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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 Schrˆdinger 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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00:16:20,000 --> 00:16:27,000
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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00:16:34,000 --> 00:16:37,000
And so it is a more complicated
problem.

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00:16:37,000 --> 00:16:41,000
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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00:16:50,000 --> 00:16:54,000
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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00:17:03,000 --> 00:17:07,000
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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00:17:39,000 --> 00:17:44,000
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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00:17:49,000 --> 00:17:54,000
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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00:18:00,000 --> 00:18:06,000
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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large, round molecule.
Actually, round,

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00:18:18,000 --> 00:18:24,000
spherical things are the most
symmetric things that we can

238
00:18:24,000 --> 00:18:29,000
think of.
Here, you can imagine that this

239
00:18:29,000 --> 00:18:35,000
is a big ball with atoms located
at various points on the surface

240
00:18:35,000 --> 00:18:39,000
of the ball.
This is just a ball and stick

241
00:18:39,000 --> 00:18:43,000
representation of the C60
molecule, also known as

242
00:18:43,000 --> 00:18:48,000
Buckminsterfullerene.
It is a geodesic dome-type of

243
00:18:48,000 --> 00:18:51,000
molecule.
This molecule has the chemical

244
00:18:51,000 --> 00:18:55,000
formula C60.
There are no hydrogens in this

245
00:18:55,000 --> 00:19:00,000
molecule.
All the atoms are displayed.

246
00:19:00,000 --> 00:19:04,000
And each carbon atom sits in a
position where it is adjacent to

247
00:19:04,000 --> 00:19:07,000
one five-membered ring and two
six-membered rings on the

248
00:19:07,000 --> 00:19:10,000
surface of this spherical
molecule.

249
00:19:10,000 --> 00:19:13,000
And that is true of every
carbon atom on the surface of

250
00:19:13,000 --> 00:19:17,000
this whole molecule.
As you go on in chemistry,

251
00:19:17,000 --> 00:19:21,000
if you go into the analysis of
molecules using nuclear magnetic

252
00:19:21,000 --> 00:19:24,000
resonance spectroscopy,
you will find that it is really

253
00:19:24,000 --> 00:19:30,000
important to be able to identify
the symmetry of a molecule.

254
00:19:30,000 --> 00:19:34,000
And you will realize that the
symmetry of the molecule is

255
00:19:34,000 --> 00:19:38,000
manifest in the nuclear magnetic
resonance spectrum of a

256
00:19:38,000 --> 00:19:41,000
molecule.
If you take the carbon-13

257
00:19:41,000 --> 00:19:45,000
nuclear magnetic resonance
spectrum of this molecule,

258
00:19:45,000 --> 00:19:49,000
a sample composed of this
molecule, you will see that

259
00:19:49,000 --> 00:19:53,000
there is only a single C13
signal in the spectrum.

260
00:19:53,000 --> 00:19:57,000
And that is because all 60
carbon atoms are in an identical

261
00:19:57,000 --> 00:20:03,000
environment in this molecule.
Every one of them feels exactly

262
00:20:03,000 --> 00:20:07,000
like everyone else.
They are all equidistant from

263
00:20:07,000 --> 00:20:11,000
the center of this molecule.
And they are all equidistant

264
00:20:11,000 --> 00:20:16,000
from their set of neighbors.
And so, in that way of looking

265
00:20:16,000 --> 00:20:20,000
at a molecule,
you would see that they are all

266
00:20:20,000 --> 00:20:23,000
equivalent.
You may have encountered this

267
00:20:23,000 --> 00:20:26,000
molecule before,
but this molecule's discovery

268
00:20:26,000 --> 00:20:30,000
--
-- was actually predicted

269
00:20:30,000 --> 00:20:34,000
initially from analysis of mass
spectrometry data by Professor

270
00:20:34,000 --> 00:20:39,000
Smalley, a Nobel Prize winner
whose efforts in this area have

271
00:20:39,000 --> 00:20:43,000
spawned off chemistries
involving not only these

272
00:20:43,000 --> 00:20:47,000
nano-sized balls of matter,
but also nanotubes made of

273
00:20:47,000 --> 00:20:52,000
carbon, and this whole area that
we think of nano technology.

274
00:20:52,000 --> 00:20:57,000
Really a lot of it is dominated
by the chemistry of new forms of

275
00:20:57,000 --> 00:21:01,000
carbon that arose with the
discovery of

276
00:21:01,000 --> 00:21:06,000
Buckminsterfullerene.
And that is just one example of

277
00:21:06,000 --> 00:21:12,000
a new allotrope of carbon that
was discovered in recent years.

278
00:21:12,000 --> 00:21:15,000
This gives you an idea for
symmetry.

