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Some of you will recognize this
molecule that I have rotating up

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on the screen because it is one
of the coordination complexes of

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Werner that we talked about last
time.

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And Werner, of course,
was well-known.

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And some of you probably read
is biography over the last few

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days on the Nobel Prize website.
Well-known for his theory that

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broke down the preconceived
notions that had been prevailing

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at that time concerning the
structure of systems that have

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unusual salt-like behavior in
some cases, but that contain the

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3D elements.
I am focusing on 3D,

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but this is also true of 4D and
5D elements, the transition

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metals.
Elements like titanium,

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vanadium, chromium,
iron, cobalt,

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nickel, etc.
This molecule here,

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that you see spinning up on the
screen, is being represented

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here on the screen in one of
these kind of arbitrary forms,

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wherein the cobalt center at
the middle of the molecule is

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just a round ball of an
arbitrary size.

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And then, this particular
molecule that contains a cobalt

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three ion is
surrounded by a compliment of

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six ligands.
We have three chloride ligands,

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that because this program chose
to do so colored those chlorides

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the same as the cobalt,
not what I would do,

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and put the three ammonia
ligands, colored the ammonia

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nitrogens in dark blue,
and the ammonia hydrogens in

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white.
And we talked about the fact

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that these transition elements,
in fact, these 3+ ions of

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metals like cobalt,
can behave simultaneously as

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Lewis acids toward a multitude
of Lewis bases,

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here six.
This is a times six Lewis acid

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in the middle and six Lewis
bases, which are three chloride

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ions and three ammonia molecules
oriented around it,

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in an arrangement that is
quasi-octahedral.

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Because the positions of the
three nitrogens of the ammonias

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that are interacting with the
cobalt center and the three

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chloride ions are located near
the vertices of a regular

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octahedron.
So we will be taking use of the

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geometry of molecules like this
in discussing electronic

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structure properties of these
molecules today.

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We will be doing that beginning
today.

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And that will be important to
understanding the magnetism and

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the color and also the reactions
of molecules like this.

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Now, one other arbitrary thing
that this program did is it

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chose not to draw lines between
the cobalt and any of the six

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ligands.
But normally,

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when you see these molecules
drawn in textbooks,

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you will see that the lines
drawn are the same as the lines

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between the nitrogens and these
hydrogens.

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So we would have to add six
more lines to this drawing to

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get the typical textbook
representation of a molecule

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like this.
But, still, that would be a

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somewhat arbitrary
representation.

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And so, I would like to show
you a less arbitrary

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representation.
And we will do that forthwith.

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And this one has to do with
looking at the same molecule,

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but represented as an electron
density isosurface.

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And that isosurface will be
colored according to this

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function, that tells us about
the propensity to pair electrons

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in 3D space.
And so you will recall that

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when we talk about coloring
electron density isosurfaces in

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this way, so this now is a
physically important kind of

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representation of this
coordination complex,

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this color scale will run from
red all the way to blue.

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And at blue is where you find
regions in space where you are

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most likely to find pairs of
electrons mapped onto the value

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here, colored mapped onto the
value of an electron density

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isosurface.
Here the value of the electron

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density that is present for
every point represented on this

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surface is 0.11 electrons per
unit volume.

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And so what you see here is
that, in fact,

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the electron density does
become low as you move from the

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cobalt in the direction of any
one of the ligands.

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But the lone pair on each
ammonia is certainly polarized

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in the direction of the cobalt
center.

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And the cobalt center is not
uniform in terms of the way that

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electron density is organized
around it.

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You will see that right on the
metal center,

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we see red.
And then sort of at the corners

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of a cube, you see this yellow
or green color on the cobalt,

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and that is very significant.
And what we are going to be

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today is we are going to try to
understand what happens when you

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put a set of ligands into an
octahedral array around a

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central metal ion that has
d-electrons and d orbitals to

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play with.
And, in fact,

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what you will see is that the
oxidation state of the cobalt

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center here is +3.
That is why I am referring to

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it as a cobalt plus three ion.

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And because cobalt is in group
nine of the periodic table,

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you then know that there are
six valance electrons on the

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cobalt center that you have to
put into orbitals.

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And so what we are really
seeking to know is how can we

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get an energy-level diagram for
a system like this,

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so we will know how to put
those six electrons into that

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diagram, the ones that are
mostly localized on the cobalt,

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and make predictions on whether
the electrons should be paired

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up or not, for example.
We saw with homonuclear

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diatomics, like the dioxygen
molecule, that electrons are not

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all paired up.
Two of the electrons are

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unpaired.
And, when we have six electrons

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to put into an energy-level
diagram for the cobalt ion,

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we are going to wonder just
what the case is.

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How many energy levels are
there, and where do we put the

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electrons?
And so we are going to need to

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very briefly review what I did
at end of lecture last time.

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And that has to do with the
properties of the five d

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orbitals --
-- because, in order to answer

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the kinds of questions that we
are posing about energy levels

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of metal ions that are
transition metals,

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we are going to really need to
know very well the nodal

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properties of these d orbitals.
And so, remember the m quantum

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number here can be zero for the
d z squared.

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It can be plus one for the
d(xz).

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And plus two for the d x
squared minus y squared.

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Minus one for the d(yz)

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orbital.
And minus two for the d(xy)

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orbital.
And let's draw those.

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d z squared is an
interesting one.

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And, because it is distinct
from the other four d orbitals,

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we are going to be spending
more time on it today than the

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others.
What it has,

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these are the x,
let's say y and z axes here,

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is it has a positive lobe along
both plus and minus z.

