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We have been talking about the
properties of transition metal

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complexes, coordination
complexes, in the context of a

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simple crystal field theory
model.

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And I just wanted to make
reference at the beginning,

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today, to state diagrams.

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00:00:40,000 --> 00:00:43,000
And, in the context of
octahedral complexes,

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you saw that an array of six
ligands surrounding a central

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metal ion would lead to a t two
g and what we are

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calling an eg star
manifold of orbitals on the

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metal in which we can populate
with the number of d electrons

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that we have present.
And the graph that I am making

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here is related to the idea that
for a system with only a single

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electron, so this could be,
for example,

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titanium OH two six times three
plus.

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This is a d one system.
And it can go from a ground

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state configuration of t two g
one eg star zero.

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And then, upon absorption of

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light, it can be promoted from
the ground state,

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represented here,
into the excited state.

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And the excited state has the
configuration t two g zero eg

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star one,
showing that that electron has

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been promoted through absorption
of a photon.

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00:02:04,000 --> 00:02:09,000
And this gap here between t(2g)
and eg star,

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of course, is our delta O
value.

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00:02:12,000 --> 00:02:19,000
And the axis on the bottom,
here, is ligand field strength.

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This is supposed to tell you
that different ligands give rise

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to different values of delta O
accounting for the fact that

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this d one titanium
three ion in a field

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00:02:44,000 --> 00:02:47,000
of water molecules has one
color, whereas,

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if you, in fact, look at
Ti Cl six three minus,

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you will be able to tell that

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that also is a titanium three
plus or d one

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system.
But they do not have the same

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color.
And the one absorption band

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00:03:07,000 --> 00:03:11,000
that you find in the visible
part of the spectrum that

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corresponds to promotion of an
electron from t two g

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00:03:16,000 --> 00:03:21,000
into eg star is just
moving around a little bit in

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energy because these ligands
behave differently with respect

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to what we call their ligand
field strength that they exert,

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that leads to a particular
energy gap between t(2g) and

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e(g)*.
I will get into this a little

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bit further in the moment,
but the point here is that d

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one or d nine
are special electron counts.

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00:04:05,000 --> 00:04:10,000
And the reason that they are
special is that these have a

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single absorption peak due to
what we call d-d transitions.

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They have a single d-d
transition.

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00:04:25,000 --> 00:04:29,000
And that means there is a
ground state and then just one

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excited state that the system
can be promoted to in the

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presence of impinging photons of
the right energy.

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And what happens is that if you
have more d electrons than just

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one, or d nine is
special because we treat that

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using a whole formalism,
like there is just one electron

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missing from a complete
manifold, a completely full d

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shell minus one.
With d^n systems,

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for example,
d two is much more

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complex.
And this is beyond the scope of

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5.112, but it is something that
they are treating in 5.04,

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for example.
If you go on in organic

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chemistry, you will learn how
come different states can arise,

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multiple states arise when you
have more than just a one

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electron picture.

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For example,
if you have a vanadium three

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plus ion,
there are three bands observed.

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And so the simple picture that
you have, just a t(2g) level and

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an e(g) level is correct and
maps onto the picture of the

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electronic states that are
available, as long as we are

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just talking about a one
electron picture.

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With more electrons in play,
depending on just which of the

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orbitals they go into,
they will have inter-electron

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repulsion terms that come into
play and give rise to more

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excited states.
Here, with three bands being

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observed in the visible,
that means there are three

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possible transitions and three
possible excited states that can

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be accessed.
So it is a very complicated

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picture.
And so we are going to be,

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for the purposes of this,
focusing on the one electron

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picture.
But I just wanted you to be

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aware that it does get a lot
more complex as soon as you go

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to many electron systems.
Now, let's talk just briefly

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about pairing energy.

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In the case of 3d system,
we typically find that the

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pairing energy is around 15
to 18,000 reciprocal

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centimeters, the typical energy
unit given for pairing energy.

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But then if we go to heavier
transition elements,

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to the 4d or the 5d elements,
then typically the pairing

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energy is about 8,000 to 12
reciprocal centimeters.

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And what that says is that it
is easier to pair up electrons

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in 4d or 5d orbitals than it is
to put two electrons in the same

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3d orbital.
And the reason for that is just

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what the radial extension is of
these orbitals.

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In effect, these orbitals are
larger than the 3d orbitals.

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And it means that if you put
two electrons into the same 4d

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or 5d orbital,
much of the time they can stay

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farther apart from each other
than if they are in a 3d

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orbital.
And so this has the practical

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consequence that 4d and 5d
systems often are low spin.

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And so this high spin/low spin
issue that we talked about last

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time is mostly focused on
complexes of the 3d ions.

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So then we are talking about
titanium, vanadium,

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chromium, etc.
And, if we come over here,

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we can ask, what is the effect
of charge on delta O?

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This is all pointing back to
the state diagram that I showed

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you over there in this kind of
sliding scale,

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where you can change things
like just what type of metal ion

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you have, 3d,
4d, or 5d.

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Or you can change the nature of
the ligands.

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Or you can even change the
charge on the metal as a way of

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changing delta O.
And so we can look at,

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for example,
chromium hexaaquo two plus

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versus chromium
hexaaquo three plus.

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The first is a d four ion.

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And then, the second is a d

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three ion.
You should practice going from

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a formula into the d^n count.
And then we can look at what

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delta O does.
This is roughly 14,000 wave

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numbers.
And then, when we oxidize from

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2+ to 3+, higher charge on the
central metal,

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delta O shoots up to about
approximately 17,000.

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And this is in reciprocal
centimeters.

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We see that putting a higher
charge on the metal center leads

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to a larger value of delta O.
And that is also true if we

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look at some other metal systems
along these lines.

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For example,
there are just a number of

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different metals for which the
hexaaquo systems can be

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generated in both the 2+ and the
3+ state of charge.

