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We are back here for some more
stimulating discussion of

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chemistry.
And I want to begin today by

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just saying that I find it
particularly ironic that on the

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day that I am all prepared to
speak to you about the molecular

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orbital energy levels of
polyatomic molecules for which I

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am using methane as an
interesting example,

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the boards are screwed up.
The reason for that apparently

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is that this screen came down
and will not go back up.

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And one of the reasons that I
have designed my presentation

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technique for you this semester
so heavily around the use of the

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blackboard is that last year,
when I taught the class,

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I had a lot of feedback from
students about how the one day

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when we had audiovisual problems
and I couldn't use my PowerPoint

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and instead gave an impromptu
chalk talk, that that was their

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favorite lecture.
Maybe next year we will

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entirely be back to PowerPoint.
I don't know.

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We will see.
But, fortunately,

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I did write out my notes this
morning in electronic format,

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so I can make use of them on
the side boards.

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And, in addition,
I want to display for you some

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molecular orbital shapes and
properties.

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And I will do that by starting
out --

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Actually, I would like to
start, if I could,

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with this one.
I will go back to the document

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camera in a moment.

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Starting out right where we
left off last time,

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you will recall that we were
looking at diatomic molecules.

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We started out with a seven
orbital problem two lectures

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ago.
We looked at the planar BH

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three molecule.
We had a molecule where the MOs

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were mixing in two dimensions.
And then last time,

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when we looked at homonuclear
and heteronuclear diatomic

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molecules the MOs were,
in fact, mixing in just along

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an axis in one dimension.
And so, when we left the

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situation last time,
we had arrived at some kind of

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an understanding of the energy
levels in the carbon monoxide

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molecule.
And these were being developed,

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especially, with reference to
CO being a poisonous molecule,

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--
-- whereas, N two,

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another ten valence electron
diatomic molecule was very,

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very inert.
Why was that?

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It turns out that you have the
very electronegative oxygen atom

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on one side of the diagram with
its very low energetically lying

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orbitals mixing with the carbon
orbitals that are higher in

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energy.
And the neat thing about this,

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for the carbon monoxide
molecule, is that it results in

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an asymmetry of the molecular
orbitals --

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-- because the coefficients are
not the same on the carbon side

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as they are on the oxygen side
as those atomic orbitals blend

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together into the four sigma and
four pi molecular orbitals that

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we have.
We have a total of eight MOs,

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and they distribute
energetically as four energy

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levels in ascending energy that
are sigma in symmetry.

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That is, cylindrically
symmetric about the internuclear

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axis.
And then we had four that were

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pi.
Two energies,

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a pi bonding with four
electrons and a pi antibonding

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with just virtual empty
orbitals.

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And because this highest
occupied molecular orbital of

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carbon monoxide is so heavily
concentrated on the carbon end

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of the molecule,
you can essentially view this

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as a carbon-based lone pair.
It has this big lobe here,

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through which the carbon can
serve as a nucleophile or as a

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base and bind to something that
has an empty orbital.

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And if that something that has
an empty orbital also has a

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filled orbital that matches the
symmetry of the LUMO of carbon

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monoxide, then you get the
possibility for multiple bonding

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between the carbon of the CO
molecule and that other entity.

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And we will see that that other
entity may, in fact,

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be a d-block transition
element, such as iron.

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And what I wanted to do was to
go ahead and show you that the

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LUMO is an interesting and
unsymmetrical one,

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as well.
Let's take a look at that.

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And so, in this picture I have
gone ahead and included a little

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ball and stick diagram
underneath the representation of

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the orbital lobes.
Here is the carbon,

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and the oxygen is colored red.
And you can see that the lobes

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on the LUMO, the lowest
unoccupied molecular orbital in

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this system, are also large on
carbon relative to oxygen.

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The same as we had found for
the highest occupied MO.

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What this means is that that
carbon atom in the CO molecule

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is simultaneously a base in the
sigma system.

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And because this is our lowest
unoccupied molecular orbital,

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our carbon atom is actually an
acid in the pi system.

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And that makes it
electronically very

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complimentary to d-block
elements, as we will see when we

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get into the chemistry of
elements that do have valence

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d-orbitals in addition to s and
p, coming up in a couple of

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lectures.
And so what can happen is that,

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let's say that a carbon
monoxide molecule somehow

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erroneously finds its way into
your blood stream,

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you have there hemoglobin,
the iron atom at the center of

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which acts reversibly to bind
the dioxygen molecule.

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But it irreversibly binds to
the carbon monoxide molecule

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because the iron center,
which is sitting at the center

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of a heme unit that I mentioned,
and we will talk about the

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structure of hemoglobin and the
way that it functions a little

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bit later.
But if CO gets in there,

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it binds to the iron through
the HOMO, acting as a base to

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the metal acting as a sigma
acid, and also through

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d-electrons being able to act as
a pi base from the metal binding

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with this LUMO of the carbon
monoxide molecule.

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And this LUMO is doubly
degenerate.

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That means there are two of
them at the same energy.

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And we will come back to this
issue of orbital degeneracy

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later in today's lecture,
but that electronic asymmetry

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of carbon monoxide and the fact
that the HOMO and LUMO are both

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centered on the carbon end of
the molecule are what make CO

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such an interesting molecule and
such a reactive molecule and

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therefore a poison.
And so now let us please go

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back to the document camera.

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What I have shown here are the
five platonic solids.

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Actually, I spent a lot of time
over the weekend practicing

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drawing these on the chalkboard
so that I would be able to do

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that very nicely for you.
And here it comes that we do

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not have that option today,
so I will show them to you this

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way.
And the platonic solids are the

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five regular polyhedra.
And that means that they are

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polyhedra that have,
as their faces,

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regular polygons that are all
identical.

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And the reason I am bringing
this up today is that we had

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talked about issues of symmetry
with reference to molecular

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orbital theory.
And these are all very highly

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symmetric shapes,
as you can immediately

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recognize.
The buckyball that we looked at

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earlier is not here because it
has both five-membered rings and

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six-membered rings on the faces.
And so it doesn't have all the

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faces identical.
And those that do have all the

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faces identical are the platonic
solids.

