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PROFESSOR: -- that this is not a quick clicker question, so
you actually need to maybe write something down to figure
out the answer.
So if you haven't clicked in yet, or if you want to change
your answer, keep in mind that you might need to jot down,
for example, a Lewis structure before you can answer this
question.
-- and get in your final answer
to the clicker question.
For those of you just walking in now, you might not have a
chance to get all of the thought process that you need
in on this clicker question, because it is based on a Lewis
structure, so we will go over it.
But for those of you that have been here for 30 seconds or
longer, see if you can get the right answer in here.
All right, so OK, it looks like we have a very mixed
response in terms of the answer to the
clicker question today.
Raise your hand if you didn't have time to figure out the
Lewis structure.
OK, so that accounts for some of you.
Let's go over the correct answer this question.
So, who got the right answer --
31% of us.
This is not the kind of percentages we're looking for,
so let's go over this.
In class on Monday, we did go over the geometries, and the
geometries themselves are very straightforward, once you know
what the Lewis structure is, but remember, you can't just
always look at a molecule and automatically
know the Lewis structure.
We actually need to think about those valence electrons.
So, here we're dealing with selenium hydride, so s e h 2.
I told you that there's 6 valence electrons in selenium,
so we have 6 there, plus 1 for each of the hydrogen.
So what we should have is 8 valence electrons
in our Lewis structure.
How many electrons do we need to have full valence shells?
STUDENT: [INAUDIBLE]
PROFESSOR: 12, that's right.
We need 8 plus 4 is 12 for full shells.
That means that we need 12 minus 8, or 4 bonding
electrons in our structure.
So I tried to give you one that you could draw pretty
quickly, it just has hydrogens in It.
So we have s e h 2.
We can do our 4 bonding electrons.
How many valence electrons do we have left?
STUDENT: [INAUDIBLE]
PROFESSOR: 4, that's right.
So, we need to fill our octet for selenium, so 1, 2, 3, 4.
So, this is our Lewis structure here, hopefully you
can see why it's not linear.
If it were linear, which 32% of you seem have thought, that
would have meant that our Lewis structure had no lone
pairs in it, right, and that's not the case.
We have two lone pairs, so if we thought about what the
bonds were everywhere, it would be 109 .
5, but it's bent because we're only looking at the bonds,
we're not counting the Lewis structures in naming our
geometry, but they do affect the angles.
And it turns out that it's actually less than 109 .
5, because those lone pairs are pushing the bonds even
further away.
So, does this make sense to everyone if you think
about it this way?
Pretty much, okay.
So, we'll try another clicker question like this later.
Some of these questions that look very straightforward just
naming the geometry, you have to remember to do the first
step before you jump in and go ahead with a naming.
So you'll got plenty of practice with this on your
problem-set if you haven't already.
All right.
So before we start in with today's notes, I do want to
mention that this morning the Nobel Prize in
chemistry was announced.
This is an exciting week in science in general because we
got to hear another Nobel Prize every
morning pretty much.
So, let's settle down for a second and start listening.
All right.
So, today we're going to be talking about molecular
orbital theory, but first I wanted to just mention, in
case some of you didn't hear what the Nobel Prize was this
morning, and this was in chemistry, it went to three
different chemists.
Osamu Shimomura, who's a Japanese chemist, and then
Martin Chalfi who's at Columbia, and Robert Chen
who's at U. C. San Diego.
The three of these chemists split the Nobel Prize this
morning for their discovery and/or their application of
using green fluorescent protein, which
is also called GFP.
How many of you have heard of GFP before?
Oh, that's so great.
OK.
Many of you have heard of GFP.
For some of you that haven't I'll just say that it's a
protein, it's 238 amino acids, which means that it's about
1,000, actually more than 1,000 atoms in size, and this
protein is fluorescent.
The protein was first discovered and first isolated
from this jellyfish here.
This was done by Shimomura, and he did this in
the 60's and 70's.
He actually recognized and, of course, many people recognized
that this jellyfish was fluorescent, and he isolated
the actual protein, and determined and proved that
this protein was sufficient to cause this fluorescence.
So, in terms of thinking about applications of why it's so
exciting that you have this fluorescent protein, other
than it's always really fun to look at things that are
fluorescent, we can think about in terms of biology why
it's so exciting, you can actually tag this protein to
any other protein that you're studying and now you have a
visual handle on what's going on.
So, for example, if you were interested in some protein
involved in cancer, you could tag it with GFP.
You could watch where it localizes in the cell, there
are fluorescent assays you could use to determine what
other proteins it interacts with.
You could see, for example, when it's expressed in a
cancer cell, and where it's expressed.
There are all sorts of things you can do once you can
visualize something with fluorescence.
The reason it is so exciting that it's a protein, and it's
a protein, this is the structure here, it's a ribbon
structure so you can kind of see what it looks like, it's
made up of all natural amino acids.
So this means we can code for it in DNA, you don't have to
worry how am I going to get into the cell.
All you have to do is mutate the DNA, which is very
straightforward to do in molecular biology, and now you
can tag absolutely any protein that you're interested in.
