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Over on the side board here,
I want to show you a few

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representation of the ethylene
molecule that we were talking

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about at the end of last hour.
When you see representations of

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molecules that appear throughout
your textbook,

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as well as in this class,
I want you to see if you can

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recognize just what property of
the molecule it is that is being

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represented.
In this case here,

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we have arbitrarily sized
spheres that represent the two

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carbon atoms and the four
peripheral hydrogen atoms.

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And then, what I have plotted
around it is a see-through

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representation of the electron
density at a particular

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isosurface value of 0.1
electrons per unit volume.

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So, that is a representation.
And we are going to be seeing a

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number of different kinds of
representations,

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and you will need to be able to
distinguish between them.

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And, in particular,
we were talking last time about

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different kinds of bonds.
We talked about sigma bonds

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that were cylindrically
symmetric about the internuclear

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axis of the two atoms that are
bonded together.

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And we also talked about pi
bonds.

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And I brought up the pi bond in
connection with the ethylene

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molecule that we were just
looking at a different

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representation of.
And so now, let me show you

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what a pi bond can look like.
Here is a representation of the

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ethylene molecule.
Again, I have arbitrarily sized

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spheres that show us just where
the nuclear positions are in

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three-dimensional space.
So, we are talking about the

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equilibrium geometry of the
ethylene molecule.

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And what you see here is that
if the molecule is lying in the

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x,y-plane, then each of these
carbons has a 2pz orbital that

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has a positive lobe and a
negative lobe,

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respectively,
above and below the plane of

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the molecule.
And so, normally,

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when we talk about atomic
orbitals, we can reference their

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phase as either positive or
negative.

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And we often,
in representations like this

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one, associate the positive
phase with one color,

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here blue, and the negative
phase with a different color,

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here yellow.
I can choose whatever colors I

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want for those,
but this is also a type of

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isosurface.
And it represents the in-phase

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side-to-side overlap of the
positive above the plane lobes

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of the carbon 2pz orbitals and
the negative lobes below the

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plane of the carbon 2pz orbitals
experiencing side-to-side

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overlap.
And so you see the two

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electrons that could be paired
up, spin paired,

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one spin up,
one spin down in this pi bond,

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are smeared out both above and
below the plane

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But there is no contribution at
all to the electron density in

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the plane from an orbital like
this.

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When we looked at electron
density isosurfaces,

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as we have done a few times in
this class, what we are looking

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at there is an isosurface
corresponding to all of the

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electrons in the molecule.
And here, what we are looking

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at is an orbital that can house
just two electrons,

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one spin up and one spin down,
so we are focusing our

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attention on a function that
involves just two electrons

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bonding in a pi bond that has,
as a nodal surface,

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the plane of the molecule.
The plane that contains all six

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of these nuclei,
the two carbons and the four

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hydrogens.
We talked about nodal planes

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last time.
And the issue of nodal planes,

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nodal surfaces,
is going to be very important

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in today's lecture.
Let's go over here and start

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talking about hybridization.

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On this board last time,
I gave you kind of a timeline,

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where we talked about
contributions to different

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aspects of electronic structure
theory for molecules by

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different people,
some of who were recognized

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with Nobel Prizes,
others who were not.

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And one of those people,
namely Linus Pauling,

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who won two Nobel Prizes,
not both in chemistry,

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came up with this idea of
hybridization to solve one of

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the problems associated with
atomic orbital nodes.

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And let me work on this and
illustrate this to you by taking

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a pretty simple molecule that
has only four atoms.

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And this molecule is one that,
in fact, we have seen before.

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It is an example of a Lewis
acid, so any discussion of

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electronic structure of this BH
three molecule better

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come out with a description of
why it is a Lewis acid.

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But let me now point something
out.

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That is, if I make the
direction that coincides with

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this top B-H bond,
the x-axis.

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And in the plane of the board,
the y-axis perpendicular to it,

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as shown.
And then, of course,

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the z-axis coming
perpendicularly out of the plane

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of the board.
Linus Pauling recognized a

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problem.
And let me give these hydrogens

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descriptors.
I will make this one A,

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this one labeled B,
and this one labeled C.

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And the problem is that we have
a pz orbital with one positive

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lobe coming out of the plane of
the board, up here.

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This would be where the blue
lobe is.

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And the yellow lobe would be
back behind the plane of the

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board, analogous to the pi bond
that we just looked at up on the

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screen.
And what you will see is I can

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draw it this way.
I am rotating this planar BH

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three molecule.

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That allows me to draw the pz
orbital in a way that you can

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visualize.
And the observation that I want

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you to take away from this
drawing is it has to do with the

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relationship in space of the
three hydrogen nuclei as an

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important property of the 2pz
orbital.

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What we can say is that these
three hydrogens,

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A, B, and C reside in the nodal
plane of the 2pz.

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They lie in the 2pz nodal
plane.

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That is, the 2pz on the boron.
And the wave functions that

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these hydrogens bring into the
problem of the electronic

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structure of the BH three
molecule are spherically

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symmetric.
They are simply 1s orbitals.

