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PROFESSOR: So, moving
to today's handout,
we're going to talk about
crystal field theory.
And we're going
to talk about how
the shapes of those
orbitals can explain
the special properties
of transition metals,
like color and magnetism.
And really, as I said,
the transition metals
have amazing properties that
can be absolutely gorgeous.
They can do incredible things.
They're very, very
useful elements.
I like to think of them
as the super heroes
of the periodic table.
They're doing all the
really spectacular stuff.
So, today, we're going
to talk about colors.
And we're going to talk about
whether things are magnetic,
if they're paramagnetic,
or diamagnetic.
So, transition metals.
I talked before about how
they're useful in biology.
They're also useful in terms
of their beautiful colors.
So, in the old days before
you generated floor plans
and drawings for
buildings, on a computer
you did blueprints
for something.
And the blue in the blueprint
is actually a transition metal
complex.
And also, currently, it's
an active area of research
to design new imaging agents
to use in MRIs, or to image
the heart, or other things.
And a lot of these also involve
transition metal complexes.
And the reason why they're
useful, say, in an MRI
is because they can have
magnetic properties.
So, scientists see
these beautiful colors.
It's kind of obvious that
the transition metals often
have these incredible colors.
And they can put the
transition metals in something
with a magnet and see that
they have magnetic properties.
So these things were obvious.
And then chemists want to try
to understand why it's true.
Why this particular metal
behaves this way, or this metal
has this particular color.
And they like to try to
categorize them and come up
with theories that
help to explain what
they're observing in nature.
And so, today, we're
going to talk about one
of those theories, which
is Ligand field theory.
And I'm also going to
differentiate that-- or Crystal
field theory and differentiate
it from Ligand field theory.
So both of these
theories, again,
are an attempt to explain the
properties that we observe.
And the idea behind
it is pretty simple.
You have a metal ion and a
set of ligands around it.
And if you compare the energy
levels of that free metal ion--
so here's a free metal
ion-- with that metal
ion in a coordination
complex-- here
is quite a pretty
coordination complex--
the energy levels are different.
That's really all these
theories are getting at.
That there's a change
when you bring ligands
in around a transition metal.
It changes the energy
levels of those d orbitals.
And if we understand how
those de-energy levels change,
we can understand a lot
about the properties
of that particular metal in
that coordination environment.
So energy levels are altered.
So there's these two
theories: Crystal field theory
and Ligand field theory.
Crystal field theory
is the more simple one.
It just considers the ionic
description of the metal ligand
bond.
And so we talked
about ionic bonds.
And mostly you're just
kind of thinking about them
as these point charges.
And you have plus,
you have minus,
and they attract each other.
Covalent bonding is a
bit more complicated.
Start talking about
hybridization and other things
like that.
So Crystal field theory
is really simple.
It just considers
the ionic nature.
And it does pretty well.
It can explain a lot,
but not everything.
Ligand field theory has
the covalent and the ionic,
so it's a more
complete description
of that metal ligand bond.
But this is more of
an advanced topic.
I just want you to
know that that exists.
So if you take 503 here
in organic chemistry,
we'll talk more about
Ligand field theory.
In this class we're just going
to do Crystal field theory.
But I like Crystal field
theory because I like it
when something that's
pretty simple theory
can explain a lot.
I think a lot of nature
can be explained by this.
So if you really
want to understand
particular properties
of a specific compound,
you probably want a
more complicated way
to think about it.
But if in general
you just want to say,
why can some cobalt complexes
can be so many colors?
Crystal field theory works
really well for that.
All right, so
Crystal field theory.
Very simple idea behind this.
It's almost
disappointing, I think.
But it's OK.
You can go, yeah we're
studying, you know,
Crystal field theory in class.
Yeah, you don't have to
tell people that it's
a really, very simple idea.
So the idea is that the
ligands are negative charges.
They're like this big
blob of negativity.
And those ligands,
with their lone pairs,
their negativeness
or negative charge--
they got those extra lone pair
electrons-- are going to be
repulsive to those d orbitals.
Negative point
charges, you bring
a d orbital, And a lone pair,
and there's some repulsion.
That's basically it.
Negative point charge repulsion.
That's all that you
have to think about
with Crystal field theory.
So it sounds really impressive
to tell people you're
doing Crystal field theory.
And you can neglect to
mention that it's just
a really very, very simple idea
that, sort of, negative charges
repel each other.
That's really all it is.
All right, so we're
going to talk first
about the octahedral case.
And then we're going to get
into some other geometries
on Friday, so that's
going to be exciting.
But here's our
octahedral molecule.
And here we have it, here.
And I can think about the metal
in the middle, M to the N plus.
