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CATHERINE DRENNAN: We gotta move up to the next lowest energy
orbital, which we'll also talk about today.
All right.
So that kind of completes our one electron systems,
and now we're gonna talk about multi-electrons.
But we're not going to move far away from the topics
that we've discussed because with multi-electrons there
are several things that are the same and some things that
are different.
But a lot of things are the same,
so we're going to come back to radial probability
distributions and energy levels and things like that.
So we'll move on to today's handout.
So similarities and differences.
Similarities.
We have the same shapes of orbitals,
whether we're talking about a one electron system or more.
So again, we're gonna have our s orbital.
So this is my version of an s orbital
that you can hold in your hand, where you have it spherically
symmetric and you have probability out
in very direction of r that you might find an electron,
except at our radial nodes.
And then we have our p orbitals, which
can be in three different directions, where
you have nodal planes.
So it doesn't matter what you're talking about-- one electron
system or more-- you still are going
to have those same shapes of the wave functions.
Other similarities again include the nodal structure,
which I just mentioned.
Whether s is always going to be spherically symmetrical.
p is gonna have those nodal planes.
The angular nodal planes.
So that's the same.
Differences, though, have to do with the energy levels.
So orbitals in a multi-electron atoms
are lower-- more negative-- in energy
than their corresponding orbitals in the H atom.
So let's take a look at this and think about why this is true.
So here we have our friend the hydrogen atom.
It has a small Z. Z equals 1.
Doesn't get smaller than that.
We have our 1s.
Then we have our n equals 2.
And we talked about the fact that in a 1 electron
system like hydrogen-- these are all
degenerate in energy with respect to each other,
so they're all equal in energy.
And then we have our n equals 3.
Again, degenerate energy levels and n equals 4 up here.
But when you go to a multi-electron system,
the first thing that I'll point out
is that the energy level is lower.
So we have 1s over here.
Now 1s is a more negative lower number.
And we can think about this in multi-electron systems-- you
have a bigger Z. So you have more positive charge
and it's kind of hauling all of those energy levels closer.
So it's a lower energy for all of those.
So 1s, again, lower, and we have 2s.
Again, that's lower in energy.
And now importantly, 2s and 2p are no longer
degenerate with respect to each other in energy.
So now the 2p system is higher in energy than the 1s
and so on.
So we have 3s down here and then the 3p's.
So compared to hydrogen in a multi-electron atom,
n is not the sole determinate of what the energy levels are.
Now, instead of just n we have n and l.
So let's review.
This is good.
This is all getting ready for the test.
Some equations that you've seen before.
And again, equations will be on the equations sheet.
You don't have to memorize them.
And the equation sheet for the exam
is handed out today, so you can take a look a it
and see where things are.
If I forgot anything you can ask me questions
and maybe we'll add some more if I forgot any ones.
So for a one electron system, binding energy
equals minus the ionization energy equals minus Z squared
RH, the Rydberg constant, over N squared.
And so for hydrogens Z is 1, but there are other 1 electron
at least ions and then you have to worry about Z.
What about a multi-electron system?
Looks pretty much the same, but now instead of just having n we
have n and l because l matters with a multi-electron system.
So the binding energy for the electron
is minus the ionization energy-- that's always
going to [INAUDIBLE] true.
The binding energy is always going
to be equal to minus the ionization energy, which
is equal to minus Z. But now we have
a different Z. We have Z effective,
which is abbreviated Z eff.
So this is the effective charge, not the actual charge.
And that's squared.
And then we have Rydberg constant and n again.
So Z effective is not the same as Z,
and it's not the same Z because of shielding.
So we talked about shielding a little bit,
and I made this point that you need
to think about shielding a little differently.
It's not just about the average size of the orbital.
It's more about the likelihood-- the probability--
that they're going to be electrons close
to the nucleus that will participate in-- that
will be affected by shielding.
So let's look at some extreme cases
now and think about what's happening
in terms of this shielding.
So extreme case one.
Extreme shielding.
Maximum shielding.
So we have the case.
We have the helium nucleus.
We have moved far in the periodic table.
We've left hydrogen. And so what is the charge going
to be on helium nucleus?