So, as I said, this was first discovered and isolated from
the jellyfish.
This was done by Shimomura -- that was kind of the first
step in this process of having it become such a useful tool.
And then, many years later, not until 1994 did Martin
Chalfi, at Columbia, show that yes, I can, in fact, take the
DNA and put it into a different organism, and he put
it into e coli, a bacteria.
And what he could show was that it could be expressed,
this is a picture from his 1994 science paper, in that e
coli, and it is going to fluoresce green.
Now, the first application tends to not be quite as
exciting as, for example, all the other organisms people
have put it in since then -- you can have flies, you see
transgenic mice that are glowing green with this GFP.
Of course, that's not the useful application for it,
it's more of a proof of principle, but it does show
you that you can put it in for studies in any organisms.
And the field was really pushed forward by the
discoveries of Robert Chen at UC San Diego, and what he did
was he actually figured out how it was that this protein
fluoresced, what caused the actual fluorescence.
And once he did that, both he and many other scientists,
could then, once they understood what caused the
fluorescence, make little changes to the actual protein,
and tune what the properties of that
fluorescent protein were.
So now, for example, there are a whole range, just a whole
rainbow of fluorescent proteins that can be used.
And I'm sure you can imagine that if you want to label onw
protein green and one red and one yellow, now you can start
looking at really complex biological processes.
So, it's pretty rare that chemistry makes
the every day news.
So hopefully you'll all look in the normal papers today,
not just the scientific journals, and get to read
something about chemistry.
It's always fun to see how its described in The New York
Times or in The Boston Globe.
The other thing I wanted to mention and I'm not sure if
the exhibit's still there.
But there was an exhibit of jellyfish, I know at least
until last year, at the Boston Museum of Science, and all of
you guys can get in there free.
And it's neat to see the glowing jellyfish and think
about the fluorescent protein that's in them.
So, I encourage you to do that the next time you have some
free time on your hands, maybe at IAP or some time like that.
All right, so let's move into today's notes.
Today we're talking about molecular orbital theory.
This is a shift, this is a new topic that we're starting.
So far we've exclusively been using Lewis structures any
time we've tried to describe bonding within molecules.
Lewis structures are really useful, we use them all the
time in chemistry.
And they're useful because, first of all, they're easy to
depict, they're easy to draw -- relatively easy once we get
all the rules down.
And also they're accurate over 90% of the time.
But they're not accurate all the time in predicting bonding
within molecules, and the reason for this is because
Lewis structures are not, in fact,
based on quantum mechanics.
So, molecular orbital theory, on the other hand, is based on
quantum mechanics.
And specifically, MO theory is the quantum mechanical
description of wave functions within molecules.
So, saying wave functions within molecules might sound a
little confusing, but remember we spent a lot of time talking
about wave functions within atoms, and we know how to
describe that, we know that a wave function just means an
atomic orbital.
It's the same thing with molecules -- a molecular wave
function just means a molecular orbital.
So, we'll start today talking about the two kinds of
molecular orbitals, we can talk about bonding or
anti-bonding orbitals.
Then we're going to actually use MO theory to describe
bonding within these molecules, and we'll start
with homonuclear diatomic molecules.
Diatomic mean it's di atomic, it's made up of two atoms, and
homonuclear means that those two are the same atoms. Then
at the end, we'll look at an example with a heteronuclear
diatomic molecules.
So, again, the same thing, but now two different atoms.
So, I will point out, in terms of MO theory, because it
rigorously does take into account quantum mechanics, it
starts to become complicated once we go
beyond diatomic molecules.
So we're going to limit in our discussion in 511-1 for
molecular orbital theory to diatomic molecules.
However, on Friday we will use a different approach so we can
talk about bonding within atoms that have more than two
atoms, molecules with more than two atoms.
All right, so one thing that I first want to point out about
MO theory that is a big difference from Lewis
structures, is that in MO theory valence electrons are
de-localized over the entire molecule.
So, when we talked about Lewis structures, we actually
assigned electrons to individual atoms or to
individual bonds.
Whereas in molecular orbital theory, what I'm telling you
is instead we understand that the electrons are spread all
over the molecule, they're not just associated with a single
atom or a single bond.
So specifically, what we do associate them instead is
within molecular orbitals, and what we say is that they can
be either in bonding or anti-bonding orbitals.
And again, I want to point out that a molecular orbital, we
can also call that a wave function,
they're the same thing.
And these orbitals arise from the combination of individual
atomic orbital.
So, if we have two atomic orbitals coming together from
two different atoms and they combine, what we end up
forming is a molecular orbital.
The reason that we can talk about this is remember that
we're talking about wave functions, we're talking about
waves, so we can have constructive interference in
which two different orbitals can constructively interfere,
we can also have destructive interference.
So, we'll start by taking a look at constructive
interference, and another way to explain this is just to say
again, molecular orbitals are a linear
combination of atomic orbitals.
So, let's start our discussion of a bonding orbital.