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So it looks like a p orbital so
far, except that we have the

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same sign both in plus or minus
z.

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And then what we have is a very
interesting toroidal shape that

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goes around in the x,y-plane all
the way around in a

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cylindrically symmetric manner.
And so, if you were to look

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down z onto this orbital,
it might look something like

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this.

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Like that.
It would be cylindrically

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symmetric about z.
And that was another feature

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that I didn't get to add to this
diagram at the end of last hour,

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which is that because of this
property, this m equals zero

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that means this is a sigma
orbital with respect to the

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z-axis.
Cylindrically symmetric about

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z.
And then, let's go to the xz,

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since I am making this x,
y, and z over here.

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The d(xz) has four lobes,
and they are in between the x

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and z axes, as I am trying to
represent here.

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These two are kind of coming
out in front here,

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and these two back behind.
And then, the phases go as

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follows.
Unshaded is,

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again, plus.
And then we have the x squared

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minus y squared
orbital, which has four lobes.

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In fact, the shape of xz,
x squared minus y squared,

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yz, and xy are all the
same.

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It's just that they point in
different directions in space

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with respect to the Cartesian
coordinate axes.

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x squared minus y squared,
like d z squared,

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is an orbital whose
lobes point along the coordinate

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axes, like this.
And it is minus along y,

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as the name suggests,
and plus along x,

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as the name suggests.
That is our d x squared minus y

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squared orbital.

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And then yz is like xz,
but let me finish this part

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here.
If d z squared is

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sigma with respect to z,
and you imagine looking down z

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onto d(xz), what would that be
with respect to z?

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Anyone.
Pi.

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Thank you.
That would be pi with respect

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to z, because the yz plane is a
nodal surface for this d

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orbital, as is the x,y-plane.
And that is what we are trying

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to do here, is become familiar
with these orbital surfaces.

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And, accordingly,
now, if you look down z onto

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the d x squared minus y squared
orbital,

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you are now going to see two
nodal surfaces when you look

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down it that way.
Because you are going to see

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that there is one located over
here.

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That is a plane that contains
the z-axis, but it bisects the x

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and y axes.
And then there is another one

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over here, but that is 90
degrees to the first one.

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And that makes this one delta
with respect to z.

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Delta is when you have two
nodes that contain the z-axis

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and if we are looking down that
z-axis.

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And then, over here,
the d(yz) orbital has its four

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lobes between y and z,
like that.

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And it is positive between y
and z, and negative over here,

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as I am shading.
And this one,

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like d(xz), is pi with respect
to z.

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And then, over here,
we have a d(xy) orbital as our

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final orbital.
And what you might be able to

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guess, we have a sigma with
respect to z,

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we have a pair of pi with
respect to z for m equals plus

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and minus one.
We also must have a pair of

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delta with respect to z for m
equals plus and minus two.

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And that means that the d(xy)
orbital, like d x squared minus

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y squared,
must lie in the x,y-plane.

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And, in order to be orthogonal
to d x squared minus y squared,

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we are going to have to rotate
it such that its four lobes

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point now between the x and y
Cartesian coordinate axes like

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this, although I am trying to
improve on that with my

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coloration.
That is like that.

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Again, this is an orbital
perpendicular to z and which has

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delta symmetry.
And now, the two nodes are,

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in fact, the xz and yz
coordinate axes.

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Indicating those nodal planes
there.

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This is certainly the basics
for what you need to know about

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the d orbitals.
And just briefly,

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I would like to switch to
Athena terminal,

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here, to show you that there
are ways for you to go ahead and

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visualize the orbitals.
And I am going to make this

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information available to you,
so that you can go ahead and do

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this yourself in order to
visualize these in a way that

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will take you right from the
equations for the orbitals to

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their graphical representation.
I think that is really

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important to get a good
understanding of orbitals.

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Let's zoom this a little bit,
so that you can begin to see

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it.

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This is a worksheet put
together that,

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in fact, contains all the
functional forms for the d

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orbitals.
Let me see.

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Where is that?
Here we go.

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And you can look at this with
Maple on Athena.

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And then, you can look at the
equation that represents the

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angular part of the wave
function for the orbital that

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you are interested in.
And then, you can go ahead and

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plot it.
And you can plot it in such a

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way that the function is
animated.

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And, rather than just seeing it
projected on a board as well as

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I can draw, you will be able to
see it drawn up graphically.

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In fact, these come from the
solution for the Schrödinger

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equation for the hydrogen atom.
And the angular part of these

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wave functions that is going to
be oh so important to us is

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something known as the set of
spherical harmonic equations.

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And that should reference you
to this issue of standing waves,

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that we have discussed.
Let's just see what we can do,

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here.
Sometimes, I am not so good at

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using Maple up in front of the
class.

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What you are seeing is we are
getting representations for d z

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squared.
Right here is a way of writing

240
00:16:17 --> 00:16:19
d z squared.
You are going to see an

241
00:16:19 --> 00:16:22
important term here,
three cosine squared theta

242
00:16:22 --> 00:16:25
minus one.
And we will come back to that

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in a moment.
That is the angular part of the

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00:16:27 --> 00:16:31
d z squared wave
function.

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00:16:31 --> 00:16:34
And then we can look at some of
these other ones.

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You will see that some of the d
orbitals come as combinations of

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real and imaginary functions
that are the solutions to the

248
00:16:43 --> 00:16:46
differential form of the
Schrödinger equation.

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00:16:46 --> 00:16:50
And then we take linear
combinations of these to get

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00:16:50 --> 00:16:54
real forms, so that we can get
plots that we can look at.

