The following content is provided by MIT OpenCourseWare
under a Creative Commons license.
Additional information about our license and MIT
OpenCourseWare in general is available at ocw.mit.edu.
We have been talking about the properties of transition metal
complexes, coordination complexes, in the context of a
simple crystal field theory model.
And I just wanted to make reference at the beginning,
today, to state diagrams.
And, in the context of octahedral complexes,
you saw that an array of six ligands surrounding a central
metal ion would lead to a t two g and what we are
calling an eg star manifold of orbitals on the
metal in which we can populate with the number of d electrons
that we have present. And the graph that I am making
here is related to the idea that for a system with only a single
electron, so this could be, for example,
titanium OH two six times three plus.
This is a d one system. And it can go from a ground
state configuration of t two g one eg star zero.
And then, upon absorption of
light, it can be promoted from the ground state,
represented here, into the excited state.
And the excited state has the configuration t two g zero eg
star one, showing that that electron has
been promoted through absorption of a photon.
And this gap here between t(2g) and eg star,
of course, is our delta O value.
And the axis on the bottom, here, is ligand field strength.
This is supposed to tell you that different ligands give rise
to different values of delta O accounting for the fact that
this d one titanium three ion in a field
of water molecules has one color, whereas,
if you, in fact, look at Ti Cl six three minus,
you will be able to tell that
that also is a titanium three plus or d one
system. But they do not have the same
color. And the one absorption band
that you find in the visible part of the spectrum that
corresponds to promotion of an electron from t two g
into eg star is just moving around a little bit in
energy because these ligands behave differently with respect
to what we call their ligand field strength that they exert,
that leads to a particular energy gap between t(2g) and
e(g)*. I will get into this a little
bit further in the moment, but the point here is that d
one or d nine are special electron counts.
And the reason that they are special is that these have a
single absorption peak due to what we call d-d transitions.
They have a single d-d transition.
And that means there is a ground state and then just one
excited state that the system can be promoted to in the
presence of impinging photons of the right energy.
And what happens is that if you have more d electrons than just
one, or d nine is special because we treat that
using a whole formalism, like there is just one electron
missing from a complete manifold, a completely full d
shell minus one. With d^n systems,
for example, d two is much more
complex. And this is beyond the scope of
5.112, but it is something that they are treating in 5.04,
for example. If you go on in organic
chemistry, you will learn how come different states can arise,
multiple states arise when you have more than just a one
electron picture.
For example, if you have a vanadium three
plus ion, there are three bands observed.
And so the simple picture that you have, just a t(2g) level and
an e(g) level is correct and maps onto the picture of the
electronic states that are available, as long as we are
just talking about a one electron picture.
With more electrons in play, depending on just which of the
orbitals they go into, they will have inter-electron
repulsion terms that come into play and give rise to more
excited states. Here, with three bands being
observed in the visible, that means there are three
possible transitions and three possible excited states that can
be accessed. So it is a very complicated
picture. And so we are going to be,
for the purposes of this, focusing on the one electron
picture. But I just wanted you to be
aware that it does get a lot more complex as soon as you go
to many electron systems. Now, let's talk just briefly
about pairing energy.
In the case of 3d system, we typically find that the
pairing energy is around 15 to 18,000 reciprocal
centimeters, the typical energy unit given for pairing energy.
But then if we go to heavier transition elements,
to the 4d or the 5d elements, then typically the pairing
energy is about 8,000 to 12 reciprocal centimeters.
And what that says is that it is easier to pair up electrons
in 4d or 5d orbitals than it is to put two electrons in the same
3d orbital. And the reason for that is just
what the radial extension is of these orbitals.
In effect, these orbitals are larger than the 3d orbitals.
And it means that if you put two electrons into the same 4d
or 5d orbital, much of the time they can stay
farther apart from each other than if they are in a 3d
orbital. And so this has the practical
consequence that 4d and 5d systems often are low spin.
And so this high spin/low spin issue that we talked about last
time is mostly focused on complexes of the 3d ions.
So then we are talking about titanium, vanadium,
chromium, etc. And, if we come over here,
we can ask, what is the effect of charge on delta O?