Our simplest case that we can look at would be if we had two
1 s orbitals coming together.
So let's say, for example, in a hydrogen atom.
So in hydrogen atom a, I'll depict that here where the
nucleus is this dot, and then the circle is what I'm
depicting as the wave function.
It makes sense to draw the wave function as a circle,
because we do know that 1 s orbitals
are spherically symmetric.
So, we can say that a circle is a good approximation for a
1 s wave function.
Similarly, with the second hydrogen atom, we've got the
nucleus in the middle, and the 1 s b wave function around it.
So these are atomic orbitals.
What we're going to do in forming a molecule is just
bring these two orbitals close together such that now we have
their nucleus, the two nuclei, at a distance apart that's
equal to the bond length.
And what we end up forming is a molecular orbital, because
as we bring these two atomic orbitals close together, the
part between them, that wave function, constructively
interferes such that in our molecular orbital, we actually
have a lot of wave function in between the two nuclei.
So we can go ahead and name our molecular orbital, just
like we know how to name our atomic orbitals.
And I'm going to name this sigma 1 s.
The 1 s just comes from the fact that the molecular
orbital is a combination of two 1 s atomic orbitals.
And the sigma tells us something about the symmetry
of this molecular orbital, specifically that it's
cylindrically symmetric about the bond axis.
So that is the bond axis -- it's just the axis between the
two nuclei.
Sometimes it's also called the internuclear axis.
So any time you have two atoms bonding, the bond axis is just
the axis that they're bonding along.
Another thing I want to point out about every sigma orbital
that you see, and it will make more sense when we contrast it
with pi orbitals later.
But in sigma orbitals, you have no nodal planes along the
bond axis, so if we had a nodal plane here, we'd see an
area where the wave function was equal to zero.
We don't see that.
It will make more sense when we can show you one where it
does have that area.
But keep in mind sigma orbitals have no nodal planes
along the bond axis.
All right.
So, let's look at this in another way, sometimes it's
hard to picture these waves combining.
So let's think of them a little bit more by graphing
the amplitude of the wave, and seeing how we can have this
constructive interference.
So again, if we think of a graph of the wave function, we
had the wave function is at its highest amplitude when
it's lined up with the nucleus, and then as we got
further away from the nucleus, the amplitude of the wave
function ends up tapering off until -- it never hits zero
exactly, but it goes down very low.
So we can draw that for 1 s a, we can also draw it for 1 s b,
and what I'm saying for the molecular wave function is
that we have the interference between the two, and we have a
constructive interference, so we end up adding these two
wave functions together.
So, we're talking about wave functions and we know that
means orbitals, but this is -- probably the better way to
think about is the physical
interpretation of the wave function.
So what is the wave function squared going to be equal to?
STUDENT: [INAUDIBLE]
PROFESSOR: Probability density, yes.
Probability density of finding an electron within that
molecule in some given volume.
So let's think about that instead, let's think about
probability density.
So if we're talking about probability density that's the
wave function squared.
But now we're talking not about an atomic wave function,
we're talking about a molecular wave function.
So to talk about it's squared, we're going to say it's sigma
1 s squared.
Oh, and actually before you skip your page in the notes, I
realized I should write out for you what the addition is
to start with.
So, when we're combining two waves, what we have is 1 s a
that we're adding together with 1 s for
b, the second atom.
And what we end up for our molecular wave
function is sigma 1 s.
So this is what we call our molecular orbital.
All right, so that will now allow you to turn the page, I
think, and we can take a look at the probability.
So the probability again, that's just the orbital
squared, the wave function squared.
So when we write that out, we just write sigma 1 s squared,
or we can break it up into its individual parts, there's no
reason we can't do that as well.
So just to say that it's 1 s squared plus 1 s b, all of
that together squared.
So if we write out every term individually, what we end up
with is essentially just the probability density for the
first atom, then the probability density for the
second atom, and then we have this last term here, and this
is what ends up being the interference term.
So in this case where we're adding it together, this is
going to be constructive interference.
So in this case the cross term represents constructive
interference between the two 1 s atomic wave functions.
And this again is what we're going to
call a bonding orbital.
So it can often make a lot more sense if we think about
things in terms of energy.
We've been discussing energy diagrams a lot in this class,
it's a very good way to visualize exactly
what's going on.
So, let's think of the energy of interaction when we're
comparing atomic orbitals to molecular bonding orbitals.
And what you find is when you have a bonding orbital, the
energy decreases compared to the atomic orbitals.
So you can see here in this slide we have the atomic
orbitals for the two hydrogen atoms, each of them have one
electron in them, hydrogen has one electron in a 1 s orbital.
So, if we look at the molecular orbital, that's
actually going to be lower in energy than either of the two
atomic orbitals.
So it's going to be favorable for the electrons instead to
go to that lower energy state and be within
the molecular orbital.
So, let's draw in our electrons there, so we have
our two electrons now in the molecular orbital.
So any time that you're drawing these molecular
orbital diagrams, you want to keep in mind that the number
of electrons that you have in atomic orbitals, you need to
add those together and put that many
electrons into your molecule.