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And here, we go from about
12,000 and then oxidize,

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and we get up to about 18,000,
so even a larger increase in

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the value of delta O on going
from the 2+ to the 3+ oxidation

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state of the metal.
And then in the notes you will

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find that I put in another
example that behaves similarly

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based on cobalt.
If you go from cobalt two to

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cobalt three hexaaquo,
you go from about 9,000 up to

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18,000 reciprocal centimeters.
So almost doubling the value of

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delta O by removing one electron
from the system.

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What is going on here is that
the greater charge --

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-- pulls the ligands in closer
to the metal.

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And if the ligands are in
closer to the metal,

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then what that ends up saying
is you will see in a moment

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because we are going to get to
the MO picture for octahedral

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coordination complexes,
is that putting electrons in e

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g star becomes more
difficult.

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Delta O is larger.
The ligands come in closer,

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and that is effectively
shrinking that value of r when

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we talked about our spherical
coordinates as applied to the

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angular parts of the d-orbital
wavefunctions and looking at

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that.
So we are changing r.

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The ligands are coming in
closer to the more positively

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charged metal ion in that case.
And then we can also look at

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the effect of changing the
ligand.

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In the system on the left,
we were looking at pairs of

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metal complexes with the same
ligand and changing the charge.

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And now we are just going to
say, what happens if you change

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the ligand?
And this will be a series of

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vanadium complexes.
We have five of them.

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We have the vanadium hexaaquo.

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This one is 3+.
We have the vanadium with six

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urea molecules.
And this is also 3+.

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We have vanadium hexafluoride
three minus.

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And these are all vanadium in
the +3 oxidation state.

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Then we have vanadium
hexachloride three minus,

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and down here the fifth one

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will be vanadium with six
cyanide ligands and three minus.

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So these are all d two

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systems with vanadium in the +3
oxidation state.

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And here, we are going from
18,000 reciprocal centimeters to

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the hexafluoride.
And we drop down to 16,000.

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And this is all approximate.
And six urea on vanadium,

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we go up to about 17,0000.
And then, with the

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hexachloride,
we are down to about a delta O

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00:14:35,000 --> 00:14:37,000
of 12,000 reciprocal
centimeters.

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And then we put cyanides on,
and delta O pops up to about

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23,000 wave numbers.
So this is the same metal ion

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but in five different ligand
environments.

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It has five very different
values of delta O.

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00:14:54,000 --> 00:14:57,000
And from this,
we can derive a spectrochemical

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00:14:57,000 --> 00:15:00,000
series --

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-- for organizing ligands with
respect to the magnitude of

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delta O that they exert for a
given metal ion.

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00:15:11,000 --> 00:15:14,000
And this one just gives us a
series that contains five

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ligands.
But, as you can imagine,

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many more ligands than just
these five simple ligands have

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been looked at with a variety of
different metal ions to make a

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pretty big overall comprehensive
spectrochemical series that sort

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of allows you to pick out which
ligand you want when you are

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interested in generating a
particular value of delta O.

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And here what you see is that
the weakest one is chloride.

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We have chloride.
And then next was fluoride.

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And then next was urea followed
by water.

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00:16:00,000 --> 00:16:05,000
And then the strongest in this
small series of five was

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00:16:05,000 --> 00:16:09,000
cyanide.
And when we try to understand

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the order that these different
ligands appear in,

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in the spectrochemical series,
we are going to find out,

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00:16:19,000 --> 00:16:25,000
as we study the properties of
the molecular orbitals of these

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systems, that cyanide is a pi
acid.

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00:16:30,000 --> 00:16:33,000
And that down here,
chloride is a pi donor.

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00:16:33,000 --> 00:16:37,000
And in the middle,
you have systems like water

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which is more or less sigma
only.

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00:16:40,000 --> 00:16:45,000
So we can understand where
specific ligands appear in the

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00:16:45,000 --> 00:16:50,000
spectrochemical series on the
basis of their orbital

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00:16:50,000 --> 00:16:54,000
considerations.
And that will be the focus of

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00:16:54,000 --> 00:16:58,000
the next part of the lecture.

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00:17:15,000 --> 00:17:18,000
I would like to talk a little
bit about sigma-only ligands.

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00:17:18,000 --> 00:17:23,000
The process in which we are now
about to engage is very similar

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00:17:23,000 --> 00:17:27,000
to that which we used in class
for generating the molecular

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00:17:27,000 --> 00:17:32,000
orbital energy level diagram for
the BH three molecule.

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00:17:32,000 --> 00:17:36,000
In that case we had a central
boron and three hydrogens around

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00:17:36,000 --> 00:17:40,000
it, and we made some hydrogen
linear combinations.

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00:17:40,000 --> 00:17:43,000
And we saw how they would
interact with the atomic

213
00:17:43,000 --> 00:17:47,000
orbitals of the boron.
And now we are going to do the

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00:17:47,000 --> 00:17:52,000
same thing, except we have a lot
more orbitals in the system as a

215
00:17:52,000 --> 00:17:55,000
whole, so we are approximating
some of them.

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00:17:55,000 --> 00:18:00,000
Here is a typical sigma-only
ligand, NH three.

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00:18:00,000 --> 00:18:05,000
And for NH three,
we have a lone pair on the

218
00:18:05,000 --> 00:18:09,000
nitrogen here.
And I am drawing it in very

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00:18:09,000 --> 00:18:13,000
simplified fashion,
as you will no doubt

220
00:18:13,000 --> 00:18:16,000
appreciate.
There is our lone pair.

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00:18:16,000 --> 00:18:21,000
And, when this points directly
at the metal,

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00:18:21,000 --> 00:18:27,000
the lone pair can make a sigma
bond to the metal.

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00:18:39,000 --> 00:18:43,000
I think you will appreciate
that if we have a metal out here

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00:18:43,000 --> 00:18:47,000
such that this lone pair of
electrons is directed right at

225
00:18:47,000 --> 00:18:51,000
the metal, then that results in
a cylindrically symmetric lone

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00:18:51,000 --> 00:18:55,000
pair about that metal-nitrogen
internuclear axis.

