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On Monday, we went through and
looked at the functional forms

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for sp two hybrid
orbitals, as found in the case

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of the BH three molecule.

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Now, you should recognize that
there are other hybridization

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schemes that go along with
different geometries.

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So, BH three was
trigonal planar.

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And we are going to talk some
more about BH three today.

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If you had been considering
last time instead a tetrahedral

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carbon atom, then the
hybridization scheme we would

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have needed to develop would
have been the sp three

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hybridization scheme.
That would be associated with a

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tetrahedral carbon atom.
On the other hand,

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if we had a linearly
coordinated carbon atom or some

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other main group atom,
boron or beryllium,

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for example,
then we would have had to

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develop the sp hybridization
scheme.

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You should not miss the forest
for the trees in putting into

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context from my previous lecture
on the sp two specific

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case for that type of geometry.
But now, rather than going

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through more examples of valance
bond theory treatments of

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molecules, instead,
it is time for me to introduce

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you to the tenants of the
molecular orbital theory for

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treatment of molecular
electronic structure.

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I mentioned valance bond theory
as having been introduced by

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Linus Pauling.
And then, I also named Robert

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S.
Mulliken, MIT undergraduate,

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as the father of molecular
orbital theory.

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And these are both two
different perspectives on

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viewing electron structure of
molecules that arises from the

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results of quantum mechanics.
In the case of valance bond

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theory, we have a situation
where you have localized

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electron pairs.

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In the case of molecular
orbital theory,

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electron pairs are still going
to be important,

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but they are going to be able
to be delocalized over the

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entire molecule in some cases.

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So, this is very different.
This is a major difference

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between MO theory and valance
bond theory.

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In the case of the BH three
molecule that we

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considered last time,
you had four positively charged

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nuclei.
And you arrange those four plus

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charges at the points in space
that correspond to the

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equilibrium geometry of the
molecule.

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And then you sprinkle in your
six valance electrons,

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and you want to understand how
those six electrons become

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organized in space in response
to that electric field set up by

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those four positively charged
nuclei.

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And we took the approach last
time that we are going to

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localize electron pairs in
between nuclei.

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And, because structurally the
BH three molecule is

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symmetric and the three
hydrogens are indistinguishable

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from one another,
we decided that we were going

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to make hybrids.
And so we talk about

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hybridization.

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The idea behind hybridization
was to change the atomic

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orbitals of boron by mixing
them, so that we would have one

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that would be able to point at
each of the three hydrogens in

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space to form three nicely,
perfectly directed electron

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sigma bonds between boron and
hydrogen.

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That was the scheme that we
adopted, this hybridization

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scheme, but in MO theory we are
not going to do that.

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We are not going to interfere
with the intrinsic atomic

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orbital structure of a boron
atom in order to make bonds.

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And we are going to see that
there are some predictions that

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come out of the MO treatment for
the molecule that differ from

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those that came out of the
valance bond treatment for the

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molecule.
And so here,

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no hybrids.

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In terms of just the vernacular
of chemical structure,

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you will hear sp three
as being used interchangeably

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with the notion of a
tetrahedron.

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But in valance bond theory,
it refers to a particular

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hybridization scheme in which we
actually mix s and p as a

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preparative to bond formation in
a molecule.

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And we mix the atomic orbitals
on that central atom in a

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hybridization scheme.
In MO theory,

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we are not going to do that.
It is very important that you

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keep these theories and the
language associated with the

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theories separate in your minds
so you can see the difference

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between these theories.
And then one of the

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consequences of this valance
bond theory and the

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hybridization scheme is that it
is not so good for excited

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states.

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And what that means is that we
were developing a scheme to

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describe the bonding in the
molecule in its ground

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electronic state.
Molecules can have excited

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states, just like atoms can have
electronic states.

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And over here,
in molecular orbital theory,

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we are going to find that we do
a much better job with excited

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states.
And that is important for

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understanding the range of
properties associated with

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molecular systems.
And you are going to see,

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indeed, that the energy level
scheme for the six valance

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electrons in the BH three
molecule is different,

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depending on whether you use
valence bond theory or molecular

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orbital theory.
Now, for molecular orbital

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theory, we are going to need,
again, like we do for valance

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bonds, to have some kind of a
procedure for forming molecular

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orbitals, conceptually.

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And the first step in such a
procedure is that you are going

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to want to analyze the
three-dimensional shape of the

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molecule.
And we do this,

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of course, when we talk about
the valance-shell electron-pair

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repulsion theory for predicting
molecular structure.

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We are going to look at the
structure.

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And we want to identify,
usually by inspection,

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sets of symmetry-related --

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-- atoms or orbitals.
And here, I am talking about

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atomic orbitals.
I will come back in a moment

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and talk about what I mean by
symmetry-related.

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This is a concept that can be
put on a very nice,

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firm, mathematical footing.
And, in fact,

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if you find this type of
analysis interesting and would

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like to see more of the math
that can help you to organize

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the results of quantum mechanics
in terms of symmetry,

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then you will want to put the
5.04 subject on your calendar

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for the future.
That subject is devoted,

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in large part,
to the applications of group

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theory to chemistry and chemical
problems.

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And symmetry plays a big role
in that.

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And then, two,
we are going to form

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combinations.
And let me just further

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quantify this by saying we are
going to form linear

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combinations --

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-- of symmetry-related
orbitals.

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One of the big approximations
that we usually kind of take for

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granted in the molecular orbital
theory of electronic structure

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is the LCAO approximation.
I am just going to mention that

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parenthetically,
here.

