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Some of you will recognize this molecule that I have rotating up
on the screen because it is one of the coordination complexes of
Werner that we talked about last time.
And Werner, of course, was well-known.
And some of you probably read is biography over the last few
days on the Nobel Prize website. Well-known for his theory that
broke down the preconceived notions that had been prevailing
at that time concerning the structure of systems that have
unusual salt-like behavior in some cases, but that contain the
3D elements. I am focusing on 3D,
but this is also true of 4D and 5D elements, the transition
metals. Elements like titanium,
vanadium, chromium, iron, cobalt,
nickel, etc. This molecule here,
that you see spinning up on the screen, is being represented
here on the screen in one of these kind of arbitrary forms,
wherein the cobalt center at the middle of the molecule is
just a round ball of an arbitrary size.
And then, this particular molecule that contains a cobalt
three ion is surrounded by a compliment of
six ligands. We have three chloride ligands,
that because this program chose to do so colored those chlorides
the same as the cobalt, not what I would do,
and put the three ammonia ligands, colored the ammonia
nitrogens in dark blue, and the ammonia hydrogens in
white. And we talked about the fact
that these transition elements, in fact, these 3+ ions of
metals like cobalt, can behave simultaneously as
Lewis acids toward a multitude of Lewis bases,
here six. This is a times six Lewis acid
in the middle and six Lewis bases, which are three chloride
ions and three ammonia molecules oriented around it,
in an arrangement that is quasi-octahedral.
Because the positions of the three nitrogens of the ammonias
that are interacting with the cobalt center and the three
chloride ions are located near the vertices of a regular
octahedron. So we will be taking use of the
geometry of molecules like this in discussing electronic
structure properties of these molecules today.
We will be doing that beginning today.
And that will be important to understanding the magnetism and
the color and also the reactions of molecules like this.
Now, one other arbitrary thing that this program did is it
chose not to draw lines between the cobalt and any of the six
ligands. But normally,
when you see these molecules drawn in textbooks,
you will see that the lines drawn are the same as the lines
between the nitrogens and these hydrogens.
So we would have to add six more lines to this drawing to
get the typical textbook representation of a molecule
like this. But, still, that would be a
somewhat arbitrary representation.
And so, I would like to show you a less arbitrary
representation. And we will do that forthwith.
And this one has to do with looking at the same molecule,
but represented as an electron density isosurface.
And that isosurface will be colored according to this
function, that tells us about the propensity to pair electrons
in 3D space. And so you will recall that
when we talk about coloring electron density isosurfaces in
this way, so this now is a physically important kind of
representation of this coordination complex,
this color scale will run from red all the way to blue.
And at blue is where you find regions in space where you are
most likely to find pairs of electrons mapped onto the value
here, colored mapped onto the value of an electron density
isosurface. Here the value of the electron
density that is present for every point represented on this
surface is 0.11 electrons per unit volume.
And so what you see here is that, in fact,
the electron density does become low as you move from the
cobalt in the direction of any one of the ligands.
But the lone pair on each ammonia is certainly polarized
in the direction of the cobalt center.
And the cobalt center is not uniform in terms of the way that
electron density is organized around it.
You will see that right on the metal center,
we see red. And then sort of at the corners
of a cube, you see this yellow or green color on the cobalt,
and that is very significant. And what we are going to be
today is we are going to try to understand what happens when you
put a set of ligands into an octahedral array around a
central metal ion that has d-electrons and d orbitals to
play with. And, in fact,
what you will see is that the oxidation state of the cobalt
center here is +3. That is why I am referring to
it as a cobalt plus three ion.
And because cobalt is in group nine of the periodic table,
you then know that there are six valance electrons on the
cobalt center that you have to put into orbitals.
And so what we are really seeking to know is how can we
get an energy-level diagram for a system like this,
so we will know how to put those six electrons into that
diagram, the ones that are mostly localized on the cobalt,
and make predictions on whether the electrons should be paired
up or not, for example. We saw with homonuclear
diatomics, like the dioxygen molecule, that electrons are not
all paired up. Two of the electrons are
unpaired. And, when we have six electrons
to put into an energy-level diagram for the cobalt ion,
we are going to wonder just what the case is.
