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PROFESSOR: OK.
Let's get started.
Can you go ahead and take 10 more seconds on this first
clicker question here?
OK.
So it looks like most of you got that the electron
configuration that we're writing here is for copper.
So I'm actually going to give the benefit of the doubt that
the people that didn't get it right were rushing to get out
their clickers and didn't have time to think it all
the way through.
Remember that when we're talking about 4 s 1, 3 d 10,
that's one of those exceptions where a completely filled d
orbital is more stable than we would expect.
So, that's actually the electron configuration we have
when we're talking about copper and some other exceptions in
the periodic table that you're going to be looking at.
So, hopefully, if you were to go back and look you could see
that this is, in fact, copper.
We're actually going to do one more clicker question to get
started with today, and as we do, I'll explain something
we're going to be trying today, which is a little bit of a
friendly competition in terms of answering the clicker
questions correctly.
So we've tagged each of your numbers to your actual
recitation, so we can see today which recitation actually is
going to be doing the best in terms of clicker questions,
who's going to get the most correct today.
So, you may or may not know this about your TAs, but this
is a pretty competitive group of TAs we have this year, and
they like to brag about how smart their recitation is, how
good questions they're getting in the recitation section.
So, do your TA proud today and see if you can be part of the
recitation that gets the most correct in terms
of a percentage.
And at the end of class we'll announce which recitation that
is, we'll also make sure to give you a little bit of a
prize if you are, in fact, in that recitation.
So we have extra incentive to get these clicker
questions right.
So, in this one we're selecting the correct electronic
configuration for an ion.
So, why don't you go ahead and take 10 more seconds on this
second clicker question for our intro.
OK.
So, it looks like we have a little bit of a
mixed consensus here.
Let's go over this question.
And I know there's a lot to talk about about this
competition, but let's just get into listening mode here and
talk about how we can figure out what the correct electron
configuration is for this ion.
Remember, ions are a little bit different.
The first thing we need to do is write the electron
configuration for the atom itself, and then we need
to take an electron away.
So here we're talking about v plus 1, so if we were to write
it just for the neutral electron itself, we would find
that the electron configuration is argon, that's the filled
shell in front of it.
Then 4 s 2 and 3 d 3.
So this would be for the actual filled, the
completely neutral atom.
But remember what we said, which was when we talked about,
this is at the end of class on Friday, we said that it turns
out that even though 3 d is higher in energy when it's not
filled, once we fill it with an electron, these 2 orbitals
actually switch place in terms of energy.
So if we were to write this in terms of energy, we would
actually have to rewrite it has 3 d 3, and then 4 s 2.
So, which orbital would we take an electron out of if we were
ionizing this atom here?
The s.
So, we would actually take an electron out of the s, which
gives us 3 d 3 and then 4 s 1.
So, it's a little bit of a trick when you're
dealing with ions.
The best suggestion is just to write it out completely for the
neutral atom, and then you want to take an electron out
of the highest orbital.
It makes sense that it's going to come out of the highest
occupied atomic orbital, because that's going to be the
lowest amount of energy that's required to actually
eject an electron.
All right.
So let's go to today's notes.
And actually before we start into today's topics, I want to
remind everyone and hopefully you all do remember that our
first exam is coming up and it's coming up in exactly a
week, so it'll be a week from today, next Wednesday.
And on Friday in class, at the beginning of class, I'll go
over just in all the detail you could possibly imagine
everything you need to know logistically for the exam --
where it is, what you do, what kind of calculators you can
bring, which by the way are any calculator.
So you'll get all of that information on Friday.
So don't worry if you have some questions right now.
I just want to let you know that.
The other thing I want to let you know is that instead of
having a new problem-set that you'll be assigned this Friday,
what we'll do instead is we'll give you some practice
problems, and these will be just more of the same type of
problems that you saw before but that's another chance
to try them out more.
These won't be graded, you don't have to turn them in,
it's just to give you some extra practice if you want
while you're studying for the exam.
We'll also post an exam from a previous year so you can
actually see exactly what the format's going to look like.
So when you go into the exam a week from today, it'll all look
really familiar, you'll be comfortable with the format and
you can just dive right in and start answering the questions.
So you'll have all that information and we'll get
it to you on Friday.
The other quick thing I want to say is that I do have office
hours today from 3 to 5, so feel free to stop by if you
have questions about problem-set 3 that
you're finishing up.
And also, for those of you that did sign up for the pizza forum
tonight, that's going to be at 5 o'clock, it's in room 56-502,
so we'll see some of you tonight for that as well.
All right.
So, let's move on to what we're talking about today.
What we're going to start with is discussing photoelectron
spectroscopy, which is a spectroscopy technique that
will give us some information about energy levels in
multielectron atoms.
We'll then take a turn to talking about the periodic
table, we'll look at a bunch of periodic trends, including
ionization energy, electron affinity, electronegativity
and atomic radius.
And then, if we have time at the end, we'll introduce
one last topic, which is isoelectronic atoms and ions.