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This is that we can form
molecular orbitals that will be

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linear combinations of atomic
orbitals.

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This LCAO approximation arises
from the fact that we can solve

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the Schrödinger equation exactly
for the hydrogen atom,

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but for big molecules and many
electrons systems,

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we cannot.
And what we like to do is to

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take the atomic orbital wave
functions, that is,

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atomic orbital here in LCAO,
and use those wave functions in

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our approximation of molecular
orbitals.

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We are saying that we can
combine these atomic orbitals on

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the atoms that are in a molecule
to form the molecular orbitals

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that will be able to spread out
and delocalize over the entire

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molecule.
This is inherent in the

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development of the theory that I
am working on here.

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And you should see that this
will pop up in a number of

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cases, but it is an inherent
approximation that we are

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accepting.
And then, three,

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we are going to combine the
linear combinations from part

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two --

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-- with central atom atomic
orbitals.

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And that is what we will do.
And if we have done that

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properly, we will have arrived
at molecular orbitals for the

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system in question.
Now, I want to take a little

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bit of time to go through each
one of these steps in order to

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define the problem that we have
for this alternative way of

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viewing the electronic structure
of the BH three molecule.

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Now, why am I choosing the BH

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three molecule for this?
Well, I am choosing it because

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it is an easy problem and one
that is illustrative of the

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steps that go into forming
molecular orbitals for a system.

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In your textbook,
you will see that before you

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get to something like BH three,
first diatomic

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molecules are considered.
Those, I will show you on

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Friday, are actually a little
trickier to understand than is

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the case with BH three.
That is because in molecular

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orbital theory we have the same
number of MOs as AOs.

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If we start out with a certain
number of atomic orbitals that

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come into play by virtue of the
atoms that are in the molecule,

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then that number of orbitals
will be the same as the number

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of molecular orbitals that we
will get at the end of the

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problem.
They won't all be filled.

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We will have some that are
empty.

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But we are going to have the
same number of molecular

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orbitals as atomic orbitals that
go into the problem.

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And so what that means is that
the complexity of the problem is

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related to the number of MOs
and, hence, the number of AOs.

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The complexity of the problem
scales with the number of atomic

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orbitals in the problem.
We actually call these our

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basis functions.
And one of the things that John

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Pople talks about,
if you have gone and looked at

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his video that I pointed you to
in the problem set,

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he talks about the wonderful
fact that our computational

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power has gotten so great,
and will grow so much greater

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in the future,
that we are able to

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computationally handle problems
of the calculations of the

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properties of enormous molecules
that bring into a problem an

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enormous number,
a vast number of atomic

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orbitals.
And so, a great computational

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power is necessary to apply
molecular orbital theory,

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or even the more recent density
functional theory,

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to problems of electronic
structure that we need to

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grapple with in order to predict
properties of molecular systems.

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And in the BH three
molecule, which is,

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as I said, a simple problem of
electronic structure and a nice

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illustrative one for today's
purposes, we have,

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this is a seven orbital
problem.

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Why is this a seven orbital
problem?

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It is a seven orbital problem
because boron has four valance

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orbitals, an s,
a px, a py, a pz.

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And we have three hydrogens,
each with their 1s orbital.

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We have our valance orbitals on
boron.

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We don't count the 1s orbital
on boron because that is filled

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and is a core orbital,
and it doesn't get involved in

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chemical bonding.
The valance shell for boron is

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the n equals two shell.
We have 2s, 2px,

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2py and 2pz.
And we have three hydrogens,

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each with their 1s orbital.
If we were to tackle right now

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the problem of the molecular
orbital energy level diagram for

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a diatomic molecule like O two,
dioxygen,

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we would find immediately that
this analysis of how many atomic

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orbitals we have available would
give us the value eight.

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And so it is a more complicated
problem.

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It has one more atomic orbital
than this problem,

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even though this one has four
atoms because three of these

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atoms happen to be hydrogens,
which only bring in one orbital

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to the problem.
That is why I am choosing this.

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And then, in addition,
I am choosing this because the

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consequences of the symmetry of
the BH three molecule

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are that the orbital
interactions that we are going

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to identify occur in nice
pair-wise sets.

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And that makes things
especially easy to see in terms

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of how chemical bonds arise in
the context of MO theory for a

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problem like this.
And so, to now go ahead and

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carry out step one for this,
we are going to draw the

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molecule and try to identify
sets of symmetry-related

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atoms/orbitals.
And, before you do that,

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I just would like to show you
an example of a highly symmetric

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molecule, because this notion of
symmetry is something that,

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at this point in time,
I really only want you to gain

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an intuitive grasp of,
I am not going to quantify it.

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But, if you look at a molecule
like the one I have placed on

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the left and right-hand screens,
you will see that it is a

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00:18:13 --> 00:18:18
large, round molecule.
Actually, round,

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00:18:18 --> 00:18:24
spherical things are the most
symmetric things that we can

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think of.
Here, you can imagine that this

252
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is a big ball with atoms located
at various points on the surface

253
00:18:35 --> 00:18:39
of the ball.
This is just a ball and stick

254
00:18:39 --> 00:18:43
representation of the C60
molecule, also known as

255
00:18:43 --> 00:18:48
Buckminsterfullerene.
It is a geodesic dome-type of

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00:18:48 --> 00:18:51
molecule.
This molecule has the chemical

257
00:18:51 --> 00:18:55
formula C60.
There are no hydrogens in this

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00:18:55 --> 00:19:00
molecule.
All the atoms are displayed.

