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CATHERINE DRENNAN: If you haven't clicked in yet,
please pay attention to the clicker question,
and I'll give you a little bit more time to click in.
And I'll give you a 10-second warning.
All right.
Let's just go ahead and take 10 more seconds.
All right.
Does someone want to say how they got this one right?
Anyone willing to say how they got it right?
We have a American Chemical Society,
whatever these are called that hang around your neck and clip
things to them.
Yes.
There's the word.
No?
Oh, there we go.
AUDIENCE: Thank you.
So this molecule is the phosphorus
has five valence electrons and then each hydrogen has three
so it's per total of eight.
So it has three bonding and then one lone pair,
which makes it a tetrahedral, but then the lone pair
pushed the other bond, so it's less than 109.5 degrees.
CATHERINE DRENNAN: Excellent.
Yeah.
So I see that some people just decided that there
were three things bound to it.
And so then they decided either 120 or less than 120,
but you really need to do the Lewis structure
and see how many loan pairs there are first.
So once you do the Lewis structure,
then you figure out the parent geometry,
Sn forces sum of tetrahedral system.
All right.
So more practice on these coming up.
All right.
So we're continuing on now with molecular orbital theory,
and we started the course talking about atomic structure
and then we talked about atomic orbitals or wave functions.
And then we moved on to bonding, that atoms can come together
and bond and talked about the structure of molecules,
and now we're going to molecular orbitals.
And then we'll also talk more about the structure
of molecules based on those molecular orbitals
on Wednesday.
So we're really coming all the way around,
we're using a lot of the ideas that you've seen before,
but now we're applying them to molecules.
And then, to me, this is really an exciting part.
I love getting up to the molecules
and talking about structures and molecules
and how orbitals really play a role in those properties that
molecules have.
And then, to me, the really exciting-- I like reactions.
So after we finish with the structures,
we're going to talk more about how molecules react.
And so on Friday, we're going to be
starting that, reactions of molecules,
and getting into thermodynamics.
So we're sort of winding our way away from orbitals for a while.
We will come back to d orbitals around Thanksgiving time,
but we'll have a long time before that where
we're talking about reactions.
So we're about to do a transition
in the type of material.
But before we do that, more orbitals.
But these are super cool because these are molecular orbitals.
So we're going to talk today about MO theory, MO
for molecular orbitals.
So molecular orbital theory presents the idea
that these valence electrons are really
going to be delocalized around these molecules
and not just sitting on individual atoms.
So to think about this electron do localization,
we need to think about molecular orbitals.
Molecular orbitals, as we've learned,
another word for orbital is also wave functions.
Wave functions are orbitals.
Orbitals are wave functions.
You need to consider the wavelike properties
of electrons to think about where the electrons are going
to be, what is their probability density,
how are they going to be arranged
with respect to the nucleus.
And so when we take atomic orbitals
and we bring them together as atoms come together
to form bonds, atomic orbitals come together
to form molecular orbitals.
So we're going to be adding superimposing, atomic orbitals
to form these molecular orbitals.
And this is called a linear combination of atomic orbitals,
or LCAO.
And so we're going to bring those atomic orbitals together
and create molecular orbitals.
And we're going to create two types of molecular orbitals.
We're going to create bonding and antibonding.
And some basic math principles apply here,
and that is that you can create N molecular orbitals
from N atomic orbitals.
All right.
So that's really the basis of molecular orbital theory,
and now let's apply it to our friends, the s orbitals.
All right.
So we're going to think about really simple molecules,
bringing together two atoms that are identical with each other
and what happens to their s orbitals when this happens.
So first we'll talk about bonding orbitals.
So bonding orbitals, again, arise
from this linear combination of atomic orbitals, the LCAO.
And if it's a bonding orbital, it's
going to arise from constructive interference.
So we talked before about the properties of waves, and one
of the great properties of waves is
that they can constructively interfere, destructively
interfere.
And orbitals are wave functions so they can constructively
interfere and destructively interfere.
Bonding orbitals are generated by
the constructive interference.
So let's look at two atomic orbitals.
And so here we have an orbital.
The nucleus is in the middle.
It's a little dot.
It's a nucleus.
And these two atomic orbitals are going to come together.
There's going to be a bonding event.
And so we have a 1s orbital from atom a
and a 1s orbital from atom b, and they're
going to come together, and they're
going to form a molecular orbital,
an ab molecular orbital, and this is called sigma 1s.
So a sigma orbital is symmetric around the bond axis,
so the bond axis here would be just a vertical, a line
between these two nuclei here.
And so this molecular orbital is symmetric around that bond
axis.
There are no nodal planes for something that is symmetric.
There are no nodal planes for our s orbitals,
and so there's none for the molecular orbital either.
And we can also write this as 1sa plus 1sb.
So the atomic orbital from 1sa, the atomic orbital 1sb coming
together to form sigma orbital 1s.
