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PROFESSOR: OK.
As you're settling into your seats, why don't we take 10
more seconds on the clicker question here.
All right, so this is a question that you saw on your
problem-set, so this is how many electrons would we expect
to see in a single atom in the 2 p state.
So, let's see what you said here.
Six.
And the correct answer is, in fact, six.
And most of you got that, about 75% of
you got that right.
So, let's consider some people got it wrong, however, and
let's see where that wrong answer might have come from,
or actually, more importantly, let's see how we can all get
to the correct answer.
So if we say that we have a 2 p orbital here, that means
that we can have how many different complete orbitals
have a 2 for an n, and a p as its l value? three.
So, we can have the 2 p x, 2 p y, and 2 p z orbitals.
Each of these orbitals can have two electrons in them, so
we get two electrons here, here, and here.
So, we end up with a total of six electrons that are
possible that have that 2 p orbital value.
So this is a question that, hopefully, if we see another
one like this we'll get a 100% on, because you've already
seen this in your problem-set as much as you're going to see
it, and you're seen it in class as much as you're
going to see it.
So if you're still having trouble with this, this is
something you want to bring up in your recitation.
And the idea behind this, of course, is that we know that
every electron has to have its own distinct set of four
quantum numbers.
So that means that if we have three orbitals, we can only
have six electrons in those complete three orbitals.
All right, so today we're going to fully have our
discussion focused on multi-electron atoms. We
started talking about these on Wednesday, and what we're
going to start with is considering specifically the
wave functions for multi-electron atoms. So, the
wave functions for multi-electron atoms. Then
we'll move on to talking about the binding energies, and
we'll specifically talk about how that differs from the
binding energies we saw of hydrogen atoms. We talked
about that quite in depth, but there are some differences now
that we have more than one electron in the atom.
Then something that you probably have a lot of
experience with is talking about electron configuration
and writing out the electron configuration.
But we'll go over that, particularly some exceptions,
when we're filling in electron configurations, and how we
would go about doing that for positive ions, which follow a
little bit of a different procedure.
And if we have time today, we'll start in on the
photo-electron spectroscopy, if not, that's where we'll
start when we come back on Wednesday.
So, what we saw just on Wednesday, in particular, but
also as we have been discussing the Schrodinger
equation for the hydrogen atom, is that this equation
can be used to correctly predict the atomic structure
of hydrogen, and also all of the energy levels of the
different orbitals in hydrogen, which matched up
with what we observed, for example, when we looked at the
hydrogen atom emission spectra.
And what we can do is we can also use the Schrodinger
equation to make these accurate predictions for any
other atom that we want to talk about in
the periodic table.
The one problem that we run into is as we go to more and
more atoms on the table, as we add on electrons, the
Schrodinger equation is going to get more complicated.
So here I've written for the hydrogen atom that deceptively
simple form of the Schrodinger equation, where we don't
actually write out the Hamiltonian operator, but you
remember that's a series of second derivatives, so we have
a differential equation that were actually dealing with.
If you think about what happens when we go from
hydrogen to helium, now instead of one electron, so
three position variables, we have to describe two
electrons, so now we have six position variables that we
need to plug into our Schrodinger equation.
So similarly, as we now move up only one more atom in the
table, so to an atomic number of three or lithium, now we're
going from six variables all the way to nine variables.
So you can see that we're starting to have a very
complicated equation, and it turns out that it's
mathematically impossible to even solve the exact
Schrodinger equation as we move up to
higher numbers of electrons.
So, what we say here is we need to take a step back here
and come up with an approximation that's going to
allow us to think about using the Schrodinger equation when
we're not just talking about hydrogen or one electron, but
when we have these multi-electron atoms.
The most straightforward way to do this is to make what's
called a one electron orbital approximation, and when you do
you get out what are called Hartree orbitals, and what
this means is that instead of considering the wave function
as a function, for example, for helium as six different
variables, what we do is we break it up and treat each
electron has a separate wave function and say that our
assumption is that the total wave function is equal to the
product of the two individual wave functions.
So, for example, for helium, we can break it up into wave
function for it the r, theta, and phi value for electron
one, and multiply that by the wave function for the r,
theta, and phi value for electron number two.
So essentially what we're saying is we have a wave
function for electron one, and a wave function
for electron two.
We know how to write that in terms of the state numbers, so
it would be 1, 0, 0, because we're talking
about the ground state.
We're always talking about the ground state unless we specify
that we're talking about an excited state.
And we have the spin quantum number as plus 1/2 for
electron one, and minus 1/2 for the electron two.
It's arbitrary which one I assigned to which, but we know
that we have to have each of those two magnetic spin
quantum numbers in order to have the distinct four letter
description of an electron.
We know that it's not enough just to describe the orbital
by three quantum numbers, we need that fourth number to
fully describe an electron.
