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At the end of last hour,
we had just gotten to the point

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of having developed the
molecular orbital energy level

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diagram for the BH three
molecule, this trigonal planar

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entity.
And I had not had time last

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hour to show you what these
molecular orbitals look like in

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their calculated form.
And so, I am going to start off

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today by doing that,
and then will proceed to answer

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a quadrangle of questions that
we can attack regarding diatomic

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molecules using molecular
orbital theory.

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Remember that the BH three
MO problem is a seven

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orbital problem.
And so, we will go through and

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just look at these seven
molecular orbitals in ascending

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energy from the lowest energy on
up.

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This one is the lowest energy
molecular orbital for BH three

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as calculated using a
modern quantum chemical package.

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And what you can see is that
since we are indicating the

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phase of the wave function by
color, there is only one phase

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to be seen.
This, of course,

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is a characteristic property of
an s orbital.

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Remember that?
And what we have here,

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at the center of the molecule,
is the 2s orbital on boron.

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And in this molecular orbital
theory, effectively what is

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happening is that 2s orbital on
boron is reaching out and

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simultaneously overlapping with
the three 1s orbitals on each of

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the hydrogens.
And the way that we have

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orthogonalized it and normalized
it amounts to a single molecular

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orbital that can,
if you will,

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house a pair of electrons in
this orbital that is

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simultaneously bonding between
boron and the three hydrogens.

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You have one electron pair in
here.

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Two of the six electrons in the
valance shell of this molecule

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are in this lowest lying,
most tightly held,

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most electronegative molecular
orbital.

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Let's look now at the next one.

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You should be thinking in your
mind just what was the next

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highest lying orbital that we
had, and you will realize that

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this is an orbital formed from
one of the boron p orbitals and

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a linear combination of two of
the hydrogen wave functions.

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And let's see if you can
recognize it based on what we

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did last time.
Maybe I can reorient it to help

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in that regard.
Here is a molecular orbital

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that can house a pair of
electrons, one spin-up and one

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spin-down, and that is
simultaneously bonding from that

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boron to this hydrogen,
up here.

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And then there is a nodal
surface, here,

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that is coincident with the
nodal surfaces of the boron 2px

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orbital, using the same
coordinate system that we were

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using last time.
And so, the wave function

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changes sign as you pass through
the boron.

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And then the opposite phase
lobe of the boron 2px orbital is

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able to overlap with these two
hydrogens, here.

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If you will remember,
when I wrote down the

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normalization constants for this
molecular orbital,

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we had double the coefficient
on the hydrogen up here as

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compared with the two hydrogens
down here.

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And they were opposite in sign.
And that leads to a really nice

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overlap, as you can see here.
This is BH bonding here,

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and BH bonding,
down here.

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And it involves that boron 2px
orbital.

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One way you can think about
this molecular orbital theory is

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that this 2px orbital on the
central atom is reaching out

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with its intrinsic plus-minus
phase combination that is

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intrinsic to a 2px orbital
simultaneously reaching out and

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bonding with all three hydrogens
in the only way that it can,

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being a 2px orbital.
And so, we created that linear

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combination with that idea in
mind.

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And then, if we go up one more
in energy, strike that.

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Not one up more in energy,
but to the other orbital in the

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molecule that is at the same
energy.

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What we will find is that we
are involving,

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now, the 2py orbital.
And it will reach out and

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interact with hydrogens in the
only way it can,

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given its intrinsic nodal
properties as a 2py orbital.

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Remember that a 2py orbital has
the x,z-plane as a nodal

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surface.
And there is a hydrogen up here

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on that x,z-plane,
namely the one that lies on the

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plus x-axis.
And because it lies on the

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intrinsic nodal surface of this
2py orbital, it can contribute

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nothing to this molecular
orbital.

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All we are seeing up here at
the top is an arbitrarily sized

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sphere, just to show the
location of that nucleus,

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the hydrogen nucleus that is on
the plus x-axis.

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And then here,
what you see is the 2py orbital

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oriented along y and overlapping
simultaneously with a plus 1s

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wave function here and minus 1s
wave function here so that you

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see a beautiful bonding
molecular orbital that is at the

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same energy at the one we just
looked at.

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And so, as you step up in
energy, looking at the energy

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level diagram for BH three
from last time,

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that is what you see.
We have now six electrons in

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the molecule,
two in this orbital and two

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each in the two orbitals we just
looked at that provide the three

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BH bonds that we represent in
the valance bond theory as

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electron pair bonds.
But this is a totally different

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way of looking at the electronic
structure that is very appearing

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because it just takes advantage
of the atomic orbital properties

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that you would calculate for
your central atom.

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Now, let's go up one more.
The very next one,

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I will skip showing to you for
purposes of saving time,

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here, but you know that the
next orbital up is simply the

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2pz orbital on the boron,
which is our lowest unoccupied

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molecular orbital,
as well as being an atomic

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orbital in this system and
responsible for the Lewis acid

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characteristics of this species.
And one up above that,

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one higher in energy than the
LUMO is the orbital that I did

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show you last time,
the out-of-phase combination of

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the boron 2s with the three
hydrogens all with the same

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sign.
It is sort of a round nodal

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surface intersection each of the
BH bonds.

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And then, finally,
orbitals number six and seven

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are the antibonding counterparts
to the ones that I just showed

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you involving boron's 2px and
2py wave functions.

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Here is what these look like.
And this one,

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you will see,
involves the 2px orbital.

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Here is our x-axis pointing up.
And the blue lobe of 2px,

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the positive lobe here,
as you can see,

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is squished out and down.
These pictures can be a little

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bit more complicated than what
we are used to looking at from

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just sketching it out on the
board or the qualitative

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pictures that you sometimes see
in your textbook.

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But what we do find here that
is the important feature is that

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as we go from the 1s
contribution on this hydrogen

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over here to the positive lobe
of 2px we are going through a

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nodal surface.
And that is antibonding in

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character.
And you can see that this

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deformation of this lobe of the
p orbital, of course,

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looks like a high energy type
of phenomena.

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And indeed it is.
This is one of the two highest

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energy orbitals in the energy
level diagram for the BH three

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molecule.
And then, similarly,

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here is the other lobe of the
2px orbital, the minus lobe.

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And, as you try to go from it
to the adjacent hydrogens,

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remember, in the bonding one we
were allowing this to overlap

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with the same sign on the
hydrogens.

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Now we have the opposite sign
on the hydrogens,

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so that we have a beautiful
antibonding node indicated here

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and here between the minus phase
lobe of 2px and the adjacent

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hydrogens.
So, this antibonding orbital

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has lots of internuclear nodes,
nodes that appear between

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nuclei and indicate that we are
not getting overlapped and

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bonding, but rather we are
getting a high frequency orbital

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that is likewise high in energy.
Finally, I will show you the

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counterpart to this one,
that involves 2py.