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And the complete list of those
is shown here.

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With triangles,
you can put four equilateral

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triangles together to make a
tetrahedron.

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There is actually a journal
called Tetrahedron.

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And there is another one called
Tetrahedron Letters.

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And there is another one called
Tetrahedron Reports.

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And the reason is that organic
chemistry is carbon-based,

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and carbon so often gets
involved in forming tetrahedra.

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And so the tetrahedron is a
very important symbol of organic

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chemistry.
And we will get back to that.

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And with triangles,
if instead of putting four of

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them together,
you put eight of them together

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this way, you get the
octahedron.

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And the octahedron is equally
important as a symbol for

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inorganic chemistry.
And the reason for that is that

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when we have metal ions in
solution, they often sit at the

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center of a coordination
octahedron in which six ligands

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bind to that metal center.
And we are going to get into

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how that was discovered pretty
shortly by Alfred Werner,

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another Nobel laureate in
chemistry.

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And with triangles you can also
make the icosahedron shown here,

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and this has 20 faces.
And the icosahedron is

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important in chemistry.
And certain elements,

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for example boron,
actually have structures in the

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solid state in which the boron
atoms are arranged at the

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vertices of a regular
icosahedron.

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So this is a chemically
important shape as well.

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And if you go from triangles to
squares you can,

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of course, make a cube.
And probably the most familiar

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of a platonic solid to us.
And then, if we go up to

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pentagons and put those
together, as you can see here,

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we can put 12 pentagons
together in a regular way,

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that gives rise to what we call
the pentagonal dodecahedron.

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And you can imagine that each
of these vertices could be a C-H

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unit, carbon-hydrogen.
And then each carbon would have

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its valence of four satisfied.
And there would be all single

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bonds at the parameter of this
round molecule.

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And, in fact,
that dodecahedron molecule was

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the objective of organic
synthesis for quite a long time

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before it was finally
successfully synthesized by Leo

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Paquette.
You would be surprised with the

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kinds of challenges,
maybe, sometimes that do drive

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chemists to do what they do.
And that is the platonic

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solids.
I am going to be focusing our

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attention on methane.
And, if I could have the side

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board go back to the computer
for a moment,

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that would be great.
These are the notes for today's

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lecture that I wrote up this
morning that you will be able to

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download later today,
or maybe you already can.

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I don't know.
In any event,

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when we approach the methane
problem in molecular orbital

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theory, what we do is recognize
that the tetrahedron can be

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conveniently inscribed into a
cube by placing hydrogens at

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alternating corners of the cube,
as shown here.

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And then, when we do that,
the carbon atom of the methane

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molecule lies at the very center
of this cube.

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And this way of arranging the
system in space will help us to

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simplify the problem of the
molecular orbital energy levels

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for the methane molecule.
And we are going to be

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ultimately working toward a
comparison of the molecular

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orbital description of the
methane molecule with that

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afforded by valance bond theory.
Now, some of you approached me

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after lecture last time and
said, what happened to

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hybridization?
And, if you wanted to talk

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about the methane molecule and
the language of valence bond

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theory, you would say that this
carbon atom is sp three

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hybridized in order that it may
form simultaneously four two

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electron bonds between the
central carbon and the four

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peripheral hydrogens.
Well, where hybridization went

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is that it stayed with valence
bond theory.

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Let me reemphasize the
distinction between the

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molecular orbital treatment of
molecule electronic structure as

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contrasted with the valence bond
theory in which hybridization

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plays a role.
It does not play a role in the

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language of MO theory,
so please keep those quite

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distinct.
Now, we need to choose a

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coordinate system for our
problem.

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And I won't spend too much time
discussing how we choose the

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coordinate system,
but what we have done in the

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present case with methane is to
choose the x,

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y, and z axes each to come out
of the faces of the cube.

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And we do this because that
choice of coordinate system will

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lead to the equivalence of the
carbons' 2px,

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2py, and 2pz orbitals in 3D
space.

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We could choose a different
coordinate system,

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and that would make the problem
more difficult.

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So, what we have done is chosen
a convenient set of coordinates

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where the x, y,
and z axes each come out of the

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faces of our cube.
Now, we recognize that the

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methane problem,
like the homonuclear and

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heteronuclear diatomic molecule
problems we talked about last

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time, is an eight orbital
problem.

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The diatomics N two and
CO each had ten electrons to go

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into their eight molecular
orbitals.

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And here, with methane,
we have eight electrons,

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four coming in as the valence
orbitals from the carbon atom,

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valence electrons from the
carbon atom, rather,

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and four coming in as the four
valence 1s electrons for the

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four hydrogens.
And our eight orbitals are the

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carbons 2s, 2px,
2py, and 2pz orbitals.

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And we have four more orbitals,
the 1s from each of the four

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hydrogens.
We know how many electrons are

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going to be able to go into our
MO energy level diagram.

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00:15:18 --> 00:15:22
We know how many MOs we are
going to have because we know

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how many valance AOs we have.
And those numbers are equal at

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eight.
The choice of this problem is

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predicated on my desire to show
you how things progress as we go

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from a planar system to a linear
system to now a system that has

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molecular orbitals that stretch
out in three-dimensional space.

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Now, we are going to take the
following approach.

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We are going to say,
how can we generate linear

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combinations of the four
hydrogen 1s orbitals that will

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be so constructed as to match
the nodal properties of the

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00:16:01 --> 00:16:06
carbon atomic orbitals that are
the valence orbitals of that

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carbon?
Okay.

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So that is our goal,
and that is our strategy.

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And let me say also,
that in this tetrahedral

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symmetry, you can recognize that
you cannot distinguish any one

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00:16:23 --> 00:16:28
of the four vertices of a
tetrahedron from the other

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00:16:28 --> 00:16:31
three.
And that is important.

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00:16:31 --> 00:16:36
And that is also true from the
way that we have decided to

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situate the hydrogen nuclei on
this cube relative to our

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00:16:40 --> 00:16:43
coordinate system.
In other words,

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none of these four hydrogens
lie on any of the three

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Cartesian axes.
And their relationship to the

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coordinate system and to the
shape of the cube each is

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indistinguishable from that of
the other three hydrogens.

