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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.

Design & Developed By ThemeShopy

@ediyakobov

This is Drennan. 🙂

She is my saviour (:

This is great! Honestly a huge help towards me 😀

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 .

I think she is much better than Dr. Tylor

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.

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

Where can I find videos of ligand field theory?

https://youtu.be/RzMrxXQyEoE

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

Is EDTA a strong or weak field ligand?

Come on guys, we are already on 30.