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Over on the side board here, I want to show you a few
representation of the ethylene molecule that we were talking
about at the end of last hour. When you see representations of
molecules that appear throughout your textbook,
as well as in this class, I want you to see if you can
recognize just what property of the molecule it is that is being
represented. In this case here,
we have arbitrarily sized spheres that represent the two
carbon atoms and the four peripheral hydrogen atoms.
And then, what I have plotted around it is a see-through
representation of the electron density at a particular
isosurface value of 0.1 electrons per unit volume.
So, that is a representation. And we are going to be seeing a
number of different kinds of representations,
and you will need to be able to distinguish between them.
And, in particular, we were talking last time about
different kinds of bonds. We talked about sigma bonds
that were cylindrically symmetric about the internuclear
axis of the two atoms that are bonded together.
And we also talked about pi bonds.
And I brought up the pi bond in connection with the ethylene
molecule that we were just looking at a different
representation of. And so now, let me show you
what a pi bond can look like. Here is a representation of the
ethylene molecule. Again, I have arbitrarily sized
spheres that show us just where the nuclear positions are in
three-dimensional space. So, we are talking about the
equilibrium geometry of the ethylene molecule.
And what you see here is that if the molecule is lying in the
x,y-plane, then each of these carbons has a 2pz orbital that
has a positive lobe and a negative lobe,
respectively, above and below the plane of
the molecule. And so, normally,
when we talk about atomic orbitals, we can reference their
phase as either positive or negative.
And we often, in representations like this
one, associate the positive phase with one color,
here blue, and the negative phase with a different color,
here yellow. I can choose whatever colors I
want for those, but this is also a type of
isosurface. And it represents the in-phase
side-to-side overlap of the positive above the plane lobes
of the carbon 2pz orbitals and the negative lobes below the
plane of the carbon 2pz orbitals experiencing side-to-side
overlap. And so you see the two
electrons that could be paired up, spin paired,
one spin up, one spin down in this pi bond,
are smeared out both above and below the plane
But there is no contribution at all to the electron density in
the plane from an orbital like this.
When we looked at electron density isosurfaces,
as we have done a few times in this class, what we are looking
at there is an isosurface corresponding to all of the
electrons in the molecule. And here, what we are looking
at is an orbital that can house just two electrons,
one spin up and one spin down, so we are focusing our
attention on a function that involves just two electrons
bonding in a pi bond that has, as a nodal surface,
the plane of the molecule. The plane that contains all six
of these nuclei, the two carbons and the four
hydrogens. We talked about nodal planes
last time. And the issue of nodal planes,
nodal surfaces, is going to be very important
in today's lecture. Let's go over here and start
talking about hybridization.
On this board last time, I gave you kind of a timeline,
where we talked about contributions to different
aspects of electronic structure theory for molecules by
different people, some of who were recognized
with Nobel Prizes, others who were not.
And one of those people, namely Linus Pauling,
who won two Nobel Prizes, not both in chemistry,
came up with this idea of hybridization to solve one of
the problems associated with atomic orbital nodes.
And let me work on this and illustrate this to you by taking
a pretty simple molecule that has only four atoms.
And this molecule is one that, in fact, we have seen before.
It is an example of a Lewis acid, so any discussion of
electronic structure of this BH three molecule better
come out with a description of why it is a Lewis acid.
But let me now point something out.
That is, if I make the direction that coincides with
this top B-H bond, the x-axis.
And in the plane of the board, the y-axis perpendicular to it,
as shown. And then, of course,
the z-axis coming perpendicularly out of the plane
of the board. Linus Pauling recognized a
problem. And let me give these hydrogens
descriptors. I will make this one A,
this one labeled B, and this one labeled C.
And the problem is that we have a pz orbital with one positive
lobe coming out of the plane of the board, up here.
This would be where the blue lobe is.
And the yellow lobe would be back behind the plane of the
board, analogous to the pi bond that we just looked at up on the
screen. And what you will see is I can
draw it this way. I am rotating this planar BH
three molecule.
That allows me to draw the pz orbital in a way that you can
visualize. And the observation that I want
you to take away from this drawing is it has to do with the
relationship in space of the three hydrogen nuclei as an
important property of the 2pz orbital.