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00:18:55,000 --> 00:18:58,000
And so that would be a
sigma-type of interaction.

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00:18:58,000 --> 00:19:02,000
So this is called a sigma-only
ligand.

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00:19:02,000 --> 00:19:08,000
And other ligands,
like cyanide or like chloride,

230
00:19:08,000 --> 00:19:16,000
we are going to have to add in
pi effects, but this is the

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00:19:16,000 --> 00:19:23,000
simple sigma-only case.
And so if you take a molecule

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00:19:23,000 --> 00:19:30,000
such as this cobalt hexamine two
plus, --

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00:19:30,000 --> 00:19:34,000
-- or, actually,
let's say three plus for

234
00:19:34,000 --> 00:19:38,000
simplicity here.
Cobalt three plus.

235
00:19:38,000 --> 00:19:44,000
And we have these NH three
ligands at each of the

236
00:19:44,000 --> 00:19:47,000
six positions.
And, in each case,

237
00:19:47,000 --> 00:19:52,000
the orientation of the
hydrogens on these nitrogens is

238
00:19:52,000 --> 00:19:58,000
such as to promote a sigma
directing effect of that lone

239
00:19:58,000 --> 00:20:04,000
pair toward the metal.
And so that lone pair of

240
00:20:04,000 --> 00:20:09,000
electrons on the nitrogen for
each of these ammonia ligands

241
00:20:09,000 --> 00:20:15,000
directed right at the metal is
providing this repulsion that

242
00:20:15,000 --> 00:20:19,000
pumps e(g) up in energy.
And you will see that,

243
00:20:19,000 --> 00:20:25,000
hopefully, very clearly now.
And let me point out that this

244
00:20:25,000 --> 00:20:29,000
orbital, here,
that I am focusing on is the

245
00:20:29,000 --> 00:20:34,000
HOMO of NH three.
What we are doing is saying you

246
00:20:34,000 --> 00:20:37,000
have a central metal ion and
then you have six ammonia

247
00:20:37,000 --> 00:20:40,000
ligands pointing at it,
each of them pointing their

248
00:20:40,000 --> 00:20:43,000
HOMO at the metal,
their highest occupied

249
00:20:43,000 --> 00:20:45,000
molecular orbital.
What does that do?

250
00:20:45,000 --> 00:20:48,000
What kinds of bonds arise as a
consequence of that between the

251
00:20:48,000 --> 00:20:52,000
metal ion and these six ammonia
molecules?

252
00:20:52,000 --> 00:20:56,000
And so we will now see what
that is on the basis of some of

253
00:20:56,000 --> 00:21:01,000
the linear combinations that we
can generate between the six

254
00:21:01,000 --> 00:21:06,000
HOMOs of the NH three's in this
kind of a geometry.

255
00:21:22,000 --> 00:21:24,000
I will do the first one over
here.

256
00:21:32,000 --> 00:21:36,000
I would like to use a different
color for that.

257
00:21:36,000 --> 00:21:40,000
I am going to use my blue.
All right.

258
00:21:40,000 --> 00:21:46,000
I am going to draw for you some
linear combinations of ammonia

259
00:21:46,000 --> 00:21:49,000
lone pairs.
These are simplistically

260
00:21:49,000 --> 00:21:55,000
representing the homo of each NH
three directed at the

261
00:21:55,000 --> 00:21:59,000
metal.
That one has the symmetry of a

262
00:21:59,000 --> 00:22:03,000
cobalt s orbital.
It has the same sign

263
00:22:03,000 --> 00:22:06,000
everywhere.
And then, we are going to see

264
00:22:06,000 --> 00:22:10,000
that you can draw three of
these.

265
00:22:15,000 --> 00:22:17,000
And I will draw them all right
here --

266
00:22:24,000 --> 00:22:30,000
-- in pairs.
And this one has the symmetry

267
00:22:30,000 --> 00:22:37,000
of a cobalt pz atomic orbital.
You see it has positive up

268
00:22:37,000 --> 00:22:39,000
there.
And down here,

269
00:22:39,000 --> 00:22:42,000
we are taking this to be x,
this to be y,

270
00:22:42,000 --> 00:22:46,000
and this to be z for each of
our diagrams,

271
00:22:46,000 --> 00:22:52,000
so that the phase properties of
this combination is that of pz.

272
00:22:52,000 --> 00:22:57,000
This one is like px.
This one is like py.

273
00:22:57,000 --> 00:23:01,000
And the only reason we have to
take these linear combinations

274
00:23:01,000 --> 00:23:06,000
is because we have multiple
equivalent ammonia molecules in

275
00:23:06,000 --> 00:23:09,000
this problem.
And then, we have two more that

276
00:23:09,000 --> 00:23:14,000
are of interest to us.
One of these will have a couple

277
00:23:14,000 --> 00:23:18,000
of big lobes located on z,
pointed at the metal and then

278
00:23:18,000 --> 00:23:21,000
four small lobes,
each with negative phase.

279
00:23:21,000 --> 00:23:25,000
I will shade them to indicate
that negative phase.

280
00:23:25,000 --> 00:23:30,000
And that one will have the same
phase properties as a cobalt d z

281
00:23:30,000 --> 00:23:35,000
squared orbital.
Plus above, plus below.

282
00:23:35,000 --> 00:23:40,000
And then, to match up with the
torus in the center of d z

283
00:23:40,000 --> 00:23:45,000
squared in the
x,y-plane, the negative phase

284
00:23:45,000 --> 00:23:47,000
shown there.
And then, finally,

285
00:23:47,000 --> 00:23:52,000
over here we have one where all
four contributions to this one

286
00:23:52,000 --> 00:23:57,000
are in the x,y-plane,
along the x and y axes and with

287
00:23:57,000 --> 00:24:01,000
shading as shown there.
And that is of the same

288
00:24:01,000 --> 00:24:07,000
symmetry as our d x squared
minus y squared

289
00:24:07,000 --> 00:24:10,000
orbital.
Having written down this set of

290
00:24:10,000 --> 00:24:15,000
six linear combinations that we
can generate from our six

291
00:24:15,000 --> 00:24:18,000
ammonia HOMOs,
we are done because the cobalt

292
00:24:18,000 --> 00:24:22,000
has atomic orbitals of these
symmetries.