This is all pointing back to the state diagram that I showed
you over there in this kind of sliding scale,
where you can change things like just what type of metal ion
you have, 3d, 4d, or 5d.
Or you can change the nature of the ligands.
Or you can even change the charge on the metal as a way of
changing delta O. And so we can look at,
for example, chromium hexaaquo two plus
versus chromium hexaaquo three plus.
The first is a d four ion.
And then, the second is a d
three ion. You should practice going from
a formula into the d^n count. And then we can look at what
delta O does. This is roughly 14,000 wave
numbers. And then, when we oxidize from
2+ to 3+, higher charge on the central metal,
delta O shoots up to about approximately 17,000.
And this is in reciprocal centimeters.
We see that putting a higher charge on the metal center leads
to a larger value of delta O. And that is also true if we
look at some other metal systems along these lines.
For example, there are just a number of
different metals for which the hexaaquo systems can be
generated in both the 2+ and the 3+ state of charge.
And here, we go from about 12,000 and then oxidize,
and we get up to about 18,000, so even a larger increase in
the value of delta O on going from the 2+ to the 3+ oxidation
state of the metal. And then in the notes you will
find that I put in another example that behaves similarly
based on cobalt. If you go from cobalt two to
cobalt three hexaaquo, you go from about 9,000 up to
18,000 reciprocal centimeters. So almost doubling the value of
delta O by removing one electron from the system.
What is going on here is that the greater charge --
-- pulls the ligands in closer to the metal.
And if the ligands are in closer to the metal,
then what that ends up saying is you will see in a moment
because we are going to get to the MO picture for octahedral
coordination complexes, is that putting electrons in e
g star becomes more difficult.
Delta O is larger. The ligands come in closer,
and that is effectively shrinking that value of r when
we talked about our spherical coordinates as applied to the
angular parts of the d-orbital wavefunctions and looking at
that. So we are changing r.
The ligands are coming in closer to the more positively
charged metal ion in that case. And then we can also look at
the effect of changing the ligand.
In the system on the left, we were looking at pairs of
metal complexes with the same ligand and changing the charge.
And now we are just going to say, what happens if you change
the ligand? And this will be a series of
vanadium complexes. We have five of them.
We have the vanadium hexaaquo.
This one is 3+. We have the vanadium with six
urea molecules. And this is also 3+.
We have vanadium hexafluoride three minus.
And these are all vanadium in the +3 oxidation state.
Then we have vanadium hexachloride three minus,
and down here the fifth one
will be vanadium with six cyanide ligands and three minus.
So these are all d two
systems with vanadium in the +3 oxidation state.
And here, we are going from 18,000 reciprocal centimeters to
the hexafluoride. And we drop down to 16,000.
And this is all approximate. And six urea on vanadium,
we go up to about 17,0000. And then, with the
hexachloride, we are down to about a delta O
of 12,000 reciprocal centimeters.
And then we put cyanides on, and delta O pops up to about
23,000 wave numbers. So this is the same metal ion
but in five different ligand environments.
It has five very different values of delta O.
And from this, we can derive a spectrochemical
series --
-- for organizing ligands with respect to the magnitude of
delta O that they exert for a given metal ion.
And this one just gives us a series that contains five
ligands. But, as you can imagine,
many more ligands than just these five simple ligands have
been looked at with a variety of different metal ions to make a
pretty big overall comprehensive spectrochemical series that sort
of allows you to pick out which ligand you want when you are
interested in generating a particular value of delta O.
And here what you see is that the weakest one is chloride.
We have chloride. And then next was fluoride.
And then next was urea followed by water.
And then the strongest in this small series of five was
cyanide. And when we try to understand
the order that these different ligands appear in,
in the spectrochemical series, we are going to find out,
as we study the properties of the molecular orbitals of these
systems, that cyanide is a pi acid.
And that down here, chloride is a pi donor.
And in the middle, you have systems like water
which is more or less sigma only.
So we can understand where specific ligands appear in the
spectrochemical series on the basis of their orbital
considerations. And that will be the focus of
the next part of the lecture.