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And let's see if we can get d z
squared,

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00:16:58 --> 00:17:00
here.
There it is.

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00:17:00 --> 00:17:04
There is a picture of d z
squared.

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00:17:04 --> 00:17:07
And you see,
if you are using this Maple

255
00:17:07 --> 00:17:12
worksheet, that you can actually
rotate that around and animate

256
00:17:12 --> 00:17:16
it a little bit.
You see that we have this torus

257
00:17:16 --> 00:17:20
that is in the x,y-plane.
And you have the two large

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00:17:20 --> 00:17:24
lobes that extend up along plus
and minus z.

259
00:17:24 --> 00:17:29
And so I am going to encourage
you to go ahead and look at that

260
00:17:29 --> 00:17:35
worksheet, which will be
available from our website.

261
00:17:35 --> 00:17:37
Go ahead and look at some of
the functions.

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00:17:37 --> 00:17:41
And if some of you are
interested in higher orbitals,

263
00:17:41 --> 00:17:44
the f orbitals are also
available in this worksheet.

264
00:17:44 --> 00:17:48
So you can visualize the f
orbitals, that are important for

265
00:17:48 --> 00:17:51
understanding the chemistry of
elements like,

266
00:17:51 --> 00:17:54
for example,
uranium, which is a little bit

267
00:17:54 --> 00:17:57
beyond the scope of 5.112.
Now let's switch to the

268
00:17:57 --> 00:18:01
document camera.
And, if you could,

269
00:18:01 --> 00:18:06
I would like you to make this
part big.

270
00:18:06 --> 00:18:11
This is a table in your
textbook that has the angular

271
00:18:11 --> 00:18:18
part of the wave functions for
various hydrogen-like orbitals.

272
00:18:18 --> 00:18:23
And this is the part that I am
most interested in,

273
00:18:23 --> 00:18:28
over here.
If you could just focus in on

274
00:18:28 --> 00:18:34
the d orbitals over here.
When I was talking about the d

275
00:18:34 --> 00:18:38
z squared orbital a
moment ago, I was focusing on

276
00:18:38 --> 00:18:41
this term here,
this cosine squared theta minus

277
00:18:41 --> 00:18:43
one term.
And here are the other

278
00:18:43 --> 00:18:46
d-orbitals.
These are the descriptors for

279
00:18:46 --> 00:18:49
the d-orbitals,
zy, yz, xz, x squared minus y

280
00:18:49 --> 00:18:52
squared,
and z squared.

281
00:18:52 --> 00:18:55
And I will need to refer to
this in a moment,

282
00:18:55 --> 00:18:59
so we will leave this up.
The reason why I have this

283
00:18:59 --> 00:19:03
arrow written into my book here
is because these are backwards

284
00:19:03 --> 00:19:06
in the text.
This one is actually x squared

285
00:19:06 --> 00:19:10
minus y squared,
and this one is actually the xy

286
00:19:10 --> 00:19:13
orbital.
We figured that out last year

287
00:19:13 --> 00:19:17
when we were doing this lecture.
And so your book isn't always

288
00:19:17 --> 00:19:19
right.
You should make sure you check.

289
00:19:19 --> 00:19:22
The same is certainly true of
your instructor,

290
00:19:22 --> 00:19:24
but we will try not to mislead
you.

291
00:19:24 --> 00:19:30
And so now here is the approach
that we are going to take.

292
00:19:30 --> 00:19:40


293
00:19:40 --> 00:19:42
Let's say that we have our
coordinate system,

294
00:19:42 --> 00:19:46
and we want to know how to
evaluate one of these d orbital

295
00:19:46 --> 00:19:49
wave functions for a particular
point in space.

296
00:19:49 --> 00:19:53
Actually, we are going to want
to evaluate the square of the

297
00:19:53 --> 00:19:56
wave function.
And so we are going to make use

298
00:19:56 --> 00:19:58
of a Cartesian coordinate
system.

299
00:19:58 --> 00:20:02
And we are going to express
things in terms of polar

300
00:20:02 --> 00:20:04
coordinates.

301
00:20:04 --> 00:20:10


302
00:20:10 --> 00:20:15
Here is a sphere projected onto
our Cartesian coordinate system.

303
00:20:15 --> 00:20:20
And some of you will be very
familiar with this.

304
00:20:20 --> 00:20:25
We are going to say if you are
at a point here in space,

305
00:20:25 --> 00:20:31
then we can describe that point
in space by a set of variables

306
00:20:31 --> 00:20:35
which will be r,
theta, and phi.

307
00:20:35 --> 00:20:42
And here is our angle theta.
And if we drop down to a

308
00:20:42 --> 00:20:49
perpendicular on the x,y-plane
from our point,

309
00:20:49 --> 00:20:58
then we are going to define phi
as being from the x-axis and

310
00:20:58 --> 00:21:04
going over in the direction of
y.

311
00:21:04 --> 00:21:08
And so, when you look at the d
orbital wave functions,

312
00:21:08 --> 00:21:13
you see that they are all
written here in terms of just

313
00:21:13 --> 00:21:16
theta and phi,
and not in terms of r,

314
00:21:16 --> 00:21:20
which is this distance here
from the nucleus,

315
00:21:20 --> 00:21:26
from the center of this metal
ion that we are talking about.

