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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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00:13:57,000 --> 00:14:02,000
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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00:14:06,000 --> 00:14:09,000
2py, and 2pz orbitals in 3D
space.

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00:14:09,000 --> 00:14:13,000
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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00:14:34,000 --> 00:14:38,000
heteronuclear diatomic molecule
problems we talked about last

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time, is an eight orbital
problem.

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00:14:40,000 --> 00:14:44,000
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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00:14:47,000 --> 00:14:50,000
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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00:15:01,000 --> 00:15:06,000
four hydrogens.
And our eight orbitals are the

233
00:15:06,000 --> 00:15:08,000
carbons 2s, 2px,
2py, and 2pz orbitals.

234
00:15:08,000 --> 00:15:12,000
And we have four more orbitals,
the 1s from each of the four

235
00:15:12,000 --> 00:15:15,000
hydrogens.
We know how many electrons are

236
00:15:15,000 --> 00:15:18,000
going to be able to go into our
MO energy level diagram.

237
00:15:18,000 --> 00:15:22,000
We know how many MOs we are
going to have because we know

238
00:15:22,000 --> 00:15:26,000
how many valance AOs we have.
And those numbers are equal at

239
00:15:26,000 --> 00:15:30,000
eight.
The choice of this problem is

240
00:15:30,000 --> 00:15:35,000
predicated on my desire to show
you how things progress as we go

241
00:15:35,000 --> 00:15:40,000
from a planar system to a linear
system to now a system that has

242
00:15:40,000 --> 00:15:45,000
molecular orbitals that stretch
out in three-dimensional space.

243
00:15:45,000 --> 00:15:49,000
Now, we are going to take the
following approach.

244
00:15:49,000 --> 00:15:52,000
We are going to say,
how can we generate linear

245
00:15:52,000 --> 00:15:56,000
combinations of the four
hydrogen 1s orbitals that will

246
00:15:56,000 --> 00:16:01,000
be so constructed as to match
the nodal properties of the

247
00:16:01,000 --> 00:16:06,000
carbon atomic orbitals that are
the valence orbitals of that

248
00:16:06,000 --> 00:16:08,000
carbon?
Okay.

249
00:16:08,000 --> 00:16:12,000
So that is our goal,
and that is our strategy.

250
00:16:12,000 --> 00:16:17,000
And let me say also,
that in this tetrahedral

251
00:16:17,000 --> 00:16:23,000
symmetry, you can recognize that
you cannot distinguish any one

252
00:16:23,000 --> 00:16:28,000
of the four vertices of a
tetrahedron from the other

253
00:16:28,000 --> 00:16:31,000
three.
And that is important.

254
00:16:31,000 --> 00:16:36,000
And that is also true from the
way that we have decided to

255
00:16:36,000 --> 00:16:40,000
situate the hydrogen nuclei on
this cube relative to our

256
00:16:40,000 --> 00:16:43,000
coordinate system.
In other words,

257
00:16:43,000 --> 00:16:47,000
none of these four hydrogens
lie on any of the three

258
00:16:47,000 --> 00:16:51,000
Cartesian axes.
And their relationship to the

259
00:16:51,000 --> 00:16:55,000
coordinate system and to the
shape of the cube each is

260
00:16:55,000 --> 00:17:01,000
indistinguishable from that of
the other three hydrogens.

261
00:17:01,000 --> 00:17:04,000
And that is an important aspect
of the way that we set up the

262
00:17:04,000 --> 00:17:07,000
problem.
And why do I mention that?

263
00:17:07,000 --> 00:17:10,000
Because I mentioned that when
constructing MOs in general,

264
00:17:10,000 --> 00:17:15,000
one of the first things you can
do, if you are going to be able

265
00:17:15,000 --> 00:17:19,000
to take advantage of symmetry to
help you solve the problem of

266
00:17:19,000 --> 00:17:23,000
these MO energy levels and how
they are constructed from linear

267
00:17:23,000 --> 00:17:27,000
combinations of atomic orbitals
is to identify symmetry-related

268
00:17:27,000 --> 00:17:32,000
sets of atoms in orbitals.
If you are keeping that in

269
00:17:32,000 --> 00:17:37,000
mind, you will realize that we
are treating all four hydrogens

270
00:17:37,000 --> 00:17:41,000
together because they are
indistinguishable from one

271
00:17:41,000 --> 00:17:46,000
another in the tetrahedral shape
of the methane molecule.

272
00:17:46,000 --> 00:17:50,000
So, that is important.
If we had a more complicated

273
00:17:50,000 --> 00:17:55,000
molecule, such as if we were
looking at --

274
00:17:55,000 --> 00:17:58,000
I get to use a little bit of
board space today.

275
00:17:58,000 --> 00:18:02,000
Here is a slightly more
complicated hydrocarbon

276
00:18:02,000 --> 00:18:04,000
molecule.
Here is a slightly more

277
00:18:04,000 --> 00:18:07,000
complicated hydrocarbon
molecule.

278
00:18:07,000 --> 00:18:11,000
And in a molecule like this,
this is cyclohexane drawn in a

279
00:18:11,000 --> 00:18:15,000
chair confirmation.
Sir Derek Barton actually was

280
00:18:15,000 --> 00:18:19,000
the Nobel laureate who got his
prize for discussing

281
00:18:19,000 --> 00:18:22,000
conformational aspects of
hydrocarbons.