I also want to note that the end of the material today, so
this last topic here, that's the end of the material that's
going to be on this first exam.
So whether we finish it today, or more likely when we finish
it up on Friday, once we get passed isoelectronic atoms,
that's it, that's all you need to study for this first exam.
So from that point on it'll be exam 2 material, so depending
on how you like to come compartmentalize your
information, you can separate that in your brain in terms of
what you're trying to learn right now versus what you
can put off until a little bit later.
So, let's start with talking about photoelectron
spectroscopy.
This actually relates very closely to what we discussed in
class on Friday before the long weekend, and what we were
talking about is the energy levels of multielectron atoms.
So what we'll start with today is talking about the technique
that's primarily used to actually experimentally figure
out what these different energy levels are.
And this is called photoelectron spectroscopy, and
essentially what it is is very similar conceptually to what we
were talking about way back in the first couple lectures when
we were talking about the photoelectric effect.
Because here what we have is some atom that we're studying,
in the case, it's going to be a gas, and we hit it with a
photon that has some incident energy.
So e sub i, some energy that the photon comes in with, and
if it has sufficient energy to eject an electron, it will do
that, and our electron will be ejected with a certain kinetic
energy, which is going to be whatever energy is left over
from the initial energy we put in minus what was taken up in
order to actually ionize or eject the electron.
So, you can see how this can directly give us different
ionization energies for any atom that we're
interested in studying.
For example, with neon we can think about all of the
different orbital energies we could be looking at.
In the first case, so here is the electron
configuration of neon.
So we can think about what is our most loosely-bound
electron, what's that highest energy orbital, and it's going
to be the 2 p orbital, that's going to be what's
highest in energy.
So if we're going to eject an electron using a minimum amount
of energy, that's where it's going to come from.
So, you can imagine, that we'll actually probably have a lot of
kinetic energy left over if we put a lot of energy in
in the first place.
We're only using up a little bit to eject the electron, then
we'll have a lot left over.
So, one difference between photoelectron spectroscopy and,
for example, the photoelectric effect is that in this case,
we're not just looking at one energy level, which is what we
were looking at from the surface of a metal, now we're
talking about this gaseous atom.
So we can actually pop an electron or eject an electron
from any single orbital that is occupied within the atom.
So, for example, it's not just the 2 p that we could actually
take an electron from, we could also think about ejecting an
electron from the 2 s orbital.
Now this, of course, is going to take more energy because a 2
s is lower, it has has a more negative binding energy than
the 2 p, but that's OK as long as we put in enough energy, but
what we're going to find is the kinetic energy coming out with
the electron is actually going to be a little bit less, right,
because we had to use up more energy to eject the electron,
so we don't have as much left over.
There's actually one more orbital that we could talk
about if we're talking about this sample case of neon, which
is a 1 s orbital, and if we're talking about a 1 s orbital,
now we're going to be even lower in energy still, so that
means the minimum energy required to eject an electron
is going to be at its highest, so that means the energy that
we have left over that turns into kinetic energy for the
electron, is now going to be really quite small.
And what happens when you irradiate one of these atoms
that you're studying with this light is in photoelectron
spectroscopy, you want to make sure that you put in enough
energy to actually ionize any single electron that
you have in the atom.
So the way that we really make sure this is done
is that we use x-rays.
So you know that x-rays are higher frequency than UV light,
for example, that means it's also higher energy than UV
light, and if you think back to our photoelectric effect
experiments, do you remember what type of light we were
usually using for those?
Does anyone remember?
Yeah.
It was UV light that we used.
Well, we can't guarantee with UV light we'll have enough
energy to eject every single electron, so that's why when we
use x-rays, they're higher energy, you can pretty much be
guaranteed we're going to eject all of those electrons there.
So I said that this technique was used to experimentally
determine what the different binding energies or the
different ionization energies are for the different states
in a multielectron atom.
Another way to say states is to talk about different orbitals.
So we can do this directly as long as we have certain
types of information.
The first that we need to know the energy of the photon that's
incident on our gaseous atom.
The second piece of information we need to know is what
actually the kinetic energy is of the ejected electron, and
that's something we can just measure by measuring
its velocity.
So we can use an equation to relate the incident energy and
the kinetic energy to the ionization energy, or the
energy that's required to eject an electron.
This should all sound incredibly familiar, like I'm
just repeating myself in terms of photoelectric effect,
because essentially that's what I'm doing, and that's one
reason we spent so much time and did so many problem-set
problems on the photoelectric effect.
So what we're saying here is the incident energy, so the
energy coming in, is just equal to the minimum energy that's
required to a eject an electron.
When we talked about the photoelectric effect, that was
called the work function.
In this case, it's called the ionization energy, plus
whatever kinetic energy we have left over in the electron.
So if we want to solve for ionization energy, we can just
rearrange this equation.
Our ionization energy is going to be equal to the incident
energy coming in, minus the kinetic energy of the electron.