279
00:21:15,000 --> 00:21:19,000
I could show you pictures,
actually, of enormous

280
00:21:19,000 --> 00:21:23,000
biomolecules.
There are large viruses that

281
00:21:23,000 --> 00:21:28,000
are composed of biopolymers,
macromolecules that pack in a

282
00:21:28,000 --> 00:21:33,000
way that is symmetrical,
so that you can see these

283
00:21:33,000 --> 00:21:37,000
things.
If you view them in the right

284
00:21:37,000 --> 00:21:40,000
kind of representation,
you will be able to see the

285
00:21:40,000 --> 00:21:44,000
symmetry in them.
And one definition of symmetry

286
00:21:44,000 --> 00:21:48,000
that I would like you to take
away from this picture is just

287
00:21:48,000 --> 00:21:51,000
that each of the atoms that are
symmetry-related is

288
00:21:51,000 --> 00:21:55,000
indistinguishable.
If you turn the molecule around

289
00:21:55,000 --> 00:21:59,000
and look at the atoms,
those that are symmetry-related

290
00:21:59,000 --> 00:22:03,000
are indistinguishable from one
another.

291
00:22:03,000 --> 00:22:07,000
And so, when we have a trigonal
planer BH three

292
00:22:07,000 --> 00:22:11,000
molecule, is the boron symmetry
related to any of the other

293
00:22:11,000 --> 00:22:13,000
atoms?
No, it is not.

294
00:22:13,000 --> 00:22:17,000
What about this hydrogen?
I labeled them last time A,

295
00:22:17,000 --> 00:22:20,000
B, and C.
This hydrogen labeled C,

296
00:22:20,000 --> 00:22:23,000
is it symmetry-related to other
hydrogens?

297
00:22:23,000 --> 00:22:25,000
Yes.
And that is because of the

298
00:22:25,000 --> 00:22:30,000
symmetry of this molecule with
these 120 degree bond angles on

299
00:22:30,000 --> 00:22:36,000
the planarity of the molecule.
So, you have a set of three

300
00:22:36,000 --> 00:22:39,000
hydrogens.
And their 1s orbitals are in

301
00:22:39,000 --> 00:22:42,000
space indistinguishable from one
another.

302
00:22:42,000 --> 00:22:47,000
They are related by symmetry.
And so, in this problem here,

303
00:22:47,000 --> 00:22:51,000
we have four orbitals here on
the boron that are not

304
00:22:51,000 --> 00:22:55,000
symmetry-related.
And then also incidentally let

305
00:22:55,000 --> 00:23:00,000
me point out that the boron atom
is located at the center of

306
00:23:00,000 --> 00:23:04,000
gravity of this system.
If atoms are going to be

307
00:23:04,000 --> 00:23:08,000
symmetry-related,
they must not be located at the

308
00:23:08,000 --> 00:23:13,000
center of gravity of the system.
Let's go over here and expand

309
00:23:13,000 --> 00:23:15,000
on these ideas.

310
00:23:35,000 --> 00:23:39,000
The method that I am going to
develop here for forming the

311
00:23:39,000 --> 00:23:44,000
linear combinations has to do
with thinking ahead,

312
00:23:44,000 --> 00:23:48,000
in this problem,
to the fact that we are going

313
00:23:48,000 --> 00:23:52,000
to want to make linear
combinations that have the

314
00:23:52,000 --> 00:23:57,000
correct symmetry to bond to
atomic orbitals on the central

315
00:23:57,000 --> 00:24:00,000
boron atom.
Let me draw the ones that are

316
00:24:00,000 --> 00:24:03,000
going to be relevant to this
part of the problem.

317
00:24:24,000 --> 00:24:28,000
This one over here would be the
boron pz orbital,

318
00:24:28,000 --> 00:24:33,000
so it has one positive lobe
coming out of the board,

319
00:24:33,000 --> 00:24:37,000
negative lobe going back.
Over here we would have the

320
00:24:37,000 --> 00:24:42,000
boron py orbital using the
coordinate system that I had

321
00:24:42,000 --> 00:24:46,000
chosen last time,
which is x up and y to the

322
00:24:46,000 --> 00:24:49,000
left.
And then here we have the

323
00:24:49,000 --> 00:24:53,000
boron's px orbital.
And then over here we have the

324
00:24:53,000 --> 00:24:59,000
boron's 2s orbital.
And so, our challenge now will

325
00:24:59,000 --> 00:25:02,000
be to construct linear
combinations,

326
00:25:02,000 --> 00:25:06,000
we are at part two,
of the set of three hydrogen 1s

327
00:25:06,000 --> 00:25:10,000
orbitals that can match in
symmetry the boron central

328
00:25:10,000 --> 00:25:13,000
atom's atomic orbitals.
Last time, remember,

329
00:25:13,000 --> 00:25:17,000
for hybridization,
we were making these orbitals

330
00:25:17,000 --> 00:25:21,000
mix with each other in order to
point at the hydrogens.