282
00:18:22,000 --> 00:18:26,000
We would find that the
hydrogens split up into two

283
00:18:26,000 --> 00:18:30,000
sets.
They are not all equivalent in

284
00:18:30,000 --> 00:18:34,000
the cyclohexane molecule in this
frozen out chair confirmation

285
00:18:34,000 --> 00:18:38,000
because we have one type that is
equatorial and one type that is

286
00:18:38,000 --> 00:18:42,000
axial on each of these carbons
as we go around this

287
00:18:42,000 --> 00:18:45,000
six-membered ring.
And so there are different sets

288
00:18:45,000 --> 00:18:48,000
that you would have to treat
separately.

289
00:18:48,000 --> 00:18:50,000
But methane is more
symmetrical.

290
00:18:50,000 --> 00:18:54,000
And all four hydrogens in the
methane molecule are identical,

291
00:18:54,000 --> 00:18:58,000
so we are going to treat them
together in developing our

292
00:18:58,000 --> 00:19:02,000
linear combinations.
And, coming back to this,

293
00:19:02,000 --> 00:19:06,000
what we are going to do,
once again, to generate these

294
00:19:06,000 --> 00:19:10,000
things, is to imagine the
central atom atomic orbital as

295
00:19:10,000 --> 00:19:13,000
growing out in space out to
where the hydrogen nuclei are.

296
00:19:13,000 --> 00:19:17,000
And then we are going to take
the sign of the wave function

297
00:19:17,000 --> 00:19:21,000
that we find out there and apply
it to those hydrogens that are

298
00:19:21,000 --> 00:19:26,000
out there in space in order to
see what our linear combination

299
00:19:26,000 --> 00:19:29,000
must look like.
If you imagine the carbon 2s

300
00:19:29,000 --> 00:19:32,000
orbital, which has the same
sign, positive,

301
00:19:32,000 --> 00:19:35,000
everywhere, and imagine it
growing out to where the

302
00:19:35,000 --> 00:19:38,000
hydrogens are,
and we are labeling the

303
00:19:38,000 --> 00:19:40,000
hydrogen A, B,
C, and D, as shown here.

304
00:19:40,000 --> 00:19:44,000
And we will keep that same
labeling scheme throughout the

305
00:19:44,000 --> 00:19:47,000
development of this problem.
You would see that the carbon

306
00:19:47,000 --> 00:19:51,000
2s positive wave function,
as we imagine it growing out to

307
00:19:51,000 --> 00:19:54,000
where the hydrogens are,
would touch each of these

308
00:19:54,000 --> 00:19:58,000
hydrogens at the same time.
And, thus, we confer the sign

309
00:19:58,000 --> 00:20:02,000
positive on all four of the
contributors to this first of

310
00:20:02,000 --> 00:20:07,000
our linear combinations.
And what that will represent is

311
00:20:07,000 --> 00:20:12,000
simultaneous bonding of the
carbon 2s orbital with all four

312
00:20:12,000 --> 00:20:15,000
hydrogens and with the same sign
everywhere.

313
00:20:15,000 --> 00:20:19,000
And so, we can write down
one-half A plus B plus C plus D,

314
00:20:19,000 --> 00:20:23,000
where I am using A here to

315
00:20:23,000 --> 00:20:28,000
represent the hydrogen 1s
orbital associated with hydrogen

316
00:20:28,000 --> 00:20:32,000
labeled A.
So I have just abbreviated that

317
00:20:32,000 --> 00:20:33,000
in A.
And, similarly,

318
00:20:33,000 --> 00:20:37,000
B is the hydrogen 1s orbital
associated with hydrogen B and

319
00:20:37,000 --> 00:20:39,000
so on.
So, this is a linear

320
00:20:39,000 --> 00:20:42,000
combination of these four
hydrogen orbitals.

321
00:20:42,000 --> 00:20:44,000
And it is normalized with a
half, here.

322
00:20:44,000 --> 00:20:48,000
And we have the same
coefficient on each of the four

323
00:20:48,000 --> 00:20:52,000
hydrogens, so the size of the
lobe that you would draw at each

324
00:20:52,000 --> 00:20:55,000
four of these positions would be
the same.

325
00:20:55,000 --> 00:20:59,000
That is number one.
We have built a linear

326
00:20:59,000 --> 00:21:03,000
combination based on the nodal
properties of the carbon 2s

327
00:21:03,000 --> 00:21:06,000
atomic orbital.
And now, let's do two more.

328
00:21:06,000 --> 00:21:11,000
And we are going to base these
on the nodal properties of the

329
00:21:11,000 --> 00:21:15,000
central carbon 2px orbital and
the central carbon 2py orbital.

330
00:21:15,000 --> 00:21:19,000
I am using the same coordinate
system as I started with,

331
00:21:19,000 --> 00:21:24,000
namely, that the positive
x-axis comes out of the face of

332
00:21:24,000 --> 00:21:28,000
the cube and comes toward us,
and the positive y-axis comes

333
00:21:28,000 --> 00:21:33,000
out of this face of the cube and
goes that way.

334
00:21:33,000 --> 00:21:35,000
And we will look at z in a
moment.