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And so, if you draw one of them
here, remember that a 1s orbital

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is spherically symmetric.
You can see that any positive

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reinforcement that would occur
by overlap here is equal and

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opposite to the negative
reinforcement that will occur

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down here.
And this is something that is

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true whenever a hydrogen s
orbital lies in a nodal plane.

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There is no net overlap--

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--between the boron's 2pz and
the hydrogen 1s orbitals.

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When we try to describe the
bonding in this molecule,

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we cannot have any bonds that
involve 2pz and any of the three

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hydrogen 1s orbitals because
there is no net overlap between

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them --
-- because those spherically

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symmetric orbitals lie in the
nodal plane of 2pz.

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And that means,
further, that we need to form

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three in-plane bonds.

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And that is the xy plane in
plane bonds.

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And these three bonds must be
equivalent.

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And what is interesting about
the fact that they must be

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equivalent is that what are the
three orbitals that the boron

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has that are in the x,y-plane,
the three valence orbitals?

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There is the 2s,
the 2px, and the 2py.

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Three orbitals that are not
identical to one another,

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yet we need to make three
identical bonds because all the

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studies that we do on the BH
three molecule tells us

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that each of these bonds are the
same length, the molecule has

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the perfect symmetry of a
Mercedes Benz symbol,

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it's trigonal planar.
It has this three-fold

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rotational symmetry to it.
And so somehow,

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using a 2s, a 2px,
and a 2py, we need to make

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three equivalent bonds.
This is an energy level

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diagram.
And so, Pauling came up with

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the idea of promotion.
And promotion will correspond

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to the following.
I can draw 2px,

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2py, 2pz, and 2s.
And I can populate with

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electrons.
The boron atom has three.

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Now I have put in three
electrons into this energy level

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diagram for the boron atom.

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And now, this concept of
promotion kicks in --

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-- and reorganizes us
electronically,

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as follows.

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According to our discussion,
we need three equivalent

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orbitals down here with the
three electrons in to pair up

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with the three electrons that
come into the problem from the

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three hydrogen atoms.
At the end, we will show how

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those electrons form the
hydrogen atoms come in and pair

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up with these three electrons on
a promoted boron atom.

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Up here, we have the boron 2pz
orbital.

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That stays by itself.
And down here,

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we are going to have a set of
sp two hybrids.

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The idea is now somehow we are
going to get three orbitals down

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here by mixing one part s and
two parts p.

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And, if we do this correctly
mathematically,

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these will have the right
orientation in space to be

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hybrids that have proper
directionality for good overlap

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with our hydrogen 1s oribtals.
These are going to be three

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equivalent hybrids.

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How do we do that?
Well, we are going to get our

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equations for hybridization.

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And we are going to develop
these using some of the results

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from quantum mechanics,
which is what Pauling did.

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And essentially,
what we are going to do is seek

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equations that have one part s
and two parts p,

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00:14:17,000 --> 00:14:20,000
each of them,
but which point in the correct

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directions in space for making
three separate two electron

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bonds between the boron and the
three hydrogens.

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00:14:30,000 --> 00:14:35,000
And so, I will just call them
sp two of A,

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because we are going to need
one that points toward that A

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00:14:41,000 --> 00:14:46,000
hydrogen, which is along the
positive x-axis,

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00:14:46,000 --> 00:14:52,000
we are going to need one that
points toward the B hydrogen,

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and we are going to need one
that points toward the C

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hydrogen.
And the idea here is that we

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are going to have some amount of
the boron 2s,

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00:15:07,000 --> 00:15:11,000
px and py --
-- because pz is not

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contributing to these three
hybrids.

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And each one of these is going
to have some quantity of the

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00:15:19,000 --> 00:15:22,000
three orbitals that we have
available.

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And what you need to keep in
mind here in setting up

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equations like this is that px
has a definite spatial

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orientation with respect to the
coordinate system that we have

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chosen.
And so does py.

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And the way that we will make
this come out using our blue

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chalk is to note that we have a
coefficient C1,

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00:15:52,000 --> 00:15:57,000
C2 and C3.
Our job is going to be to find

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the values of these
coefficients,

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00:16:01,000 --> 00:16:06,000
C4, C5, C6, C7,
C8, and C9.

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00:16:06,000 --> 00:16:12,000
We can use the rules of quantum
mechanics, and I am going to

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summarize a few of these rules
for you before this lecture

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00:16:17,000 --> 00:16:22,000
ends, to find the values of
these constants.

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00:16:22,000 --> 00:16:26,000
Let's go over here and begin.

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00:16:50,000 --> 00:16:53,000
Let's make note of the dictates
of symmetry.