293
00:24:22,000 --> 00:24:26,000
We can now use these to make
bonds between the ammonia

294
00:24:26,000 --> 00:24:32,000
molecules and the cobalt.
And we can use them to make

295
00:24:32,000 --> 00:24:34,000
corresponding antibonds.

296
00:24:39,000 --> 00:24:43,000
Here is the MO diagram for the
sigma-only case.

297
00:25:00,000 --> 00:25:07,000
What we have is an energy level
diagram that will represent the

298
00:25:07,000 --> 00:25:13,000
interaction of these linear
combinations with the metal

299
00:25:13,000 --> 00:25:18,000
valance orbitals.
And so I am going to draw the

300
00:25:18,000 --> 00:25:23,000
metal valance orbitals over here
as follows.

301
00:25:23,000 --> 00:25:27,000
We have, on the metal,
the 3d, the 4s,

302
00:25:27,000 --> 00:25:32,000
and the 4p.
And some of you will notice

303
00:25:32,000 --> 00:25:37,000
that this ordering of energies
is a little different than what

304
00:25:37,000 --> 00:25:41,000
you saw when you just built up
atomic configurations,

305
00:25:41,000 --> 00:25:45,000
because what I have drawn here
is the set of five 3d atomic

306
00:25:45,000 --> 00:25:51,000
orbitals is down lower in energy
than the cobalt 4s and lower in

307
00:25:51,000 --> 00:25:54,000
energy, that is,
even still, than the cobalt 4p.

308
00:25:54,000 --> 00:25:59,000
This is our cobalt atom here.
And the reason this energy

309
00:25:59,000 --> 00:26:03,000
ordering for the valance
orbitals is difference is

310
00:26:03,000 --> 00:26:06,000
because we have an ionized
cobalt.

311
00:26:06,000 --> 00:26:10,000
And the energies change around
when you remove an electron from

312
00:26:10,000 --> 00:26:14,000
the system as compared to the
neutral atom.

313
00:26:14,000 --> 00:26:18,000
But one thing that will help
you remember this ordering of

314
00:26:18,000 --> 00:26:22,000
energy levels is that it goes in
the order of the principle

315
00:26:22,000 --> 00:26:25,000
quantum number,
which kind of does make a

316
00:26:25,000 --> 00:26:29,000
little sense.
In fact, these energies switch

317
00:26:29,000 --> 00:26:33,000
around, depending on which of
the metals you are talking

318
00:26:33,000 --> 00:26:35,000
about.
But generally 3d and 4s,

319
00:26:35,000 --> 00:26:39,000
or alternatively 4d and 5s,
are pretty close in energy to

320
00:26:39,000 --> 00:26:43,000
each other, depending on the ion
that you are talking about.

321
00:26:43,000 --> 00:26:46,000
And then 4p is usually
energetically a bit separated

322
00:26:46,000 --> 00:26:49,000
from that.
And then the thing that makes

323
00:26:49,000 --> 00:26:52,000
us focus our attention so
strongly on the d orbitals is

324
00:26:52,000 --> 00:26:57,000
that those are the ones closest
in energy to the ligand orbitals

325
00:26:57,000 --> 00:27:03,000
that can interact with them.
And so let me draw over here a

326
00:27:03,000 --> 00:27:10,000
bar that represents the six LCs
of NH three HOMOs.

327
00:27:10,000 --> 00:27:16,000
In essence, this represents
these six drawings that I drew

328
00:27:16,000 --> 00:27:23,000
over here, linear combinations
that have the right symmetry to

329
00:27:23,000 --> 00:27:30,000
interact with certain of valance
orbitals on the cobalt,

330
00:27:30,000 --> 00:27:36,000
here on the left.
And what we will find is that--

331
00:27:36,000 --> 00:27:42,000
And we will start with the 3d,
since that is what we are going

332
00:27:42,000 --> 00:27:45,000
to be principally interested in
here.

333
00:27:45,000 --> 00:27:51,000
We come over and find that
three of the d orbitals do not

334
00:27:51,000 --> 00:27:57,000
match any of these because these
NH three HOMOs all lie

335
00:27:57,000 --> 00:28:02,000
in nodal surfaces of them.
And that will be,

336
00:28:02,000 --> 00:28:06,000
of course, our t(2g) set.

337
00:28:12,000 --> 00:28:15,000
Those have the same energy as
in the ion itself.

338
00:28:15,000 --> 00:28:20,000
They come straight over and
don't get involved in bonding or

339
00:28:20,000 --> 00:28:24,000
antibonding interactions.
These are nonbonding.

340
00:28:30,000 --> 00:28:33,000
That is t(2g).
And then, what we find is that

341
00:28:33,000 --> 00:28:35,000
of these six linear
combinations,

342
00:28:35,000 --> 00:28:38,000
two of them,
shown right here,

343
00:28:38,000 --> 00:28:42,000
have the correct phase
properties to make bonding and

344
00:28:42,000 --> 00:28:46,000
antibonding combinations by
interacting with d z squared

345
00:28:46,000 --> 00:28:51,000
and d x squared minus
y squared.

346
00:28:51,000 --> 00:28:54,000
We show that down here.

347
00:29:00,000 --> 00:29:06,000
Here is a bonding combination.
This has e(g) symmetry,

348
00:29:06,000 --> 00:29:14,000
and it represents the formation
of two bonds between the metal

349
00:29:14,000 --> 00:29:18,000
and the ligands.
And, of course,

350
00:29:18,000 --> 00:29:24,000
we also make an antibonding
combination of the same

351
00:29:24,000 --> 00:29:29,000
symmetry.
This should just remind you of

352
00:29:29,000 --> 00:29:33,000
the hydrogen H two
molecule molecular orbital

353
00:29:33,000 --> 00:29:36,000
problem built into this much
larger structure.