335
00:21:35,000 --> 00:21:40,000
That one is going straight up
out of the top face of the cube

336
00:21:40,000 --> 00:21:44,000
for positive z.
What you can see is that if you

337
00:21:44,000 --> 00:21:48,000
imagine the carbon 2px atomic
orbital as just growing out to

338
00:21:48,000 --> 00:21:53,000
where the four hydrogens are,
we will see that the hydrogens

339
00:21:53,000 --> 00:21:57,000
in positions A and B will
experience the positive lobe of

340
00:21:57,000 --> 00:22:02,000
the carbon 2px orbital at the
same moment as the two hydrogens

341
00:22:02,000 --> 00:22:06,000
in back labeled C and D
experience the negative lobe of

342
00:22:06,000 --> 00:22:12,000
the carbon 2px orbital.
And so, we give positive phase

343
00:22:12,000 --> 00:22:16,000
to A and B and negative phase to
C and D.

344
00:22:16,000 --> 00:22:20,000
And we give them the same
coefficient everywhere such that

345
00:22:20,000 --> 00:22:25,000
this linear combination is
normalized as one-half A plus B

346
00:22:25,000 --> 00:22:30,000
minus C minus D.

347
00:22:30,000 --> 00:22:33,000
That is our second linear
combination of hydrogen

348
00:22:33,000 --> 00:22:36,000
orbitals.
And this one has been so

349
00:22:36,000 --> 00:22:40,000
generated as to match the nodal
property of the carbon 2px

350
00:22:40,000 --> 00:22:44,000
orbital, the key feature of
which is that it is everywhere

351
00:22:44,000 --> 00:22:48,000
positive along positive x,
and everywhere along negative x

352
00:22:48,000 --> 00:22:51,000
it is negative.
And the 2py orbital of carbon

353
00:22:51,000 --> 00:22:56,000
has the property that everywhere
in the plus y region of space,

354
00:22:56,000 --> 00:22:59,000
it is positive,
so that C and B will be

355
00:22:59,000 --> 00:23:05,000
associated with a positive sign
for this linear combination.

356
00:23:05,000 --> 00:23:09,000
And the 2py orbital is negative
everywhere along negative y,

357
00:23:09,000 --> 00:23:12,000
back here.
And that is where hydrogens A

358
00:23:12,000 --> 00:23:15,000
and D are located.
So, A and D will carry a

359
00:23:15,000 --> 00:23:18,000
negative sign for this new
linear combination,

360
00:23:18,000 --> 00:23:22,000
which is one-half of minus A
plus B plus C minus D.

361
00:23:22,000 --> 00:23:26,000
This is just one way
of writing down

362
00:23:26,000 --> 00:23:32,000
the picture that we see here.
And there is only one left to

363
00:23:32,000 --> 00:23:37,000
be generated from the nodal
properties of the carbon 2pz

364
00:23:37,000 --> 00:23:42,000
orbital, which has its positive
node oriented along positive z

365
00:23:42,000 --> 00:23:46,000
and its negative lobe oriented
along negative z,

366
00:23:46,000 --> 00:23:50,000
and it has the x,y-plane as a
nodal surface.

367
00:23:50,000 --> 00:23:54,000
And so, therefore,
hydrogens A and C will carry a

368
00:23:54,000 --> 00:24:00,000
positive sign because they are
oriented and located in the plus

369
00:24:00,000 --> 00:24:05,000
z region of space.
Whereas, hydrogens B and D are

370
00:24:05,000 --> 00:24:09,000
located in the negative z region
of space, so they will carry a

371
00:24:09,000 --> 00:24:13,000
negative sign.
And that one can be normalized

372
00:24:13,000 --> 00:24:16,000
also with a normalization
coefficient of one-half.

373
00:24:16,000 --> 00:24:19,000
And so it is A minus B plus C
minus D.

374
00:24:19,000 --> 00:24:23,000
And just don't forget
that these letters

375
00:24:23,000 --> 00:24:27,000
refer to the 1s orbitals
associated with the hydrogens

376
00:24:27,000 --> 00:24:32,000
that are so labeled.
And we started with identifying

377
00:24:32,000 --> 00:24:37,000
the fact that the four hydrogens
in this problem are all

378
00:24:37,000 --> 00:24:40,000
equivalent.
And, thus, their 1s orbitals

379
00:24:40,000 --> 00:24:44,000
are all equivalent,
indistinguishable from each

380
00:24:44,000 --> 00:24:47,000
other with respect to their
spatial orientation.

381
00:24:47,000 --> 00:24:51,000
And, so we took four atomic
orbitals, and we are

382
00:24:51,000 --> 00:24:56,000
constructing from them four
linear combinations of those

383
00:24:56,000 --> 00:25:01,000
four atomic orbitals.
I want to emphasize what I have

384
00:25:01,000 --> 00:25:04,000
written here,
which is that with the

385
00:25:04,000 --> 00:25:08,000
coordinate system we have chosen
and in this high tetrahedral

386
00:25:08,000 --> 00:25:13,000
symmetry, the carbons 2p x,
y and z orbitals are identical.

387
00:25:13,000 --> 00:25:17,000
And, just like in a carbon atom
floating free in space,

388
00:25:17,000 --> 00:25:21,000
they are degenerate.
That means they have the same

389
00:25:21,000 --> 00:25:24,000
energy.
And so we will talk about

390
00:25:24,000 --> 00:25:29,000
orbitals in atoms being
degenerate in a moment.

391
00:25:29,000 --> 00:25:33,000
But we find that the 2p x,
y and z orbitals on that carbon

392
00:25:33,000 --> 00:25:37,000
are going to have the same
energy, and they are degenerate.

393
00:25:37,000 --> 00:25:41,000
And the molecular orbitals
constructed from them in this

394
00:25:41,000 --> 00:25:44,000
tetrahedral system will have the
same property.

