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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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00:06:01 --> 00:06:05
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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00:06:25 --> 00:06:29
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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00:06:36 --> 00:06:40
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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00:07:40 --> 00:07:50


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They lie in the 2pz nodal
plane.

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00:07:55 --> 00:08:00


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00:08:00 --> 00:08:05
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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00:10:11 --> 00:10:18


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00:10:18 --> 00:10:22
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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00:10:31 --> 00:10:34
There is the 2s,
the 2px, and the 2py.

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00:10:34 --> 00:10:38
Three orbitals that are not
identical to one another,

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00:10:38 --> 00:10:43
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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00:10:53 --> 00:10:57
the perfect symmetry of a
Mercedes Benz symbol,

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00:10:57 --> 00:11:02
it's trigonal planar.
It has this three-fold

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00:11:02 --> 00:11:06
rotational symmetry to it.
And so somehow,

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00:11:06 --> 00:11:10
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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00:14:00 --> 00:14:10


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

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00:14:46 --> 00:14:52
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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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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three orbitals that we have
available.

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00:15:22 --> 00:15:26
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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00:15:38 --> 00:15:46
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,

212
00:15:52 --> 00:15:57
C2 and C3.
Our job is going to be to find

213
00:15:57 --> 00:16:01
the values of these
coefficients,

214
00:16:01 --> 00:16:06
C4, C5, C6, C7,
C8, and C9.

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

216
00:16:12 --> 00:16:17
summarize a few of these rules
for you before this lecture

217
00:16:17 --> 00:16:22
ends, to find the values of
these constants.

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

219
00:16:26 --> 00:16:50


220
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Let's make note of the dictates
of symmetry.

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

222
00:16:58 --> 00:17:03
in working on this problem of
finding those coefficients,

223
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one through nine.
The neat thing is that the 2s

224
00:17:07 --> 00:17:11
orbital centered on that boron,
which is going to be

225
00:17:11 --> 00:17:16
participating in these hybrids
that are going to point toward

226
00:17:16 --> 00:17:19
hydrogens A, B,
and C, that 2s orbital is

227
00:17:19 --> 00:17:23
spherically symmetric.
Its wave function is the same

228
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as you go radially out from the
boron center in any direction

229
00:17:28 --> 00:17:31
into space.
What that symmetry

230
00:17:31 --> 00:17:38
consideration dictates is that
there will be the same amount of

231
00:17:38 --> 00:17:43
boron 2s in the hybrids for A,
as for B, as for C.

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

233
00:17:50 --> 00:17:58


234
00:17:58 --> 00:18:01
And we are going to need a
value for this.

235
00:18:01 --> 00:18:06
And we can get a value for that
by taking into consideration

236
00:18:06 --> 00:18:11
something called unit orbital
contribution.

237
00:18:11 --> 00:18:19


238
00:18:19 --> 00:18:26
What that says is that we have
to use the boron's 2s orbital

239
00:18:26 --> 00:18:35
completely and identically once
in that set of three equations.

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

241
00:18:38 --> 00:18:40
evenly among those three
hybrids.

242
00:18:40 --> 00:18:44
And we have to distribute all
of it evenly among those three

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

244
00:18:47 --> 00:18:50
interested in the square of the
coefficients.

245
00:18:50 --> 00:18:54
And that is because when we
talk about atomic orbitals,

246
00:18:54 --> 00:18:58
we know that the probability of
finding an electron in that

247
00:18:58 --> 00:19:04
orbital is related to the square
of the wave function.

248
00:19:04 --> 00:19:07
And so, normally when we plot
orbitals, like I just did with

249
00:19:07 --> 00:19:11
the pi bond of ethylene a few
minutes ago, we are making an

250
00:19:11 --> 00:19:14
isosurface that corresponds to
some percent probability of

251
00:19:14 --> 00:19:17
finding the electron at that
point in space.

252
00:19:17 --> 00:19:20
So, we are interested in
probability distributions.

253
00:19:20 --> 00:19:24
And these are related to the
square of the wave function.