354
00:29:36,000 --> 00:29:39,000
And this one,
this antibonding case is e g

355
00:29:39,000 --> 00:29:42,000
star.
And let's just quickly look at

356
00:29:42,000 --> 00:29:44,000
them.

357
00:29:55,000 --> 00:30:00,000
Here, what we have is a d z
squared orbital with

358
00:30:00,000 --> 00:30:05,000
its torus interacting in an
out-of-phase manner with these

359
00:30:05,000 --> 00:30:09,000
ammonia lone pairs.
That creates antibonding

360
00:30:09,000 --> 00:30:14,000
character here and here,
making this a high-energy

361
00:30:14,000 --> 00:30:18,000
orbital.
And then, we have antibonding

362
00:30:18,000 --> 00:30:22,000
character, too,
in between each of the ammonias

363
00:30:22,000 --> 00:30:27,000
located on the x,y-plane and the
torus of the d z squared.

364
00:30:27,000 --> 00:30:33,000
So that is one of our two e g

365
00:30:33,000 --> 00:30:38,000
star orbitals.
And over here,

366
00:30:38,000 --> 00:30:46,000
we have our d x squared minus y
squared orbital

367
00:30:46,000 --> 00:30:52,000
making antibonding interactions
with this other linear

368
00:30:52,000 --> 00:30:59,000
combination, as shown here,
such that we get antibonding

369
00:30:59,000 --> 00:31:05,000
all the way around.
That is e g star.

370
00:31:05,000 --> 00:31:10,000
That is the reason why the e(g)
set, derived from our set of d

371
00:31:10,000 --> 00:31:12,000
orbitals, is marked with an
asterisk.

372
00:31:12,000 --> 00:31:16,000
It is antibonding with respect
to these bonds.

373
00:31:16,000 --> 00:31:20,000
If you start to populate e(g)*
with electrons,

374
00:31:20,000 --> 00:31:23,000
as you will do in certain cases
that are high-spin,

375
00:31:23,000 --> 00:31:27,000
for example,
those electrons that go into

376
00:31:27,000 --> 00:31:31,000
e(g)*, if you reflect back on
the way that we calculate bond

377
00:31:31,000 --> 00:31:36,000
order, they go in there and
weaken these two metal ligand

378
00:31:36,000 --> 00:31:40,000
bonds.
And these two bonds look just

379
00:31:40,000 --> 00:31:43,000
identical to the ones I drew
here for e(g)*,

380
00:31:43,000 --> 00:31:47,000
just reversing the phase and
making them bonding everywhere.

381
00:31:47,000 --> 00:31:50,000
Let's just look at those.

382
00:31:58,000 --> 00:32:03,000
Here, we have d z squared
making a nice bonding

383
00:32:03,000 --> 00:32:05,000
combination.

384
00:32:10,000 --> 00:32:14,000
It is interacting in an
in-phase manner with that linear

385
00:32:14,000 --> 00:32:19,000
combination, so we have bonding
both in the x,y-plane and

386
00:32:19,000 --> 00:32:25,000
strongly along the z-axis.
That is one of these two bonds.

387
00:32:25,000 --> 00:32:29,000
And then over here,
we will have x squared minus y

388
00:32:29,000 --> 00:32:34,000
squared,
making bonds also --

389
00:32:39,000 --> 00:32:45,000
-- like that.
Down here, this is bonding,

390
00:32:45,000 --> 00:32:50,000
and this is antibonding.

391
00:32:55,000 --> 00:33:00,000
Now, that is not the entire
diagram.

392
00:33:00,000 --> 00:33:04,000
One runs out of space pretty
quickly on these

393
00:33:04,000 --> 00:33:09,000
horizontally-oriented boards for
putting together tall MO

394
00:33:09,000 --> 00:33:12,000
diagrams, but there were six LCs
over here.

395
00:33:12,000 --> 00:33:18,000
And you will see that there
will be another one that makes a

396
00:33:18,000 --> 00:33:22,000
bond with the 4s orbital.
And so, accordingly,

397
00:33:22,000 --> 00:33:27,000
the 4s orbital comes up here.
There will be an antibond

398
00:33:27,000 --> 00:33:32,000
derived from 4s.
There will be also a set of

399
00:33:32,000 --> 00:33:39,000
three that are antibonding
derived from interaction of 4p

400
00:33:39,000 --> 00:33:41,000
with the other ones,
here.

401
00:33:41,000 --> 00:33:47,000
And there is a big energy
mismatch between 4p and these,

402
00:33:47,000 --> 00:33:55,000
so we will draw these as the
least stabilized of the set.

403
00:33:55,000 --> 00:34:00,000
But now, what you can see down
here is that we have found a way

404
00:34:00,000 --> 00:34:04,000
to form six bonds,
six pairs of electrons,

405
00:34:04,000 --> 00:34:09,000
each coming from the highest
occupied molecular orbital of

406
00:34:09,000 --> 00:34:14,000
the six ammonia molecules.
We have formed six bonds using

407
00:34:14,000 --> 00:34:17,000
these valance orbitals of the
metal.

408
00:34:17,000 --> 00:34:22,000
And how many more electrons do
we have to put into the diagram?

409
00:34:22,000 --> 00:34:27,000
We have six because this,
in the case of our cobalt (NH

410
00:34:27,000 --> 00:34:33,000
three) six three plus --

411
00:34:33,000 --> 00:34:37,000
This is Group 9.
Minus three for the 3+ charge.

412
00:34:37,000 --> 00:34:41,000
All the ammonia ligands are
neutral, equals six.

413
00:34:41,000 --> 00:34:43,000
So this is a d six case.

414
00:34:43,000 --> 00:34:48,000
The orbitals here that we
called t(2g) and e g star

415
00:34:48,000 --> 00:34:52,000
can now take up six
more electrons.