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00:19:00 --> 00:19:04
And each carbon atom sits in a
position where it is adjacent to

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00:19:04 --> 00:19:07
one five-membered ring and two
six-membered rings on the

261
00:19:07 --> 00:19:10
surface of this spherical
molecule.

262
00:19:10 --> 00:19:13
And that is true of every
carbon atom on the surface of

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00:19:13 --> 00:19:17
this whole molecule.
As you go on in chemistry,

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00:19:17 --> 00:19:21
if you go into the analysis of
molecules using nuclear magnetic

265
00:19:21 --> 00:19:24
resonance spectroscopy,
you will find that it is really

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00:19:24 --> 00:19:30
important to be able to identify
the symmetry of a molecule.

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00:19:30 --> 00:19:34
And you will realize that the
symmetry of the molecule is

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00:19:34 --> 00:19:38
manifest in the nuclear magnetic
resonance spectrum of a

269
00:19:38 --> 00:19:41
molecule.
If you take the carbon-13

270
00:19:41 --> 00:19:45
nuclear magnetic resonance
spectrum of this molecule,

271
00:19:45 --> 00:19:49
a sample composed of this
molecule, you will see that

272
00:19:49 --> 00:19:53
there is only a single C13
signal in the spectrum.

273
00:19:53 --> 00:19:57
And that is because all 60
carbon atoms are in an identical

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00:19:57 --> 00:20:03
environment in this molecule.
Every one of them feels exactly

275
00:20:03 --> 00:20:07
like everyone else.
They are all equidistant from

276
00:20:07 --> 00:20:11
the center of this molecule.
And they are all equidistant

277
00:20:11 --> 00:20:16
from their set of neighbors.
And so, in that way of looking

278
00:20:16 --> 00:20:20
at a molecule,
you would see that they are all

279
00:20:20 --> 00:20:23
equivalent.
You may have encountered this

280
00:20:23 --> 00:20:26
molecule before,
but this molecule's discovery

281
00:20:26 --> 00:20:30
--
-- was actually predicted

282
00:20:30 --> 00:20:34
initially from analysis of mass
spectrometry data by Professor

283
00:20:34 --> 00:20:39
Smalley, a Nobel Prize winner
whose efforts in this area have

284
00:20:39 --> 00:20:43
spawned off chemistries
involving not only these

285
00:20:43 --> 00:20:47
nano-sized balls of matter,
but also nanotubes made of

286
00:20:47 --> 00:20:52
carbon, and this whole area that
we think of nano technology.

287
00:20:52 --> 00:20:57
Really a lot of it is dominated
by the chemistry of new forms of

288
00:20:57 --> 00:21:01
carbon that arose with the
discovery of

289
00:21:01 --> 00:21:06
Buckminsterfullerene.
And that is just one example of

290
00:21:06 --> 00:21:12
a new allotrope of carbon that
was discovered in recent years.

291
00:21:12 --> 00:21:15
This gives you an idea for
symmetry.

292
00:21:15 --> 00:21:19
I could show you pictures,
actually, of enormous

293
00:21:19 --> 00:21:23
biomolecules.
There are large viruses that

294
00:21:23 --> 00:21:28
are composed of biopolymers,
macromolecules that pack in a

295
00:21:28 --> 00:21:33
way that is symmetrical,
so that you can see these

296
00:21:33 --> 00:21:37
things.
If you view them in the right

297
00:21:37 --> 00:21:40
kind of representation,
you will be able to see the

298
00:21:40 --> 00:21:44
symmetry in them.
And one definition of symmetry

299
00:21:44 --> 00:21:48
that I would like you to take
away from this picture is just

300
00:21:48 --> 00:21:51
that each of the atoms that are
symmetry-related is

301
00:21:51 --> 00:21:55
indistinguishable.
If you turn the molecule around

302
00:21:55 --> 00:21:59
and look at the atoms,
those that are symmetry-related

303
00:21:59 --> 00:22:03
are indistinguishable from one
another.

304
00:22:03 --> 00:22:07
And so, when we have a trigonal
planer BH three

305
00:22:07 --> 00:22:11
molecule, is the boron symmetry
related to any of the other

306
00:22:11 --> 00:22:13
atoms?
No, it is not.

307
00:22:13 --> 00:22:17
What about this hydrogen?
I labeled them last time A,

308
00:22:17 --> 00:22:20
B, and C.
This hydrogen labeled C,

309
00:22:20 --> 00:22:23
is it symmetry-related to other
hydrogens?

310
00:22:23 --> 00:22:25
Yes.
And that is because of the

311
00:22:25 --> 00:22:30
symmetry of this molecule with
these 120 degree bond angles on

312
00:22:30 --> 00:22:36
the planarity of the molecule.
So, you have a set of three

313
00:22:36 --> 00:22:39
hydrogens.
And their 1s orbitals are in

314
00:22:39 --> 00:22:42
space indistinguishable from one
another.

315
00:22:42 --> 00:22:47
They are related by symmetry.
And so, in this problem here,

316
00:22:47 --> 00:22:51
we have four orbitals here on
the boron that are not

317
00:22:51 --> 00:22:55
symmetry-related.
And then also incidentally let

318
00:22:55 --> 00:23:00
me point out that the boron atom
is located at the center of

319
00:23:00 --> 00:23:04
gravity of this system.
If atoms are going to be

320
00:23:04 --> 00:23:08
symmetry-related,
they must not be located at the

321
00:23:08 --> 00:23:13
center of gravity of the system.
Let's go over here and expand

322
00:23:13 --> 00:23:15
on these ideas.