Plus 2.
So now we have this electron one.
We're interested in how much this electron is
shielded by electron two because we've got two electrons.
Now say electron two is close to the nucleus
and it's maximally shielding electron one
from this positive charge down here.
So think about this electron.
I like to think about this electron two
as kind of the elasti-girl of electrons.
So elasti-girl electron is shielding everywhere,
stretching everywhere around that nucleus.
Just completely shielding it from electron
one, using the full negative charge to really shield.
So in that kind of exaggerated case, the Z effective is not 2.
It's 1.
So this electron basically cancels
the whole positive charge of this nucleus.
Really shields that away.
And so this is the effective charge with total shielding.
Maximal shielding that you can get.
So now we can calculate what the binding energy-- or if we
wanted, what the ionization energy would
be for this case, given that particular effective charge.
So again, the binding energy of electron one
is equal to minus its ionization energy equal to minus the Z
effective squared R H over n squared.
And we can plug in those numbers.
Don't forget the minus.
We can have Z effective 1 over 1 and we calculate this value,
which is, of course, just the Rydberg constant
or minus the Rydberg constant minus 2.18 times 10
to the minus 18th joules.
So this is just like it was a hydrogen atom.
A one electron system.
So it had two electrons but one of the electron shielded
so completely it was like a one electron system.
Again, this is an exaggerated case of total shielding.
So now let's go to the other extreme
and consider zero shielding.
No shielding at all.
So here we have the helium nucleus.
We have electron one is now close
and electron two is pretty far away.
So again, we're asking, how much is this electron one going
to shielded by electron two?
And in this extreme scenario, electron two
is not participating in shielding at all.
It's far away.
I'm not sure what it's doing.
Maybe it lost its super suit and can't find it.
It might be at the dry cleaner's.
We don't know what's going on.
In any case, it is out of the game.
It is not involved.
It is not shielding at all.
So in this extreme case, electron one
feels that full force of the nucleus.
So its effective charge is the full complete plus 2.
So we can plug that in now and calculate
what the binding energy is going to be.
Again, binding energy for electron one
equals minus the ionization energy for this electron.
Same equation.
We put this in but now the Z effective is 2
and we can calculate that.
And now we get a value of minus 8.72 times 10
to the minus 18th joules.
And this is actually the same as you
would get for the scenario of helium plus, which
is a one electron system.
Then, in a one electron system you
can use a formula of just Z where Z is 2
and get your value.
So here are the two extreme cases for helium.
One it's like a hydrogen atom one electron system, and one
it's like helium plus.
And one case is like it just has one electron,
and in the other case you have zero shielding.
So extreme case one the Z effective is 1,
and we have the binding energy that's
very similar to hydrogen. And so this is a total shielding case.
It's shields so much it's like a one electron case.
Extreme case two you have the full force.
So there's zero shielding.
No shielding at all.
And then, this is like helium plus case
where you've actually lost that other electron.
It's not shielding.
It's not even there.
So no shielding.
And the reality in most cases is that you're in between,
and you can determine this experimentally.
You can measure ionization energies.
So if we measured the ionization energy for helium
we would find that it's 3.94 times 10 to the minus 18th.
So it's greater than the 2.18 and less than the 8.72.
So it's in between, and that's what you find most of the time.
The Z effective is in between zero shielding
and total shielding.
So we can calculate what the Z effective is actually
in this case if we know the ionization energy.
So if you know the ionization energy
you can calculate the Z effective,
or if you're given a Z effective you
can calculate what the ionization
energy should be for that case.
And so I told you that it was measured at 3.94 times 10
to the minus 18th.
And so in that case the Z effective,
if you plug the numbers in would be 1.34.
And so this number is in fact in between the two cases.
With total shielding you would have a Z effective of 1
and with no shielding you have a Z effective of 2.
And so in reality, we're somewhere
in between in this case.
So thinking about this now, let's try your hand
at a clicker question and see if you can tell me
which of these is a possible Z effective for an element
with a Z equals 3.
I'm doing good time wise.
OK.
10 more seconds.
So does someone want to say why the other ones are not correct?
Why don't you run up there and you can give them this pen?
American Chemical Society pen.