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00:16:53,000 --> 00:16:58,000
I will reference symmetry a
couple of times today in class

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00:16:58,000 --> 00:17:03,000
in working on this problem of
finding those coefficients,

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00:17:03,000 --> 00:17:07,000
one through nine.
The neat thing is that the 2s

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00:17:07,000 --> 00:17:11,000
orbital centered on that boron,
which is going to be

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00:17:11,000 --> 00:17:16,000
participating in these hybrids
that are going to point toward

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00:17:16,000 --> 00:17:19,000
hydrogens A, B,
and C, that 2s orbital is

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00:17:19,000 --> 00:17:23,000
spherically symmetric.
Its wave function is the same

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00:17:23,000 --> 00:17:28,000
as you go radially out from the
boron center in any direction

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00:17:28,000 --> 00:17:31,000
into space.
What that symmetry

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00:17:31,000 --> 00:17:38,000
consideration dictates is that
there will be the same amount of

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00:17:38,000 --> 00:17:43,000
boron 2s in the hybrids for A,
as for B, as for C.

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00:17:43,000 --> 00:17:50,000
And what that means is that C1
is equal to C4 is equal to C7.

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00:17:58,000 --> 00:18:01,000
And we are going to need a
value for this.

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00:18:01,000 --> 00:18:06,000
And we can get a value for that
by taking into consideration

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00:18:06,000 --> 00:18:11,000
something called unit orbital
contribution.

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00:18:19,000 --> 00:18:26,000
What that says is that we have
to use the boron's 2s orbital

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00:18:26,000 --> 00:18:35,000
completely and identically once
in that set of three equations.

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00:18:35,000 --> 00:18:38,000
In other words,
we are going to distribute it

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evenly among those three
hybrids.

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00:18:40,000 --> 00:18:44,000
And we have to distribute all
of it evenly among those three

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00:18:44,000 --> 00:18:47,000
hybrids.
And here we are going to be

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00:18:47,000 --> 00:18:50,000
interested in the square of the
coefficients.

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00:18:50,000 --> 00:18:54,000
And that is because when we
talk about atomic orbitals,

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00:18:54,000 --> 00:18:58,000
we know that the probability of
finding an electron in that

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00:18:58,000 --> 00:19:04,000
orbital is related to the square
of the wave function.

230
00:19:04,000 --> 00:19:07,000
And so, normally when we plot
orbitals, like I just did with

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00:19:07,000 --> 00:19:11,000
the pi bond of ethylene a few
minutes ago, we are making an

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00:19:11,000 --> 00:19:14,000
isosurface that corresponds to
some percent probability of

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00:19:14,000 --> 00:19:17,000
finding the electron at that
point in space.

234
00:19:17,000 --> 00:19:20,000
So, we are interested in
probability distributions.

235
00:19:20,000 --> 00:19:24,000
And these are related to the
square of the wave function.

236
00:19:24,000 --> 00:19:27,000
Here, for unit orbital,
we are going to say that C1

237
00:19:27,000 --> 00:19:32,000
squared equals C4 squared.
Sorry, plus C4 squared plus C7

238
00:19:32,000 --> 00:19:38,000
squared is equal to one. And,

239
00:19:38,000 --> 00:19:44,000
therefore, C1 is equal to C4 is
equal to C7 is equal to one over

240
00:19:44,000 --> 00:19:48,000
root three.

241
00:19:48,000 --> 00:19:51,000
Now, we have found C1,
C4, and C7.

242
00:19:51,000 --> 00:19:56,000
And, if we are going to
complete sp two of A,

243
00:19:56,000 --> 00:20:01,000
get this first hybrid of the
three, I am going to go across

244
00:20:01,000 --> 00:20:05,000
this way.
I just went down one,

245
00:20:05,000 --> 00:20:09,000
four, and seven and pointed out
that by the symmetry of the s

246
00:20:09,000 --> 00:20:13,000
orbital is due to the fact that
the s orbital contributes once,

247
00:20:13,000 --> 00:20:16,000
unit orbital contribution,
we could get C1,

248
00:20:16,000 --> 00:20:19,000
C4, and C7 equal to one over
root three.

249
00:20:19,000 --> 00:20:23,000
Now I am interested in C2 and

250
00:20:23,000 --> 00:20:27,000
C3, so let's draw a little
picture.

251
00:20:35,000 --> 00:20:38,000
This, remember,
is our 1s orbital on HA.

252
00:20:38,000 --> 00:20:44,000
And what we are trying to do is
to construct a hybrid focusing

253
00:20:44,000 --> 00:20:49,000
on that top equation that will
be directed at HA.

254
00:20:49,000 --> 00:20:53,000
And we are keeping our axes the
same as before,

255
00:20:53,000 --> 00:20:59,000
so that x points up and y goes
to the left, as I am drawing it

256
00:20:59,000 --> 00:21:04,000
here.
And z would be coming out at

257
00:21:04,000 --> 00:21:08,000
us.
What I am going to draw now is

258
00:21:08,000 --> 00:21:13,000
a picture of the boron's 2py
orbital.

259
00:21:13,000 --> 00:21:20,000
And, as I usually do,
I am taking the convention that

260
00:21:20,000 --> 00:21:28,000
the negative phase lobe of the
boron's 2py orbital--

261
00:21:28,000 --> 00:21:32,000
Negative is shaded.
That is my convention.