316
00:21:26 --> 00:21:30
This would be the r.
And what we are going to do,

317
00:21:30 --> 00:21:35
is if this point on the surface
of our sphere represents one of

318
00:21:35 --> 00:21:38
our ligand atoms,
so think back to that complex

319
00:21:38 --> 00:21:43
we were describing a few moments
ago, cobalt with three ammonia

320
00:21:43 --> 00:21:46
ligands and three chloride
ligands, we are going to

321
00:21:46 --> 00:21:50
approximate each of those six
ligands by a point on the

322
00:21:50 --> 00:21:54
surface of our sphere.
And then, we are going to say,

323
00:21:54 --> 00:21:57
if there is a metal at the
center, that cobalt ion in

324
00:21:57 --> 00:22:01
particular, what is the
probability of finding a d

325
00:22:01 --> 00:22:05
electron where that ligand atom
is?

326
00:22:05 --> 00:22:08
And, in order to do that,
we are going to need to be able

327
00:22:08 --> 00:22:11
to evaluate the wave function.
And we are making the

328
00:22:11 --> 00:22:14
assumption, for simplicity,
that this is a perfect sphere,

329
00:22:14 --> 00:22:17
and also that our coordination
geometry is a perfect

330
00:22:17 --> 00:22:19
octahedron.
And so we are not going to

331
00:22:19 --> 00:22:22
consider r, because r is going
to be the same everywhere.

332
00:22:22 --> 00:22:25
It is just a perfect sphere
with one value of r,

333
00:22:25 --> 00:22:29
no matter which ligand position
we are looking at.

334
00:22:29 --> 00:22:33
And so that means we can focus
in just on these which are the

335
00:22:33 --> 00:22:36
angular part of the wave
function for the molecule in

336
00:22:36 --> 00:22:39
question.
And so let's see what this

337
00:22:39 --> 00:22:43
means with respect to d z
squared.

338
00:22:43 --> 00:22:54


339
00:22:54 --> 00:22:57
The probability of finding an
electron at some point in space

340
00:22:57 --> 00:23:01
in a particular atomic orbital
is proportional to the square of

341
00:23:01 --> 00:23:05
that atomic orbital at that
point in space.

342
00:23:05 --> 00:23:07
That is our probability
density.

343
00:23:07 --> 00:23:10
And we encountered that very
early in the semester.

344
00:23:10 --> 00:23:14
Now, we are going to make use
of that to make a plot.

345
00:23:14 --> 00:23:18
We are going to plot this
probability density of finding

346
00:23:18 --> 00:23:22
electron in d z squared
as a function of

347
00:23:22 --> 00:23:25
theta.
Why can I do that ignoring phi?

348
00:23:25 --> 00:23:29
Well, that is because if you
look at d z squared,

349
00:23:29 --> 00:23:33
there is no phi in that
equation.

350
00:23:33 --> 00:23:35
Why is that?
That is because d z squared

351
00:23:35 --> 00:23:39
is cylindrically
symmetric about z.

352
00:23:39 --> 00:23:42
And look at how we defined phi.
It is as you go in the

353
00:23:42 --> 00:23:45
x,y-plane around starting from
x.

354
00:23:45 --> 00:23:49
Because of the sigma symmetry
of d z squared with respect to

355
00:23:49 --> 00:23:53
z, there is no phi dependence of
this wave function.

356
00:23:53 --> 00:23:57
And that should appear here in
the angular form of the

357
00:23:57 --> 00:24:01
description of the d z squared
orbital.

358
00:24:01 --> 00:24:06
But we do know that d z squared
does depend on theta,

359
00:24:06 --> 00:24:10
because theta starts out
somewhere here along,

360
00:24:10 --> 00:24:15
let's say initially theta
equals zero would be right on

361
00:24:15 --> 00:24:19
the positive z-axis.
And then, as we keep a constant

362
00:24:19 --> 00:24:23
r and we sweep down here toward
the x,y-plane,

363
00:24:23 --> 00:24:28
the d z squared
probability density is dropping

364
00:24:28 --> 00:24:32
off.
And then at some point,

365
00:24:32 --> 00:24:36
when we get to the node here,
and we are going to be

366
00:24:36 --> 00:24:39
interested in that,
it goes to zero,

367
00:24:39 --> 00:24:44
because that is what happens on
nodes, as we continue down

368
00:24:44 --> 00:24:50
toward the x,y-plane past that
node, it is going to be nonzero

369
00:24:50 --> 00:24:55
again and rise up as we approach
this smaller torus in the

370
00:24:55 --> 00:24:57
x,y-plane.
Smaller, that is,

371
00:24:57 --> 00:25:03
than the big lobes that extend
up along z and down along minus

372
00:25:03 --> 00:25:05
z.
Let's represent that

373
00:25:05 --> 00:25:08
graphically.

374
00:25:08 --> 00:25:14


375
00:25:14 --> 00:25:20
Coming down in theta from theta
equals zero to some value here,

376
00:25:20 --> 00:25:26
we are going to be interested
in just what that value is.

377
00:25:26 --> 00:25:32
And then, rising up again to
theta is equal to pi over two.

378
00:25:32 --> 00:25:36
Do you see that?
This is another way of

379
00:25:36 --> 00:25:43
displaying this property.
This is at constant r --

380
00:25:43 --> 00:25:48


381
00:25:48 --> 00:25:50
-- and varying theta.

382
00:25:50 --> 00:25:56


383
00:25:56 --> 00:25:59
First of all,
how can we find out at what

384
00:25:59 --> 00:26:04
value theta goes to zero?
Well, we look up here at the

385
00:26:04 --> 00:26:08
functional form of the d z
squared.

386
00:26:08 --> 00:26:12
We get nodal properties of d z
squared.