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It is a little easier to
understand from inspecting it.

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And it also evinces the
internuclear nodes that you

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would expect for an antibonding
molecular orbital.

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And I am orienting it so that
you will be able to understand

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it with reference to our
coordinate system as introduced

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last time.
And so, you have the boron and

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the hydrogen along x,
up here.

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Here is our 2py.
It is sort of bending away from

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these two hydrogens whose wave
functions have opposite sign to

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the lobes that are directed into
their vicinity.

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You can see that we have here a
nodal surface between boron and

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this hydrogen wherein the
bonding counterpart that we

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looked at had the same sign and
a nice bond.

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And then over here,
once again, the mirror image.

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So, this is one of our final
orbitals in looking at the BH

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three energy level
diagram.

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It is the case that if you
start to put electrons into

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antibonding orbitals,
they will cancel the bonding

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properties of the bonding
counterparts.

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And we will explore that in
quite a bit more detail here

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shortly.
Let me leave that up there for

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now and tell you what quadrangle
of problems it is that I hope to

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answer for you today using more
MO theory ideas.

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First of these will be why He
two is so stable?

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And so, today's lecture is
predominantly devoted to

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diatomic molecules.
And He two will be the

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first of these.
Let's make this unstable.

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Let's do stable here in just a
moment, but first let's say,

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next, why does O two
have unpaired electrons?

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Certainly, if we draw the
valance bond picture for O two,

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like that,
it does not give us any

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indication that this diatomic
molecule would have unpaired

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electrons.
And so, we are going to see how

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MO theory can shed some light on
this issue.

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And then, the final two
questions that I will fold into

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one line here given the way I am
using up space on this board is,

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why is N two so stable
--

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-- and CO poison?

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MO theory can shed very clear
light on all of these questions.

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And so, that is what we will
work on for the rest of the

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hour.
In MO theory,

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there are a couple things you
should keep in mind when trying

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to grasp all the subtleties of a
given energy level diagram with

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00:13:11,000 --> 00:13:15,000
which you may be confronted in
your textbook or on an exam or

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just through reading papers in
the Journal of the American

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Chemical Society,
wherever you may find these.

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You should keep in mind that,
number one, interaction is

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strong.
That is, when atomic orbitals

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interact to form molecular
orbitals, the AO interaction is

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strong --

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-- when there is good spatial
overlap.

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Overlap is a central concept in
molecular orbital theory.

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The idea is that if you have
atoms that are far apart from

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each other in space,
then their wave functions have

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dropped off exponentially and
are not interacting much and

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00:14:27,000 --> 00:14:32,000
there is not good overlap.
But when orbitals are close in

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00:14:32,000 --> 00:14:36,000
space and if their orbitals are
directed in space toward one

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another, then there may be
overlap that leads to good

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00:14:40,000 --> 00:14:43,000
bonding.
And also I can echo this first

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part, AO interaction is strong
when, and say two,

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AOs are close in energy.
The interacting orbitals are

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00:14:50,000 --> 00:14:53,000
close in energy.

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00:15:13,000 --> 00:15:17,000
When I talk about where atomic
orbitals are in energy,

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00:15:17,000 --> 00:15:21,000
I am going to really want you
to think about your periodic

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table and the properties that
appear with periodicity and are

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00:15:26,000 --> 00:15:30,000
organized and collated in the
periodic table.

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00:15:30,000 --> 00:15:33,000
We are going to be talking
about properties like

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00:15:33,000 --> 00:15:37,000
electronegativity as it relates
to the energy.

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00:15:37,000 --> 00:15:42,000
And I will get to that a little
bit on the next board,

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but let's first make a small MO
diagram here.

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00:15:45,000 --> 00:15:51,000
And this will be a diagram that
we may use both for H two

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00:15:51,000 --> 00:15:55,000
or He two.
This is the simplest of all MO

220
00:15:55,000 --> 00:16:01,000
diagrams and one with which you
should be quite familiar.

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00:16:01,000 --> 00:16:07,000
We have a bonding molecular
orbital at low-end energy.

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00:16:07,000 --> 00:16:14,000
And notice that if this is a 1s
orbital over here and this is a

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00:16:14,000 --> 00:16:20,000
1s orbital over here for H two
or He two

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00:16:20,000 --> 00:16:25,000
systems, then when these come
together and bond,

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00:16:25,000 --> 00:16:30,000
here are your two nuclei,
--

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00:16:30,000 --> 00:16:33,000
-- and here is a probability
density isosurface that I am

227
00:16:33,000 --> 00:16:37,000
drawing around that shows
in-phase combination of those

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00:16:37,000 --> 00:16:42,000
two 1s orbitals merging with
each other, giving good overlap,

229
00:16:42,000 --> 00:16:45,000
and the interacting AOs being
of the same energy.

230
00:16:45,000 --> 00:16:49,000
Because the H two and He
two molecules have such

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00:16:49,000 --> 00:16:52,000
symmetry we have a good bonding
orbital.

232
00:16:52,000 --> 00:16:55,000
And then, up here,
we must find there to be an

233
00:16:55,000 --> 00:16:59,000
internuclear node in that
corresponding antibonding

234
00:16:59,000 --> 00:17:03,000
orbital.
And so, it will look something

235
00:17:03,000 --> 00:17:05,000
like this.
And, therefore,

236
00:17:05,000 --> 00:17:10,000
I am going to associate with
this high-lying orbital an

237
00:17:10,000 --> 00:17:14,000
asterisk to indicate the
antibonding character,

238
00:17:14,000 --> 00:17:18,000
the presence of this
internuclear node,

239
00:17:18,000 --> 00:17:23,000
the change of sign as we go
from one nucleus to the other,

240
00:17:23,000 --> 00:17:27,000
the absence of bonding
character.

241
00:17:27,000 --> 00:17:30,000
Here we have the buildup of
electron density in the

242
00:17:30,000 --> 00:17:34,000
internuclear region,
electrons being stabilized by

243
00:17:34,000 --> 00:17:38,000
being held simultaneously by
more than one nucleus that is

244
00:17:38,000 --> 00:17:42,000
positively charged.
And here, the exact inverse of

245
00:17:42,000 --> 00:17:47,000
that, constructive interference
and destructive interference in

246
00:17:47,000 --> 00:17:50,000
the H two or He two
systems.