So this just indicates it's
any metal and to the oxidation
number.
And the black balls,
here, are the ligands
and we can think of those
as negative charges.
So here, we have drawn sort
of this octohedral shape
around the metal in the middle.
And we can think
about these NH3 groups
as these negative point charges.
So these dots here are
negative point charges.
And so now if we think about
the shapes of the d orbitals
again, that we just looked at,
we can look over here and say,
wow, some of those ligands
that are those negative point
charges are basically
right there up
against that d orbital.
Pointing directly
toward the d orbital.
So that's going to
be pretty repulsive.
The point charge is right
there next to the d orbital.
This also has a point charge
kind of right toward it
along z.
But these, the point
charges, are on axis.
Octahedral, you have
your ligands along z,
along x, and along y.
But these sets of orbitals,
the negative point charges,
aren't pointing directly
at the d orbitals.
So one could think
about the fact
that these are not going to
be not as much repulsion.
This is a lot of repulsion.
All right, so let's go through
orbital set by orbital set
and consider how
repulsive it's going to be
compared to the other orbitals.
OK, so if we look over here
at the octahedral case--
and I have both, these
drawings were on the first page
of your handout.
This is on the second page.
But you can think about them.
Here it's drawn more
where they're all
drawn kind of the same way.
Here it's sort of moved a
little so you can see better.
So we can keep both
of these in mind.
So, again, the negative point
charges around the orbitals
here.
And on these drawings, I have
these little negative signs
pointing in to give you a sense
of where those negative point
charge ligands are coming.
So we have our dz2 and we can
see the negative point charge
right toward this maximum
amplitude along z.
So that's going to be repulsive.
Dx2, y2, same thing, point
charges right toward these.
So the ligands point directly
at dz2 and dx2 minus y2.
So there's a large
repulsion there.
And, in this particular
case, in the octahedral case,
what turns out to be true,
both of these are stabilized.
And they're destabilized,
actually, by the same amount.
And so that means that they
have the same energy, which
we learned before,
means that there-- you
could say they're degenerate
with respect to each other.
So both of these, big repulsion.
Both are destabilized by
those negative charges,
by the same amount.
And you have much more
repulsion, and much more
destabilization, for these
two orbitals compared
to the other set of orbitals.
The ones that are
45 degrees off axis.
So now, let's look
at these and look
at where the point charges are.
So this is the picture
in your handout.
I just kind of put
these up there,
too, so you can think about
where those point charges are
for octahedral geometry.
So the ligands,
again now, are not
pointing right at the orbitals.
They're off axis.
Because of ligands are on axis,
and the orbitals are off axis,
and they're not directly
pointing toward each other.
And so this is a
lot less repulsive
than the other
situation where you
have the ligands
and the orbitals,
both on axis, pointing
right toward each other.
So now, this is why it's
important to know which
orbitals are 45 degree
off axis because we, now,
can think about the
fact that there will
be less repulsion in this case.
So, we can say then, that
the xy, xz, and yz orbitals
are stabilized compared to
dz2 and dx2 minus y2 orbitals.
And they're also stabilized
by the same amount.
So these are also degenerate.
With respect to each other,
they have the same energy.
All right, so we have these
now two sets of orbital types.
Two orbitals that
are destabilized,
and three orbitals that are
stabilized, compared to them.
So now let's draw what's known
as an Octahedral Crystal Field
Splitting Diagram and think
about what's happening.
So, I have some over there.
Ignore those for now.
I'm going to draw one over here.
So first, we have
energy going up.
And we have a situation we
first want to think about.
What would be the case, say,
if you had ligands everywhere?
Not just on the axis,
not octahedral geometry,
but just everywhere.
So if you had
ligands everywhere,
you would have
your 5 d orbitals,
all would have the same energy.
Because all of them--
there's ligands everywhere--
all of them would
be experiencing
the same amount of repulsion.
So this would be the case for
a Spherical Crystal Field.
And I like to think about
this with this prop.
I had this before
as an s orbital.
It's convenient
for that as well.
But, really, what it is, is a
Hypothetical Spherical Crystal
Field.
Because these are
all little ligands.
And you can see these
little ligands are
completely symmetric around.
And if you feel
sort of the inside,
there's something in there.
Well that, of course,
is a metal ion.
And these ligands are
completely around.
So all the d
orbitals are equally
feeling the repulsive
effects of these ligands.
And so they all
have the same energy
in this hypothetical case.
I was so excited when
I found this at CVS.
And I went up and
I was, like, wow,
you sell hypothetical
spherical crystal fields here.
And the person was not amused.
OK, so now what happens--
because this is not the case.
This doesn't exist.
This is just hypothetical.
It looks like it
exists, but it doesn't.