254
00:19:24 --> 00:19:27
Here, for unit orbital,
we are going to say that C1

255
00:19:27 --> 00:19:32
squared equals C4 squared.
Sorry, plus C4 squared plus C7

256
00:19:32 --> 00:19:38
squared is equal to one. And,

257
00:19:38 --> 00:19:44
therefore, C1 is equal to C4 is
equal to C7 is equal to one over

258
00:19:44 --> 00:19:48
root three.

259
00:19:48 --> 00:19:51
Now, we have found C1,
C4, and C7.

260
00:19:51 --> 00:19:56
And, if we are going to
complete sp two of A,

261
00:19:56 --> 00:20:01
get this first hybrid of the
three, I am going to go across

262
00:20:01 --> 00:20:05
this way.
I just went down one,

263
00:20:05 --> 00:20:09
four, and seven and pointed out
that by the symmetry of the s

264
00:20:09 --> 00:20:13
orbital is due to the fact that
the s orbital contributes once,

265
00:20:13 --> 00:20:16
unit orbital contribution,
we could get C1,

266
00:20:16 --> 00:20:19
C4, and C7 equal to one over
root three.

267
00:20:19 --> 00:20:23
Now I am interested in C2 and

268
00:20:23 --> 00:20:27
C3, so let's draw a little
picture.

269
00:20:27 --> 00:20:35


270
00:20:35 --> 00:20:38
This, remember,
is our 1s orbital on HA.

271
00:20:38 --> 00:20:44
And what we are trying to do is
to construct a hybrid focusing

272
00:20:44 --> 00:20:49
on that top equation that will
be directed at HA.

273
00:20:49 --> 00:20:53
And we are keeping our axes the
same as before,

274
00:20:53 --> 00:20:59
so that x points up and y goes
to the left, as I am drawing it

275
00:20:59 --> 00:21:04
here.
And z would be coming out at

276
00:21:04 --> 00:21:08
us.
What I am going to draw now is

277
00:21:08 --> 00:21:13
a picture of the boron's 2py
orbital.

278
00:21:13 --> 00:21:20
And, as I usually do,
I am taking the convention that

279
00:21:20 --> 00:21:28
the negative phase lobe of the
boron's 2py orbital--

280
00:21:28 --> 00:21:32
Negative is shaded.
That is my convention.

281
00:21:32 --> 00:21:38
That 2py orbital lies in the xy
plane and points along y.

282
00:21:38 --> 00:21:43
Its positive lobe,
in fact, points along positive

283
00:21:43 --> 00:21:46
y.
Its negative lobe points along

284
00:21:46 --> 00:21:50
negative y.
And what can we say about the

285
00:21:50 --> 00:21:56
relationship of the HA 1s
orbital to the 2py orbital of

286
00:21:56 --> 00:22:00
that central boron atom?

287
00:22:00 --> 00:22:05


288
00:22:05 --> 00:22:11
Let me phrase it this way.
The 2py orbital of our central

289
00:22:11 --> 00:22:16
boron atom has a nodal plane.
What is that nodal plane?

290
00:22:16 --> 00:22:22
xz is a nodal plane of the 2py
orbital, right here.

291
00:22:22 --> 00:22:27
Just like this.
And so, what you see is that

292
00:22:27 --> 00:22:31
this HA is not only in the nodal
plane of 2pz,

293
00:22:31 --> 00:22:37
it is also in the nodal plane
of 2py.

294
00:22:37 --> 00:22:42
And that means any positive
interference by bonding between

295
00:22:42 --> 00:22:47
HA's 1s and the positive lobe of
2py would be exactly canceled by

296
00:22:47 --> 00:22:52
the equal and opposite negative
interference between a

297
00:22:52 --> 00:22:57
hydrogen's 1s orbital and the
negative phase lobe on 2py.

298
00:22:57 --> 00:23:02
And that means that we know the
value of C3.

299
00:23:02 --> 00:23:08
We can say, therefore,
that C3 is equal to zero.

