The following content is provided under a Creative
Commons license.
Your support will help MIT OpenCourseWare continue to
offer high quality educational resources for free.
To make a donation or view additional materials from
hundreds of MIT courses, visit MIT OpenCourseWare at
ocw.mit.edu.
PROFESSOR: OK, everyone, pay attention
to the clicker question.
If you haven't responded, now's a good time to click in
your response.
All right, let's just take 10 more seconds.
OK, we can do better than this.
So, tetrahedral complexes.
Do you recall tetrahedral complexes with angles of what
between the ligands?
109 .
5.
The ligands' negative point charges aren't facing any of
the d orbitals perfectly.
They're a little bit closer to the orbitals that are 45
degrees off-axis, so those three are the most repelled.
But they're not really directly hitting any of them.
So that's in contrast with the octahedral system or square
planar where the ligands negative point charges are
headed directly toward some of the d orbitals.
So because the ligands in a tetrahedral case are not
headed directly toward any of the d orbitals, there is not a
huge amount of crystal fields splitting, so that's small.
And so, when the splitting is small, then you tend to have
high spin systems. So you put in all of the electrons singly
to the fullest extent of the orbital as
possible before you pair.
And so, since you put them in singly for the fullest extent
possible before you pair them up, that will lead to a high
spin system, which is the maximum
number of unpaired electrons.
So today is then the last lecture on transition metals,
and we've been talking about crystal field theory, and
today we're going to talk about colors and
crystal field theory.
So, colors, there are a lot of beautiful colors in nature,
and some of the beautiful colors you find in nature have
to do with transition metals or other liganded states.
So, we're going to start with an example of how colors can
change, how a molecule's color can depend on its oxidation
state, how a molecule's color can depend on
its liganded state.
So, Dr. Taylor is going to be doing this for you -- should
we do it the demo and then look at the questions or do
the questions first?
Demo first?
PROFESSOR: OK.
PROFESSOR: So, here are some of the reactions up on the
Powerpoint that you're going to be looking at, and as the
reaction proceeds, there's going to be changes in
oxidation state, and also changes in ligand in state,
and that will lead to changes in color.
[DEMONSTRATION]
PROFESSOR: So, how would you describe that first color?
Any predictions?
Have any of you seen this before, what's
going to happen next?
[DEMONSTRATION]
PROFESSOR: So this is called an oscillating clock reaction,
so as it runs through, it cycles between
the different colors.
So, that should happily keep going, and we can consider
what's happening.
So here's the overall reaction here and it can be divided
into two components.
So, why don't you tell me, what is happening to iodide in
that first reaction, and there it is again.
All right, so let's take 10 more seconds.
Very good.
So, if you looked up here, you see that you have minus 2 for
the oxygen, three of them, minus 6, and it needs to equal
minus 1, so you have plus 5.
And then over here, we again have plus 1 minus 2, and so
that has to be plus 1.
So that was actually a quiz question, but you score three
points if you answered, even if you got it wrong, but we
could have had that four points, most people got that
right anyway.
But if you answered, you get an extra few
points on that one.
All right, so very good.
That's what's happening to iodide.
Now you notice that in this reaction h o i is being
produced, and then in the second
step it's being consumed.
This later reaction can be divided, then, into two
additional reactions.
And one of the reactions, you can tell me again what is
happening, what is being oxidized and what is being
reduced in this part of the reaction.
OK, let's take 10 more seconds.
Excellent.
Even a little higher than before.
All right, so you figured out that this was the plus 1, this
is minus 1, and they're both going to 0.
So you have oxidation and reduction going on all
involving iodide in that reaction.
OK, so as the reaction was proceeding, you could see that
it started out with clear and then it went to sort of amber
color, and that was the i 2, the clear was i minus, and the
sort of darker blue color that you saw is the complex with
starch in the reaction.
So that both of these guys have colors independent, but
then when they're liganded to something else,
the color is different.
And that's very true with coordination complexes that
the metal by itself will be strongly influenced by what
the ligands are.
And depending on what type of ligands it has, it can have a
really entirely different color than it had before.
So, today we're going to talk about why that's true, and how
you can predict colors based on what type of ligand is
bound to a transition metal.
