1
00:00:00 --> 00:00:04
The following content is
provided by MIT Open Courseware
2
00:00:04 --> 00:00:06
under a Creative Commons
license.
3
00:00:06 --> 00:00:10
Additional information about
our license and MIT Open
4
00:00:10 --> 00:00:15
Courseware in general is
available at ocw.mit.edu.
5
00:00:15 --> 00:00:20
Some of you will recognize this
molecule that I have rotating up
6
00:00:20 --> 00:00:26
on the screen because it is one
of the coordination complexes of
7
00:00:26 --> 00:00:30
Werner that we talked about last
time.
8
00:00:30 --> 00:00:33
And Werner, of course,
was well-known.
9
00:00:33 --> 00:00:38
And some of you probably read
is biography over the last few
10
00:00:38 --> 00:00:43
days on the Nobel Prize website.
Well-known for his theory that
11
00:00:43 --> 00:00:48
broke down the preconceived
notions that had been prevailing
12
00:00:48 --> 00:00:52
at that time concerning the
structure of systems that have
13
00:00:52 --> 00:00:57
unusual salt-like behavior in
some cases, but that contain the
14
00:00:57 --> 00:01:01
3D elements.
I am focusing on 3D,
15
00:01:01 --> 00:01:06
but this is also true of 4D and
5D elements, the transition
16
00:01:06 --> 00:01:09
metals.
Elements like titanium,
17
00:01:09 --> 00:01:11
vanadium, chromium,
iron, cobalt,
18
00:01:11 --> 00:01:14
nickel, etc.
This molecule here,
19
00:01:14 --> 00:01:19
that you see spinning up on the
screen, is being represented
20
00:01:19 --> 00:01:24
here on the screen in one of
these kind of arbitrary forms,
21
00:01:24 --> 00:01:29
wherein the cobalt center at
the middle of the molecule is
22
00:01:29 --> 00:01:34
just a round ball of an
arbitrary size.
23
00:01:34 --> 00:01:38
And then, this particular
molecule that contains a cobalt
24
00:01:38 --> 00:01:43
three ion is
surrounded by a compliment of
25
00:01:43 --> 00:01:46
six ligands.
We have three chloride ligands,
26
00:01:46 --> 00:01:52
that because this program chose
to do so colored those chlorides
27
00:01:52 --> 00:01:55
the same as the cobalt,
not what I would do,
28
00:01:55 --> 00:02:00
and put the three ammonia
ligands, colored the ammonia
29
00:02:00 --> 00:02:04
nitrogens in dark blue,
and the ammonia hydrogens in
30
00:02:04 --> 00:02:08
white.
And we talked about the fact
31
00:02:08 --> 00:02:12
that these transition elements,
in fact, these 3+ ions of
32
00:02:12 --> 00:02:16
metals like cobalt,
can behave simultaneously as
33
00:02:16 --> 00:02:20
Lewis acids toward a multitude
of Lewis bases,
34
00:02:20 --> 00:02:23
here six.
This is a times six Lewis acid
35
00:02:23 --> 00:02:28
in the middle and six Lewis
bases, which are three chloride
36
00:02:28 --> 00:02:32
ions and three ammonia molecules
oriented around it,
37
00:02:32 --> 00:02:37
in an arrangement that is
quasi-octahedral.
38
00:02:37 --> 00:02:41
Because the positions of the
three nitrogens of the ammonias
39
00:02:41 --> 00:02:45
that are interacting with the
cobalt center and the three
40
00:02:45 --> 00:02:49
chloride ions are located near
the vertices of a regular
41
00:02:49 --> 00:02:52
octahedron.
So we will be taking use of the
42
00:02:52 --> 00:02:56
geometry of molecules like this
in discussing electronic
43
00:02:56 --> 00:03:00
structure properties of these
molecules today.
44
00:03:00 --> 00:03:02
We will be doing that beginning
today.
45
00:03:02 --> 00:03:06
And that will be important to
understanding the magnetism and
46
00:03:06 --> 00:03:09
the color and also the reactions
of molecules like this.
47
00:03:09 --> 00:03:13
Now, one other arbitrary thing
that this program did is it
48
00:03:13 --> 00:03:17
chose not to draw lines between
the cobalt and any of the six
49
00:03:17 --> 00:03:18
ligands.