247
00:17:50,000 --> 00:17:54,000
And then we can ask,
how many electrons do we have

248
00:17:54,000 --> 00:18:00,000
to put into the diagram?
If it is the H two case,

249
00:18:00,000 --> 00:18:04,000
then, to figure out the bond
order, --

250
00:18:14,000 --> 00:18:18,000
-- what we do is populate the
diagram according to the same

251
00:18:18,000 --> 00:18:23,000
kind of Aufbau principle that
you would for generating the

252
00:18:23,000 --> 00:18:27,000
energy level diagram for an
atom, and taking into account

253
00:18:27,000 --> 00:18:32,000
things like Hund's rule.
If it is H two,

254
00:18:32,000 --> 00:18:37,000
we have two electrons that we
can put in, which in here with

255
00:18:37,000 --> 00:18:39,000
opposing spin,
like that.

256
00:18:39,000 --> 00:18:43,000
That is our bonding molecular
orbital.

257
00:18:43,000 --> 00:18:48,000
We have two bonding electrons.
Zero antibonding electrons

258
00:18:48,000 --> 00:18:52,000
divided by two equals a bond
order of one.

259
00:18:52,000 --> 00:18:57,000
We have an H-H single bond in
the case of the H two

260
00:18:57,000 --> 00:19:02,000
molecule.
But if we have He two,

261
00:19:02,000 --> 00:19:07,000
then in the He two system,
we have two more electrons,

262
00:19:07,000 --> 00:19:12,000
and the only place they can go
is up here in the antibonding

263
00:19:12,000 --> 00:19:16,000
molecular orbital.
We have two bonding minus two

264
00:19:16,000 --> 00:19:22,000
antibonding over two is equal to
zero for a bond order of zero.

265
00:19:22,000 --> 00:19:27,000
And so, the idea is that if two
He atoms collide with each

266
00:19:27,000 --> 00:19:31,000
other, when they do so,
their orbitals may overlap and

267
00:19:31,000 --> 00:19:36,000
give rise to bonding.
But, simultaneously,

268
00:19:36,000 --> 00:19:40,000
you get antibonding.
And the net bond order is zero,

269
00:19:40,000 --> 00:19:44,000
and so He two is not
bound, but these atoms just

270
00:19:44,000 --> 00:19:47,000
bounce off each other.
There is no stabilization.

271
00:19:47,000 --> 00:19:51,000
And you can certainly
understand that if we were in

272
00:19:51,000 --> 00:19:55,000
the middle somewhere,
as we would be if we had He two

273
00:19:55,000 --> 00:19:58,000
plus,
missing one electron up here,

274
00:19:58,000 --> 00:20:02,000
then our bond order would be
one-half.

275
00:20:02,000 --> 00:20:06,000
It is going to get a lot more
complicated than this,

276
00:20:06,000 --> 00:20:11,000
but this you have to know
because this is a really nice

277
00:20:11,000 --> 00:20:16,000
starting point for thinking
about bonding in all kinds of

278
00:20:16,000 --> 00:20:20,000
scenarios where you may need to
consider it.

279
00:20:20,000 --> 00:20:24,000
And so, let's look at some
periodic properties.

280
00:20:24,000 --> 00:20:30,000
This will be with reference,
again, to energy.

281
00:20:30,000 --> 00:20:36,000
And I am interested in how some
of the properties of the atoms

282
00:20:36,000 --> 00:20:41,000
vary as we go across the
periodic table,

283
00:20:41,000 --> 00:20:48,000
lithium to beryllium to boron
to carbon to nitrogen to oxygen

284
00:20:48,000 --> 00:20:52,000
and over to fluorine.
And I will stop there.

285
00:20:52,000 --> 00:21:00,000
We won't worry about any more
of the noble gases today.

286
00:21:00,000 --> 00:21:03,000
As we go from left to right,
one of the things that you know

287
00:21:03,000 --> 00:21:07,000
from your study of the periodic
table is that electronegativity

288
00:21:07,000 --> 00:21:08,000
increases.

289
00:21:21,000 --> 00:21:24,000
This increases from left to
right across the Periodic Table

290
00:21:24,000 --> 00:21:27,000
in concert with increasing Z.

291
00:21:34,000 --> 00:21:36,000
That is, the increasing atomic
number.

292
00:21:36,000 --> 00:21:41,000
That is, the increasing number
of positively charged protons in

293
00:21:41,000 --> 00:21:46,000
our nucleus, as we go across.
Each time we are adding more

294
00:21:46,000 --> 00:21:50,000
protons to that nucleus.
The nucleus is becoming more

295
00:21:50,000 --> 00:21:53,000
and more positively charged as
we go across.

296
00:21:53,000 --> 00:21:58,000
And an important consequence of
that has to do with the fact

297
00:21:58,000 --> 00:22:03,000
that s orbitals don't have a
node at the nucleus.

298
00:22:03,000 --> 00:22:08,000
Whereas, p orbitals do.
Since these elements are in the

299
00:22:08,000 --> 00:22:12,000
so-called p block of our
periodic table,

300
00:22:12,000 --> 00:22:18,000
well, lithium in the s block,
but we will consider the

301
00:22:18,000 --> 00:22:24,000
properties all the way across.
And the energy issue up here

302
00:22:24,000 --> 00:22:29,000
that I am going to be concerned
with is the energy of the atomic

303
00:22:29,000 --> 00:22:33,000
2s versus 2p orbitals as we go
from left to right across the

304
00:22:33,000 --> 00:22:37,000
periodic table,
systematically adding to the

305
00:22:37,000 --> 00:22:42,000
number of protons in our nucleus
in a way that does not affect 2s

306
00:22:42,000 --> 00:22:46,000
orbitals the same way that it
affects 2p orbitals.

307
00:22:46,000 --> 00:22:50,000
It turns out that by the time
we get over to fluorine,

308
00:22:50,000 --> 00:22:55,000
where we have essentially our
most electronegative element,

309
00:22:55,000 --> 00:23:00,000
the 2s orbital has sunk to a
very low energy.

310
00:23:00,000 --> 00:23:05,000
And the 2p orbital has gone
down in energy as well,

311
00:23:05,000 --> 00:23:12,000
but just not by nearly as much.
And so, it is nearly an

312
00:23:12,000 --> 00:23:15,000
isotonic function as we go
across.

313
00:23:15,000 --> 00:23:22,000
2p is decreasing in energy in
response to this increased Z,

314
00:23:22,000 --> 00:23:27,000
but not as much as 2s is
decreasing in energy,

315
00:23:27,000 --> 00:23:34,000
having to do with the fact that
2s electrons see much more of

316
00:23:34,000 --> 00:23:41,000
the nucleus than 2p electrons do
because 2s electrons don't have

317
00:23:41,000 --> 00:23:47,000
a node there.
They spend a lot more time in

318
00:23:47,000 --> 00:23:52,000
close to the nucleus than do
your 2p electrons.