300
00:23:08 --> 00:23:17
That is really cool because now
we are going to be able to bring

301
00:23:17 --> 00:23:26
in one more consideration and
complete the formula for sp two

302
00:23:26 --> 00:23:29
of A.

303
00:23:29 --> 00:23:41


304
00:23:41 --> 00:23:43
We are going to be able to use
the normalization condition,

305
00:23:43 --> 00:23:44
--

306
00:23:44 --> 00:23:54


307
00:23:54 --> 00:24:01
-- which tells us that C1
squared plus C2 squared plus C3

308
00:24:01 --> 00:24:08
squared is equal to one.

309
00:24:08 --> 00:24:11
Once we formed this new hybrid
by mixing together s and p wave

310
00:24:11 --> 00:24:15
functions, we are going to find
that this normalization

311
00:24:15 --> 00:24:19
condition must be satisfied for
the hybrid orbital just as it is

312
00:24:19 --> 00:24:21
satisfied for these atomic
orbitals.

313
00:24:21 --> 00:24:25
That is so that the probability
of finding an electron somewhere

314
00:24:25 --> 00:24:29
is space associated with this
hybrid wave function is one,

315
00:24:29 --> 00:24:33
if there is an electron in this
orbital.

316
00:24:33 --> 00:24:39
That is where this comes from.
And we know that C3 squared is

317
00:24:39 --> 00:24:42
zero. C1 squared is one

318
00:24:42 --> 00:24:46
over root three. So,

319
00:24:46 --> 00:24:52
we know that sp squared of A is
equal to one over root three.

320
00:24:52 --> 00:24:57
And then we can say that,

321
00:24:57 --> 00:25:02
by virtue of this and the fact
that C3 is zero,

322
00:25:02 --> 00:25:08
C2 could be equal to either
plus or minus the square root of

323
00:25:08 --> 00:25:14
two over three.

324
00:25:14 --> 00:25:18
And we have to figure out which
one of these to use.

325
00:25:18 --> 00:25:22
Remember that C2 is the
coefficient on px.

326
00:25:22 --> 00:25:26
Let's reference our diagram on
our coordinate system.

327
00:25:26 --> 00:25:30
We have our x-axis and our
y-axis.

328
00:25:30 --> 00:25:37
And we are looking for the sign
of the coefficient on the boron

329
00:25:37 --> 00:25:42
2px orbital.
That is our C2 coefficient.

330
00:25:42 --> 00:25:49
And I am shading negative here.
And, in order to overlap

331
00:25:49 --> 00:25:56
constructively and form a bond
to the 1s orbital of HA,

332
00:25:56 --> 00:26:02
we want positive as our
coefficient.

333
00:26:02 --> 00:26:08
Because that is going to make
the same phase lobes interact in

334
00:26:08 --> 00:26:13
a sigma fashion.
We know that sp squared of A is

335
00:26:13 --> 00:26:19
equal to one over root three
times the boron's 2s orbital

336
00:26:19 --> 00:26:25
plus the square root of two over
three 2px orbital.

337
00:26:25 --> 00:26:31


338
00:26:31 --> 00:26:36
And now that we have the
coefficients on that hybrid

339
00:26:36 --> 00:26:41
orbital, we can draw a picture
of what that is.

340
00:26:41 --> 00:26:48
That is equal to taking the
boron 1s orbital here and adding

341
00:26:48 --> 00:26:53
to it the 2px orbital that looks
like this.

342
00:26:53 --> 00:27:00
This is the conceptual
breakthrough that Pauling made.

343
00:27:00 --> 00:27:03
You could mix atomic orbital
wave functions located on a

344
00:27:03 --> 00:27:07
central atom in order to
generate these hybrids that have

345
00:27:07 --> 00:27:12
the proper directionality for
good overlap with the peripheral

346
00:27:12 --> 00:27:14
atoms that bond to the central
atom.