262
00:21:32,000 --> 00:21:38,000
That 2py orbital lies in the xy
plane and points along y.

263
00:21:38,000 --> 00:21:43,000
Its positive lobe,
in fact, points along positive

264
00:21:43,000 --> 00:21:46,000
y.
Its negative lobe points along

265
00:21:46,000 --> 00:21:50,000
negative y.
And what can we say about the

266
00:21:50,000 --> 00:21:56,000
relationship of the HA 1s
orbital to the 2py orbital of

267
00:21:56,000 --> 00:22:00,000
that central boron atom?

268
00:22:05,000 --> 00:22:11,000
Let me phrase it this way.
The 2py orbital of our central

269
00:22:11,000 --> 00:22:16,000
boron atom has a nodal plane.
What is that nodal plane?

270
00:22:16,000 --> 00:22:22,000
xz is a nodal plane of the 2py
orbital, right here.

271
00:22:22,000 --> 00:22:27,000
Just like this.
And so, what you see is that

272
00:22:27,000 --> 00:22:31,000
this HA is not only in the nodal
plane of 2pz,

273
00:22:31,000 --> 00:22:37,000
it is also in the nodal plane
of 2py.

274
00:22:37,000 --> 00:22:42,000
And that means any positive
interference by bonding between

275
00:22:42,000 --> 00:22:47,000
HA's 1s and the positive lobe of
2py would be exactly canceled by

276
00:22:47,000 --> 00:22:52,000
the equal and opposite negative
interference between a

277
00:22:52,000 --> 00:22:57,000
hydrogen's 1s orbital and the
negative phase lobe on 2py.

278
00:22:57,000 --> 00:23:02,000
And that means that we know the
value of C3.

279
00:23:02,000 --> 00:23:08,000
We can say, therefore,
that C3 is equal to zero.

280
00:23:08,000 --> 00:23:17,000
That is really cool because now
we are going to be able to bring

281
00:23:17,000 --> 00:23:26,000
in one more consideration and
complete the formula for sp two

282
00:23:26,000 --> 00:23:29,000
of A.

283
00:23:41,000 --> 00:23:43,000
We are going to be able to use
the normalization condition,

284
00:23:43,000 --> 00:23:44,000
--

285
00:23:54,000 --> 00:24:01,000
-- which tells us that C1
squared plus C2 squared plus C3

286
00:24:01,000 --> 00:24:08,000
squared is equal to one.

287
00:24:08,000 --> 00:24:11,000
Once we formed this new hybrid
by mixing together s and p wave

288
00:24:11,000 --> 00:24:15,000
functions, we are going to find
that this normalization

289
00:24:15,000 --> 00:24:19,000
condition must be satisfied for
the hybrid orbital just as it is

290
00:24:19,000 --> 00:24:21,000
satisfied for these atomic
orbitals.

291
00:24:21,000 --> 00:24:25,000
That is so that the probability
of finding an electron somewhere

292
00:24:25,000 --> 00:24:29,000
is space associated with this
hybrid wave function is one,

293
00:24:29,000 --> 00:24:33,000
if there is an electron in this
orbital.

294
00:24:33,000 --> 00:24:39,000
That is where this comes from.
And we know that C3 squared is

295
00:24:39,000 --> 00:24:42,000
zero. C1 squared is one

296
00:24:42,000 --> 00:24:46,000
over root three. So,

297
00:24:46,000 --> 00:24:52,000
we know that sp squared of A is
equal to one over root three.

298
00:24:52,000 --> 00:24:57,000
And then we can say that,

299
00:24:57,000 --> 00:25:02,000
by virtue of this and the fact
that C3 is zero,

300
00:25:02,000 --> 00:25:08,000
C2 could be equal to either
plus or minus the square root of

301
00:25:08,000 --> 00:25:14,000
two over three.

302
00:25:14,000 --> 00:25:18,000
And we have to figure out which
one of these to use.

303
00:25:18,000 --> 00:25:22,000
Remember that C2 is the
coefficient on px.

304
00:25:22,000 --> 00:25:26,000
Let's reference our diagram on
our coordinate system.

305
00:25:26,000 --> 00:25:30,000
We have our x-axis and our
y-axis.

306
00:25:30,000 --> 00:25:37,000
And we are looking for the sign
of the coefficient on the boron

307
00:25:37,000 --> 00:25:42,000
2px orbital.
That is our C2 coefficient.

308
00:25:42,000 --> 00:25:49,000
And I am shading negative here.
And, in order to overlap

309
00:25:49,000 --> 00:25:56,000
constructively and form a bond
to the 1s orbital of HA,

310
00:25:56,000 --> 00:26:02,000
we want positive as our
coefficient.

311
00:26:02,000 --> 00:26:08,000
Because that is going to make
the same phase lobes interact in

312
00:26:08,000 --> 00:26:13,000
a sigma fashion.
We know that sp squared of A is

313
00:26:13,000 --> 00:26:19,000
equal to one over root three
times the boron's 2s orbital

314
00:26:19,000 --> 00:26:25,000
plus the square root of two over
three 2px orbital.