This is what really exists.
You have certain geometries.
The ligands come
together with the metal
and form particular geometries.
And we already talked about
what those geometries are.
There's not a whole
lot of variation.
We get the same
kinds of geometries.
And octahedral is the
most common geometry
for a transition metal complex.
So what's going to happen here?
So, the ligands will now split
the energy of the d orbitals.
And three of them are
going to go down in energy.
They're going to be stabilized
compared to the two that
go up in energy.
So the three that go--
are stabilized again,
are dxy, dyz, and dxz.
If you can't read this, it
all should be in your handout.
So it's OK, you don't have
to read my handwriting.
And just kind of-- just
follow along in your notes.
And then the ones that
have the most repulsion are
dx2 minus y2 and dz2 up here.
All right, so two
go up and energy.
There's more repulsion.
Three go down in energy.
And instead of writing
all of those d's
and x's and y's, they're little
abbreviations that people use.
So they call these two
orbitals-- the eg orbitals.
So those are the ones
that are destabilized.
And the ones that
are stabilized are
called t2g orbitals,
which is going
to make our nomenclature
easier later on.
OK, these two again.
Same energy I tried to draw
a straight line between them.
They're degenerate with
respect to each other.
These three of the same energy,
again, degenerate with respect
to each other.
And then the difference of
how much it's being split.
This energy difference
has a special name.
It's sort of this delta sub o.
And this is the octahedral.
And that's the o.
The o is octahedral.
Crystal field splitting
energy because it's
the energy that shows the
splitting of the crystal field.
So that's at least
a good name for it.
All right, now,
overall this energy
is going to be conserved.
So the orbitals that go--
that are destabilized--
and two go up and three go down.
So we can also put that the two
that go up, go up by 3/5 times
the octahedral splitting energy.
And the three that go
down-- are stabilized--
go down by minus 2/5 times
the octahedral crystal
field, splitting energy.
So this is why I have
it in your notes,
because you can't really
read my writing too well.
So two go-- are
destabilized, up by 3/5.
Three, or stabilized,
down by 2/5 in energy.
All right, so now what
controls the overall size
of this splitting energy?
How do you know?
Is it a small splitting?
Is this, you know,
should I have drawn this,
that they're just a tiny bit?
Or is it, like, could it be
a really large splitting?
What determines the splitting?
Or The magnitude of
this splitting energy.
So what determines that
are the nature of ligands.
So some ligands are
really repulsive.
Big splitting.
Others-- eh-- you barely
notice they're there.
Very little splitting.
There's always some splitting.
But it can be different.
So the relative
ability of a ligand
to split the crystal
field gives rise
to what's known as the
Spectrochemical Series.
So it's the relative
ability of common ligands
to split the energies
of these d orbitals.
And this is what gives rise
to these beautiful colors.
So all of those colors are
possible from cobalt compounds.
Every single one of them.
Because different ligands
will split the energy
different amounts, giving rise
to those different colors.
So we have what's known
as Strong field ligands.
And as that name would
suggest, Strong field ligands
are going to have
a large splitting.
They're very strong.
They split to a large degree.
And we have Weak field ligands.
Weak field don't cause
much of a splitting.
It's a small delta sub o,
small octahedral crystal
field splitting energy.
All right, so which
common ligands
fall into which categories?
Here is the list that we
talk about pretty much
in this class.
I don't think there
are any other ones.
So what I will put on
your equation sheet--
your equation
sheet for exam four
is lots of cool,
interesting things
that were not in the
other equation sheets.
And I will tell you
which are weak field,
and which are strong
field, but I will not
tell you that weak field means
that it's a small splitting.
That's something that
you need to know.
Or that strong field means
it's a large splitting.
But I will tell you that
your halides down here
are weak field ligands.
Water and hydroxide are
sort of in the middle.
And they can have all
sorts of different colors
when they're bound to
coordination complexes.
So they're sort of intermediate.
And then your strong field
ligands-- with your strongest
being cyanide.
Very strong field ligand.
All right, so, if your
metal has a weak field,
intermediate field, or
strong field ligand,
it can have very
different properties.
Especially very
different colors.
Also, it could be
paramagnetic or diamagnetic,
depending on this.
So let's take a look at two
different iron complexes.
We have iron with six
waters, with a plus 3
charge, and iron with
cyanide, with six cyanides
and a minus 3.
So the first thing we
have to do is figure out
the oxidation number of
the iron and the d count.
And that's our first
clicker question.
All right, let's just
take 10 more seconds.
OK.
There you go.
There's the right one.
A little more people say that.
All right.
So, let's take a look at this.
And you will get
used to thinking
about what the charges are
on the various ligands.