416
00:34:52,000 --> 00:34:57,000
And I am drawing them in here
with the assumption that this is

417
00:34:57,000 --> 00:35:02,000
a low-spin case.
And that is reasonable based on

418
00:35:02,000 --> 00:35:06,000
the 3+ charge of this ion.
We have already discussed the

419
00:35:06,000 --> 00:35:09,000
effective charge.
It draws the ligands in.

420
00:35:09,000 --> 00:35:14,000
It tends to increase the value
of delta O, as we have seen.

421
00:35:14,000 --> 00:35:17,000
And so this is how I would
populate that diagram,

422
00:35:17,000 --> 00:35:21,000
with the 6d electrons going
into the nonbonding t(2g)

423
00:35:21,000 --> 00:35:26,000
constituting the highest
occupied molecular orbital of

424
00:35:26,000 --> 00:35:30,000
this system.
And then, stabilize the lone

425
00:35:30,000 --> 00:35:32,000
pairs.
Down here what is really going

426
00:35:32,000 --> 00:35:37,000
on, you can think about these
six ammonia molecules acting

427
00:35:37,000 --> 00:35:42,000
simultaneously as six Lewis
bases to a metal that has enough

428
00:35:42,000 --> 00:35:46,000
empty orbitals to accept six
lone pair donations to that same

429
00:35:46,000 --> 00:35:49,000
metal.
So it is a Lewis acid times

430
00:35:49,000 --> 00:35:51,000
six.
And it is positively charged,

431
00:35:51,000 --> 00:35:57,000
which of course tends to draw
electron density to it.

432
00:35:57,000 --> 00:36:00,000
And that is our simplified
molecular orbital diagram for

433
00:36:00,000 --> 00:36:04,000
the sigma-only case.
Now, there are a lot of other

434
00:36:04,000 --> 00:36:07,000
interesting cases.
As you can see here from this

435
00:36:07,000 --> 00:36:11,000
spectrochemical series where the
ligands in the middle are

436
00:36:11,000 --> 00:36:13,000
basically sigma-only urea,
water.

437
00:36:13,000 --> 00:36:17,000
But then at either end of the
spectrochemical series,

438
00:36:17,000 --> 00:36:21,000
you get into systems where you
must take pi effects into

439
00:36:21,000 --> 00:36:23,000
account.

440
00:36:39,000 --> 00:36:43,000
And we will treat,
first, the case of a pi donor.

441
00:36:43,000 --> 00:36:48,000
Remember that if we are talking
about a donor,

442
00:36:48,000 --> 00:36:53,000
that means we are effectively
talking about a Lewis base.

443
00:36:53,000 --> 00:36:59,000
And you have to ask yourself
the question in a system that

444
00:36:59,000 --> 00:37:05,000
has a ligand that can make pi
bonds, are the orbitals on the

445
00:37:05,000 --> 00:37:12,000
metal that have pi symmetry
filled, or are they empty?

446
00:37:12,000 --> 00:37:14,000
And, similarly,
are the pi symmetry orbitals on

447
00:37:14,000 --> 00:37:17,000
the ligand in question filled or
are they empty?

448
00:37:17,000 --> 00:37:20,000
And, if you can get all that
right, you will really

449
00:37:20,000 --> 00:37:24,000
understand what happens in terms
of attenuating t(2g) when you

450
00:37:24,000 --> 00:37:26,000
add in pi effects.
We have seen that the e(g)

451
00:37:26,000 --> 00:37:30,000
orbitals of a metal are the d z
squared and d x

452
00:37:30,000 --> 00:37:33,000
squared minus y squared
that point along the

453
00:37:33,000 --> 00:37:38,000
axes.
The xz, yz, and xy that point

454
00:37:38,000 --> 00:37:45,000
between the axes have the
potential to make pi bonds with

455
00:37:45,000 --> 00:37:49,000
ligands.
And so here is an example of a

456
00:37:49,000 --> 00:37:55,000
type of ligand that can make pi
bonds to a metal.

457
00:37:55,000 --> 00:38:01,000
This ligand is NH two minus.

458
00:38:01,000 --> 00:38:04,000
And it can make both sigma and
pi bonds to a metal.

459
00:38:04,000 --> 00:38:09,000
It is usually the case that if
a ligand is capable of making pi

460
00:38:09,000 --> 00:38:12,000
bonds to a metal,
it is also capable of making

461
00:38:12,000 --> 00:38:15,000
sigma bonds.
And so the pi effects are sort

462
00:38:15,000 --> 00:38:17,000
of mapped onto the sigma
framework.

463
00:38:17,000 --> 00:38:21,000
And so what we found over here
and here about the sigma only

464
00:38:21,000 --> 00:38:26,000
case, that will also be true for
pi systems.

465
00:38:26,000 --> 00:38:31,000
But then we are superposing on
that diagram the pi bonds that

466
00:38:31,000 --> 00:38:35,000
can form in the system.
And the way that NH two minus

467
00:38:35,000 --> 00:38:41,000
can make both sigma
and pi bonds with a metal is,

468
00:38:41,000 --> 00:38:44,000
first of all,
it has a lone pair that is

469
00:38:44,000 --> 00:38:48,000
directed at the metal,
like this.

470
00:38:53,000 --> 00:38:57,000
For forming sigma contacts.
And, if you have six of these

471
00:38:57,000 --> 00:39:00,000
around a metal,
the situation would be just the

472
00:39:00,000 --> 00:39:05,000
same as what we derived over
here for the cobalt hexamine.

473
00:39:05,000 --> 00:39:10,000
But this is a planar system.
The sum of the bond angles

474
00:39:10,000 --> 00:39:15,000
around the nitrogen in systems
like this are 360 degrees

475
00:39:15,000 --> 00:39:18,000
meaning that,
unlike ammonia itself,

476
00:39:18,000 --> 00:39:23,000
which is a trigonal pyramid,
this is a planar system.

477
00:39:23,000 --> 00:39:29,000
And so it has a lone pair
perpendicular to that plane.