That is a bonding orbital because it's
constructive interference.
It's a bonding MO, or molecular orbital.
So now let's consider the wavelike properties
and think about these atomic orbitals coming together
as waves in what is happening.
So here we have the same equation up here.
We're bringing together 1sa and 1sb,
but now let's think about this as a wave function.
So there is an amplitude associated
with the wave function for 1sa, and there
is an amplitude associated with the wave function for 1sb.
These wave functions come together.
Here is one nuclei, here's the other nuclei.
And for a bonding orbital, it will
be constructive interference, and so the amplitude
where these atomic wave functions overlap
will be increased when you have constructive interference.
So our sigma 1s now has increased
amplitude between these two nuclei
due to that constructive interference.
So an increased amplitude between these two nuclei,
again, this is the bonding axis-- here is one nuclei, here
is another-- so an increased amplitude here
corresponds to an increased or enhanced probability density.
Remember our wave function squared is probability density.
It suggests the likelihood of finding
an electron in a certain region of volume
and if it's enhanced by this constructive interference.
So if that probability density is enhanced,
you have a greater chance of finding
an electron between these two nuclei, which will
be attracted by both nuclei.
So now why don't you tell me what
you think is going to happen to an electron that
is in this region of constructive interference,
this increased area of probability density.
OK, 10 more seconds.
Interesting.
OK.
So we should have probably not put up the answer there
and re-pole.
Yes.
So here, if we have an electron that's
attracted to both nuclei, then we
want to think about whether that's
going to be lower or higher in energy
than something in an atomic orbital.
So if it's attracted to both, it's
going to be more stably bound to those,
It will be harder to remove that electron, which means
that it's lower down in energy.
And so we should look at that, and we're going to.
So the answer is it should be lower energy, more stable,
harder to remove that electron.
It's in a sweet spot.
It has two positively-charged nuclei,
and it's hanging out right in the middle.
It's very happy.
It's going to be more stable, and that means lower in energy.
All right.
So let's take a look at that.
So the electron is lower in energy,
and the bonding orbital energy is also
going to be decreased compared to the atomic orbitals,
and that has to be true if that's where the electron is.
So let's look at the atomic orbital
from a and the atomic orbital from b.
And now, I'm going to put the bonding orbital at a lower
energy level.
First, I was going to put the electrons on,
and now, I'm going to put the orbital at a lower energy.
So remember, this is increasing energy here,
so the atomic orbitals would be up here,
whereas the molecular orbital is down here.
The molecular bonding orbital will be lower in energy.
And now we'll put our electrons there.
So we have one electron up here and one here,
and so when they come together, we're
going to have two electrons in this molecular orbital.
So when you have these two electrons
and they both occupy the bonding orbital,
and this is the case for H2, each H atom
is bringing one electron, H2 has these two electrons,
and that's going to make H2 more stable.
And we saw that before that to associate the H2 bonds,
you have to add energy into the system.
H2 is more stable than free H plus H.
So when you bring atomic orbitals together
and you have constructive interference, an increased
probability of electrons between those two nuclei,
that's a sweet spot.
Those electrons are going to be very happy there,
and that will result in a more stable, lower energy structure.
That's bonding.
But whenever there is a positive event like this,
there's always a negative event because that's
just how life works.
So we have bonding orbitals, but we also
have antibonding orbitals.
So antibonding arise from the linear combination
of atomic orbitals, LCAO, through
destructive interference.
So here these are going to be destructively interfering,
and that will generate a molecular orbital
that is an antibonding orbital.
So here are our little nuclei again,
and this antibonding orbital is called sigma 1s star.
So we can write an equation for this
as 1sa minus 1sb equals sigma 1s star.
That is an antibonding molecular orbital.
And let's just think about how this kind of shape
arises considering the wavelike properties
of these atomic orbitals.
So I'm going to now move this equation up to the top.
And now, I have my 1sa here, my 1sb here,
and now it's destructive interference.
So we have opposite phase of the wave function.
And we'll put up a wave function there.
Now, the next one has the opposite phase,
so they're going to destructively interfere.
There is overlap over here, but when
you have destructive interference,
then the amplitude is going to decrease.
So now, when we consider this destructive interference
between these two orbitals, you see that you
have what arises between them.
Instead of enhanced probability of finding an electron,
you actually get a node.
So you have decreased amplitude translates
to decreased probability density between these two nuclei-- one
here, one here-- and that results in a node
between the two nuclei.
So in the antibonding orbital, there
is a much lower probability that it
will be in this sweet spot between the two nuclei
that have this nice, positive charge for it's
little negative charge, so there's
pretty much no shot at being in that nice, sweet spot.
And so what that ends up meaning is
that an electron and an antibonding orbital
is pretty much excluded from that internuclear
region, that region between those two nuclei,
and that results in a molecular orbital
that has higher energy than the atomic orbital.
There's just no chance of being in that wonderful spot.