315
00:26:31,000 --> 00:26:36,000
And now that we have the
coefficients on that hybrid

316
00:26:36,000 --> 00:26:41,000
orbital, we can draw a picture
of what that is.

317
00:26:41,000 --> 00:26:48,000
That is equal to taking the
boron 1s orbital here and adding

318
00:26:48,000 --> 00:26:53,000
to it the 2px orbital that looks
like this.

319
00:26:53,000 --> 00:27:00,000
This is the conceptual
breakthrough that Pauling made.

320
00:27:00,000 --> 00:27:03,000
You could mix atomic orbital
wave functions located on a

321
00:27:03,000 --> 00:27:07,000
central atom in order to
generate these hybrids that have

322
00:27:07,000 --> 00:27:12,000
the proper directionality for
good overlap with the peripheral

323
00:27:12,000 --> 00:27:14,000
atoms that bond to the central
atom.

324
00:27:14,000 --> 00:27:18,000
And when I do this,
what you can see is that we are

325
00:27:18,000 --> 00:27:21,000
going to get constructive
interference with the positive

326
00:27:21,000 --> 00:27:25,000
lobe and destructive
interference with the negative

327
00:27:25,000 --> 00:27:29,000
lobe when we add this boron 2s
orbital to the boron 2px

328
00:27:29,000 --> 00:27:33,000
orbital.
And that should give us

329
00:27:33,000 --> 00:27:38,000
something that looks like this,
with an enlarged lobe pointing

330
00:27:38,000 --> 00:27:42,000
along positive x and a
diminished negative node

331
00:27:42,000 --> 00:27:46,000
pointing along negative x.
That is our hybrid.

332
00:27:46,000 --> 00:27:49,000
We have developed some
coefficients here.

333
00:27:49,000 --> 00:27:54,000
We just got it pictorially,
and here we show you the

334
00:27:54,000 --> 00:28:00,000
picture of the hybrid.
And then, we need to continue.

335
00:28:09,000 --> 00:28:14,000
We need to use the remainder of
our 2px orbital.

336
00:28:14,000 --> 00:28:20,000
The one thing we cannot do in
constructing a set of hybrid

337
00:28:20,000 --> 00:28:26,000
orbitals from a set of atomic
orbitals is forgetting to use

338
00:28:26,000 --> 00:28:31,000
part of our orbital.
We need to use all of the

339
00:28:31,000 --> 00:28:38,000
orbital that we are given to
construct these hybrids with.

340
00:28:38,000 --> 00:28:44,000
And what that means is we have
now a relationship between C5

341
00:28:44,000 --> 00:28:51,000
and C8, because we now know C2
is plus root two over three.

342
00:28:51,000 --> 00:28:57,000
Let's use
this unit orbital idea and

343
00:28:57,000 --> 00:29:03,000
reference 2px --
-- and say that C2 squared plus

344
00:29:03,000 --> 00:29:10,000
C5 squared plus C8 squared,
these are all the coefficients

345
00:29:10,000 --> 00:29:16,000
that deal with the px
contributions to these three

346
00:29:16,000 --> 00:29:22,000
hybrids, is equal to one. And

347
00:29:22,000 --> 00:29:27,000
we know C2 squared is equal to
two-thirds.

348
00:29:27,000 --> 00:29:33,000
And now let's draw a picture to

349
00:29:33,000 --> 00:29:39,000
help us figure out something
else about C5 and C8.

350
00:29:39,000 --> 00:29:46,000
Here I am simply representing,
with the same coordinate system

351
00:29:46,000 --> 00:29:52,000
that I am using throughout,
the boron atomic orbital that

352
00:29:52,000 --> 00:29:58,000
is the 2px orbital.
And, if I consider,

353
00:29:58,000 --> 00:30:04,000
down here, the s orbitals on HA
and HB, I can say something

354
00:30:04,000 --> 00:30:10,000
about the way in which 2px
interacts with HA as compared

355
00:30:10,000 --> 00:30:15,000
with the way in which it
interacts with HB.

356
00:30:15,000 --> 00:30:21,000
Notice that these HA and HB
atomic positions are mirror

357
00:30:21,000 --> 00:30:27,000
images of each other with
respect to this 2px orbital in

358
00:30:27,000 --> 00:30:33,000
the middle.
That tells us that C5 must be

359
00:30:33,000 --> 00:30:37,000
equal to C8. We are using

360
00:30:37,000 --> 00:30:43,000
symmetry as a mathematical
argument here to say that C5 is

361
00:30:43,000 --> 00:30:48,000
equal to C8.
Therefore, we can figure out

362
00:30:48,000 --> 00:30:54,000
exactly what C5 and C8 are
because we already know that C2

363
00:30:54,000 --> 00:31:00,000
squared is two-thirds.