And when we describe this in terms of talking about
chemistry terminology, we would call the first one the 1
s, and 1 is in parentheses because we're talking about
the first electron there, and we would multiply it by the
wave function for the second one, which is also 1 s, but
now we are talking about that second electron.
We can do the exact same thing when we talk about lithium,
but now instead of breaking it up into two wave functions,
we're breaking it up into three wave functions because
we have three electrons.
So, the first again is the 1 s 1 electron.
We then have the 1 s 2 electron, and what is our
third electron going to be?
Yeah.
So it's going to be the 2 s 1 electron.
So we can do this essentially for any atom we want, we just
have more and more wave functions that we're breaking
it up to as we get to more and more electrons.
And we can also write this in an even simpler form, which is
what's called electron configuration, and this is
just a shorthand notation for these electron wave functions.
So, for example, again we see hydrogen is 1 s 1, helium we
say is 1 s 2, or 1 s squared, so instead of writing out the
1 s 1 and the 1 s 2, we just combine it as 1 s squared,
lithium is 1 s 2, 2 s 1.
So writing out electron configurations I realize is
something that a lot of you had experience with in high
school, you're probably -- many of you are very
comfortable doing it, especially for the more
straightforward atoms. But what's neat to kind of think
about is if you think about what a question might have
been in high school, which is please write the electron
configuration for lithium, now we can also answer what sounds
like a much more impressive, and a much more complicated
question, which would be write the shorthand notation for the
one electron orbital approximation to solve the
Schrodinger equation for lithium.
So essentially, that is the exact same thing.
The electronic configuration, all it is is the shorthand
notation for that one electron approximation for the
Schrodinger equation for lithium.
So, if you're at hanging your exam from high school on the
fridge and you want to make it look more impressive, you
could just rewrite the question as that, and
essentially you're answering the same thing.
But now, hopefully, we understand where that comes
from, why it is that we use the shorthand notation.
So, let's write this one electron orbital approximation
for berylium, that sounds like a pretty complicated question,
but hopefully we know that it's not at all, it's just 1 s
2, and then 2 s 2.
And we can go on and on down the table.
So, for example, for boron, now we're dealing with 1 s 2,
then 2 s 2, and now we have to move into the p orbital so we
go to 2 p 1.
So that's a little bit of an introduction into electron
configuration.
We'll get into some spots where it gets a little
trickier, a little bit more complicated later in class.
But that's an idea of what it actually means to talk about
electron configuration.
So now that we can do this, we can compare and think about,
we know how to consider wave functions for individual
electrons in multi-electron atoms using those Hartree
orbitals or the one electron wave approximations.
So let's compare what some of the similarities and
differences are between hydrogen atom orbitals, which
we spent a lot of time studying, and now these one
electron orbital approximations for these
multi-electron atoms.
So, as an example, let's take argon, I've written up the
electron configuration here, and let's think about what
some of the similarities might be between wave functions in
argon and wave functions for hydrogen.
So the first is that the orbitals are similar in shape.
So for example, if you know how to draw an s orbital for a
hydrogen atom, then you already know how to draw the
shape of an s orbital or a p orbital for argon.
Similarly, if we were to look at the radial probability
distributions, what we would find is that there's an
identical nodal structure.
So, for example, if we look at the 2 s orbital of argon, it's
going to have the same amount of nodes and the same type of
nodes that the 2 s orbital for hydrogen has.
So how many nodes does the 2 s orbital for hydrogen have?
It has one node, right, because if we're talking about
nodes it's just n minus 1 is total nodes, so you would just
say 2 minus 1 equals 1 node for the 2 s orbital.
And how many of those nodes are angular nodes? zero.
L equals 0, so we have zero angular nodes, that means that
they're all radial nodes.
So what we end up with is one radial node for the 2 s
orbital of hydrogen, and we can apply that for argon or
any other multi-electron atom here, we also have one radial
node for the 2 s orbital of argon.
But there's also some differences that we need to
keep in mind, and that will be the focus of a lot of the
lecture today.
One of the main difference is is that when you're talking
about multi-electron orbitals, they're actually smaller than
the corresponding orbital for the hydrogen atom.
We can think about why that would be.
Let's consider again an s orbital for argon, so let's
say we're looking at the 1 s orbital for argon.
What is the pull from the nucleus from argon
going to be equal to?
What is the charge of the nucleus?
Does anyone know, it's a quick addition problem here.
Yeah, so it's 18.
So z equals 18, so the nucleus is going to be pulling at the
electron with a Coulombic attraction that has a charge
of plus 18, if we're talking about the 1 s electron or the
1 s orbital in argon.
It turns out, and we're going to get the idea of shielding,
so it's not going to actually feel that full plus 18, but
it'll feel a whole lot more than it will just feel in
terms of a hydrogen atom where we only have a
nuclear charge of one.