It's really very sad for the poor electron
that has to occupy an antibonding orbital.
So now, let's put this on our energy scale.
So we'll go back to our energy scale.
And we saw before that when we had 1sa and 1sb
and you had a sigma 1s, a bonding orbital,
that was lower in energy.
Electrons that occupy it are more stable compared
to their positions in the atomic orbital, but we also
now have an antibonding orbital from destructive interference
between the wave functions of the atomic orbitals,
and that's higher in energy.
And so that's up here.
So this is what our diagram is going
to look like that brings two atomic orbitals together
to create two molecular orbitals.
So the antibonding orbital up here
is raised in energy by the same amount
that the bonding energy is lowered.
And so that gives rise to this diagram.
And importantly, as I mentioned, we
have N atomic orbitals forming N molecular orbitals.
So if we have two atomic orbitals,
we generate two molecular orbitals,
one is bonding, lower in energy, and one is antibonding, higher
in energy.
All right.
So let's take a look at some examples.
So always start with hydrogen. So hydrogen has how many
electrons?
One hydrogen, hydrogen atom.
AUDIENCE: One.
CATHERINE DRENNAN: One.
So we have two hydrogen atoms, and so we
have two 1s orbitals, 1sa, 1sb.
And now where do we want to put our electrons?
In the highest energy possible, lower energy?
Where do we want to put them?
AUDIENCE: Lower.
CATHERINE DRENNAN: Always start with lower energy.
So we're going to put them down here.
And so this is now the MO diagram for H2.
And we can write the electron configuration
for the MO diagram, which you'll note is a different electron
configuration than you were writing when you were looking
at the periodic table because now we're not writing it
in terms of 1s 2, 2s 2, we're writing it
in terms of molecular orbitals.
And there are some of these on the problem set,
and I tried to indicate example to make
sure you know what kind of electron configurations
we're talking about.
So for MO diagrams, when it says electron configuration
we're talking about where the electrons are with respect
to the molecular orbitals.
So the answer to this would be sigma 1s 2.
There are two electrons in the molecular orbital, sigma 1s.
Good.
So that means that you can do the same thing for dihelium,
and that's a clicker question.
So let's just take 10 more seconds.
OK, you can vote.
All right.
So does someone want to tell me-- that is correct--
what's number two for?
For hydrogen. What about number three, what was wrong there?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: OK, everyone's doing really well.
Yeah.
So all of the electrons were parallel in there,
and what does that violate?
What would be true in that case?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: Yeah.
And so they'd have the same four principle quantum numbers.
This is not allowed.
So that one's not good.
And what's wrong with four?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: Yeah, star was on the bottom.
So they were flipped around.
Antibonding was lower in energy.
Great.
So we'll just put those in over here,
and so we had two electrons.
Helium brings two.
So two went into the bonding, and two
went into the antibonding with opposite spins.
And we have the bonding, lower energy;
antibonding, higher energy.
All right.
So then we can put in the electron configuration here,
and we have sigma 1s 2, sigma 1s star 2,
so that's the electron configuration.
Now interestingly, you see, you have two
at lower energy and two at higher energy,
so there's no net loss or gain in energy of H2
compared to just two helium atoms.
And so that raises the question, does helium 2 exist?
So what would molecular orbital theory
tell you about whether it exists, and it would actually
predict that it does not exist.
And the way that molecular orbital theory gives you
these predictions is through the calculation of something
called bond order.
So bond order is half of the number
of bonding electrons minus the number of antibonding
electrons.
So let's just write out what the bond order for helium would be.
So helium 2, our dihelium molecule.
So we have a bond order equals 1/2.
There's always 1/2.
And now, the number of bonding electrons, so
how many bonding electrons do we have for helium?
AUDIENCE: Two.
CATHERINE DRENNAN: Two.
How many antibonding electrons do we have for dihelium?
AUDIENCE: Two
CATHERINE DRENNAN: Two.
And can someone do this math for me?
AUDIENCE: Zero.
CATHERINE DRENNAN: Zero.
Right.
So that would suggest a bond order
of 0, which means no bond.
And let's just compare that to hydrogen, H2,
which should have a bond order equal to 1/2.
It's always a 1/2.
In the hydrogen, how many bonding electrons did we have?
AUDIENCE: Two.
CATHERINE DRENNAN: We had two.
How many antibonding electrons did we have?
AUDIENCE: Zero.
CATHERINE DRENNAN: Zero.
And the math?
Bond order is?
AUDIENCE: 1.
CATHERINE DRENNAN: 1.
So that means 1 bond or a single bond.
So MO theory would predict that dihelium, has no bond, i.e.
it's not really a molecule without a bond,
but H2 should exist.
So let's take a look at what experiment tells us,
and it does exist, but only really not that much.
So it was not discovered until 1993, which for some of you
might seem like quite a long time ago,
but since we've been mostly talking
about discoveries that were made in the 1800s,
it took a long time before someone could
prove that dihelium existed.