How many energy levels are there, and where do we put the
electrons? And so we are going to need to
very briefly review what I did at end of lecture last time.
And that has to do with the properties of the five d
orbitals -- -- because, in order to answer
the kinds of questions that we are posing about energy levels
of metal ions that are transition metals,
we are going to really need to know very well the nodal
properties of these d orbitals. And so, remember the m quantum
number here can be zero for the d z squared.
It can be plus one for the d(xz).
And plus two for the d x squared minus y squared.
Minus one for the d(yz)
orbital. And minus two for the d(xy)
orbital. And let's draw those.
d z squared is an interesting one.
And, because it is distinct from the other four d orbitals,
we are going to be spending more time on it today than the
others. What it has,
these are the x, let's say y and z axes here,
is it has a positive lobe along both plus and minus z.
So it looks like a p orbital so far, except that we have the
same sign both in plus or minus z.
And then what we have is a very interesting toroidal shape that
goes around in the x,y-plane all the way around in a
cylindrically symmetric manner. And so, if you were to look
down z onto this orbital, it might look something like
this.
Like that. It would be cylindrically
symmetric about z. And that was another feature
that I didn't get to add to this diagram at the end of last hour,
which is that because of this property, this m equals zero
that means this is a sigma orbital with respect to the
z-axis. Cylindrically symmetric about
z. And then, let's go to the xz,
since I am making this x, y, and z over here.
The d(xz) has four lobes, and they are in between the x
and z axes, as I am trying to represent here.
These two are kind of coming out in front here,
and these two back behind. And then, the phases go as
follows. Unshaded is,
again, plus. And then we have the x squared
minus y squared orbital, which has four lobes.
In fact, the shape of xz, x squared minus y squared,
yz, and xy are all the same.
It's just that they point in different directions in space
with respect to the Cartesian coordinate axes.
x squared minus y squared, like d z squared,
is an orbital whose lobes point along the coordinate
axes, like this. And it is minus along y,
as the name suggests, and plus along x,
as the name suggests. That is our d x squared minus y
squared orbital.
And then yz is like xz, but let me finish this part
here. If d z squared is
sigma with respect to z, and you imagine looking down z
onto d(xz), what would that be with respect to z?
Anyone. Pi. Thank you. That would be pi with respect
to z, because the yz plane is a nodal surface for this d
orbital, as is the x,y-plane. And that is what we are trying
to do here, is become familiar with these orbital surfaces.
And, accordingly, now, if you look down z onto
the d x squared minus y squared orbital,
you are now going to see two nodal surfaces when you look
down it that way. Because you are going to see
that there is one located over here.
That is a plane that contains the z-axis, but it bisects the x
and y axes. And then there is another one
over here, but that is 90 degrees to the first one.
And that makes this one delta with respect to z.
Delta is when you have two nodes that contain the z-axis
and if we are looking down that z-axis.
And then, over here, the d(yz) orbital has its four
lobes between y and z, like that.
And it is positive between y and z, and negative over here,
as I am shading. And this one,
like d(xz), is pi with respect to z.
And then, over here, we have a d(xy) orbital as our
final orbital. And what you might be able to
guess, we have a sigma with respect to z,
we have a pair of pi with respect to z for m equals plus
and minus one. We also must have a pair of
delta with respect to z for m equals plus and minus two.
And that means that the d(xy) orbital, like d x squared minus
y squared, must lie in the x,y-plane.
And, in order to be orthogonal to d x squared minus y squared,
we are going to have to rotate it such that its four lobes
point now between the x and y Cartesian coordinate axes like
this, although I am trying to improve on that with my
coloration. That is like that.
Again, this is an orbital perpendicular to z and which has
delta symmetry. And now, the two nodes are,
in fact, the xz and yz coordinate axes.
Indicating those nodal planes there.
This is certainly the basics for what you need to know about
the d orbitals. And just briefly,
I would like to switch to Athena terminal,
here, to show you that there are ways for you to go ahead and
visualize the orbitals. And I am going to make this
information available to you, so that you can go ahead and do
this yourself in order to visualize these in a way that
will take you right from the equations for the orbitals to
their graphical representation. I think that is really
important to get a good understanding of orbitals.