So, a lot of transition metals have really beautiful colors,
and my laboratory studies metals bound to proteins, and
often the proteins will have really beautiful colors
because of the metal cofactor involved, and that's one of
the things that I liked about that particular area of study
is just how beautiful these proteins can be.
So, the color given off depends on the nature of the
metal, and depends on the nature of the ligand.
And so we can use crystal field theory, which again, is
a very simplified theory, to try to predict or explain the
observed colors.
And so again, this is not always very precise, but
given, if you're told information out of color, you
can rationalize why that would be true, and you can also
predict at least a range of color that you would have
under certain circumstances.
So, let's take a look at this more.
So the ligands again, have the ability to
split those d orbitals.
And when we're talking about for metals, it's all about the
d orbitals.
And we talked already about strong field ligands and weak
field ligands, and we're going to talk more about that today,
but this time in terms of color.
So a strong field ligand, as we discussed, creates a big
splitting in the d orbital energy, whereas weak field
ligand, like the first question you had today with
tetrahedral complexes, there's usually a weak field there, so
we have a small energy separation
between the d orbitals.
So here is something that you actually have to memorize --
there's not much memorization in this course, but you do
need to memorize these six ligands in terms of their
ability to split d orbitals.
So, on the side, we have three that are strong field ligands,
cyanide, c o, and ammonia, and so those are going to be
strong field, so they're going to have big splitting energy,
and so they'll tend to be low spin.
Then we have three that are sort of in between that are
intermediate field -- water, hydroxide and f minus.
And so, in comparing, those are intermediate, so you're
going to be asked questions such as how does that compare
to a weak field, how does that compare to a strong field.
And then our weak field ligands or a lot of our
halides down here, i minus, b r minus, c l minus.
And so those are weak field ligands, so you'll have a
small splitting, so they'll tend to be
in high spin complexes.
So let's take a look at some examples.
So we talked about iron before in complexes, and now we can
consider two cases where we have iron plus 3, so the same
metal in the same oxidation state, but it
has different ligands.
It So in one case you have a high spin system with six
water ligands, and then the other in a low spin system
with six cyanide ligands.
So first, before you do anything else with this, you
always have to think about what the d count is.
So, to do the d count, we're going to look at where iron is
in the periodic table, and we're going to see
it's in group 8.
And then, we'll have 8 minus 3, the oxidation
number is d 5 system.
So now we have two diagrams here, one has a big splitting,
one has a small splitting, and why don't you fill in for me
in a clicker question, what the high spin
system would look like.
OK, let's just take 10 more seconds.
Very good.
I think that's one of our highest numbers in a while.
That's right, so we're going to fill to the fullest extent
possible before we pair any of the electrons.
So again, here we put in electrons down here, and then
go up here, because the splitting is small, so it
doesn't take that much energy to put an electron in the
upper orbitals, it takes more energy to pair the electrons
for this weak field system.
And so, this is a high spin case, we have a maximum number
of unpaired electrons.
So, over here when we have a much bigger splitting energy,
it's going to take a lot more energy to put electrons up
there, and so we're going to fill up all of these orbitals
down here until we have to put an electron up there.
So, if we do that, put in the three, and now we're going to
pair, because it takes less energy to pair than it does to
put an electron up there, so we do four and five.
And so then here is our system where we have a strong field,
and so here's a weak field and it's going to be high spin,
maximum number of unpaired electrons, here we have the
strong field and it'll be low spin, the minimum number of
unpaired electrons.
And we're doing this, again, because we know that cyanide
is a strong field ligand, whereas water is an
intermediate field ligand, it's a
lot weaker than cyanide.
OK, now we can continue doing some of the things that we've
done before.
So just to review this material, we can write the d n
electron configurations.
What are the orbitals called that are down here?
Yup, t 2 g.
And how many electrons do we have there? three.
And then the e g system up here with two electrons.
And over here what do we have?
Yup, t 2 g to the 5.
So, a review, electron configurations, these are just
shorthand notations, which tell people what these
diagrams look like.
Then, we also talked about this before.
So, what does this term stand for?
Crystal field stabilization energy, right.
So now why don't you tell me what it is for
the high spin system.
OK, let's just take 10 more seconds.
Yup, zero.
So if we look at that, we have three electrons down here, so
that's minus 2/5, two electrons up here, which is
plus 3/5, so 3 times minus 2/5 is minus 6/5, plus
6/5 gives you 0.