Hand up there.
AUDIENCE: Who had their hand up?
CATHERINE DRENNAN: I think it's on.
AUDIENCE: So in the case of no shielding
the charge should be 3, and in the case of maximum shielding
there's only two electrons in the 1s so it would be 1.
So anything below 1 would be beyond maximum shielding,
so it's gotta be between 1 and 3.
CATHERINE DRENNAN: Yeah, that's right.
So now we're going to talk more about
why shielding is important.
And shielding really has to do with this order
of orbital energy that we all know and are very fond of.
So when I show you this we want to ask the question,
why is 2s lower than 2p or 3s lower than 3p?
So let's take a look at this, and we're
going to go back to our radial probability distribution.
I told you I wasn't going to leave it for very long
and we're back again.
So here, again, if we're considering 2s and 2p--
so we'll consider this case here--
the maximum probable radius is longer.
It's greater for 2s than for 2p.
But what we really care about when
we're talking about shielding is this part right here.
And so there are different ways you can express this.
You can say that the electrons in the orbital with lower
values of l, like 2s is lower than 2p--
those electrons penetrate closer to the nucleus,
even though we have this trend where the radius decreases
with increasing l.
So despite this size difference, when
you compare this dotted line with this line here,
there's more probability that they're
going to be electrons near the nucleus with 2s.
And that results in this lower energy.
So there's less shielding for the s orbitals
than for the p orbitals.
And as a result of that, with less shielding
because their probability is that they're closer,
that they're bound more tightly, you have this lower energy.
So that explains this energy difference.
So we can look at this now for the three system
again for n equals 3.
And here we see that p electrons are also less shielded
than the d electrons, despite the fact if you look
that the most probable radius over here is longer for p
than for d.
But now, if we look near the nucleus at the probability
that they're going to be electrons near the nucleus,
there's a higher probability that they'll
be electrons near the nucleus with p than with d.
So these are going to be less shielded and lower in energy.
And if I throw s on there now we see
s has the most probability here of being closer.
Then p then d.
So for a multi-electron atom, the order of energy--
we have s lower than p, p lower than d, d lower than f due
to this shielding.
So again, we want to be able to draw
some version of these diagrams with appropriate features
to explain answers.
And this all leads into what we're
doing next, which is electron configurations.
So we're going to think about how
we're going to write the electron configurations
and this indicates how you build up in energy.
So electron configurations.
So first, we're going to fill our electrons in to the energy
states that are our lowest.
Nature doesn't want to do a lot of work,
so it's going to put them in the lower states.
And again, where those energy levels are
depends on, for multi-electron atoms, both n and l
And we're going to put them in one electron at a time,
starting with the lowest energy state
and heading the following rules.
So there are some rules in doing this, and most of you
have probably heard some of these before.
And if you haven't I'm sure you'll like them.
I know you heard this because I already just told you
about that a few minutes ago.
We have Pauli's Exclusion Principle,
which says that you can't have the same four quantum numbers.
So if you put one electron in and it's spin up the next one
has to be spin down.
And Hund's Rule, which is one of my favorite rules in chemistry,
and that is when you're adding electrons to the same state
you're going to put them in singly, when there's
degenerate energy orbitals, before a second one enters
the same orbital.
And the spins remain parallel as you're adding them across.
So let's consider these and put some electrons into these.
And we'll do this for oxygen, which is a Z equals 8 system.
So first, I want to put them in the lowest energy state.
So that's 1s.
So I'll put one electron in there.
And then I'm going to put the second electron in there
because it's the lowest energy state,
so I'm going to fill it up.
But I'm going to pay attention to Pauli's Exclusion Principle
and put it in spin down.
One electron spin up.
One electron spin down.
I can't put them both spin up because they would
have the same four quantum numbers,
and that would violate Pauli's Exclusion Principle.
So next I'm going to put electrons in 2s because that's
the next lowest energy state.
I'll put one in spin up.
And then, because of Pauli, I'll put the other one in spin down.
Next, we come to the 2p system.
And I'm going to put the first electron in,
but I'm not going to pair the second one.
I'm going to put electron in the second orbital, both
being spin up.