But normally,
50
00:03:18 --> 00:03:21
when you see these molecules
drawn in textbooks,
51
00:03:21 --> 00:03:25
you will see that the lines
drawn are the same as the lines
52
00:03:25 --> 00:03:29
between the nitrogens and these
hydrogens.
53
00:03:29 --> 00:03:33
So we would have to add six
more lines to this drawing to
54
00:03:33 --> 00:03:37
get the typical textbook
representation of a molecule
55
00:03:37 --> 00:03:40
like this.
But, still, that would be a
56
00:03:40 --> 00:03:43
somewhat arbitrary
representation.
57
00:03:43 --> 00:03:47
And so, I would like to show
you a less arbitrary
58
00:03:47 --> 00:03:51
representation.
And we will do that forthwith.
59
00:03:51 --> 00:03:55
And this one has to do with
looking at the same molecule,
60
00:03:55 --> 00:04:01
but represented as an electron
density isosurface.
61
00:04:01 --> 00:04:05
And that isosurface will be
colored according to this
62
00:04:05 --> 00:04:10
function, that tells us about
the propensity to pair electrons
63
00:04:10 --> 00:04:13
in 3D space.
And so you will recall that
64
00:04:13 --> 00:04:17
when we talk about coloring
electron density isosurfaces in
65
00:04:17 --> 00:04:22
this way, so this now is a
physically important kind of
66
00:04:22 --> 00:04:25
representation of this
coordination complex,
67
00:04:25 --> 00:04:31
this color scale will run from
red all the way to blue.
68
00:04:31 --> 00:04:34
And at blue is where you find
regions in space where you are
69
00:04:34 --> 00:04:38
most likely to find pairs of
electrons mapped onto the value
70
00:04:38 --> 00:04:42
here, colored mapped onto the
value of an electron density
71
00:04:42 --> 00:04:45
isosurface.
Here the value of the electron
72
00:04:45 --> 00:04:48
density that is present for
every point represented on this
73
00:04:48 --> 00:04:51
surface is 0.11 electrons per
unit volume.
74
00:04:51 --> 00:04:54
And so what you see here is
that, in fact,
75
00:04:54 --> 00:04:57
the electron density does
become low as you move from the
76
00:04:57 --> 00:05:02
cobalt in the direction of any
one of the ligands.
77
00:05:02 --> 00:05:05
But the lone pair on each
ammonia is certainly polarized
78
00:05:05 --> 00:05:08
in the direction of the cobalt
center.
79
00:05:08 --> 00:05:12
And the cobalt center is not
uniform in terms of the way that
80
00:05:12 --> 00:05:14
electron density is organized
around it.
81
00:05:14 --> 00:05:17
You will see that right on the
metal center,
82
00:05:17 --> 00:05:20
we see red.
And then sort of at the corners
83
00:05:20 --> 00:05:24
of a cube, you see this yellow
or green color on the cobalt,
84
00:05:24 --> 00:05:28
and that is very significant.
And what we are going to be
85
00:05:28 --> 00:05:33
today is we are going to try to
understand what happens when you
86
00:05:33 --> 00:05:37
put a set of ligands into an
octahedral array around a
87
00:05:37 --> 00:05:41
central metal ion that has
d-electrons and d orbitals to
88
00:05:41 --> 00:05:42
play with.
And, in fact,
89
00:05:42 --> 00:05:46
what you will see is that the
oxidation state of the cobalt
90
00:05:46 --> 00:05:50
center here is +3.
That is why I am referring to
91
00:05:50 --> 00:05:53
it as a cobalt plus three ion.
92
00:05:53 --> 00:05:57
And because cobalt is in group
nine of the periodic table,
93
00:05:57 --> 00:06:01
you then know that there are
six valance electrons on the
94
00:06:01 --> 00:06:06
cobalt center that you have to
put into orbitals.
95
00:06:06 --> 00:06:09
And so what we are really
seeking to know is how can we
96
00:06:09 --> 00:06:12
get an energy-level diagram for
a system like this,
97
00:06:12 --> 00:06:15
so we will know how to put
those six electrons into that
98
00:06:15 --> 00:06:19
diagram, the ones that are
mostly localized on the cobalt,
99
00:06:19 --> 00:06:22
and make predictions on whether
the electrons should be paired
100
00:06:22 --> 00:06:25
up or not, for example.