323
00:23:15 --> 00:23:35


324
00:23:35 --> 00:23:39
The method that I am going to
develop here for forming the

325
00:23:39 --> 00:23:44
linear combinations has to do
with thinking ahead,

326
00:23:44 --> 00:23:48
in this problem,
to the fact that we are going

327
00:23:48 --> 00:23:52
to want to make linear
combinations that have the

328
00:23:52 --> 00:23:57
correct symmetry to bond to
atomic orbitals on the central

329
00:23:57 --> 00:24:00
boron atom.
Let me draw the ones that are

330
00:24:00 --> 00:24:03
going to be relevant to this
part of the problem.

331
00:24:03 --> 00:24:24


332
00:24:24 --> 00:24:28
This one over here would be the
boron pz orbital,

333
00:24:28 --> 00:24:33
so it has one positive lobe
coming out of the board,

334
00:24:33 --> 00:24:37
negative lobe going back.
Over here we would have the

335
00:24:37 --> 00:24:42
boron py orbital using the
coordinate system that I had

336
00:24:42 --> 00:24:46
chosen last time,
which is x up and y to the

337
00:24:46 --> 00:24:49
left.
And then here we have the

338
00:24:49 --> 00:24:53
boron's px orbital.
And then over here we have the

339
00:24:53 --> 00:24:59
boron's 2s orbital.
And so, our challenge now will

340
00:24:59 --> 00:25:02
be to construct linear
combinations,

341
00:25:02 --> 00:25:06
we are at part two,
of the set of three hydrogen 1s

342
00:25:06 --> 00:25:10
orbitals that can match in
symmetry the boron central

343
00:25:10 --> 00:25:13
atom's atomic orbitals.
Last time, remember,

344
00:25:13 --> 00:25:17
for hybridization,
we were making these orbitals

345
00:25:17 --> 00:25:21
mix with each other in order to
point at the hydrogens.

346
00:25:21 --> 00:25:27
Now what we are doing is kind
of an inverse concept.

347
00:25:27 --> 00:25:30
We are going to mix the
hydrogen orbitals so that they

348
00:25:30 --> 00:25:34
have the right symmetry to
interact with the central atom

349
00:25:34 --> 00:25:37
atomic orbitals.
There is a nice parallelism

350
00:25:37 --> 00:25:40
here.
Here is going to be our LCs.

351
00:25:40 --> 00:25:50


352
00:25:50 --> 00:25:53
They key feature of the boron's
2s orbital is that it does not

353
00:25:53 --> 00:25:56
have any nodes.
Remember nodes are surfaces.

354
00:25:56 --> 00:25:59
When you pass from one side of
a node to the other,

355
00:25:59 --> 00:26:03
you get a change in sign of the
wave function.

356
00:26:03 --> 00:26:08
And we indicate that change in
sign by differential shading.

357
00:26:08 --> 00:26:11
The 2s orbital has no nodes
whatsoever.

358
00:26:11 --> 00:26:16
And a way that we can construct
a linear combination that has

359
00:26:16 --> 00:26:20
the same spatial,
nodal properties as that boron

360
00:26:20 --> 00:26:25
2s atomic orbital is as follows.
We can involve contributions

361
00:26:25 --> 00:26:30
from each of the three hydrogen
1s orbitals.

362
00:26:30 --> 00:26:33
Remember this one is A,
this one is B,

363
00:26:33 --> 00:26:37
and this one is C.
This is going to be a linear

364
00:26:37 --> 00:26:40
combination of the three
hydrogen 1s orbitals.

365
00:26:40 --> 00:26:43
I will write this one as
follows.

366
00:26:43 --> 00:26:47
This one will be written as A
plus B plus C.

367
00:26:47 --> 00:26:52
That indicates the 1s orbital
on A, plus the 1s orbital on B,

368
00:26:52 --> 00:26:57
plus the 1s orbital on C.
And, as we talked about last

369
00:26:57 --> 00:27:03
time, wave functions that we
write should be normalized.

370
00:27:03 --> 00:27:07
And they should satisfy the
unit orbital contribution rule.

371
00:27:07 --> 00:27:10
For normalization,
here, I give a factor of one

372
00:27:10 --> 00:27:15
over root three for this linear
combination formed as a symmetry

373
00:27:15 --> 00:27:18
match with the boron 2s orbital.
You can see,

374
00:27:18 --> 00:27:22
I hope, what we are doing.
We are projecting the nodal

375
00:27:22 --> 00:27:27
properties of the central atom
atomic orbitals onto the linear

376
00:27:27 --> 00:27:31
combinations that we are
forming.

377
00:27:31 --> 00:27:36
And we are going to form a
complete set of three linear

378
00:27:36 --> 00:27:40
combinations in this way.
Let's make one that has

379
00:27:40 --> 00:27:45
symmetry properties that remind
us of this px orbital.

380
00:27:45 --> 00:27:50
On HA we are going to have a
positive contribution to match

381
00:27:50 --> 00:27:56
the positive contribution of
this lobe of the px orbital that

382
00:27:56 --> 00:28:02
points along the plus x axis.
And then down here we are going

383
00:28:02 --> 00:28:05
to have contributions from
hydrogens B and C.

384
00:28:05 --> 00:28:09
And they are going to be
smaller, and they are going to

385
00:28:09 --> 00:28:13
be opposite in-phase.
They are going to be opposite

386
00:28:13 --> 00:28:17
in-phase because we are building
a linear combination that has a

387
00:28:17 --> 00:28:21
node approximately at the center
of the system here,

388
00:28:21 --> 00:28:25
so as you go from positive x
into the negative x region of

389
00:28:25 --> 00:28:30
space, the wave function changes
sign to match the change in sign

390
00:28:30 --> 00:28:36
associated with the px orbital.
We are projecting the nodal

391
00:28:36 --> 00:28:40
properties of px onto the linear
combination of hydrogen orbitals

392
00:28:40 --> 00:28:43
that we are forming here.
And this one,

393
00:28:43 --> 00:28:48
written in normalized fashion,
will be root two over three,

394
00:28:48 --> 00:28:50
A minus one-half B minus
one-half C.