395
00:25:44,000 --> 00:25:47,000
They will be degenerate.
So, something that is

396
00:25:47,000 --> 00:25:51,000
occasionally associated with
high symmetry is multiple MO

397
00:25:51,000 --> 00:25:55,000
degeneracy.
Next, knowing what our carbon

398
00:25:55,000 --> 00:25:59,000
atom atomic orbitals are and
knowing also what our four

399
00:25:59,000 --> 00:26:03,000
linear combinations are of the
hydrogen 1s orbitals,

400
00:26:03,000 --> 00:26:07,000
we can now allow them to
interact in ways that are either

401
00:26:07,000 --> 00:26:11,000
bonding or antibonding.
And they interact,

402
00:26:11,000 --> 00:26:14,000
of course, according to their
nodal properties.

403
00:26:14,000 --> 00:26:19,000
Those linear combinations that
have the same nodal properties

404
00:26:19,000 --> 00:26:23,000
as a particular carbon atom
atomic orbital will lead to a

405
00:26:23,000 --> 00:26:29,000
bonding interaction and an
antibonding interaction.

406
00:26:29,000 --> 00:26:33,000
And we have an eight orbital
problem, so we better see eight

407
00:26:33,000 --> 00:26:37,000
energy levels appearing in our
molecular orbital energy level

408
00:26:37,000 --> 00:26:39,000
diagram for this methane
molecule.

409
00:26:39,000 --> 00:26:43,000
And, when we go down here,
starting at the lowest energy

410
00:26:43,000 --> 00:26:46,000
molecular orbital,
we find that it can be

411
00:26:46,000 --> 00:26:49,000
constructed by adding to the
carbon 2s orbital,

412
00:26:49,000 --> 00:26:53,000
which is, once again,
spherically symmetric,

413
00:26:53,000 --> 00:26:57,000
this linear combination that is
A plus B plus C plus D.

414
00:26:57,000 --> 00:27:00,000
And we can draw this with a

415
00:27:00,000 --> 00:27:03,000
surface like this,
which envelopes all five of our

416
00:27:03,000 --> 00:27:08,000
nuclei which has the same sign
everywhere.

417
00:27:08,000 --> 00:27:10,000
And so our lowest lying
molecular orbital,

418
00:27:10,000 --> 00:27:14,000
our most bonding molecular
orbital, our most stabilized MO,

419
00:27:14,000 --> 00:27:17,000
which can house a pair of
electrons because it is singly

420
00:27:17,000 --> 00:27:21,000
degenerate, is one that will
look something like this.

421
00:27:21,000 --> 00:27:23,000
It envelopes all five of our
nuclei.

422
00:27:23,000 --> 00:27:26,000
And, effectively,
the pair of electrons is

423
00:27:26,000 --> 00:27:30,000
associated simultaneously with
all five nuclei.

424
00:27:30,000 --> 00:27:33,000
And the wave function has the
same sign everywhere,

425
00:27:33,000 --> 00:27:38,000
just like the carbon 2s orbital
from which this is partially

426
00:27:38,000 --> 00:27:41,000
constructed does.
And then, next we have three

427
00:27:41,000 --> 00:27:45,000
molecular orbitals that are each
built by forming bonding

428
00:27:45,000 --> 00:27:48,000
interactions with the carbon's
p-orbitals, px,

429
00:27:48,000 --> 00:27:53,000
py, and pz with the linear
combination that matches that

430
00:27:53,000 --> 00:27:56,000
carbon orbital in terms of the
nodal properties.

431
00:27:56,000 --> 00:28:00,000
And we built them that way so
that becomes particularly

432
00:28:00,000 --> 00:28:04,000
straightforward.
This one here is coming out,

433
00:28:04,000 --> 00:28:07,000
the face that points to the
right.

434
00:28:07,000 --> 00:28:11,000
And so you can see that this
one is based on the carbon 2py

435
00:28:11,000 --> 00:28:13,000
orbital.
And, if that carbon 2py orbital

436
00:28:13,000 --> 00:28:18,000
makes a bonding interaction with
the linear combination denoted

437
00:28:18,000 --> 00:28:21,000
as minus A plus B plus
C minus D,

438
00:28:21,000 --> 00:28:25,000
the bonding interaction here is
signified by this plus sign.

439
00:28:25,000 --> 00:28:30,000
So, we have a plus sign here
and a plus sign here.

440
00:28:30,000 --> 00:28:34,000
And, thus, we have a molecular
orbital that has the same nodal

441
00:28:34,000 --> 00:28:37,000
properties as a carbon 2py
orbital does,

442
00:28:37,000 --> 00:28:41,000
which means that the x,z-plane
is a nodal surface for this

443
00:28:41,000 --> 00:28:45,000
molecular orbital.
And what it looks like is kind

444
00:28:45,000 --> 00:28:49,000
of a two-bladed propeller.
You have a plus phase that is

445
00:28:49,000 --> 00:28:54,000
distributed, enveloping the two
hydrogens that we labeled B and

446
00:28:54,000 --> 00:29:00,000
C, along with that positive lobe
of the carbon 2py orbital.

447
00:29:00,000 --> 00:29:04,000
So that you have that forming
one blade of our two-bladed

448
00:29:04,000 --> 00:29:08,000
propeller.
And the blade with the other

449
00:29:08,000 --> 00:29:12,000
sign is rotated relative to the
first by 45 degrees.