So, let's take a look at the different kinetic energies that
would be observed in a spectrum for neon where we had this
incident energy here.
And it turns out that the first kinetic energy that we would
see or the highest kinetic energy, would be 12
32 electron volts.
So if that's the case doing a quick little calculation, what
would the ionization energy be for a 2 p electron in neon?
Yup, 22.
So, basically all we did was take 12 54, subtract 12 32,
and we got 22 electron volts.
We can do the same thing for the other observed
kinetic energy.
So, for example, in the second case, we say that we see 12 06
in terms of the kinetic energy.
Same sort of subtraction problem, what do we have
for the ionization energy of the 2 s electron?
Good, quick math.
All right, so 48 electron volts.
And let's look at the final kinetic energy that we'd
observe in this spectrum, which is 384 electron volts, so what
is that third corresponding ionization energy?
I couldn't quite hear, but I have a feeling everyone
said 870 electron volts.
So, we can actually kind of visualize what we would see
if we were looking at a photoelectron spectrum.
And what we would see if we were graphing, for example,
increasing kinetic energy, is we would see 1 line
corresponding to each of these energies of electrons
that we see coming out.
And, of course, each of those electrons correspond to an
electron coming out of a particular orbital.
So in the case of 12 32, that is our highest kinetic energy,
that means it's our lowest ionization energy -- it's the
smallest amount of energy it takes to pop an electron
out of that orbital.
So that's why we see the 2 p here, the 2 s is 12 06, and it
makes sense that what we see as the greatest ionization energy,
which is also the smallest kinetic energy is
that 1 s orbital.
Remember, because that 1 s orbital is all the way down in
terms of if we're thinking about an energy diagram, we're
all the way down here, so we have a huge amount of energy we
have to put into the system in order to eject an electron.
So what I want to point out when you're kind of looking at
these numbers here, what the significance is, look at that
huge difference between what the ionization energies are for
what we call those valence electrons, those outer shell
electrons, versus the ionization energy for the 1
s orbital -- those are core electrons there.
So we can think about something I mentioned last time, which is
when we're thinking about chemistry and what's really
interesting in terms of chemical reactions, it's mostly
valence electrons we're talking about, those are the ones that
tend to be involved in chemical reactions.
It makes a lot of sense when we look at it energetically,
because if we think about a 1 s core electron, that's going
to be held really, really tightly to the nucleus.
We see that we have to put this huge energy in to actually get
a 1 s electron ejected, so it makes a lot of sense that we
wouldn't want to pay that energy cost in a normal
chemical reaction.
And we don't -- we very rarely would see these core electrons
actually being involved in any type of a reaction.
All right, so one thing that I want to point out, which I said
many, many times on Friday, and this is perhaps the last time
I'll say it, but one last time is we can think about why we
only see a line for the 2 p orbital, versus we don't see
separate lines for a 2 p x, a 2 p y, and a 2 p z.
Remember, we need those three quantum numbers to completely
describe the orbital.
Why do we just see one for all the p's?
And the reason is that the energy of the orbitals, depend
on two quantum numbers, and that's quantum number n,
and quantum number l.
M does not actually have an effect, in this case, on
the energy of the orbital.
So that's why we're not seeing separate lines
in this spectrum.
All right.
So let's go ahead and try an example here in thinking about
photoelectron spectroscopy.
So, let's say we're looking at an element and we have an
emission spectra, and we know that it has five distinct
different kinetic energies in that spectrum.
We might be asked, for example, to determine what all of the
different elements could be that would produce a spectrum
that gave us 5 different lines.
So the first thing that we want to do, if we're thinking about
something like this, is just to determine exactly what orbitals
are causing the five different lines that we're seeing
in the spectrum.
So, if we're talking about five different orbitals and we're
talking about a ground state atom, we know that we just
need to start at the bottom and work our way out up.
So, our first orbital that an electron must be
coming from is the 1 s.
What comes after that?
2 s.
All right, then what?
2 p.
After that?
3 s.
Next?
3 p, and that's 1, 2, 3, 4 -- that gives us five different
options, five different orbitals, five different
energies right there.
So, then all we need to do to determine which elements that
corresponds to is take a look at our periodic table.
So we want to look at any element that has a 3 p orbital
filled, but that does not then go on and have a 4 s, because
if it had the 4 s filled then we would actually see six
lines in the spectrum.
So that is relevant for all of these atoms here, so
we actually have several different possibilities.
It could be aluminum, silicone, phosphorous, sulfur,
chlorine or argon.
Any one of these different elements could actually produce
a photoelectron spectroscopy spectrum that has
five distinct lines.
If I went on and told you what the different incident light
was, and what the electrons were ejected with, and then you
could look up the ionization energy for the particular
different elements, you should be able to actually determine
exactly which element it is, but just with the information
given, we can only narrow it down to these choices here.
So let's actually let you try another example of solving a
problem that has to do with one of the spectrums.
So, let's turn to another clicker question here.