And if we look at the bond association energy for H2,
0.01, and compare that to H2, 432,
dihelium really doesn't exist very much.
0 is a much better approximation of its bond than 1 would be.
It really is not a very good molecule.
So I would call this a win for molecular orbital theory.
It correctly predicts that our helium 2 is not
going to be a very good molecule,
but H2 will be a good molecule, and that works.
All right.
Now, let's consider 2s.
So 2s orbitals are analogous to 1s
except that you have to remember that they're bigger.
So we have our 1s and our 2s, but for the purposes of this,
it doesn't really matter.
So let's look at a diagram now that has 1s and 2s.
So we have lithium.
So how many electrons does lithium have?
AUDIENCE: Three.
CATHERINE DRENNAN: Yep.
So we'll put on lithium, dilithium.
We'll put on lithiums has two in 1s and one in 2s,
and so we have one lithium here and one lithium over here.
And our 1s orbitals are going to be lower energy,
so they're down here.
Our 2s orbitals are higher in energy.
So that goes for both the atomic and the molecular orbitals
that are generated.
So bringing together 1s with 1s, we get sigma 1s and also
sigma star 1s.
And so we can start putting our electrons in.
We're going to start with the lowest energy and move on up.
So we have four electrons that we need to put in,
so we fill up everything here.
And now we go up to our 2s.
The 2s orbitals will generate signal 2s and sigma 2s star.
We have two electrons, so they both can go down here.
So here is what our MO diagram looks like.
We can write out the electron configuration again,
based on this diagram.
So we have two electrons in sigma 1s, so we have a 2 there.
We have two electrons in sigma 1s star, our antibonding
orbital, and we have two electrons in sigma 2s.
And now we can calculate the bond order, which
is another clicker question.
OK, just 10 more seconds.
Excellent.
84%.
So if we do out the math here, we always have our 1/2.
We have four bonding electrons.
We have two down here and two up here.
We have two antibonding electrons in our sigma 1s star,
and so that gives us a bond order of 1.
And in fact, the dissociation energy
does suggest there is a bond.
It's not a great bond.
It's 105 kilojoules per mole, not necessarily enough to power
a starship, but still this molecule does exist.
All right.
So let's look at beryllium now, do another example.
So how many electrons is beryllium going to have?
It will have four.
So we'll put those up.
Now, we're going to start with our lowest energy orbital,
and we're going to put some in the antibonding as well.
And then we'll do this, and then we'll do that.
And we filled up everything.
And so we can write out our configuration.
We have two in our sigma 1s orbital.
We have two in our sigma 1s star.
We have two in our sigma 2s, and two in our sigma 2s star
antibonding orbital there.
And now there are two different ways
that I can calculate the bond order here,
and we'll do both of them and show you that they come out
the same way.
One way that we can calculate this
is to consider all electrons.
So if we consider all electrons, our bond order,
and that's often just said B.O., is 1/2 of our bonding.
And now we can count up both the 1s and the 2s.
So how many bonding electrons do we have?
We should have four, so we have two here, two here, so four.
How many antibonding?
Also four, so that suggest a bond order of 0.
Or we could just consider our valence electrons,
so that would be the electrons in 2s,
and if we do that, bond order equals 1/2.
It's always a 1/2, so how many valence electrons
do we have in bonding orbitals?
Two.
How many in antibonding orbitals?
Two.
And that gives the answer of 0.
So it should always work unless you do something very strange.
You should be able to do it both ways and get the same answer.
So if this is a complicated problem,
you might want to only consider the valence electrons.
In fact, on a test, you may only be
asked to draw the molecular orbital diagrams
for the valence electrons, and you don't
have to do the other ones.
So that should work both ways.
And we get a bond order of 0, and in fact,
the dissociation energy is only 9 kilojoules per mole.
It's a little stronger than dihelium,
but this is an exceedingly weak molecule.
So when you have the same number of electrons and bonding
as antibonding, it doesn't lead to a very strong molecule.
All right.
That's the orbitals.
Now, it's time for molecules that have p orbitals as well.
So in this case, now I have my p orbital and another p orbital.
And we have our nuclei in the middle,
and we're going to bring these together,
and we will have constructive interference.
And so here we're bringing together
2p x of a and 2p xb or 2p ya and 2p yb.
So just x and y we're considering right now.
And if we bring those together with constructive interference,
then we're going to form a bonding orbital that
has enhanced probability density in both cases
and nodal plane along the bond axis
because we had nodal planes along there to begin with.
So if we think about this and we have both of these,
we're bringing them together and they're
going to interfere constructively, enhance
probability density here, enhance probability density
down here, but we still have our nodal plane because we
started out with one.
And so if we have a nodal plane, this cannot be a sigma orbital.
It has to be a pi orbital, because sigma orbitals are not
going to have a nodal plane along the bonding access.