264
00:17:01 --> 00:17:04
And that is an important aspect
of the way that we set up the

265
00:17:04 --> 00:17:07
problem.
And why do I mention that?

266
00:17:07 --> 00:17:10
Because I mentioned that when
constructing MOs in general,

267
00:17:10 --> 00:17:15
one of the first things you can
do, if you are going to be able

268
00:17:15 --> 00:17:19
to take advantage of symmetry to
help you solve the problem of

269
00:17:19 --> 00:17:23
these MO energy levels and how
they are constructed from linear

270
00:17:23 --> 00:17:27
combinations of atomic orbitals
is to identify symmetry-related

271
00:17:27 --> 00:17:32
sets of atoms in orbitals.
If you are keeping that in

272
00:17:32 --> 00:17:37
mind, you will realize that we
are treating all four hydrogens

273
00:17:37 --> 00:17:41
together because they are
indistinguishable from one

274
00:17:41 --> 00:17:46
another in the tetrahedral shape
of the methane molecule.

275
00:17:46 --> 00:17:50
So, that is important.
If we had a more complicated

276
00:17:50 --> 00:17:55
molecule, such as if we were
looking at --

277
00:17:55 --> 00:17:58
I get to use a little bit of
board space today.

278
00:17:58 --> 00:18:02
Here is a slightly more
complicated hydrocarbon

279
00:18:02 --> 00:18:04
molecule.
Here is a slightly more

280
00:18:04 --> 00:18:07
complicated hydrocarbon
molecule.

281
00:18:07 --> 00:18:11
And in a molecule like this,
this is cyclohexane drawn in a

282
00:18:11 --> 00:18:15
chair confirmation.
Sir Derek Barton actually was

283
00:18:15 --> 00:18:19
the Nobel laureate who got his
prize for discussing

284
00:18:19 --> 00:18:22
conformational aspects of
hydrocarbons.

285
00:18:22 --> 00:18:26
We would find that the
hydrogens split up into two

286
00:18:26 --> 00:18:30
sets.
They are not all equivalent in

287
00:18:30 --> 00:18:34
the cyclohexane molecule in this
frozen out chair confirmation

288
00:18:34 --> 00:18:38
because we have one type that is
equatorial and one type that is

289
00:18:38 --> 00:18:42
axial on each of these carbons
as we go around this

290
00:18:42 --> 00:18:45
six-membered ring.
And so there are different sets

291
00:18:45 --> 00:18:48
that you would have to treat
separately.

292
00:18:48 --> 00:18:50
But methane is more
symmetrical.

293
00:18:50 --> 00:18:54
And all four hydrogens in the
methane molecule are identical,

294
00:18:54 --> 00:18:58
so we are going to treat them
together in developing our

295
00:18:58 --> 00:19:02
linear combinations.
And, coming back to this,

296
00:19:02 --> 00:19:06
what we are going to do,
once again, to generate these

297
00:19:06 --> 00:19:10
things, is to imagine the
central atom atomic orbital as

298
00:19:10 --> 00:19:13
growing out in space out to
where the hydrogen nuclei are.

299
00:19:13 --> 00:19:17
And then we are going to take
the sign of the wave function

300
00:19:17 --> 00:19:21
that we find out there and apply
it to those hydrogens that are

301
00:19:21 --> 00:19:26
out there in space in order to
see what our linear combination

302
00:19:26 --> 00:19:29
must look like.
If you imagine the carbon 2s

303
00:19:29 --> 00:19:32
orbital, which has the same
sign, positive,

304
00:19:32 --> 00:19:35
everywhere, and imagine it
growing out to where the

305
00:19:35 --> 00:19:38
hydrogens are,
and we are labeling the

306
00:19:38 --> 00:19:40
hydrogen A, B,
C, and D, as shown here.

307
00:19:40 --> 00:19:44
And we will keep that same
labeling scheme throughout the

308
00:19:44 --> 00:19:47
development of this problem.
You would see that the carbon

309
00:19:47 --> 00:19:51
2s positive wave function,
as we imagine it growing out to

310
00:19:51 --> 00:19:54
where the hydrogens are,
would touch each of these

311
00:19:54 --> 00:19:58
hydrogens at the same time.
And, thus, we confer the sign

312
00:19:58 --> 00:20:02
positive on all four of the
contributors to this first of

313
00:20:02 --> 00:20:07
our linear combinations.
And what that will represent is

314
00:20:07 --> 00:20:12
simultaneous bonding of the
carbon 2s orbital with all four

315
00:20:12 --> 00:20:15
hydrogens and with the same sign
everywhere.

316
00:20:15 --> 00:20:19
And so, we can write down
one-half A plus B plus C plus D,

317
00:20:19 --> 00:20:23
where I am using A here to

318
00:20:23 --> 00:20:28
represent the hydrogen 1s
orbital associated with hydrogen

319
00:20:28 --> 00:20:32
labeled A.
So I have just abbreviated that

320
00:20:32 --> 00:20:33
in A.
And, similarly,

321
00:20:33 --> 00:20:37
B is the hydrogen 1s orbital
associated with hydrogen B and

322
00:20:37 --> 00:20:39
so on.
So, this is a linear

323
00:20:39 --> 00:20:42
combination of these four
hydrogen orbitals.

324
00:20:42 --> 00:20:44
And it is normalized with a
half, here.

325
00:20:44 --> 00:20:48
And we have the same
coefficient on each of the four

326
00:20:48 --> 00:20:52
hydrogens, so the size of the
lobe that you would draw at each

327
00:20:52 --> 00:20:55
four of these positions would be
the same.

328
00:20:55 --> 00:20:59
That is number one.
We have built a linear

329
00:20:59 --> 00:21:03
combination based on the nodal
properties of the carbon 2s

330
00:21:03 --> 00:21:06
atomic orbital.
And now, let's do two more.

331
00:21:06 --> 00:21:11
And we are going to base these
on the nodal properties of the

332
00:21:11 --> 00:21:15
central carbon 2px orbital and
the central carbon 2py orbital.