Remember, your answer holds great weight in terms
of the state of the TA bragging for next week.
So, how many distinct, so again, we're talking about
distinct kinetic energies, would be displayed if you're
talking about a spectrum for the element hafnium, and I'll
tell you here that it has a z of 72, so you don't have to
spend two minutes searching your periodic table.
The period of table's on the back page of your notes if
you don't see that there.
All right.
It looks like a lot of you are done, so let's take
10 more seconds here.
Part of the challenge is speed, too, how quickly you can get
these answers in terms of getting them in on time.
So let's see what we say.
All right.
So I think I can safely say that most people had the right
idea and were counting quickly, though I have a feeling that
some people who wrote 13 might have forgotten about those
4 f, the 4 f electrons.
So, remember when you're looking at your periodic table,
don't forget about the lanthinides, sometimes
they come into play.
So it's actually 14, and the way that we got that answer was
we just wrote out or just looked at your period table,
figured out all of the different orbitals that you
could have in terms of the principle quantum number, and
then the l quantum number, and then write them all down -- it
turns out to be 14, so that's what the answer is.
So, it looks like this is good, because we'll have some
separation in terms of not everyone's going to get 100% in
terms of recitations here, which is what we're going for.
All right.
So let's turn our attention to a new topic, which is thinking
a little bit about the periodic table, and also talking
about periodic trends.
And there's a lot we can explain by talking about what
we see in the periodic table in terms of what different trends
are in grouping different elements in different spots
within the periodic table.
So, here we have a picture of Dmitri Mendeleev, who is one of
the scientists responsible for first compiling the
periodic table.
You'll notice I have what's a very flattering picture of him
up here, and if you haven't done the reading yet you might
not think this is particularly flattering, but if you look at
the picture of him in the book, you'll notice I chose a very
flattering picture of Dmitri up here, and here he's pondering
putting these elements together in a periodic table.
And he actually did this in the late 1800's, back before even
all of the elements that we know today were discovered,
really only about 60% or so, 70% were discovered then
that we now know today.
But still, he was able to put together a periodic table.
And what he did what he actually grouped things
in terms of their chemical properties.
So the way that we like to think of things now is in terms
of electron configurations, right, but at the time that
wasn't really understood.
So, instead, it was amazing he was able to group things in
terms of the properties that he saw.
So, for example, if he was talking about the group one
metals, lithium, sodium, potassium -- he noticed these
were all very soft reactive metals, those were
grouped together.
Versus looking at, for example, helium or neon or argon, these
are all inert gases, inert meaning essentially do not
react, those were grouped together in the periodic table.
So basically, at the time he was just going on size and then
traits, but what we actually know today is that we can also
order things in the periodic table by electron
configuration.
In fact, that is the most logical way for
us to look at it now.
So, for example, if we're actually thinking about
electron configuration and we look at lithium, sodium and
potassium, these all have one valence electron.
So basically, that means one electron in an s orbital in
their outer-most most shell.
So that explains why they're so reactive, they're all very
willing to give up that 1 s orbital and then drop to
a lower energy level.
In contrast, helium, neon, and argon all have filled shells.
That also explains why they're very stable.
They're not going to want to add on another electron,
because then it'll have to jump a very large energy level and
start filling in another shell -- go from n equals 2,
to n equals 3, and n equals 4, and so on.
So it turns out that we can really know a lot just by
looking at the periodic table.
You will never in this class have to memorize anything
about the periodic table.
Depending on what kind of chemistry you go in to, you
might accidentally memorize parts of the table, which is
fine, but what you really want to know how to do is know how
to use the periodic table.
But you actually need to keep a few caveats in mind as you do
this, which is the fact that trends predict a lot of
chemical properties, but they can't predict everything in
terms of biological properties.
And after the periodic table was developed in the late
1800's, people didn't understand this quite as
well, they took things a little more literally.
They thought, for example, if you could do something with one
element, if you looked at an element very close to it, it
would be similar enough that you could maybe
replace it with that.
Today we know, for example, if you can put one certain kind of
element in your mouth or eat that, it doesn't necessarily
mean you want to put the element next to it and your
mouth as well, that might not be safe.
But this is things we've learned as the years
have gone past.
So, let's just take a quick example to show how not
completely you can use these periodic trends, that
there are limits.
So if we consider lithium, potassium, and sodium, they're
all together in the same group on the periodic table, knowing
what we do about biology we can immediately think of sodium and
potassium, or even just knowing what you know about table salt,
for example, that these are two elements that we find, and
particularly in the ion form in very high concentrations
in our body.
For example, sodium in our blood plasma is almost to
the point sometimes of 100 millimol or that's very,
very concentrated.
Similarly, we find it in table salt, we're taking it in all
the time, the same with potassium, think of bananas,
were always eating potassium.
Not so with lithium.
I don't think too many people and here are probably
taking lithium.
It turns out there's actually no natural function known
in the body for lithium.
So there's nothing naturally going on unless we were to
introduce it ourselves in our body that we know of, at
least, that involves lithium.