All right.
So we could generate pi 2px or pi 2py this way.
So a pi orbital is a molecular orbital
that has a nodal plane through the bond axis or maybe I
should say along the bond axis, so here is our nodal plane
right through the bond axis.
We can also have antibonding, which
means destructive interference.
So now, I'm going to be subtracting
one of these from one of these, and I'm
going to get something that looks like this,
and it will be pi 2px star or pi 2py star,
and it will have two nodal planes.
So let's look at this.
So this is destructive interference.
I'm subtracting one of these from one of these,
so the phase has to change.
So I'm going to change the phase on one of them
and then bring them together.
Now, we're not going to have that awesome, constructive
interference increased probability density.
These are negatively interacting with each other.
And this generates a nodal plane between the molecules.
They really look much more like this now,
so we have still our nodal plane through the bond.
We had that before.
We're always going to have that.
But now we have an additional nodal plane between the nuclei.
So in one case, we have enhanced density, probability density
again.
And the other case, in antibonding,
we have another nodal plane.
All right.
So now, let's look at what happens
to the energies of these pi orbitals.
And the diagrams that I'm about to show you,
we're only talking about px and py now.
We have for the moment forgotten 2pz.
So these diagrams are rated I for incomplete.
Warning to the viewer, people come to me
and go where are the 2pz?
Yes, they're not in these diagrams,
but when you're asked for a complete diagram,
you will always have to put those orbitals in.
And in fact, completing these diagrams
could be a question you get later.
But for now, we're going to have 2pz here,
but it's not forming a molecular orbital in this diagram.
This diagram is, thus, incomplete,
but we're going to start simple and build more complicated.
So only 2p orbitals first, and then we'll add the third one.
Because this compound doesn't actually need that orbital,
so we're good for now.
All right.
So we have moved on to the first element that has
a p electron in a p orbital.
We have boron, and we have two of them.
So now I dropped off 1s to simplify this.
Now, we just have our valence electrons,
so it's a good thing we can calculate bond order just using
our valence electrons.
All right.
So we have two in 2s and one in 2px or 2py.
I could have put it in either place.
So let's put in where they would go.
So we have 2s orbitals.
They'll go into bonding sigma 2s first, then
into antibonding, sigma 2s star, and now we have p electrons,
and we're going to put them into our pi orbitals,
our molecular orbitals.
I'll put one in, and where am I going to put the other one?
Am I going to put it next to it or over here?
What do you think?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: Yeah.
So we're going to do that because, again, when you're
going to sit on a bus, you want to have if they're degenerate
in energy levels as they are here,
you're always going to put one electron in each orbital
of the same energy first before you pair them up,
and they'll have parallel spins.
So we're reviewing things we learned before.
I love doing that.
So now, let's see what our electron configuration is,
and this is just the valence electron configuration.
We're simplifying.
We're not going to consider the 1s orbital,
and we can write this down.
So we have two electrons in sigma 2s,
and we have two electrons in sigma 2s star, our antibonding
orbital, and we have one in pi 2px and one in pi 2py.
And I can put a 1 or not put a 1.
If I don't put anything, for a 1, 1 is assumed.
And we can calculate our bond order as well.
So we have 1/2, and again, we're just
using our valence electrons, but that's OK.
We have four now who are bonding, two down here, two
up here, these are bonding orbitals,
and we have two that are antibonding.
And notice for our pi orbitals, this
is what we saw before bonding are lower in energy,
antibonding are higher in energy.
With the bonding orbitals, we had constructive interference,
enhanced probability of the electrons near the nuclei,
and so that's lower in energy.
But in our antibonding ones, we have a nodal plane
in between our nuclei.
So we don't have any probability that electrons
are right in between there because there's a nodal plane,
so those are higher in energy.
So here is our B2 diagram.
So now let's try the same thing for carbon, C2,
and that is a clicker question.
All right.
Let's just take 10 more seconds.
So let's take a look at that over here.
The easiest thing to do to answer this question
was to fill in the diagram in your handout.
And so if you did that, you would have put two down here
and you would have put two up here.
Then you would have put one here, one there,
another one there, and another one there.
So now we have used these up, and so our configuration
is sigma 2s 2, sigma star 2, pi 2px 2, pi 2py 2.
And the bond order is 1/2.
There are six bonding electrons-- 1, 2, 3, 4, 5, 6--
and two antibonding electrons, and so that adds up
to a bond order of 2.
And so sometimes on a test, they'll
be a simple question what is the bond order, but to get there
you have to draw your whole molecular orbital diagram
and figure out how many bonding and how many antibonding,
so these are not really that fast questions.
And it's nice.
Sometimes we give you like a little space,
and you see this whole little molecular orbital diagram
fit in there to answer the question.
All right.
So let's just compare these two diagrams for a minute.
So in both cases, we had 2s orbitals, two atomic orbitals
for 2s, and they both generated two molecular orbitals,
a bonding and an antibonding.