So here, we know that
the overall charge
has to be the same as the
overall charge on the complex.
Water is neutral.
And so that means this
iron has to be plus 3.
Cyanide has a minus
charge, and you
may have noticed that
in the list of ligands,
that it was cn minus.
That when I just showed you
the strong field ligands.
So there's six minus ones.
And so it's also
plus 3, to give you
an overall charge of minus 3.
And as you do these
problems, and problem sets,
you'll become very
familiar with what
the charges of the
different ligands are.
So overall, plus
3, 8 minus 3, is 5.
Is a d5 system.
So the next thing
we're going to do
is we're going to draw our
octahedral crystal field
splitting diagrams
and place electrons.
And I already have, over here,
a starting place for those.
So we're doing
here our d5 system.
And we have a diagram
with a small splitting
and one with a large splitting.
So with the small
splitting, is that
going to be a weak
field or a strong field?
That will be a weak field.
And when we have
a large splitting,
that is a strong field
OK, so which of
our two compounds
is going to be the
weak field compound?
And which will be the strong?
Let's do strong first.
Which one has a
strong field ligand?
Cyanide one.
So this one is going to be
our cyanide complex because it
has the strong field.
And so this one
would be the water.
Water's intermediate.
So you kind of have to see
what the comparison is first
before you decide.
Oops, six bracket 3 plus.
OK, so here we have water.
There, we have cyanide.
So a weak, or intermediate
field, versus a strong field.
All right, now we need
to put electrons in here.
So I have to decide.
I have five electrons
I'm going to play with.
And I could put them in here.
Let's just start here.
Do one, two, three.
Now the question is, where
am I going to put four?
Am I going to put
four down here?
Or put it up here?
And so, here, it depends on, if
I put it down here on pairing--
and, you know, the seats
on the bus-- people
don't really want
to sit together
if there are empty seats.
Now there are empty
seats up here.
So, the question is, is it going
to take more energy to pair?
Or more energy to
put it up here?
Now, if this is a weak
field, which it is,
then that's not
all that far away.
You get on a bus and you
see a seat in the back.
You'd rather sit up
front, but you're
like, than you have to sit with
someone, so you go to the back.
So that's what happens
with a weak field.
The other energy levels
are just not that far away,
so you go up there.
And if I kind of put
this up-- move it up
for just a second-- and then
I'll say, so this rule here,
and I'm going to kind
of cover this up again.
But maybe I'll just
put it down here.
This is in your notes.
So here, the octahedral
crystal field splitting energy
is less than we call PE,
which is pairing energy.
So, it's not that far away.
Just put them singly,
to the full extent
possible, before you pair.
So that's what happens
with a weak field.
And we also can think
about how we're putting
in our electrons parallel.
We're not putting them in all
sorts of different directions.
And we're not-- if we do pair--
we're going to not pair--
we're going to pair.
So one's up, one's down.
So all the rules that
we learned earlier
are going to apply here.
So this is good, good review.
So now, in this case, we can put
in our first three electrons.
And what do you think?
Do you think I'm going to pair?
Or do you think I'm
going to go up there?
What do you think?
I'm going pair.
Yeah, I'm definitely
going to pair.
So I'm going to put my
other two over here.
So in this case, the pair--
the octahedral crystal field
splitting energy--
it's so much fun
to say that so many
times-- is greater
than the pairing energy.
And, so, in this case,
there might be empty seats.
But maybe it's a long train.
And there is-- you can sit
in car one with someone else,
or carry all your luggage.
Why did you bring so much?
You're just going
home for Thanksgiving.
All the way, 23 cars later,
to an empty seat in the back.
You're just not going to do it.
You don't have enough energy.
You'd rather sit with
somebody up front.
So how you put in
electrons depends if it's
a weak field or strong field.
It depends on whether it takes
just a little bit of energy,
and it's worse to pair.
Or takes a huge
amount of energy.
There's a big splitting,
a really strong field.
And in that case, you're
going to pair first.
And you're only going to
put electrons up here when
you're completely done here.
All right, so now we have
placed our electrons.
We've talked about
pairing energy.
And we've talked about the
weak and strong fields.
And now we can talk
about some notation.
Because there's always notation.
So this is what
we've done already.
And now we're going to do step
E, which is to our d to the n
electron configurations.
Every diagram, there's different
configurations every time.
So this one we do using those
little kind of cool terms
I told you about.
So instead of writing, I
have one electron in dxy,
one electron in dxz,
and one electron in dyz,
I can just say, I have three
in the t2g set of orbitals
that is a lot more convenient.
So you would just say t2g3.
And I have two up here.
So EG2.
And that's how
you would do this.