331
00:25:21,000 --> 00:25:27,000
Now what we are doing is kind
of an inverse concept.

332
00:25:27,000 --> 00:25:30,000
We are going to mix the
hydrogen orbitals so that they

333
00:25:30,000 --> 00:25:34,000
have the right symmetry to
interact with the central atom

334
00:25:34,000 --> 00:25:37,000
atomic orbitals.
There is a nice parallelism

335
00:25:37,000 --> 00:25:40,000
here.
Here is going to be our LCs.

336
00:25:50,000 --> 00:25:53,000
They key feature of the boron's
2s orbital is that it does not

337
00:25:53,000 --> 00:25:56,000
have any nodes.
Remember nodes are surfaces.

338
00:25:56,000 --> 00:25:59,000
When you pass from one side of
a node to the other,

339
00:25:59,000 --> 00:26:03,000
you get a change in sign of the
wave function.

340
00:26:03,000 --> 00:26:08,000
And we indicate that change in
sign by differential shading.

341
00:26:08,000 --> 00:26:11,000
The 2s orbital has no nodes
whatsoever.

342
00:26:11,000 --> 00:26:16,000
And a way that we can construct
a linear combination that has

343
00:26:16,000 --> 00:26:20,000
the same spatial,
nodal properties as that boron

344
00:26:20,000 --> 00:26:25,000
2s atomic orbital is as follows.
We can involve contributions

345
00:26:25,000 --> 00:26:30,000
from each of the three hydrogen
1s orbitals.

346
00:26:30,000 --> 00:26:33,000
Remember this one is A,
this one is B,

347
00:26:33,000 --> 00:26:37,000
and this one is C.
This is going to be a linear

348
00:26:37,000 --> 00:26:40,000
combination of the three
hydrogen 1s orbitals.

349
00:26:40,000 --> 00:26:43,000
I will write this one as
follows.

350
00:26:43,000 --> 00:26:47,000
This one will be written as A
plus B plus C.

351
00:26:47,000 --> 00:26:52,000
That indicates the 1s orbital
on A, plus the 1s orbital on B,

352
00:26:52,000 --> 00:26:57,000
plus the 1s orbital on C.
And, as we talked about last

353
00:26:57,000 --> 00:27:03,000
time, wave functions that we
write should be normalized.

354
00:27:03,000 --> 00:27:07,000
And they should satisfy the
unit orbital contribution rule.

355
00:27:07,000 --> 00:27:10,000
For normalization,
here, I give a factor of one

356
00:27:10,000 --> 00:27:15,000
over root three for this linear
combination formed as a symmetry

357
00:27:15,000 --> 00:27:18,000
match with the boron 2s orbital.
You can see,

358
00:27:18,000 --> 00:27:22,000
I hope, what we are doing.
We are projecting the nodal

359
00:27:22,000 --> 00:27:27,000
properties of the central atom
atomic orbitals onto the linear

360
00:27:27,000 --> 00:27:31,000
combinations that we are
forming.

361
00:27:31,000 --> 00:27:36,000
And we are going to form a
complete set of three linear

362
00:27:36,000 --> 00:27:40,000
combinations in this way.
Let's make one that has

363
00:27:40,000 --> 00:27:45,000
symmetry properties that remind
us of this px orbital.

364
00:27:45,000 --> 00:27:50,000
On HA we are going to have a
positive contribution to match

365
00:27:50,000 --> 00:27:56,000
the positive contribution of
this lobe of the px orbital that

366
00:27:56,000 --> 00:28:02,000
points along the plus x axis.
And then down here we are going

367
00:28:02,000 --> 00:28:05,000
to have contributions from
hydrogens B and C.

368
00:28:05,000 --> 00:28:09,000
And they are going to be
smaller, and they are going to

369
00:28:09,000 --> 00:28:13,000
be opposite in-phase.
They are going to be opposite

370
00:28:13,000 --> 00:28:17,000
in-phase because we are building
a linear combination that has a

371
00:28:17,000 --> 00:28:21,000
node approximately at the center
of the system here,

372
00:28:21,000 --> 00:28:25,000
so as you go from positive x
into the negative x region of

373
00:28:25,000 --> 00:28:30,000
space, the wave function changes
sign to match the change in sign

374
00:28:30,000 --> 00:28:36,000
associated with the px orbital.
We are projecting the nodal

375
00:28:36,000 --> 00:28:40,000
properties of px onto the linear
combination of hydrogen orbitals

376
00:28:40,000 --> 00:28:43,000
that we are forming here.
And this one,

377
00:28:43,000 --> 00:28:48,000
written in normalized fashion,
will be root two over three,

378
00:28:48,000 --> 00:28:50,000
A minus one-half B minus
one-half C.