347
00:27:14 --> 00:27:18
And when I do this,
what you can see is that we are

348
00:27:18 --> 00:27:21
going to get constructive
interference with the positive

349
00:27:21 --> 00:27:25
lobe and destructive
interference with the negative

350
00:27:25 --> 00:27:29
lobe when we add this boron 2s
orbital to the boron 2px

351
00:27:29 --> 00:27:33
orbital.
And that should give us

352
00:27:33 --> 00:27:38
something that looks like this,
with an enlarged lobe pointing

353
00:27:38 --> 00:27:42
along positive x and a
diminished negative node

354
00:27:42 --> 00:27:46
pointing along negative x.
That is our hybrid.

355
00:27:46 --> 00:27:49
We have developed some
coefficients here.

356
00:27:49 --> 00:27:54
We just got it pictorially,
and here we show you the

357
00:27:54 --> 00:28:00
picture of the hybrid.
And then, we need to continue.

358
00:28:00 --> 00:28:09


359
00:28:09 --> 00:28:14
We need to use the remainder of
our 2px orbital.

360
00:28:14 --> 00:28:20
The one thing we cannot do in
constructing a set of hybrid

361
00:28:20 --> 00:28:26
orbitals from a set of atomic
orbitals is forgetting to use

362
00:28:26 --> 00:28:31
part of our orbital.
We need to use all of the

363
00:28:31 --> 00:28:38
orbital that we are given to
construct these hybrids with.

364
00:28:38 --> 00:28:44
And what that means is we have
now a relationship between C5

365
00:28:44 --> 00:28:51
and C8, because we now know C2
is plus root two over three.

366
00:28:51 --> 00:28:57
Let's use
this unit orbital idea and

367
00:28:57 --> 00:29:03
reference 2px --
-- and say that C2 squared plus

368
00:29:03 --> 00:29:10
C5 squared plus C8 squared,
these are all the coefficients

369
00:29:10 --> 00:29:16
that deal with the px
contributions to these three

370
00:29:16 --> 00:29:22
hybrids, is equal to one. And

371
00:29:22 --> 00:29:27
we know C2 squared is equal to
two-thirds.

372
00:29:27 --> 00:29:33
And now let's draw a picture to

373
00:29:33 --> 00:29:39
help us figure out something
else about C5 and C8.

374
00:29:39 --> 00:29:46
Here I am simply representing,
with the same coordinate system

375
00:29:46 --> 00:29:52
that I am using throughout,
the boron atomic orbital that

376
00:29:52 --> 00:29:58
is the 2px orbital.
And, if I consider,

377
00:29:58 --> 00:30:04
down here, the s orbitals on HA
and HB, I can say something

378
00:30:04 --> 00:30:10
about the way in which 2px
interacts with HA as compared

379
00:30:10 --> 00:30:15
with the way in which it
interacts with HB.

380
00:30:15 --> 00:30:21
Notice that these HA and HB
atomic positions are mirror

381
00:30:21 --> 00:30:27
images of each other with
respect to this 2px orbital in

382
00:30:27 --> 00:30:33
the middle.
That tells us that C5 must be

383
00:30:33 --> 00:30:37
equal to C8. We are using

384
00:30:37 --> 00:30:43
symmetry as a mathematical
argument here to say that C5 is

385
00:30:43 --> 00:30:48
equal to C8.
Therefore, we can figure out

386
00:30:48 --> 00:30:54
exactly what C5 and C8 are
because we already know that C2

387
00:30:54 --> 00:31:00
squared is two-thirds.

388
00:31:00 --> 00:31:08
We know that C5 squared plus C8
squared has got to be equal to

389
00:31:08 --> 00:31:13
one-third. And that

390
00:31:13 --> 00:31:21
means that C5 is equal to C6,
sorry, C8, which is equal to

391
00:31:21 --> 00:31:30
one over the square root of six.

392
00:31:30 --> 00:31:34
Because we put this in here,
we will get a sixth plus a

393
00:31:34 --> 00:31:37
sixth, two-sixths,
which is one-third,

394
00:31:37 --> 00:31:39
plus the two-thirds,
which will be one.

395
00:31:39 --> 00:31:42
And so, we know it is one over
root six.

396
00:31:42 --> 00:31:46
However, it could be a positive
or a negative.