Right, we had one from each atom, so that means we need a
total of two in our molecular orbital.
And what you can see directly from looking at this energy
level diagram, is that the molecule that we have is now
more stable in the individual atoms. That makes sense
because we're lower in energy, the electrons are
now lower in energy.
So that's the idea of a bonding molecular orbital.
Since we're talking about wave functions, since we're talking
about the properties of waves, we don't only have
constructive interference, we can also imagine a case where
we would have destructive interference.
Just like we see destructive interference with water waves
or with light waves, we can also see destructive
interference with orbitals.
So, let's think about what that would look like.
So in this case we would have 1 s a and 1 s b, and instead
we would subtract one from the other, and what we would see
is that instead of having additional, more wave function
in the middle here, we've actually cancelled out the
wave function and we end up with a node.
So we can also name this orbital, and this orbital
we're going to call sigma 1 s star.
So if we name this orbital, this is an anti-bonding
molecular orbital.
So we had bonding and now we're talking about
anti-bonding.
When we talk about anti-bonding, essentially
we're taking 1 s a and now we're subtracting 1 s b, and
what we end up with again is sigma 1 s, and the important
thing to remember is to write this star here.
So any time you see a star that means an
anti-bonding orbital.
Again we can look at this in terms of thinking about a
picture this way, in terms of drawing the wave function out
on an axis.
So we have 1 s a, and we're drawing this as having a
positive amplitude, but since we have destructive
interference we're going to draw 1 s b as having the
opposite sign, so we have a plus and a
minus in terms of signs.
So that should make it very easy to picture that this is
being cancelled out in the middle.
If we overlay what the actual molecular orbital is on top of
it, what you see is that in the center you end up
cancelling out the wave function entirely.
So this is the 1 s star, sigma 1 s star orbital, and what you
have in the center here is a node, right in the center
between the two nuclei.
So again if we look at this in terms of its physical
interpretation or probability density, what we need to do is
square the wave function.
So if we square sigma 1 s star, we flip the amplitude so
it's all positive now, but again we still have this node
right in the middle.
So if we talk about the probability density and we
write that in, it's going to be sigma 1 s star squared, so
now we're talking about 1 s a minus 1 s b, all
of that being squared.
And again, if we write out what all the terms are, we
again have 1 s a squared plus 1 s b squared, but now what
we're doing is we're actually subtracting the
interference term.
So if we're subtracting the interference term, what we
have here now is destructive interference.
So let's think about the energy of interaction here.
When we were talking about constructive interference, we
had more electron density in between the 2 nuclei.
So that lowered the energy of the molecular orbital.
So, bonding orbitals are down here.
But when we think about where anti-bonding orbitals should
be, it should be higher in energy.
It's increased compared to the atomic orbitals.
So we would label our anti-bonding orbital higher in
energy than our 1 s atomic orbitals.
So, let's think a little bit about what this means.
First of all, again to repeat, any time we see the star,
sigma 1 s star, that's an anti-bonding orbital.
When we talk about that, basically what we're saying,
and you can see that because of that negative interference,
we actually have less electron density between the nuclei
than we did when they were two separate atoms. So you can see
that this is non-bonding, this is even worse than
non-bonding, it's anti-bonding, because we're
actually getting rid of electron density between the
two nuclei.
And we know that it's electron density between the nuclei
that holds two atoms together in a bond.
So what I want to point out is that it creates an effect that
is exactly opposite of a bond.
You might have thought before we started talking about
molecular orbital theory that non-bonding was the opposite
of bonding, it's not, anti-bonding is the opposite
of bonding, and anti-bonding is not non-bonding.
We can see that if we just look at this picture here.
Here is bonding, and here is non-bonding.
Anti-bonding is even higher in energy than non-bonding.
And the other thing to point out is that the energy that an
anti-bonding orbital is raised by, is the same amount as a
bonding orbital is lowered by.
So any time I draw these molecular orbitals, I do my
best, and I'm not always perfect, yet trying to make
this energy different exactly the same for the anti-bonding
orbital being raised, versus the bonding
orbital being lowered.
So those should be raised and lowered by the same energy.
So now let's take a look at some of -- is there
a question up there?
STUDENT: Why would two atoms decide to [INAUDIBLE].
PROFESSOR: Well, they don't want to.
It's a higher energy situation.
So we'll start to look at molecules and we'll see if we
take two atoms and we fill in our molecular orbital and it
turns out that they have more anti-bonding orbitals than
bonding, that's -- a diatomic molecule we'll never see.
So it helps us predict, will we see this, for example, h 2,
which we're going to be about to do, we'll see is stabilized
because it has more bonding than anti-bonding.
So we'll predict, yes, there's a bond here.
That's a really good question.
So, let's go ahead and do this and take a look at some of the
actual atoms that we can think about and think
about them in molecules.
So our simplest case that we started talking about was
molecular hydrogen.
I want to finish this discussion by including the
anti-bonding orbital, and this is a tip for you when you're
drawing your molecular orbital diagrams, any time you draw a
bonding orbital, there is also an anti-bonding
orbital that exists.