So because we're feeling a stronger attractive force from
the nucleus, we're actually pulling that electron in
closer, which means that the probability squared of where
the electron is going to be is actually a smaller radius.
So when we talk about the size of multi-electron orbitals,
they're actually going to be smaller because they're being
pulled in closer to the nucleus because of that
stronger attraction because of the higher charge of the
nucleus in a multi-electron atom compared
to a hydrogen atom.
The other main difference that we're really going to get to
today is that in multi-electron atoms, orbital
energies depend not just on the shell, which is what we
saw before, not just on the value of n, but also on the
angular momentum quantum number.
So they also depend on the sub-shell or l.
And we'll really get to see a picture of that, and I'll be
repeating that again and again today, because this is
something I really want everyone to get firmly into
their heads.
So, let's now take a look at the energies.
We looked at the wave functions, we know the other
part of solving the Schrodinger equation is to
solve for the binding energy of electrons to the nucleus,
so let's take a look at those.
And there again is another difference between
multi-electron atom and the hydrogen atoms. So when we
talk about orbitals in multi-electron atoms, they're
actually lower in energy than the
corresponding h atom orbitals.
And when we say lower in energy, of course, what we
mean is more negative.
Right, because when we think of an energy diagram, that
lowest spot there is going to have the lowest value of the
binding energy or the most negative value of binding.
So, let's take a look here at an example of an energy
diagram for the hydrogen atom, and we can also look at a
energy diagram for a multi-electron atom, and this
is just a generic one here, so I haven't actually listed
energy numbers, but I want you to see the trend.
So for example, if you look at the 1 s orbital here, you can
see that actually it is lower in the case of the
multi-electron atom than it is for the hydrogen atom.
You see the same thing regardless of which orbital
you're looking at.
For example, for the 2 s, again what you see is that the
multi-electron atom, its 2 s orbital is lower in energy
than it is for the hydrogen.
The same thing we see for the 2 p.
Again the 2 p orbitals for the multi-electron atom, lower in
energy than for the hydrogen atom.
But there's something you'll note here also when I point
out the case of the 2 s versus the 2 p, which is what I
mentioned that I would be saying again and again, which
is when we look at the hydrogen atom, the energy of
all of the n equals 2 orbitals are exactly the same.
That's what we call degenerate orbitals,
they're the same energy.
But when we get to the multi-electron atoms, we see
that actually the p orbitals are higher in energy than the
s orbitals.
So we'll see specifically why it is that the s orbitals are
lower in energy.
We'll get to discussing that, but what I want to point out
here again is the fact that instead of just being
dependent on n, the energy level is dependent
on both n and l.
And is no longer that sole determining factor for energy,
energy also depends both on n and on l.
And we can look at precisely why that is by looking at the
equations for the energy levels for a hydrogen atom
versus the multi-electron atom.
So, for a hydrogen atom, and actually for any one electron
atom at all, this is our energy or our binding energy.
This is what came out of solving the Schrodinger
equation, we've seen this several times before that the
energy is equal to negative z squared times the Rydberg
constant over n squared.
Remember the z squared, that's just the atomic number or the
charge on the nucleus, and we can figure that out for any
one electron atom at all.
And an important thing to note is in terms of what that
physically means, so physically the binding energy
is just the negative of the ionization energy.
So if we can figure out the binding energy, we can also
figure out how much energy we have to put into our atom in
order to a eject or ionize an electron.
We can also look at the energy equation now for a
multi-electron atom.
And the big difference is right here in this term.
So instead of being equal to negative z squared, now we're
equal to negative z effective squared times r
h all over n squared.
So when we say z effective, what we're talking about is
instead of z, the charge on the nucleus, we're talking
about the effective charge on a nucleus.
So for an example, even if a nucleus has a charge of 7, but
the electron we're interested in only feels the charge as if
it were a 5, then what we would say is that the z
effective for the nucleus is 5 for that electron.
And we'll talk about this more, so if this is not
completely intuitive, we'll see why in a second.
So the main idea here is z effective is not z, so don't
try to plug one in for the other, they're absolutely
different quantities in any case when we're not talking
about a 1 electron atom.
And the point that I also want to make is the way that they
differ, z effective actually differs from the total charge
in the nucleus due to an idea called shielding.
So, shielding happens when you have more than one electron in
an atom, and the reason that it's happening is because
you're actually canceling out some of that positive charge
from the nucleus or that attractive force with a
repulsive force between two electrons.
So if you have some charge in the nucleus, but you also have
repulsion with another electron, the net attractive
charge that a given electron going to feel is actually less
than that total charge in the nucleus.
And shielding is a little bit of a misnomer because it's not
actually that's the electron's blocking the charge from
another electron, it's more like you're canceling out a
positive attractive force with a negative repulsive force.
But shielding is a good way to think about it, and actually,
that's what we'll use in this class to sort of visualize
what's happening when we have many electrons in an atom and
they're shielding each other.