I would like to talk a little bit about sigma-only ligands.
The process in which we are now about to engage is very similar
to that which we used in class for generating the molecular
orbital energy level diagram for the BH three molecule.
In that case we had a central boron and three hydrogens around
it, and we made some hydrogen linear combinations.
And we saw how they would interact with the atomic
orbitals of the boron. And now we are going to do the
same thing, except we have a lot more orbitals in the system as a
whole, so we are approximating some of them.
Here is a typical sigma-only ligand, NH three.
And for NH three, we have a lone pair on the
nitrogen here. And I am drawing it in very
simplified fashion, as you will no doubt
appreciate. There is our lone pair.
And, when this points directly at the metal,
the lone pair can make a sigma bond to the metal.
I think you will appreciate that if we have a metal out here
such that this lone pair of electrons is directed right at
the metal, then that results in a cylindrically symmetric lone
pair about that metal-nitrogen internuclear axis.
And so that would be a sigma-type of interaction.
So this is called a sigma-only ligand.
And other ligands, like cyanide or like chloride,
we are going to have to add in pi effects, but this is the
simple sigma-only case. And so if you take a molecule
such as this cobalt hexamine two plus, --
-- or, actually, let's say three plus for
simplicity here. Cobalt three plus.
And we have these NH three ligands at each of the
six positions. And, in each case,
the orientation of the hydrogens on these nitrogens is
such as to promote a sigma directing effect of that lone
pair toward the metal. And so that lone pair of
electrons on the nitrogen for each of these ammonia ligands
directed right at the metal is providing this repulsion that
pumps e(g) up in energy. And you will see that,
hopefully, very clearly now. And let me point out that this
orbital, here, that I am focusing on is the
HOMO of NH three. What we are doing is saying you
have a central metal ion and then you have six ammonia
ligands pointing at it, each of them pointing their
HOMO at the metal, their highest occupied
molecular orbital. What does that do?
What kinds of bonds arise as a consequence of that between the
metal ion and these six ammonia molecules?
And so we will now see what that is on the basis of some of
the linear combinations that we can generate between the six
HOMOs of the NH three's in this kind of a geometry.
I will do the first one over here.
I would like to use a different color for that.
I am going to use my blue. All right.
I am going to draw for you some linear combinations of ammonia
lone pairs. These are simplistically
representing the homo of each NH three directed at the
metal. That one has the symmetry of a
cobalt s orbital. It has the same sign
everywhere. And then, we are going to see
that you can draw three of these.
And I will draw them all right here --
-- in pairs. And this one has the symmetry
of a cobalt pz atomic orbital. You see it has positive up
there. And down here,
we are taking this to be x, this to be y,
and this to be z for each of our diagrams,
so that the phase properties of this combination is that of pz.
This one is like px. This one is like py.
And the only reason we have to take these linear combinations
is because we have multiple equivalent ammonia molecules in
this problem. And then, we have two more that
are of interest to us. One of these will have a couple
of big lobes located on z, pointed at the metal and then
four small lobes, each with negative phase.
I will shade them to indicate that negative phase.
And that one will have the same phase properties as a cobalt d z
squared orbital. Plus above, plus below.
And then, to match up with the torus in the center of d z
squared in the x,y-plane, the negative phase
shown there. And then, finally,
over here we have one where all four contributions to this one
are in the x,y-plane, along the x and y axes and with
shading as shown there. And that is of the same
symmetry as our d x squared minus y squared
orbital. Having written down this set of
six linear combinations that we can generate from our six
ammonia HOMOs, we are done because the cobalt
has atomic orbitals of these symmetries.
We can now use these to make bonds between the ammonia
molecules and the cobalt. And we can use them to make
corresponding antibonds.
Here is the MO diagram for the sigma-only case.
What we have is an energy level diagram that will represent the
interaction of these linear combinations with the metal
valance orbitals. And so I am going to draw the
metal valance orbitals over here as follows.