But this did not stop people, for example, in the late
1800's, early 1900's, and, in fact, in 1927 a new soft drink
was put on to the market and they wanted to make a
lemon-lime soft drink, these were very popular in the early
1900's, and to get sort of that lemony flavor, they decided to
use citric acid, so that's a good idea, that gives
that soury taste.
And they wanted to use a soluble salt of citric acid, so
they could have used sodium, they could have used potassium.
But, you know why not do something a little special,
little different, and they decided instead to use lithium.
So, here we have this soda with lithium citrate, some of
you might be familiar with this, soda is called 7-Up.
So, 7-Up no longer has lithium in it, but from 1927 to 1950 it
did, and, in fact, not only did they not try to hide the fact
that there's lithium in the soda, this they used as a
really special marketing technique, they really pointed
out this is something that stands out about our soda,
this is something special.
There's a lot of good things about lithium.
I don't know if you can see, probably not, what's written
on here, so let me point out to you a few things.
Lithium, slenderizing, that's great to see in a soda.
Other nice things about lithium in your soda, it dispells
hangovers, takes the ouch out of grouch.
That's very nice.
So basically, you get a lot of benefit supposedly from this
7-Up soda from the 1920's or so.
And this went on and was unregulated for some time, but
at some point the Food and Drug Administration did take a step
in, so here's a case where they did do something important --
that's not what I mean at all -- where they did take the
step, they do many things that are important, often
not quickly enough.
Here's a -- actually here it did take 25 years, but they
did, they did eventually step on before we started
drinking 7-Up.
And what they said was, look, you can't put this in, we're
starting to notice it does some strange things.
Because it was in the 1950's or so, maybe the late 1940's, that
people started to discover lithium, even though it had no
natural function, it did do something in our bodies.
Does anyone know what was lithium's used for?
Yeah, it's an anti-psychotic drug, so, for example, some
people with bipolar disorder even today still take it, it
works really well for some people, for other people
it doesn't work so well.
But anyway, this isn't really something you want to have
in your soda, so they did take it out eventually.
Another side effect if you take too much lithium is death, so
that's no good to have in sodas either, and it might not have
been as big a deal back in the 1920's, but you can imagine
with supersizing today, this might be a bigger problem.
So anyway, when we talk about periodic trends, it
doesn't always match up.
This was eventually taken out, and actually just for your
interest, there was no overlap between the time when cocaine
was in Coca Cola and lithium was in 7-Up, so there was a few
years difference between those two times, but it's amazing to
think about what does go into processed foods.
And the other thing to point out, which I don't know if this
is true or not, but does anyone know -- well that's part's
true, does anyone know what the atomic mass of lithium is?
Yes, it's 7.
So, I don't know if this is true or not, but I wonder if
that's where the actual name 7-Up came from.
So, even though we don't have the lithium anymore, we still
keep that atomic number 7 around.
All right.
So that is an anti-example of using periodic trends.
So let's go to some actual real examples, which might come
more in handy for this class.
So it's going to keep in mind the limitations, so let's
start off with talking about ionization energy.
Now this is a good place to start, because we are very
familiar with ionization energy, we've been talking
about it in a lot of different forms for quite a while -- it's
that minimum energy required to remove an electron
from an atom.
And specifically, when we talk about ionization energy, it's
assumed that what we mean is actually the first
ionization energy.
So, you can imagine, we could talk about any of the different
electrons, or we could talk about taking out an electron
and taking out second electron.
Whenever you hear the term ionization energy, make sure
you keep in mind that unless we say otherwise, we're
talking about that first ionization energy.
And we know what that's equal to, this is something we've
been over and over, ionization energy is simply equal to the
negative of the binding energy.
So negative e, which is sub n l, because it's a function of
n and l in terms of quantum numbers.
So, let's think about kind of differentiating, however,
between first ionization energy or just ionization energy, and
other types such as second or third ionization energy, and
let's take boron as an example here.
So, if we want to think about what the first ionization
energy is of boron, what you want to do is write out the
electron configuration, because then you can think about where
it is that the electron's coming out of.
The electron's going to come out of that highest occupied
atomic orbital, that one that's the highest in energy, because
that's going to be the at least amount of energy it needs
to eject something.
So what we'll end up with is boron plus, 1 s 2, 2 s 2, and
what we say is the delta energy or the change in energy as the
same thing as saying the energy of the products minus the
energy of our reactant here, and we just call that the
ionization energy -- that's how much energy we have to put into
the system to eject an electron.
And again, this is just the negative, the binding
energy, when we're talking about the 2 p orbital.
So, this is first ionization energy, let's think about
second ionization energy.
So, second ionization energy simply means you've already
taken one electron out, now how much energy does it take for
you to take a second electron out.
So in the case of boron here, what we're starting with is the
ion, boron 1 s 2, 2 s 2, and now we're going to pull
one more electron out.