What we can say is that these three hydrogens,
A, B, and C reside in the nodal plane of the 2pz.
They lie in the 2pz nodal plane.
That is, the 2pz on the boron. And the wave functions that
these hydrogens bring into the problem of the electronic
structure of the BH three molecule are spherically
symmetric. They are simply 1s orbitals.
And so, if you draw one of them here, remember that a 1s orbital
is spherically symmetric. You can see that any positive
reinforcement that would occur by overlap here is equal and
opposite to the negative reinforcement that will occur
down here. And this is something that is
true whenever a hydrogen s orbital lies in a nodal plane.
There is no net overlap--
--between the boron's 2pz and the hydrogen 1s orbitals.
When we try to describe the bonding in this molecule,
we cannot have any bonds that involve 2pz and any of the three
hydrogen 1s orbitals because there is no net overlap between
them -- -- because those spherically
symmetric orbitals lie in the nodal plane of 2pz.
And that means, further, that we need to form
three in-plane bonds.
And that is the xy plane in plane bonds.
And these three bonds must be equivalent.
And what is interesting about the fact that they must be
equivalent is that what are the three orbitals that the boron
has that are in the x,y-plane, the three valence orbitals?
There is the 2s, the 2px, and the 2py.
Three orbitals that are not identical to one another,
yet we need to make three identical bonds because all the
studies that we do on the BH three molecule tells us
that each of these bonds are the same length, the molecule has
the perfect symmetry of a Mercedes Benz symbol,
it's trigonal planar. It has this three-fold
rotational symmetry to it. And so somehow,
using a 2s, a 2px, and a 2py, we need to make
three equivalent bonds. This is an energy level
diagram. And so, Pauling came up with
the idea of promotion. And promotion will correspond
to the following. I can draw 2px,
2py, 2pz, and 2s. And I can populate with
electrons. The boron atom has three.
Now I have put in three electrons into this energy level
diagram for the boron atom.
And now, this concept of promotion kicks in --
-- and reorganizes us electronically,
as follows.
According to our discussion, we need three equivalent
orbitals down here with the three electrons in to pair up
with the three electrons that come into the problem from the
three hydrogen atoms. At the end, we will show how
those electrons form the hydrogen atoms come in and pair
up with these three electrons on a promoted boron atom.
Up here, we have the boron 2pz orbital.
That stays by itself. And down here,
we are going to have a set of sp two hybrids.
The idea is now somehow we are going to get three orbitals down
here by mixing one part s and two parts p.
And, if we do this correctly mathematically,
these will have the right orientation in space to be
hybrids that have proper directionality for good overlap
with our hydrogen 1s oribtals. These are going to be three
equivalent hybrids.
How do we do that? Well, we are going to get our
equations for hybridization.
And we are going to develop these using some of the results
from quantum mechanics, which is what Pauling did.
And essentially, what we are going to do is seek
equations that have one part s and two parts p,
each of them, but which point in the correct
directions in space for making three separate two electron
bonds between the boron and the three hydrogens.
And so, I will just call them sp two of A,
because we are going to need one that points toward that A
hydrogen, which is along the positive x-axis,
we are going to need one that points toward the B hydrogen,
and we are going to need one that points toward the C
hydrogen. And the idea here is that we
are going to have some amount of the boron 2s,
px and py -- -- because pz is not
contributing to these three hybrids.
And each one of these is going to have some quantity of the
three orbitals that we have available.
And what you need to keep in mind here in setting up
equations like this is that px has a definite spatial
orientation with respect to the coordinate system that we have
chosen. And so does py.
And the way that we will make this come out using our blue
chalk is to note that we have a coefficient C1,
C2 and C3. Our job is going to be to find
the values of these coefficients,
C4, C5, C6, C7, C8, and C9.
We can use the rules of quantum mechanics, and I am going to
summarize a few of these rules for you before this lecture
ends, to find the values of these constants.
Let's go over here and begin.
Let's make note of the dictates of symmetry.
I will reference symmetry a couple of times today in class
in working on this problem of finding those coefficients,
one through nine. The neat thing is that the 2s
orbital centered on that boron, which is going to be
participating in these hybrids that are going to point toward
hydrogens A, B, and C, that 2s orbital is
spherically symmetric. Its wave function is the same
as you go radially out from the boron center in any direction
into space. What that symmetry
consideration dictates is that there will be the same amount of
boron 2s in the hybrids for A, as for B, as for C.