We have, on the metal, the 3d, the 4s,
and the 4p. And some of you will notice
that this ordering of energies is a little different than what
you saw when you just built up atomic configurations,
because what I have drawn here is the set of five 3d atomic
orbitals is down lower in energy than the cobalt 4s and lower in
energy, that is, even still, than the cobalt 4p.
This is our cobalt atom here. And the reason this energy
ordering for the valance orbitals is difference is
because we have an ionized cobalt.
And the energies change around when you remove an electron from
the system as compared to the neutral atom.
But one thing that will help you remember this ordering of
energy levels is that it goes in the order of the principle
quantum number, which kind of does make a
little sense. In fact, these energies switch
around, depending on which of the metals you are talking
about. But generally 3d and 4s,
or alternatively 4d and 5s, are pretty close in energy to
each other, depending on the ion that you are talking about.
And then 4p is usually energetically a bit separated
from that. And then the thing that makes
us focus our attention so strongly on the d orbitals is
that those are the ones closest in energy to the ligand orbitals
that can interact with them. And so let me draw over here a
bar that represents the six LCs of NH three HOMOs.
In essence, this represents these six drawings that I drew
over here, linear combinations that have the right symmetry to
interact with certain of valance orbitals on the cobalt,
here on the left. And what we will find is that--
And we will start with the 3d, since that is what we are going
to be principally interested in here.
We come over and find that three of the d orbitals do not
match any of these because these NH three HOMOs all lie
in nodal surfaces of them. And that will be,
of course, our t(2g) set.
Those have the same energy as in the ion itself.
They come straight over and don't get involved in bonding or
antibonding interactions. These are nonbonding.
That is t(2g). And then, what we find is that
of these six linear combinations,
two of them, shown right here,
have the correct phase properties to make bonding and
antibonding combinations by interacting with d z squared
and d x squared minus y squared.
We show that down here.
Here is a bonding combination. This has e(g) symmetry,
and it represents the formation of two bonds between the metal
and the ligands. And, of course,
we also make an antibonding combination of the same
symmetry. This should just remind you of
the hydrogen H two molecule molecular orbital
problem built into this much larger structure.
And this one, this antibonding case is e g
star. And let's just quickly look at
them.
Here, what we have is a d z squared orbital with
its torus interacting in an out-of-phase manner with these
ammonia lone pairs. That creates antibonding
character here and here, making this a high-energy
orbital. And then, we have antibonding
character, too, in between each of the ammonias
located on the x,y-plane and the torus of the d z squared.
So that is one of our two e g
star orbitals. And over here,
we have our d x squared minus y squared orbital
making antibonding interactions with this other linear
combination, as shown here, such that we get antibonding
all the way around. That is e g star.
That is the reason why the e(g) set, derived from our set of d
orbitals, is marked with an asterisk.
It is antibonding with respect to these bonds.
If you start to populate e(g)* with electrons,
as you will do in certain cases that are high-spin,
for example, those electrons that go into
e(g)*, if you reflect back on the way that we calculate bond
order, they go in there and weaken these two metal ligand
bonds. And these two bonds look just
identical to the ones I drew here for e(g)*,
just reversing the phase and making them bonding everywhere.
Let's just look at those.
Here, we have d z squared making a nice bonding
combination.
It is interacting in an in-phase manner with that linear
combination, so we have bonding both in the x,y-plane and
strongly along the z-axis. That is one of these two bonds.
And then over here, we will have x squared minus y
squared, making bonds also --
-- like that. Down here, this is bonding,
and this is antibonding.
Now, that is not the entire diagram.
One runs out of space pretty quickly on these
horizontally-oriented boards for putting together tall MO
diagrams, but there were six LCs over here.
And you will see that there will be another one that makes a
bond with the 4s orbital. And so, accordingly,
the 4s orbital comes up here. There will be an antibond
derived from 4s. There will be also a set of
three that are antibonding derived from interaction of 4p
with the other ones, here.
And there is a big energy mismatch between 4p and these,
so we will draw these as the least stabilized of the set.
But now, what you can see down here is that we have found a way
to form six bonds, six pairs of electrons,
each coming from the highest occupied molecular orbital of
the six ammonia molecules. We have formed six bonds using
these valance orbitals of the metal.
