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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.