The highest occupied orbital is now the 2 s orbital, so we're
going to end up with boron 2 plus 1 s 2, 2 s 1, plus the
electron coming out of there.
And what we say when we talk about the delta energy is that
this is going to be equal to i e 2, or the second ionization
energy, or we could say the negative of the binding energy
of a 2 s electron in b plus. so it's important to note that
it's not in b, now we're talking about b plus, because
we've already taken an electron out here.
So, similarly if we start talking about our third
ionization energy, this is going to be going from b
plus 2, to 1 s 2, 2 s 1.
Now we're going to pull that second electron out of the 2 s,
so we end up with boron plus 3, and then the configuration is
just 1 s 2, plus our extra electron here.
So, what we call this is the third ionization energy, or the
negative of the binding energy, again of the 2 s orbital, but
now it's in boron plus 2 to we're starting with.
So, this raises kind of an interesting question in terms
of what the difference is between these two cases, and
we're talking about numbers of energy.
So let's address this by considering another example,
which should clarify what the difference is between
these ionization energies.
So let's think about the energy required now to remove a 2 s
electron, let's say we're removing it from boron plus
1 versus neutral boron.
So, in the case of boron plus 1, what we are starting with
is the ion, so we're starting with a 2 s electron, and then
we're going to 2 s 1 here.
And what we call the binding energy is negative 2 s in b
plus -- this is what we saw on the last slide.
And the second case here looks a lot more like what we saw
when we were talking about photoelectron spectroscopy,
because here we want to remove a 2 s electron, but it's
actually not the highest occupied orbital, so that's not
the one that would naturally come out first, but let's say
we're hitting it with high energy light sufficient to
knock out all the different electrons, and one that we end
up knocking out is this 2 s here.
So if we think about what that delta energy is, we call that
the ionization of the 2 s, that's different from saying
second ionization energy.
And that's going to be equal to the negative the binding energy
of 2 s in b, in neutral boron.
So, my question to you is are these two energies equal?
No.
All right, good answer.
So, we can think about why is it that these are not equal.
In both cases we're taking an electron out of
the 2 s orbital.
And it turns out that if we're talking about a 2 s orbital in
an ion, that means it doesn't have as many electrons in it,
so what we're going to see is less sheilding.
There are fewer electrons around to shield some
of that nuclear charge.
So what we're going to see is less sheilding, which means
that it will actually feel a higher z effective.
So even though they're both 2 s electrons, in one case it's
going to think its feeling more pull from the nucleus, and it,
in fact, will be, than in the other case, and if its feeling
a higher z effective, then it's actually going to require more
energy to remove that electron, right, it's being pulled in
closer and more tightly to the nucleus, you have to put in
more energy to rip it away from that very close interaction.
So, that's the difference in thinking about different types
of ionization energy, so it can get a little bit confusing with
terminology if you're just looking at something quickly,
so make sure you look really carefully about what
we're discussing here.
If you see a problem that asks you, for example, the third
ionization energy versus taking a second electron out of
the 2 s in a photoelectron spectroscopy experiment, those
are two very different things.
So, let's make sure everyone kind of has this down, let's do
another clicker question here.
And in this case we're going to look at silicone, and we'll say
if you can point out to me which requires the least
amount of energy.
So, which has the smallest energy that you have to put
in in order to eject this electron?
Will it be if you take a 3 s electron from neutral silicone,
if you take a 3 p electron from the neutral atom, or if you
take a 3 p from the ion?
So this you should be able to know pretty quickly, so let's
just take 10 seconds here.
All right, great.
So most of you see that, in fact, the energy that's going
to be the least that we need to put in is in case 2 here.
Let's compare case 2 and 3, since this where some people
seem to have gotten confused.
In case 2, we're taking it out of -- oh, it's kind of hard to
compare case 2 and 3 when we can't see it anymore.
In case 2, we're taking the 3 p out of the neutral atom,
whereas in case 3, we're taking it out of the ion.
Remember in the ion, we're going to have less electrons
around to counteract the pull from the nucleus.
So we're going to feel a higher z effective in the case of
the ion compared to the neutral atom.
If we have a higher z effective, it's pulled in
tighter, we have to put in more energy in order to eject an
electron, so it turns out that that's why case 2 is actually
the lowest energy that we need to put in.
The z effective is lower, so we have to put less energy
in to get an ion out.
So, let's take a look at this in terms of periodic trends --
that's our topic here, we're talking about periodic trends.
So as we go across the row, and this is my beautiful picture
of a periodic table here.
As we go across the row what happens is that the ionization
energy actually increases, and we can think about
logically why it is that that's happening.
As we go across the row, what we have happening is that the z
or the atomic charge -- I'm not talking about z effective here,
I'm just talking about z -- the z is increasing as we go across
a row, that's easy to see.
But we're still in the same shell, so we still have the
same n value as we go all the way across a row
in the periodic table.
So, in general what we're going to see is that what happens to
z effective if we have z increasing but we're in
the same shell here.
Would it increase or decrease z effective?
All right.