387
00:26:12 --> 00:26:22


388
00:26:22 --> 00:26:25
We can pretty quickly see that
in each of these,

389
00:26:25 --> 00:26:30
we have a factor leading out in
front, which is a normalization

390
00:26:30 --> 00:26:35
factor that assures us that the
sum integrated over all space of

391
00:26:35 --> 00:26:39
this wave function will come out
to be one.

392
00:26:39 --> 00:26:42
If there is an electron in that
orbital somewhere,

393
00:26:42 --> 00:26:45
the probability of finding that
electron somewhere in space will

394
00:26:45 --> 00:26:47
be one.
We have these normalizing

395
00:26:47 --> 00:26:51
constants out in front that
allow for that and ensure that

396
00:26:51 --> 00:26:54
that is the case.
But where we actually find the

397
00:26:54 --> 00:26:56
angular dependence is in that
second term.

398
00:26:56 --> 00:26:59
Here it is three cosine squared
theta minus one.

399
00:26:59 --> 00:27:02


400
00:27:02 --> 00:27:10


401
00:27:10 --> 00:27:14
And what we want to do is say,
when does this function go to

402
00:27:14 --> 00:27:17
zero?
Because when that goes to zero,

403
00:27:17 --> 00:27:22
we will have this angle here.
If we say, let this equal zero,

404
00:27:22 --> 00:27:26
we are looking for the value of
the angle theta which

405
00:27:26 --> 00:27:32
corresponds to the node of the d
z squared orbital.

406
00:27:32 --> 00:27:35
Remember, we mentioned this
last time, because of this

407
00:27:35 --> 00:27:39
cylindrical symmetry of the d z
squared orbital,

408
00:27:39 --> 00:27:44
this really is a conical nodal
surface that is above and below

409
00:27:44 --> 00:27:47
the plane here.
If you are anywhere on that

410
00:27:47 --> 00:27:51
cone, either in plus or minus z,
the value of that d z squared

411
00:27:51 --> 00:27:55
orbital is zero.
And so we are setting it equal

412
00:27:55 --> 00:28:02
to zero to find the angle theta.
And we can rearrange this and

413
00:28:02 --> 00:28:07
say that cosine squared theta is
equal to one over three.

414
00:28:07 --> 00:28:14
And, if we go ahead
and solve that,

415
00:28:14 --> 00:28:19
this comes out to the arccosine
of root three over three.

416
00:28:19 --> 00:28:25
And the other possibility that

417
00:28:25 --> 00:28:31
satisfies that relation is pi
minus the arccosine of root

418
00:28:31 --> 00:28:37
three over three

419
00:28:37 --> 00:28:41
And so what that means is that
this relation,

420
00:28:41 --> 00:28:48
here, gives us the angle for
that cone in the plus z axis.

421
00:28:48 --> 00:28:54
And then this one down here,
pi minus arccosine root three

422
00:28:54 --> 00:29:01
over three,
gives us the angle for

423
00:29:01 --> 00:29:05
that cone down in the minus
z-axis.

424
00:29:05 --> 00:29:09
And what is this?
This is, in degrees,

425
00:29:09 --> 00:29:14
something like 54.476 dot,
dot, dot degrees,

426
00:29:14 --> 00:29:18
approximately.
That is just how far down you

427
00:29:18 --> 00:29:27
are from the z-axis when you hit
that nodal surface of z.

428
00:29:27 --> 00:29:29
And that number,
you are going to see,

429
00:29:29 --> 00:29:32
is kind of a magical number in
chemistry.

430
00:29:32 --> 00:29:37
And it will hearken back to
some of the things that we have

431
00:29:37 --> 00:29:39
been taking about,
recently.

432
00:29:39 --> 00:29:44
And, in order to get to that
point, I am going to need to now

433
00:29:44 --> 00:29:48
talk about our ligands again
with respect to d z squared.

434
00:29:48 --> 00:29:50


435
00:29:50 --> 00:30:07


436
00:30:07 --> 00:30:10
We are considering,
now, an octahedral metal

437
00:30:10 --> 00:30:11
complex.

438
00:30:11 --> 00:30:20


439
00:30:20 --> 00:30:24
This is the type of
Werner-esque complex that we

440
00:30:24 --> 00:30:29
talked about last time.
You are going to see that we

441
00:30:29 --> 00:30:35
are going to have ligands.
We are not really specifying

442
00:30:35 --> 00:30:38
them.
We are numbering them and

443
00:30:38 --> 00:30:43
locating them at positions one,
two, three, four,

444
00:30:43 --> 00:30:50
five, and six relative to our
metal center at the middle.

445
00:30:50 --> 00:30:54
And what you might begin to
realize is that in order to find

446
00:30:54 --> 00:30:58
out what our energy-level
diagram will be that refers only

447
00:30:58 --> 00:31:02
to the five d orbitals on the
cobalt center,

448
00:31:02 --> 00:31:06
or whatever metal center is at
the middle of this ion,

449
00:31:06 --> 00:31:10
what we are going to have to do
is evaluate the square of the

450
00:31:10 --> 00:31:15
wave function at each of these
ligand positions.

451
00:31:15 --> 00:31:19
And we are assuming that all
the ligands are equivalent.