Let's zoom this a little bit, so that you can begin to see
it.
This is a worksheet put together that,
in fact, contains all the functional forms for the d
orbitals. Let me see.
Where is that? Here we go.
And you can look at this with Maple on Athena.
And then, you can look at the equation that represents the
angular part of the wave function for the orbital that
you are interested in. And then, you can go ahead and
plot it. And you can plot it in such a
way that the function is animated.
And, rather than just seeing it projected on a board as well as
I can draw, you will be able to see it drawn up graphically.
In fact, these come from the solution for the Schrˆdinger
equation for the hydrogen atom. And the angular part of these
wave functions that is going to be oh so important to us is
something known as the set of spherical harmonic equations.
And that should reference you to this issue of standing waves,
that we have discussed. Let's just see what we can do,
here. Sometimes, I am not so good at
using Maple up in front of the class.
What you are seeing is we are getting representations for d z
squared. Right here is a way of writing
d z squared. You are going to see an
important term here, three cosine squared theta
minus one. And we will come back to that
in a moment. That is the angular part of the
d z squared wave function.
And then we can look at some of these other ones.
You will see that some of the d orbitals come as combinations of
real and imaginary functions that are the solutions to the
differential form of the Schrˆdinger equation.
And then we take linear combinations of these to get
real forms, so that we can get plots that we can look at.
And let's see if we can get d z squared,
here. There it is.
There is a picture of d z squared.
And you see, if you are using this Maple
worksheet, that you can actually rotate that around and animate
it a little bit. You see that we have this torus
that is in the x,y-plane. And you have the two large
lobes that extend up along plus and minus z.
And so I am going to encourage you to go ahead and look at that
worksheet, which will be available from our website.
Go ahead and look at some of the functions.
And if some of you are interested in higher orbitals,
the f orbitals are also available in this worksheet.
So you can visualize the f orbitals, that are important for
understanding the chemistry of elements like,
for example, uranium, which is a little bit
beyond the scope of 5.112. Now let's switch to the
document camera. And, if you could,
I would like you to make this part big.
This is a table in your textbook that has the angular
part of the wave functions for various hydrogen-like orbitals.
And this is the part that I am most interested in,
over here. If you could just focus in on
the d orbitals over here. When I was talking about the d
z squared orbital a moment ago, I was focusing on
this term here, this cosine squared theta minus
one term. And here are the other
d-orbitals. These are the descriptors for
the d-orbitals, zy, yz, xz, x squared minus y
squared, and z squared.
And I will need to refer to this in a moment,
so we will leave this up. The reason why I have this
arrow written into my book here is because these are backwards
in the text. This one is actually x squared
minus y squared, and this one is actually the xy
orbital. We figured that out last year
when we were doing this lecture. And so your book isn't always
right. You should make sure you check.
The same is certainly true of your instructor,
but we will try not to mislead you.
And so now here is the approach that we are going to take.
Let's say that we have our coordinate system,
and we want to know how to evaluate one of these d orbital
wave functions for a particular point in space.
Actually, we are going to want to evaluate the square of the
wave function. And so we are going to make use
of a Cartesian coordinate system.
And we are going to express things in terms of polar
coordinates.
Here is a sphere projected onto our Cartesian coordinate system.
And some of you will be very familiar with this.
We are going to say if you are at a point here in space,
then we can describe that point in space by a set of variables
which will be r, theta, and phi.
And here is our angle theta. And if we drop down to a
perpendicular on the x,y-plane from our point,
then we are going to define phi as being from the x-axis and
going over in the direction of y.
And so, when you look at the d orbital wave functions,
you see that they are all written here in terms of just
theta and phi, and not in terms of r,
which is this distance here from the nucleus,
from the center of this metal ion that we are talking about.