And what that means is that C1 is equal to C4 is equal to C7.
And we are going to need a value for this.
And we can get a value for that by taking into consideration
something called unit orbital contribution.
What that says is that we have to use the boron's 2s orbital
completely and identically once in that set of three equations.
In other words, we are going to distribute it
evenly among those three hybrids.
And we have to distribute all of it evenly among those three
hybrids. And here we are going to be
interested in the square of the coefficients.
And that is because when we talk about atomic orbitals,
we know that the probability of finding an electron in that
orbital is related to the square of the wave function.
And so, normally when we plot orbitals, like I just did with
the pi bond of ethylene a few minutes ago, we are making an
isosurface that corresponds to some percent probability of
finding the electron at that point in space.
So, we are interested in probability distributions.
And these are related to the square of the wave function.
Here, for unit orbital, we are going to say that C1
squared equals C4 squared. Sorry, plus C4 squared plus C7
squared is equal to one. And,
therefore, C1 is equal to C4 is equal to C7 is equal to one over
root three.
Now, we have found C1, C4, and C7.
And, if we are going to complete sp two of A,
get this first hybrid of the three, I am going to go across
this way. I just went down one,
four, and seven and pointed out that by the symmetry of the s
orbital is due to the fact that the s orbital contributes once,
unit orbital contribution, we could get C1,
C4, and C7 equal to one over root three.
Now I am interested in C2 and
C3, so let's draw a little picture.
This, remember, is our 1s orbital on HA.
And what we are trying to do is to construct a hybrid focusing
on that top equation that will be directed at HA.
And we are keeping our axes the same as before,
so that x points up and y goes to the left, as I am drawing it
here. And z would be coming out at
us. What I am going to draw now is
a picture of the boron's 2py orbital.
And, as I usually do, I am taking the convention that
the negative phase lobe of the boron's 2py orbital--
Negative is shaded. That is my convention.
That 2py orbital lies in the xy plane and points along y.
Its positive lobe, in fact, points along positive
y. Its negative lobe points along
negative y. And what can we say about the
relationship of the HA 1s orbital to the 2py orbital of
that central boron atom?
Let me phrase it this way. The 2py orbital of our central
boron atom has a nodal plane. What is that nodal plane?
xz is a nodal plane of the 2py orbital, right here.
Just like this. And so, what you see is that
this HA is not only in the nodal plane of 2pz,
it is also in the nodal plane of 2py.
And that means any positive interference by bonding between
HA's 1s and the positive lobe of 2py would be exactly canceled by
the equal and opposite negative interference between a
hydrogen's 1s orbital and the negative phase lobe on 2py.
And that means that we know the value of C3.
We can say, therefore, that C3 is equal to zero.
That is really cool because now we are going to be able to bring
in one more consideration and complete the formula for sp two
of A.
We are going to be able to use the normalization condition,
--
-- which tells us that C1 squared plus C2 squared plus C3
squared is equal to one.
Once we formed this new hybrid by mixing together s and p wave
functions, we are going to find that this normalization
condition must be satisfied for the hybrid orbital just as it is
satisfied for these atomic orbitals.
That is so that the probability of finding an electron somewhere
is space associated with this hybrid wave function is one,
if there is an electron in this orbital.
That is where this comes from. And we know that C3 squared is
zero. C1 squared is one
over root three. So,
we know that sp squared of A is equal to one over root three.
And then we can say that,
by virtue of this and the fact that C3 is zero,
C2 could be equal to either plus or minus the square root of
two over three.
And we have to figure out which one of these to use.
Remember that C2 is the coefficient on px.
Let's reference our diagram on our coordinate system.
We have our x-axis and our y-axis.
And we are looking for the sign of the coefficient on the boron
2px orbital. That is our C2 coefficient.
And I am shading negative here. And, in order to overlap
constructively and form a bond to the 1s orbital of HA,
we want positive as our coefficient.
Because that is going to make the same phase lobes interact in
a sigma fashion. We know that sp squared of A is
equal to one over root three times the boron's 2s orbital
plus the square root of two over three 2px orbital.