319
00:23:52,000 --> 00:23:57,000
This has a really important
influence on that set of

320
00:23:57,000 --> 00:24:02,000
diatomic molecule MO theory
problems, eight orbital

321
00:24:02,000 --> 00:24:09,000
problems, that we call homo or
heteronuclear diatomic molecules

322
00:24:09,000 --> 00:24:14,000
of these elements.
And that has to do with the

323
00:24:14,000 --> 00:24:20,000
fact that in between nitrogen
and oxygen, a fundamental change

324
00:24:20,000 --> 00:24:24,000
takes place in the energy level
diagram.

325
00:24:24,000 --> 00:24:28,000
And that change,
that results from the atomic

326
00:24:28,000 --> 00:24:33,000
structure of the elements,
ends up with the result that

327
00:24:33,000 --> 00:24:38,000
the MO diagram is much easier to
draw for oxygen,

328
00:24:38,000 --> 00:24:41,000
O two,
and fluorine,

329
00:24:41,000 --> 00:24:45,000
F two,
than it is for the earlier

330
00:24:45,000 --> 00:24:48,000
ones.
And it is much easier to

331
00:24:48,000 --> 00:24:53,000
understand for O two and
F two than it is for N

332
00:24:53,000 --> 00:24:55,000
two, C two, B two,
Be two, Li two.

333
00:24:57,000 --> 00:25:00,000
So, let's just see what that
means.

334
00:25:11,000 --> 00:25:14,000
First of all,
when we are going to draw an

335
00:25:14,000 --> 00:25:17,000
energy level diagram for a
diatomic molecule,

336
00:25:17,000 --> 00:25:23,000
if it is a homonuclear diatomic
molecule, meaning the two atoms

337
00:25:23,000 --> 00:25:27,000
that are going to be bonded
together are the same,

338
00:25:27,000 --> 00:25:31,000
that means that the atomic
orbitals for the contributing

339
00:25:31,000 --> 00:25:36,000
atoms are the same.
Here, on the left,

340
00:25:36,000 --> 00:25:44,000
what I am drawing is,
let's say, an F or an O atom.

341
00:25:44,000 --> 00:25:49,000
We have the 2s and the 2p,
x-y-z.

342
00:25:49,000 --> 00:25:58,000
That is four of the eight
orbitals in our problem.

343
00:25:58,000 --> 00:26:03,000
And then, over here,
we have the other identical F

344
00:26:03,000 --> 00:26:10,000
or O atom, with its 2p x-y-z and
2s orbitals at exactly the same

345
00:26:10,000 --> 00:26:15,000
energy as for the counterpart,
over here.

346
00:26:15,000 --> 00:26:18,000
So, this is another F or O
atom.

347
00:26:18,000 --> 00:26:25,000
And then we are going to ask,
how can these atomic orbitals

348
00:26:25,000 --> 00:26:31,000
combine to give molecular
orbitals in these diatomic

349
00:26:31,000 --> 00:26:35,000
molecules?
This is unlike the BH three

350
00:26:35,000 --> 00:26:40,000
problem because there is
no atom at the center of gravity

351
00:26:40,000 --> 00:26:43,000
of our system,
but the ideas are the same.

352
00:26:43,000 --> 00:26:47,000
And the ideas are very much the
same as what we have here for H

353
00:26:47,000 --> 00:26:51,000
two, where also we could
make molecular orbitals despite

354
00:26:51,000 --> 00:26:56,000
the fact that there was no atom
at the center of our system.

355
00:26:56,000 --> 00:27:00,000
We can pretend that there is,
and we can use that pretend

356
00:27:00,000 --> 00:27:05,000
atom to identify what linear
combinations we should make.

357
00:27:05,000 --> 00:27:09,000
A simpler way to do it,
at this point,

358
00:27:09,000 --> 00:27:14,000
would be to say that our
molecule has the two nuclei

359
00:27:14,000 --> 00:27:20,000
oriented along the z-axis.
And then you have x and y

360
00:27:20,000 --> 00:27:25,000
perpendicular to z.
And so, if we take this choice

361
00:27:25,000 --> 00:27:32,000
of coordinate system that the
molecule is oriented along z,

362
00:27:32,000 --> 00:27:39,000
the two nuclei lie one on plus
z, one on minus z.

363
00:27:39,000 --> 00:27:47,000
Then we can immediately
classify our atomic orbitals as

364
00:27:47,000 --> 00:27:55,000
being either sigma or pi with
respect to the z-axis.

365
00:27:55,000 --> 00:28:02,000
I will say AO is sigma or pi
with respect to z,

366
00:28:02,000 --> 00:28:10,000
which is the molecular axis.
And remember what that means in

367
00:28:10,000 --> 00:28:15,000
terms of the nodal properties.
If something is sigma with

368
00:28:15,000 --> 00:28:19,000
respect to z,
that means if you view it down

369
00:28:19,000 --> 00:28:25,000
z it will look like a cylinder.
It will look like a sphere,

370
00:28:25,000 --> 00:28:28,000
actually.
It will be cylindrically

371
00:28:28,000 --> 00:28:32,000
symmetric about the z-axis.
If it is sigma then,

372
00:28:32,000 --> 00:28:37,000
if you are viewing down z,
you see that.

373
00:28:37,000 --> 00:28:43,000
That could be either an s
orbital or it could be a pz

374
00:28:43,000 --> 00:28:47,000
orbital.
They would look the same if you

375
00:28:47,000 --> 00:28:52,000
were viewing right down z.
And then pi,

376
00:28:52,000 --> 00:28:56,000
viewing down z,
would look like this.

377
00:28:56,000 --> 00:29:03,000
And that would be a px or a py
orbital perpendicular to the

378
00:29:03,000 --> 00:29:07,000
z-axis.
And this would be an s orbital

379
00:29:07,000 --> 00:29:11,000
or a pz orbital.
What I can say right away is

380
00:29:11,000 --> 00:29:15,000
that we can separate out the
molecular orbitals that we are

381
00:29:15,000 --> 00:29:20,000
going to form from these atomic
orbitals into their sigma or pi

382
00:29:20,000 --> 00:29:24,000
characteristics relative to the
z-axis.

383
00:29:24,000 --> 00:29:28,000
And because we are in the O or
F atom case, very far on the

384
00:29:28,000 --> 00:29:33,000
right-hand side the periodic
table, we have a large gap.

385
00:29:33,000 --> 00:29:36,000
This gap is large.