379
00:28:50,000 --> 00:28:55,000
And, although what we are

380
00:28:55,000 --> 00:28:58,000
working with here are linear
combinations of these

381
00:28:58,000 --> 00:29:02,000
symmetry-related hydrogen 1s
wave functions,

382
00:29:02,000 --> 00:29:06,000
you are going to find that
these coefficients on the atomic

383
00:29:06,000 --> 00:29:10,000
orbitals, that contribute
eventually to the molecular

384
00:29:10,000 --> 00:29:14,000
orbitals, are going to come out
as normalized and as unit

385
00:29:14,000 --> 00:29:20,000
orbital contributions.
So that if we started out this

386
00:29:20,000 --> 00:29:24,000
problem with a single 1s orbital
on HA, that will be entirely

387
00:29:24,000 --> 00:29:29,000
accounted for among these linear
combinations and the molecular

388
00:29:29,000 --> 00:29:33,000
orbitals that we are going to
make with them.

389
00:29:33,000 --> 00:29:37,000
Now, let's generate a linear
combination having the nodal

390
00:29:37,000 --> 00:29:41,000
properties of py.
And, in order to do that,

391
00:29:41,000 --> 00:29:46,000
we need to have a negative
coefficient out here in the

392
00:29:46,000 --> 00:29:49,000
minus y direction.
We are going to put in a

393
00:29:49,000 --> 00:29:53,000
contribution from HC as
negative, like that,

394
00:29:53,000 --> 00:29:56,000
to match that.
And then over here,

395
00:29:56,000 --> 00:30:00,000
we are going to make a
contribution for HB that is

396
00:30:00,000 --> 00:30:05,000
positive.
And then, noting that there is

397
00:30:05,000 --> 00:30:08,000
a nodal plane along the
y,z-plane, which comes out of

398
00:30:08,000 --> 00:30:12,000
the board like this,
so that we are always negative

399
00:30:12,000 --> 00:30:15,000
along minus y and we are
positive along plus y.

400
00:30:15,000 --> 00:30:19,000
We match that here.
And the coincidence of that

401
00:30:19,000 --> 00:30:23,000
nodal plane with the location of
HA dictates no contribution from

402
00:30:23,000 --> 00:30:27,000
HA to this orbital,
for reasons that actually we

403
00:30:27,000 --> 00:30:30,000
looked at last time,
--

404
00:30:30,000 --> 00:30:35,000
-- namely, that we cannot bring
a hydrogen 1s orbital in here

405
00:30:35,000 --> 00:30:41,000
and also change sign on going
half-way through that hydrogen

406
00:30:41,000 --> 00:30:46,000
1s orbital, because s orbitals
have to have the same sign

407
00:30:46,000 --> 00:30:50,000
everywhere.
It does not contribute to this

408
00:30:50,000 --> 00:30:54,000
linear combination.
And our normalized form for

409
00:30:54,000 --> 00:30:58,000
this will be one over root two B
minus C.

410
00:30:58,000 --> 00:31:04,000
And then, if I were to ask the

411
00:31:04,000 --> 00:31:07,000
question, can I make a linear
combination of the three

412
00:31:07,000 --> 00:31:11,000
hydrogen 1s orbitals that has
the same nodal properties as pz,

413
00:31:11,000 --> 00:31:15,000
the answer would be no because
they all lie in the x,y-plane

414
00:31:15,000 --> 00:31:20,000
and they are just s orbitals and
cannot change sign as you go

415
00:31:20,000 --> 00:31:23,000
through the x,y-plane from plus
z to minus z.

416
00:31:23,000 --> 00:31:26,000
And so, we are done here.
And what you are going to find

417
00:31:26,000 --> 00:31:30,000
is that we have created these
three linear combination

418
00:31:30,000 --> 00:31:35,000
according to step two.
And taking into account both

419
00:31:35,000 --> 00:31:39,000
the symmetry properties of the
molecule to identify a set of

420
00:31:39,000 --> 00:31:42,000
three symmetry-related atoms and
orbitals.

421
00:31:42,000 --> 00:31:47,000
And then, taking into account
an analysis of the nodal

422
00:31:47,000 --> 00:31:50,000
properties of the central atom
atomic orbitals,

423
00:31:50,000 --> 00:31:55,000
so that we could project those
out to help us find appropriate

424
00:31:55,000 --> 00:32:01,000
linear combinations for mixing
with the central atom orbitals.

425
00:32:10,000 --> 00:32:14,000
And, when we do that mixing,
we are going to find out that

426
00:32:14,000 --> 00:32:17,000
there are three ways that we can
do it.

427
00:32:17,000 --> 00:32:21,000
We are about to move onto step
three of this problem.

