:23
atomic orbitals.
 

389
00:18:23 --> 00:18:26
So it's going to be favorable
for the electrons instead to

390
00:18:26 --> 00:18:29
go to that lower energy
state and be within

391
00:18:29 --> 00:18:31
the molecular orbital.
 

392
00:18:31 --> 00:18:36
So, let's draw in our electrons
there, so we have our two

393
00:18:36 --> 00:18:40
electrons now in the
molecular orbital.

394
00:18:40 --> 00:18:42
So any time that you're drawing
these molecular orbital

395
00:18:42 --> 00:18:45
diagrams, you want to keep in
mind that the number of

396
00:18:45 --> 00:18:48
electrons that you have in
atomic orbitals, you need to

397
00:18:48 --> 00:18:50
add those together and put that
many electrons into

398
00:18:50 --> 00:18:51
your molecule.
 

399
00:18:51 --> 00:18:54
Right, we had one from each
atom, so that means we need

400
00:18:54 --> 00:18:57
a total of two in our
molecular orbital.

401
00:18:57 --> 00:18:59
And what you can see directly
from looking at this energy

402
00:18:59 --> 00:19:02
level diagram, is that the
molecule that we have is now

403
00:19:02 --> 00:19:04
more stable in the
individual atoms.

404
00:19:04 --> 00:19:07
That makes sense because we're
lower in energy, the electrons

405
00:19:07 --> 00:19:09
are now lower in energy.
 

406
00:19:09 --> 00:19:13
So that's the idea of a
bonding molecular orbital.

407
00:19:13 --> 00:19:16
Since we're talking about wave
functions, since we're talking

408
00:19:16 --> 00:19:19
about the properties of waves,
we don't only have constructive

409
00:19:19 --> 00:19:22
interference, we can also
imagine a case where we would

410
00:19:22 --> 00:19:24
have destructive interference.
 

411
00:19:24 --> 00:19:27
Just like we see destructive
interference with water waves

412
00:19:27 --> 00:19:29
or with light waves, we
can also see destructive

413
00:19:29 --> 00:19:30
interference with orbitals.
 

414
00:19:30 --> 00:19:33
So, let's think about what
that would look like.

415
00:19:33 --> 00:19:37
So in this case we would have 1
s a and 1 s b, and instead we

416
00:19:37 --> 00:19:41
would subtract one from the
other, and what we would see is

417
00:19:41 --> 00:19:44
that instead of having
additional, more wave function

418
00:19:44 --> 00:19:47
in the middle here, we've
actually cancelled out the wave

419
00:19:47 --> 00:19:50
function and we end
up with a node.

420
00:19:50 --> 00:19:53
So we can also name this
orbital, and this orbital we're

421
00:19:53 --> 00:19:56
going to call sigma 1 s star.
 

422
00:19:56 --> 00:20:00
So if we name this orbital,
this is an anti-bonding

423
00:20:00 --> 00:20:00
molecular orbital.
 

424
00:20:00 --> 00:20:04
So we had bonding and now we're
talking about anti-bonding.

425
00:20:04 --> 00:20:08
When we talk about
anti-bonding, essentially we're

426
00:20:08 --> 00:20:14
taking 1 s a and now we're
subtracting 1 s b, and what we

427
00:20:14 --> 00:20:19
end up with again is sigma 1 s,
and the important thing to

428
00:20:19 --> 00:20:21
remember is to write
this star here.

429
00:20:21 --> 00:20:27
So any time you see a star that
means an anti-bonding orbital.

430
00:20:27 --> 00:20:30
Again we can look at this in
terms of thinking about a

431
00:20:30 --> 00:20:32
picture this way, in terms
of drawing the wave

432
00:20:32 --> 00:20:34
function out on an axis.
 

433
00:20:34 --> 00:20:38
So we have 1 s a, and we're
drawing this as having a

434
00:20:38 --> 00:20:41
positive amplitude, but
since we have destructive

435
00:20:41 --> 00:20:45
interference we're going to
draw 1 s b as having the

436
00:20:45 --> 00:20:48
opposite sign, so we have a
plus and a minus in

437
00:20:48 --> 00:20:49
terms of signs.
 

438
00:20:49 --> 00:20:51
So that should make it very
easy to picture that this

439
00:20:51 --> 00:20:53
is being cancelled
out in the middle.

440
00:20:53 --> 00:20:57
If we overlay what the actual
molecular orbital is on top of

441
00:20:57 --> 00:21:01
it, what you see is that in the
center you end up cancelling

442
00:21:01 --> 00:21:03
out the wave function entirely.
 

443
00:21:03 --> 00:21:08
So this is the 1 s star, sigma
1 s star orbital, and what you

444
00:21:08 --> 00:21:12
have in the center here is a
node, right in the center

445
00:21:12 --> 00:21:15
between the two nuclei.
 

446
00:21:15 --> 00:21:17
So again if we look at this
in terms of its physical

447
00:21:17 --> 00:21:21
interpretation or probability
density, what we need to do is

448
00:21:21 --> 00:21:22
square the wave function.
 

449
00:21:22 --> 00:21:28
So if we square sigma 1 s star,
we flip the amplitude so it's

450
00:21:28 --> 00:21:31
all positive now, but again we
still have this node

451
00:21:31 --> 00:21:32
right in the middle.
 

452
00:21:32 --> 00:21:34
So if we talk about the
probability density and we

453
00:21:34 --> 00:21:38
write that in, it's going to be
sigma 1 s star squared, so now

454
00:21:38 --> 00:21:43
we're talking about 1 s a minus
1 s b, all of that

455
00:21:43 --> 00:21:44
being squared.
 

456
00:21:44 --> 00:21:47
And again, if we write out what
all the terms are, we again

457
00:21:47 --> 00:21:52
have 1 s a squared plus 1 s b
squared, but now what we're

458
00:21:52 --> 00:21:54
doing is we're actually
subtracting the

459
00:21:54 --> 00:21:56
interference term.
 

460
00:21:56 --> 00:21:58
So if we're subtracting the
interference term, what

461
00:21:58 --> 00:22:04
we have here now is
destructive interference.

462
00:22:04 --> 00:22:07
So let's think about the
energy of interaction here.

463
00:22:07 --> 00:22:11
When we were talking about
constructive interference, we

464
00:22:11 --> 00:22:14
had more electron density
in between the 2 nuclei.

465
00:22:14 --> 00:22:17
So that lowered the energy
of the molecular orbital.

466
00:22:17 --> 00:22:20
So, bonding orbitals
are down here.

467
00:22:20 --> 00:22:22
But when we think about
where anti-bonding orbitals

468
00:22:22 --> 00:22:25
should be, it should
be higher in energy.

469
00:22:25 --> 00:22:29
It's increased compared
to the atomic orbitals.