It might not have any electrons in it, but it still
exists, so you need to draw these into your molecular
orbital diagram.
So I wanted to make sure you have a complete set for
hydrogen in your notes.
So let's take a look at this.
Hydrogen, we can first draw in our atomic electrons.
So there's one electron in each hydrogen atom.
And then this means we'll have a total of two electrons in
our hydrogen molecule, so we can fill both of those into
the sigma 1 s orbital, the bonding orbital.
We don't have to put anything into the anti-bonding orbital,
so that's great.
What we've seen is we have a net lowering of energy of the
molecule versus the individual atoms.
So let's draw the electron configuration of hydrogen, the
molecule, molecular hydrogen.
What you saw, what we've done a lot of is drawing the
electron configurations for different atoms, we can do the
same thing for a molecule.
So, if we take h 2, and we want to draw the electron
configuration, it's very short.
All it is sigma 1 s, and then we have two electrons in it,
so it's sigma 1 s squared.
So this is our electron configuration.
Let's take a look at another example.
Let's draw the molecular diagram for h e 2 now.
So again, we can fill in our atomic orbitals here, there's
going to be two electrons in each of our atomic orbitals.
So now let's go ahead and fill in our molecular orbitals.
We need to fill in a total of four electrons.
So we have two electrons in our bonding orbital, but
because we use the same rules to fill up molecular orbitals
as we do atomic orbitals, so the Pauli exclusion principle
tells us we can't have more than two electrons per
orbital, so we have to go up to our
anti-bonding orbital here.
So this means that we have two of the electrons are lowered
in energy, but two are raised in energy.
So would this be a stabilized molecule then?
STUDENT: [INAUDIBLE]
PROFESSOR: No.
So, compared to the atoms, it should be somewhat the same
energy, we shouldn't get any extra stabilization from
forming the molecule.
So lets go ahead and write what the electron
configuration would be of h e 2.
And again, we're just filling in the different orbitals, so
we have sigma 1 s, that's going to be squared, and now
we have sigma 1 s star squared.
So we can compare the two electron configurations, and
we can actually think about -- what we figure out from them,
we see that two are lowered in energy, two electrons are
raised in energy, so we have no net gain or no net loss in
energy for h e 2.
And there's actually a way that we can make predictions
here, and what I'll tell you is molecular orbital theory
predicts that h e 2 does not exist because it's not
stabilized in terms of forming the molecule.
The way that we can figure this out is using something
called bond order, and bond order is equal to 1/2 times
the number of bonding electrons, minus the number of
anti-bonding electrons.
And the bond order you get out will either be, for example,
zero, which would mean that you have no bond, or you could
have 1, a single bond, 1 .
5, a 1 and 1/2 bond, 2, a double bond, and so on.
So let's figure out the bond order for our two molecules
here that we figured out the electron configuration for.
So I guess we'll start with helium 2.
So the bond order is going to be equal to 1/2, and then it
will be 2 minus 2.
So our bond order for h e 2 is going to be equal to 0.
So it has a 0 bond, there's no bond in h e 2.
Let's look for hydrogen.
For hydrogen our bond order is going to equal 1/2, 2 minus 0.
So we would predict a bond order of 1.
What kind of a bond is a bond order of 1?
Yeah, we'd expect to see a single bond in hydrogen.
So what actually turns out the reality is that h e 2 does
exist, but it exists as the weakest chemical bond known,
and it wasn't, in fact, even found to exist until 1993, so
I can assure you this is not a bond that you see very often
in nature, and it is a very, very weak bond.
It only has a dissociation energy of 0 .
1 kilojoules per mole.
So that should make sense, because we saw no energy
difference between the actual atoms and the molecules.
Molecular orbital theory, even at this very basic level,
allowed us to predict that no, we're not going to see a true
bond here, a strong bond.
In contrast, the dissociation energy of a bond for hydrogen,
and molecular hydrogen is everywhere around us, we see
432 kilojoules per mole.
All right, so we can now see a little bit of what the power
of molecular orbital theory is in predicting what kind of
bonds we're going to see in molecules, or whether or not
we'll see this bonding occur at all.
So let's look at another example, let's take lithium 2
and see what we can figure out here.
In lithium 2, we have two atoms of lithium, each have
three electrons in them.
So now we have to include both the 1 s orbitals and also the
2 s orbitals.
So any time in a molecular orbital diagram you draw in
orbitals, you need to draw the
corresponding molecular orbitals.
So, this means we need to have sigma 1 s, sigma 1 s star, and
now sigma 2 s and sigma 2 s star.
Something I'll also point out as you see these dashed line
that tell you where the individual molecular orbitals
are arising from, as you get to higher and higher atomic
numbers of molecules that you're making, it makes a lot
more sense to look at a diagram when you draw these
dotted lines in, because they can start to get
a little bit confusing.
So when you go ahead and draw these on your problem-sets or
on your exams, it's a good idea to put these dashed lines
in, both for you and for people reading it to see
exactly where your molecular orbitals are coming from.