386
00:29:42,000 --> 00:29:47,000
We can make the approximation
that this 2s orbital here,

387
00:29:47,000 --> 00:29:53,000
which could in principle form a
sigma bond with this 2pz orbital

388
00:29:53,000 --> 00:29:59,000
over here, that would be an
interaction looking like that,

389
00:29:59,000 --> 00:30:02,000
--
-- that that does not occur.

390
00:30:02,000 --> 00:30:07,000
Because remember here I said
you get strong interaction when

391
00:30:07,000 --> 00:30:10,000
the interacting AOs are close in
energy.

392
00:30:10,000 --> 00:30:15,000
When you are talking about
these very electronegative

393
00:30:15,000 --> 00:30:20,000
elements, the s manifold of
orbitals is very well separated

394
00:30:20,000 --> 00:30:25,000
in energy from the p manifold.
That leads to simplicity in the

395
00:30:25,000 --> 00:30:30,000
case of the diagram that we are
about to draw.

396
00:30:30,000 --> 00:30:34,000
Because we say that this
interacts with this in a sigma

397
00:30:34,000 --> 00:30:39,000
fashion, giving rise to bonding
and antibonding interactions

398
00:30:39,000 --> 00:30:43,000
that look exactly like what we
have drawn over there.

399
00:30:43,000 --> 00:30:46,000
I will draw the bonding orbital
here.

400
00:30:46,000 --> 00:30:51,000
Actually, the antibonding
orbital here and the bonding

401
00:30:51,000 --> 00:30:54,000
orbital here.
And I will show the parentage

402
00:30:54,000 --> 00:30:59,000
of this molecular orbitals in
this way.

403
00:30:59,000 --> 00:31:03,000
And you will recognize that
this is our in-phase combination

404
00:31:03,000 --> 00:31:08,000
and this is our out-of-phase
combination, analogous to the H

405
00:31:08,000 --> 00:31:12,000
two problem.
This is like the H two

406
00:31:12,000 --> 00:31:16,000
problem built into the F two
or the O two

407
00:31:16,000 --> 00:31:19,000
problem.
And next what we find is that

408
00:31:19,000 --> 00:31:24,000
we can also make a sigma bond by
interaction of pz with pz.

409
00:31:24,000 --> 00:31:27,000
This is going to be a nice
strong directed bonding

410
00:31:27,000 --> 00:31:32,000
interaction that will look like
this.

411
00:31:38,000 --> 00:31:42,000
You see we have overlap here in
the center, a nice strong

412
00:31:42,000 --> 00:31:47,000
bonding interaction in the
center with 2pz orbitals that

413
00:31:47,000 --> 00:31:52,000
are directed at each other.
And this is a sigma bond

414
00:31:52,000 --> 00:31:57,000
because you view that down z,
it is going to look like a

415
00:31:57,000 --> 00:32:02,000
round thing.
You are not going to see any

416
00:32:02,000 --> 00:32:06,000
nodal surfaces.
And then it has up here a

417
00:32:06,000 --> 00:32:11,000
corresponding antibonding
orbital right there.

418
00:32:11,000 --> 00:32:14,000
So far I have --

419
00:32:21,000 --> 00:32:27,000
-- two pair-wise combinations
that give us two sigma bonding

420
00:32:27,000 --> 00:32:34,000
orbitals and two sigma star
antibonding orbitals.

421
00:32:34,000 --> 00:32:37,000
So, that is pretty
straightforward.

422
00:32:37,000 --> 00:32:40,000
And, once again,
I can say that we have the H

423
00:32:40,000 --> 00:32:46,000
two problem built into
this energy level diagram twice,

424
00:32:46,000 --> 00:32:50,000
with the s interactions,
with the pz interactions.

425
00:32:50,000 --> 00:32:56,000
The only thing I have left is
to do my px and py interactions.

426
00:32:56,000 --> 00:33:00,000
And we can make a pair of pi
bonds and a pair of pi

427
00:33:00,000 --> 00:33:06,000
anti-bonds that corresponds to
those pi bonds.

428
00:33:06,000 --> 00:33:08,000
This one is pi.
This one is pi star.

429
00:33:08,000 --> 00:33:12,000
And what do they look like?
Well, they look like

430
00:33:12,000 --> 00:33:16,000
side-to-side bonds between pairs
of p orbitals.

431
00:33:16,000 --> 00:33:21,000
This is just like what I showed
you for the ethylene molecule.

432
00:33:21,000 --> 00:33:25,000
The pi bond,
in fact, has this nice overlap

433
00:33:25,000 --> 00:33:28,000
on one side and up here on the
other.

434
00:33:28,000 --> 00:33:33,000
There is your pi bond.
And you have one of those that

435
00:33:33,000 --> 00:33:38,000
lie in the x,z-plane and one of
those that lie in the y,z-plane.

436
00:33:38,000 --> 00:33:43,000
You have two of them,
and they are at equal energy

437
00:33:43,000 --> 00:33:44,000
here.
And then up here,

438
00:33:44,000 --> 00:33:47,000
what does the pi star look
like?

439
00:33:47,000 --> 00:33:51,000
Well, the same,
except we turn around one of

440
00:33:51,000 --> 00:33:54,000
the p orbitals,
introducing a nodal surface,

441
00:33:54,000 --> 00:33:59,000
like that.
And we have two of those.

442
00:33:59,000 --> 00:34:04,000
And here is our nodal surface.
Now it is an internuclear

443
00:34:04,000 --> 00:34:09,000
between the nuclei nodal
surface, indicating antibonding

444
00:34:09,000 --> 00:34:14,000
character there.
That is the easy case because s

445
00:34:14,000 --> 00:34:18,000
and p are very well separated.
What you can see,

446
00:34:18,000 --> 00:34:23,000
though, is that type of
interaction of a 2s on our left

447
00:34:23,000 --> 00:34:29,000
atom with a pz on the right
atom, that might start to become

448
00:34:29,000 --> 00:34:34,000
important as you are over here
in this part of the diagram,

449
00:34:34,000 --> 00:34:41,000
where s and p are pretty close
together in energy.

450
00:34:41,000 --> 00:34:43,000
And that is what happens,
in fact.

451
00:34:43,000 --> 00:34:48,000
And that is what happens as
soon as you go to the left of

452
00:34:48,000 --> 00:34:53,000
oxygen and to nitrogen or
anything lighter than nitrogen.

453
00:34:53,000 --> 00:34:57,000
Suddenly, you cannot ignore
interactions of s on one side

454
00:34:57,000 --> 00:35:00,000
with p on the other.

