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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.