The bonding is lower in energy, and the antibonding
is higher in energy.
We also had two 2px atomic orbitals.
They generated 2 pi 2 px orbitals,
one bonding, one antibonding.
And the same for our two atomic orbitals for 2 py.
We had two of those, and they generated one lower energy
bonding, pi, bonding and one pi star antibonding.
So you always have N atomic orbitals generating
N molecular orbitals.
So the stability of the resulting molecules
in these cases depend on how many of the electrons
are bonding, how many are in energy lower
as a result of formation of the molecule,
and how many are at higher energy as a result of formation
of the molecule.
And if the net result are more electrons in lower energy, more
electrons in bonding orbitals, then that molecule
is more stable.
If there's a very slight or no difference,
then that's not a very stable molecule.
So now let's just compare these two
and think about which of these is going to be more stable.
So we have our configurations again.
So in the case of B2, how many electrons
are in lower energy or bonding orbitals?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: Yeah, we have four-- 1, 2, 3, 4.
How many in higher?
AUDIENCE: Two.
CATHERINE DRENNAN: Two.
Up here.
For carbon, we had six-- 1, 2, 3, 4, 5, 6-- two in higher.
And so the bond order here was 1, the bond order here was 2.
Which is more stable?
Higher dissociation energy, which one do you think?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: Carbon.
Right.
Has a bond order of two.
It has more electrons in lower energy orbitals.
So it cannot really well out of this bonding deal,
and the dissociation energy for B2, 289, whereas, for C2, 599.
So when molecules come together such that more of the electrons
are in lower energy or bonding orbitals,
you form a nice, stable molecule.
When molecules come together such that more
of their electrons are an antibonding or higher energy,
that's not a happy molecule.
So I'll just and with one way to think about this.
In this cartoon molecular, break up lines,
sometimes two atoms just have an incompatible number
of valence electrons.
And there are just too many-- just here this molecule saying,
I'm sorry.
Too many of your electrons are in my antibonding regions.
I don't know how many times we've all heard that,
but it's time to dissociate, but our atomic orbitals, well, they
can still be friends.
OK.
See you on Wednesday.
Take a look at the clicker question.
All right.
Let's just take 10 more seconds.
All right.
Let's just go through this one.
So this is a review of where we were last time.
So the correct answer is 1.
So sigma orbitals are cylindrically symmetrical.
Let's quiet down a minute.
You can hear the answer.
So this one is true.
The second one is not true.
A bond order of zero doesn't mean that you just
have antibonding orbitals.
Whenever you bring together two atomic orbitals,
you have to make two molecular orbitals.
So it isn't that sometimes you make bonding orbitals
and sometimes you make antibonding orbitals.
Every time you bring together two atomic orbitals,
you make two molecular orbitals, one that's lower in energy,
and that's the bonding orbital, and one
that's higher in energy, and that's the antibonding orbital.
So that is not what that means.
A bond order of zero means that you
have equal numbers of electrons in your bonding and antibonding
orbitals.
So there's no net stabilization due to the formation
of these bonds.
So here bonding occurs when you bring together
two atomic orbitals to make two molecular orbitals that
are both of lower energy.
No.
Every time you make the two orbitals, one is lower energy,
one is higher that energy.
You can't make two that are both lower in energy.
And bond order of 1 means constructive interference
is generated at one bonding orbital.
That's not what a bond order of 1 is.
And again, every time you generate a bonding orbital,
you generate an antibonding orbital.
And 1 means that you have twice as many electrons
in your bonding orbitals as antibonding orbitals
because the formula is 1/2 the number of bonding
minus antibonding.
But that's good, and it's important to remember
that sigma orbitals are symmetric around the bond axis.
All right.
So we had these diagrams for boron and carbon,
just talking about the interactions of the px and py,
and so we had forgotten about our pz.
And you can't do that on a test.
You get into trouble, so I'm always
telling people for these two handouts,
you must include the molecular orbitals that
are derived from p to z.
So on a test, you need to put them even if they're empty.
Even if they don't have anything in them,
they need to be part of your molecular orbital diagram.
We didn't have them in the diagram
because we hadn't talked about them yet,
so now we're going to talk about them.
So two pz orbitals, again, this is on Monday's handout.
You have this linear combination of atomic orbitals.
And all are p orbitals.
They all look the same as each other,
they're just different in orientation.
You have one along x, one along y, one along z,
but they're the same.
So now we're going to bring our two pz orbitals together,
and we're going to do it along the bonding axis.
So we've defined this as the bonding axis in the class,
so we'll bring them together, and they'll
be constructive interference with our bonding orbitals.
There's always constructive interference that
generates bonding orbitals.
And so we're going to create a enhanced amplitude as the wave
functions come together, and it's
going to be cylindrically symmetric.
So what type of orbital do you think this is going to be,
sigma or pi?