So, this means we have a total of six electrons that we need
to put into molecular orbitals.
So again we just start filling those up -- we have two in the
1 s, two in the sigma 1 s star, and then we have two in
the sigma 2 s.
So we should be able to also calculate the bond order, just
like we did for hydrogen and helium.
First we can do that by knowing the electron
configuration, we can write it out just by going up the table
here, up the energy levels.
And you can go ahead and tell me what you think the bond
order is going to be for this molecule.
All right.
Let's take 10 more seconds on this.
OK, good, we're back on track a little bit
with our clicker answers.
So it's selection three or one is the bond order.
So let's switch back to our class notes and look at what
this means.
So we know that it's 1, because we have 1, 2, 3, 4
bonding, minus 2 anti-bonding, and 1/2 of that is a bond
order of 1.
We would predict to see a single bond between lithium,
and it turns out that's what we see.
And so you have for a reference, the dissociation
energy of lithium 2 is 105 kilojoules per mole.
So let's keep moving along the periodic table and keep
applying our molecular orbitals.
For example now, with b e 2, so beryllium 2 has four
electrons in terms of each atom.
So you can start by filling those in, and now we can fill
in our molecular orbitals as well.
This means we need a total of eight electrons in our
molecular orbitals.
So we have two in 1 s, two in the sigma 1 s star, two in the
sigma 2 s, and two in the sigma 2 s star.
So let's go ahead and figure out the bonding order for
beryllium here.
So when we figure it out for beryllium -- let's see if I
wrote in your notes what the actual -- electron
configuration, OK that's already in your notes for you.
So let's go right to the bond order for beryllium.
So for the bond order we want to take 1/2 of the total
number of bonding electrons, so that's going to be 4 minus
anti-bonding is 4, so we end up getting a bond order that's
equal to 0.
So what I want to point out with this case in beryllium is
that you don't have to use all of the electrons to figure out
the bond order, and in fact, once you get to molecules that
are from atoms with atomic numbers of 8 or 10, you're not
going to want to maybe draw out the full
molecular orbital diagram.
So what I want to tell you is we also always get the same
bond order if we instead only deal
with the valence electrons.
So let's just prove that to ourselves and figure out the
bond order just using valence electrons.
So this would mean the bond order is equal to 1/2, and in
terms of valence electrons, how many bonding valence
electrons do we have?
STUDENT: [INAUDIBLE]
PROFESSOR: All right, what about anti-bonding?
STUDENT: [INAUDIBLE]
PROFESSOR: Two.
OK, good.
So again, we're going to see that we have a
bonding order of 0.
So we would not predict to see a b e 2 bond.
So what we see is a bond order of 0, and again, the bond is
very, very weak.
Essentially we're not going to see this, it's 9
kilojoules per mole.
All right.
So, so far we've looked only at molecules that involve
atoms that have only s orbitals in them.
I'm sure you're thinking well, what do we do in the case of p
orbitals, and, in fact, we can do the same thing.
Again, we're going to take the linear combination of those p
atomic orbitals and make what are called pi or some more
sigma molecular orbitals.
So let's look at the first case where we have either the
2 p x or 2 p y type of orbitals that we're combining.
So they're the same shape, this is the shape of the
orbital or the shape of the wave function, and we can call
this either 2 p x a being combined with 2 p x b, or we
could say since it's the same shape, it's 2 p y a being
combined with 2 p y b.
And in either case if we first talk about constructive
interference, what again we're going to see is that where
these two orbitals come together, we're going to see
increased wave function in that area, so we saw
constructive interference.
So again, we can name these molecular orbitals and these
we're going to call -- also to point out there is now a bond
axis along this nodal plane, which is something we didn't
see before when we were combining the s orbitals.
So when we go ahead and name these, we're going to call
these pi orbitals.
We'll call it either pi 2 p x, if we're combining the x
orbitals, or pi 2 p y.
The reason that I wanted to point out this nodal plane
here is because this is why it is called a pi orbital.
Pi orbitals are a molecular orbital that have a nodal
plane through the bond axis.
Remember this is our bond axis here, and you can see there is
this area where the wave function is equal to zero all
along that plane, that's a nodal plane.
So that's why these are pi orbitals
instead of sigma orbitals.
So again, we can think about the probability density in
terms of squaring the wave function.
So now what it is that we're squaring is if we're talking
about x orbital, it's pi 2 p x squared, and this is just
equal to the 2 p x a plus the 2 p x b all squared, or if we
write out all of the terms we have 2 p x a squared plus 2 p
x b squared, and then this term here, and again, this is
our interference term.
In this case is it constructive or destructive
interference?
STUDENT: [INAUDIBLE]
PROFESSOR: Constructive interference.
We're seeing that the wave function's adding together and
giving us more wave function in the center here.
All right.
So we see constructive interference, of course, we
can also see destructive interference.
So I changed the colors here to show that these are 2 p
orbitals with an opposite phase or an opposite sign.