470
00:22:29 --> 00:22:32
So we would label our
anti-bonding orbital higher

471
00:22:32 --> 00:22:37
in energy than our 1
s atomic orbitals.

472
00:22:37 --> 00:22:39
So, let's think a little
bit about what this means.

473
00:22:39 --> 00:22:42
First of all, again to repeat,
any time we see the star, sigma

474
00:22:42 --> 00:22:46
1 s star, that's an
anti-bonding orbital.

475
00:22:46 --> 00:22:48
When we talk about that,
basically what we're saying,

476
00:22:48 --> 00:22:51
and you can see that because of
that negative interference, we

477
00:22:51 --> 00:22:55
actually have less electron
density between the nuclei than

478
00:22:55 --> 00:22:57
we did when they were
two separate atoms.

479
00:22:57 --> 00:23:01
So you can see that this is
non-bonding, this is even

480
00:23:01 --> 00:23:03
worse than non-bonding, it's
anti-bonding, because we're

481
00:23:03 --> 00:23:06
actually getting rid of
electron density between

482
00:23:06 --> 00:23:07
the two nuclei.
 

483
00:23:07 --> 00:23:09
And we know that it's electron
density between the nuclei

484
00:23:09 --> 00:23:13
that holds two atoms
together in a bond.

485
00:23:13 --> 00:23:15
So what I want to point out is
that it creates an effect that

486
00:23:15 --> 00:23:18
is exactly opposite of a bond.
 

487
00:23:18 --> 00:23:20
You might have thought before
we started talking about

488
00:23:20 --> 00:23:23
molecular orbital theory that
non-bonding was the opposite of

489
00:23:23 --> 00:23:27
bonding, it's not, anti-bonding
is the opposite of bonding,

490
00:23:27 --> 00:23:29
and anti-bonding is
not non-bonding.

491
00:23:29 --> 00:23:32
We can see that if we just
look at this picture here.

492
00:23:32 --> 00:23:35
Here is bonding, and
here is non-bonding.

493
00:23:35 --> 00:23:40
Anti-bonding is even higher
in energy than non-bonding.

494
00:23:40 --> 00:23:43
And the other thing to point
out is that the energy that an

495
00:23:43 --> 00:23:47
anti-bonding orbital is raised
by, is the same amount as a

496
00:23:47 --> 00:23:49
bonding orbital is lowered by.
 

497
00:23:49 --> 00:23:52
So any time I draw these
molecular orbitals, I do my

498
00:23:52 --> 00:23:55
best, and I'm not always
perfect, yet trying to make

499
00:23:55 --> 00:23:59
this energy different exactly
the same for the anti-bonding

500
00:23:59 --> 00:24:01
orbital being raised, versus
the bonding orbital

501
00:24:01 --> 00:24:02
being lowered.
 

502
00:24:02 --> 00:24:07
So those should be raised and
lowered by the same energy.

503
00:24:07 --> 00:24:09
So now let's take a look
at some of -- is there

504
00:24:09 --> 00:24:16
a question up there?
 

505
00:24:16 --> 00:24:17
STUDENT: Why would two atoms
decide to [INAUDIBLE].

506
00:24:17 --> 00:24:18
PROFESSOR: Well,
they don't want to.

507
00:24:18 --> 00:24:19
It's a higher energy situation.
 

508
00:24:19 --> 00:24:22
So we'll start to look at
molecules and we'll see if we

509
00:24:22 --> 00:24:27
take two atoms and we fill in
our molecular orbital and it

510
00:24:27 --> 00:24:29
turns out that they have more
anti-bonding orbitals than

511
00:24:29 --> 00:24:33
bonding, that's -- a diatomic
molecule we'll never see.

512
00:24:33 --> 00:24:37
So it helps us predict, will we
see this, for example, h 2,

513
00:24:37 --> 00:24:40
which we're going to be about
to do, we'll see is stabilized

514
00:24:40 --> 00:24:42
because it has more bonding
than anti-bonding.

515
00:24:42 --> 00:24:44
So we'll predict, yes,
there's a bond here.

516
00:24:44 --> 00:24:45
That's a really good question.
 

517
00:24:45 --> 00:24:51
So, let's go ahead and do this
and take a look at some of the

518
00:24:51 --> 00:24:54
actual atoms that we can think
about and think about

519
00:24:54 --> 00:24:56
them in molecules.
 

520
00:24:56 --> 00:24:58
So our simplest case that
we started talking about

521
00:24:58 --> 00:25:00
was molecular hydrogen.
 

522
00:25:00 --> 00:25:03
I want to finish this
discussion by including the

523
00:25:03 --> 00:25:05
anti-bonding orbital, and this
is a tip for you when you're

524
00:25:05 --> 00:25:09
drawing your molecular orbital
diagrams, any time you draw a

525
00:25:09 --> 00:25:12
bonding orbital, there is
also an anti-bonding

526
00:25:12 --> 00:25:13
orbital that exists.
 

527
00:25:13 --> 00:25:16
It might not have any electrons
in it, but it still exists, so

528
00:25:16 --> 00:25:20
you need to draw these into
your molecular orbital diagram.

529
00:25:20 --> 00:25:22
So I wanted to make sure you
have a complete set for

530
00:25:22 --> 00:25:23
hydrogen in your notes.
 

531
00:25:23 --> 00:25:24
So let's take a look at this.
 

532
00:25:24 --> 00:25:29
Hydrogen, we can first draw
in our atomic electrons.

533
00:25:29 --> 00:25:32
So there's one electron
in each hydrogen atom.

534
00:25:32 --> 00:25:36
And then this means we'll have
a total of two electrons in our

535
00:25:36 --> 00:25:39
hydrogen molecule, so we can
fill both of those into

536
00:25:39 --> 00:25:43
the sigma 1 s orbital,
the bonding orbital.

537
00:25:43 --> 00:25:45
We don't have to put anything
into the anti-bonding

538
00:25:45 --> 00:25:47
orbital, so that's great.
 

539
00:25:47 --> 00:25:51
What we've seen is we have a
net lowering of energy of the

540
00:25:51 --> 00:25:56
molecule versus the
individual atoms.

541
00:25:56 --> 00:26:00
So let's draw the electron
configuration of hydrogen, the

542
00:26:00 --> 00:26:02
molecule, molecular hydrogen.
 

543
00:26:02 --> 00:26:05
What you saw, what we've done a
lot of is drawing the electron

544
00:26:05 --> 00:26:08
configurations for different
atoms, we can do the same

545
00:26:08 --> 00:26:10
thing for a molecule.
 

546
00:26:10 --> 00:26:13
So, if we take h 2, and we
want to draw the electron

547
00:26:13 --> 00:26:15
configuration, it's very short.
 