395
00:28:50 --> 00:28:55
And, although what we are

396
00:28:55 --> 00:28:58
working with here are linear
combinations of these

397
00:28:58 --> 00:29:02
symmetry-related hydrogen 1s
wave functions,

398
00:29:02 --> 00:29:06
you are going to find that
these coefficients on the atomic

399
00:29:06 --> 00:29:10
orbitals, that contribute
eventually to the molecular

400
00:29:10 --> 00:29:14
orbitals, are going to come out
as normalized and as unit

401
00:29:14 --> 00:29:20
orbital contributions.
So that if we started out this

402
00:29:20 --> 00:29:24
problem with a single 1s orbital
on HA, that will be entirely

403
00:29:24 --> 00:29:29
accounted for among these linear
combinations and the molecular

404
00:29:29 --> 00:29:33
orbitals that we are going to
make with them.

405
00:29:33 --> 00:29:37
Now, let's generate a linear
combination having the nodal

406
00:29:37 --> 00:29:41
properties of py.
And, in order to do that,

407
00:29:41 --> 00:29:46
we need to have a negative
coefficient out here in the

408
00:29:46 --> 00:29:49
minus y direction.
We are going to put in a

409
00:29:49 --> 00:29:53
contribution from HC as
negative, like that,

410
00:29:53 --> 00:29:56
to match that.
And then over here,

411
00:29:56 --> 00:30:00
we are going to make a
contribution for HB that is

412
00:30:00 --> 00:30:05
positive.
And then, noting that there is

413
00:30:05 --> 00:30:08
a nodal plane along the
y,z-plane, which comes out of

414
00:30:08 --> 00:30:12
the board like this,
so that we are always negative

415
00:30:12 --> 00:30:15
along minus y and we are
positive along plus y.

416
00:30:15 --> 00:30:19
We match that here.
And the coincidence of that

417
00:30:19 --> 00:30:23
nodal plane with the location of
HA dictates no contribution from

418
00:30:23 --> 00:30:27
HA to this orbital,
for reasons that actually we

419
00:30:27 --> 00:30:30
looked at last time,
--

420
00:30:30 --> 00:30:35
-- namely, that we cannot bring
a hydrogen 1s orbital in here

421
00:30:35 --> 00:30:41
and also change sign on going
half-way through that hydrogen

422
00:30:41 --> 00:30:46
1s orbital, because s orbitals
have to have the same sign

423
00:30:46 --> 00:30:50
everywhere.
It does not contribute to this

424
00:30:50 --> 00:30:54
linear combination.
And our normalized form for

425
00:30:54 --> 00:30:58
this will be one over root two B
minus C.

426
00:30:58 --> 00:31:04
And then, if I were to ask the

427
00:31:04 --> 00:31:07
question, can I make a linear
combination of the three

428
00:31:07 --> 00:31:11
hydrogen 1s orbitals that has
the same nodal properties as pz,

429
00:31:11 --> 00:31:15
the answer would be no because
they all lie in the x,y-plane

430
00:31:15 --> 00:31:20
and they are just s orbitals and
cannot change sign as you go

431
00:31:20 --> 00:31:23
through the x,y-plane from plus
z to minus z.

432
00:31:23 --> 00:31:26
And so, we are done here.
And what you are going to find

433
00:31:26 --> 00:31:30
is that we have created these
three linear combination

434
00:31:30 --> 00:31:35
according to step two.
And taking into account both

435
00:31:35 --> 00:31:39
the symmetry properties of the
molecule to identify a set of

436
00:31:39 --> 00:31:42
three symmetry-related atoms and
orbitals.

437
00:31:42 --> 00:31:47
And then, taking into account
an analysis of the nodal

438
00:31:47 --> 00:31:50
properties of the central atom
atomic orbitals,

439
00:31:50 --> 00:31:55
so that we could project those
out to help us find appropriate

440
00:31:55 --> 00:32:01
linear combinations for mixing
with the central atom orbitals.

441
00:32:01 --> 00:32:10


442
00:32:10 --> 00:32:14
And, when we do that mixing,
we are going to find out that

443
00:32:14 --> 00:32:17
there are three ways that we can
do it.

444
00:32:17 --> 00:32:21
We are about to move onto step
three of this problem.

445
00:32:21 --> 00:32:26
We are going to need to combine
these linear combinations with

446
00:32:26 --> 00:32:31
the central atom atomic orbitals
according to the rules of MO

447
00:32:31 --> 00:32:35
theory to generate,
first, bonding molecular

448
00:32:35 --> 00:32:37
orbitals.

449
00:32:37 --> 00:32:43


450
00:32:43 --> 00:32:50
And the bonding molecular
orbitals that we will get will

451
00:32:50 --> 00:32:54
be an in-phase combination --

452
00:32:54 --> 00:33:01


453
00:33:01 --> 00:33:04
-- of our LCs,
our linear combinations of

454
00:33:04 --> 00:33:09
atomic orbitals,
with our boron atomic orbitals.

455
00:33:09 --> 00:33:14
That will describe the chemical
bonding in our system.