And then I'm going to do that again.
Most people kind of refer to the Hund's Rule
as kind of the rule of seating on a bus,
where it always seems that one person goes in--
and even though there are two seats, one person takes it.
The next person gets on the bus.
Doesn't sit next to the person who's already there.
They take another empty seat and so on and so on.
And so you fill up the bus with one person per bench
before all those seats are taken.
Then the next person comes on.
Sadly has to sit with someone else.
So if you think about that, that's the Hund's Rule.
You'll always remember to do that.
And once you put one in each, then the next person on the bus
has to sit next to someone and they're
going to do that spin down.
Because if it was spin up again you'd
have the same four quantum numbers.
So these are the rules that you need to know to put these in.
And then, after you do that you can write an electron
configuration that says what you did.
So here we can write it this way.
We'll say we have 1s two.
There are two electrons in the 1s.
2s two.
Two electrons in 2s.
And 2p four.
We have four electrons in the 2p orbitals.
And if for some reason the question
asks you to specify m sub l, you can do that too.
So then instead of just saying 2p4 then you would say 2px2.
So there were two electrons in px.
And then to 2pz1 and 2py1.
And if you don't write 1, 1 is implied.
So you will see this both ways.
You will see the 1 indicated and then
you'll see the orbital just listed with nothing.
If you do that you are implying there is one electron in there.
If you don't mean to imply there is one electron in there
don't write it that way because that's
what is implied if that's the way it is.
So now let's do a clicker question.
OK.
10 more seconds.
I think we can get at least 93% on this one.
That's not bad.
90.
That was a decent guess.
So I think the trick here was just counting.
So the bottom one that some people
liked-- there are only six electrons and that has more.
AUDIENCE: Question.
CATHERINE DRENNAN: Yep.
AUDIENCE: Are the p and sub l notations always
filled in with order x, z, y or can it
be x, y, z, or [INAUDIBLE]?
CATHERINE DRENNAN: Yeah, you don't have to worry about that.
That's kind of arbitrary what you put for that.
So if it is indicated then that's fine,
but we didn't give you an option that would be different.
So now you can imagine that if you were writing these electron
configurations and you were asked to write an electron
configuration for something way down on the periodic table
you would be writing for a very, very, very long time.
So you can use information about core electrons and valence
electrons to make your life easier.
Now, sometimes you will be asked to write the full electron
configuration.
It will say don't use the noble gas
short hand, then that's what you have to do.
Very important on the exam.
Read questions carefully.
So let's introduce this concept of core electrons and valence
electrons.
So if we look at the periodic table we have our 1s1, 1s2,
and then we come down.
We have our 2s's and our 2p's, and then we have a noble gas,
and then we come down and we have our first 3s1.
And so if we're talking about sodium,
we have core electrons, which are
the electrons that make up the noble gas element that's
on the row before.
And then, those are usually not very reactive.
They're held pretty tight.
And the valence electrons-- the valence electrons
are the electrons that do all the exciting chemistry-- lose
and gain valence electrons.
They're the fun ones.
And so those are going to be beyond that noble gas
configuration.
Those core electrons.
And so here sodium has one 3s1.
So we can also write sodium then as bracket neon,
indicating the noble gas before, and then
just put the valence electrons.
The 3s1.
And unless it's clearly specified
you're not supposed to do that this will be acceptable.
So we can go on in that row of the periodic table
in the third period.
And so the next one we would have 3s2,
and then we jump over here to the 3p's and so
on and so on until we get to our next noble gas.
So this is pretty straightforward.
There are no exceptions on the third period.
But when we get to the fourth period of the periodic table
we start to have a couple of exceptions
that you will be responsible for.
So it's looking good up here in the beginning.
We have now our noble gas, and then we
have our valence electrons 4s1, 4s2,
and then we start down in the 3d's over here.
And when we get halfway through we have an exception.
So you have exceptions halfway through and also
when you're almost at the end.
So instead of having 4s2 3d4, it really
wants to have 5 electrons.
And in this case, instead of 4s2 3d9 it wants 10.
So it like things being half full or totally full,
and so you have these exceptions there.
So you don't have a d4 and you don't have d9.