548
00:26:15 --> 00:26:20
All it is sigma 1 s, and then
we have two electrons in it,

549
00:26:20 --> 00:26:23
so it's sigma 1 s squared.
 

550
00:26:23 --> 00:26:25
So this is our electron
configuration.

551
00:26:25 --> 00:26:27
Let's take a look at
another example.

552
00:26:27 --> 00:26:30
Let's draw the molecular
diagram for h e 2 now.

553
00:26:30 --> 00:26:35
So again, we can fill in our
atomic orbitals here, there's

554
00:26:35 --> 00:26:39
going to be two electrons in
each of our atomic orbitals.

555
00:26:39 --> 00:26:42
So now let's go ahead and fill
in our molecular orbitals.

556
00:26:42 --> 00:26:45
We need to fill in a
total of four electrons.

557
00:26:45 --> 00:26:50
So we have two electrons in our
bonding orbital, but because we

558
00:26:50 --> 00:26:53
use the same rules to fill up
molecular orbitals as we do

559
00:26:53 --> 00:26:56
atomic orbitals, so the Pauli
exclusion principle tells us we

560
00:26:56 --> 00:26:59
can't have more than two
electrons per orbital, so

561
00:26:59 --> 00:27:02
we have to go up to our
anti-bonding orbital here.

562
00:27:02 --> 00:27:06
So this means that we have two
of the electrons are lowered in

563
00:27:06 --> 00:27:09
energy, but two are
raised in energy.

564
00:27:09 --> 00:27:12
So would this be a
stabilized molecule then?

565
00:27:12 --> 00:27:14
STUDENT: [INAUDIBLE]
 

566
00:27:14 --> 00:27:15
PROFESSOR: No.
 

567
00:27:15 --> 00:27:18
So, compared to the atoms, it
should be somewhat the same

568
00:27:18 --> 00:27:20
energy, we shouldn't get any
extra stabilization from

569
00:27:20 --> 00:27:21
forming the molecule.
 

570
00:27:21 --> 00:27:25
So lets go ahead and write what
the electron configuration

571
00:27:25 --> 00:27:31
would be of h e 2.
 

572
00:27:31 --> 00:27:34
And again, we're just filling
in the different orbitals, so

573
00:27:34 --> 00:27:39
we have sigma 1 s, that's going
to be squared, and now we

574
00:27:39 --> 00:27:44
have sigma 1 s star squared.
 

575
00:27:44 --> 00:27:48
So we can compare the two
electron configurations, and we

576
00:27:48 --> 00:27:52
can actually think about --
what we figure out from them,

577
00:27:52 --> 00:27:55
we see that two are lowered in
energy, two electrons are

578
00:27:55 --> 00:27:58
raised in energy, so we have no
net gain or no net loss

579
00:27:58 --> 00:28:01
in energy for h e 2.
 

580
00:28:01 --> 00:28:05
And there's actually a way that
we can make predictions here,

581
00:28:05 --> 00:28:07
and what I'll tell you is
molecular orbital theory

582
00:28:07 --> 00:28:10
predicts that h e 2 does not
exist because it's not

583
00:28:10 --> 00:28:14
stabilized in terms of
forming the molecule.

584
00:28:14 --> 00:28:17
The way that we can figure this
out is using something called

585
00:28:17 --> 00:28:21
bond order, and bond order is
equal to 1/2 times the number

586
00:28:21 --> 00:28:24
of bonding electrons, minus the
number of anti-bonding

587
00:28:24 --> 00:28:25
electrons.
 

588
00:28:25 --> 00:28:28
And the bond order you get out
will either be, for example,

589
00:28:28 --> 00:28:31
zero, which would mean that you
have no bond, or you could

590
00:28:31 --> 00:28:33
have 1, a single bond, 1 .
 

591
00:28:33 --> 00:28:37
5, a 1 and 1/2 bond, 2, a
double bond, and so on.

592
00:28:37 --> 00:28:42
So let's figure out the bond
order for our two molecules

593
00:28:42 --> 00:28:45
here that we figured out the
electron configuration for.

594
00:28:45 --> 00:28:47
So I guess we'll
start with helium 2.

595
00:28:47 --> 00:28:53
So the bond order is going to
be equal to 1/2, and then

596
00:28:53 --> 00:28:56
it will be 2 minus 2.
 

597
00:28:56 --> 00:29:00
So our bond order for h e 2
is going to be equal to 0.

598
00:29:00 --> 00:29:04
So it has a 0 bond,
there's no bond in h e 2.

599
00:29:04 --> 00:29:06
Let's look for hydrogen.
 

600
00:29:06 --> 00:29:14
For hydrogen our bond order is
going to equal 1/2, 2 minus 0.

601
00:29:14 --> 00:29:16
So we would predict
a bond order of 1.

602
00:29:16 --> 00:29:19
What kind of a bond is
a bond order of 1?

603
00:29:19 --> 00:29:23
Yeah, we'd expect to see a
single bond in hydrogen.

604
00:29:23 --> 00:29:27
So what actually turns out the
reality is that h e 2 does

605
00:29:27 --> 00:29:31
exist, but it exists as the
weakest chemical bond known,

606
00:29:31 --> 00:29:35
and it wasn't, in fact, even
found to exist until 1993, so I

607
00:29:35 --> 00:29:39
can assure you this is not a
bond that you see very often

608
00:29:39 --> 00:29:42
in nature, and it is a
very, very weak bond.

609
00:29:42 --> 00:29:45
It only has a dissociation
energy of 0 .

610
00:29:45 --> 00:29:47
1 kilojoules per mole.
 

611
00:29:47 --> 00:29:50
So that should make sense,
because we saw no energy

612
00:29:50 --> 00:29:53
difference between the actual
atoms and the molecules.

613
00:29:53 --> 00:29:57
Molecular orbital theory, even
at this very basic level,

614
00:29:57 --> 00:30:00
allowed us to predict that no,
we're not going to see a true

615
00:30:00 --> 00:30:02
bond here, a strong bond.
 

616
00:30:02 --> 00:30:06
In contrast, the dissociation
energy of a bond for hydrogen,

617
00:30:06 --> 00:30:10
and molecular hydrogen is
everywhere around us, we see

618
00:30:10 --> 00:30:12
432 kilojoules per mole.
 

619
00:30:12 --> 00:30:17
All right, so we can now see a
little bit of what the power of

620
00:30:17 --> 00:30:20
molecular orbital theory is in
predicting what kind of bonds

621
00:30:20 --> 00:30:23
we're going to see in
molecules, or whether or not

622
00:30:23 --> 00:30:25
we'll see this bonding
occur at all.

623
00:30:25 --> 00:30:27
So let's look at another
example, let's take

624
00:30:27 --> 00:30:30
lithium 2 and see what
we can figure out here.