I think this is our last
configuration for electrons
in a diagram that you're
going to be learning.
There's a lot of them.
OK, so then over
here, we would just
say we have five, all
in the lower energy
set, all in our t2g.
All right, one other thing, part
F that you can be asked about.
Which is the crystal field
stabilization energy.
Not to be confused with the
crystal field splitting energy.
They both start with an s.
So I don't know why you
wouldn't be confused by that.
But if you can try to remember
it, that would be awesome.
It's often abbreviated, CFSE.
And this is the
energy change that's
due to going from this
hypothetical spherical crystal
field, where all the d
orbitals have the same energy.
And if you put your
electrons in here,
where they'd all have the same
energy, to the diff energy
difference, when you're placing
energies-- when you're placing
the electrons in lower energy
orbitals or higher energy
orbitals.
So you're thinking about
how much more stabilized
is the system if all of the
electrons are down here.
That's going to be
stabilized compared
to where they were here.
But if a lot of
them are up here,
then there'd be less
stabilization due
to this splitting.
All right, so let's look
at how we would write that.
So we have three electrons
that are down in energy.
And they are down in
energy by minus 2/5 times
the octahedral crystal
field splitting energy.
And that's because the
overall energy is maintained.
Three go down in energy,
so that's minus 2.
Two go up in energy,
again, plus 3.
And then we have two
electrons up here.
So 2 times plus 3/5, again times
the octahedral crystal field
splitting energy.
And so overall, our
stabilization is zero.
There's no stabilization
of this system,
compared to the system here,
because three electrons are
down in energy, and two are up.
All right, so now
you might imagine
that this is going to have
some extra stabilization.
And you would be right.
So we would write
this like this.
We'd say there are
five electrons down.
Stabilized in the
lower energy orbitals.
So five times minus 2/5, times
the optic single crystal field
splitting energy, minus 10/5
times the octahedral crystal
field splitting energy.
And sometimes you'll also
see, kind of a little truth,
that there's also some energy
kind of due to this pairing.
So you might see an indication
that there's pairing energy
for two sets of electrons.
So sometimes that's included.
Sometimes it's not.
The questions were asked
to indicate, you know,
how many of sets of
electrons ended up
being paired for
that stabilization?
All right, so we
can think about,
again, we have this
nomenclature here.
And we can think about
the stabilization.
So there's a big difference
between these iron compounds.
One, there's no stabilization
due to the splitting.
And the other one,
there's quite a bit,
because all the
electrons ended down
in the lower energy orbitals.
So now it's time to try
some of these on your own.
See if you got the rules down.
And let's try some
clicker questions.
Clicker question one.
OK, 10 more seconds.
All right, yes.
So the trick here is to remind
yourself what weak field meant.
And in this case, if
it's a weak field, that
meant that you place all
the electrons singly--
there are seven of them-- to
the fullest extent possible
before you pair.
You still have to pair.
But here you put more--
you filled these up
before you paired.
And this is a
strong field diagram
where you put-- you
paired all of them
possible before you put
any of them up here.
And to make it harder, I
made-- drew them the same.
So you have to
really pay attention
to what weak field meant.
But this would be the
version of the strong field.
And this is just wrong.
So, you wouldn't
have more electrons
up in the higher energy
orbitals for really any reason
whatsoever.
OK, so let's try the next one.
All right, let's just
take 10 more seconds
and we'll mention high spin
and we'll do this problem.
Yeah.
People did very well with that.
OK, so I forgot to mention
that I ran out of room to draw.
So I decided that was enough.
So it's in your notes
though, for here.
And when you have a
very small, weak field,
and you end up putting all
the electrons in singly,
that is also known-- and
I'll move this up a little
bit-- as a high spin.
And so you have a lot
of unpaired electrons
in this case.
So that is the maximum
number of unpaired electrons.
And it's easier to
remember because, you know,
you have these spins.
There's a lot of spins.
There are a lot of single
spins and they're also high up.
And this is a low spin case.
When you have a strong field.
Because you like to pair first
before putting them singly.
So this is going to give
you the minimum number
of unpaired electrons.
So high spin, weak field.
Low spin often comes
from a strong field.
So it's about whether
you have maximum number
of unpaired electrons,
which you do
with weak field, high
spin, minimum number.
All right, so let's
look at this one then.
So it's a high field one.
So that means that you're going
to have the maximum number
of unpaired electrons.
So this would be diagram here.
And I can put that over here.
And so then you have to
figure out how to write that.
And so you would have three ones
down in energy, minus 2/5 times
the octahedral crystal
field splitting energy.
And you would have one
electron, up by 3/5 times
the octahedral crystal
field splitting energy.
So minus 6 plus 3 is minus 3/5
times the octahedral crystal
field splitting energy.