456
00:33:14 --> 00:33:19
And we will see that it
contrasts in a very interesting

457
00:33:19 --> 00:33:24
way with the hybridization
scheme developed last time.

458
00:33:24 --> 00:33:27
And the key phrase to
underline, here,

459
00:33:27 --> 00:33:32
is in-phase.
And what that in-phase means is

460
00:33:32 --> 00:33:36
that when two positive lobes of
two orbitals centered on two

461
00:33:36 --> 00:33:40
different atoms are juxtaposed
and neighbor one another and can

462
00:33:40 --> 00:33:44
have good overlap of their
atomic orbital wave functions,

463
00:33:44 --> 00:33:48
that leads to in-phase
constructive interference and

464
00:33:48 --> 00:33:52
stabilization of the electrons
associated with that newly

465
00:33:52 --> 00:33:54
formed bonding molecular
orbital.

466
00:33:54 --> 00:34:00
And that stabilization is what
we call the chemical bond.

467
00:34:00 --> 00:34:04
The analogy to that in valence
bond theory is the idea that a

468
00:34:04 --> 00:34:08
pair of electrons associated
with two nuclei in a sigma bond

469
00:34:08 --> 00:34:11
is more stable because it
experiences, simultaneously,

470
00:34:11 --> 00:34:15
two positive charges.
And here we are generalizing

471
00:34:15 --> 00:34:19
that and allowing electrons to
flow over the molecule as a

472
00:34:19 --> 00:34:21
whole.
But now we have a new concept,

473
00:34:21 --> 00:34:24
and that is antibonding.

474
00:34:24 --> 00:34:30


475
00:34:30 --> 00:34:36
Antibonding molecular orbitals
will be out-of-phase

476
00:34:36 --> 00:34:42
combinations that are repulsive
and lead to high energy

477
00:34:42 --> 00:34:45
interactions.

478
00:34:45 --> 00:34:50


479
00:34:50 --> 00:34:55
And when, as in the case of the
problem we are considering here,

480
00:34:55 --> 00:34:58
the interactions occur in
pair-wise sets,

481
00:34:58 --> 00:35:03
we will find that we get very
nicely, for every bonding

482
00:35:03 --> 00:35:07
molecular orbital,
a corresponding anti-bonding

483
00:35:07 --> 00:35:10
molecular orbital.

484
00:35:10 --> 00:35:15


485
00:35:15 --> 00:35:18
Also, one of the interesting
things is that if you start

486
00:35:18 --> 00:35:20
putting electrons into
anti-bonding orbitals,

487
00:35:20 --> 00:35:23
if your system just happens to
be so constructed as to have

488
00:35:23 --> 00:35:26
many electrons,
such that you fill up not only

489
00:35:26 --> 00:35:29
the bonding molecular orbitals
of electrons to make the

490
00:35:29 --> 00:35:31
chemical bonds,
but you continue on and you

491
00:35:31 --> 00:35:34
have enough electrons to keep
going and put them into

492
00:35:34 --> 00:35:37
anti-bonding orbitals,
those antibonds start to cancel

493
00:35:37 --> 00:35:40
the bonds.
And so, you will see a very

494
00:35:40 --> 00:35:44
nice progression of this as we
study the MO theory of the

495
00:35:44 --> 00:35:48
homonuclear diatomic molecules,
starting on Friday.

496
00:35:48 --> 00:35:52
And then, here is another
concept that arises from the MO

497
00:35:52 --> 00:35:55
analysis of a molecule,
and that is that certain

498
00:35:55 --> 00:35:58
orbitals can be non-bonding.

499
00:35:58 --> 00:36:04


500
00:36:04 --> 00:36:10
And an example of this would be
a lone pair of electrons.

501
00:36:10 --> 00:36:16
And it happens when an orbital
or a linear combination of

502
00:36:16 --> 00:36:21
orbitals finds no counterpart --

503
00:36:21 --> 00:36:27


504
00:36:27 --> 00:36:31
-- of like nodal symmetry.
These nodal properties of

505
00:36:31 --> 00:36:36
orbitals are very important,
and I will show you later how

506
00:36:36 --> 00:36:42
the nodal properties are related
to the energies of the orbitals,

507
00:36:42 --> 00:36:46
as we consider them.
Having given you this preview

508
00:36:46 --> 00:36:52
of how orbitals are going to be
able to combine in the MO

509
00:36:52 --> 00:36:57
theory, let's see how it
actually takes place in the case

510
00:36:57 --> 00:37:00
of BH three.

511
00:37:00 --> 00:37:15


512
00:37:15 --> 00:37:18
Now we are drawing another
example of an energy-level

513
00:37:18 --> 00:37:23
diagram, where the energy is low
at the bottom and rises as it

514
00:37:23 --> 00:37:26
goes up.
And these energy-level diagrams

515
00:37:26 --> 00:37:30
that we are now developing for
molecules are analogous to those

516
00:37:30 --> 00:37:35
that you studied earlier in the
semester for atoms.

517
00:37:35 --> 00:37:41
And we are just generalizing
this notion to the atoms.

518
00:37:41 --> 00:37:48
And what I am drawing over here
is our boron 2s orbital and here

519
00:37:48 --> 00:37:52
is our boron 2px,
2py, 2pz orbital.

520
00:37:52 --> 00:37:58
Those horizontal bars just
represent the energy of these

521
00:37:58 --> 00:38:04
orbitals in the molecule.
Here I am just redrawing the

522
00:38:04 --> 00:38:08
boron atom.
And then, over here on the

523
00:38:08 --> 00:38:14
right, we are going to see that
we have our linear combinations

524
00:38:14 --> 00:38:19
that we developed.
These are our three H 1s linear

525
00:38:19 --> 00:38:22
combinations.
We found that the three

526
00:38:22 --> 00:38:28
hydrogens in BH three
were symmetry equivalent,

527
00:38:28 --> 00:38:33
so we generated linear
combinations.