So what happens when we add a 2 p a and we subtract from it
a 2 p x b, or the same with a 2 p y a subtracting a 2 p y b,
is that we're actually going to cancel out the wave
function in the center, so we now have 2 nodal planes.
So again, this is an anti-bonding orbital, and what
you see is that there is now less electron density between
the two nuclei than there was when you had non-bonding.
So we're going to call this the sigma 2 p x star, or if
we're talking about the 2 p y orbitals we'll call this the
pi 2 p x star, and the pi 2 p y star.
And the pi star orbitals result from any time you have
destructive interference from 2 p orbitals that are either
the p x or the p y.
So now we can move on to an example where we do, in fact,
have to use some p orbitals, so this would be b 2.
How many electrons are in boron?
Five.
I see some hands going up.
There's five electrons So what you'll notice here is that I
only filled in 3 electrons.
What do these electrons represent?
STUDENT: Valence.
PROFESSOR: Valence electrons.
OK, sometimes you're going to be asked to draw a molecular
orbital diagram where you're asked to include all
electrons, and sometimes it will specifically say only
include valence electrons.
That happens because of space issues that you were asked to
do that, because you can always assume that all of the
core orbitals are already going to be filled.
So, in this case, we're just drawing the molecular orbital
diagram for the valence electrons, so we
have three for each.
And what we see here is now when we're combining the p, we
have our 2 p x and our 2 p y orbitals that are lower in
energy, and then our pi anti-bonding orbitals that are
higher in energy.
You might be asking where the 2 p z orbital is and we'll get
to that soon once we need it.
Let's just first fill in this for the b 2 case.
So we can start at the bottom, two electrons in sigma 2 s,
two electrons in sigma 2 s star.
Now we need to jump up to using these pi orbitals, and
what we're going to do is put one electron into each of our
pi 2 p x and 2 p y orbitals.
So again you can see as we're filling up our molecular
orbitals, we're using the exact same principle we used
to fill up atomic orbitals.
So let's think about what the valence electron
configuration is here.
So now we're looking at the case of b 2.
And what we're looking at is a valence electron molecular
orbital diagram, so let's just draw the electron
configuration for the valence orbitals, so that will be
sigma 2 s 2, sigma 2 s star 2, and now we start in with our
pi 2 p x 1, and our pi 2 p y 1.
So this is our valence electron
configuration for b 2.
All right.
So what would you expect the bonding order for b 2 to be?
Shout it out if you know.
STUDENT: one.
PROFESSOR: And I didn't write up there but it is one, and we
can see that it's 1, because it's 1/2 of 2, 4 minus 2, so
1/2 of 2, the bonding order is going to be equal to one.
So let's move on to another example, let's talk about
carbon here.
Again, we're just talking about the valence electrons.
So carbon has four valence electrons, so if we talk about
c 2, again we're going to start filling in our molecular
orbitals, and now we're going to have eight electrons to
fill into our molecular orbitals.
So, we'll put two in the sigma 2 s, two in the sigma 2 s
star, and now we're going to fill one and one into each of
our pi 2 p x and 2 p y, but we still have two electrons left,
so what we're going to do is double up in terms of our 2 p
x and our 2 p y.
So let's think about what this valence electron configuration
is for c 2.
And again, I want you to have practiced drawing these out in
the form -- you always need to start with the sigma and then
write the number of the orbital.
So, sigma 2 s has two electrons, sigma 2 s star with
two electrons, and now we have sigma 2 p x -- how many
electrons here?
STUDENT: Two.
PROFESSOR: Two.
And sigma 2 p y, two electrons here.
Oh excuse me, pi 2 p y, thank you -- pi 2 p x and pi 2 p y.
So let's talk about what the bonding order is
going to be for c 2.
So what's the bonding order for c 2?
STUDENT: Two.
PROFESSOR: two?
OK, so we have 2, 4, 6 minus 2, so we have 1/2 of 6 minus
2, so that's 1/2 half 4, so we have a bonding order of two
for carbon 2.
So we would expect to see a double bond for a c 2 where we
would expect to see a -- double bond for c 2 and a
single bond for b 2.
And that is, in fact, what we can surmise if we look at the
different dissociation energies for the two bonds.
So for b 2, which is a single bond, that's 289 kilojoules
per mole to break it, and it takes us more energy to break
this double bond for carbon, which is 599
kilojoules per mole.
So, in general what we see, and this is always true if
we're comparing the same atom, and in general, if we're
comparing different types of molecules, but we know that a
single bond is always weaker than a double bond, which is
weaker than a triple bond.
And obviously, no bond is the weakest of all is not bonding.
All right.
So let's look now at the case where we do have 2 p z
orbitals that we're talking about.
So again, what we're talking about is the linear
combination of atomic 2 p orbitals, and now we're
talking about 2 p z.
So if we have constructive interference between the two,
what we're going to see is our molecular orbital looks
something like this.
Do you predict that this will be a sigma or a pi orbital?
STUDENT: [INAUDIBLE]
PROFESSOR: All right, I'm hearing a little of both, but
I'm very encouraged to hear quite a few
people saying sigma.