So this problem required you
to know what high spin was,
then correctly identify
the weak field diagram,
and then figure out the crystal
field stabilization energy.
So that was actually quite good.
And maybe you'll remember the
high spin, low spin definitions
as well.
OK, so let's just do
one more thing and then
we'll-- because we've just
been talking about high spin.
So back to our compounds, we're
going to talk about magnetism.
And then we're going
to come back next time
and talk about color.
And we have some cool demos for
color which we'll do next time.
But before we leave from
this, our iron compounds,
we want to think about whether
they're paramagnetic, which is
attracted by a magnetic field.
Or diamagnetic, repelled
by a magnetic field.
And we've already talked
about this in this class.
So, based-- and here are
our iron diagrams again.
Based on these
diagrams, would you
expect these to be
paramagnetic, or diamagnetic?
Paramedic.
And that was because it
has what kind of electrons?
Paired, right.
So here you just have to
remember your definition
of paramagnetic.
It has unpaired electrons.
Now, normally, things with a
weak field, and a high spin,
are going to be more
likely to be paramagnetic,
have unpaired electrons.
Things that have a
very strong field.
And so you pair first,
before you fill up.
Minimum number-- low
spin, minimum number
of unpaired electrons are
more likely to be diamagnetic.
But in this case, we didn't
have enough electrons,
so they're still
both paramagnetic.
All right, so we'll stop there.
We'll come back and talk
about the colors of those iron
compounds.
And more colors and
cool demos on Friday.
Don't miss Friday.
Remember, Friday, double
clicker competition.
The team that comes in second is
automatically in the playoffs.
Are you--
AUDIENCE: We're at
166, but it stopped.
PROFESSOR: OK, let's just
take 10 more a seconds
on the clicker question.
All right, let's quiet down.
I'd like to see the
85% that's good.
Of course it doesn't distinguish
all that much for the clicker
competition today.
So if we can quiet
down a little bit.
People can just yell out.
What would this have been
the correct answer to?
What should the question
have said for that
to be the correct answer?
Low spin, right.
So low spin-- shh--
is the minimum number
of unpaired electrons.
And so this is the low spin.
High spin is the maximum
number of unpaired electrons.
So this is the correct diagram.
And C is wrong.
It's not correct for
either high or low spin.
Because we have
two electrons here
that would have the
same four quantum
numbers, which is not allowed.
OK, so let's continue
on with the lecture.
And we are talking about colors.
So, we're talking--
we're talking
about the colors of
the two iron compounds
that we had described before.
And so let's continue.
And when we talk about colors,
we need to do a little review
and think about what's happening
when a substance gives off
a color.
And so, when a substance--
whether a substance is going
to absorb a photon or not, to
excite an electron to a higher
energy level that then,
when it falls back,
will emit a beautiful light.
We can talk about whether
a substance is going
to absorb a photon or not.
And it will if the
energy of that photon
is equal to the difference
in that energy level.
So we talked about this before.
Exam one, exam two,
somewhere around there.
And so, we saw this
equation a lot.
That the energy equals Planck's
constant, times the frequency
of the light.
And now we can take that
same gorgeous equation
and add a little thing to it.
So we can add that
that energy is
going to be equal to the
octahedral crystal field
splitting energy.
And so, a substance, one
of these transition metal
complexes, will absorb
a photon of light
if the energy of that photon
is equal to this splitting
difference.
And if it is, it can promote
an electron between a lower
and a higher state.
So this is all the
same kind of thing
that we talked about before,
but now we're just applying it
to transition metals.
So, let's think about what's
happening with our iron
complex.
And if we say, had
a low frequency
of light that was
absorbed, what would
be true about the wavelength?
Would it be long or short?
It would be long.
And how do we know this?
We know this from the fact that
if you're talking about light,
the speed of light equals the
wavelength times the frequency.
So if you have a low
frequency, then you're
going to have a long wavelength.
So our long wavelengths,
then, are on this
and over here, sort of our
yellow, orange, red, with red
being our longest wavelength.
So if, then, we have a high
frequency of light absorbed,
then the wavelength
would be short.
So it'll be short
wavelength absorbed.
And so, again, our
short wavelengths
are down here, with our shortest
being the violet wavelength.
All right, and now, the
color that you actually see
is the one that is
complimentary to the color that
is of the absorbed light.
So we're going to think
about how big those energy
differences are,
whether that translates
to high frequency, low
frequency, long wavelength,
short length-- wavelength
for the absorb light.
And then, the complimentary
of the absorb light
is the light that we see,
the transmitted light.
All right, so let's think
about the iron complexes
that we talked about.
And we had a high spin, or
a higher spin, iron water
complex.