So here is a case where you really don't have much
stabilization.
It would be equivalent then, so there's zero stabilization
because you have three electrons down and two up.
All right, so what about the low spin system now?
So if we can look at that, what are we going to have for
this system?
So, we have 5 times minus 2/5 or minus 10/5, and we also
have two pairing energy terms there, which we can write in,
because there are two sets that are paired.
So this is mostly review on what we've had before, and we
haven't talked about some of these things in a little
while, so we go over it again, and now we're going to take it
a next step and think about what sort of wavelength would
be absorbed if you're going to promote some of these
electrons to unfilled orbitals.
So what about the light absorbed by these octahedral
coordination complexes?
So, if you remember back to the beginning of the course, a
physics course or high school, a substance absorbs photons of
light if the energy of the photons match the energy
required to excite those electrons to a
higher energy level.
And so now we are doing some review from the first part of
the course, which is always good.
As I mentioned, everything kind of comes together and we
need to go over everything for the final, but there's also
connections between all the different
parts of the semester.
So this should look familiar to you.
So, the energy of the absorbed light equals Planck's constant
times the frequency of that light.
But now we can make that equal to another term, a term we've
been talking about in this unit, and that is our
octahedral crystal field splitting energy.
Because the energy that's going to be required to bump
an electron from here to here is that energy,
that splitting energy.
So that's going to be equal to this term here.
OK, so what does this mean in terms of the wavelengths of
lights absorbed by different coordination complexes.
So we can think about that.
So if you have high frequency of light is absorbed, the
wavelength of the absorbed light is going to be short.
And we know that relationship -- again, think back to the
beginning of the course and also probably to physics and
to high school, we know a very handy equation for telling us
about the relationship of frequency and wavelength of
light, so we have the speed of light equals the wavelength
times the frequency.
So if you have a high frequency of light absorbed,
the absorbed wavelength is going to be short.
So, let's look at a couple of examples, the example here,
going back to our example.
So we had this high spin system with water, and now I'm
telling you that the splitting energy is 171 kilojoules per
mole, and when you have cyanide as your ligand, your
splitting energy is 392 kilojoules per mole.
Again, this was a stronger field ligand, so we have a
bigger splitting energy.
And this was an intermediate field ligand, certainly weaker
than cyanide, so this has a smaller splitting energy.
So, from these values now, we can calculate the wavelength
of absorbed light.
So for the high spin system first, we can rearrange these
equations, which you know well, to come up with the
rearranged equation, the wavelength equals Planck's
constant times the speed of light, and divided by e, and
this time our e is that crystal
field splitting energy.
So we can put in these terms. So we have Planck's constant
times the speed of light over our octahedral field splitting
energy, oh, but then we have some other terms here.
Now one thing that you have to pay attention to in this unit
is your units.
So, splitting energies are often given in kilojoules per
mole, whereas you often see Planck's constant in joules,
so we want to make sure that we convert one or the other,
and here it's set up to convert the
kilojoules to joules.
And also, we want our final unit, we're talking about
wavelengths, in meters or nanometers, so we need to get
rid of this mole term, and we use Avagadro's
number to do that.
So now we should be able to cancel our units and get the
correct units.
So we should be able to cancel the seconds over here, and we
should also be able to go in and cancel our moles.
We should be able to cancel the joules and the kilojoules
out, and that should leave us just with this over here,
which is meters.
So this is 7 .
0 0 times 10 to the minus 7 meters.
Does that make sense in terms of a wavelength of light?
Because that would convert to what nanometers?
700 nanometers.
So if you do something strange and you forget Avagadro's
number, you're going to come up with a very interesting
wavelength.
So that's a good way to check to make sure that you've done
the problem correctly.
So, 700 nanometers, anyone remember what light that is,
what color that corresponds to?
Red, so it's absorbing red light.
All right, so now let's just do the same thing for the low
spin system with the cyanide ligand, and we're going to
plug in our 392 here, and we get 305 over here.
And so, that is a much shorter wavelength of light.
So again, light absorbed for the compound with water, 700
nanometers, and the compound of iron with cyanide, 305, and
so we're absorbing a red light over with the water compound,
and sort of purple or violet light is being absorbed in the
cyanide complex.