This is, in fact, a sigma 2 p z orbital is what
this orbital is called.
The reason that it's sigma is if you look at the bonding
axis here, is that there is no nodal plane
along the bonding axis.
Also, it is cylindrically symmetric around the bonding
axis, so this is how we know that it's a sigma orbital.
So some p orbitals form pi molecular orbitals, and some
form sigma p orbitals.
Specifically, it's always the z that forms the sigma
orbital, and the reason is at least at a minimum for this
class we always define the internuclear axis as the z
axis, so this is always the z axis, so it's always going to
be the 2 p z's that are coming together head-on.
The reason that there is increased electron density
here is you can see that these two orbitals come together and
constructively interfere.
We can also talk about anti-bonding orbitals where we
have destructive interference.
So instead, these would be canceling out wave functions
between the two, so we would end up with a nodal plane down
the center.
Would this be a sigma or a pi?
It's still sigma.
So even though we see a nodal plane down the center, I just
want to really point out that it's only when we have a nodal
plane in the internuclear or the bond axis that we're
calling that a pi orbital.
So this is still a sigma -- it's a
sigma 2 p z star orbital.
We have destructive interference here.
So now let's look at an example where we talk about
using these 2 p z orbitals, so let's look at oxygen.
So the first thing I want to point out is that the 2 p z,
the sigma 2 p z is even lower in energy than the pi orbitals
here, and the anti-bonding sigma orbital is going to be
higher in energy than the pi 2 p orbitals that we have here.
So in oxygen again, this is just showing the valence
electrons, so we end up having six valence electrons from
each oxygen atom.
We can fill right up our table just like we did before, but
now we have included our 2 p z orbital here.
So we have two electrons in sigma 2 s, two in sigma 2 s
star, in sigma 2 p z we have two -- those are filled first,
and now we're going to put one into pi 2 p x, and
one into pi 2 p y.
So we can write out what the electron configuration is
here, and I think that I have already written that out for
you in your notes.
What is the bond order of o 2?
It's two.
Oh excuse me, I didn't fill in all of my electrons.
All right.
So we had a total of 2, 4, 6, 8.
So we had a total of 12.
So we filled in four here -- we need to keep going.
All right, so I did this not at all purposely, but this can
point out for you that you need to make sure that the
number of electrons that you have in your molecular orbital
does match up with the total number that you have in your
atomic orbitals.
So I did not do any counting up here, you should make sure
you do counting, I apologize.
So we need to fill all the way up to the pi 2 p x,
and the pi 2 p y.
All right, so the bonding order, you're correct, should
be 2, if we subtract the number of bonding minus
anti-bonding electrons and take that in 1/2.
And what I want to point out that we just figured out for
molecular orbital theory, is that o 2 is a biradical,
because remember, the definition of a radical is
when we have an unpaired electron.
You can see that we have two unpaired electrons in this
molecule here -- one in the pi 2 p x star, and one in the pi
2 p y star orbital.
This was something we could not predict using Lewis
structures, but we can predict using MO theory that we have a
radical species here.
So that's a really important type of an application that we
can use MO theory for that we weren't able to do with our
Lewis structures.
All right, I want to do one more at homonuclear example
here, and this is n 2.
The first thing that I need to point out is you can actually
see an n 2 versus o 2 that we flip-flopped the energy of the
sigma and the pi 2 p orbitals.
So this is a glitch, just like sometimes we had glitches in
filling up our atomic orbitals.
This is something that you need to remember.
And what you need to remember is if the z is equal to eight
or greater, such as oxygen being the cut-off point, this
sigma 2 p orbital is actually lower in energy than the pi 2
p orbitals, the molecular orbitals.
But for anything 7 or less, so what is the
atomic number for nitrogen?
STUDENT: Five.
PROFESSOR: five -- there's five valence electrons, but
the atomic number is actually seven.
So z equals 7 -- this is the cut-off where, in fact, the
sigma orbital is going to be higher in energy than the pi 2
p orbitals.
This is something that you just need to remember.
I wrote it down your notes, if you can put a big star next to
it so you don't forget this.
This is something you're going to be responsible for in
drawing out your molecular orbitals.
So let's fill it out in this way, keeping in mind that
we're going to fill out the pi 2 p's before the sigma.
So we have a total of 2, 4, 6, 8, 10 valence electrons, so
I'll make sure I count to 10 as we fill up our molecular
orbitals here.
We have two, then we have four.
Now we're going to start in with that pi 2 p orbitals,
which gives us 1 each, and then two each in those, and
then after that, we'll go up to our sigma 2 p z orbital.
So this is going to be our molecular orbital diagram.
And again, I've written for you, but you can figure out
what the electron configuration is just by
writing up in this order here.
And what is the bond order going to be n 2?
STUDENT: [INAUDIBLE]
PROFESSOR: It's going to be three.
So, you can go ahead and calculate that, if you can't
see that right away.
So we'll end here, we'll finish up with the
heteronuclear example on Friday.