30. Magnetism and spectrochemical theory


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MIT OpenCourseWare at 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

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12 thoughts on “30. Magnetism and spectrochemical theory”

  1. LemonScience33 says:

    This is Drennan. 🙂

  2. jaismeer chohan says:

    She is my saviour (:

  3. Execratos says:

    This is great! Honestly a huge help towards me 😀

  4. aoifewest says:

    such a great lecture. so smart. I could learn in 30 in what I was thinking not being able to understand. Thanks for sharing your knowledge. My teacher passed over all the explanation and background .

  5. gamealot1999 says:

    I think she is much better than Dr. Tylor

  6. Leo Chia says:

    Can coordination complexes (in aqueous solution under white light) also not have colour if the ligand bonded to the central metal atom is very high/low in the spectrochemical series? Since the absorbed light maybe won't be within the frequencies that human eyes can perceive.

  7. martaschoenle10 says:

    Such a great lecture :). We never went into the math behind the transitions at my school. It is so elegant.

  8. Alvaro Alvites says:

    Where can I find videos of ligand field theory?

  9. Athena Osborne says:

  10. Vincent Doan says:

    Amazing lecturer with absolutely on point organisation of knowledge and exercises to help students understand as she teaches.

  11. relike868p says:

    Is EDTA a strong or weak field ligand?

  12. 김언수 says:

    Come on guys, we are already on 30.

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