625
00:30:30 --> 00:30:34
In lithium 2, we have two
atoms of lithium, each have

626
00:30:34 --> 00:30:36
three electrons in them.
 

627
00:30:36 --> 00:30:40
So now we have to include
both the 1 s orbitals and

628
00:30:40 --> 00:30:42
also the 2 s orbitals.
 

629
00:30:42 --> 00:30:45
So any time in a molecular
orbital diagram you draw in

630
00:30:45 --> 00:30:48
orbitals, you need to
draw the corresponding

631
00:30:48 --> 00:30:49
molecular orbitals.
 

632
00:30:49 --> 00:30:54
So, this means we need to have
sigma 1 s, sigma 1 s star, and

633
00:30:54 --> 00:30:58
now sigma 2 s and
sigma 2 s star.

634
00:30:58 --> 00:31:01
Something I'll also point out
as you see these dashed line

635
00:31:01 --> 00:31:04
that tell you where the
individual molecular orbitals

636
00:31:04 --> 00:31:08
are arising from, as you get to
higher and higher atomic

637
00:31:08 --> 00:31:12
numbers of molecules that
you're making, it makes a lot

638
00:31:12 --> 00:31:15
more sense to look at a diagram
when you draw these dotted

639
00:31:15 --> 00:31:16
lines in, because they can
start to get a little

640
00:31:16 --> 00:31:17
bit confusing.
 

641
00:31:17 --> 00:31:20
So when you go ahead and draw
these on your problem-sets or

642
00:31:20 --> 00:31:23
on your exams, it's a good idea
to put these dashed lines in,

643
00:31:23 --> 00:31:26
both for you and for people
reading it to see exactly where

644
00:31:26 --> 00:31:29
your molecular orbitals
are coming from.

645
00:31:29 --> 00:31:32
So, this means we have a total
of six electrons that we need

646
00:31:32 --> 00:31:34
to put into molecular orbitals.
 

647
00:31:34 --> 00:31:37
So again we just start filling
those up -- we have two in the

648
00:31:37 --> 00:31:41
1 s, two in the sigma 1 s star,
and then we have two

649
00:31:41 --> 00:31:44
in the sigma 2 s.
 

650
00:31:44 --> 00:31:46
So we should be able to also
calculate the bond order,

651
00:31:46 --> 00:31:49
just like we did for
hydrogen and helium.

652
00:31:49 --> 00:31:53
First we can do that by knowing
the electron configuration, we

653
00:31:53 --> 00:31:56
can write it out just by going
up the table here, up

654
00:31:56 --> 00:31:58
the energy levels.
 

655
00:31:58 --> 00:32:01
And you can go ahead and tell
me what you think the bond

656
00:32:01 --> 00:32:22
order is going to be
for this molecule.

657
00:32:22 --> 00:32:22
All right.
 

658
00:32:22 --> 00:32:36
Let's take 10 more
seconds on this.

659
00:32:36 --> 00:32:38
OK, good, we're back on
track a little bit with

660
00:32:38 --> 00:32:39
our clicker answers.
 

661
00:32:39 --> 00:32:46
So it's selection three or
one is the bond order.

662
00:32:46 --> 00:32:48
So let's switch back to
our class notes and look

663
00:32:48 --> 00:32:49
at what this means.
 

664
00:32:49 --> 00:32:54
So we know that it's 1, because
we have 1, 2, 3, 4 bonding,

665
00:32:54 --> 00:32:58
minus 2 anti-bonding, and 1/2
of that is a bond order of 1.

666
00:32:58 --> 00:33:01
We would predict to see
a single bond between

667
00:33:01 --> 00:33:04
lithium, and it turns
out that's what we see.

668
00:33:04 --> 00:33:07
And so you have for a
reference, the dissociation

669
00:33:07 --> 00:33:10
energy of lithium 2 is
105 kilojoules per mole.

670
00:33:10 --> 00:33:13
So let's keep moving
along the periodic table

671
00:33:13 --> 00:33:16
and keep applying our
molecular orbitals.

672
00:33:16 --> 00:33:23
For example now, with b e 2, so
beryllium 2 has four electrons

673
00:33:23 --> 00:33:24
in terms of each atom.
 

674
00:33:24 --> 00:33:27
So you can start by filling
those in, and now we can

675
00:33:27 --> 00:33:30
fill in our molecular
orbitals as well.

676
00:33:30 --> 00:33:32
This means we need a total
of eight electrons in

677
00:33:32 --> 00:33:35
our molecular orbitals.
 

678
00:33:35 --> 00:33:40
So we have two in 1 s, two in
the sigma 1 s star, two in the

679
00:33:40 --> 00:33:44
sigma 2 s, and two in
the sigma 2 s star.

680
00:33:44 --> 00:33:48
So let's go ahead and figure
out the bonding order

681
00:33:48 --> 00:33:57
for beryllium here.
 

682
00:33:57 --> 00:34:00
So when we figure it out for
beryllium -- let's see if I

683
00:34:00 --> 00:34:06
wrote in your notes what
the actual -- electron

684
00:34:06 --> 00:34:08
configuration, OK that's
already in your notes for you.

685
00:34:08 --> 00:34:12
So let's go right to the
bond order for beryllium.

686
00:34:12 --> 00:34:16
So for the bond order we want
to take 1/2 of the total number

687
00:34:16 --> 00:34:20
of bonding electrons, so that's
going to be 4 minus

688
00:34:20 --> 00:34:24
anti-bonding is 4, so we end up
getting a bond order

689
00:34:24 --> 00:34:25
that's equal to 0.
 

690
00:34:25 --> 00:34:28
So what I want to point out
with this case in beryllium is

691
00:34:28 --> 00:34:31
that you don't have to use all
of the electrons to figure out

692
00:34:31 --> 00:34:35
the bond order, and in fact,
once you get to molecules that

693
00:34:35 --> 00:34:39
are from atoms with atomic
numbers of 8 or 10, you're not

694
00:34:39 --> 00:34:42
going to want to maybe draw out
the full molecular

695
00:34:42 --> 00:34:44
orbital diagram.
 

696
00:34:44 --> 00:34:46
So what I want to tell you is
we also always get the same

697
00:34:46 --> 00:34:49
bond order if we instead
only deal with the

698
00:34:49 --> 00:34:50
valence electrons.
 

699
00:34:50 --> 00:34:53
So let's just prove that to
ourselves and figure out the

700
00:34:53 --> 00:34:57
bond order just using
valence electrons.

701
00:34:57 --> 00:34:59
So this would mean the bond
order is equal to 1/2, and in

702
00:34:59 --> 00:35:02
terms of valence electrons,
how many bonding valence

703
00:35:02 --> 00:35:02
electrons do we have?
 