455
00:35:09,000 --> 00:35:14,000
And so, I am going to draw here
the diagram that you would use,

456
00:35:14,000 --> 00:35:17,000
say, for N two.
And, for simplicity,

457
00:35:17,000 --> 00:35:21,000
I am just going to draw the
middle part.

458
00:35:21,000 --> 00:35:26,000
This part here in the middle is
that which corresponds to the

459
00:35:26,000 --> 00:35:29,000
molecule.
This one is what you would use

460
00:35:29,000 --> 00:35:33,000
for O two or F two.

461
00:35:33,000 --> 00:35:39,000
And let's just go ahead and
draw that diagram for the

462
00:35:39,000 --> 00:35:44,000
molecule.
This one would be this set of

463
00:35:44,000 --> 00:35:49,000
energy levels corresponding to,
for example, N two.

464
00:35:49,000 --> 00:35:54,000
And we have increasing energy

465
00:35:54,000 --> 00:36:00,000
on the vertical axis.
We are going to find that

466
00:36:00,000 --> 00:36:07,000
again, we have a sigma and a
sigma star.

467
00:36:07,000 --> 00:36:12,000
And then, the difference comes
right here where we find that as

468
00:36:12,000 --> 00:36:17,000
we go up in energy we next get
to our pi bonds.

469
00:36:17,000 --> 00:36:22,000
And then, the next sigma
orbital is slightly higher in

470
00:36:22,000 --> 00:36:29,000
energy than the pi system rather
than slightly lower in energy.

471
00:36:29,000 --> 00:36:34,000
And then we will encounter up
here our pi star.

472
00:36:34,000 --> 00:36:38,000
And then, finally,
that highest in energy,

473
00:36:38,000 --> 00:36:41,000
sigma star.
This switch,

474
00:36:41,000 --> 00:36:48,000
this sigma going up relative to
that pi is really a consequence

475
00:36:48,000 --> 00:36:55,000
of all of the sigma orbitals
being responsive to the value of

476
00:36:55,000 --> 00:37:02,000
the electronegativity of the 2s
orbital of the atom.

477
00:37:02,000 --> 00:37:06,000
You can really almost consider
the pi system independent from

478
00:37:06,000 --> 00:37:09,000
the sigma.
These four sigma orbitals,

479
00:37:09,000 --> 00:37:12,000
sigma, sigma star,
sigma, sigma star,

480
00:37:12,000 --> 00:37:16,000
I can call them one sigma,
two, three, four,

481
00:37:16,000 --> 00:37:20,000
just to show you that we have,
in ascending order,

482
00:37:20,000 --> 00:37:22,000
four orbitals of sigma
symmetry.

483
00:37:22,000 --> 00:37:26,000
That whole manifold,
when you can start mixing a

484
00:37:26,000 --> 00:37:31,000
little bit of very low energy s
into it, sinks down relative to

485
00:37:31,000 --> 00:37:34,000
pi.
In the case over here,

486
00:37:34,000 --> 00:37:39,000
we are basically pulling down
the four sigma orbitals relative

487
00:37:39,000 --> 00:37:42,000
to pi.
And here, because our s

488
00:37:42,000 --> 00:37:47,000
orbitals are higher in energy
relative to our p orbitals,

489
00:37:47,000 --> 00:37:51,000
they go up a little bit and
bumps three sigma up over pi.

490
00:37:51,000 --> 00:37:56,000
And what it means is that these
orbitals are not so simple

491
00:37:56,000 --> 00:37:59,000
anymore.
Any one of them is a linear

492
00:37:59,000 --> 00:38:02,000
combination of four atomic
orbitals.

493
00:38:02,000 --> 00:38:06,000
This orbital here,
one sigma, will not look like

494
00:38:06,000 --> 00:38:09,000
it looks in the H two
problem, anymore.

495
00:38:09,000 --> 00:38:12,000
This one sigma will have some p
mixed into it.

496
00:38:12,000 --> 00:38:16,000
And so, one way we can
represent that through a simple

497
00:38:16,000 --> 00:38:19,000
drawing is like this.

498
00:38:25,000 --> 00:38:29,000
You see the difference between
that sigma bonding orbital and

499
00:38:29,000 --> 00:38:33,000
the one up there?
We have some extra little nodes

500
00:38:33,000 --> 00:38:37,000
in here that are an inherent
characteristic of the

501
00:38:37,000 --> 00:38:40,000
contribution of 2pz orbitals
into this.

502
00:38:40,000 --> 00:38:45,000
This is a linear combination of
the 2s and the 2pz orbitals on

503
00:38:45,000 --> 00:38:49,000
the two atoms we have.
There are four atomic orbitals

504
00:38:49,000 --> 00:38:54,000
that go in with different
coefficients to these different

505
00:38:54,000 --> 00:38:57,000
parts of the sigma manifold.
And then up here,

506
00:38:57,000 --> 00:39:03,000
the other bonding orbital,
we can draw it this way.

507
00:39:08,000 --> 00:39:11,000
And this molecular orbital,
this one here,

508
00:39:11,000 --> 00:39:14,000
three sigma,
has diminished bonding

509
00:39:14,000 --> 00:39:20,000
character between the two nuclei
and enhanced character on the

510
00:39:20,000 --> 00:39:25,000
outside, the part of the p
orbital that is contributing to

511
00:39:25,000 --> 00:39:30,000
it that points away from the
other atom.

512
00:39:30,000 --> 00:39:34,000
And so, while this has bonding
character, it also has what we

513
00:39:34,000 --> 00:39:37,000
might think of as lone pair
characteristics.

514
00:39:37,000 --> 00:39:40,000
In MO theory,
some orbitals are not that

515
00:39:40,000 --> 00:39:42,000
simple.
They might be at the same time

516
00:39:42,000 --> 00:39:46,000
a little bit bonding and a
little bit lone pair in

517
00:39:46,000 --> 00:39:49,000
character.
Let's look at a couple of

518
00:39:49,000 --> 00:39:52,000
orbitals like that,
briefly.

519
00:40:04,000 --> 00:40:08,000
And here I have done this for
the C two molecule.

520
00:40:08,000 --> 00:40:12,000
I have done the calculation for
C two,

521
00:40:12,000 --> 00:40:17,000
so now we are over here on the
left-hand side of that dividing

522
00:40:17,000 --> 00:40:20,000
line where s and p are a big
closer together.

523
00:40:20,000 --> 00:40:25,000
And I am opening up this
orbital, this one down here.