333
00:21:15 --> 00:21:19
I am using the same coordinate
system as I started with,

334
00:21:19 --> 00:21:24
namely, that the positive
x-axis comes out of the face of

335
00:21:24 --> 00:21:28
the cube and comes toward us,
and the positive y-axis comes

336
00:21:28 --> 00:21:33
out of this face of the cube and
goes that way.

337
00:21:33 --> 00:21:35
And we will look at z in a
moment.

338
00:21:35 --> 00:21:40
That one is going straight up
out of the top face of the cube

339
00:21:40 --> 00:21:44
for positive z.
What you can see is that if you

340
00:21:44 --> 00:21:48
imagine the carbon 2px atomic
orbital as just growing out to

341
00:21:48 --> 00:21:53
where the four hydrogens are,
we will see that the hydrogens

342
00:21:53 --> 00:21:57
in positions A and B will
experience the positive lobe of

343
00:21:57 --> 00:22:02
the carbon 2px orbital at the
same moment as the two hydrogens

344
00:22:02 --> 00:22:06
in back labeled C and D
experience the negative lobe of

345
00:22:06 --> 00:22:12
the carbon 2px orbital.
And so, we give positive phase

346
00:22:12 --> 00:22:16
to A and B and negative phase to
C and D.

347
00:22:16 --> 00:22:20
And we give them the same
coefficient everywhere such that

348
00:22:20 --> 00:22:25
this linear combination is
normalized as one-half A plus B

349
00:22:25 --> 00:22:30
minus C minus D.

350
00:22:30 --> 00:22:33
That is our second linear
combination of hydrogen

351
00:22:33 --> 00:22:36
orbitals.
And this one has been so

352
00:22:36 --> 00:22:40
generated as to match the nodal
property of the carbon 2px

353
00:22:40 --> 00:22:44
orbital, the key feature of
which is that it is everywhere

354
00:22:44 --> 00:22:48
positive along positive x,
and everywhere along negative x

355
00:22:48 --> 00:22:51
it is negative.
And the 2py orbital of carbon

356
00:22:51 --> 00:22:56
has the property that everywhere
in the plus y region of space,

357
00:22:56 --> 00:22:59
it is positive,
so that C and B will be

358
00:22:59 --> 00:23:05
associated with a positive sign
for this linear combination.

359
00:23:05 --> 00:23:09
And the 2py orbital is negative
everywhere along negative y,

360
00:23:09 --> 00:23:12
back here.
And that is where hydrogens A

361
00:23:12 --> 00:23:15
and D are located.
So, A and D will carry a

362
00:23:15 --> 00:23:18
negative sign for this new
linear combination,

363
00:23:18 --> 00:23:22
which is one-half of minus A
plus B plus C minus D.

364
00:23:22 --> 00:23:26
This is just one way
of writing down

365
00:23:26 --> 00:23:32
the picture that we see here.
And there is only one left to

366
00:23:32 --> 00:23:37
be generated from the nodal
properties of the carbon 2pz

367
00:23:37 --> 00:23:42
orbital, which has its positive
node oriented along positive z

368
00:23:42 --> 00:23:46
and its negative lobe oriented
along negative z,

369
00:23:46 --> 00:23:50
and it has the x,y-plane as a
nodal surface.

370
00:23:50 --> 00:23:54
And so, therefore,
hydrogens A and C will carry a

371
00:23:54 --> 00:24:00
positive sign because they are
oriented and located in the plus

372
00:24:00 --> 00:24:05
z region of space.
Whereas, hydrogens B and D are

373
00:24:05 --> 00:24:09
located in the negative z region
of space, so they will carry a

374
00:24:09 --> 00:24:13
negative sign.
And that one can be normalized

375
00:24:13 --> 00:24:16
also with a normalization
coefficient of one-half.

376
00:24:16 --> 00:24:19
And so it is A minus B plus C
minus D.

377
00:24:19 --> 00:24:23
And just don't forget
that these letters

378
00:24:23 --> 00:24:27
refer to the 1s orbitals
associated with the hydrogens

379
00:24:27 --> 00:24:32
that are so labeled.
And we started with identifying

380
00:24:32 --> 00:24:37
the fact that the four hydrogens
in this problem are all

381
00:24:37 --> 00:24:40
equivalent.
And, thus, their 1s orbitals

382
00:24:40 --> 00:24:44
are all equivalent,
indistinguishable from each

383
00:24:44 --> 00:24:47
other with respect to their
spatial orientation.

384
00:24:47 --> 00:24:51
And, so we took four atomic
orbitals, and we are

385
00:24:51 --> 00:24:56
constructing from them four
linear combinations of those

386
00:24:56 --> 00:25:01
four atomic orbitals.
I want to emphasize what I have

387
00:25:01 --> 00:25:04
written here,
which is that with the

388
00:25:04 --> 00:25:08
coordinate system we have chosen
and in this high tetrahedral

389
00:25:08 --> 00:25:13
symmetry, the carbons 2p x,
y and z orbitals are identical.

390
00:25:13 --> 00:25:17
And, just like in a carbon atom
floating free in space,

391
00:25:17 --> 00:25:21
they are degenerate.
That means they have the same

392
00:25:21 --> 00:25:24
energy.
And so we will talk about

393
00:25:24 --> 00:25:29
orbitals in atoms being
degenerate in a moment.

394
00:25:29 --> 00:25:33
But we find that the 2p x,
y and z orbitals on that carbon

395
00:25:33 --> 00:25:37
are going to have the same
energy, and they are degenerate.

396
00:25:37 --> 00:25:41
And the molecular orbitals
constructed from them in this

397
00:25:41 --> 00:25:44
tetrahedral system will have the
same property.

398
00:25:44 --> 00:25:47
They will be degenerate.
So, something that is

399
00:25:47 --> 00:25:51
occasionally associated with
high symmetry is multiple MO

400
00:25:51 --> 00:25:55
degeneracy.
Next, knowing what our carbon

401
00:25:55 --> 00:25:59
atom atomic orbitals are and
knowing also what our four

402
00:25:59 --> 00:26:03
linear combinations are of the
hydrogen 1s orbitals,

403
00:26:03 --> 00:26:07
we can now allow them to
interact in ways that are either

404
00:26:07 --> 00:26:11
bonding or antibonding.
And they interact,

405
00:26:11 --> 00:26:14
of course, according to their
nodal properties.