And now that we have the coefficients on that hybrid
orbital, we can draw a picture of what that is.
That is equal to taking the boron 1s orbital here and adding
to it the 2px orbital that looks like this.
This is the conceptual breakthrough that Pauling made.
You could mix atomic orbital wave functions located on a
central atom in order to generate these hybrids that have
the proper directionality for good overlap with the peripheral
atoms that bond to the central atom.
And when I do this, what you can see is that we are
going to get constructive interference with the positive
lobe and destructive interference with the negative
lobe when we add this boron 2s orbital to the boron 2px
orbital. And that should give us
something that looks like this, with an enlarged lobe pointing
along positive x and a diminished negative node
pointing along negative x. That is our hybrid.
We have developed some coefficients here.
We just got it pictorially, and here we show you the
picture of the hybrid. And then, we need to continue.
We need to use the remainder of our 2px orbital.
The one thing we cannot do in constructing a set of hybrid
orbitals from a set of atomic orbitals is forgetting to use
part of our orbital. We need to use all of the
orbital that we are given to construct these hybrids with.
And what that means is we have now a relationship between C5
and C8, because we now know C2 is plus root two over three.
Let's use this unit orbital idea and
reference 2px -- -- and say that C2 squared plus
C5 squared plus C8 squared, these are all the coefficients
that deal with the px contributions to these three
hybrids, is equal to one. And
we know C2 squared is equal to two-thirds.
And now let's draw a picture to
help us figure out something else about C5 and C8.
Here I am simply representing, with the same coordinate system
that I am using throughout, the boron atomic orbital that
is the 2px orbital. And, if I consider,
down here, the s orbitals on HA and HB, I can say something
about the way in which 2px interacts with HA as compared
with the way in which it interacts with HB.
Notice that these HA and HB atomic positions are mirror
images of each other with respect to this 2px orbital in
the middle. That tells us that C5 must be
equal to C8. We are using
symmetry as a mathematical argument here to say that C5 is
equal to C8. Therefore, we can figure out
exactly what C5 and C8 are because we already know that C2
squared is two-thirds.
We know that C5 squared plus C8 squared has got to be equal to
one-third. And that
means that C5 is equal to C6, sorry, C8, which is equal to
one over the square root of six.
Because we put this in here, we will get a sixth plus a
sixth, two-sixths, which is one-third,
plus the two-thirds, which will be one.
And so, we know it is one over root six.
However, it could be a positive or a negative.
And we need to figure out just which it is.
Is it positive or is it negative here?
And I will come back to that issue just in a moment,
when we draw these functions. But notice that these are the
coefficients on 2px and the positive lobe points up toward
HA. We are going to need it to be
flipped around and be pointing down if we are going to form
bonds to HA and HB. So, that will dictate the sign
here as negative one over root six.
We do not have, yet, the complete equations for
sp two of B and sp two of C because we
need C6 and C9. Let's see how we can get those.
C6 and C9 relate to the py orbital that is oriented like
this with reference to our coordinate system,
with the negative phase shaded as follows.
And specifically, C6 relates to the interaction
with this HB down here. And C9 relates to the
interaction with the 1s orbital on HC down here,
on the lower right. And because this py orbital
charges phase as we pass through the origin, and we want the
positive phase to match up both with HB and with HC in the
contributions to the hybrid wave functions, we are going to know,
and by the symmetry of this problem, that these are negative
mirror images of each other. So, we know that C6 is equal to
negative C9. And if we put that
together with unit orbital contribution,
we can see, because we know that we have our sp two of B
and sp two of C,
both of them have one over root three of the 2s contributing.
And then we are looking for six and nine.
We found out over there that both of them have negative one
over root six as the coefficient on px.
And then, we know that since this is the one that points to
B, we do not need to flip the phase on py to get a bond
forming between py and HB. This is going to be a positive,
and then this one is going to be a negative,
for the reasons I just mentioned, that we are going to
need to flip that py contribution to sp squared of C,
as compared to sp squared of B,
in order to get this positive
lobe over here interacting with HC in that function.
I am going to draw these functions in just a moment.
But now, when we note that normalization requires that the
sum of the squares of the coefficients be equal to one,
and we know that C3 was zero, so that there is no
contribution at all of py to sp squared of A,
and all of it must be here and here, that means that our
coefficient is one over root two of py.