452
00:31:19 --> 00:31:25
If you think of the ligand as
an electron or as a point charge

453
00:31:25 --> 00:31:30
in space, then you can imagine
that if we have an electron in d

454
00:31:30 --> 00:31:35
z squared up here,
that it is going to interact

455
00:31:35 --> 00:31:40
strongly and very repulsively
with a point charge that would

456
00:31:40 --> 00:31:45
be located at position one.
Whereas, if we instead had a

457
00:31:45 --> 00:31:49
point charge located at that
theta angle of 54 point whatever

458
00:31:49 --> 00:31:53
over there that we solved for,
since that is on the node of d

459
00:31:53 --> 00:31:56
z squared,
that would be the least

460
00:31:56 --> 00:32:00
possible repulsive interaction
that you could get between an

461
00:32:00 --> 00:32:04
electron and an electron in d z
squared because it is

462
00:32:04 --> 00:32:07
on the node.
And so what we would like to do

463
00:32:07 --> 00:32:10
is to go ahead and solve for
some of these things.

464
00:32:10 --> 00:32:13
Essentially,
what we seek --

465
00:32:13 --> 00:32:25


466
00:32:25 --> 00:32:28
-- is an energy-level diagram.

467
00:32:28 --> 00:32:33


468
00:32:33 --> 00:32:35
And so let's write this,
up here.

469
00:32:35 --> 00:32:39
This is five over 16pi.
This is d z squared

470
00:32:39 --> 00:32:43
that I am writing up.
Three cosine squared theta

471
00:32:43 --> 00:32:45
minus one.

472
00:32:45 --> 00:32:50
And because we are talking
about the probability density of

473
00:32:50 --> 00:32:54
finding an electron at a
particular point in space,

474
00:32:54 --> 00:32:58
which will mean a particular
theta value in our case for d z

475
00:32:58 --> 00:33:02
squared,
we are talking about that wave

476
00:33:02 --> 00:33:07
function squared.
What we need to do is evaluate

477
00:33:07 --> 00:33:10
it.
We have already evaluated it

478
00:33:10 --> 00:33:14
where the node is,
but we would like to evaluate

479
00:33:14 --> 00:33:18
it at position one.
Because we have a ligand at

480
00:33:18 --> 00:33:21
position one,
and we need to know what the

481
00:33:21 --> 00:33:26
relative response will be of d z
squared to a ligand

482
00:33:26 --> 00:33:32
along position one versus the
other five positions.

483
00:33:32 --> 00:33:35
We can say something about that
by symmetry, already.

484
00:33:35 --> 00:33:38
And so we are going to find out
that the value that you get for

485
00:33:38 --> 00:33:42
evaluating d z squared
at position one is

486
00:33:42 --> 00:33:45
the same as you would get for
evaluating it at position six.

487
00:33:45 --> 00:33:49
And that is because the big
lobes of d z squared

488
00:33:49 --> 00:33:52
are along plus and minus z.
That is where ligands one and

489
00:33:52 --> 00:33:54
six are.
And then, because the torus is

490
00:33:54 --> 00:33:58
also cylindrically symmetric,
two, three, four and five have

491
00:33:58 --> 00:34:02
the same value when you evaluate
d z squared at those

492
00:34:02 --> 00:34:06
positions.
But that value is smaller than

493
00:34:06 --> 00:34:10
along z, as illustrated by this
graph over here.

494
00:34:10 --> 00:34:14
But we just want to know,
how much smaller.

495
00:34:14 --> 00:34:30


496
00:34:30 --> 00:34:33
And so this is position one.
We get a value of

497
00:34:33 --> 00:34:36
five-quarters.
And down here,

498
00:34:36 --> 00:34:39
position two,
a value of five-sixteenths.

499
00:34:39 --> 00:34:42
I am leaving off a factor of
pi.

500
00:34:42 --> 00:34:45
Please don't be concerned by
that.

501
00:34:45 --> 00:34:50
But these are the relative
values that you get when you

502
00:34:50 --> 00:34:54
evaluate this function at ligand
positions one and two,

503
00:34:54 --> 00:34:59
which is all we need to do
because of the symmetry of this

504
00:34:59 --> 00:35:05
because that value for position
one is also true for position

505
00:35:05 --> 00:35:09
six.
So we have positions one and

506
00:35:09 --> 00:35:12
six.
And then, this is also three,

507
00:35:12 --> 00:35:17
four, and five.
By doing two quick evaluations

508
00:35:17 --> 00:35:21
at two different theta
positions, position one,

509
00:35:21 --> 00:35:25
of course, theta is equal to
zero, up here.

510
00:35:25 --> 00:35:32
And we evaluate that squared
function for theta equals zero.

511
00:35:32 --> 00:35:36
And we get five-fourths pi,
but I am leaving off the pi.

512
00:35:36 --> 00:35:38
And down here,
at position two,

513
00:35:38 --> 00:35:42
we evaluate this for theta
equals pi over two because two,

514
00:35:42 --> 00:35:45
three, four,
and five are all in the

515
00:35:45 --> 00:35:49
x,y-plane at 90 degrees to z.
So the value of theta anywhere

516
00:35:49 --> 00:35:53
for those four ligands would be
pi over two.

517
00:35:53 --> 00:35:57
You evaluate this function for
pi over two, and you will get

518
00:35:57 --> 00:36:04
five-sixteenths pi.
And I have just left off the

519
00:36:04 --> 00:36:06
pi.
That is useful.

520
00:36:06 --> 00:36:12
Let's take this one up to the
top.

521
00:36:12 --> 00:36:23


522
00:36:23 --> 00:36:27
Here we have done one of the d
orbitals.

523
00:36:27 --> 00:36:35
This is d z squared
for all six ligand positions.

524
00:36:35 --> 00:36:40


525
00:36:40 --> 00:36:49
Now, let's do d x squared minus
y squared for

526
00:36:49 --> 00:36:54
all six ligand positions.