364
00:31:00,000 --> 00:31:08,000
We know that C5 squared plus C8
squared has got to be equal to

365
00:31:08,000 --> 00:31:13,000
one-third. And that

366
00:31:13,000 --> 00:31:21,000
means that C5 is equal to C6,
sorry, C8, which is equal to

367
00:31:21,000 --> 00:31:30,000
one over the square root of six.

368
00:31:30,000 --> 00:31:34,000
Because we put this in here,
we will get a sixth plus a

369
00:31:34,000 --> 00:31:37,000
sixth, two-sixths,
which is one-third,

370
00:31:37,000 --> 00:31:39,000
plus the two-thirds,
which will be one.

371
00:31:39,000 --> 00:31:42,000
And so, we know it is one over
root six.

372
00:31:42,000 --> 00:31:46,000
However, it could be a positive
or a negative.

373
00:31:46,000 --> 00:31:49,000
And we need to figure out just
which it is.

374
00:31:49,000 --> 00:31:52,000
Is it positive or is it
negative here?

375
00:31:52,000 --> 00:31:56,000
And I will come back to that
issue just in a moment,

376
00:31:56,000 --> 00:32:02,000
when we draw these functions.
But notice that these are the

377
00:32:02,000 --> 00:32:06,000
coefficients on 2px and the
positive lobe points up toward

378
00:32:06,000 --> 00:32:08,000
HA.
We are going to need it to be

379
00:32:08,000 --> 00:32:13,000
flipped around and be pointing
down if we are going to form

380
00:32:13,000 --> 00:32:16,000
bonds to HA and HB.
So, that will dictate the sign

381
00:32:16,000 --> 00:32:19,000
here as negative one over root
six.

382
00:32:19,000 --> 00:32:22,000
We do not have,
yet, the complete equations for

383
00:32:22,000 --> 00:32:27,000
sp two of B and sp
two of C because we

384
00:32:27,000 --> 00:32:34,000
need C6 and C9.
Let's see how we can get those.

385
00:33:01,000 --> 00:33:08,000
C6 and C9 relate to the py
orbital that is oriented like

386
00:33:08,000 --> 00:33:14,000
this with reference to our
coordinate system,

387
00:33:14,000 --> 00:33:19,000
with the negative phase shaded
as follows.

388
00:33:19,000 --> 00:33:25,000
And specifically,
C6 relates to the interaction

389
00:33:25,000 --> 00:33:31,000
with this HB down here.
And C9 relates to the

390
00:33:31,000 --> 00:33:35,000
interaction with the 1s orbital
on HC down here,

391
00:33:35,000 --> 00:33:39,000
on the lower right.
And because this py orbital

392
00:33:39,000 --> 00:33:44,000
charges phase as we pass through
the origin, and we want the

393
00:33:44,000 --> 00:33:48,000
positive phase to match up both
with HB and with HC in the

394
00:33:48,000 --> 00:33:53,000
contributions to the hybrid wave
functions, we are going to know,

395
00:33:53,000 --> 00:33:58,000
and by the symmetry of this
problem, that these are negative

396
00:33:58,000 --> 00:34:06,000
mirror images of each other.
So, we know that C6 is equal to

397
00:34:06,000 --> 00:34:12,000
negative C9. And if we put that

398
00:34:12,000 --> 00:34:18,000
together with unit orbital
contribution,

399
00:34:18,000 --> 00:34:26,000
we can see, because we know
that we have our sp two of B

400
00:34:26,000 --> 00:34:31,000
and sp two of C,

401
00:34:31,000 --> 00:34:42,000
both of them have one over root
three of the 2s contributing.

402
00:34:42,000 --> 00:34:46,000
And then we are looking for six
and nine.

403
00:34:46,000 --> 00:34:53,000
We found out over there that
both of them have negative one

404
00:34:53,000 --> 00:34:59,000
over root six as the coefficient
on px.

405
00:34:59,000 --> 00:35:03,000
And then, we know that since
this is the one that points to

406
00:35:03,000 --> 00:35:06,000
B, we do not need to flip the
phase on py to get a bond

407
00:35:06,000 --> 00:35:10,000
forming between py and HB.
This is going to be a positive,

408
00:35:10,000 --> 00:35:13,000
and then this one is going to
be a negative,

409
00:35:13,000 --> 00:35:17,000
for the reasons I just
mentioned, that we are going to

410
00:35:17,000 --> 00:35:21,000
need to flip that py
contribution to sp squared of C,

411
00:35:21,000 --> 00:35:24,000
as compared to sp squared of B,

412
00:35:24,000 --> 00:35:27,000
in order to get this positive

413
00:35:27,000 --> 00:35:32,000
lobe over here interacting with
HC in that function.

414
00:35:32,000 --> 00:35:36,000
I am going to draw these
functions in just a moment.

415
00:35:36,000 --> 00:35:41,000
But now, when we note that
normalization requires that the

416
00:35:41,000 --> 00:35:45,000
sum of the squares of the
coefficients be equal to one,

417
00:35:45,000 --> 00:35:50,000
and we know that C3 was zero,
so that there is no

418
00:35:50,000 --> 00:35:55,000
contribution at all of py to sp
squared of A,

419
00:35:55,000 --> 00:35:59,000
and all of it must be here and
here, that means that our

420
00:35:59,000 --> 00:36:06,000
coefficient is one over root two
of py.