Shielding is the term that's used, it brings up a certain
image in our mind, and even though that's not precisely
what's going on, it's a very good way to visualize what
we're trying to think about here.
So let's take two cases of shielding if we're talking
about, for example, the helium, a helium nucleus or a
helium atom.
So what is the charge on a helium nucleus?
What is z?
Yup, so it's plus 2.
So the charge is actually just equal to z, we can write plus
2, or you can write plus 2 e, e just means the absolute
value of the charge on an electron.
When we plug it into equations we just use the number, the e
is assumed there.
So, let's think of what we could have if we have two
electrons in a helium atom that are shielded in two
extreme ways.
So, in the first extreme way, let's consider that our first
electron is at some distance very far away from the
nucleus, we'll call this electron one, and our second
electron is, in fact, much, much closer to the nucleus,
and let's think of the idea of shielding in more of the
classical sense where we're actually blocking some of that
positive charge.
So if we have total and complete shielding where that
can actually negate a full positive charge, because
remember our nucleus is plus 2, one of the electrons is
minus 1, so if it totally blocks it, all we would have
left from the nucleus is an effective charge of plus 1.
So in our first case, our first extreme case, would be
that the z effective that is felt by electron number 1, is
going to be plus 1.
So, what we can do is figure out what we would expect the
binding energy of that electron to be in the case of
this total shielding.
And remember again, the binding energy physically is
the negative of the ionization energy, and that's actually
how you can experimentally check to see if this is
actually correct.
And that's going to be equal to negative z effective
squared times r h over n squared.
So, let's plug in these values and see what we would expect
to see for the energy.
So it would be negative 1 squared times r h all over 1
squared, since our z effective we're saying is 1, and n is
also equal to 1, because we're in the ground state here so
we're talking about a 1 s orbital.
So if we have a look at what the answer would be, this
looks very familiar.
We would expect our binding energy to be a negative 2 .
1 8 times 10 to the negative 18 joules.
This is actually what the binding energy is for hydrogen
atom, and in fact, that makes sense because in our extreme
case where we have total shielding by the second
electron of the electron of interest, it's essentially
seeing the same nuclear force that an electron in a hydrogen
atom would see.
All right.
Let's consider now the second extreme case, or extreme case
b, for our helium atom.
Again we have the charge of the nucleus on plus 2, but
let's say this time the electron now is going to be
very, very close to the nucleus.
And let's say our second electron now is really far
away, such that it's actually not going to shield any of the
nuclear charge at all from that first electron.
So what we end up saying is that the z effective or the
effective charge that that first electron feels is now
going to be plus 2.
Again, we can just plug this into our equation, so if we
write in our numbers now saying that z effective is
equal to 2, we find that we get negative 2 squared r h,
all divided again by 1 squared -- we're still talking about a
1 s orbital here.
And if we do that calculation, what we find out is that the
binding energy, in this case where we have no shielding, is
negative 8 .
7 2 times 10 to the negative 18 joules.
So, let's compare what we've just seen as our two extremes.
So in extreme case a, we saw that z effective was 1.
This is what we call total shielding.
The electron completely canceled out it's equivalent
of charge from the nucleus, such that we only saw in a z
effective of 1.
In an extreme case b, we had a z effective of 2, so
essentially what we had was no shielding at all.
We said that that second electron was so far out of the
picture, that it had absolutely no affect on what
the charge was felt by that first electron.
So, we can actually think about now, we know the extreme
cases, but what is the reality, and the reality is if
we think about the ionization energy, and we measure it
experimentally, we find that it's 3 .
9 4 times 10 to the negative 18 joules, and what you can
see is that falls right in the middle between the two
ionization energies that we would expect
for the extreme cases.
And this is absolutely confirming that what is
happening is what we would expect to happen, because we
would expect the case of reality is that, in fact, some
shielding is going on, but it's not going to be total
shielding, but at the same time it's not going to be no
shielding at all.
And if we experimentally know what the ionization energy is,
we actually have a way to find out what the z effective will
be equal to.
And we can use this equation here, this is just the
equation for the ionization energy, which is the same
thing as saying the negative of the binding energy that's
equal to z effective squared r h over n squared.
So, what we can do instead of talking about the ionization
energy, because that's one of our known quantities, is we
can instead solve so that we can find z effective.
So, if we just rearrange this equation, what we find is that
z effective is equal to n squared times the ionization
energy, all over the Rydberg constant and the
square root of this.
So the square root of n squared r e over r h.
So what's our value for n here? one.
Yup, that's right.
And then what's our value for ionization energy?
Yup.
So it's just that ionization energy that we have
experimentally measured, 3 .
9 4 times 10 to the negative 18 joules.
We put all of this over the Rydberg constant, which is 2 .