428
00:32:21,000 --> 00:32:26,000
We are going to need to combine
these linear combinations with

429
00:32:26,000 --> 00:32:31,000
the central atom atomic orbitals
according to the rules of MO

430
00:32:31,000 --> 00:32:35,000
theory to generate,
first, bonding molecular

431
00:32:35,000 --> 00:32:37,000
orbitals.

432
00:32:43,000 --> 00:32:50,000
And the bonding molecular
orbitals that we will get will

433
00:32:50,000 --> 00:32:54,000
be an in-phase combination --

434
00:33:01,000 --> 00:33:04,000
-- of our LCs,
our linear combinations of

435
00:33:04,000 --> 00:33:09,000
atomic orbitals,
with our boron atomic orbitals.

436
00:33:09,000 --> 00:33:14,000
That will describe the chemical
bonding in our system.

437
00:33:14,000 --> 00:33:19,000
And we will see that it
contrasts in a very interesting

438
00:33:19,000 --> 00:33:24,000
way with the hybridization
scheme developed last time.

439
00:33:24,000 --> 00:33:27,000
And the key phrase to
underline, here,

440
00:33:27,000 --> 00:33:32,000
is in-phase.
And what that in-phase means is

441
00:33:32,000 --> 00:33:36,000
that when two positive lobes of
two orbitals centered on two

442
00:33:36,000 --> 00:33:40,000
different atoms are juxtaposed
and neighbor one another and can

443
00:33:40,000 --> 00:33:44,000
have good overlap of their
atomic orbital wave functions,

444
00:33:44,000 --> 00:33:48,000
that leads to in-phase
constructive interference and

445
00:33:48,000 --> 00:33:52,000
stabilization of the electrons
associated with that newly

446
00:33:52,000 --> 00:33:54,000
formed bonding molecular
orbital.

447
00:33:54,000 --> 00:34:00,000
And that stabilization is what
we call the chemical bond.

448
00:34:00,000 --> 00:34:04,000
The analogy to that in valence
bond theory is the idea that a

449
00:34:04,000 --> 00:34:08,000
pair of electrons associated
with two nuclei in a sigma bond

450
00:34:08,000 --> 00:34:11,000
is more stable because it
experiences, simultaneously,

451
00:34:11,000 --> 00:34:15,000
two positive charges.
And here we are generalizing

452
00:34:15,000 --> 00:34:19,000
that and allowing electrons to
flow over the molecule as a

453
00:34:19,000 --> 00:34:21,000
whole.
But now we have a new concept,

454
00:34:21,000 --> 00:34:24,000
and that is antibonding.

455
00:34:30,000 --> 00:34:36,000
Antibonding molecular orbitals
will be out-of-phase

456
00:34:36,000 --> 00:34:42,000
combinations that are repulsive
and lead to high energy

457
00:34:42,000 --> 00:34:45,000
interactions.

458
00:34:50,000 --> 00:34:55,000
And when, as in the case of the
problem we are considering here,

459
00:34:55,000 --> 00:34:58,000
the interactions occur in
pair-wise sets,

460
00:34:58,000 --> 00:35:03,000
we will find that we get very
nicely, for every bonding

461
00:35:03,000 --> 00:35:07,000
molecular orbital,
a corresponding anti-bonding

462
00:35:07,000 --> 00:35:10,000
molecular orbital.

463
00:35:15,000 --> 00:35:18,000
Also, one of the interesting
things is that if you start

464
00:35:18,000 --> 00:35:20,000
putting electrons into
anti-bonding orbitals,

465
00:35:20,000 --> 00:35:23,000
if your system just happens to
be so constructed as to have

466
00:35:23,000 --> 00:35:26,000
many electrons,
such that you fill up not only

467
00:35:26,000 --> 00:35:29,000
the bonding molecular orbitals
of electrons to make the

468
00:35:29,000 --> 00:35:31,000
chemical bonds,
but you continue on and you

469
00:35:31,000 --> 00:35:34,000
have enough electrons to keep
going and put them into

470
00:35:34,000 --> 00:35:37,000
anti-bonding orbitals,
those antibonds start to cancel

471
00:35:37,000 --> 00:35:40,000
the bonds.
And so, you will see a very

472
00:35:40,000 --> 00:35:44,000
nice progression of this as we
study the MO theory of the

473
00:35:44,000 --> 00:35:48,000
homonuclear diatomic molecules,
starting on Friday.

474
00:35:48,000 --> 00:35:52,000
And then, here is another
concept that arises from the MO

475
00:35:52,000 --> 00:35:55,000
analysis of a molecule,
and that is that certain

476
00:35:55,000 --> 00:35:58,000
orbitals can be non-bonding.

477
00:36:04,000 --> 00:36:10,000
And an example of this would be
a lone pair of electrons.