450
00:29:12,000 --> 00:29:16,000
Sorry, by 90 degrees,
in fact, and will involve the

451
00:29:16,000 --> 00:29:21,000
distribution of bonding
character over carbon's negative

452
00:29:21,000 --> 00:29:26,000
2py lobe interacting with these
two hydrogens back here,

453
00:29:26,000 --> 00:29:32,000
which are hydrogens A and D.
And so we will visualize these

454
00:29:32,000 --> 00:29:34,000
in a moment to make that more
clear.

455
00:29:34,000 --> 00:29:37,000
But these three bonds,
one oriented along x,

456
00:29:37,000 --> 00:29:41,000
one oriented along y,
and one oriented along z all

457
00:29:41,000 --> 00:29:44,000
look exactly like this.
They are just rotated 90

458
00:29:44,000 --> 00:29:47,000
degrees to each other in space,
just like the x,

459
00:29:47,000 --> 00:29:51,000
y, and z Cartesian axes are.
I have only drawn one of them,

460
00:29:51,000 --> 00:29:55,000
but we have three of them
because there is a three-fold

461
00:29:55,000 --> 00:30:00,000
orbital degeneracy in this
molecular orbital.

462
00:30:00,000 --> 00:30:03,000
So, singly degenerate and then
triply degenerate.

463
00:30:03,000 --> 00:30:07,000
And then, as we go up in
energy, we will find that our

464
00:30:07,000 --> 00:30:10,000
lowest unoccupied molecular
orbital is constructed by

465
00:30:10,000 --> 00:30:13,000
subtracting.
It is the same as I have

466
00:30:13,000 --> 00:30:15,000
written here,
but with a minus sign,

467
00:30:15,000 --> 00:30:19,000
the C 2s subtracting A plus B
plus C plus D.

468
00:30:19,000 --> 00:30:21,000
A, B, C and D hydrogens all

469
00:30:21,000 --> 00:30:25,000
have the same sign to their wave
function as each other,

470
00:30:25,000 --> 00:30:29,000
but they have the opposite sign
as the contribution from the

471
00:30:29,000 --> 00:30:34,000
carbon 2s orbital to this
molecular orbital.

472
00:30:34,000 --> 00:30:37,000
And what that results in,
as I have tried to indicate

473
00:30:37,000 --> 00:30:41,000
with some red dashes here,
is that you generate a nodal

474
00:30:41,000 --> 00:30:46,000
surface in between the carbon
and hydrogen nuclei that will go

475
00:30:46,000 --> 00:30:50,000
completely around and will
intersect perpendicular to each

476
00:30:50,000 --> 00:30:54,000
of these carbon-hydrogen
inter-nuclear vectors.

477
00:30:54,000 --> 00:30:58,000
You have a change of sign as
you are traversing the path from

478
00:30:58,000 --> 00:31:03,000
carbon to any one of the four
hydrogen nuclei.

479
00:31:03,000 --> 00:31:06,000
And that is what we get by just
reversing the sign relative to

480
00:31:06,000 --> 00:31:09,000
the bonding counterpart of that
orbital.

481
00:31:09,000 --> 00:31:11,000
And that makes this one
anti-bonding.

482
00:31:11,000 --> 00:31:15,000
We will give it an asterisk
here to denote antibonding

483
00:31:15,000 --> 00:31:17,000
character to this molecular
orbital.

484
00:31:17,000 --> 00:31:21,000
And now, ascending in energy,
we will find that for each one

485
00:31:21,000 --> 00:31:25,000
of these orbitals that is bonded
and oriented with respect to x,

486
00:31:25,000 --> 00:31:28,000
y or z in Cartesian space,
we find that there is an

487
00:31:28,000 --> 00:31:32,000
antibonding counterpart.
Again, what we are doing is

488
00:31:32,000 --> 00:31:36,000
taking, for example,
carbons 2py and subtracting the

489
00:31:36,000 --> 00:31:40,000
linear combination that
corresponds to carbon 2py in

490
00:31:40,000 --> 00:31:42,000
terms of its nodal symmetry
properties.

491
00:31:42,000 --> 00:31:47,000
We are subtracting minus A plus
B plus C minus D.

492
00:31:47,000 --> 00:31:49,000
The carbon 2py has the same

493
00:31:49,000 --> 00:31:53,000
plus, its positive lobe is
oriented along positive y,

494
00:31:53,000 --> 00:31:57,000
just as it is down here,
but now we have reversed the

495
00:31:57,000 --> 00:32:02,000
sign of the four hydrogens.
And what this introduces will

496
00:32:02,000 --> 00:32:07,000
be antibonding nodal surfaces in
between the carbon and the

497
00:32:07,000 --> 00:32:11,000
hydrogens.
And they are strong antibonding

498
00:32:11,000 --> 00:32:16,000
interactions because of the good
directional overlap of,

499
00:32:16,000 --> 00:32:20,000
say, the positive lobe of
carbon's 2py with these

500
00:32:20,000 --> 00:32:26,000
s-orbital contributions in the
negative from hydrogens C and B,

501
00:32:26,000 --> 00:32:28,000
here.
And back here,

502
00:32:28,000 --> 00:32:33,000
A and D rotated,
of course, by 90 degrees.

503
00:32:33,000 --> 00:32:37,000
One of the points that I was
illustrating last time and

504
00:32:37,000 --> 00:32:41,000
really trying to emphasize in
the context of MO theory is

505
00:32:41,000 --> 00:32:45,000
that, in general,
we expect orbitals to be higher

506
00:32:45,000 --> 00:32:48,000
in energy if they have more
internuclear nodes.