And how many more electrons do we have to put into the diagram?
We have six because this, in the case of our cobalt (NH
three) six three plus --
This is Group 9. Minus three for the 3+ charge.
All the ammonia ligands are neutral, equals six.
So this is a d six case.
The orbitals here that we called t(2g) and e g star
can now take up six more electrons.
And I am drawing them in here with the assumption that this is
a low-spin case. And that is reasonable based on
the 3+ charge of this ion. We have already discussed the
effective charge. It draws the ligands in.
It tends to increase the value of delta O, as we have seen.
And so this is how I would populate that diagram,
with the 6d electrons going into the nonbonding t(2g)
constituting the highest occupied molecular orbital of
this system. And then, stabilize the lone
pairs. Down here what is really going
on, you can think about these six ammonia molecules acting
simultaneously as six Lewis bases to a metal that has enough
empty orbitals to accept six lone pair donations to that same
metal. So it is a Lewis acid times
six. And it is positively charged,
which of course tends to draw electron density to it.
And that is our simplified molecular orbital diagram for
the sigma-only case. Now, there are a lot of other
interesting cases. As you can see here from this
spectrochemical series where the ligands in the middle are
basically sigma-only urea, water.
But then at either end of the spectrochemical series,
you get into systems where you must take pi effects into
account.
And we will treat, first, the case of a pi donor.
Remember that if we are talking about a donor,
that means we are effectively talking about a Lewis base.
And you have to ask yourself the question in a system that
has a ligand that can make pi bonds, are the orbitals on the
metal that have pi symmetry filled, or are they empty?
And, similarly, are the pi symmetry orbitals on
the ligand in question filled or are they empty?
And, if you can get all that right, you will really
understand what happens in terms of attenuating t(2g) when you
add in pi effects. We have seen that the e(g)
orbitals of a metal are the d z squared and d x
squared minus y squared that point along the
axes. The xz, yz, and xy that point
between the axes have the potential to make pi bonds with
ligands. And so here is an example of a
type of ligand that can make pi bonds to a metal.
This ligand is NH two minus.
And it can make both sigma and pi bonds to a metal.
It is usually the case that if a ligand is capable of making pi
bonds to a metal, it is also capable of making
sigma bonds. And so the pi effects are sort
of mapped onto the sigma framework.
And so what we found over here and here about the sigma only
case, that will also be true for pi systems.
But then we are superposing on that diagram the pi bonds that
can form in the system. And the way that NH two minus
can make both sigma and pi bonds with a metal is,
first of all, it has a lone pair that is
directed at the metal, like this.
For forming sigma contacts. And, if you have six of these
around a metal, the situation would be just the
same as what we derived over here for the cobalt hexamine.
But this is a planar system. The sum of the bond angles
around the nitrogen in systems like this are 360 degrees
meaning that, unlike ammonia itself,
which is a trigonal pyramid, this is a planar system.
And so it has a lone pair perpendicular to that plane.
I will draw it that way. It is just a pure p-orbital,
perpendicular to the plane of the substituents on that
nitrogen. And the way that could make a
pi bond with the metal is through simultaneous donation
above and below the plane of its substituents on that nitrogen,
so that this ligand can interact as a double Lewis base
with a metal if the metal has appropriate orbitals available.
And so a situation in which you might find this would be the
following.
An example of such a molecule, this would be a neutral one.
And what you see is that the planes of these NH two
units are lining up with the coordinate xz,
yz, and xy Cartesian planes in the molecule.
Since each of these ligands is negatively charged and the
molecule is neutral, the chromium is therefore in
the +6 oxidation state to balance the six negative charges
on the amido ligands, we call them.
This is amido. And so this one,
in fact, is a d zero case.
And d zero is an electron count that is pretty
typical for the formation of pi bonds from ligands that can
donate into the metal center. If you have d electrons present
in a system like this, then you won't have the empty
d-orbitals on the metal that are necessary to accept bonds formed
by lone pair donation of this sort.