We saw with homonuclear
101
00:06:25 --> 00:06:29
diatomics, like the dioxygen
molecule, that electrons are not
102
00:06:29 --> 00:06:32
all paired up.
Two of the electrons are
103
00:06:32 --> 00:06:35
unpaired.
And, when we have six electrons
104
00:06:35 --> 00:06:39
to put into an energy-level
diagram for the cobalt ion,
105
00:06:39 --> 00:06:42
we are going to wonder just
what the case is.
106
00:06:42 --> 00:06:47
How many energy levels are
there, and where do we put the
107
00:06:47 --> 00:06:50
electrons?
And so we are going to need to
108
00:06:50 --> 00:06:54
very briefly review what I did
at end of lecture last time.
109
00:06:54 --> 00:06:58
And that has to do with the
properties of the five d
110
00:06:58 --> 00:07:02
orbitals --
-- because, in order to answer
111
00:07:02 --> 00:07:08
the kinds of questions that we
are posing about energy levels
112
00:07:08 --> 00:07:11
of metal ions that are
transition metals,
113
00:07:11 --> 00:07:16
we are going to really need to
know very well the nodal
114
00:07:16 --> 00:07:22
properties of these d orbitals.
And so, remember the m quantum
115
00:07:22 --> 00:07:27
number here can be zero for the
d z squared.
116
00:07:27 --> 00:07:31
It can be plus one for the
d(xz).
117
00:07:31 --> 00:07:38
And plus two for the d x
squared minus y squared.
118
00:07:38 --> 00:07:43
Minus one for the d(yz)
119
00:07:43 --> 00:07:48
orbital.
And minus two for the d(xy)
120
00:07:48 --> 00:07:52
orbital.
And let's draw those.
121
00:07:52 --> 00:08:00
d z squared is an
interesting one.
122
00:08:00 --> 00:08:04
And, because it is distinct
from the other four d orbitals,
123
00:08:04 --> 00:08:09
we are going to be spending
more time on it today than the
124
00:08:09 --> 00:08:11
others.
What it has,
125
00:08:11 --> 00:08:14
these are the x,
let's say y and z axes here,
126
00:08:14 --> 00:08:19
is it has a positive lobe along
both plus and minus z.
127
00:08:19 --> 00:08:24
So it looks like a p orbital so
far, except that we have the
128
00:08:24 --> 00:08:28
same sign both in plus or minus
z.
129
00:08:28 --> 00:08:32
And then what we have is a very
interesting toroidal shape that
130
00:08:32 --> 00:08:36
goes around in the x,y-plane all
the way around in a
131
00:08:36 --> 00:08:40
cylindrically symmetric manner.
And so, if you were to look
132
00:08:40 --> 00:08:44
down z onto this orbital,
it might look something like
133
00:08:44 --> 00:08:45
this.
134
00:08:45 --> 00:08:52
135
00:08:52 --> 00:08:54
Like that.
It would be cylindrically
136
00:08:54 --> 00:08:58
symmetric about z.
And that was another feature
137
00:08:58 --> 00:09:03
that I didn't get to add to this
diagram at the end of last hour,
138
00:09:03 --> 00:09:08
which is that because of this
property, this m equals zero
139
00:09:08 --> 00:09:12
that means this is a sigma
orbital with respect to the
140
00:09:12 --> 00:09:16
z-axis.
Cylindrically symmetric about
141
00:09:16 --> 00:09:19
z.
And then, let's go to the xz,
142
00:09:19 --> 00:09:22
since I am making this x,
y, and z over here.
143
00:09:22 --> 00:09:27
The d(xz) has four lobes,
and they are in between the x
144
00:09:27 --> 00:09:32
and z axes, as I am trying to
represent here.
145
00:09:32 --> 00:09:37
146
00:09:37 --> 00:09:41
These two are kind of coming
out in front here,
147
00:09:41 --> 00:09:46
and these two back behind.
And then, the phases go as
148
00:09:46 --> 00:09:48
follows.
Unshaded is,
149
00:09:48 --> 00:09:52
again, plus.