478
00:36:10,000 --> 00:36:16,000
And it happens when an orbital
or a linear combination of

479
00:36:16,000 --> 00:36:21,000
orbitals finds no counterpart --

480
00:36:27,000 --> 00:36:31,000
-- of like nodal symmetry.
These nodal properties of

481
00:36:31,000 --> 00:36:36,000
orbitals are very important,
and I will show you later how

482
00:36:36,000 --> 00:36:42,000
the nodal properties are related
to the energies of the orbitals,

483
00:36:42,000 --> 00:36:46,000
as we consider them.
Having given you this preview

484
00:36:46,000 --> 00:36:52,000
of how orbitals are going to be
able to combine in the MO

485
00:36:52,000 --> 00:36:57,000
theory, let's see how it
actually takes place in the case

486
00:36:57,000 --> 00:37:00,000
of BH three.

487
00:37:15,000 --> 00:37:18,000
Now we are drawing another
example of an energy-level

488
00:37:18,000 --> 00:37:23,000
diagram, where the energy is low
at the bottom and rises as it

489
00:37:23,000 --> 00:37:26,000
goes up.
And these energy-level diagrams

490
00:37:26,000 --> 00:37:30,000
that we are now developing for
molecules are analogous to those

491
00:37:30,000 --> 00:37:35,000
that you studied earlier in the
semester for atoms.

492
00:37:35,000 --> 00:37:41,000
And we are just generalizing
this notion to the atoms.

493
00:37:41,000 --> 00:37:48,000
And what I am drawing over here
is our boron 2s orbital and here

494
00:37:48,000 --> 00:37:52,000
is our boron 2px,
2py, 2pz orbital.

495
00:37:52,000 --> 00:37:58,000
Those horizontal bars just
represent the energy of these

496
00:37:58,000 --> 00:38:04,000
orbitals in the molecule.
Here I am just redrawing the

497
00:38:04,000 --> 00:38:08,000
boron atom.
And then, over here on the

498
00:38:08,000 --> 00:38:14,000
right, we are going to see that
we have our linear combinations

499
00:38:14,000 --> 00:38:19,000
that we developed.
These are our three H 1s linear

500
00:38:19,000 --> 00:38:22,000
combinations.
We found that the three

501
00:38:22,000 --> 00:38:28,000
hydrogens in BH three
were symmetry equivalent,

502
00:38:28,000 --> 00:38:33,000
so we generated linear
combinations.

503
00:38:33,000 --> 00:38:38,000
The pictures of them are over
there.

504
00:38:38,000 --> 00:38:45,000
I can give them names.
Why don't I call them D,

505
00:38:45,000 --> 00:38:48,000
E, and F.

506
00:39:00,000 --> 00:39:05,000
We have D constructed to match
the nodal properties of the

507
00:39:05,000 --> 00:39:09,000
boron's 2s orbital.
E and F were respectively

508
00:39:09,000 --> 00:39:14,000
constructed to match the nodal
properties of the boron's 2s,

509
00:39:14,000 --> 00:39:18,000
2px, and 2py orbitals.
And I am showing you their

510
00:39:18,000 --> 00:39:22,000
relative energies.
And now we need to do this

511
00:39:22,000 --> 00:39:26,000
issue referenced here as point
three.

512
00:39:26,000 --> 00:39:30,000
We need to combine these
things.

513
00:39:30,000 --> 00:39:33,000
I have these on the one hand
and these on the other.

514
00:39:33,000 --> 00:39:38,000
Our seven atomic orbitals have
been changed into four atomic

515
00:39:38,000 --> 00:39:40,000
orbitals and three linear
combinations,

516
00:39:40,000 --> 00:39:43,000
so I still have seven orbitals
total.

517
00:39:43,000 --> 00:39:47,000
And now I am going to combine
these with these according to

518
00:39:47,000 --> 00:39:51,000
their nodal properties to
generate seven molecular

519
00:39:51,000 --> 00:39:55,000
orbitals.
And let's do it this way.

520
00:40:03,000 --> 00:40:08,000
We are going to take linear
combination D that has been

521
00:40:08,000 --> 00:40:13,000
constructed to match the boron
2s orbital in terms of nodal

522
00:40:13,000 --> 00:40:18,000
symmetry properties,
and we are going to make a

523
00:40:18,000 --> 00:40:22,000
bonding combination.
I am going to draw these

524
00:40:22,000 --> 00:40:26,000
pictorially in a moment.
And, in this case,

525
00:40:26,000 --> 00:40:32,000
those are the only two orbitals
in my seven orbital system here

526
00:40:32,000 --> 00:40:38,000
that have that set of nodal
symmetry properties.

527
00:40:38,000 --> 00:40:42,000
And for every bonding MO,
I must have an antibonding MO.