Water is actually kind of an
intermediate field ligand.
And it absorbs low frequency,
or a longer wavelength of light.
And so it's going to transmit
on the shorter end of things.
Again, water sort of
an intermediate field.
But in this case, it was less
strong than our other one,
which was cyanide,
which is very strong.
And, actually, these
iron water complexes
can appear actually
a variety of colors.
If they're solid, they're
more of this pale violet.
So definitely a very short color
wavelength that we're seeing.
But in solution, and
depending on the pH,
they can be sort of
yellowish, brownish color.
But the strong field ligand,
remember the other compound
had cyanide on it.
That's a very
strong field ligand.
And so that's going to absorb,
then, a high frequency.
So a strong field, so you
have a big energy difference.
So you have a big energy, high
frequency, and therefore short
wavelength.
So it will transmit on
the longer end of things.
And this is actually a
bright orange red color.
So you have a very
long, long wavelength
that you're observing here.
This beautiful red orange.
It's a really brilliant color.
Now interestingly, I mentioned
in terms of applications,
as some transition metal
complexes were used to color
things-- and I mentioned
the blue in blueprint--
and if you take this compound
and actually add iron
in a different
oxidation state to it,
and, form this complex
that has iron plus 3,
and iron plus true with
cyanide, you actually
get the blueprint blue.
So the colors can be just
dramatically different
depending on what gets
added to the system.
All right, so now let's think
about-- it's a little sad--
but things without color.
So what coordination
complexes would be colorless?
What would be true
about the d orbitals
If you had a colorless thing?
Yeah, what would be true?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Yeah, they
could all be degenerate,
so there's no splitting at all.
Which usually doesn't happen if
there are any ligands around.
That's their hypothetical
spherical crystal field,
which I brought with me again.
Because I just
love carrying this
through the infinite corridor
and have people look at me.
But, if you do have the
ligands around, it will happen.
If all of your d
orbitals are filled,
or if the energy
levels are basically
out of the visible range.
So the transitions are not
in the visible range at all.
So, if they're all
filled-- so, there's
no way that you can
move an electron.
Or, if you can.
But it's outside
the visible region
All right, so now let's
think about some examples
of what transition metals
would fit into this.
And I'll bring up
my periodic table.
And we'll also bring
up a clicker question.
All right, let's just take
10 more seconds on this.
All right.
So we can take a look at this.
And with our periodic table,
we have nickel and palladium,
group 10 minus 2 is 8.
Copper plus 2, gold
plus 2, 11 minus 2 is 9.
Zinc plus 2, cadmium plus
2, twelve minus 2 is 10.
And so 10 would be the
correct number for our field d
orbitals.
They can hold 10 electrons.
S if we, then, go and fill this
in, examples would include:
zinc plus 2, and cadmium
plus 2, with our d10 system.
And in fact, zinc plus 2 is
a very common oxidation state
for zinc.
Many proteins require zinc.
Many of you have zinc
in vitamin tablets.
Some people take
extra zinc to make
sure you have enough
zinc plus 2 in your body.
And a lot of times people
who are studying proteins
do not realize it's a zinc
containing protein because they
isolate the protein.
And it's clear, so
they don't think it
has a transition metal in it.
But zinc is hiding
in that protein
because it's colorless so
you don't know it's there.
If you purify a protein with
color with other metals,
it's really obvious
that the metal is there.
But zinc can be sneaky.
Cadmium can also be sneaky.
Cadmium, for the most
part, is a poison to us.
And they used to use it
to coat barbecue grills.
Which is not-- you
don't want to put
a poisonous substance
on a barbecue grill
and then heat it up.
That's a really bad idea.
So if you go to a
barbecue and you think,
wow, that grill looks like
it's, like, 70 or 80 years old,
or something like that.
It looks ancient.
Maybe you don't want to eat
from that barbecue grill.
They don't do this anymore,
but old barbecue grills
had cadmium on it.
And I know someone who
actually had cadmium poisoning.
And it was a really pretty
terrible thing because it's
hard to diagnose that.
But they finally got
the right diagnosis.
All right, so again,
colorless things,
you don't know that they're
there, but they sometimes are.
All right, so what
about cobalt plus 3?
You can just yell this out.
Would this be a
colorless compound?
No.
So we have in our group 9, minus
3 would be sync -- would be 6.
So it's not.
Which vitamin contains cobalt?
And you probably all
know because we've
been talking about it.
Vitamin B-12.
And I just thought I
would share with you
the colors of vitamin B-12.
And so this is crystals
that contain vitamin B-12.
And these are their actual
colors of the crystals.
So it's really fun to
work with vitamin B12.
It's absolutely brilliant.