1 8 times 10 to the negative 18 joules, and we want to
raise this all to the 1/2.
So what we end up seeing is that the z effective is equal
to positive 1 .
3 4.
So, this is what we find the actual z effective is for an
electron in the helium atom.
Does this seem like a reasonable number?
Yeah?
Who says yes, raise your hand if this seems reasonable.
Does anyone think this seems not reasonable?
OK.
How can we check, for example, if it does or if it doesn't
seem reasonable.
Well, the reason, the way that we can check it is just to see
if it's in between our two extreme cases.
We know that it has to be more than 1, because even if we had
total shielding, we would at least feel is the
effective of 1.
We know that it has to be equal to less than 2, because
even if we had absolutely no shielding at all, the highest
z effective we could have is 2, so it makes perfect sense
that we have a z effective that falls somewhere in the
middle of those two.
So, let's look at another example of thinking about
whether we get an answer out that's reasonable.
So we should be able to calculate a z effective for
any atom that we want to talk about, as long as we know what
that ionization energy is.
And I'm not expecting you to do that calculation here,
because it involves the calculator, among maybe a
piece of paper as well.
But what you should be able to do is take a look at a list of
answers for what we're saying z effective might be, and
determining which ones are possible versus which ones are
not possible.
So, why don't you take a look at this and tell me which are
possible for a 2 s electron in a lithium atom where z is
going to be equal to three?
Let's do 10 more seconds on that.
OK, great.
So, the majority of you got it right.
There are some people that are a little bit confused still on
where this make sense, so, let's just think about this a
little bit more.
So now we're saying that z is equal to 3, so if, for
example, we had total shielding by the other two
electrons, if they totally canceled out one unit of
positive charge each in the nucleus, what we would end up
with is we started with 3 and then we would subtract a
charge of 2, so we would end up with a plus 1 z effective
from the nucleus.
So our minimum that we're going to see is that the
smallest we can have for a z effective is going
to be equal to 1.
So any of the answers that said a z effective of .
3 9 or .
8 7 are possible, they actually aren't possible
because even if we saw a total shielding, the minimum z
effective we would see is 1.
And then I think it looks like most people understood that
four was not a possibility.
Of course, if we saw no shielding at all what we would
end up with is a z effective of 3.
So again, when we check these, what we want to see is that
our z effective falls in between the two extreme cases
that we could envision for shielding.
And again, just go back and look at this and think about
this, this should make sense if you kind of look at those
two extreme examples, so even if it doesn't make entire
sense in the 10 seconds you have to answer a clicker
question right now, make sure this weekend you can go over
it and be able to predict if you saw a list of answers or
if you calculate your own answer on the p-set, whether
or not it's right or it's wrong, you should be able to
qualitatively confirm whether you have a reasonable or a not
reasonable answer after you do the calculation part.
All right.
So now that we have a general idea of what we're talking
about with shielding, we can now go back and think about
why it is that the orbitals are ordered in the
order that they are.
We know that the orbitals for multi-electron atoms depend
both on n and on l.
But we haven't yet addressed why, for example, a 2 s
orbital is lower in energy than the 2 p orbital, or why,
for example, a 3 s orbital is lower in energy than a 3 p,
which in turn is lower than a 3 d orbital.
So let's think about shielding in trying to answer why, in
fact, it's those s orbitals that are the lowest in energy.
And when we make these comparisons, one thing I want
to point out is that we need to keep the constant principle
quantum number constant, so we're talking about a certain
state, so we could talk about the n equals 2 state, or the n
equals 3 state.
And when we're talking about orbitals in the same state,
what we find is that orbitals that have lower values of l
can actually penetrate closer to the nucleus.
This is an idea we introduced on Wednesday when we were
looking at the radial probability distributions of p
orbitals versus s orbitals versus d orbitals.
But now it's going to make more sense because in that
case we were just talking about single electron atoms,
and now we're talking about a case where we
actually can see shielding.
So what is actually going to matter is how closely that
electron can penetrate to the nucleus, and what I mean by
penetrate to the nucleus is is there probability density a
decent amount that's very close to the nucleus.
So, if we superimpose, for example, the 2 s radial
probability distribution over the 2 p, what we see is
there's this little bit of probability density in the 2
s, but it is significant, and that's closer to the nucleus
than it is for the 2 p.
And remember, this is in complete opposition to what we
call the size of the orbitals, because we know that the 2 p
is actually a smaller orbital.
For example, when we're talking about radial
probability distributions, the most probable radius is closer
into the nucleus than it is for the s orbital.
But what's important is not where that most probable
radius is when we're talking about the z effective it
feels, what's more important is how close the electron
actually can get the nucleus.
And for the s electron, since it can get closer, what we're
going to see is that s electrons are actually less
shielded than the corresponding p electrons.
They're less shielded because they're closer to the nucleus,
they feel a greater z effective.