524
00:40:25,000 --> 00:40:30,000
It is a little more complicated
than the way I drew it on the

525
00:40:30,000 --> 00:40:35,000
board, but what you should be
able to see is --

526
00:40:35,000 --> 00:40:37,000
Think back, now,
to the first orbital I showed

527
00:40:37,000 --> 00:40:41,000
you today, which was the lowest
lying orbital in the BH three

528
00:40:41,000 --> 00:40:43,000
molecule.
You remember,

529
00:40:43,000 --> 00:40:46,000
no matter which way we turned
it, we only saw blue,

530
00:40:46,000 --> 00:40:48,000
which was the positive wave
function.

531
00:40:48,000 --> 00:40:50,000
It looked like an s orbital,
everywhere.

532
00:40:50,000 --> 00:40:53,000
It had the same sign
everywhere, and you could not

533
00:40:53,000 --> 00:40:56,000
see any nodal properties.
Well, that would be true also

534
00:40:56,000 --> 00:41:00,000
for this one down here in the
case of O two or F two

535
00:41:00,000 --> 00:41:04,000
because it is not mixing
in p character.

536
00:41:04,000 --> 00:41:07,000
Because the atomic s and p
orbitals are so far away in

537
00:41:07,000 --> 00:41:09,000
energy.
But now, when they are closer

538
00:41:09,000 --> 00:41:13,000
in energy, this orbital down
here, the s over here sees the p

539
00:41:13,000 --> 00:41:15,000
over here and they mix in,
and vice versa.

540
00:41:15,000 --> 00:41:19,000
And, if I turn this around,
what you are going to see are

541
00:41:19,000 --> 00:41:22,000
these little negative lobes here
that I am indicating that are

542
00:41:22,000 --> 00:41:24,000
mixing in.
But this is hugely bonding

543
00:41:24,000 --> 00:41:27,000
orbital, still,
because look at how much of the

544
00:41:27,000 --> 00:41:32,000
orbital is centered in the
region between the nuclei.

545
00:41:32,000 --> 00:41:37,000
This is a very strongly bonding
orbital in this system.

546
00:41:37,000 --> 00:41:44,000
Let's now visualize the three
sigma, the third sigma symmetry

547
00:41:44,000 --> 00:41:50,000
orbital, as we ascend in energy.
And it will look hopefully

548
00:41:50,000 --> 00:41:56,000
something like what I drew here.
There it is.

549
00:42:04,000 --> 00:42:07,000
You see this is a sigma
orbital, because as we look down

550
00:42:07,000 --> 00:42:11,000
the z-axis, it looks just round.
But it certainly has a lot of p

551
00:42:11,000 --> 00:42:15,000
orbital character because as we
go from outside the molecule

552
00:42:15,000 --> 00:42:18,000
through the nucleus,
we get a change in sign.

553
00:42:18,000 --> 00:42:21,000
And then there is overlap in
here, but the amount of the

554
00:42:21,000 --> 00:42:25,000
overlap in here is small
compared to the amount of this

555
00:42:25,000 --> 00:42:29,000
lobe out here that is just
nonbonding in character because

556
00:42:29,000 --> 00:42:33,000
it is pointing out into space.
It is like a lone pair of

557
00:42:33,000 --> 00:42:37,000
electrons at the same time as it
is a bonding orbital.

558
00:42:37,000 --> 00:42:41,000
If this molecule actually had
enough electrons to fill up

559
00:42:41,000 --> 00:42:46,000
through three sigma and no
higher, than you would recognize

560
00:42:46,000 --> 00:42:50,000
that this orbital that we are
looking at would be the highest

561
00:42:50,000 --> 00:42:55,000
occupied molecular orbital.
And it would be responsible for

562
00:42:55,000 --> 00:42:58,000
the basicity and the
nucleophilicity of the molecule

563
00:42:58,000 --> 00:43:02,000
--
-- because any Lewis acid that

564
00:43:02,000 --> 00:43:07,000
came in would want to come in on
the z-axis to interact with this

565
00:43:07,000 --> 00:43:10,000
lone pair electron density out
there.

566
00:43:10,000 --> 00:43:15,000
So, that is how our three sigma
orbital now looks when we are on

567
00:43:15,000 --> 00:43:20,000
the left side of that red dotted
line to the lighter elements.

568
00:43:20,000 --> 00:43:25,000
I will leave that up there for
a moment while we consider a

569
00:43:25,000 --> 00:43:30,000
couple of interesting points.
One of the things that you will

570
00:43:30,000 --> 00:43:34,000
often be asked to do is to write
down the configuration.

571
00:43:40,000 --> 00:43:45,000
And I want to finish these two
questions here of O two,

572
00:43:45,000 --> 00:43:49,000
why does it have unpaired
electrons?

573
00:43:49,000 --> 00:43:54,000
In the case of O two,
we are over on this diagram,

574
00:43:54,000 --> 00:43:58,000
we have 12 valance electrons.
We can go two,

575
00:43:58,000 --> 00:44:02,000
four, six, eight,
ten, and then 11 and 12,

576
00:44:02,000 --> 00:44:08,000
two electrons up here in the pi
star manifold.

577
00:44:08,000 --> 00:44:11,000
We are completely filled up
through pi.

578
00:44:11,000 --> 00:44:16,000
And then, we have a half-filled
pi star with 12 electrons

579
00:44:16,000 --> 00:44:19,000
populating the O two diagram.

580
00:44:19,000 --> 00:44:24,000
Let's represent that this way.
We have our one sigma with two

581
00:44:24,000 --> 00:44:27,000
electrons in.
We have our second sigma,

582
00:44:27,000 --> 00:44:32,000
which is antibonding with two
electrons.

583
00:44:32,000 --> 00:44:36,000
Then we have our third sigma
with two electrons.

584
00:44:36,000 --> 00:44:40,000
And then, as we go up in
energy, the next is our first pi

585
00:44:40,000 --> 00:44:44,000
orbital bonding.
That accommodates four

586
00:44:44,000 --> 00:44:47,000
electrons.
And then we have next our

587
00:44:47,000 --> 00:44:51,000
second pi type orbital.
It is antibonding and has two

588
00:44:51,000 --> 00:44:55,000
electrons in it.
And then all the other orbitals

589
00:44:55,000 --> 00:45:00,000
are empty, so I won't indicate
them.

590
00:45:00,000 --> 00:45:04,000
This is the configuration for
the O two molecule.

591
00:45:04,000 --> 00:45:08,000
That means, we come back over
here, by Hund's rule of maximum

592
00:45:08,000 --> 00:45:11,000
multiplicity,
we have two spin-up electrons

593
00:45:11,000 --> 00:45:16,000
in the O two molecule
with all these other orbitals

594
00:45:16,000 --> 00:45:20,000
down here being filled.
That configuration that I wrote

595
00:45:20,000 --> 00:45:24,000
down for the O two molecule over
there is a very shorthand way of

596
00:45:24,000 --> 00:45:29,000
representing the MO energy level
diagram --

597
00:45:29,000 --> 00:45:32,000
-- because I started here at
the lowest energy,

598
00:45:32,000 --> 00:45:35,000
and I went up in energy like
this.