397
00:31:46 --> 00:31:49
And we need to figure out just
which it is.

398
00:31:49 --> 00:31:52
Is it positive or is it
negative here?

399
00:31:52 --> 00:31:56
And I will come back to that
issue just in a moment,

400
00:31:56 --> 00:32:02
when we draw these functions.
But notice that these are the

401
00:32:02 --> 00:32:06
coefficients on 2px and the
positive lobe points up toward

402
00:32:06 --> 00:32:08
HA.
We are going to need it to be

403
00:32:08 --> 00:32:13
flipped around and be pointing
down if we are going to form

404
00:32:13 --> 00:32:16
bonds to HA and HB.
So, that will dictate the sign

405
00:32:16 --> 00:32:19
here as negative one over root
six.

406
00:32:19 --> 00:32:22
We do not have,
yet, the complete equations for

407
00:32:22 --> 00:32:27
sp two of B and sp
two of C because we

408
00:32:27 --> 00:32:34
need C6 and C9.
Let's see how we can get those.

409
00:32:34 --> 00:33:01


410
00:33:01 --> 00:33:08
C6 and C9 relate to the py
orbital that is oriented like

411
00:33:08 --> 00:33:14
this with reference to our
coordinate system,

412
00:33:14 --> 00:33:19
with the negative phase shaded
as follows.

413
00:33:19 --> 00:33:25
And specifically,
C6 relates to the interaction

414
00:33:25 --> 00:33:31
with this HB down here.
And C9 relates to the

415
00:33:31 --> 00:33:35
interaction with the 1s orbital
on HC down here,

416
00:33:35 --> 00:33:39
on the lower right.
And because this py orbital

417
00:33:39 --> 00:33:44
charges phase as we pass through
the origin, and we want the

418
00:33:44 --> 00:33:48
positive phase to match up both
with HB and with HC in the

419
00:33:48 --> 00:33:53
contributions to the hybrid wave
functions, we are going to know,

420
00:33:53 --> 00:33:58
and by the symmetry of this
problem, that these are negative

421
00:33:58 --> 00:34:06
mirror images of each other.
So, we know that C6 is equal to

422
00:34:06 --> 00:34:12
negative C9. And if we put that

423
00:34:12 --> 00:34:18
together with unit orbital
contribution,

424
00:34:18 --> 00:34:26
we can see, because we know
that we have our sp two of B

425
00:34:26 --> 00:34:31
and sp two of C,

426
00:34:31 --> 00:34:42
both of them have one over root
three of the 2s contributing.

427
00:34:42 --> 00:34:46
And then we are looking for six
and nine.

428
00:34:46 --> 00:34:53
We found out over there that
both of them have negative one

429
00:34:53 --> 00:34:59
over root six as the coefficient
on px.

430
00:34:59 --> 00:35:03
And then, we know that since
this is the one that points to

431
00:35:03 --> 00:35:06
B, we do not need to flip the
phase on py to get a bond

432
00:35:06 --> 00:35:10
forming between py and HB.
This is going to be a positive,

433
00:35:10 --> 00:35:13
and then this one is going to
be a negative,

434
00:35:13 --> 00:35:17
for the reasons I just
mentioned, that we are going to

435
00:35:17 --> 00:35:21
need to flip that py
contribution to sp squared of C,

436
00:35:21 --> 00:35:24
as compared to sp squared of B,

437
00:35:24 --> 00:35:27
in order to get this positive

438
00:35:27 --> 00:35:32
lobe over here interacting with
HC in that function.

439
00:35:32 --> 00:35:36
I am going to draw these
functions in just a moment.

440
00:35:36 --> 00:35:41
But now, when we note that
normalization requires that the

441
00:35:41 --> 00:35:45
sum of the squares of the
coefficients be equal to one,

442
00:35:45 --> 00:35:50
and we know that C3 was zero,
so that there is no

443
00:35:50 --> 00:35:55
contribution at all of py to sp
squared of A,

444
00:35:55 --> 00:35:59
and all of it must be here and
here, that means that our

445
00:35:59 --> 00:36:06
coefficient is one over root two
of py.