704
00:35:02 --> 00:35:03
STUDENT: [INAUDIBLE]
 

705
00:35:03 --> 00:35:06
PROFESSOR: All right,
what about anti-bonding?

706
00:35:06 --> 00:35:07
STUDENT: [INAUDIBLE]
 

707
00:35:07 --> 00:35:08
PROFESSOR: Two.
 

708
00:35:08 --> 00:35:08
OK, good.
 

709
00:35:08 --> 00:35:11
So again, we're going
to see that we have a

710
00:35:11 --> 00:35:13
bonding order of 0.
 

711
00:35:13 --> 00:35:18
So we would not predict
to see a b e 2 bond.

712
00:35:18 --> 00:35:22
So what we see is a bond
order of 0, and again, the

713
00:35:22 --> 00:35:24
bond is very, very weak.
 

714
00:35:24 --> 00:35:25
Essentially we're not
going to see this, it's

715
00:35:25 --> 00:35:27
9 kilojoules per mole.
 

716
00:35:27 --> 00:35:29
All right.
 

717
00:35:29 --> 00:35:33
So, so far we've looked only
at molecules that involve

718
00:35:33 --> 00:35:36
atoms that have only
s orbitals in them.

719
00:35:36 --> 00:35:38
I'm sure you're thinking well,
what do we do in the case of p

720
00:35:38 --> 00:35:40
orbitals, and, in fact, we
can do the same thing.

721
00:35:40 --> 00:35:43
Again, we're going to take the
linear combination of those p

722
00:35:43 --> 00:35:47
atomic orbitals and make what
are called pi or some more

723
00:35:47 --> 00:35:49
sigma molecular orbitals.
 

724
00:35:49 --> 00:35:53
So let's look at the first case
where we have either the 2 p x

725
00:35:53 --> 00:35:57
or 2 p y type of orbitals
that we're combining.

726
00:35:57 --> 00:36:00
So they're the same shape, this
is the shape of the orbital or

727
00:36:00 --> 00:36:04
the shape of the wave function,
and we can call this either 2 p

728
00:36:04 --> 00:36:08
x a being combined with 2 p x
b, or we could say since it's

729
00:36:08 --> 00:36:13
the same shape, it's 2 p y a
being combined with 2 p y b.

730
00:36:13 --> 00:36:16
And in either case if we first
talk about constructive

731
00:36:16 --> 00:36:19
interference, what again we're
going to see is that where

732
00:36:19 --> 00:36:22
these two orbitals come
together, we're going to see

733
00:36:22 --> 00:36:25
increased wave function
in that area, so we saw

734
00:36:25 --> 00:36:27
constructive interference.
 

735
00:36:27 --> 00:36:31
So again, we can name these
molecular orbitals and these

736
00:36:31 --> 00:36:35
we're going to call -- also to
point out there is now a bond

737
00:36:35 --> 00:36:38
axis along this nodal plane,
which is something we didn't

738
00:36:38 --> 00:36:41
see before when we were
combining the s orbitals.

739
00:36:41 --> 00:36:43
So when we go ahead and
name these, we're going to

740
00:36:43 --> 00:36:45
call these pi orbitals.
 

741
00:36:45 --> 00:36:49
We'll call it either pi 2 p
x, if we're combining the

742
00:36:49 --> 00:36:53
x orbitals, or pi 2 p y.
 

743
00:36:53 --> 00:36:57
The reason that I wanted to
point out this nodal plane here

744
00:36:57 --> 00:37:00
is because this is why it
is called a pi orbital.

745
00:37:00 --> 00:37:04
Pi orbitals are a molecular
orbital that have a nodal

746
00:37:04 --> 00:37:07
plane through the bond axis.
 

747
00:37:07 --> 00:37:10
Remember this is our bond axis
here, and you can see there is

748
00:37:10 --> 00:37:13
this area where the wave
function is equal to zero

749
00:37:13 --> 00:37:16
all along that plane,
that's a nodal plane.

750
00:37:16 --> 00:37:18
So that's why these are
pi orbitals instead

751
00:37:18 --> 00:37:20
of sigma orbitals.
 

752
00:37:20 --> 00:37:23
So again, we can think about
the probability density

753
00:37:23 --> 00:37:25
in terms of squaring
the wave function.

754
00:37:25 --> 00:37:28
So now what it is that we're
squaring is if we're talking

755
00:37:28 --> 00:37:34
about x orbital, it's pi 2 p x
squared, and this is just equal

756
00:37:34 --> 00:37:43
to the 2 p x a plus the 2 p x b
all squared, or if we write out

757
00:37:43 --> 00:37:49
all of the terms we have 2 p x
a squared plus 2 p x b squared,

758
00:37:49 --> 00:37:52
and then this term here, and
again, this is our

759
00:37:52 --> 00:37:53
interference term.
 

760
00:37:53 --> 00:37:57
In this case is it constructive
or destructive interference?

761
00:37:57 --> 00:37:57
STUDENT: [INAUDIBLE]
 

762
00:37:57 --> 00:37:59
PROFESSOR: Constructive
interference.

763
00:37:59 --> 00:38:02
We're seeing that the wave
function's adding together and

764
00:38:02 --> 00:38:06
giving us more wave function
in the center here.

765
00:38:06 --> 00:38:07
All right.
 

766
00:38:07 --> 00:38:09
So we see constructive
interference, of course, we

767
00:38:09 --> 00:38:12
can also see destructive
interference.

768
00:38:12 --> 00:38:15
So I changed the colors here
to show that these are 2 p

769
00:38:15 --> 00:38:19
orbitals with an opposite
phase or an opposite sign.

770
00:38:19 --> 00:38:25
So what happens when we add a 2
p a and we subtract from it a 2

771
00:38:25 --> 00:38:30
p x b, or the same with a 2 p y
a subtracting a 2 p y b, is

772
00:38:30 --> 00:38:33
that we're actually going to
cancel out the wave function in

773
00:38:33 --> 00:38:36
the center, so we now
have 2 nodal planes.

774
00:38:36 --> 00:38:39
So again, this is an
anti-bonding orbital, and what

775
00:38:39 --> 00:38:43
you see is that there is now
less electron density between

776
00:38:43 --> 00:38:49
the two nuclei than there was
when you had non-bonding.

777
00:38:49 --> 00:38:53
So we're going to call this the
sigma 2 p x star, or if we're

778
00:38:53 --> 00:38:58
talking about the 2 p y
orbitals we'll call this the pi

779
00:38:58 --> 00:39:05
2 p x star, and the
pi 2 p y star.

780
00:39:05 --> 00:39:09
And the pi star orbitals result
from any time you have

781
00:39:09 --> 00:39:12
destructive interference from 2
p orbitals that are either

782
00:39:12 --> 00:39:16
the p x or the p y.
 