406
00:26:14 --> 00:26:19
Those linear combinations that
have the same nodal properties

407
00:26:19 --> 00:26:23
as a particular carbon atom
atomic orbital will lead to a

408
00:26:23 --> 00:26:29
bonding interaction and an
antibonding interaction.

409
00:26:29 --> 00:26:33
And we have an eight orbital
problem, so we better see eight

410
00:26:33 --> 00:26:37
energy levels appearing in our
molecular orbital energy level

411
00:26:37 --> 00:26:39
diagram for this methane
molecule.

412
00:26:39 --> 00:26:43
And, when we go down here,
starting at the lowest energy

413
00:26:43 --> 00:26:46
molecular orbital,
we find that it can be

414
00:26:46 --> 00:26:49
constructed by adding to the
carbon 2s orbital,

415
00:26:49 --> 00:26:53
which is, once again,
spherically symmetric,

416
00:26:53 --> 00:26:57
this linear combination that is
A plus B plus C plus D.

417
00:26:57 --> 00:27:00
And we can draw this with a

418
00:27:00 --> 00:27:03
surface like this,
which envelopes all five of our

419
00:27:03 --> 00:27:08
nuclei which has the same sign
everywhere.

420
00:27:08 --> 00:27:10
And so our lowest lying
molecular orbital,

421
00:27:10 --> 00:27:14
our most bonding molecular
orbital, our most stabilized MO,

422
00:27:14 --> 00:27:17
which can house a pair of
electrons because it is singly

423
00:27:17 --> 00:27:21
degenerate, is one that will
look something like this.

424
00:27:21 --> 00:27:23
It envelopes all five of our
nuclei.

425
00:27:23 --> 00:27:26
And, effectively,
the pair of electrons is

426
00:27:26 --> 00:27:30
associated simultaneously with
all five nuclei.

427
00:27:30 --> 00:27:33
And the wave function has the
same sign everywhere,

428
00:27:33 --> 00:27:38
just like the carbon 2s orbital
from which this is partially

429
00:27:38 --> 00:27:41
constructed does.
And then, next we have three

430
00:27:41 --> 00:27:45
molecular orbitals that are each
built by forming bonding

431
00:27:45 --> 00:27:48
interactions with the carbon's
p-orbitals, px,

432
00:27:48 --> 00:27:53
py, and pz with the linear
combination that matches that

433
00:27:53 --> 00:27:56
carbon orbital in terms of the
nodal properties.

434
00:27:56 --> 00:28:00
And we built them that way so
that becomes particularly

435
00:28:00 --> 00:28:04
straightforward.
This one here is coming out,

436
00:28:04 --> 00:28:07
the face that points to the
right.

437
00:28:07 --> 00:28:11
And so you can see that this
one is based on the carbon 2py

438
00:28:11 --> 00:28:13
orbital.
And, if that carbon 2py orbital

439
00:28:13 --> 00:28:18
makes a bonding interaction with
the linear combination denoted

440
00:28:18 --> 00:28:21
as minus A plus B plus
C minus D,

441
00:28:21 --> 00:28:25
the bonding interaction here is
signified by this plus sign.

442
00:28:25 --> 00:28:30
So, we have a plus sign here
and a plus sign here.

443
00:28:30 --> 00:28:34
And, thus, we have a molecular
orbital that has the same nodal

444
00:28:34 --> 00:28:37
properties as a carbon 2py
orbital does,

445
00:28:37 --> 00:28:41
which means that the x,z-plane
is a nodal surface for this

446
00:28:41 --> 00:28:45
molecular orbital.
And what it looks like is kind

447
00:28:45 --> 00:28:49
of a two-bladed propeller.
You have a plus phase that is

448
00:28:49 --> 00:28:54
distributed, enveloping the two
hydrogens that we labeled B and

449
00:28:54 --> 00:29:00
C, along with that positive lobe
of the carbon 2py orbital.

450
00:29:00 --> 00:29:04
So that you have that forming
one blade of our two-bladed

451
00:29:04 --> 00:29:08
propeller.
And the blade with the other

452
00:29:08 --> 00:29:12
sign is rotated relative to the
first by 45 degrees.

453
00:29:12 --> 00:29:16
Sorry, by 90 degrees,
in fact, and will involve the

454
00:29:16 --> 00:29:21
distribution of bonding
character over carbon's negative

455
00:29:21 --> 00:29:26
2py lobe interacting with these
two hydrogens back here,

456
00:29:26 --> 00:29:32
which are hydrogens A and D.
And so we will visualize these

457
00:29:32 --> 00:29:34
in a moment to make that more
clear.

458
00:29:34 --> 00:29:37
But these three bonds,
one oriented along x,

459
00:29:37 --> 00:29:41
one oriented along y,
and one oriented along z all

460
00:29:41 --> 00:29:44
look exactly like this.
They are just rotated 90

461
00:29:44 --> 00:29:47
degrees to each other in space,
just like the x,

462
00:29:47 --> 00:29:51
y, and z Cartesian axes are.
I have only drawn one of them,

463
00:29:51 --> 00:29:55
but we have three of them
because there is a three-fold

464
00:29:55 --> 00:30:00
orbital degeneracy in this
molecular orbital.

465
00:30:00 --> 00:30:03
So, singly degenerate and then
triply degenerate.

466
00:30:03 --> 00:30:07
And then, as we go up in
energy, we will find that our

467
00:30:07 --> 00:30:10
lowest unoccupied molecular
orbital is constructed by

468
00:30:10 --> 00:30:13
subtracting.
It is the same as I have

469
00:30:13 --> 00:30:15
written here,
but with a minus sign,

470
00:30:15 --> 00:30:19
the C 2s subtracting A plus B
plus C plus D.

471
00:30:19 --> 00:30:21
A, B, C and D hydrogens all

472
00:30:21 --> 00:30:25
have the same sign to their wave
function as each other,

473
00:30:25 --> 00:30:29
but they have the opposite sign
as the contribution from the

474
00:30:29 --> 00:30:34
carbon 2s orbital to this
molecular orbital.