446
00:36:06 --> 00:36:10
Using these restrictions from
quantum mechanics,

447
00:36:10 --> 00:36:15
we actually have coefficients
on the spx and py orbitals that

448
00:36:15 --> 00:36:20
show how they go together.
And the way in which these add

449
00:36:20 --> 00:36:24
up is similar to what we have
seen over here.

450
00:36:24 --> 00:36:29
It is a little more complicated
because both px and py

451
00:36:29 --> 00:36:34
contribute.
Let me show you what they look

452
00:36:34 --> 00:36:37
like.
This mathematical manipulation

453
00:36:37 --> 00:36:43
of these functions produces
linear combinations that all

454
00:36:43 --> 00:36:49
three look the same as the
symmetry of this problem would

455
00:36:49 --> 00:36:53
dictate.
It is a three-fold symmetry

456
00:36:53 --> 00:36:58
type of problem.
And this is our sp two

457
00:36:58 --> 00:37:05
orbital pointing toward B with
this mathematical form.

458
00:37:05 --> 00:37:09
We have to flip around px so
that the positive lobe of px

459
00:37:09 --> 00:37:13
contributes down here.
We do not have to flip py.

460
00:37:13 --> 00:37:18
And we have differing amounts
of px and py in this hybrid wave

461
00:37:18 --> 00:37:21
function.
But the math of the atomic

462
00:37:21 --> 00:37:25
orbitals guarantees that this
form would be physically

463
00:37:25 --> 00:37:30
indistinguishable from this one
over here.

464
00:37:30 --> 00:37:34
They would look exactly
identical if I plotted a

465
00:37:34 --> 00:37:39
probability density isosurface.
These two-dimensional pictures

466
00:37:39 --> 00:37:43
that we draw on the board are
like slices through

467
00:37:43 --> 00:37:47
three-dimensional probability
density isosurfaces.

468
00:37:47 --> 00:37:50
That is what these
representations are.

469
00:37:50 --> 00:37:55
And then, we note that the py
orbital is flipped around when

470
00:37:55 --> 00:38:00
we form sp squared of C.

471
00:38:00 --> 00:38:04
And that leads to this
appearance.

472
00:38:04 --> 00:38:10


473
00:38:10 --> 00:38:15
I would like to show you what
one of these might look like if

474
00:38:15 --> 00:38:21
you calculated it from the math
that we have developed,

475
00:38:21 --> 00:38:22
here.

476
00:38:22 --> 00:38:50


477
00:38:50 --> 00:38:54
I have actually got more work
to do here in the next five

478
00:38:54 --> 00:38:57
minutes, so let's keep our
concentration.

479
00:38:57 --> 00:39:01
Look at this orbital.
This thing is a beautiful

480
00:39:01 --> 00:39:04
hybrid orbital.
And I have calculated this with

481
00:39:04 --> 00:39:06
reference to a particular
arbitrary molecule.

482
00:39:06 --> 00:39:10
This was calculated with
reference to the water molecule.

483
00:39:10 --> 00:39:13
The positions of the nuclei in
the water molecular are

484
00:39:13 --> 00:39:16
indicated by these arbitrarily
sized spheres.

485
00:39:16 --> 00:39:19
That is a typical ball and
stick representation of a

486
00:39:19 --> 00:39:21
molecule in 3D.
You can see that the oxygen of

487
00:39:21 --> 00:39:24
the water molecule is colored
red, in there.

488
00:39:24 --> 00:39:27
That is arbitrary.
And then we have the hydrogens

489
00:39:27 --> 00:39:30
in white.
And then I have a big positive

490
00:39:30 --> 00:39:34
lobe in blue here and then a
smaller yellow lobe that is our

491
00:39:34 --> 00:39:38
negative lobe on this orbital,
here, which is an added mixture

492
00:39:38 --> 00:39:40
of an s orbital with a p
orbital.