Using these restrictions from quantum mechanics,
we actually have coefficients on the spx and py orbitals that
show how they go together. And the way in which these add
up is similar to what we have seen over here.
It is a little more complicated because both px and py
contribute. Let me show you what they look
like. This mathematical manipulation
of these functions produces linear combinations that all
three look the same as the symmetry of this problem would
dictate. It is a three-fold symmetry
type of problem. And this is our sp two
orbital pointing toward B with this mathematical form.
We have to flip around px so that the positive lobe of px
contributes down here. We do not have to flip py.
And we have differing amounts of px and py in this hybrid wave
function. But the math of the atomic
orbitals guarantees that this form would be physically
indistinguishable from this one over here.
They would look exactly identical if I plotted a
probability density isosurface. These two-dimensional pictures
that we draw on the board are like slices through
three-dimensional probability density isosurfaces.
That is what these representations are.
And then, we note that the py orbital is flipped around when
we form sp squared of C.
And that leads to this appearance.
I would like to show you what one of these might look like if
you calculated it from the math that we have developed,
here.
I have actually got more work to do here in the next five
minutes, so let's keep our concentration.
Look at this orbital. This thing is a beautiful
hybrid orbital. And I have calculated this with
reference to a particular arbitrary molecule.
This was calculated with reference to the water molecule.
The positions of the nuclei in the water molecular are
indicated by these arbitrarily sized spheres.
That is a typical ball and stick representation of a
molecule in 3D. You can see that the oxygen of
the water molecule is colored red, in there.
That is arbitrary. And then we have the hydrogens
in white. And then I have a big positive
lobe in blue here and then a smaller yellow lobe that is our
negative lobe on this orbital, here, which is an added mixture
of an s orbital with a p orbital.
I am not telling you the actual coefficients that went into this
particular hybrid, but you can see how if I were
to take a two-dimensional slice along the long axis of this
hybrid orbital, it might look something like
what we have drawn over there. Namely, with one nice large
directed lobe, that can have good overlap with
a hydrogen 1s orbital in order to give rise to a good two
electron bond between hydrogen and boron.
And let me illustrate how that impacts on how we draw our
energy-level diagram for the BH three molecule.
What we have now is a situation in which, as you go up in
energy, you find three orbitals that are all at the same energy.
These lines indicate places where we can put electrons.
And, in fact, they look like this.
They represent positive interference up along x.
Positive interference between a sp two hybrid that
points down at atom B, right here, overlapping nicely
with the hydrogen 1s wave function that is sitting out
there, surrounding that nucleus. And then over here is our
hybrid orbital sp squared of C that points down
here and interacts with the hydrogen 1s orbital on atom C.
And so, the idea is now this system, BH three,
has a total of six valance electrons.
Valance shell electron pair repulsion theory could be used
by you to predict the geometry that we started with for this
problem. But now, we can populate these
bonding orbitals with electron pairs, as shown here.
And then, this problem has one more aspect to it,
namely that up here somewhere higher in energy there is a 2pz
orbital on the boron that is empty.
And this empty orbital --
-- is our feature that gives this molecule Lewis acid
character. We know that this is an
electron deficient system. It does not satisfy the octet
rule. And, specifically,
where it is missing electrons are in that boron 2pz orbital.
That allows you to predict something about the
stereochemistry of reactions, for example.
It says that if some nucleophile or Lewis base,
a source of a pair of electrons is going to come in and approach
that planar BH three molecule, it will do so either
along positive z or along negative z because that is where
the big lobes are of that empty pz orbital point.
And, this valance bond theory --
-- satisfies this problem, that Pauling noticed,
of how do we make three identical bonds in a plane that
contains three dissimilar atomic orbital wave functions,
an s orbital px and a py? Well, he found that
mathematically you can mix them together in ways that gives you
three identical hybrids that have the needed directionality
for the forming of these three two electron bonds.
There is an alternative solution to this problem,
and we will call that molecular orbital theory.
Don't forget about Robert S. Mulliken, MIT undergraduate
thesis 1917. And MO theory is a different
perspective on electronic structure theory for molecules.
And we will launch into that on Wednesday.