783
00:39:16 --> 00:39:19
So now we can move on to an
example where we do, in fact,

784
00:39:19 --> 00:39:23
have to use some p orbitals,
so this would be b 2.

785
00:39:23 --> 00:39:26
How many electrons
are in boron?

786
00:39:26 --> 00:39:27
Five.
 

787
00:39:27 --> 00:39:29
I see some hands going up.
 

788
00:39:29 --> 00:39:32
There's five electrons So what
you'll notice here is that I

789
00:39:32 --> 00:39:35
only filled in 3 electrons.
 

790
00:39:35 --> 00:39:37
What do these
electrons represent?

791
00:39:37 --> 00:39:38
STUDENT: Valence.
 

792
00:39:38 --> 00:39:39
PROFESSOR: Valence electrons.
 

793
00:39:39 --> 00:39:41
OK, sometimes you're going to
be asked to draw a molecular

794
00:39:41 --> 00:39:45
orbital diagram where you're
asked to include all electrons,

795
00:39:45 --> 00:39:48
and sometimes it will
specifically say only

796
00:39:48 --> 00:39:50
include valence electrons.
 

797
00:39:50 --> 00:39:53
That happens because of space
issues that you were asked to

798
00:39:53 --> 00:39:56
do that, because you can always
assume that all of the core

799
00:39:56 --> 00:39:59
orbitals are already
going to be filled.

800
00:39:59 --> 00:40:01
So, in this case, we're just
drawing the molecular orbital

801
00:40:01 --> 00:40:04
diagram for the valence
electrons, so we have

802
00:40:04 --> 00:40:05
three for each.
 

803
00:40:05 --> 00:40:09
And what we see here is now
when we're combining the p, we

804
00:40:09 --> 00:40:13
have our 2 p x and our 2 p y
orbitals that are lower in

805
00:40:13 --> 00:40:16
energy, and then our pi
anti-bonding orbitals that

806
00:40:16 --> 00:40:18
are higher in energy.
 

807
00:40:18 --> 00:40:22
You might be asking where the 2
p z orbital is and we'll get

808
00:40:22 --> 00:40:24
to that soon once we need it.
 

809
00:40:24 --> 00:40:28
Let's just first fill in
this for the b 2 case.

810
00:40:28 --> 00:40:32
So we can start at the bottom,
two electrons in sigma 2 s, two

811
00:40:32 --> 00:40:35
electrons in sigma 2 s star.
 

812
00:40:35 --> 00:40:40
Now we need to jump up to using
these pi orbitals, and what

813
00:40:40 --> 00:40:44
we're going to do is put one
electron into each of our pi

814
00:40:44 --> 00:40:47
2 p x and 2 p y orbitals.
 

815
00:40:47 --> 00:40:50
So again you can see as we're
filling up our molecular

816
00:40:50 --> 00:40:53
orbitals, we're using the exact
same principle we used to

817
00:40:53 --> 00:40:55
fill up atomic orbitals.
 

818
00:40:55 --> 00:41:00
So let's think about what
the valence electron

819
00:41:00 --> 00:41:03
configuration is here.
 

820
00:41:03 --> 00:41:07
So now we're looking
at the case of b 2.

821
00:41:07 --> 00:41:11
And what we're looking at is a
valence electron molecular

822
00:41:11 --> 00:41:14
orbital diagram, so let's just
draw the electron configuration

823
00:41:14 --> 00:41:20
for the valence orbitals, so
that will be sigma 2 s 2, sigma

824
00:41:20 --> 00:41:30
2 s star 2, and now we start in
with our pi 2 p x 1,

825
00:41:30 --> 00:41:34
and our pi 2 p y 1.
 

826
00:41:34 --> 00:41:38
So this is our valence electron
configuration for b 2.

827
00:41:38 --> 00:41:41
All right.
 

828
00:41:41 --> 00:41:48
So what would you expect the
bonding order for b 2 to be?

829
00:41:48 --> 00:41:50
Shout it out if you know.
 

830
00:41:50 --> 00:41:51
STUDENT: one.
 

831
00:41:51 --> 00:41:55
PROFESSOR: And I didn't write
up there but it is one, and we

832
00:41:55 --> 00:42:00
can see that it's 1, because
it's 1/2 of 2, 4 minus 2, so

833
00:42:00 --> 00:42:04
1/2 of 2, the bonding order
is going to be equal to one.

834
00:42:04 --> 00:42:06
So let's move on to another
example, let's talk

835
00:42:06 --> 00:42:07
about carbon here.
 

836
00:42:07 --> 00:42:10
Again, we're just talking
about the valence electrons.

837
00:42:10 --> 00:42:13
So carbon has four valence
electrons, so if we talk about

838
00:42:13 --> 00:42:17
c 2, again we're going to start
filling in our molecular

839
00:42:17 --> 00:42:21
orbitals, and now we're going
to have eight electrons to fill

840
00:42:21 --> 00:42:22
into our molecular orbitals.
 

841
00:42:22 --> 00:42:27
So, we'll put two in the sigma
2 s, two in the sigma 2 s star,

842
00:42:27 --> 00:42:31
and now we're going to fill one
and one into each of our pi 2 p

843
00:42:31 --> 00:42:35
x and 2 p y, but we still have
two electrons left, so what

844
00:42:35 --> 00:42:38
we're going to do is double up
in terms of our 2 p

845
00:42:38 --> 00:42:40
x and our 2 p y.
 

846
00:42:40 --> 00:42:42
So let's think about what
this valence electron

847
00:42:42 --> 00:42:48
configuration is for c 2.
 

848
00:42:48 --> 00:42:51
And again, I want you to have
practiced drawing these out in

849
00:42:51 --> 00:42:54
the form -- you always need to
start with the sigma and then

850
00:42:54 --> 00:42:56
write the number
of the orbital.

851
00:42:56 --> 00:43:02
So, sigma 2 s has two
electrons, sigma 2 s star with

852
00:43:02 --> 00:43:06
two electrons, and now we
have sigma 2 p x -- how

853
00:43:06 --> 00:43:07
many electrons here?
 

854
00:43:07 --> 00:43:09
STUDENT: Two.
 

855
00:43:09 --> 00:43:10
PROFESSOR: Two.
 

856
00:43:10 --> 00:43:15
And sigma 2 p y, two
electrons here.

857
00:43:15 --> 00:43:20
Oh excuse me, pi 2 p y, thank
you -- pi 2 p x and pi 2 p y.

858
00:43:20 --> 00:43:23
So let's talk about what
the bonding order is

859
00:43:23 --> 00:43:25
going to be for c 2.
 