507
00:32:48,000 --> 00:32:53,000
And the energy of the orbitals
is also affected by which atomic

508
00:32:53,000 --> 00:32:58,000
orbitals are close to the MO in
energy and also by factors of

509
00:32:58,000 --> 00:33:01,000
overlap.
And normally,

510
00:33:01,000 --> 00:33:04,000
for example,
a sigma bond is typically

511
00:33:04,000 --> 00:33:09,000
associated with greater overlap
than a pi bond because of the

512
00:33:09,000 --> 00:33:13,000
directionality of the overlap.
And here, you have an

513
00:33:13,000 --> 00:33:17,000
interesting case,
where this p-orbital is not

514
00:33:17,000 --> 00:33:23,000
directed right at the hydrogen.
And it is not a pi bond either.

515
00:33:23,000 --> 00:33:27,000
It is something in between sort
of sigma and a pi type of

516
00:33:27,000 --> 00:33:32,000
orientation.
It is an oblique interaction of

517
00:33:32,000 --> 00:33:36,000
the hydrogen 1s wavefunction
with that lobe of the p-orbital.

518
00:33:36,000 --> 00:33:40,000
But, still, it gives rise to
very good overlap down here,

519
00:33:40,000 --> 00:33:44,000
and consequently a very strong
antibonding character up here in

520
00:33:44,000 --> 00:33:47,000
counterpart.
And, from calculation,

521
00:33:47,000 --> 00:33:51,000
this is the order of the energy
levels that come out of that.

522
00:33:51,000 --> 00:33:55,000
And so we have only four energy
levels, two of which are triply

523
00:33:55,000 --> 00:33:59,000
degenerate, so we indeed have
eight wave functions that

524
00:33:59,000 --> 00:34:04,000
correspond to the molecular
orbitals of this system.

525
00:34:04,000 --> 00:34:25,000
Question down here?
Could you say that louder?

526
00:34:25,000 --> 00:34:26,000
Why did this not have the
one-half?

527
00:34:26,000 --> 00:34:29,000
The one-half is missing there
because I forgot to put it

528
00:34:29,000 --> 00:34:32,000
there.
Yes, to be normalized it should

529
00:34:32,000 --> 00:34:35,000
have a one-half there.
That is right.

530
00:34:35,000 --> 00:34:39,000
But then there is a further
point that that brings up,

531
00:34:39,000 --> 00:34:43,000
which is the following.
What I am implying here is that

532
00:34:43,000 --> 00:34:48,000
there are equal amounts of
carbon 2s in the bonding orbital

533
00:34:48,000 --> 00:34:51,000
and as in the antibonding
counterpart.

534
00:34:51,000 --> 00:34:55,000
But that need not be the case.
It could be that this is 0.6

535
00:34:55,000 --> 00:35:01,000
here and that this 0.4 up here.
And that is a greater level of

536
00:35:01,000 --> 00:35:06,000
detail than I am really
interested in going into for the

537
00:35:06,000 --> 00:35:09,000
purposes of this problem.
But your point,

538
00:35:09,000 --> 00:35:13,000
with respect to the
coefficient, is well taken.

539
00:35:13,000 --> 00:35:16,000
Thanks.
Christine, maybe we can fix

540
00:35:16,000 --> 00:35:20,000
that on the one that gets posted
on the web.

541
00:35:20,000 --> 00:35:26,000
Let's now look at these things
using our VMD program.

542
00:35:37,000 --> 00:35:41,000
Starting from the lowest energy
and we will work our way up.

543
00:35:48,000 --> 00:35:52,000
Which orbital is this?
This is a molecular orbital of

544
00:35:52,000 --> 00:35:54,000
the methane molecule.

545
00:36:00,000 --> 00:36:04,000
One way you can identify
orbitals is where they are with

546
00:36:04,000 --> 00:36:06,000
reference to the HOMO or the
LUMO.

547
00:36:06,000 --> 00:36:09,000
In this case,
the HOMO is triply degenerate.

548
00:36:09,000 --> 00:36:13,000
And there is one orbital lower
in energy than the HOMO.

549
00:36:13,000 --> 00:36:17,000
This would be then the HOMO
minus one because it is one

550
00:36:17,000 --> 00:36:21,000
below the HOMO in energy.
That is the one which is the

551
00:36:21,000 --> 00:36:24,000
carbon 2s, plus A plus B plus C
plus D.

552
00:36:24,000 --> 00:36:27,000
That is that MO there.

553
00:36:33,000 --> 00:36:39,000
Actually, let's go ahead and
take that one away.

554
00:36:49,000 --> 00:36:50,000
Sorry about that.

555
00:36:55,000 --> 00:37:00,000
Let's see if you can figure out
which one this is.

556
00:37:14,000 --> 00:37:18,000
In order to figure out which
one it is, you need to be able

557
00:37:18,000 --> 00:37:23,000
to realize where the nuclei are
and where your internuclear

558
00:37:23,000 --> 00:37:26,000
nodes are, if any.
And, if so, how many.

559
00:37:26,000 --> 00:37:30,000
This is one of our hydrogen
contributions.

560
00:37:30,000 --> 00:37:33,000
Here is a hydrogen contribution
with opposite sign.

561
00:37:33,000 --> 00:37:38,000
And there is a piece of our
carbon 2p orbital in the center.

562
00:37:38,000 --> 00:37:43,000
Now, this is just one of three
identical counterparts that are

563
00:37:43,000 --> 00:37:47,000
rotated 90 degrees with respect
to each other in space,

564
00:37:47,000 --> 00:37:51,000
so I am not going to show you
all three of them.