528
00:40:42,000 --> 00:40:44,000
This is one of our molecular
orbitals.

529
00:40:44,000 --> 00:40:49,000
And there is going to be a
corresponding molecular orbital

530
00:40:49,000 --> 00:40:53,000
up here, high in energy.
And this will be an antibonding

531
00:40:53,000 --> 00:40:57,000
molecular orbital that will be
the out-of-phase combination of

532
00:40:57,000 --> 00:41:03,000
the boron 2s with this linear
combination that I labeled D.

533
00:41:03,000 --> 00:41:09,000
Antibonding molecular orbitals
are usually denoted with a star.

534
00:41:09,000 --> 00:41:15,000
We have a bonding combination
and an anti-bonding combination.

535
00:41:15,000 --> 00:41:21,000
Now, we can form two more bonds
that will spread out over the

536
00:41:21,000 --> 00:41:24,000
molecule because,
if you recall,

537
00:41:24,000 --> 00:41:31,000
we had our px and py pair that
served as the nodal template for

538
00:41:31,000 --> 00:41:37,000
our construction of linear
combinations E and F.

539
00:41:37,000 --> 00:41:42,000
We are going to be able to
match up those to form two more

540
00:41:42,000 --> 00:41:48,000
bonding molecular orbitals.
And these will be found to be

541
00:41:48,000 --> 00:41:53,000
higher in energy than the first
one that we formed from D.

542
00:41:53,000 --> 00:41:59,000
Here is a pair of bonding
molecular orbitals that derive

543
00:41:59,000 --> 00:42:05,000
from linear combinations E and F
combining in-phase with boron's

544
00:42:05,000 --> 00:42:10,000
px and py atomic orbitals.
And there will be a

545
00:42:10,000 --> 00:42:14,000
corresponding antibonding
combination, where we allow

546
00:42:14,000 --> 00:42:19,000
those orbitals to interact in an
out-of-phase manner.

547
00:42:19,000 --> 00:42:22,000
And you will see what that
means shortly.

548
00:42:22,000 --> 00:42:27,000
Let me put that up there and
add a star to indicate that this

549
00:42:27,000 --> 00:42:32,000
high energy pair of molecular
orbitals is an antibonding pair

550
00:42:32,000 --> 00:42:36,000
of orbitals.
I have six orbitals now in my

551
00:42:36,000 --> 00:42:40,000
molecular orbital energy-level
diagram for BH three.

552
00:42:40,000 --> 00:42:44,000
And that means I am not done
because I have to have seven,

553
00:42:44,000 --> 00:42:47,000
and I started out with seven
atomic orbitals.

554
00:42:47,000 --> 00:42:50,000
Look over here.
2pz was an atomic orbital on

555
00:42:50,000 --> 00:42:55,000
boron that did not find any way
of serving as a template for

556
00:42:55,000 --> 00:43:00,000
making a linear combination
involving the three hydrogens.

557
00:43:00,000 --> 00:43:05,000
And so, it comes over here as
nonbonding.

558
00:43:05,000 --> 00:43:12,000
It has no counterpart of like
nodal symmetry because of the

559
00:43:12,000 --> 00:43:18,000
location of those three
hydrogens in the x,y-plane,

560
00:43:18,000 --> 00:43:25,000
which is a nodal plane for the
boron pz orbital.

561
00:43:25,000 --> 00:43:32,000
And so, this one is nonbonding.
These three orbitals up here

562
00:43:32,000 --> 00:43:35,000
are anti-bonding.
And the ones down at the

563
00:43:35,000 --> 00:43:40,000
bottom, which are the lowest in
energy, corresponding to being

564
00:43:40,000 --> 00:43:45,000
able to most tightly hold onto
electrons in them are bonding

565
00:43:45,000 --> 00:43:49,000
molecular orbitals.
And so our electrons can fill

566
00:43:49,000 --> 00:43:53,000
into this MO energy-level
diagram in that way.

567
00:43:53,000 --> 00:43:59,000
We have our six electrons that
come into this problem.

568
00:43:59,000 --> 00:44:03,000
We have boron bringing in three
valance electrons and three

569
00:44:03,000 --> 00:44:07,000
hydrogens each bringing in one
valance electron.

570
00:44:07,000 --> 00:44:10,000
There are six electrons to put
into the diagram,

571
00:44:10,000 --> 00:44:14,000
filling up three of the
molecular orbitals and then

572
00:44:14,000 --> 00:44:18,000
leaving empty pz.
Let me introduce a little bit

573
00:44:18,000 --> 00:44:22,000
more MO language right now.
This one here will be called

574
00:44:22,000 --> 00:44:27,000
the highest occupied molecular
orbital, and this one here will

575
00:44:27,000 --> 00:44:32,000
be called the lowest unoccupied
molecular orbital.