This would be the r. And what we are going to do,
is if this point on the surface of our sphere represents one of
our ligand atoms, so think back to that complex
we were describing a few moments ago, cobalt with three ammonia
ligands and three chloride ligands, we are going to
approximate each of those six ligands by a point on the
surface of our sphere. And then, we are going to say,
if there is a metal at the center, that cobalt ion in
particular, what is the probability of finding a d
electron where that ligand atom is?
And, in order to do that, we are going to need to be able
to evaluate the wave function. And we are making the
assumption, for simplicity, that this is a perfect sphere,
and also that our coordination geometry is a perfect
octahedron. And so we are not going to
consider r, because r is going to be the same everywhere.
It is just a perfect sphere with one value of r,
no matter which ligand position we are looking at.
And so that means we can focus in just on these which are the
angular part of the wave function for the molecule in
question. And so let's see what this
means with respect to d z squared.
The probability of finding an electron at some point in space
in a particular atomic orbital is proportional to the square of
that atomic orbital at that point in space.
That is our probability density.
And we encountered that very early in the semester.
Now, we are going to make use of that to make a plot.
We are going to plot this probability density of finding
electron in d z squared as a function of
theta. Why can I do that ignoring phi?
Well, that is because if you look at d z squared,
there is no phi in that equation.
Why is that? That is because d z squared
is cylindrically symmetric about z.
And look at how we defined phi. It is as you go in the
x,y-plane around starting from x.
Because of the sigma symmetry of d z squared with respect to
z, there is no phi dependence of this wave function.
And that should appear here in the angular form of the
description of the d z squared orbital.
But we do know that d z squared does depend on theta,
because theta starts out somewhere here along,
let's say initially theta equals zero would be right on
the positive z-axis. And then, as we keep a constant
r and we sweep down here toward the x,y-plane,
the d z squared probability density is dropping
off. And then at some point,
when we get to the node here, and we are going to be
interested in that, it goes to zero,
because that is what happens on nodes, as we continue down
toward the x,y-plane past that node, it is going to be nonzero
again and rise up as we approach this smaller torus in the
x,y-plane. Smaller, that is,
than the big lobes that extend up along z and down along minus
z. Let's represent that
graphically.
Coming down in theta from theta equals zero to some value here,
we are going to be interested in just what that value is.
And then, rising up again to theta is equal to pi over two.
Do you see that? This is another way of
displaying this property. This is at constant r --
-- and varying theta.
First of all, how can we find out at what
value theta goes to zero? Well, we look up here at the
functional form of the d z squared.
We get nodal properties of d z squared.
We can pretty quickly see that in each of these,
we have a factor leading out in front, which is a normalization
factor that assures us that the sum integrated over all space of
this wave function will come out to be one.
If there is an electron in that orbital somewhere,
the probability of finding that electron somewhere in space will
be one. We have these normalizing
constants out in front that allow for that and ensure that
that is the case. But where we actually find the
angular dependence is in that second term.
Here it is three cosine squared theta minus one.
And what we want to do is say, when does this function go to
zero? Because when that goes to zero,
we will have this angle here. If we say, let this equal zero,
we are looking for the value of the angle theta which
corresponds to the node of the d z squared orbital.
Remember, we mentioned this last time, because of this
cylindrical symmetry of the d z squared orbital,
this really is a conical nodal surface that is above and below
the plane here. If you are anywhere on that
cone, either in plus or minus z, the value of that d z squared
orbital is zero. And so we are setting it equal
to zero to find the angle theta. And we can rearrange this and
say that cosine squared theta is equal to one over three.
And, if we go ahead and solve that,
this comes out to the arccosine of root three over three.
And the other possibility that
satisfies that relation is pi minus the arccosine of root
three over three
And so what that means is that this relation,
here, gives us the angle for that cone in the plus z axis.
And then this one down here, pi minus arccosine root three
over three, gives us the angle for
that cone down in the minus z-axis.
And what is this? This is, in degrees,
something like 54.476 dot, dot, dot degrees,
approximately. That is just how far down you
are from the z-axis when you hit that nodal surface of z.
And that number, you are going to see,
is kind of a magical number in chemistry.
And it will hearken back to some of the things that we have
been taking about, recently.
And, in order to get to that point, I am going to need to now
talk about our ligands again with respect to d z squared.