And we're talk in a minute about the light that is being
transmitted, which is complementary to the color of
the light absorbed.
So by knowing something about splitting energies, by knowing
something about the types of ligands, then you can know
something about colors.
So now, for another example, we're going to look at the
different colors of two chromium complexes.
So first, what is the oxidation number of chromium
in the water complex here?
What is it?
And what about over here with n h 3 ligands.
Plus 3.
So, plus 3 and plus 3.
And what is our d count?
You know where chromium is, in what group?
Six.
6 minus 3 is 3, so we have a d 3 system.
What does c n mean again?
Coordination number, so what is that for both of these?
Six.
So there's Six things coordinated to the chromium,
and so we have, again, an octahedral system.
And what type of ligand is water?
Intermediate.
And what about n h 3?
Strong.
So this is strong, and water is and intermediate ligand,
and certainly, it is weaker than n h 3.
So, we expect one system over here for a strong ligand, and
then something that's weaker than that strong ligand.
All right, so here are two diagrams, one with a big
splitting energy, and one with a smaller splitting energy.
Are the diagrams going to look same or different?
The same.
Because we only have three electrons, so they're going to
be the same.
In both cases, we put in the three electrons in the lowest
orbitals, and then there's no decision to be made, because
there isn't that fourth electron, so we don't have to
decide which place to put this.
So these diagrams are going to look the same, and before when
we were doing this in this unit, we said OK, we're done,
they look the same.
But now, we realize that these are really not the same
compounds and that they're going to have different
properties, even though their diagrams are
going to look the same.
Because the energy that it's going to take to excite an
electron here is much smaller than over here, and that's
going to result in a different wavelength of light being
absorbed in these two different cases, which will
mean a different wavelength of light being transmitted.
So, again, here we have a weaker field and here we have
a stronger field.
So again, we can go through and think
about these two cases.
So when we have a smaller term here, that means a lower
energy -- this is a splitting energy, and so we'll have a
lower frequency.
When we have a larger case or higher energy, we have a
higher frequency.
Again, if you have a lower frequency absorbed, we have a
longer wavelength absorbed, and in this case, the higher
frequency translates into a shorter wavelength absorbed.
The color of the transmitted light is complementary to the
color of the absorbed light.
So now we can think about --
I always want to ask, when do people learn about
complementary colors?
Is that 6th grade, earlier?
I don't really remember, but I think it's pretty early on.
And people always ask me, are you going to have that on the
equation sheet, or do I have to actually remember my
complementary colors?
And I tell people that I will put some version of this on,
so you don't have to review your kindergarten notes for
this class, if that's when you learned it.
It's pretty early, I don't know when it is exactly.
So, the color of the light is going to be complementary
about -- this is very approximate.
So here, if the transmitted light is shorter because we
have this weaker splitting, then we are expecting we're
going to have this shorter wavelength, the transmitted
light, and experimentally, if you make this compound you'll
see it's violet.
In this other case, we have a stronger field ligand, and so
you have a larger energy, higher frequency absorbed,
shorter wavelength absorbed, then you're going to have a
longer wavelength from your transmitted light.
And that this compound, if you make it, is actually yellow.
So if we go back to our colors for a minute, you see that
when you have the shorter wavelength, we have a violet,
and that is a short wavelength.
And in this case for the strong field ligand, we are
going to have transmitted light of a longer wavelength
and it's yellow.
So it's the same oxidation state of chromium, it's the
same -- it's an octahedral complex, six ligands in both
cases, same octahedral crystal field diagrams, but yet one
compound has a violet color, and the other one
has a yellow color.
All right, so one can also be asked to calculate a crystal
field splitting energy in kilojoules per mole, given the
appropriate information.
So we've looked at when a splitting energy is given, and
we've been asked to calculate wavelength absorbed, you can
also be asked to go in the other direction.
And so here we have another chromium complex to work with.
And we're told that the wavelength of the most
intensely absorbed light is 740, and so what would you
predict the color of this to be?
It would be greenish.
So that would be what you would predict.
Again, chemistry is an experimental science, but
based on having a complementary color to the one
absorbed, that would be a guess.
All right, so we can actually calculate the frequency of the
light absorbed.
So we were given the wavelength, and use speed of
light, and plug in your wavelength and you can come up
with a frequency, 4.05 times 10 to the 14 per second.