We can see the same thing when we compare p electrons to d
electrons, or p and d orbitals.
I've drawn the 3 p and the 3 d orbital here, and again, what
you can see is that the p electron are going to be able
to penetrate closer to the nucleus because of the fact
that there's this bit of probability density that's in
significantly closer to the nucleus than it
is for the 3 d orbital.
And if we go ahead and superimpose the 3 s on top of
the 3 p, you can see that the 3 s actually has some bit of
probability density that gets even closer to the nucleus
than the 3 p did.
So that's where that trend comes from where the s orbital
is lower than the d orbital, which is
lower than the d orbital.
So now that we have this idea of shielding and we can talk
about the differences in the radial probability
distributions, we can consider more completely why, for
example, if we're talking about lithium, we write the
electron configuration as 1 s 2, 2 s 1, and we don't instead
jump from the 1 s 2 all the way to a p orbital.
So the most basic answer that doesn't explain why is just to
say well, the s orbital is lower in energy than the p
orbital, but we now have a more complete answer, so we
can actually describe why that is.
And what we're actually talking about
again is the z effective.
So that z effective felt by the 2 p is going to be less
than the z effective felt by the 2 s.
And another way to say this, I think it's easiest to look at
just the fact that there's some probability density very
close the nucleus, but what we can actually do is average the
z effective over this entire radial probability
distribution, and when we find that, we find that it does
turn out that the average of the z effective over the 2 p
is going to be less than that of the 2 s.
So we know that we can relate to z effective to the actual
energy level of each of those orbitals, and we can do that
using this equation here where it's negative z effective
squared r h over n squared, we're going to see
that again and again.
And it turns out that if we have a, for example, for s, a
very large z effective or larger z effective than for 2
p, and we plug in a large value here in the numerator,
that means we're going to end up with a very
large negative number.
So in other words a very low energy is what we're going to
have when we talk about the orbitals -- the energy of the
2 s orbital is going to be less than the
energy of the 2 p orbital.
Another way to say that it's going to be less, so you don't
get confused with that the fact this is in the numerator
here, there is that negative sign so it's less energy but
it's a bigger negative number that gives us
that less energy there.
All right, so let's go back to electrons configurations now
that we have an idea of why the orbitals are listed in the
energy that they are listed under, why, for example, the 2
s is lower than the 2 p.
So now we can go back and think about filling in these
electron configurations for any atom.
I think most and you are familiar with the Aufbau or
the building up principle, you probably have seen it quite a
bit in high school, and this is the idea that we're filling
up our energy states, again, which depend on both n and l,
one electron at a time starting with that lowest
energy and then working our way up into
higher and higher orbitals.
And when we follow the Aufbau principle, we have to follow
two other rules.
One is the Pauli exclusion principal, we discussed this
on Wednesday.
So this is just the idea that the most electrons that you
can have in a single orbital is two electrons.
That makes sense because we know that every single
electron has to have its own distinct set of four quantum
numbers, the only way that we can do that is to have a
maximum of two spins in any single orbital or two
electrons per orbital.
We also need to follow Hund's rule, this is that a single
electron enters each state before it
enters a second state.
And by state we just mean orbital, so if we're looking
at the p orbitals here, that means that a single electron
goes in x, and then it will go in the z orbital before a
second one goes in the x orbital.
This intuitively should make a lot of sense, because we know
we're trying to minimize electron repulsions to keep
things in as low an energy state as possible, so it makes
sense that we would put one electron in each orbital first
before we double up in any orbital.
And the third fact that we need to keep in mind is that
spins remain parallel prior to adding a second electron in
any of the orbitals.
So by parallel we mean they're either both spin up or they're
both spin down -- remember that's our spin quantum
number, that fourth quantum number.
And the reason for this comes out of solving the
relativistic version of the Schrodinger equation, so
unfortunately it's not as intuitive as knowing that we
want to fill separate before we double up a degenerate
orbital, but you just need to keep this in mind and you need
to just memorize the fact that you need to be parallel before
you double up in the orbital.
So, we'll see how this works in a second.
So let's do this considering, for example, what it would
look like if we were to write out the electron configuration
for oxygen where z is going to be equal to 8.
So what we're doing is filling in those eight electrons
following the Aufbau principle, so our first
electron is going to go in the 1 s, and then we have no other
options for other orbitals that are at that same energy,
so we put the second electron in the 1 s as well.
Then we go up to the 2 s, and we have two electrons that we
can fill in the 2 s.
And now we get the p orbitals, remember we want to fill up 1
orbital at a time before we double up, so we'll put one in
the 2 p x, then one in the 2 p z, and then one in the 2 p y.
At this point, we have no other choice but to double up
before going to the next energy level, so we'll put a
second one in the 2 p x.
And I arbitrarily chose to put it in the 2 p x, we also could
have put it in the 2 p y or the 2 p z, it doesn't matter
where you double up, they're all the same energy.