Kind of mixed thoughts here.
So it turns out that it increases, and the reason is
because the predominant thing that's going on here is
that z is increasing.
So the z is increasing, and what goes along with it is that
the z effective is increasing, because it turns out that while
we're in the same n, even though we know that energy
depends on both the n and the l in terms of quantum numbers,
while we're in the same n, the distance from the nucleus, it's
pretty close, it's not hugely different.
So the factor that predominates is that we're actually
increasing z.
That's why we see z effective increase, and that's why we see
ionization energy increase.
As we go down a column, what happens is that the
ionization energy decreases.
We can also think about this in terms of z effective.
This is because even though z, the atomic number is still
increasing, we are also getting further away from the nucleus.
So, remember when we talk about Coulomb force, what's holding
the nucleus and the electron together, there's 2 things
we need to think about.
The first is this the z effective, or how much charge
is actually in the nucleus that's felt, or the I guess we
would say the z, how much the charge is on the nucleus that
holds it close together.
But the second factor is how far away we
are from the nucleus.
So, if we're really close to the nucleus, that's when z
effective is high, but if we get really far away, then z
effective is going to get low, because even though we have the
same charge in the nucleus, the z effective gets lower.
So this is not even thinking about the other electron
shielding, just if we think of the fact, all we need to think
about is that the effect of going to a further away n
actually dominates as we go down the table.
We're getting further away from the nucleus because we're
jumping, for example, from the n equals 2 to the n equals 3
shell, or from the n equals 3 to the n equals 4 shell.
And when you're switching n's, you're actually getting
quite a bit farther away.
That's why in the earlier models of the atom, they're not
horrible to sometimes think about just each n value
as a little ring around.
It's not complete and it's not accurate, but it's OK to kind
of think of it in terms of how far we're getting away
from the nucleus.
So, as we go down a column, we see ionization energy's
going to decrease.
So, this means we have the general trends down, so we
should be able to look at actual atoms in our periodic
table and graph them and see that they match up
with our trends.
So here we have that graphed here, we have atomic number z
graphed against ionization energy, so, let's fill in what
the actual atoms are here, and we can see in general, yes,
we're following the trend.
For row one, we're increasing, as we should, across the row.
Let's look at row two also.
Well, we're generally increasing here, we'll talk
about that more in a minute.
And also in a row three, yeah, we're generally increasing,
there's some glitches here, but the general trend holds true.
Similarly we see as we go down the table, so as we're going
from one row to the next row, so, for example, between helium
and lithium, we see a drop; the same with neon to sodium,
we see a drop here.
So it looks like we're generally following our
trend, that's a good thing.
But hopefully, you will not be satisfied to just make a
general statement here when we do have these glitches.
So you can see, for example, we have several places where
instead of going up as we go across the row, we actually
go down in ionization energy a little bit.
So between b e, and b, between n and o, magesium and aluminum,
and then phosphorous and sulfur, what we see here is
that we're kind of going down, or quite specifically,
we are going down.
So, let's take a look at one of these rows in more detail to
think about why this might be happening, and it turns out the
reason that these glitches occur are because the sub shell
structure predominates in certain instances, and that's
where these glitches take place.
So I've sort of just spread out what we have as the second
row here, graphed against the ionization energy.
So, let's consider specifically where these glitches
are taking place.
So, let's look at the first one between beryllium
and boron here.
And the glitch that doesn't make sense just through
periodic trends, is that it turns out that the ionization
energy of boron is actually less than the ionization
energy up beryllium.
So I put the electron configurations and actually
drew it on an energy diagram here, so we can actually
think about why this might be happening.
So what is this, which element did I chart here?
Which one is that, the boron or the beryllium?
I couldn't tell what you said, sorry.
So, I'm going to assume that was beryllium, and then it
turns out that if that's beryllium, the other
one must be boron.
So, we have beryllium in the first case here, it has four
electrons, that's how we know it's beryllium, boron
has five electrons.
So, just looking at putting in the electrons, filling up the
energy diagram here, we should be able to see a little bit
why this is happening.
And the reason is simply because the energy that we gain
in terms of moving up in z, so from going to z equals 4 to z
equals 5, is actually outweighed by the energy it
takes to go into this new shell, to go into
this new sub shell.
So to jump from the 2 s to the 2 p, takes more energy than we
can actually compensate with by increasing the pull
from the nucleus.
So, it turns out that in this case, and any time that we see
we're going from a 2 s to 2 p, filling in of electrons, we
actually see that little bit of glitch in ionization energy.
So it's shown here for the second row, but it's actually
also going to be true for the third row.
The same thing when you're going from filling in the 2
s to putting that first electron into the 2 p.
So that explains one of our glitches here, but we have
another glitch, and that second glitch comes between
nitrogen and oxygen.
So, these sound more different, so I think I'll
be able to distinguish.
Which element is shown here?
Yeah, nitrogen.
So, nitrogen is shown here, we know that because
it has 7 electrons.