Then we can calculate from that the crystal field
splitting energy, and so we use Planck's constant, and we
have our frequency, and we calculate 2 .
6 8 times 10 to the minus 19 joules.
Am I done with the problem?
What does the problem ask for?
It asks for it in kilojoules per mole, and so, we're not
done, we need to convert to kilojoules per mole.
And I'm making this point, because often this is where
people lose points on the final exam, and that's not
where you want to lose points.
You want to lose them on a really hard problem, not on
something like this.
So, most of the time you're asked for kilojoules per mole,
so make sure that if that's what asked for, that's what
you provide.
So here, we can just do the conversion of units, and then
we're going to use Avagadro's number to
give us that per mole.
And so, this translates into a 160 kilojoules per mole, which
you might recognize is more similar to the other numbers
that you saw for octahedral crystal
field splitting energy.
All right.
Sadly, there are some coordination complexes that do
not have colors.
Why would that be?
Why would something not have a color?
It has d orbitals, it's a transition metal.
So what would be true about all of the d orbitals?
Yeah, so one example, is if they're all full, and that is
the most common thing that we see, so it's not possible to
have a d to d transition in the visible range.
So there are a number of examples of metals who have
this situation.
Zinc and cadmium are two of the most common that give you
problems in biological systems. So why is this?
Well, that's because they're over here in group twelve.
But their most common oxidation states are plus 2.
So, if you have 12 minus 2 you have a d 10 system, and that's
the case for both of these systems here.
And so all the d orbitals are filled.
Now, zinc is a really important metal in biological
systems, and because it has all these d orbitals filled,
it doesn't have a color.
And so it's very hard to tell if an enzyme molecule has
zinc. And I think one of the problems that you have on this
problem-set talks about how zinc is important in a
biological system by altering the p k a of a residue that
coordinates to it.
And that's often its job, and so biochemists are often faced
with the problem of trying to figure out if their protein
has zinc, but they have no color of the protein, also
they might try to look for a paramagnetic
or diamagnetic system.
They're not -- you know if they see a paramagnetic
system, they say oh, unpaired electrons, we know we must
have metal involved, but there's no sort of
spectroscopic probe for zinc.
And so, often someone will determine a crystal structure,
and it'll be a huge surprise that there's zinc associated
with this protein.
Do you think there's a lot of proteins that use cadmium as
of part of their mechanism?
What do you know about cadmium?
Yeah, cadmium is poisonous.
Old barbecue grills were sometimes, they used to coat
things with cadmium on a barbecue grill.
Yeah, that was not very smart.
So, cadmium poisoning is a problem, and people have been
trying to figure out the mechanism of that, but again,
it's hard to study cadmium because it has no
spectroscopic signal.
All right, so getting back now to just kind of review over
what we've talked about in terms of colors.
So we have our weak field ligands, again, you need to
memorize what they are.
You have your intermediate field ligands, which you need
to memorize, and also your strong field ligands.
So these weak field ligands are going to have a small
splitting energy, and that means that in terms of how the
complex absorbs, low energy, low frequency, long
wavelength, and that the color transmitted will be
complementary.
And usually what this means is that it'll be sort of in the
end of the spectra, so it's often hard to say well, red
will definitely be green.
So it's not a perfect agreement, but you can usually
say well, it's probably going to be in the
blue-violet or green end.
So in sort of one part of the spectra.
Strong field ligands, again, have a huge splitting energy.
So you're going to have big energy, high frequency, short
wavelength, and so it's going to transmit, then, in the
complementary, so it should be in the yellow,
orange or red end.
And you will be asked in some of the problems, in again,
problem-set 9 due Wednesday, and you should be able to
finish that up pretty quickly tonight after this lecture,
and there are some problems on this.
So, for cobalt complexes, you get pretty much the entire
range of colors.
So I'm just going to end with one more biological example.
And here are some pictures of actual colors, and so this is
cobalt coordinated to vitamin B12.
So one ligand gives you this brilliant red color, another
gives you an orange color, and a third ligand gives you a
pink color.
So you can tell the oxidation state of vitamin B12 by the
colors of the molecule.
All right.
Now you have all the information to finish your
problem-set, and that's the end of transition metals.
On Wednesday we start kinetics.