478
00:39:34,000 --> 00:39:38,000
I will draw it that way.
It is just a pure p-orbital,

479
00:39:38,000 --> 00:39:42,000
perpendicular to the plane of
the substituents on that

480
00:39:42,000 --> 00:39:46,000
nitrogen.
And the way that could make a

481
00:39:46,000 --> 00:39:50,000
pi bond with the metal is
through simultaneous donation

482
00:39:50,000 --> 00:39:56,000
above and below the plane of its
substituents on that nitrogen,

483
00:39:56,000 --> 00:40:00,000
so that this ligand can
interact as a double Lewis base

484
00:40:00,000 --> 00:40:07,000
with a metal if the metal has
appropriate orbitals available.

485
00:40:07,000 --> 00:40:13,000
And so a situation in which you
might find this would be the

486
00:40:13,000 --> 00:40:15,000
following.

487
00:40:30,000 --> 00:40:34,000
An example of such a molecule,
this would be a neutral one.

488
00:41:00,000 --> 00:41:05,000
And what you see is that the
planes of these NH two

489
00:41:05,000 --> 00:41:10,000
units are lining up with the
coordinate xz,

490
00:41:10,000 --> 00:41:14,000
yz, and xy Cartesian planes in
the molecule.

491
00:41:14,000 --> 00:41:20,000
Since each of these ligands is
negatively charged and the

492
00:41:20,000 --> 00:41:25,000
molecule is neutral,
the chromium is therefore in

493
00:41:25,000 --> 00:41:31,000
the +6 oxidation state to
balance the six negative charges

494
00:41:31,000 --> 00:41:36,000
on the amido ligands,
we call them.

495
00:41:36,000 --> 00:41:38,000
This is amido.
And so this one,

496
00:41:38,000 --> 00:41:41,000
in fact, is a d zero case.

497
00:41:41,000 --> 00:41:45,000
And d zero is an
electron count that is pretty

498
00:41:45,000 --> 00:41:50,000
typical for the formation of pi
bonds from ligands that can

499
00:41:50,000 --> 00:41:55,000
donate into the metal center.
If you have d electrons present

500
00:41:55,000 --> 00:41:59,000
in a system like this,
then you won't have the empty

501
00:41:59,000 --> 00:42:04,000
d-orbitals on the metal that are
necessary to accept bonds formed

502
00:42:04,000 --> 00:42:09,000
by lone pair donation of this
sort.

503
00:42:09,000 --> 00:42:13,000
The system will become choked
up on itself with too many

504
00:42:13,000 --> 00:42:15,000
electrons.
Let's see what those

505
00:42:15,000 --> 00:42:20,000
perturbations will lead to for
our MO diagram.

506
00:42:40,000 --> 00:42:44,000
I am going to focus in on
linear combinations of amide

507
00:42:44,000 --> 00:42:48,000
lone pair orbitals that,
in fact, have the correct

508
00:42:48,000 --> 00:42:52,000
symmetry to interact with our
t(2g) set, since that is our pi

509
00:42:52,000 --> 00:42:56,000
set of orbitals from the d
manifold on the metal.

510
00:42:56,000 --> 00:42:59,000
So let's look at this.

511
00:43:09,000 --> 00:43:12,000
This would be d(yz).

512
00:43:19,000 --> 00:43:23,000
And then, let me choose --

513
00:43:41,000 --> 00:43:45,000
We can make a pi bond of this
sort using a linear combination

514
00:43:45,000 --> 00:43:50,000
that I am mating up together
with the d(yz) orbital.

515
00:43:50,000 --> 00:43:54,000
And while I have drawn here the
bonding counterpart there,

516
00:43:54,000 --> 00:43:58,000
of course, will be an
antibonding counterpart for

517
00:43:58,000 --> 00:44:03,000
that.
Let's draw the one that belongs

518
00:44:03,000 --> 00:44:04,000
to d(xz).

519
00:44:10,000 --> 00:44:13,000
And that will be the one up
top, here.

520
00:44:31,000 --> 00:44:34,000
So what we have is pi bonding
happening here,

521
00:44:34,000 --> 00:44:39,000
here, here, and here that
corresponds to pi lone paired

522
00:44:39,000 --> 00:44:42,000
donation into that metal
orbital.

523
00:44:42,000 --> 00:44:46,000
And then, finally,
we have one that involves the

524
00:44:46,000 --> 00:44:48,000
d(xy).

525
00:44:55,000 --> 00:45:01,000
And this would be with our
hydrogens up and down,

526
00:45:01,000 --> 00:45:04,000
here and here.
Sorry.

527
00:45:04,000 --> 00:45:10,000
Up and down in front and in
back like that,

528
00:45:10,000 --> 00:45:18,000
so that we can make two pi
bonding interactions over here

529
00:45:18,000 --> 00:45:22,000
with this.
There is pi bonding,

530
00:45:22,000 --> 00:45:28,000
pi bonding, bonding,
bonding.

531
00:45:28,000 --> 00:45:31,000
At this point,
we have gotten to where we can

532
00:45:31,000 --> 00:45:35,000
recognize how there are three
linear combinations of the pi

533
00:45:35,000 --> 00:45:39,000
lone pairs from these amido
ligands that will have the

534
00:45:39,000 --> 00:45:43,000
correct symmetry to interact
with the t(2g) set from our

535
00:45:43,000 --> 00:45:46,000
metal.
There are three more where you

536
00:45:46,000 --> 00:45:51,000
would just flip one of the two
on each, and those would have

537
00:45:51,000 --> 00:45:54,000
the correct symmetry to interact
with the px, py,

538
00:45:54,000 --> 00:45:59,000
and pz orbitals on the metal.
I am not going to draw those

539
00:45:59,000 --> 00:46:03,000
out since we are focusing on the
d part of our diagram at the

540
00:46:03,000 --> 00:46:04,000
moment.

541
00:46:10,000 --> 00:46:15,000
And what we will find then,
in the molecular orbital

542
00:46:15,000 --> 00:46:21,000
diagram for a system of this
sort, is that our d manifold,

543
00:46:21,000 --> 00:46:25,000
this is our 3d,
4s, 4p, is such that,

544
00:46:25,000 --> 00:46:33,000
you remember previously t two g
came straight over.