860
00:43:25 --> 00:43:30
So what's the bonding
order for c 2?

861
00:43:30 --> 00:43:31
STUDENT: Two.
 

862
00:43:31 --> 00:43:32
PROFESSOR: two?
 

863
00:43:32 --> 00:43:41
OK, so we have 2, 4, 6 minus 2,
so we have 1/2 of 6 minus 2, so

864
00:43:41 --> 00:43:44
that's 1/2 half 4, so we have
a bonding order of

865
00:43:44 --> 00:43:46
two for carbon 2.
 

866
00:43:46 --> 00:43:50
So we would expect to see a
double bond for a c 2 where

867
00:43:50 --> 00:43:54
we would expect to see a --
double bond for c 2 and

868
00:43:54 --> 00:43:58
a single bond for b 2.
 

869
00:43:58 --> 00:44:01
And that is, in fact, what we
can surmise if we look at

870
00:44:01 --> 00:44:04
the different dissociation
energies for the two bonds.

871
00:44:04 --> 00:44:08
So for b 2, which is a single
bond, that's 289 kilojoules per

872
00:44:08 --> 00:44:12
mole to break it, and it takes
us more energy to break this

873
00:44:12 --> 00:44:17
double bond for carbon, which
is 599 kilojoules per mole.

874
00:44:17 --> 00:44:20
So, in general what we see, and
this is always true if we're

875
00:44:20 --> 00:44:23
comparing the same atom, and in
general, if we're comparing

876
00:44:23 --> 00:44:26
different types of molecules,
but we know that a single bond

877
00:44:26 --> 00:44:28
is always weaker than a double
bond, which is weaker

878
00:44:28 --> 00:44:29
than a triple bond.
 

879
00:44:29 --> 00:44:33
And obviously, no bond is the
weakest of all is not bonding.

880
00:44:33 --> 00:44:34
All right.
 

881
00:44:34 --> 00:44:38
So let's look now at the case
where we do have 2 p z orbitals

882
00:44:38 --> 00:44:39
that we're talking about.
 

883
00:44:39 --> 00:44:42
So again, what we're talking
about is the linear combination

884
00:44:42 --> 00:44:47
of atomic 2 p orbitals, and now
we're talking about 2 p z.

885
00:44:47 --> 00:44:50
So if we have constructive
interference between the two,

886
00:44:50 --> 00:44:54
what we're going to see is our
molecular orbital looks

887
00:44:54 --> 00:44:56
something like this.
 

888
00:44:56 --> 00:44:59
Do you predict that this will
be a sigma or a pi orbital?

889
00:44:59 --> 00:45:02
STUDENT: [INAUDIBLE]
 

890
00:45:02 --> 00:45:03
PROFESSOR: All right, I'm
hearing a little of both, but

891
00:45:03 --> 00:45:05
I'm very encouraged to hear
quite a few people

892
00:45:05 --> 00:45:07
saying sigma.
 

893
00:45:07 --> 00:45:10
This is, in fact, a sigma
2 p z orbital is what

894
00:45:10 --> 00:45:12
this orbital is called.
 

895
00:45:12 --> 00:45:15
The reason that it's sigma is
if you look at the bonding axis

896
00:45:15 --> 00:45:20
here, is that there is no nodal
plane along the bonding axis.

897
00:45:20 --> 00:45:25
Also, it is cylindrically
symmetric around the bonding

898
00:45:25 --> 00:45:28
axis, so this is how we know
that it's a sigma orbital.

899
00:45:28 --> 00:45:32
So some p orbitals form pi
molecular orbitals, and some

900
00:45:32 --> 00:45:34
form sigma p orbitals.
 

901
00:45:34 --> 00:45:38
Specifically, it's always the z
that forms the sigma orbital,

902
00:45:38 --> 00:45:42
and the reason is at least at a
minimum for this class we

903
00:45:42 --> 00:45:47
always define the internuclear
axis as the z axis, so this is

904
00:45:47 --> 00:45:51
always the z axis, so it's
always going to be the 2 p

905
00:45:51 --> 00:45:53
z's that are coming
together head-on.

906
00:45:53 --> 00:45:56
The reason that there is
increased electron density here

907
00:45:56 --> 00:45:59
is you can see that these two
orbitals come together and

908
00:45:59 --> 00:46:02
constructively interfere.
 

909
00:46:02 --> 00:46:05
We can also talk about
anti-bonding orbitals where we

910
00:46:05 --> 00:46:07
have destructive interference.
 

911
00:46:07 --> 00:46:10
So instead, these would be
canceling out wave functions

912
00:46:10 --> 00:46:13
between the two, so we
would end up with a nodal

913
00:46:13 --> 00:46:15
plane down the center.
 

914
00:46:15 --> 00:46:18
Would this be a sigma or a pi?
 

915
00:46:18 --> 00:46:19
It's still sigma.
 

916
00:46:19 --> 00:46:22
So even though we see a nodal
plane down the center, I just

917
00:46:22 --> 00:46:25
want to really point out that
it's only when we have a nodal

918
00:46:25 --> 00:46:27
plane in the internuclear or
the bond axis that we're

919
00:46:27 --> 00:46:30
calling that a pi orbital.
 

920
00:46:30 --> 00:46:32
So this is still a sigma
-- it's a sigma 2

921
00:46:32 --> 00:46:35
p z star orbital.
 

922
00:46:35 --> 00:46:38
We have destructive
interference here.

923
00:46:38 --> 00:46:41
So now let's look at an example
where we talk about using

924
00:46:41 --> 00:46:45
these 2 p z orbitals, so
let's look at oxygen.

925
00:46:45 --> 00:46:48
So the first thing I want to
point out is that the 2 p z,

926
00:46:48 --> 00:46:53
the sigma 2 p z is even lower
in energy than the pi orbitals

927
00:46:53 --> 00:46:57
here, and the anti-bonding
sigma orbital is going to be

928
00:46:57 --> 00:47:03
higher in energy than the pi 2
p orbitals that we have here.

929
00:47:03 --> 00:47:05
So in oxygen again, this is
just showing the valence

930
00:47:05 --> 00:47:09
electrons, so we end up
having six valence electrons

931
00:47:09 --> 00:47:11
from each oxygen atom.
 

932
00:47:11 --> 00:47:14
We can fill right up our table
just like we did before, but

933
00:47:14 --> 00:47:18
now we have included our
2 p z orbital here.

934
00:47:18 --> 00:47:23
So we have two electrons in
sigma 2 s, two in sigma 2 s

935
00:47:23 --> 00:47:29
star, in sigma 2 p z we have
two -- those are filled first,

936
00:47:29 --> 00:47:33
and now we're going to put one
into pi 2 p x, and

937
00:47:33 --> 00:47:37
one into pi 2 p y.
 