565
00:37:51,000 --> 00:37:55,000
They all look exactly the same.
Is this one bonding or

566
00:37:55,000 --> 00:38:00,000
antibonding?
Yeah, this one is antibonding.

567
00:38:00,000 --> 00:38:03,000
And it does not have the same
sign on all the hydrogens,

568
00:38:03,000 --> 00:38:07,000
so it has to be part of our
LUMO, which is triply degenerate

569
00:38:07,000 --> 00:38:09,000
and involves bonding or,
in this case,

570
00:38:09,000 --> 00:38:13,000
the antibonding between a
carbon 2p orbital with a linear

571
00:38:13,000 --> 00:38:17,000
combination of four hydrogens.
You can see that when you have

572
00:38:17,000 --> 00:38:20,000
an antibonding orbital,
you are depleting the electron

573
00:38:20,000 --> 00:38:25,000
density in the space between the
nuclei rather than increasing it

574
00:38:25,000 --> 00:38:30,000
as you do in a bonding orbital.
Let's go ahead and look at the

575
00:38:30,000 --> 00:38:34,000
bonding counterpart of this one.
And we will do that.

576
00:38:34,000 --> 00:38:38,000
And let's delete it.
One thing that you will get,

577
00:38:38,000 --> 00:38:43,000
if you haven't already gotten
it, is you are going to get a

578
00:38:43,000 --> 00:38:48,000
link to be able to download all
the files that I used to show

579
00:38:48,000 --> 00:38:51,000
you molecular orbitals with in
class.

580
00:38:51,000 --> 00:38:55,000
Because this program I am
using, this VMD program,

581
00:38:55,000 --> 00:39:01,000
is available on Athena.
And you can also download it

582
00:39:01,000 --> 00:39:06,000
yourself for free onto your own
computer if you want to.

583
00:39:06,000 --> 00:39:10,000
And so this is going to give
you the opportunity to study

584
00:39:10,000 --> 00:39:16,000
these orbitals and rotate them
around at your leisure on your

585
00:39:16,000 --> 00:39:19,000
desktop.
And now, let's look at the

586
00:39:19,000 --> 00:39:24,000
bonding counterpart to that
antibonding orbital that we just

587
00:39:24,000 --> 00:39:27,000
looked at.
This one is the two-bladed

588
00:39:27,000 --> 00:39:33,000
propeller to which I referred.
And you can see that this

589
00:39:33,000 --> 00:39:37,000
orbital has the same symmetry
with respect to its nodal

590
00:39:37,000 --> 00:39:41,000
properties as a carbon 2p
orbital because it has a single

591
00:39:41,000 --> 00:39:45,000
nodal plane.
And one side all the signs are

592
00:39:45,000 --> 00:39:49,000
positive, and on the other side
all the signs are reversed.

593
00:39:49,000 --> 00:39:54,000
And we have a nice piece of
bonding going on there and there

594
00:39:54,000 --> 00:39:59,000
when we have this oblique
interaction of one of the lobes

595
00:39:59,000 --> 00:40:04,000
of a p-orbital simultaneously
with two of the hydrogens that

596
00:40:04,000 --> 00:40:08,000
it sits directly between in
space.

597
00:40:08,000 --> 00:40:13,000
There is one of our triply
degenerate components of our

598
00:40:13,000 --> 00:40:17,000
highest occupied molecular
orbital.

599
00:40:17,000 --> 00:40:23,000
And then, there is just one
left to visualize.

600
00:40:36,000 --> 00:40:40,000
All right.
We will delete that one.

601
00:40:50,000 --> 00:40:53,000
We looked at our most symmetric
and most bonding molecular

602
00:40:53,000 --> 00:40:55,000
orbital.
And now, we are going to go

603
00:40:55,000 --> 00:40:59,000
ahead and look at the
antibonding counterpart of that

604
00:40:59,000 --> 00:41:00,000
one.

605
00:41:13,000 --> 00:41:17,000
And so this is clearly a
molecular orbital that is formed

606
00:41:17,000 --> 00:41:20,000
using some piece of the carbon
2s orbital.

607
00:41:20,000 --> 00:41:24,000
And you see that you get a
change in sign as you go from

608
00:41:24,000 --> 00:41:28,000
carbon to each of the hydrogens,
that accordingly the electron

609
00:41:28,000 --> 00:41:32,000
density is depleted in the
region of space between the

610
00:41:32,000 --> 00:41:36,000
nuclei.
And that is a characteristic of

611
00:41:36,000 --> 00:41:39,000
an antibonding orbital.
And this one,

612
00:41:39,000 --> 00:41:42,000
according to our diagram,
in fact, is the LUMO.

613
00:41:42,000 --> 00:41:47,000
So, as you will see over here,
that was what we predicted our

614
00:41:47,000 --> 00:41:51,000
lowest unoccupied molecular
orbital would look like.

615
00:41:51,000 --> 00:41:54,000
And then, up here,
we had our triply degenerate

616
00:41:54,000 --> 00:42:00,000
LUMO plus one based on the
carbon 2p orbital interactions.

617
00:42:00,000 --> 00:42:04,000
Now, you can see what these
four types of molecular orbital

618
00:42:04,000 --> 00:42:06,000
would look like.
And now, you wonder,

619
00:42:06,000 --> 00:42:10,000
just what kind of prediction
does this make regarding

620
00:42:10,000 --> 00:42:13,000
experimental observables.
And so let's look at this

621
00:42:13,000 --> 00:42:15,000
picture.