599
00:45:35,000 --> 00:45:38,000
And what you can see is sigma
star cancels sigma,

600
00:45:38,000 --> 00:45:42,000
so no bonding there.
We have here three electron

601
00:45:42,000 --> 00:45:45,000
pairs in bonding orbitals,
sigma and 2pi,

602
00:45:45,000 --> 00:45:49,000
but we have two electrons that
are unpaired electrons,

603
00:45:49,000 --> 00:45:53,000
responsible for the
paramagnetism of the O two

604
00:45:53,000 --> 00:45:56,000
molecule,
that cancel some of the bonding

605
00:45:56,000 --> 00:46:01,000
down here.
When we want to write that down

606
00:46:01,000 --> 00:46:04,000
in terms of our bond order
convention, we have six bonding

607
00:46:04,000 --> 00:46:08,000
electrons minus two antibonding
electrons over two,

608
00:46:08,000 --> 00:46:11,000
which is going to be equal to a
bond order of two.

609
00:46:11,000 --> 00:46:14,000
That is our bond order for O
two.

610
00:46:14,000 --> 00:46:18,000
While the bond order is what we
expect to find based on that

611
00:46:18,000 --> 00:46:21,000
valance bond representation of O
two up there,

612
00:46:21,000 --> 00:46:25,000
what we find out is that the
representation of O two up there

613
00:46:25,000 --> 00:46:29,000
in the valance bond terminology
would not tell us that the

614
00:46:29,000 --> 00:46:32,000
molecule should actually have
two electrons that are not

615
00:46:32,000 --> 00:46:36,000
paired in the molecule,
--

616
00:46:36,000 --> 00:46:39,000
-- that it should have
paramagnetism due to the lack of

617
00:46:39,000 --> 00:46:42,000
pairing up of all the electrons,
spin pairing in the molecule.

618
00:46:42,000 --> 00:46:46,000
And it has to do with this high
symmetry of the O two molecule.

619
00:46:46,000 --> 00:46:48,000
And it cannot distort from high

620
00:46:48,000 --> 00:46:51,000
symmetry because it has just two
atoms.

621
00:46:51,000 --> 00:46:53,000
It has very few degrees of
freedom.

622
00:46:53,000 --> 00:46:56,000
It cannot bend or something to
get those electrons to pair up,

623
00:46:56,000 --> 00:47:01,000
and so they are unpaired.
And that is why dioxegen has

624
00:47:01,000 --> 00:47:04,000
unpaired electrons.
And then, if we go to nitrogen,

625
00:47:04,000 --> 00:47:09,000
the order that we write these
things down in is a little

626
00:47:09,000 --> 00:47:13,000
different because the energy
level diagram is different

627
00:47:13,000 --> 00:47:16,000
because now s and p are closer
together.

628
00:47:16,000 --> 00:47:19,000
And what we have is one sigma
with two electrons,

629
00:47:19,000 --> 00:47:22,000
or two sigma star with two
electrons.

630
00:47:22,000 --> 00:47:25,000
And then, next,
we have our one pi with its

631
00:47:25,000 --> 00:47:29,000
four electrons.
And then next,

632
00:47:29,000 --> 00:47:32,000
we have our three sigma with
its two electrons.

633
00:47:32,000 --> 00:47:35,000
And that is it.
We have two less electrons in

634
00:47:35,000 --> 00:47:38,000
the system than we do in
O two.

635
00:47:38,000 --> 00:47:41,000
That is N two.
And our bond order now is

636
00:47:41,000 --> 00:47:43,000
three.
That is in accord with our

637
00:47:43,000 --> 00:47:48,000
Lewis structure of N two,
but it is a triple bond.

638
00:47:48,000 --> 00:47:52,000
And it is a triple bond in a
molecule that is completely

639
00:47:52,000 --> 00:47:54,000
non-polar.
And it is one of the strongest

640
00:47:54,000 --> 00:47:58,000
chemical bonds known.
That triple bond is worth about

641
00:47:58,000 --> 00:48:03,000
226 kilocalories per mole.
And it makes N two a

642
00:48:03,000 --> 00:48:06,000
very fascinating molecule.
And, of course,

643
00:48:06,000 --> 00:48:09,000
it is the major constituent of
our atmosphere.

644
00:48:09,000 --> 00:48:13,000
And it is very inert,
due to the stability associated

645
00:48:13,000 --> 00:48:16,000
with having this diagram
populated up all through the

646
00:48:16,000 --> 00:48:19,000
bonding orbitals,
and then no more antibonding

647
00:48:19,000 --> 00:48:23,000
electrons present in the system.
So, that is pretty interesting.

648
00:48:23,000 --> 00:48:27,000
And given the time,
I think I will leave you after

649
00:48:27,000 --> 00:48:31,000
the following.
I just want to show you one

650
00:48:31,000 --> 00:48:36,000
more picture as a prelude to
where I will start next time.

651
00:48:36,000 --> 00:48:41,000
And that has to do with what
happens when the two atoms that

652
00:48:41,000 --> 00:48:45,000
are bonding are different.
Let's look at the three sigma

653
00:48:45,000 --> 00:48:49,000
orbital for the carbon monoxide
molecule.

654
00:48:49,000 --> 00:48:52,000
It would look like it does for
N two.

655
00:48:52,000 --> 00:48:55,000
It is the same number of
electrons.

656
00:48:55,000 --> 00:49:00,000
But look here.
It is very asymmetric.

657
00:49:00,000 --> 00:49:04,000
And that has to do with the
fact that carbon and oxygen have

658
00:49:04,000 --> 00:49:06,000
very different
electronegativity.

659
00:49:06,000 --> 00:49:10,000
And so, this highest occupied
molecular orbital for carbon

660
00:49:10,000 --> 00:49:13,000
monoxide, again,
an eight orbital problem,

661
00:49:13,000 --> 00:49:16,000
and again, a ten electron
system, is one where we have a

662
00:49:16,000 --> 00:49:21,000
larger coefficient on the lone
pair piece that is on the carbon

663
00:49:21,000 --> 00:49:24,000
atom of the molecule.
And I will start off with this

664
00:49:24,000 --> 00:49:30,000
next time, but that is one of
two reasons why CO is a poison.

665
00:49:30,000 --> 00:49:33,000
Have a nice weekend.