475
00:30:34 --> 00:30:37
And what that results in,
as I have tried to indicate

476
00:30:37 --> 00:30:41
with some red dashes here,
is that you generate a nodal

477
00:30:41 --> 00:30:46
surface in between the carbon
and hydrogen nuclei that will go

478
00:30:46 --> 00:30:50
completely around and will
intersect perpendicular to each

479
00:30:50 --> 00:30:54
of these carbon-hydrogen
inter-nuclear vectors.

480
00:30:54 --> 00:30:58
You have a change of sign as
you are traversing the path from

481
00:30:58 --> 00:31:03
carbon to any one of the four
hydrogen nuclei.

482
00:31:03 --> 00:31:06
And that is what we get by just
reversing the sign relative to

483
00:31:06 --> 00:31:09
the bonding counterpart of that
orbital.

484
00:31:09 --> 00:31:11
And that makes this one
anti-bonding.

485
00:31:11 --> 00:31:15
We will give it an asterisk
here to denote antibonding

486
00:31:15 --> 00:31:17
character to this molecular
orbital.

487
00:31:17 --> 00:31:21
And now, ascending in energy,
we will find that for each one

488
00:31:21 --> 00:31:25
of these orbitals that is bonded
and oriented with respect to x,

489
00:31:25 --> 00:31:28
y or z in Cartesian space,
we find that there is an

490
00:31:28 --> 00:31:32
antibonding counterpart.
Again, what we are doing is

491
00:31:32 --> 00:31:36
taking, for example,
carbons 2py and subtracting the

492
00:31:36 --> 00:31:40
linear combination that
corresponds to carbon 2py in

493
00:31:40 --> 00:31:42
terms of its nodal symmetry
properties.

494
00:31:42 --> 00:31:47
We are subtracting minus A plus
B plus C minus D.

495
00:31:47 --> 00:31:49
The carbon 2py has the same

496
00:31:49 --> 00:31:53
plus, its positive lobe is
oriented along positive y,

497
00:31:53 --> 00:31:57
just as it is down here,
but now we have reversed the

498
00:31:57 --> 00:32:02
sign of the four hydrogens.
And what this introduces will

499
00:32:02 --> 00:32:07
be antibonding nodal surfaces in
between the carbon and the

500
00:32:07 --> 00:32:11
hydrogens.
And they are strong antibonding

501
00:32:11 --> 00:32:16
interactions because of the good
directional overlap of,

502
00:32:16 --> 00:32:20
say, the positive lobe of
carbon's 2py with these

503
00:32:20 --> 00:32:26
s-orbital contributions in the
negative from hydrogens C and B,

504
00:32:26 --> 00:32:28
here.
And back here,

505
00:32:28 --> 00:32:33
A and D rotated,
of course, by 90 degrees.

506
00:32:33 --> 00:32:37
One of the points that I was
illustrating last time and

507
00:32:37 --> 00:32:41
really trying to emphasize in
the context of MO theory is

508
00:32:41 --> 00:32:45
that, in general,
we expect orbitals to be higher

509
00:32:45 --> 00:32:48
in energy if they have more
internuclear nodes.

510
00:32:48 --> 00:32:53
And the energy of the orbitals
is also affected by which atomic

511
00:32:53 --> 00:32:58
orbitals are close to the MO in
energy and also by factors of

512
00:32:58 --> 00:33:01
overlap.
And normally,

513
00:33:01 --> 00:33:04
for example,
a sigma bond is typically

514
00:33:04 --> 00:33:09
associated with greater overlap
than a pi bond because of the

515
00:33:09 --> 00:33:13
directionality of the overlap.
And here, you have an

516
00:33:13 --> 00:33:17
interesting case,
where this p-orbital is not

517
00:33:17 --> 00:33:23
directed right at the hydrogen.
And it is not a pi bond either.

518
00:33:23 --> 00:33:27
It is something in between sort
of sigma and a pi type of

519
00:33:27 --> 00:33:32
orientation.
It is an oblique interaction of

520
00:33:32 --> 00:33:36
the hydrogen 1s wavefunction
with that lobe of the p-orbital.

521
00:33:36 --> 00:33:40
But, still, it gives rise to
very good overlap down here,

522
00:33:40 --> 00:33:44
and consequently a very strong
antibonding character up here in

523
00:33:44 --> 00:33:47
counterpart.
And, from calculation,

524
00:33:47 --> 00:33:51
this is the order of the energy
levels that come out of that.

525
00:33:51 --> 00:33:55
And so we have only four energy
levels, two of which are triply

526
00:33:55 --> 00:33:59
degenerate, so we indeed have
eight wave functions that

527
00:33:59 --> 00:34:04
correspond to the molecular
orbitals of this system.

528
00:34:04 --> 00:34:25
Question down here?
Could you say that louder?

529
00:34:25 --> 00:34:26
Why did this not have the
one-half?

530
00:34:26 --> 00:34:29
The one-half is missing there
because I forgot to put it

531
00:34:29 --> 00:34:32
there.
Yes, to be normalized it should

532
00:34:32 --> 00:34:35
have a one-half there.
That is right.

533
00:34:35 --> 00:34:39
But then there is a further
point that that brings up,

534
00:34:39 --> 00:34:43
which is the following.
What I am implying here is that

535
00:34:43 --> 00:34:48
there are equal amounts of
carbon 2s in the bonding orbital

536
00:34:48 --> 00:34:51
and as in the antibonding
counterpart.

537
00:34:51 --> 00:34:55
But that need not be the case.
It could be that this is 0.6

538
00:34:55 --> 00:35:01
here and that this 0.4 up here.
And that is a greater level of

539
00:35:01 --> 00:35:06
detail than I am really
interested in going into for the

540
00:35:06 --> 00:35:09
purposes of this problem.
But your point,

541
00:35:09 --> 00:35:13
with respect to the
coefficient, is well taken.

542
00:35:13 --> 00:35:16
Thanks.
Christine, maybe we can fix

543
00:35:16 --> 00:35:20
that on the one that gets posted
on the web.