Except that it's
also light sensitive.
So you have to work in
the dark, under red light.
So everything, then, is red
because you're under red light.
But if you bring them out,
and you expose them to light,
they're really, really pretty.
All right, so we're going to
continue on this cobalt theme.
Because cobalt is one of the
most spectacular transition
metals when it comes to color.
And I'm going to get you
ready for a little demo.
And you're going
to help me first
figure out what colors you
should observe in this demo.
All right, so, we're going to
have a cobalt compound that
has six waters with it.
And you're given the octahedral
crystal field splitting energy.
So now we want to
predict the color.
And when we predict
the color, we're
asking about what
sort of wavelength
is going to be absorbed.
So we can think
about the wavelength
that will be transmitted.
So we need to think
about our equations.
And we can combine these just
like we did in many problems
in the earlier
part of the course.
And so wavelength equals
Planck's constant times
the speed of light.
Now, instead of just
divided by any energy,
we're dividing by the energy
that's the octahedral crystal
field splitting energy.
And so we can put in
our Planck's constant
and our speed of light.
And we can put in the octahedral
crystal field splitting energy
that we were given.
But we want units in meters.
And so we need our
joules to cancel out.
And in the bottom we're
given the splitting energy
in kilojoules.
So we need to do
some conversions.
So first we need to get
rid of this kilojoules.
So we're going to
convert it to joules.
Then we can get
rid of our joules.
We also have seconds.
We don't want seconds
in our wavelength.
But that is going to cancel out
here and here, so we're good.
But now our answer, we
have meters at the top,
but we have moles on the bottom.
So we need to use Avogadro's
number to cancel out our moles.
And if you don't do
that, you're going
to get a really weird number
for your wavelength that's
not going to make sense.
It's going to be
off by something,
like, a factor of 10 to the 23.
That should remind you, you want
to use Avogadro's number here
to get rid of your per mole.
All right, so now we
have a wavelength.
And yes, on your equation
sheet, we will give you this.
We will give you
the color spectrums.
You don't have to memorize this.
So about 500 nanometers is in
our green region over here.
So the color that
should be absorbed,
given this octahedral crystal
field splitting energy,
is green.
So now, for green, what is the
complementary color of green?
Yup, so it's going
to be reddish.
We have our little
drawing over here.
So the predicted
color would be red.
So let me just now tell
you about this demo.
And we're actually going
to see some red color.
But we're also going to see
another color which is blue.
So, in this demo, if you have,
start with some copper chloride
and add a lot of water.
A lot, a lot of water.
Hydrate it really well.
You'll get this
octahedral system
that you just told me was red.
But if you don't
add a lot of water,
just a little bit
of water, you'll
only display some
of the chlorides.
And then you're going
to have a blue system.
So if you have a
lot of water, red.
And you're hydrated, red.
If you're more
dehydrated, you get blue.
So we're going to now try
this out and see if it works.
GUEST SPEAKER: Is it on?
Yeah, it's on.
Great.
OK, so that's the cobalt flower.
And as Cathy said, it's-- oh,
we're going to put it under
there.
As Cathy said, it's
got some cobalt.
And it's got water ligands.
And it's also got chlorine.
And Eric's going to
sprinkle it with water now.
PROFESSOR: So we're
going to hydrate it.
GUEST SPEAKER:
Hydrate it, as you do,
when you have to water a flower.
And it's turned, like, it's a
pinkish, reddish color, right?
Can they see that up there?
Yeah.
PROFESSOR: It looks
better with it.
Hold it against
this, too, I think.
GUEST SPEAKER: Right.
So to show you that it
was, in fact, the water
that led to this
color change, we're
going to try to
dry out the flower.
Is it working?
Am I on hot?
Yes, I am.
I know it's really slow.
I'd just rather it go faster.
And, well--
PROFESSOR: Yeah.
Why don't you hold it
under the document camera?
I think you can kind
of see it happening.
GUEST SPEAKER: So
as-- there we go.
That's working.
As Eric continues to
warm this thing up,
the water is evaporating.
And as that happens, it starts
forming that hydrous chloride
complex instead of up
here, hydrous complex,
and it's going back to emitting
blue color instead of red.
And hopefully we can see.
Oh, there.
It's working.
Yea.
PROFESSOR: So some people
give each other roses.
But if your significant
other is a geek, what's
better than a flower that
changes color on hydration?
I don't know.
I think this is a
pretty good gift.
Valentine's Day isn't
quite coming up,
but just keep it in mind.
All right, so we'll
leave the flower here
and we'll keep an eye
on it as we go along.
It will change back eventually.
It depends a lot on the weather,
but it's pretty dry right now
in this time of year.