So if we think about what we would do to actually write out
this configuration, we just write the energy levels that
we see here or the orbital approximations.
So if we're talking about oxygen, we would say that it's
1 s 2, then we have 2 s 2, and then we have 2 p, and our
total number of electrons in the p orbitals are four.
So it's OK to not specify.
I want to point out, whether you're in the p x, the p y, or
the p z, unless a question specifically asks you to
specify the m sub l, which occasionally will happen, but
if it doesn't happen you just write it like this.
But if, in fact, you are asked to specify the m sub l's, then
we would have to write it out more completely, which would
be the 1 s 2, the 2 s 2, and then we would say 2 p x 2, 2 p
z 1, and 2 p y 1.
So again, in general, just go ahead and write it out like
this, but if we do ask you to specify you should be able to
know that the p orbital separates into these three --
the p sub-shell separates into these three orbitals.
So let's do a clicker question on assigning electron
configurations using the Aufbau principle.
So why don't you go ahead and identify the correct electron
configuration for carbon, and I'll tell you that z
is equal to 6 here.
And in terms of doing this for your homework, I actually want
to mention that in the back page of your notes I attached
a periodic table that does not have electron
configurations on them.
It's better to practice doing electron configurations when
you cannot actually see the electron configurations.
And this is the same periodic table that you're going to get
in your exams, so it's good to practice doing your
problem-sets with that periodic table so you're not
relying on having the double check right there of seeing
what the electron configuration is.
So, let's do 10 seconds on this problem here.
OK, great.
So this might be our best clicker question yet.
Most people were able to identify the correct electron
configuration here.
Some people, the next most popular answer with 5%, which
is a nice low number, wanted to put two in the 2 p x before
they moved on.
Remember we have to put one in each degenerate orbital before
we double up on any orbital, so just keep that rule in mind
that we would fill one in each p orbital before we a to the
second one.
But it looks like you guys are all experts here on doing
these electron configurations.
So, let's move on to some more complicated electron
configurations.
So, for example, we can move to the next periods in the
periodic table.
When we talk about a period, we're just talking about that
principle quantum number, so period 2 means that we're
talking about starting with the 2 s orbitals, period 3
starts with, what we're now filling into the
3 s orbitals here.
So if we're talking about the third period, that starts with
sodium and it goes all the way up to argon.
So if we write the electron configuration for sodium,
which you can try later -- hopefully you would all get it
correctly -- you see that this is the electron configuration
here, 1 s 2, 2 s 2, 2 p 6, and now we're going into that
third shell, 3 s 1.
And I want to point out the difference between core
electrons and valence electrons here.
If we look at this configuration, what we say is
all of the electrons in these inner shells are what we call
core electrons.
The core electrons tend not to be involved in much chemistry
in bonding or in reactions.
They're very deep and held very tightly to the nucleus,
so we can often lump them together, and instead of
writing them all out separately, we can just write
the equivalent noble gas that has that configuration.
So, for example, for sodium, we can instead write neon and
then 3 s 1.
So the 3 s 1, or any of the other electrons that are in
the outer-most shell, those are what we call our valence
electrons, and those are where all the excitement happens.
That's what we see are involved in bonding.
It makes sense, right, because they're the furthest away from
the nucleus, they're the ones that are most willing to be
involved in some chemistry or in some bonding, or those are
the orbitals that are most likely to accept an electron
from another atom, for example.
So the valence electrons, those are the exciting ones.
We want to make sure we have a full picture of
what's going on there.
So, no matter whether or not you write out the full form
here, or the noble gas configuration where you write
ne first or whatever the corresponding noble gas is to
the core electrons, we always write out the valence
electrons here.
So for sodium, again, we can write n e and then 3 s 1.
We can go all the way down, magnesium, aluminum, all the
way to this noble gas, argon, which would be n e and then 3
s 2, 3 p 6.
Now we can think about the fourth period, and the fourth
period is where we start to run into some exceptions, so
this is where things get a teeny bit more complicated,
but you just need to remember the exceptions and then you
should be OK, no matter what you're asked to write.
So for the fourth period, now we're into the 4 s 1 for
potassium here.
And what we notice when we get to the third element in and
the fourth period is that we go 4 s 2 and then we're back
to the 3 d's.
So if you look at the energy diagram, what we see is that
the 4 s orbitals are actually just a teeny bit lower in
energy -- they're just ever so slightly lower in energy than
the 3 d orbitals.
You can see that as you fill up your periodic table, it's
very clear.
But also we'll tell you a pneumonic device to keep that
in mind, so you always remember and get the orbital
energy straight.
But it just turns out that the 4 s is so low in energy that
it actually surpasses the 3 d, because we know the 3 d is
going to be pretty high in terms of the three shell, and
the 4 s is going to be the lowest in terms of the 4
shell, and it turns out that we need to fill up the 4 s
before we fill in the 3 d.