544
00:35:20 --> 00:35:26
Let's now look at these things
using our VMD program.

545
00:35:26 --> 00:35:37


546
00:35:37 --> 00:35:41
Starting from the lowest energy
and we will work our way up.

547
00:35:41 --> 00:35:48


548
00:35:48 --> 00:35:52
Which orbital is this?
This is a molecular orbital of

549
00:35:52 --> 00:35:54
the methane molecule.

550
00:35:54 --> 00:36:00


551
00:36:00 --> 00:36:04
One way you can identify
orbitals is where they are with

552
00:36:04 --> 00:36:06
reference to the HOMO or the
LUMO.

553
00:36:06 --> 00:36:09
In this case,
the HOMO is triply degenerate.

554
00:36:09 --> 00:36:13
And there is one orbital lower
in energy than the HOMO.

555
00:36:13 --> 00:36:17
This would be then the HOMO
minus one because it is one

556
00:36:17 --> 00:36:21
below the HOMO in energy.
That is the one which is the

557
00:36:21 --> 00:36:24
carbon 2s, plus A plus B plus C
plus D.

558
00:36:24 --> 00:36:27
That is that MO there.

559
00:36:27 --> 00:36:33


560
00:36:33 --> 00:36:39
Actually, let's go ahead and
take that one away.

561
00:36:39 --> 00:36:49


562
00:36:49 --> 00:36:50
Sorry about that.

563
00:36:50 --> 00:36:55


564
00:36:55 --> 00:37:00
Let's see if you can figure out
which one this is.

565
00:37:00 --> 00:37:14


566
00:37:14 --> 00:37:18
In order to figure out which
one it is, you need to be able

567
00:37:18 --> 00:37:23
to realize where the nuclei are
and where your internuclear

568
00:37:23 --> 00:37:26
nodes are, if any.
And, if so, how many.

569
00:37:26 --> 00:37:30
This is one of our hydrogen
contributions.

570
00:37:30 --> 00:37:33
Here is a hydrogen contribution
with opposite sign.

571
00:37:33 --> 00:37:38
And there is a piece of our
carbon 2p orbital in the center.

572
00:37:38 --> 00:37:43
Now, this is just one of three
identical counterparts that are

573
00:37:43 --> 00:37:47
rotated 90 degrees with respect
to each other in space,

574
00:37:47 --> 00:37:51
so I am not going to show you
all three of them.

575
00:37:51 --> 00:37:55
They all look exactly the same.
Is this one bonding or

576
00:37:55 --> 00:38:00
antibonding?
Yeah, this one is antibonding.

577
00:38:00 --> 00:38:03
And it does not have the same
sign on all the hydrogens,

578
00:38:03 --> 00:38:07
so it has to be part of our
LUMO, which is triply degenerate

579
00:38:07 --> 00:38:09
and involves bonding or,
in this case,

580
00:38:09 --> 00:38:13
the antibonding between a
carbon 2p orbital with a linear

581
00:38:13 --> 00:38:17
combination of four hydrogens.
You can see that when you have

582
00:38:17 --> 00:38:20
an antibonding orbital,
you are depleting the electron

583
00:38:20 --> 00:38:25
density in the space between the
nuclei rather than increasing it

584
00:38:25 --> 00:38:30
as you do in a bonding orbital.
Let's go ahead and look at the

585
00:38:30 --> 00:38:34
bonding counterpart of this one.
And we will do that.

586
00:38:34 --> 00:38:38
And let's delete it.
One thing that you will get,

587
00:38:38 --> 00:38:43
if you haven't already gotten
it, is you are going to get a

588
00:38:43 --> 00:38:48
link to be able to download all
the files that I used to show

589
00:38:48 --> 00:38:51
you molecular orbitals with in
class.

590
00:38:51 --> 00:38:55
Because this program I am
using, this VMD program,

591
00:38:55 --> 00:39:01
is available on Athena.
And you can also download it

592
00:39:01 --> 00:39:06
yourself for free onto your own
computer if you want to.

593
00:39:06 --> 00:39:10
And so this is going to give
you the opportunity to study

594
00:39:10 --> 00:39:16
these orbitals and rotate them
around at your leisure on your

595
00:39:16 --> 00:39:19
desktop.
And now, let's look at the

596
00:39:19 --> 00:39:24
bonding counterpart to that
antibonding orbital that we just

597
00:39:24 --> 00:39:27
looked at.
This one is the two-bladed

598
00:39:27 --> 00:39:33
propeller to which I referred.
And you can see that this

599
00:39:33 --> 00:39:37
orbital has the same symmetry
with respect to its nodal

600
00:39:37 --> 00:39:41
properties as a carbon 2p
orbital because it has a single

601
00:39:41 --> 00:39:45
nodal plane.
And one side all the signs are

602
00:39:45 --> 00:39:49
positive, and on the other side
all the signs are reversed.

603
00:39:49 --> 00:39:54
And we have a nice piece of
bonding going on there and there

604
00:39:54 --> 00:39:59
when we have this oblique
interaction of one of the lobes

605
00:39:59 --> 00:40:04
of a p-orbital simultaneously
with two of the hydrogens that

606
00:40:04 --> 00:40:08
it sits directly between in
space.

607
00:40:08 --> 00:40:13
There is one of our triply
degenerate components of our

608
00:40:13 --> 00:40:17
highest occupied molecular
orbital.

609
00:40:17 --> 00:40:23
And then, there is just one
left to visualize.

610
00:40:23 --> 00:40:36


611
00:40:36 --> 00:40:40
All right.
We will delete that one.

612
00:40:40 --> 00:40:50


613
00:40:50 --> 00:40:53
We looked at our most symmetric
and most bonding molecular

614
00:40:53 --> 00:40:55
orbital.
And now, we are going to go

615
00:40:55 --> 00:40:59
ahead and look at the
antibonding counterpart of that

616
00:40:59 --> 00:41:00
one.

617
00:41:00 --> 00:41:13


618
00:41:13 --> 00:41:17
And so this is clearly a
molecular orbital that is formed

619
00:41:17 --> 00:41:20
using some piece of the carbon
2s orbital.