527
00:36:54 --> 00:36:58


528
00:36:58 --> 00:37:00
First of all,
d x squared minus y squared

529
00:37:00 --> 00:37:05
has a pretty
interesting relationship with

530
00:37:05 --> 00:37:09
ligands one and six.
What is that relationship?

531
00:37:09 --> 00:37:17


532
00:37:17 --> 00:37:19
Zero.
Because x squared minus y

533
00:37:19 --> 00:37:23
squared has two
nodal surfaces that intersect

534
00:37:23 --> 00:37:26
along the z-axis.
And so those ligands,

535
00:37:26 --> 00:37:30
four and six,
lie on a nodal surface.

536
00:37:30 --> 00:37:35
And so, we know that that is
going to be equal to zero.

537
00:37:35 --> 00:37:38
If, however,
we go ahead and evaluate the

538
00:37:38 --> 00:37:44
square of d x squared minus y
squared, let me just write it

539
00:37:44 --> 00:37:46
up.
And I have to switch it,

540
00:37:46 --> 00:37:49
since they are wrong in the
book.

541
00:37:49 --> 00:37:55
15 over 16pi square root sine
squared theta cosine two phi.

542
00:37:55 --> 00:38:01


543
00:38:01 --> 00:38:03
We have that.
Now, just as we do,

544
00:38:03 --> 00:38:07
we are converting an orbital
angular property into a

545
00:38:07 --> 00:38:09
probability density by squaring
it.

546
00:38:09 --> 00:38:13
This is what we normally do.
When you see pictures of

547
00:38:13 --> 00:38:18
orbitals, they are representing
the square of the wave function

548
00:38:18 --> 00:38:21
in space.
We need to evaluate this as a

549
00:38:21 --> 00:38:26
function of theta and phi in
order to find out what --

550
00:38:26 --> 00:38:28
The first two,
one and six,

551
00:38:28 --> 00:38:31
we did by inspection,
but what about positions two,

552
00:38:31 --> 00:38:33
three, four,
and five?

553
00:38:33 --> 00:38:37
In the case of those four,
you can see that where they lie

554
00:38:37 --> 00:38:42
in the x,y-plane with respect to
the x squared minus y squared

555
00:38:42 --> 00:38:46
orbital is all
identical to each other by

556
00:38:46 --> 00:38:49
symmetry.
Because x squared minus y

557
00:38:49 --> 00:38:53
squared has lobes that extend
along x and plus x and y and

558
00:38:53 --> 00:38:56
minus y.
And that is where all these

559
00:38:56 --> 00:38:59
ligands lie, at positions two,
three, four,

560
00:38:59 --> 00:39:06
and five.
We know that theta is equal to

561
00:39:06 --> 00:39:11
what?
It is going to be pi over two.

562
00:39:11 --> 00:39:17
And phi, of course,
can be zero pi over two pi,

563
00:39:17 --> 00:39:24
and three pi over two for any
of those positions.

564
00:39:24 --> 00:39:32
And what we will find is that
this evaluates --

565
00:39:32 --> 00:39:37
For any of those,
let's just use phi equals zero.

566
00:39:37 --> 00:39:43
This evaluates as 15 over 16pi.
And I am dropping the pi.

567
00:39:43 --> 00:39:49
Now we have d z squared
and d x squared minus

568
00:39:49 --> 00:39:54
y squared
evaluated for all six ligand

569
00:39:54 --> 00:39:59
positions.
And then, let's make a table of

570
00:39:59 --> 00:40:01
this.

571
00:40:01 --> 00:40:23


572
00:40:23 --> 00:40:27
z squared.
x squared minus y squared.

573
00:40:27 --> 00:40:29
Let's do xz,

574
00:40:29 --> 00:40:32
yz, and xy.
And here are our ligand

575
00:40:32 --> 00:40:35
positions.
And in the table here,

576
00:40:35 --> 00:40:41
we are going to put what these
relative evaluated squared wave

577
00:40:41 --> 00:40:45
functions are.
And the biggest one of all is

578
00:40:45 --> 00:40:51
this one here that we got at
position one for d z squared,

579
00:40:51 --> 00:40:56
which is
five-fourths.

580
00:40:56 --> 00:41:00
I am going to divide everybody
through by five-fourths,

581
00:41:00 --> 00:41:03
so that I can make our biggest
value equal to one for

582
00:41:03 --> 00:41:07
simplicity here.
And then, when we do that,

583
00:41:07 --> 00:41:11
you are going to see that two
gives a value of a quarter.

584
00:41:11 --> 00:41:14
That is position two,
down in the torus.

585
00:41:14 --> 00:41:18
Three, down in the torus,
is one-quarter relative to that

586
00:41:18 --> 00:41:20
one.
Four is one-quarter.

587
00:41:20 --> 00:41:23
Five is one-quarter.
And six, down in minus z,

588
00:41:23 --> 00:41:26
is one.
So we have evaluated the d z

589
00:41:26 --> 00:41:30
squared orbital
squared at the ligand positions

590
00:41:30 --> 00:41:35
one through six.
And these are the relative

591
00:41:35 --> 00:41:40
values that we got for the
probability of finding an

592
00:41:40 --> 00:41:46
electron at that point in space,
given a constant value of r.

593
00:41:46 --> 00:41:51
And x squared minus y squared,

594
00:41:51 --> 00:41:56
we got for positions one and
six zero, we just said that.

595
00:41:56 --> 00:42:02
And then, on the same scale
here, we get three-quarter,

596
00:42:02 --> 00:42:04
three-quarter,
three-quarter,

597
00:42:04 --> 00:42:09
and three-quarter.
And then for xz,

598
00:42:09 --> 00:42:15
yz, and xy, we have gone
through these two steps to do z

599
00:42:15 --> 00:42:21
squared and x squared
minus y squared explicitly.