528
00:38:33 --> 00:38:38
The pictures of them are over
there.

529
00:38:38 --> 00:38:45
I can give them names.
Why don't I call them D,

530
00:38:45 --> 00:38:48
E, and F.

531
00:38:48 --> 00:39:00


532
00:39:00 --> 00:39:05
We have D constructed to match
the nodal properties of the

533
00:39:05 --> 00:39:09
boron's 2s orbital.
E and F were respectively

534
00:39:09 --> 00:39:14
constructed to match the nodal
properties of the boron's 2s,

535
00:39:14 --> 00:39:18
2px, and 2py orbitals.
And I am showing you their

536
00:39:18 --> 00:39:22
relative energies.
And now we need to do this

537
00:39:22 --> 00:39:26
issue referenced here as point
three.

538
00:39:26 --> 00:39:30
We need to combine these
things.

539
00:39:30 --> 00:39:33
I have these on the one hand
and these on the other.

540
00:39:33 --> 00:39:38
Our seven atomic orbitals have
been changed into four atomic

541
00:39:38 --> 00:39:40
orbitals and three linear
combinations,

542
00:39:40 --> 00:39:43
so I still have seven orbitals
total.

543
00:39:43 --> 00:39:47
And now I am going to combine
these with these according to

544
00:39:47 --> 00:39:51
their nodal properties to
generate seven molecular

545
00:39:51 --> 00:39:55
orbitals.
And let's do it this way.

546
00:39:55 --> 00:40:03


547
00:40:03 --> 00:40:08
We are going to take linear
combination D that has been

548
00:40:08 --> 00:40:13
constructed to match the boron
2s orbital in terms of nodal

549
00:40:13 --> 00:40:18
symmetry properties,
and we are going to make a

550
00:40:18 --> 00:40:22
bonding combination.
I am going to draw these

551
00:40:22 --> 00:40:26
pictorially in a moment.
And, in this case,

552
00:40:26 --> 00:40:32
those are the only two orbitals
in my seven orbital system here

553
00:40:32 --> 00:40:38
that have that set of nodal
symmetry properties.

554
00:40:38 --> 00:40:42
And for every bonding MO,
I must have an antibonding MO.

555
00:40:42 --> 00:40:44
This is one of our molecular
orbitals.

556
00:40:44 --> 00:40:49
And there is going to be a
corresponding molecular orbital

557
00:40:49 --> 00:40:53
up here, high in energy.
And this will be an antibonding

558
00:40:53 --> 00:40:57
molecular orbital that will be
the out-of-phase combination of

559
00:40:57 --> 00:41:03
the boron 2s with this linear
combination that I labeled D.

560
00:41:03 --> 00:41:09
Antibonding molecular orbitals
are usually denoted with a star.

561
00:41:09 --> 00:41:15
We have a bonding combination
and an anti-bonding combination.

562
00:41:15 --> 00:41:21
Now, we can form two more bonds
that will spread out over the

563
00:41:21 --> 00:41:24
molecule because,
if you recall,

564
00:41:24 --> 00:41:31
we had our px and py pair that
served as the nodal template for

565
00:41:31 --> 00:41:37
our construction of linear
combinations E and F.

566
00:41:37 --> 00:41:42
We are going to be able to
match up those to form two more

567
00:41:42 --> 00:41:48
bonding molecular orbitals.
And these will be found to be

568
00:41:48 --> 00:41:53
higher in energy than the first
one that we formed from D.

569
00:41:53 --> 00:41:59
Here is a pair of bonding
molecular orbitals that derive

570
00:41:59 --> 00:42:05
from linear combinations E and F
combining in-phase with boron's

571
00:42:05 --> 00:42:10
px and py atomic orbitals.
And there will be a

572
00:42:10 --> 00:42:14
corresponding antibonding
combination, where we allow

573
00:42:14 --> 00:42:19
those orbitals to interact in an
out-of-phase manner.

574
00:42:19 --> 00:42:22
And you will see what that
means shortly.

575
00:42:22 --> 00:42:27
Let me put that up there and
add a star to indicate that this

576
00:42:27 --> 00:42:32
high energy pair of molecular
orbitals is an antibonding pair

577
00:42:32 --> 00:42:36
of orbitals.
I have six orbitals now in my

578
00:42:36 --> 00:42:40
molecular orbital energy-level
diagram for BH three.

579
00:42:40 --> 00:42:44
And that means I am not done
because I have to have seven,

580
00:42:44 --> 00:42:47
and I started out with seven
atomic orbitals.

581
00:42:47 --> 00:42:50
Look over here.
2pz was an atomic orbital on

582
00:42:50 --> 00:42:55
boron that did not find any way
of serving as a template for

583
00:42:55 --> 00:43:00
making a linear combination
involving the three hydrogens.

584
00:43:00 --> 00:43:05
And so, it comes over here as
nonbonding.

585
00:43:05 --> 00:43:12
It has no counterpart of like
nodal symmetry because of the

586
00:43:12 --> 00:43:18
location of those three
hydrogens in the x,y-plane,

587
00:43:18 --> 00:43:25
which is a nodal plane for the
boron pz orbital.