545
00:46:33,000 --> 00:46:36,000
And now, instead,
we are going to see that it

546
00:46:36,000 --> 00:46:40,000
goes up in energy because what
t(2g) now is,

547
00:46:40,000 --> 00:46:44,000
it still has a lot of d(xz),
d(yz) and d(xy).

548
00:46:44,000 --> 00:46:49,000
But now, t two g has
acquired a pi star

549
00:46:49,000 --> 00:46:54,000
character due to the fact that
we have these linear

550
00:46:54,000 --> 00:46:58,000
combinations over here.
These are our pi LCs that we

551
00:46:58,000 --> 00:47:04,000
drew over there that are of the
appropriate symmetry to interact

552
00:47:04,000 --> 00:47:09,000
with d(xz), d(yz) and d(xy).
And, therefore,

553
00:47:09,000 --> 00:47:14,000
these that are filled are
stabilized, come down here and

554
00:47:14,000 --> 00:47:19,000
form a t two g set
that is bonding because those,

555
00:47:19,000 --> 00:47:24,000
of course, are filled in,
NH two minus.

556
00:47:24,000 --> 00:47:28,000
And, if the metal has an empty
t two g set,

557
00:47:28,000 --> 00:47:33,000
you see that we do get three pi
bonds.

558
00:47:33,000 --> 00:47:36,000
t two g now gives us
pi antibonds.

559
00:47:36,000 --> 00:47:40,000
This is being mapped on the
sigma framework that is the same

560
00:47:40,000 --> 00:47:44,000
as what we have right here for
the cobalt hexamine,

561
00:47:44,000 --> 00:47:47,000
so that above t(2g) we will
have the e(g)*.

562
00:47:47,000 --> 00:47:51,000
And will now recognize that
e(g) is sigma star

563
00:47:51,000 --> 00:47:55,000
rather than pi star,
so it is more strongly

564
00:47:55,000 --> 00:47:59,000
destabilized due to the greater
overlap considerations of

565
00:47:59,000 --> 00:48:03,000
forming sigma bonds versus the
side to side overlap of pi

566
00:48:03,000 --> 00:48:10,000
bonds.
And the key observation here is

567
00:48:10,000 --> 00:48:14,000
that pi donors --

568
00:48:20,000 --> 00:48:26,000
-- decrease the magnitude of
delta O by raising the energy of

569
00:48:26,000 --> 00:48:30,000
t(2g), by making t(2g) pi* in
character.

570
00:48:30,000 --> 00:48:34,000
Antibonding.
We can see, not only that t(2g)

571
00:48:34,000 --> 00:48:39,000
should be raised up,
but that mapping that onto our

572
00:48:39,000 --> 00:48:44,000
full sigma framework diagram,
delta O is shrinking.

573
00:48:44,000 --> 00:48:51,000
We still have our six sigma
bonds between the NH two

574
00:48:51,000 --> 00:48:55,000
ligands and the metal.
They are down here,

575
00:48:55,000 --> 00:49:01,000
just not drawn.
It is the same as up here.

576
00:49:01,000 --> 00:49:03,000
And now, we have three
additional pi bonds.

577
00:49:03,000 --> 00:49:07,000
There are another three
additional pi bonds that you can

578
00:49:07,000 --> 00:49:11,000
draw because we do have six of
these NH two minus

579
00:49:11,000 --> 00:49:13,000
ligands.
And then, the corresponding

580
00:49:13,000 --> 00:49:17,000
antibonding orbitals involve 4p.
And they are way up here

581
00:49:17,000 --> 00:49:19,000
somewhere in energy and very
empty.

582
00:49:19,000 --> 00:49:23,000
Since this was a d zero
case, what we have is that you

583
00:49:23,000 --> 00:49:27,000
have we pairs of electrons on
six NH two minus's

584
00:49:27,000 --> 00:49:30,000
that are down here in this
manifold of metal ligand sigma

585
00:49:30,000 --> 00:49:33,000
and pi bonding.

586
00:49:38,000 --> 00:49:40,000
And so we have 12 pairs of
electrons down here that

587
00:49:40,000 --> 00:49:44,000
describe the sigma plus pi
bonding between the metal and

588
00:49:44,000 --> 00:49:46,000
the ligands.
And then you come here to your

589
00:49:46,000 --> 00:49:49,000
d manifold, these are empty
because this is a d zero

590
00:49:49,000 --> 00:49:53,000
case, because I happened to
choose chromium in the six plus

591
00:49:53,000 --> 00:49:56,000
oxidation here.
So it is a d zero case,

592
00:49:56,000 --> 00:50:00,000
and we wouldn't have any
electrons to put in here.

593
00:50:00,000 --> 00:50:04,000
But, under these circumstances,
in order to get color in a

594
00:50:04,000 --> 00:50:07,000
system like this,
you would have to be promoting

595
00:50:07,000 --> 00:50:12,000
an electron from down here from
these pi bonds probably into the

596
00:50:12,000 --> 00:50:16,000
d manifold, for example.
And what we have seen here,

597
00:50:16,130 --> 00:50:20,000
namely that pi donors decrease
delta O should give you a clue

598
00:50:20,000 --> 00:50:24,000
as to what is going on with
chloride and fluoride.

599
00:50:24,000 --> 00:50:28,000
Since those have small values
of delta O, they lie at a

600
00:50:28,190 --> 00:50:33,000
position of weakness in the
spectrochemical series.

601
00:50:33,000 --> 00:50:36,000
And in the converse,
that ligands like cyanide and

602
00:50:36,000 --> 00:50:41,000
especially carbon monoxide are
very high in the spectrochemical

603
00:50:41,000 --> 00:50:44,000
series because they are pi acid
ligands.

604
00:50:44,000 --> 00:50:48,000
And this is something you will
be getting more information

605
00:50:48,382 --> 00:50:51,000
about in recitation this week.
See you all on Wednesday.