We are considering, now, an octahedral metal
complex.
This is the type of Werner-esque complex that we
talked about last time. You are going to see that we
are going to have ligands. We are not really specifying
them. We are numbering them and
locating them at positions one, two, three, four,
five, and six relative to our metal center at the middle.
And what you might begin to realize is that in order to find
out what our energy-level diagram will be that refers only
to the five d orbitals on the cobalt center,
or whatever metal center is at the middle of this ion,
what we are going to have to do is evaluate the square of the
wave function at each of these ligand positions.
And we are assuming that all the ligands are equivalent.
If you think of the ligand as an electron or as a point charge
in space, then you can imagine that if we have an electron in d
z squared up here, that it is going to interact
strongly and very repulsively with a point charge that would
be located at position one. Whereas, if we instead had a
point charge located at that theta angle of 54 point whatever
over there that we solved for, since that is on the node of d
z squared, that would be the least
possible repulsive interaction that you could get between an
electron and an electron in d z squared because it is
on the node. And so what we would like to do
is to go ahead and solve for some of these things.
Essentially, what we seek --
-- is an energy-level diagram.
And so let's write this, up here.
This is five over 16pi. This is d z squared
that I am writing up. Three cosine squared theta
minus one.
And because we are talking about the probability density of
finding an electron at a particular point in space,
which will mean a particular theta value in our case for d z
squared, we are talking about that wave
function squared. What we need to do is evaluate
it. We have already evaluated it
where the node is, but we would like to evaluate
it at position one. Because we have a ligand at
position one, and we need to know what the
relative response will be of d z squared to a ligand
along position one versus the other five positions.
We can say something about that by symmetry, already.
And so we are going to find out that the value that you get for
evaluating d z squared at position one is
the same as you would get for evaluating it at position six.
And that is because the big lobes of d z squared
are along plus and minus z. That is where ligands one and
six are. And then, because the torus is
also cylindrically symmetric, two, three, four and five have
the same value when you evaluate d z squared at those
positions. But that value is smaller than
along z, as illustrated by this graph over here.
But we just want to know, how much smaller.
And so this is position one. We get a value of
five-quarters. And down here,
position two, a value of five-sixteenths.
I am leaving off a factor of pi.
Please don't be concerned by that.
But these are the relative values that you get when you
evaluate this function at ligand positions one and two,
which is all we need to do because of the symmetry of this
because that value for position one is also true for position
six. So we have positions one and
six. And then, this is also three,
four, and five. By doing two quick evaluations
at two different theta positions, position one,
of course, theta is equal to zero, up here.
And we evaluate that squared function for theta equals zero.
And we get five-fourths pi, but I am leaving off the pi.
And down here, at position two,
we evaluate this for theta equals pi over two because two,
three, four, and five are all in the
x,y-plane at 90 degrees to z. So the value of theta anywhere
for those four ligands would be pi over two.
You evaluate this function for pi over two, and you will get
five-sixteenths pi. And I have just left off the
pi. That is useful.
Let's take this one up to the top.
Here we have done one of the d orbitals.
This is d z squared for all six ligand positions.
Now, let's do d x squared minus y squared for
all six ligand positions.
First of all, d x squared minus y squared
has a pretty interesting relationship with
ligands one and six. What is that relationship?
Zero. Because x squared minus y
squared has two nodal surfaces that intersect
along the z-axis. And so those ligands,
four and six, lie on a nodal surface.
And so, we know that that is going to be equal to zero.
If, however, we go ahead and evaluate the
square of d x squared minus y squared, let me just write it
up. And I have to switch it,
since they are wrong in the book.
15 over 16pi square root sine squared theta cosine two phi.
We have that. Now, just as we do,
we are converting an orbital angular property into a
probability density by squaring it.
This is what we normally do. When you see pictures of
orbitals, they are representing the square of the wave function
in space. We need to evaluate this as a
function of theta and phi in order to find out what --
The first two, one and six,
we did by inspection, but what about positions two,
three, four, and five?