588
00:43:25 --> 00:43:32
And so, this one is nonbonding.
These three orbitals up here

589
00:43:32 --> 00:43:35
are anti-bonding.
And the ones down at the

590
00:43:35 --> 00:43:40
bottom, which are the lowest in
energy, corresponding to being

591
00:43:40 --> 00:43:45
able to most tightly hold onto
electrons in them are bonding

592
00:43:45 --> 00:43:49
molecular orbitals.
And so our electrons can fill

593
00:43:49 --> 00:43:53
into this MO energy-level
diagram in that way.

594
00:43:53 --> 00:43:59
We have our six electrons that
come into this problem.

595
00:43:59 --> 00:44:03
We have boron bringing in three
valance electrons and three

596
00:44:03 --> 00:44:07
hydrogens each bringing in one
valance electron.

597
00:44:07 --> 00:44:10
There are six electrons to put
into the diagram,

598
00:44:10 --> 00:44:14
filling up three of the
molecular orbitals and then

599
00:44:14 --> 00:44:18
leaving empty pz.
Let me introduce a little bit

600
00:44:18 --> 00:44:22
more MO language right now.
This one here will be called

601
00:44:22 --> 00:44:27
the highest occupied molecular
orbital, and this one here will

602
00:44:27 --> 00:44:32
be called the lowest unoccupied
molecular orbital.

603
00:44:32 --> 00:44:37
The reason why I am drawing
attention to these orbitals is

604
00:44:37 --> 00:44:41
that in chemistry,
the chemical properties derive

605
00:44:41 --> 00:44:47
oftentimes from those orbitals
that are in what is called the

606
00:44:47 --> 00:44:52
frontier orbital region.
And the frontier orbitals are

607
00:44:52 --> 00:44:56
those close in energy to the
HOMO-LUMO gap.

608
00:44:56 --> 00:45:02
And I will come back to this.
But those highest energetically

609
00:45:02 --> 00:45:05
lying electrons are going to be
the ones responsible for

610
00:45:05 --> 00:45:10
nucleophilic properties of the
molecule and basic properties of

611
00:45:10 --> 00:45:13
the molecule and reducing
properties of the molecule.

612
00:45:13 --> 00:45:17
Whereas, low-lying empty
orbitals are going to be the

613
00:45:17 --> 00:45:20
ones responsible for acidic
properties of the molecule or

614
00:45:20 --> 00:45:23
oxidizing properties of the
molecule.

615
00:45:23 --> 00:45:27
And we will come back to that
in a moment.

616
00:45:27 --> 00:45:31
But that is something pretty
general and very useful that

617
00:45:31 --> 00:45:36
comes out of studying molecular
orbital energy-level diagrams.

618
00:45:36 --> 00:45:41


619
00:45:41 --> 00:45:46
Now that we have the diagram,
let's see what the orbitals in

620
00:45:46 --> 00:45:50
the diagram look like.
And I will try to do this

621
00:45:50 --> 00:45:54
relatively quickly.
Here, let's start with the

622
00:45:54 --> 00:45:59
lowest lying molecular orbital
in the system.

623
00:45:59 --> 00:46:04
This is a representation of an
in-phase combination of the

624
00:46:04 --> 00:46:09
boron 2s plus D,
where D is defined up here as

625
00:46:09 --> 00:46:14
that linear combination.
This will be a molecular

626
00:46:14 --> 00:46:19
orbital having the nodal
properties of a 2s orbital

627
00:46:19 --> 00:46:25
centered on that central atom.
That is where our lowest-lying

628
00:46:25 --> 00:46:31
two electrons reside.
Now, if you look at the LUMO

629
00:46:31 --> 00:46:35
over there and then go up one
orbital in energy,

630
00:46:35 --> 00:46:41
you will be looking at this
orbital, which is the boron 2s

631
00:46:41 --> 00:46:43
orbital minus D.

632
00:46:43 --> 00:46:48


633
00:46:48 --> 00:46:54
And the thing that makes this
out-of-phase linear combination

634
00:46:54 --> 00:47:01
so much higher in energy than
its in-phase counterpart is the

635
00:47:01 --> 00:47:06
appearance, now,
of a nodal surface.

636
00:47:06 --> 00:47:08
And this node is between the
nuclei.

637
00:47:08 --> 00:47:13
It goes all the way around and
is between the central atom s

638
00:47:13 --> 00:47:16
orbital and those peripheral
hydrogens.

639
00:47:16 --> 00:47:20
And I will show you a picture
of it.

640
00:47:20 --> 00:47:37


641
00:47:37 --> 00:47:42
This is our BH three
LUMO plus one molecular orbital.

642
00:47:42 --> 00:47:52


643
00:47:52 --> 00:47:56
You can see that we have a wave
function in the center of one

644
00:47:56 --> 00:47:58
sign.
And then, as we go along any

645
00:47:58 --> 00:48:02
one of the B-H bond vectors from
boron to hydrogen,

646
00:48:02 --> 00:48:06
we change phase midway along
the bond from positive to

647
00:48:06 --> 00:48:09
negative.
And that is true,

648
00:48:09 --> 00:48:14
no matter which of the three
B-H bonds we pick to traverse

649
00:48:14 --> 00:48:17
along.
That one has the nodal

650
00:48:17 --> 00:48:22
properties as drawn down there
for the boron 2s interacting in

651
00:48:22 --> 00:48:28
an out-of-phase manner with
linear combination letter D.

652
00:48:28 --> 00:48:32
And so, next time,
I will finish up and show you

653
00:48:32 --> 00:48:36
what these other orbitals look
like as calculated.

654
00:48:36 --> 00:48:41
I hope you have enjoyed this.
We will see more MO theory on

655
00:48:41.751 --> 48:44
Friday.