In this case, we're talking about 8 electrons,
which is oxygen.
So if we're comparing the difference between these 2 now,
what you'll notice is that in nitrogen we have all
half-filled 2 p orbitals, and now, once we move into oxygen,
we actually have to add 1 more electron into 1 of
the 2 p orbitals.
There's no more 2 p orbitals to put it into, so we're going to
actually have to double up.
So now we're putting 2 electrons into the same p
orbital, that's not a problem, we can do it, it's not a huge
energy cost to do that.
But actually there is a little bit of an energy cost into
doubling up into a single orbital, because, of course, it
takes energy when you create more electron repulsion, that's
not something we want to do, but we have to do it here, and
it turns out that that effect predominates over, again, the
energy that we gain by increasing the atomic
number by one.
So, our two glitches we see when we go from the 2 p, or
from 2 s to start filling the 2 p, and then we also get another
glitch when we've half-filled the 2 p, and now we're adding
and having to double up in one of those p orbitals.
Again, we see the same effect as we go into different
rows as well.
So let's talk about another periodic trend, this one is
called electron affinity.
Electron affinity is actually the ability of an atom, or we
could also talk about an ion to gain electrons.
So it's the affinity it has for electrons, it's how much it
likes to get an electron.
We can write out what it is for any certain atom or ion
x, so it's just x plus an electron gives us x minus.
We have the minus because we're adding a negative charge
from the electron.
So, basically any time we have a really high positive number
of electron affinity, it means that that atom or ion really
wants to gain another electron, and it will be very stable
and happy if it does so.
So let's look at an example of chlorine here.
So chlorine, if we talk about it in terms of electron
affinity, we would be writing that we're actually gaining an
electron here, and getting the ion, c l minus.
And the change in energy for this reaction is negative
349 kilojoules per mole.
So if we have a negative change in energy for any reaction as
it's written, what that actually means is we're
giving off energy as the reaction proceeds.
So, in other words, this c l minus is actually lower in
energy than the reactants were.
So that's why we're giving off extra energy.
We saw a similar thing as we saw electrons move
from different levels.
We can think of it in the same type of way when
we're talking about actual reactions happening.
So, if we have energy that's released, would you say that
the chlorine ion is more or less stable than
the chlorine atom?
Who thinks it's more stable, show of hands?
All right, who thinks it's less stable?
Very good.
So it turns out it is, in fact, more stable.
It's more stable because you actually -- this happens
spontaneously, you actually get energy out of the
reaction as this happens.
And we haven't talked about reactions at all yet, so you
don't need to worry about the specifics of that exactly, but
just that if you have this negative change in energy, you
have a more stable product than you do reactant.
So, we were talking, however, about energy in terms of
electron affinity, so we can actually relate electron
affinity to any reaction by saying if we have this reaction
written as here where we're gaining an electron, we say
that electron affinity is just equal to the negative of
that change in energy.
So, for example, for the chlorine case, we would say
that the electron affinity for chlorine is actually positive
349 kilojoules per mole.
That's a very large number, it's all relative, so you don't
necessarily know it's large without me telling you or
giving you other ions to compare to, but chlorine does
have a very large affinity, meaning it's really likes
getting an electron and becoming a chlorine ion.
One major difference between electron affinity and
ionization energy is that when we talked about ionization
energy, remember ionization energy always has
to be positive.
We will never have a case where ionization energy is negative.
Electron affinity, however, can be either negative
or it can be positive.
So let's look at a case where it's actually
going to be negative.
So, if we took the case of nitrogen, if we add an electron
to nitrogen and go to n minus, we find that the
change in energy is 7 kilojoules per mole.
This means in order to do that we actually have to put 7
kilojoules per mole of energy into the reaction
to make it happen.
So this is not going to be a favorable process, we're going
to find that the electron affinity is actually a
negative 7 kilojoules per mole for nitrogen.
So this means nitrogen has low electron affinity, it
doesn't actually want to gain an electron.
So, that also tells us that the n minus ion is less stable
than the neutral atom itself.
So, we can think about trends in electron affinity just like
we did for ionization energy, and what we see is
a similar trend.
Electron affinity increases as we go across a row in the
periodic table, and it decreases as we go
down a column.
I left out the noble gases here because they do something a
little bit special, and actually, I'm going to give you
one last clicker question today to see if you can tell me what
you think noble gases do.
To answer this question you just really want to think
about what does electron affinity means.
It means how much a certain atom actually wants
to get an electron.
So do you think noble gases would have a high positive
electron affinity, a low positive, or negative
electron affinity?
So, let's take 10 seconds on that.
All right.
Great.
So most of you recognize, if we switch back to the notes, that
they do have a negative electron affinity.
This should make sense to you, because they don't, in fact,
want to gain another electron, because that would mean that
electron would have to go into a new value of n, a new shell,
and that's really going to increase the energy
of the system.
So they'd much rather just stay the way they are and not have
another electron come on, and it turns out that halogens have
the highest electron affinities.