938
00:47:37 --> 00:47:42
So we can write out what the
electron configuration is here,

939
00:47:42 --> 00:47:44
and I think that I have already
written that out for

940
00:47:44 --> 00:47:45
you in your notes.
 

941
00:47:45 --> 00:47:50
What is the bond order of o 2?
 

942
00:47:50 --> 00:47:54
It's two.
 

943
00:47:54 --> 00:47:56
Oh excuse me, I didn't fill
in all of my electrons.

944
00:47:56 --> 00:47:57
All right.
 

945
00:47:57 --> 00:48:00
So we had a total
of 2, 4, 6, 8.

946
00:48:00 --> 00:48:01
So we had a total of 12.
 

947
00:48:01 --> 00:48:07
So we filled in four here
-- we need to keep going.

948
00:48:07 --> 00:48:10
All right, so I did this not at
all purposely, but this can

949
00:48:10 --> 00:48:14
point out for you that you need
to make sure that the number of

950
00:48:14 --> 00:48:18
electrons that you have in your
molecular orbital does match up

951
00:48:18 --> 00:48:21
with the total number that you
have in your atomic orbitals.

952
00:48:21 --> 00:48:24
So I did not do any counting
up here, you should make sure

953
00:48:24 --> 00:48:25
you do counting, I apologize.
 

954
00:48:25 --> 00:48:29
So we need to fill all
the way up to the pi 2

955
00:48:29 --> 00:48:31
p x, and the pi 2 p y.
 

956
00:48:31 --> 00:48:34
All right, so the bonding
order, you're correct, should

957
00:48:34 --> 00:48:38
be 2, if we subtract the number
of bonding minus anti-bonding

958
00:48:38 --> 00:48:42
electrons and take that in 1/2.
 

959
00:48:42 --> 00:48:45
And what I want to point out
that we just figured out for

960
00:48:45 --> 00:48:49
molecular orbital theory, is
that o 2 is a biradical,

961
00:48:49 --> 00:48:51
because remember, the
definition of a radical is when

962
00:48:51 --> 00:48:53
we have an unpaired electron.
 

963
00:48:53 --> 00:48:57
You can see that we have two
unpaired electrons in this

964
00:48:57 --> 00:49:02
molecule here -- one in the pi
2 p x star, and one in the

965
00:49:02 --> 00:49:05
pi 2 p y star orbital.
 

966
00:49:05 --> 00:49:08
This was something we could not
predict using Lewis structures,

967
00:49:08 --> 00:49:11
but we can predict using MO
theory that we have a

968
00:49:11 --> 00:49:13
radical species here.
 

969
00:49:13 --> 00:49:15
So that's a really important
type of an application that we

970
00:49:15 --> 00:49:18
can use MO theory for that we
weren't able to do with

971
00:49:18 --> 00:49:20
our Lewis structures.
 

972
00:49:20 --> 00:49:23
All right, I want to do one
more at homonuclear example

973
00:49:23 --> 00:49:25
here, and this is n 2.
 

974
00:49:25 --> 00:49:28
The first thing that I need to
point out is you can actually

975
00:49:28 --> 00:49:33
see an n 2 versus o 2 that we
flip-flopped the energy of the

976
00:49:33 --> 00:49:36
sigma and the pi 2 p orbitals.
 

977
00:49:36 --> 00:49:39
So this is a glitch, just like
sometimes we had glitches in

978
00:49:39 --> 00:49:41
filling up our atomic orbitals.
 

979
00:49:41 --> 00:49:43
This is something that
you need to remember.

980
00:49:43 --> 00:49:47
And what you need to remember
is if the z is equal to eight

981
00:49:47 --> 00:49:51
or greater, such as oxygen
being the cut-off point, this

982
00:49:51 --> 00:49:55
sigma 2 p orbital is actually
lower in energy than the

983
00:49:55 --> 00:49:57
pi 2 p orbitals, the
molecular orbitals.

984
00:49:57 --> 00:50:02
But for anything 7 or less,
so what is the atomic

985
00:50:02 --> 00:50:03
number for nitrogen?
 

986
00:50:03 --> 00:50:05
STUDENT: Five.
 

987
00:50:05 --> 00:50:07
PROFESSOR: five -- there's five
valence electrons, but the

988
00:50:07 --> 00:50:09
atomic number is
actually seven.

989
00:50:09 --> 00:50:13
So z equals 7 -- this is the
cut-off where, in fact, the

990
00:50:13 --> 00:50:17
sigma orbital is going to be
higher in energy than

991
00:50:17 --> 00:50:19
the pi 2 p orbitals.
 

992
00:50:19 --> 00:50:21
This is something that you
just need to remember.

993
00:50:21 --> 00:50:24
I wrote it down your notes, if
you can put a big star next to

994
00:50:24 --> 00:50:25
it so you don't forget this.
 

995
00:50:25 --> 00:50:27
This is something you're
going to be responsible

996
00:50:27 --> 00:50:29
for in drawing out your
molecular orbitals.

997
00:50:29 --> 00:50:32
So let's fill it out in this
way, keeping in mind that we're

998
00:50:32 --> 00:50:36
going to fill out the pi
2 p's before the sigma.

999
00:50:36 --> 00:50:42
So we have a total of 2, 4, 6,
8, 10 valence electrons, so

1000
00:50:42 --> 00:50:44
I'll make sure I count to 10 as
we fill up our molecular

1001
00:50:44 --> 00:50:45
orbitals here.
 

1002
00:50:45 --> 00:50:48
We have two, then we have four.
 

1003
00:50:48 --> 00:50:51
Now we're going to start in
with that pi 2 p orbitals,

1004
00:50:51 --> 00:50:56
which gives us 1 each, and then
two each in those, and then

1005
00:50:56 --> 00:51:00
after that, we'll go up to
our sigma 2 p z orbital.

1006
00:51:00 --> 00:51:04
So this is going to be our
molecular orbital diagram.

1007
00:51:04 --> 00:51:06
And again, I've written for
you, but you can figure out

1008
00:51:06 --> 00:51:09
what the electron configuration
is just by writing up

1009
00:51:09 --> 00:51:10
in this order here.
 

1010
00:51:10 --> 00:51:13
And what is the bond
order going to be n 2?

1011
00:51:13 --> 00:51:15
STUDENT: [INAUDIBLE]
 

1012
00:51:15 --> 00:51:16
PROFESSOR: It's
going to be three.

1013
00:51:16 --> 00:51:18
So, you can go ahead and
calculate that, if you

1014
00:51:18 --> 00:51:20
can't see that right away.
 

1015
00:51:20 --> 00:51:23
So we'll end here, we'll finish
up with the heteronuclear

1016
00:51:23 --> 00:51:25
example on Friday.