And then we have the x squared
150
00:09:52 --> 00:09:59
minus y squared
orbital, which has four lobes.
151
00:09:59 --> 00:10:03
In fact, the shape of xz,
x squared minus y squared,
152
00:10:03 --> 00:10:06
yz, and xy are all the
same.
153
00:10:06 --> 00:10:10
It's just that they point in
different directions in space
154
00:10:10 --> 00:10:13
with respect to the Cartesian
coordinate axes.
155
00:10:13 --> 00:10:18
x squared minus y squared,
like d z squared,
156
00:10:18 --> 00:10:23
is an orbital whose
lobes point along the coordinate
157
00:10:23 --> 00:10:26
axes, like this.
And it is minus along y,
158
00:10:26 --> 00:10:29
as the name suggests,
and plus along x,
159
00:10:29 --> 00:10:34
as the name suggests.
That is our d x squared minus y
160
00:10:34 --> 00:10:37
squared orbital.
161
00:10:37 --> 00:10:42
And then yz is like xz,
but let me finish this part
162
00:10:42 --> 00:10:45
here.
If d z squared is
163
00:10:45 --> 00:10:50
sigma with respect to z,
and you imagine looking down z
164
00:10:50 --> 00:10:56
onto d(xz), what would that be
with respect to z?
165
00:10:56 --> 00:11:02
166
00:11:02 --> 00:11:02
Anyone.
Pi.
167
00:11:02 --> 00:11:06
Thank you.
That would be pi with respect
168
00:11:06 --> 00:11:10
to z, because the yz plane is a
nodal surface for this d
169
00:11:10 --> 00:11:15
orbital, as is the x,y-plane.
And that is what we are trying
170
00:11:15 --> 00:11:20
to do here, is become familiar
with these orbital surfaces.
171
00:11:20 --> 00:11:24
And, accordingly,
now, if you look down z onto
172
00:11:24 --> 00:11:29
the d x squared minus y squared
orbital,
173
00:11:29 --> 00:11:34
you are now going to see two
nodal surfaces when you look
174
00:11:34 --> 00:11:39
down it that way.
Because you are going to see
175
00:11:39 --> 00:11:42
that there is one located over
here.
176
00:11:42 --> 00:11:47
That is a plane that contains
the z-axis, but it bisects the x
177
00:11:47 --> 00:11:50
and y axes.
And then there is another one
178
00:11:50 --> 00:11:55
over here, but that is 90
degrees to the first one.
179
00:11:55 --> 00:12:00
And that makes this one delta
with respect to z.
180
00:12:00 --> 00:12:06
Delta is when you have two
nodes that contain the z-axis
181
00:12:06 --> 00:12:10
and if we are looking down that
z-axis.
182
00:12:10 --> 00:12:16
And then, over here,
the d(yz) orbital has its four
183
00:12:16 --> 00:12:20
lobes between y and z,
like that.
184
00:12:20 --> 00:12:26
And it is positive between y
and z, and negative over here,
185
00:12:26 --> 00:12:31
as I am shading.
And this one,
186
00:12:31 --> 00:12:34
like d(xz), is pi with respect
to z.
187
00:12:34 --> 00:12:38
And then, over here,
we have a d(xy) orbital as our
188
00:12:38 --> 00:12:41
final orbital.
And what you might be able to
189
00:12:41 --> 00:12:45
guess, we have a sigma with
respect to z,
190
00:12:45 --> 00:12:49
we have a pair of pi with
respect to z for m equals plus
191
00:12:49 --> 00:12:53
and minus one.
We also must have a pair of
192
00:12:53 --> 00:12:59
delta with respect to z for m
equals plus and minus two.
193
00:12:59 --> 00:13:03
And that means that the d(xy)
orbital, like d x squared minus
194
00:13:03 --> 00:13:08
y squared,
must lie in the x,y-plane.
195
00:13:08 --> 00:13:13
And, in order to be orthogonal
to d x squared minus y squared,
196
00:13:13 --> 00:13:17
we are going to have to rotate
it such that its four lobes
197
00:13:17 --> 00:13:22
point now between the x and y
Cartesian coordinate axes like
198
00:13:22 --> 00:13:26
this, although I am trying to
improve on that with my
199
00:13:26 --> 00:13:30
coloration.