And we can do that just going along, 3 d 1, 2 3, and the
problem comes when we get to chromium here, which is
instead of what we would expect, we might expect to see
4 s 2, 3 d 4.
What we see is that instead it's 4 s 1, and 3 d 5.
So this is the first exception that you need
to the Aufbau principle.
The reason this an exception is because it turns out that
half filled d orbitals are more stable than
we could even predict.
You wouldn't be expected to be able to guess that this would
happen, because using any kind of simple theory, we would, in
fact, predict that this would not be the case, but what we
find experimentally is that it's more stable to have half
filled d orbital than to have a 4 s 2, and a 3 d 4.
So you're going to need to remember, so this is an
exception, you have to memorize.
Another exception in the fourth period is in copper
here, we see that again, we have 4 s 1 instead of 4 s 2.
This is 4 s 1, 3 d 10, we might expect 4 s 2, 3 d 9, but
again, this exception comes out of experimental
observation, which is the fact that full d orbitals also are
lower in energy then we could theoretically predict using
simple calculations.
So again, you need to memorize these two exceptions, and the
exception in general is that filled d 10, or half-filled d
5 orbitals are lower in energy than would be expected, so we
got this flip-flip where if we can get to that half filled
orbital by only removing one s electron, then we're going to
do it, and the same with the filled d orbital.
And actually, when we get to the fifth period of the
periodic table, that again takes place, so when you get
to a half filled, or a filled d orbital, again you want to
do it, so those exceptions would be with molybdenum and
silver would be the corresponding elements in the
fifth period where you're going to see the same case
here where it's lower in energy to have the half filled
or the filled d orbitals.
So here's the pneumonic I mentioned for writing the
electron configuration and getting those orbital energies
in the right order.
All you do is just write out all the orbitals, the 1 s,
then the 2 s 2 p 3, 3 s 3 p d, just write them in a straight
line like this, and then if you draw diagonals down them,
what you'll get is the correct order in
terms of orbital energies.
So if we go down the diagonal, we start with 1 s, then we get
2 s, then 2 p and 3 s, then 3 p, and 4 s, and then that's
how we see here that 4 s is actually lower in energy than
3 d, then 4 p, 5 s and so on.
So if you want to on an exam, you can just write this down
quickly at the beginning and refer to it as you're filling
up your electron configurations, but also if
you look at the periodic table it's very clear as you try to
fill it up that way that the same order comes out of that.
So, whichever works best for you can do in terms of
figuring out electron configurations.
So the last thing I want to mention today is how we can
think about electron configurations for ions.
It turns out that it's going to be a little bit different
when we're talking about positive ions here.
We need to change our rules just slightly.
So what we know is that these 3 d orbitals are higher in
energy than 4 s orbitals, so I've written the energy of the
orbital here for potassium and for calcium.
But what happens is that once a d orbital is filled, I said
the two are very close in energy, and once a d orbital
is filled, it actually drops to become lower in energy than
the 4 s orbital.
So once we move past, we fill the 4 s first, but once we
fill in the d orbital, now that's going
to be lower in energy.
So that doesn't make a difference for us when we're
talking about neutral atoms, because we would fill up the 4
s first, because that's lower in energy until we fill it,
and then we just keep going with the d orbitals.
So, for example, if we needed to figure out the electron
configuration for titanium, it would just be argon then 4 s
2, and then we would fill in the 3 d 2.
So, actually we don't have to worry about this fact any time
we're dealing with neutrals.
The problem comes when instead we're dealing with ions.
So what I want to point out is what we said now is that the 3
d 2 is actually lower in energy, so if we were to
rewrite this in terms of what the actual energy order is, we
should instead write it 3 d 2, 4 s 2.
So you might ask in terms of when you're writing electron
configurations, which way should you write it.
And we'll absolutely accept both answers
for a neutral atom.
They're both correct.
In one case you decided to order in terms of energy and
in one case you decided to order in terms of
how it fills up.
I don't care how you do it on exams or on problem sets, but
you do need to be aware that the 3 d once filled is lower
in energy than the 4 s, and the reason you need to be
aware of that is if you're asked for the electron
configuration now of the titanium ion.
So, let's say we're asked for the plus two ion.
So a plus two ion means that we're removing two electrons
from the atom and the electrons that we're going to
remove are always going to be the highest energy electrons.
So it's good to write it like this because this illustrates
the fact that in fact the 4 s electrons are the ones that
are higher in energy.
So the correct answer for titanium plus two is going to
be argon 3 d 2, whereas if we did not rearrange our order
here we might have been tempted to write as 4 s 2 so
keep that in mind when you're doing the positive ions of
corresponding atoms. Alright, so we'll pick up with
photoelectron spectroscopy on Wednesday.
Have a great weekend.