620
00:41:20 --> 00:41:24
And you see that you get a
change in sign as you go from

621
00:41:24 --> 00:41:28
carbon to each of the hydrogens,
that accordingly the electron

622
00:41:28 --> 00:41:32
density is depleted in the
region of space between the

623
00:41:32 --> 00:41:36
nuclei.
And that is a characteristic of

624
00:41:36 --> 00:41:39
an antibonding orbital.
And this one,

625
00:41:39 --> 00:41:42
according to our diagram,
in fact, is the LUMO.

626
00:41:42 --> 00:41:47
So, as you will see over here,
that was what we predicted our

627
00:41:47 --> 00:41:51
lowest unoccupied molecular
orbital would look like.

628
00:41:51 --> 00:41:54
And then, up here,
we had our triply degenerate

629
00:41:54 --> 00:42:00
LUMO plus one based on the
carbon 2p orbital interactions.

630
00:42:00 --> 00:42:04
Now, you can see what these
four types of molecular orbital

631
00:42:04 --> 00:42:06
would look like.
And now, you wonder,

632
00:42:06 --> 00:42:10
just what kind of prediction
does this make regarding

633
00:42:10 --> 00:42:13
experimental observables.
And so let's look at this

634
00:42:13 --> 00:42:15
picture.

635
00:42:15 --> 00:42:25


636
00:42:25 --> 00:42:30
This diagram contains a number
of interesting features.

637
00:42:30 --> 00:42:34
It is a diagram that plots
ionization energy here on the

638
00:42:34 --> 00:42:39
x-axis against a number of
different molecules on the

639
00:42:39 --> 00:42:42
y-axis.
At the top you have simply a

640
00:42:42 --> 00:42:46
neon atom.
Neon is our noble gas atom that

641
00:42:46 --> 00:42:50
lies just immediately to the
right of fluorine in the

642
00:42:50 --> 00:42:53
periodic table.
If anything is more

643
00:42:53 --> 00:43:00
electronegative than fluorine,
it would actually be neon.

644
00:43:00 --> 00:43:04
This inert gas that is
extremely difficult to ionize

645
00:43:04 --> 00:43:09
because it holds so tightly onto
all eight of its electrons.

646
00:43:09 --> 00:43:14
Does it hold equally tightly
onto all eight of its electrons?

647
00:43:14 --> 00:43:19
The answer is no because,
as you scroll across here in

648
00:43:19 --> 00:43:23
ionization energy,
you see that its 2p electrons

649
00:43:23 --> 00:43:29
ionize here, and its 2s
electrons ionize way over here.

650
00:43:29 --> 00:43:31
The type of diagram I showed
you last time,

651
00:43:31 --> 00:43:35
where the 2s orbital,
as you go across the periodic

652
00:43:35 --> 00:43:38
table, decreases in energy
faster than does the set of 2p

653
00:43:38 --> 00:43:42
orbitals associated with an
atom, is expressed in this

654
00:43:42 --> 00:43:44
diagram here because we have
neon.

655
00:43:44 --> 00:43:47
And then we are going across
the periodic table.

656
00:43:47 --> 00:43:51
This next molecule is HF,
an eight electron species,

657
00:43:51 --> 00:43:54
and H two O,
an eight electron species,

658
00:43:54 --> 00:43:57
and then NH three,
ammonia, the other eight

659
00:43:57 --> 00:44:02
electron species.
And then over here we get to CH

660
00:44:02 --> 00:44:05
four, the molecule
that we have just

661
00:44:05 --> 00:44:08
discussed today.
And what is very interesting

662
00:44:08 --> 00:44:12
here is that the ionization
energies of the molecular

663
00:44:12 --> 00:44:16
orbitals, in the case of the
molecules, and of the atom in

664
00:44:16 --> 00:44:18
the case of neon track very
nicely.

665
00:44:18 --> 00:44:22
Here is the molecular orbital
that has 2s character of HF,

666
00:44:22 --> 00:44:25
of H two O,
of NH three,

667
00:44:25 --> 00:44:29
and of CH four.
And it is rising in energy as

668
00:44:29 --> 00:44:34
we go from right to left across
the periodic table.

669
00:44:34 --> 00:44:38
And that is because the carbon
2s orbital is much higher than

670
00:44:38 --> 00:44:41
the neon 2s orbital is,
for example.

671
00:44:41 --> 00:44:46
And the carbon 2s orbital is
contributing to that most

672
00:44:46 --> 00:44:50
bonding of molecular orbitals of
the methane molecule.

673
00:44:50 --> 00:44:55
And then there is this manifold
here, that corresponds to

674
00:44:55 --> 00:45:00
ionizing an electron out of the
HOMO of methane.

675
00:45:00 --> 00:45:04
And it is a broadened peak
distribution here because of the

676
00:45:04 --> 00:45:08
different vibrational states
that the methane can be found in

677
00:45:08 --> 00:45:11
when, in fact,
it experiences an incoming high

678
00:45:11 --> 00:45:15
energy photon that is capable of
knocking the electron out of the

679
00:45:15 --> 00:45:17
methane molecule and ionizing
it.

680
00:45:17 --> 00:45:20
This is called photoelectron
spectroscopy.

681
00:45:20 --> 00:45:24
And you will be reading about
it also in the context of your

682
00:45:24 --> 00:45:30
problem set because we have a
problem that deals with this.

683
00:45:30 --> 00:45:33
And I point you to a piece of
your text that covers

684
00:45:33 --> 00:45:37
photoelectron spectroscopy.
But the key point here is that

685
00:45:37 --> 00:45:40
if you had all your eight
electrons at one energy,

686
00:45:40 --> 00:45:44
as valance bond theory might
suggest for a molecule like

687
00:45:44 --> 00:45:49
this, you might think you should
only see one peak in the PES of

688
00:45:49 --> 00:45:51
methane.
But you see two because

689
00:45:51 --> 00:45:55
methane, the molecule has two
molecular orbital energy levels,

690
00:45:55 --> 00:45:57
one of them being triply
degenerate.

691
00:45:57.907 --> 46:00
See you on Wednesday.