That is like that.
200
00:13:30 --> 00:13:35
Again, this is an orbital
perpendicular to z and which has
201
00:13:35 --> 00:13:38
delta symmetry.
And now, the two nodes are,
202
00:13:38 --> 00:13:42
in fact, the xz and yz
coordinate axes.
203
00:13:42 --> 00:13:45
Indicating those nodal planes
there.
204
00:13:45 --> 00:13:50
This is certainly the basics
for what you need to know about
205
00:13:50 --> 00:13:53
the d orbitals.
And just briefly,
206
00:13:53 --> 00:13:57
I would like to switch to
Athena terminal,
207
00:13:57 --> 00:14:02
here, to show you that there
are ways for you to go ahead and
208
00:14:02 --> 00:14:08
visualize the orbitals.
And I am going to make this
209
00:14:08 --> 00:14:12
information available to you,
so that you can go ahead and do
210
00:14:12 --> 00:14:16
this yourself in order to
visualize these in a way that
211
00:14:16 --> 00:14:21
will take you right from the
equations for the orbitals to
212
00:14:21 --> 00:14:25
their graphical representation.
I think that is really
213
00:14:25 --> 00:14:30
important to get a good
understanding of orbitals.
214
00:14:30 --> 00:14:38
Let's zoom this a little bit,
so that you can begin to see
215
00:14:38 --> 00:14:39
it.
216
00:14:39 --> 00:14:46
217
00:14:46 --> 00:14:49
This is a worksheet put
together that,
218
00:14:49 --> 00:14:54
in fact, contains all the
functional forms for the d
219
00:14:54 --> 00:14:56
orbitals.
Let me see.
220
00:14:56 --> 00:15:00
Where is that?
Here we go.
221
00:15:00 --> 00:15:02
And you can look at this with
Maple on Athena.
222
00:15:02 --> 00:15:06
And then, you can look at the
equation that represents the
223
00:15:06 --> 00:15:09
angular part of the wave
function for the orbital that
224
00:15:09 --> 00:15:12
you are interested in.
And then, you can go ahead and
225
00:15:12 --> 00:15:14
plot it.
And you can plot it in such a
226
00:15:14 --> 00:15:16
way that the function is
animated.
227
00:15:16 --> 00:15:20
And, rather than just seeing it
projected on a board as well as
228
00:15:20 --> 00:15:25
I can draw, you will be able to
see it drawn up graphically.
229
00:15:25 --> 00:15:30
In fact, these come from the
solution for the Schrödinger
230
00:15:30 --> 00:15:36
equation for the hydrogen atom.
And the angular part of these
231
00:15:36 --> 00:15:42
wave functions that is going to
be oh so important to us is
232
00:15:42 --> 00:15:48
something known as the set of
spherical harmonic equations.
233
00:15:48 --> 00:15:54
And that should reference you
to this issue of standing waves,
234
00:15:54 --> 00:16:00
that we have discussed.
Let's just see what we can do,
235
00:16:00 --> 00:16:03
here.
Sometimes, I am not so good at
236
00:16:03 --> 00:16:06
using Maple up in front of the
class.
237
00:16:06 --> 00:16:11
238
00:16:11 --> 00:16:14
What you are seeing is we are
getting representations for d z
239
00:16:14 --> 00:16:17
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
243
00:16:25 --> 00:16:27
in a moment.
That is the angular part of the
244
00:16:27 --> 00:16:31
d z squared wave
function.
245
00:16:31 --> 00:16:34
And then we can look at some of
these other ones.
246
00:16:34 --> 00:16:39
You will see that some of the d
orbitals come as combinations of
247
00:16:39 --> 00:16:43
real and imaginary functions
that are the solutions to the
248
00:16:43 --> 00:16:46
differential form of the
Schrödinger equation.
249
00:16:46 --> 00:16:50
And then we take linear
combinations of these to get
250
00:16:50 --> 00:16:54
real forms, so that we can get
plots that we can look at.
251
00:16:54 --> 00:16:58
And let's see if we can get d z
squared,
252
00:16:58 --> 00:17:00
here.
There it is.
253
00:17:00 --> 00:17:04
There is a picture of d z
squared.
254
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
258
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.
262
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.