600
00:42:21 --> 00:42:25
Now, we have to look at where

601
00:42:25 --> 00:42:31
ligands one through six are
relative to the nodes of xz or

602
00:42:31 --> 00:42:37
yz and xy.
And what we will find is that,

603
00:42:37 --> 00:42:44
in each case,
these ligands lie on nodal

604
00:42:44 --> 00:42:48
surfaces of xz,
yz, and xy.

605
00:42:48 --> 00:42:53
This is zero,
zero, zero, zero,

606
00:42:53 --> 00:42:57
zero, zero, zero,
and so on.

607
00:42:57 --> 00:43:01
Okay?
All the ligands,

608
00:43:01 --> 00:43:04
one through six,
lie on nodal planes of xz,

609
00:43:04 --> 00:43:06
yz, and xy.
Only four of the ligands

610
00:43:06 --> 00:43:10
interact with x squared minus y
squared,

611
00:43:10 --> 00:43:14
ligands two through five,
because those are the ones that

612
00:43:14 --> 00:43:17
lie in the x,y-plane.
And they interact strongly,

613
00:43:17 --> 00:43:20
this relative value of
three-quarters,

614
00:43:20 --> 00:43:25
but not as strongly as the two
ligands in positions one and six

615
00:43:25 --> 00:43:30
interact with the big lobes of d
z squared.

616
00:43:30 --> 00:43:32
And then, also,
ligands two through five,

617
00:43:32 --> 00:43:36
which lie in the x,y-plane,
interact with the torus of d z

618
00:43:36 --> 00:43:40
squared,
but to a much smaller extent

619
00:43:40 --> 00:43:44
because the torus does not have
as great a radial extent.

620
00:43:44 --> 00:43:48
And these ligands are all at
the same radius out from the

621
00:43:48 --> 00:43:51
metal center.
And so what this corresponds to

622
00:43:51 --> 00:43:54
is now an energy-level diagram
as follows.

623
00:43:54 --> 00:43:58
And this is for an octahedral
complex, --

624
00:43:58 --> 00:44:03


625
00:44:03 --> 00:44:07
-- where we have relative
energy units of zero,

626
00:44:07 --> 00:44:11
one, two, and three.
And what we have to do is say

627
00:44:11 --> 00:44:17
that the amount an electron in d
z squared would be

628
00:44:17 --> 00:44:22
repelled simultaneously by
electrons in positions one

629
00:44:22 --> 00:44:27
through six would be the sum of
these values.

630
00:44:27 --> 00:44:31
And so we add that up and get,
in fact for d z squared,

631
00:44:31 --> 00:44:34
up here,
a three.

632
00:44:34 --> 00:44:38
And, interestingly,
for d x squared minus y

633
00:44:38 --> 00:44:42
squared,
if we take a sum of the four

634
00:44:42 --> 00:44:46
interactions that we found that
are non-zero,

635
00:44:46 --> 00:44:50
we also get a three for d x
squared minus y squared.

636
00:44:50 --> 00:44:55
So then the net of all their
interactions is the same for x

637
00:44:55 --> 00:45:00
squared minus y squared and z
squared.

638
00:45:00 --> 00:45:03
And then down here,
we found that these three

639
00:45:03 --> 00:45:07
orbitals, xz,
yz, and xy, where the ligands

640
00:45:07 --> 00:45:11
at positions one through six lie
on their nodal surfaces,

641
00:45:11 --> 00:45:14
lie right on their nodal
surfaces.

642
00:45:14 --> 00:45:17
And so the number evaluates to
zero.

643
00:45:17 --> 00:45:22
And so a wave function squared
will evaluate to zero any time

644
00:45:22 --> 00:45:26
you are looking at a position
that is on one of its nodal

645
00:45:26 --> 00:45:29
surfaces.
It evaluates to zero.

646
00:45:29 --> 00:45:34
And so we have d(xz),
d(yz), and d(xy).

647
00:45:34 --> 00:45:39
And then, a diagram like this,
which is a d-orbital splitting

648
00:45:39 --> 00:45:40
diagram --

649
00:45:40 --> 00:45:53


650
00:45:53 --> 00:45:56
-- is associated with a couple
of different parameters.

651
00:45:56 --> 00:46:00
One is, down here we have a
triply degenerate energy level.

652
00:46:00 --> 00:46:04
Previously, we had only seen
doubly degenerate energy levels.

653
00:46:04 --> 00:46:08
Now we have a triply degenerate
on composed of xz,

654
00:46:08 --> 00:46:11
yz, and xy.
And that level we are going to

655
00:46:11 --> 00:46:14
be calling t(2g).
And then, up here,

656
00:46:14 --> 00:46:19
we have a doubly degenerate
level that will get the label

657
00:46:19 --> 00:46:22
e(g).
And then, whatever the value of

658
00:46:22 --> 00:46:25
the splitting of these two
energy levels,

659
00:46:25 --> 00:46:29
one triply and one doubly
degenerate, we are going to give

660
00:46:29 --> 00:46:34
that the label delta O for
octahedral.

661
00:46:34 --> 00:46:38
And at the beginning of next
hour, I will say more about

662
00:46:38 --> 00:46:42
tables like this.
And I will show you how you can

663
00:46:42 --> 00:46:46
actually figure out d-orbital
splitting diagrams for other

664
00:46:46 --> 00:46:51
coordination geometries and how
they compare for a couple of the

665
00:46:51 --> 46:54
most popular coordination
geometries.