And you're responsible for these two,
and then in the period right below same position
same exception.
So you can think about this-- I don't
know if it's four exceptions or really just two in two places,
but these are the exceptions you need to know.
So in the fifth it's the ones right below are the same.
You're not going to have d4 or a d9 here.
So you bump one up because it's just happier
when it has 5 or 10 electrons in the d orbitals.
You're also responsible for knowing the orders.
We have a question way up there.
Who wants to run?
Get some exercise.
I don't always see questions so yell out my name
if I don't see people.
I am wearing my glasses.
That's good.
Yeah?
AUDIENCE: So what happens when we have an ion?
Like a titanium [INAUDIBLE]--
CATHERINE DRENNAN: Good question.
I'm getting to that.
AUDIENCE: All right.
Cool.
Thanks.
CATHERINE DRENNAN: Hold on a minute.
Yes.
We're getting to ions.
They're right at the end.
So first, let's consider what happens
when we're not doing an ion.
And there are a couple different ways that we can look at this.
This is one way to remember.
So you just write out 1s, 2s, 2p, threes, your fours,
your fives, your six, your sevens, and then draw a line.
First you fill 1s, then 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, et
cetera, et cetera, et cetera.
That works.
That's one way to do it.
It's kind of time consuming.
Not sure that's the easiest way.
The other way that you can do-- this
is the periodic table you'll be getting on the exam,
and you can just remember what's happening
at these different cases.
So over here we're filling up our s's.
Over here we're filling up our p's, except for helium.
Over here we're doing our d's in the transition metals.
Over here we have f.
And then in terms of the period numbers-- so
we have one over here.
So we do 1s2, 2s2, or 1s1, 1s3, 2s.
Then we come over here.
We have the two p's, three s's, three p's, four s's.
Then we have four d's and go to four p's, five s's, four d's,
five p's, and then we can't forget when we
get down here that we have f.
What number goes here?
Four.
So when you get to principal quantum
number four is when you start having f orbitals.
So if you just remember this, it's
going to help you think about what
the energy levels-- how you're going to write those
configurations.
So maybe I'll leave this up and we'll do a clicker question now
on this.
And you have your periodic table available to look at.
And then we'll get to ions.
This should be 93%.
Yeah.
OK.
10 seconds.
Well, 85.
That's still pretty good.
So this one was one of our exceptions over here.
And we wanted to have no d9.
So it just is happier as d10 system,
so that's our exception.
One of the four that you need to know for this.
So now we have our two methods of remembering this.
And we did the clicker question, and so I'm
going to jump to ions.
There's one question on problem sets
and people have been asking about ions.
So when we're just here filling up our 4s ones.
4s we're putting in first and then
we're going to our 3d, which is what we just learned about.
But what actually happens when you start putting electrons
in your d orbitals is that the orbital energy
of the d orbitals drops below the 4s.
So once they start becoming filled you
have this change that happens.
And this is really only important
when you're talking about ions.
So if we were asking here about this one--
so we would normally say, OK, we put our two 4s in
and then we have two 3d electrons.
But now, because we've started to fill the 3d,
if we're going to really do this in terms of energy level
we could reverse the order.
And we would accept both of these for the neutral ion.
You could do either here.
But importantly, if you start ionizing it and losing
electrons, you need to think where are
those electrons coming from?
Which are the electrons that are easiest to remove?
And what happens is that you end up losing the 4s2 electrons.
And so the configuration for titanium plus 2 is just 3d2.
And so this is different.
Ions behave differently, and now you
can go run and finish problem set two.
That last question.
And this is because now we're at the end of exam one material.
So one thing I didn't get to-- it said in the beginning.
Number 12 on the problem set we're not going
to get to until next week.
So if you've done number 12 already you can turn it in.
It won't be graded and you can turn it in again on problem set
3 so you won't have wasted time.
But that won't be on exam one material.
So this is the end of exam one material.
And read your instructions for the exams very carefully.
And there's a couple questions, but I think
it's too noisy to answer them.
Note that not everyone is taking the exam in this room,
and you need to go to the appropriate place.
So please look at the instructions.
All right.
And who won the clicker competition for today?
Recitation 12.