421
00:36:06,000 --> 00:36:10,000
Using these restrictions from
quantum mechanics,

422
00:36:10,000 --> 00:36:15,000
we actually have coefficients
on the spx and py orbitals that

423
00:36:15,000 --> 00:36:20,000
show how they go together.
And the way in which these add

424
00:36:20,000 --> 00:36:24,000
up is similar to what we have
seen over here.

425
00:36:24,000 --> 00:36:29,000
It is a little more complicated
because both px and py

426
00:36:29,000 --> 00:36:34,000
contribute.
Let me show you what they look

427
00:36:34,000 --> 00:36:37,000
like.
This mathematical manipulation

428
00:36:37,000 --> 00:36:43,000
of these functions produces
linear combinations that all

429
00:36:43,000 --> 00:36:49,000
three look the same as the
symmetry of this problem would

430
00:36:49,000 --> 00:36:53,000
dictate.
It is a three-fold symmetry

431
00:36:53,000 --> 00:36:58,000
type of problem.
And this is our sp two

432
00:36:58,000 --> 00:37:05,000
orbital pointing toward B with
this mathematical form.

433
00:37:05,000 --> 00:37:09,000
We have to flip around px so
that the positive lobe of px

434
00:37:09,000 --> 00:37:13,000
contributes down here.
We do not have to flip py.

435
00:37:13,000 --> 00:37:18,000
And we have differing amounts
of px and py in this hybrid wave

436
00:37:18,000 --> 00:37:21,000
function.
But the math of the atomic

437
00:37:21,000 --> 00:37:25,000
orbitals guarantees that this
form would be physically

438
00:37:25,000 --> 00:37:30,000
indistinguishable from this one
over here.

439
00:37:30,000 --> 00:37:34,000
They would look exactly
identical if I plotted a

440
00:37:34,000 --> 00:37:39,000
probability density isosurface.
These two-dimensional pictures

441
00:37:39,000 --> 00:37:43,000
that we draw on the board are
like slices through

442
00:37:43,000 --> 00:37:47,000
three-dimensional probability
density isosurfaces.

443
00:37:47,000 --> 00:37:50,000
That is what these
representations are.

444
00:37:50,000 --> 00:37:55,000
And then, we note that the py
orbital is flipped around when

445
00:37:55,000 --> 00:38:00,000
we form sp squared of C.

446
00:38:00,000 --> 00:38:04,000
And that leads to this
appearance.

447
00:38:10,000 --> 00:38:15,000
I would like to show you what
one of these might look like if

448
00:38:15,000 --> 00:38:21,000
you calculated it from the math
that we have developed,

449
00:38:21,000 --> 00:38:22,000
here.

450
00:38:50,000 --> 00:38:54,000
I have actually got more work
to do here in the next five

451
00:38:54,000 --> 00:38:57,000
minutes, so let's keep our
concentration.

452
00:38:57,000 --> 00:39:01,000
Look at this orbital.
This thing is a beautiful

453
00:39:01,000 --> 00:39:04,000
hybrid orbital.
And I have calculated this with

454
00:39:04,000 --> 00:39:06,000
reference to a particular
arbitrary molecule.

455
00:39:06,000 --> 00:39:10,000
This was calculated with
reference to the water molecule.

456
00:39:10,000 --> 00:39:13,000
The positions of the nuclei in
the water molecular are

457
00:39:13,000 --> 00:39:16,000
indicated by these arbitrarily
sized spheres.

458
00:39:16,000 --> 00:39:19,000
That is a typical ball and
stick representation of a

459
00:39:19,000 --> 00:39:21,000
molecule in 3D.
You can see that the oxygen of

460
00:39:21,000 --> 00:39:24,000
the water molecule is colored
red, in there.

461
00:39:24,000 --> 00:39:27,000
That is arbitrary.
And then we have the hydrogens

462
00:39:27,000 --> 00:39:30,000
in white.
And then I have a big positive

463
00:39:30,000 --> 00:39:34,000
lobe in blue here and then a
smaller yellow lobe that is our

464
00:39:34,000 --> 00:39:38,000
negative lobe on this orbital,
here, which is an added mixture

465
00:39:38,000 --> 00:39:40,000
of an s orbital with a p
orbital.

466
00:39:40,000 --> 00:39:44,000
I am not telling you the actual
coefficients that went into this

467
00:39:44,000 --> 00:39:47,000
particular hybrid,
but you can see how if I were

468
00:39:47,000 --> 00:39:51,000
to take a two-dimensional slice
along the long axis of this

469
00:39:51,000 --> 00:39:54,000
hybrid orbital,
it might look something like

470
00:39:54,000 --> 00:39:59,000
what we have drawn over there.
Namely, with one nice large

471
00:39:59,000 --> 00:40:03,000
directed lobe,
that can have good overlap with

472
00:40:03,000 --> 00:40:08,000
a hydrogen 1s orbital in order
to give rise to a good two

473
00:40:08,000 --> 00:40:12,000
electron bond between hydrogen
and boron.