The system will become choked up on itself with too many
electrons. Let's see what those
perturbations will lead to for our MO diagram.
I am going to focus in on linear combinations of amide
lone pair orbitals that, in fact, have the correct
symmetry to interact with our t(2g) set, since that is our pi
set of orbitals from the d manifold on the metal.
So let's look at this.
This would be d(yz).
And then, let me choose --
We can make a pi bond of this sort using a linear combination
that I am mating up together with the d(yz) orbital.
And while I have drawn here the bonding counterpart there,
of course, will be an antibonding counterpart for
that. Let's draw the one that belongs
to d(xz).
And that will be the one up top, here.
So what we have is pi bonding happening here,
here, here, and here that corresponds to pi lone paired
donation into that metal orbital.
And then, finally, we have one that involves the
d(xy).
And this would be with our hydrogens up and down,
here and here. Sorry.
Up and down in front and in back like that,
so that we can make two pi bonding interactions over here
with this. There is pi bonding,
pi bonding, bonding, bonding.
At this point, we have gotten to where we can
recognize how there are three linear combinations of the pi
lone pairs from these amido ligands that will have the
correct symmetry to interact with the t(2g) set from our
metal. There are three more where you
would just flip one of the two on each, and those would have
the correct symmetry to interact with the px, py,
and pz orbitals on the metal. I am not going to draw those
out since we are focusing on the d part of our diagram at the
moment.
And what we will find then, in the molecular orbital
diagram for a system of this sort, is that our d manifold,
this is our 3d, 4s, 4p, is such that,
you remember previously t two g came straight over.
And now, instead, we are going to see that it
goes up in energy because what t(2g) now is,
it still has a lot of d(xz), d(yz) and d(xy).
But now, t two g has acquired a pi star
character due to the fact that we have these linear
combinations over here. These are our pi LCs that we
drew over there that are of the appropriate symmetry to interact
with d(xz), d(yz) and d(xy). And, therefore,
these that are filled are stabilized, come down here and
form a t two g set that is bonding because those,
of course, are filled in, NH two minus.
And, if the metal has an empty t two g set,
you see that we do get three pi bonds.
t two g now gives us pi antibonds.
This is being mapped on the sigma framework that is the same
as what we have right here for the cobalt hexamine,
so that above t(2g) we will have the e(g)*.
And will now recognize that e(g) is sigma star
rather than pi star, so it is more strongly
destabilized due to the greater overlap considerations of
forming sigma bonds versus the side to side overlap of pi
bonds. And the key observation here is
that pi donors --
-- decrease the magnitude of delta O by raising the energy of
t(2g), by making t(2g) pi* in character.
Antibonding. We can see, not only that t(2g)
should be raised up, but that mapping that onto our
full sigma framework diagram, delta O is shrinking.
We still have our six sigma bonds between the NH two
ligands and the metal. They are down here,
just not drawn. It is the same as up here.
And now, we have three additional pi bonds.
There are another three additional pi bonds that you can
draw because we do have six of these NH two minus
ligands. And then, the corresponding
antibonding orbitals involve 4p. And they are way up here
somewhere in energy and very empty.
Since this was a d zero case, what we have is that you
have we pairs of electrons on six NH two minus's
that are down here in this manifold of metal ligand sigma
and pi bonding.
And so we have 12 pairs of electrons down here that
describe the sigma plus pi bonding between the metal and
the ligands. And then you come here to your
d manifold, these are empty because this is a d zero
case, because I happened to choose chromium in the six plus
oxidation here. So it is a d zero case,
and we wouldn't have any electrons to put in here.
But, under these circumstances, in order to get color in a
system like this, you would have to be promoting
an electron from down here from these pi bonds probably into the
d manifold, for example. And what we have seen here,
namely that pi donors decrease delta O should give you a clue
as to what is going on with chloride and fluoride.
Since those have small values of delta O, they lie at a
position of weakness in the spectrochemical series.
And in the converse, that ligands like cyanide and
especially carbon monoxide are very high in the spectrochemical
series because they are pi acid ligands.
And this is something you will be getting more information
about in recitation this week. See you all on Wednesday.