AUDIENCE: Sigma.
CATHERINE DRENNAN: It'll be sigma
because it's cylindrically symmetric.
So we do not have any bonding plane along the bond axis,
and it's symmetric around.
And we have enhanced probability density,
and we have the wave function squared
enhanced probability of having an electron between the two
nuclei.
And so this is a sigma 2pz.
So p orbitals can form sigma molecular orbitals.
So we do have nodes passing through our nuclei.
Here are our nuclei again.
We do have them.
They were here before in our p orbitals.
There's the nodal plane in our p orbitals, but we do not have,
in this case, a node along the bond axis.
So that is our bonding.
So whenever you generate a bonding orbital, which
is going to be lower energy, we're
going to have our increased amplitude between the nuclei,
again, our increased probability density and therefore,
lower energy.
So whenever you have constructive interference
generating a molecular orbital of lower energy,
you got to create something of higher energy.
That's just how life works.
So we also are going to have antibonding
orbitals, which are generated by destructive interference.
And again, these orbitals can be thought
about as wave functions, and a property of waves
is that they constructively interfere and destructively
interfere.
So now we can subtract our two orbitals,
which we're going to switch the sign
and they're going to be out of phase.
So they'll destructively interfere,
and that's going to look like this.
So now, you're going to generate a nodal plane between the two
nuclei, but it's still symmetric around the bond axis.
So this is a sigma 2pz star, so it's an antibonding orbital.
So again, it's sigma, so it's still
cylindrically symmetrical with no nodal plane along that bond
axis, but you do have a new nodal plane that's generated.
So nodes pass, again, through the nuclei,
but also now between these two orbitals.
So we have a new nodal plane that's
generated that's between these nuclei,
and that's a result of destructive interference.
Generates that nodal plane.
You have decreased probability density
for an electron being found there.
And so that poor electron is shut out of that sweet spot.
The electrons like to be between those two nuclei
where they have the two positive charges of the nuclei and then
their little negative charge, and they can sit right there
and be very happy in a low-energy state.
But here there's really lower probability density,
lower likelihood the electron will be found here,
and that generates a molecular orbital
that's antibonding or higher in energy.
All right.
So now, we have to go back to our MO diagrams
and figure out where to put these new molecular orbitals
onto our nice diagrams.
And it's not as simple as it was before because where
we put these new sigma 2pz molecular orbitals depends
on what z is.
So it depends on the value of z.
So if z is less than the magic number of 8,
then we have our pi 2px and 2py orbitals are lower in energy
than our sigma 2pz molecular orbital.
But if we are equal to or greater than 8,
then the sigma 2pz orbital is lower energy
than the pi 2px and 2py.
So less than 8, pi is first.
It's lower in energy as you go up your energy scale.
And if z is equal to or greater than 8, pi is second
and sigma is first.
So how are you going to remember this?
There could be many different ways one can remember it,
but I'll tell you how I remember it.
And my life revolves around my daughter and my dog.
And so at Thanksgiving, we always have the question,
can I eat pie first?
So if you are under the age of 8,
you always want your pie first.
So if z is less than 8, pi comes first.
Pi is lower in energy and sigma is higher.
However, when you mature to the grand age of 9,
say, or 10, if you were a kid, 10
is like the oldest you can possibly
imagine being, very mature.
And you can eat your Thanksgiving dinner
and wait for pie.
So that is how I would remember it,
under 8 pi is first, equal to 8 or greater,
you can wait till after dinner to have your pie,
pi comes second.
Note that the ordering of the antibonding orbitals
is the same, so all you have to remember
is down here depends on z, is pi first or is pi second?
So let's take a look at an example.
Let's look at our friend molecular oxygen that has a z
equal to 8.
So oxygen is at the old, mature age of 8,
and so it's going to have its sigma 2pz first.
It can wait for its pi orbitals until later.
So let's start putting in our electrons.
So we have each oxygen making up molecular oxygen, or O2.
Brings two 2s orbitals to the Thanksgiving dinner table,
and two of them go down in energy
into the bonding sigma 2s orbital and two
go into our antibonding, sigma star 2s orbital.
Now, we have four electrons in our atomic pz
orbitals, four from each molecule,
so we need to put all of those in.
So we always start with the lowest energy orbital.
So we'll put two in there, then we'll go up.
We have two more here, two more here.
We'll put them in singly with their spins parallel,
and then we'll pair them up in the lowest energy orbitals.
And then we have two more left, so we're
going to have to put those up in our antibonding orbitals.
So they go into pi 2px and pi 2py star orbitals up here.
All right.
So now we can calculate the bond order for oxygen,
and that's a clicker question.
All right.
Let's just take 10 more seconds.
All right.
So let's take a look over here.
So bond order, again, is 1/2 the number
of bonding electrons minus the number of antibonding
electrons.
And we have eight bonding-- 1, 2, 3, 4, 5, 6, 7, 8.