622
00:42:25,000 --> 00:42:30,000
This diagram contains a number
of interesting features.

623
00:42:30,000 --> 00:42:34,000
It is a diagram that plots
ionization energy here on the

624
00:42:34,000 --> 00:42:39,000
x-axis against a number of
different molecules on the

625
00:42:39,000 --> 00:42:42,000
y-axis.
At the top you have simply a

626
00:42:42,000 --> 00:42:46,000
neon atom.
Neon is our noble gas atom that

627
00:42:46,000 --> 00:42:50,000
lies just immediately to the
right of fluorine in the

628
00:42:50,000 --> 00:42:53,000
periodic table.
If anything is more

629
00:42:53,000 --> 00:43:00,000
electronegative than fluorine,
it would actually be neon.

630
00:43:00,000 --> 00:43:04,000
This inert gas that is
extremely difficult to ionize

631
00:43:04,000 --> 00:43:09,000
because it holds so tightly onto
all eight of its electrons.

632
00:43:09,000 --> 00:43:14,000
Does it hold equally tightly
onto all eight of its electrons?

633
00:43:14,000 --> 00:43:19,000
The answer is no because,
as you scroll across here in

634
00:43:19,000 --> 00:43:23,000
ionization energy,
you see that its 2p electrons

635
00:43:23,000 --> 00:43:29,000
ionize here, and its 2s
electrons ionize way over here.

636
00:43:29,000 --> 00:43:31,000
The type of diagram I showed
you last time,

637
00:43:31,000 --> 00:43:35,000
where the 2s orbital,
as you go across the periodic

638
00:43:35,000 --> 00:43:38,000
table, decreases in energy
faster than does the set of 2p

639
00:43:38,000 --> 00:43:42,000
orbitals associated with an
atom, is expressed in this

640
00:43:42,000 --> 00:43:44,000
diagram here because we have
neon.

641
00:43:44,000 --> 00:43:47,000
And then we are going across
the periodic table.

642
00:43:47,000 --> 00:43:51,000
This next molecule is HF,
an eight electron species,

643
00:43:51,000 --> 00:43:54,000
and H two O,
an eight electron species,

644
00:43:54,000 --> 00:43:57,000
and then NH three,
ammonia, the other eight

645
00:43:57,000 --> 00:44:02,000
electron species.
And then over here we get to CH

646
00:44:02,000 --> 00:44:05,000
four, the molecule
that we have just

647
00:44:05,000 --> 00:44:08,000
discussed today.
And what is very interesting

648
00:44:08,000 --> 00:44:12,000
here is that the ionization
energies of the molecular

649
00:44:12,000 --> 00:44:16,000
orbitals, in the case of the
molecules, and of the atom in

650
00:44:16,000 --> 00:44:18,000
the case of neon track very
nicely.

651
00:44:18,000 --> 00:44:22,000
Here is the molecular orbital
that has 2s character of HF,

652
00:44:22,000 --> 00:44:25,000
of H two O,
of NH three,

653
00:44:25,000 --> 00:44:29,000
and of CH four.
And it is rising in energy as

654
00:44:29,000 --> 00:44:34,000
we go from right to left across
the periodic table.

655
00:44:34,000 --> 00:44:38,000
And that is because the carbon
2s orbital is much higher than

656
00:44:38,000 --> 00:44:41,000
the neon 2s orbital is,
for example.

657
00:44:41,000 --> 00:44:46,000
And the carbon 2s orbital is
contributing to that most

658
00:44:46,000 --> 00:44:50,000
bonding of molecular orbitals of
the methane molecule.

659
00:44:50,000 --> 00:44:55,000
And then there is this manifold
here, that corresponds to

660
00:44:55,000 --> 00:45:00,000
ionizing an electron out of the
HOMO of methane.

661
00:45:00,000 --> 00:45:04,000
And it is a broadened peak
distribution here because of the

662
00:45:04,000 --> 00:45:08,000
different vibrational states
that the methane can be found in

663
00:45:08,000 --> 00:45:11,000
when, in fact,
it experiences an incoming high

664
00:45:11,000 --> 00:45:15,000
energy photon that is capable of
knocking the electron out of the

665
00:45:15,000 --> 00:45:17,000
methane molecule and ionizing
it.

666
00:45:17,000 --> 00:45:20,000
This is called photoelectron
spectroscopy.

667
00:45:20,000 --> 00:45:24,000
And you will be reading about
it also in the context of your

668
00:45:24,000 --> 00:45:30,000
problem set because we have a
problem that deals with this.

669
00:45:30,000 --> 00:45:33,000
And I point you to a piece of
your text that covers

670
00:45:33,000 --> 00:45:37,000
photoelectron spectroscopy.
But the key point here is that

671
00:45:37,000 --> 00:45:40,000
if you had all your eight
electrons at one energy,

672
00:45:40,000 --> 00:45:44,000
as valance bond theory might
suggest for a molecule like

673
00:45:44,000 --> 00:45:49,000
this, you might think you should
only see one peak in the PES of

674
00:45:49,000 --> 00:45:51,000
methane.
But you see two because

675
00:45:51,000 --> 00:45:55,000
methane, the molecule has two
molecular orbital energy levels,

676
00:45:55,000 --> 00:45:57,000
one of them being triply
degenerate.

677
00:45:57,907 --> 00:46:00,000
See you on Wednesday.