576
00:44:32,000 --> 00:44:37,000
The reason why I am drawing
attention to these orbitals is

577
00:44:37,000 --> 00:44:41,000
that in chemistry,
the chemical properties derive

578
00:44:41,000 --> 00:44:47,000
oftentimes from those orbitals
that are in what is called the

579
00:44:47,000 --> 00:44:52,000
frontier orbital region.
And the frontier orbitals are

580
00:44:52,000 --> 00:44:56,000
those close in energy to the
HOMO-LUMO gap.

581
00:44:56,000 --> 00:45:02,000
And I will come back to this.
But those highest energetically

582
00:45:02,000 --> 00:45:05,000
lying electrons are going to be
the ones responsible for

583
00:45:05,000 --> 00:45:10,000
nucleophilic properties of the
molecule and basic properties of

584
00:45:10,000 --> 00:45:13,000
the molecule and reducing
properties of the molecule.

585
00:45:13,000 --> 00:45:17,000
Whereas, low-lying empty
orbitals are going to be the

586
00:45:17,000 --> 00:45:20,000
ones responsible for acidic
properties of the molecule or

587
00:45:20,000 --> 00:45:23,000
oxidizing properties of the
molecule.

588
00:45:23,000 --> 00:45:27,000
And we will come back to that
in a moment.

589
00:45:27,000 --> 00:45:31,000
But that is something pretty
general and very useful that

590
00:45:31,000 --> 00:45:36,000
comes out of studying molecular
orbital energy-level diagrams.

591
00:45:41,000 --> 00:45:46,000
Now that we have the diagram,
let's see what the orbitals in

592
00:45:46,000 --> 00:45:50,000
the diagram look like.
And I will try to do this

593
00:45:50,000 --> 00:45:54,000
relatively quickly.
Here, let's start with the

594
00:45:54,000 --> 00:45:59,000
lowest lying molecular orbital
in the system.

595
00:45:59,000 --> 00:46:04,000
This is a representation of an
in-phase combination of the

596
00:46:04,000 --> 00:46:09,000
boron 2s plus D,
where D is defined up here as

597
00:46:09,000 --> 00:46:14,000
that linear combination.
This will be a molecular

598
00:46:14,000 --> 00:46:19,000
orbital having the nodal
properties of a 2s orbital

599
00:46:19,000 --> 00:46:25,000
centered on that central atom.
That is where our lowest-lying

600
00:46:25,000 --> 00:46:31,000
two electrons reside.
Now, if you look at the LUMO

601
00:46:31,000 --> 00:46:35,000
over there and then go up one
orbital in energy,

602
00:46:35,000 --> 00:46:41,000
you will be looking at this
orbital, which is the boron 2s

603
00:46:41,000 --> 00:46:43,000
orbital minus D.

604
00:46:48,000 --> 00:46:54,000
And the thing that makes this
out-of-phase linear combination

605
00:46:54,000 --> 00:47:01,000
so much higher in energy than
its in-phase counterpart is the

606
00:47:01,000 --> 00:47:06,000
appearance, now,
of a nodal surface.

607
00:47:06,000 --> 00:47:08,000
And this node is between the
nuclei.

608
00:47:08,000 --> 00:47:13,000
It goes all the way around and
is between the central atom s

609
00:47:13,000 --> 00:47:16,000
orbital and those peripheral
hydrogens.

610
00:47:16,000 --> 00:47:20,000
And I will show you a picture
of it.

611
00:47:37,000 --> 00:47:42,000
This is our BH three
LUMO plus one molecular orbital.

612
00:47:52,000 --> 00:47:56,000
You can see that we have a wave
function in the center of one

613
00:47:56,000 --> 00:47:58,000
sign.
And then, as we go along any

614
00:47:58,000 --> 00:48:02,000
one of the B-H bond vectors from
boron to hydrogen,

615
00:48:02,000 --> 00:48:06,000
we change phase midway along
the bond from positive to

616
00:48:06,000 --> 00:48:09,000
negative.
And that is true,

617
00:48:09,000 --> 00:48:14,000
no matter which of the three
B-H bonds we pick to traverse

618
00:48:14,000 --> 00:48:17,000
along.
That one has the nodal

619
00:48:17,000 --> 00:48:22,000
properties as drawn down there
for the boron 2s interacting in

620
00:48:22,000 --> 00:48:28,000
an out-of-phase manner with
linear combination letter D.

621
00:48:28,000 --> 00:48:32,000
And so, next time,
I will finish up and show you

622
00:48:32,000 --> 00:48:36,000
what these other orbitals look
like as calculated.

623
00:48:36,000 --> 00:48:41,000
I hope you have enjoyed this.
We will see more MO theory on

624
00:48:41,751 --> 00:48:44,000
Friday.