And we have four antibonding-- 1, 2, 3, 4.
So that gives us a bond order of 2.
And the bond order equation is one
that you do have to memorize.
That will not be given to you on an exam.
And with a bond order of 2, we have a pretty big number
for dissociation energy.
Again, that's energy you have to put into a bond to break it,
to dissociate it, and that means it's a pretty strong bond
if it's a big number.
If you need a lot of energy, that's a strong bond.
Another thing that you can see from this diagram
is that O2 is a biradical.
It has two lone pair electrons, so two unpaired electrons,
which also makes it paramagnetic,
or attracted to a magnetic field.
So whenever you have unpaired electrons,
you will have a paramagnetic species,
and diamagnetic means they're all paired.
And so there are some questions on problem sets and on exams,
so you need to know the definitions of those.
All right.
So we talked about this when we were doing Lewis structures,
if you recall.
And we drew a beautiful Lewis structure of molecular oxygen
that had two lone pairs on each oxygen and a double bond.
But I told you that that was not really
a complete description of molecular oxygen,
that molecular oxygen was actually a biradical,
but you would not get a hint of that from the Lewis structure.
So if you draw the Lewis structure as a biradical,
then you have a single electron here
and a single electron there and a single bond.
But we also see the bond orders 2,
so this one describes the biradical nature,
but doesn't really describe that double bond
character that it has.
So neither of these Lewis structures really completely
describe molecular oxygen. And we
need our molecular orbital diagram to really help
us understand the properties of molecular oxygen,
that it's pretty strong bond between it,
it has double bond character, but it also is very reactive.
It's a biradical.
So this is just a bizarre molecule.
And really that is why this diagram right here really tells
us why our life, our planet is what it is,
right here, that describes it.
It's because of oxygen that we have
the life forms that we did.
Life was very different on this planet
before molecular oxygen came about.
You did have microbes that lived anaerobically without oxygen.
But when oxygen came, everything changed.
And so if I say, can you explain life to me?
You can draw this, and there it is.
This explains life as we know it because life
as we know it exists because of this crazy molecule that is O2,
nothing else really like it.
It is an amazing molecule that allows us to live.
So there you go.
I was going to say you don't learn anything in chemistry.
This diagram I just explains life
as we know it to you, right there.
All right.
It's not just oxygen, there's a few other elements that, yeah,
are pretty important, and one of them
is nitrogen. We wouldn't really be much anywhere
without nitrogen either.
Oxygen is, no, nitrogen is pretty special too.
This also helps.
Maybe these two molecular orbital
diagrams really sum up life as we know it.
All right.
So let's look at molecular nitrogen, N2.
So we have two electrons in our 2s orbitals,
so we're going to bring them together.
We're going to put two in the lower energy orbital, our sigma
2s, and two in our antibonding orbital.
And then we have three in our p orbitals
from this nitrogen, three electrons from this nitrogen,
and so we'll put them down here.
We'll put in 2, 3, 4, 5, 6.
So we didn't need to use any of these antibonding orbitals,
and this is a z less than 8 case.
So here we have our pi orbitals first
because it is z less than 8.
So let's look at the bond order here.
And the bond order is 1/2 of 8-- 1, 2, 3, 4, 5, 6, 7,
8-- eight bonding electrons and just
these two antibonding electrons, so 1/2 of 8 minus 2 is 3.
And you have a really big number for your dissociation energy,
941.
This is a very stable molecule, and you
can draw the Lewis structure of this without much difficulty.
This one works quite well, and you get a triple bond
and two lone pairs.
So this would be a diamagnetic molecule,
no unpaired electrons, no radicals here.
But this is a crazy, strong bond.
It's really hard to split nitrogen
because of this triple bond.
And molecular orbital theory tells you
it should be its triple bond.
It should be really, really strong interactions
between these nitrogens.
And so this is a very hard thing.
We want to do is industrially, and we'll
talk more about this when we get a chemical equilibrium.
How do you split the nitrogen bond.
This is actually, currently, a big area of research,
how you break that bond because we need nitrogen for life.
And so how do you split it?
There's lots of nitrogen N2 in the atmosphere,
but we need it here, and we need it usable,
so we need to break that bond to use it.
So we'll come back to this in chemical equilibrium
and how people are able to split the nitrogen bond.
So forget man of steel.
We should have man of nitrogen. That's strong.
Nitrogen is strong.
Forget steel.
Man of nitrogen.
One more thing to do before we move on to today's handout
and this is really fast, you can also
be asked to draw molecular orbital diagrams where
both atoms are not the same.
So if you were asked to do that, you
can use the following rules.
If z is less than 8 for both atoms, pi is first.
If z is not less than 8 for both atoms,
it's hard to know what to do, so you
don't need to know what to do.
And that's it.
So we might tell you something about it
and then you can do it, but otherwise, just
worry about things when z is less than 8.
So that's molecular orbital theory.