In the case of those four, you can see that where they lie
in the x,y-plane with respect to the x squared minus y squared
orbital is all identical to each other by
symmetry. Because x squared minus y
squared has lobes that extend along x and plus x and y and
minus y. And that is where all these
ligands lie, at positions two, three, four,
and five. We know that theta is equal to
what? It is going to be pi over two.
And phi, of course, can be zero pi over two pi,
and three pi over two for any of those positions.
And what we will find is that this evaluates --
For any of those, let's just use phi equals zero.
This evaluates as 15 over 16pi. And I am dropping the pi.
Now we have d z squared and d x squared minus
y squared evaluated for all six ligand
positions. And then, let's make a table of
this.
z squared. x squared minus y squared.
Let's do xz,
yz, and xy. And here are our ligand
positions. And in the table here,
we are going to put what these relative evaluated squared wave
functions are. And the biggest one of all is
this one here that we got at position one for d z squared,
which is five-fourths.
I am going to divide everybody through by five-fourths,
so that I can make our biggest value equal to one for
simplicity here. And then, when we do that,
you are going to see that two gives a value of a quarter.
That is position two, down in the torus.
Three, down in the torus, is one-quarter relative to that
one. Four is one-quarter.
Five is one-quarter. And six, down in minus z,
is one. So we have evaluated the d z
squared orbital squared at the ligand positions
one through six. And these are the relative
values that we got for the probability of finding an
electron at that point in space, given a constant value of r.
And x squared minus y squared,
we got for positions one and six zero, we just said that.
And then, on the same scale here, we get three-quarter,
three-quarter, three-quarter,
and three-quarter. And then for xz,
yz, and xy, we have gone through these two steps to do z
squared and x squared minus y squared explicitly.
Now, we have to look at where
ligands one through six are relative to the nodes of xz or
yz and xy. And what we will find is that,
in each case, these ligands lie on nodal
surfaces of xz, yz, and xy.
This is zero, zero, zero, zero,
and so on.
Okay? All the ligands,
one through six, lie on nodal planes of xz,
yz, and xy. Only four of the ligands
interact with x squared minus y squared,
ligands two through five, because those are the ones that
lie in the x,y-plane. And they interact strongly,
this relative value of three-quarters,
but not as strongly as the two ligands in positions one and six
interact with the big lobes of d z squared.
And then, also, ligands two through five,
which lie in the x,y-plane, interact with the torus of d z
squared, but to a much smaller extent
because the torus does not have as great a radial extent.
And these ligands are all at the same radius out from the
metal center. And so what this corresponds to
is now an energy-level diagram as follows.
And this is for an octahedral complex, --
-- where we have relative energy units of zero,
one, two, and three. And what we have to do is say
that the amount an electron in d z squared would be
repelled simultaneously by electrons in positions one
through six would be the sum of these values.
And so we add that up and get, in fact for d z squared,
up here, a three.
And, interestingly, for d x squared minus y
squared, if we take a sum of the four
interactions that we found that are non-zero,
we also get a three for d x squared minus y squared.
So then the net of all their interactions is the same for x
squared minus y squared and z squared.
And then down here, we found that these three
orbitals, xz, yz, and xy, where the ligands
at positions one through six lie on their nodal surfaces,
lie right on their nodal surfaces.
And so the number evaluates to zero.
And so a wave function squared will evaluate to zero any time
you are looking at a position that is on one of its nodal
surfaces. It evaluates to zero.
And so we have d(xz), d(yz), and d(xy).
And then, a diagram like this, which is a d-orbital splitting
diagram --
-- is associated with a couple of different parameters.
One is, down here we have a triply degenerate energy level.
Previously, we had only seen doubly degenerate energy levels.
Now we have a triply degenerate on composed of xz,
yz, and xy. And that level we are going to
be calling t(2g). And then, up here,
we have a doubly degenerate level that will get the label
e(g). And then, whatever the value of
the splitting of these two energy levels,
one triply and one doubly degenerate, we are going to give
that the label delta O for octahedral.
And at the beginning of next hour, I will say more about
tables like this. And I will show you how you can
actually figure out d-orbital splitting diagrams for other
coordination geometries and how they compare for a couple of the
most popular coordination geometries.