493
00:39:40 --> 00:39:44
I am not telling you the actual
coefficients that went into this

494
00:39:44 --> 00:39:47
particular hybrid,
but you can see how if I were

495
00:39:47 --> 00:39:51
to take a two-dimensional slice
along the long axis of this

496
00:39:51 --> 00:39:54
hybrid orbital,
it might look something like

497
00:39:54 --> 00:39:59
what we have drawn over there.
Namely, with one nice large

498
00:39:59 --> 00:40:03
directed lobe,
that can have good overlap with

499
00:40:03 --> 00:40:08
a hydrogen 1s orbital in order
to give rise to a good two

500
00:40:08 --> 00:40:12
electron bond between hydrogen
and boron.

501
00:40:12 --> 00:40:17
And let me illustrate how that
impacts on how we draw our

502
00:40:17 --> 00:40:22
energy-level diagram for the BH
three molecule.

503
00:40:22 --> 00:40:27
What we have now is a situation
in which, as you go up in

504
00:40:27 --> 00:40:35
energy, you find three orbitals
that are all at the same energy.

505
00:40:35 --> 00:40:40
These lines indicate places
where we can put electrons.

506
00:40:40 --> 00:40:44
And, in fact,
they look like this.

507
00:40:44 --> 00:40:52


508
00:40:52 --> 00:41:00
They represent positive
interference up along x.

509
00:41:00 --> 00:41:04
Positive interference between a
sp two hybrid that

510
00:41:04 --> 00:41:08
points down at atom B,
right here, overlapping nicely

511
00:41:08 --> 00:41:12
with the hydrogen 1s wave
function that is sitting out

512
00:41:12 --> 00:41:17
there, surrounding that nucleus.
And then over here is our

513
00:41:17 --> 00:41:21
hybrid orbital sp squared of C
that points down

514
00:41:21 --> 00:41:27
here and interacts with the
hydrogen 1s orbital on atom C.

515
00:41:27 --> 00:41:31
And so, the idea is now this
system, BH three,

516
00:41:31 --> 00:41:34
has a total of six valance
electrons.

517
00:41:34 --> 00:41:39
Valance shell electron pair
repulsion theory could be used

518
00:41:39 --> 00:41:44
by you to predict the geometry
that we started with for this

519
00:41:44 --> 00:41:48
problem.
But now, we can populate these

520
00:41:48 --> 00:41:52
bonding orbitals with electron
pairs, as shown here.

521
00:41:52 --> 00:41:56
And then, this problem has one
more aspect to it,

522
00:41:56 --> 00:42:02
namely that up here somewhere
higher in energy there is a 2pz

523
00:42:02 --> 00:42:05
orbital on the boron that is
empty.

524
00:42:05 --> 00:42:09
And this empty orbital --

525
00:42:09 --> 00:42:17


526
00:42:17 --> 00:42:21
-- is our feature that gives
this molecule Lewis acid

527
00:42:21 --> 00:42:24
character.
We know that this is an

528
00:42:24 --> 00:42:29
electron deficient system.
It does not satisfy the octet

529
00:42:29 --> 00:42:31
rule.
And, specifically,

530
00:42:31 --> 00:42:35
where it is missing electrons
are in that boron 2pz orbital.

531
00:42:35 --> 00:42:38
That allows you to predict
something about the

532
00:42:38 --> 00:42:40
stereochemistry of reactions,
for example.

533
00:42:40 --> 00:42:43
It says that if some
nucleophile or Lewis base,

534
00:42:43 --> 00:42:48
a source of a pair of electrons
is going to come in and approach

535
00:42:48 --> 00:42:51
that planar BH three
molecule, it will do so either

536
00:42:51 --> 00:42:55
along positive z or along
negative z because that is where

537
00:42:55 --> 00:43:00
the big lobes are of that empty
pz orbital point.

538
00:43:00 --> 00:43:03
And, this valance bond theory
--

539
00:43:03 --> 00:43:11


540
00:43:11 --> 00:43:15
-- satisfies this problem,
that Pauling noticed,

541
00:43:15 --> 00:43:20
of how do we make three
identical bonds in a plane that

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

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

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

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

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

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

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

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

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

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

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