474
00:40:12,000 --> 00:40:17,000
And let me illustrate how that
impacts on how we draw our

475
00:40:17,000 --> 00:40:22,000
energy-level diagram for the BH
three molecule.

476
00:40:22,000 --> 00:40:27,000
What we have now is a situation
in which, as you go up in

477
00:40:27,000 --> 00:40:35,000
energy, you find three orbitals
that are all at the same energy.

478
00:40:35,000 --> 00:40:40,000
These lines indicate places
where we can put electrons.

479
00:40:40,000 --> 00:40:44,000
And, in fact,
they look like this.

480
00:40:52,000 --> 00:41:00,000
They represent positive
interference up along x.

481
00:41:00,000 --> 00:41:04,000
Positive interference between a
sp two hybrid that

482
00:41:04,000 --> 00:41:08,000
points down at atom B,
right here, overlapping nicely

483
00:41:08,000 --> 00:41:12,000
with the hydrogen 1s wave
function that is sitting out

484
00:41:12,000 --> 00:41:17,000
there, surrounding that nucleus.
And then over here is our

485
00:41:17,000 --> 00:41:21,000
hybrid orbital sp squared of C
that points down

486
00:41:21,000 --> 00:41:27,000
here and interacts with the
hydrogen 1s orbital on atom C.

487
00:41:27,000 --> 00:41:31,000
And so, the idea is now this
system, BH three,

488
00:41:31,000 --> 00:41:34,000
has a total of six valance
electrons.

489
00:41:34,000 --> 00:41:39,000
Valance shell electron pair
repulsion theory could be used

490
00:41:39,000 --> 00:41:44,000
by you to predict the geometry
that we started with for this

491
00:41:44,000 --> 00:41:48,000
problem.
But now, we can populate these

492
00:41:48,000 --> 00:41:52,000
bonding orbitals with electron
pairs, as shown here.

493
00:41:52,000 --> 00:41:56,000
And then, this problem has one
more aspect to it,

494
00:41:56,000 --> 00:42:02,000
namely that up here somewhere
higher in energy there is a 2pz

495
00:42:02,000 --> 00:42:05,000
orbital on the boron that is
empty.

496
00:42:05,000 --> 00:42:09,000
And this empty orbital --

497
00:42:17,000 --> 00:42:21,000
-- is our feature that gives
this molecule Lewis acid

498
00:42:21,000 --> 00:42:24,000
character.
We know that this is an

499
00:42:24,000 --> 00:42:29,000
electron deficient system.
It does not satisfy the octet

500
00:42:29,000 --> 00:42:31,000
rule.
And, specifically,

501
00:42:31,000 --> 00:42:35,000
where it is missing electrons
are in that boron 2pz orbital.

502
00:42:35,000 --> 00:42:38,000
That allows you to predict
something about the

503
00:42:38,000 --> 00:42:40,000
stereochemistry of reactions,
for example.

504
00:42:40,000 --> 00:42:43,000
It says that if some
nucleophile or Lewis base,

505
00:42:43,000 --> 00:42:48,000
a source of a pair of electrons
is going to come in and approach

506
00:42:48,000 --> 00:42:51,000
that planar BH three
molecule, it will do so either

507
00:42:51,000 --> 00:42:55,000
along positive z or along
negative z because that is where

508
00:42:55,000 --> 00:43:00,000
the big lobes are of that empty
pz orbital point.

509
00:43:00,000 --> 00:43:03,000
And, this valance bond theory
--

510
00:43:11,000 --> 00:43:15,000
-- satisfies this problem,
that Pauling noticed,

511
00:43:15,000 --> 00:43:20,000
of how do we make three
identical bonds in a plane that

512
00:43:20,000 --> 00:43:25,000
contains three dissimilar atomic
orbital wave functions,

513
00:43:25,000 --> 00:43:29,000
an s orbital px and a py?
Well, he found that

514
00:43:29,000 --> 00:43:33,000
mathematically you can mix them
together in ways that gives you

515
00:43:33,000 --> 00:43:37,000
three identical hybrids that
have the needed directionality

516
00:43:37,000 --> 00:43:41,000
for the forming of these three
two electron bonds.

517
00:43:41,000 --> 00:43:44,000
There is an alternative
solution to this problem,

518
00:43:44,000 --> 00:43:48,000
and we will call that molecular
orbital theory.

519
00:43:48,000 --> 00:43:52,000
Don't forget about Robert S.
Mulliken, MIT undergraduate

520
00:43:52,000 --> 00:43:54,000
thesis 1917.
And MO theory is a different

521
00:43:54,000 --> 00:44:00,000
perspective on electronic
structure theory for molecules.

522
00:44:00,000 --> 00:44:03,000
And we will launch into that on
Wednesday.