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PROFESSOR: OK, let's just
take 10 more seconds on

the clicker question.

OK, 76, I think that says,
%, which is not bad, but

we should be at 100%.

So, when you're past the
equivalence point, so you've

converted all of your weak, in
this case, acid to its

conjugate base, and because it
was a weak acid, the conjugate

base is going to be a weak
based and so it's not

contributing a whole lot it'll
make the solution basic, but

it's nothing compared to adding
strong base in there.

So even though you have
the weak base around, at

this point it's really
a strong base problem.

So you would calculate this by
looking at how many mils of the

strong base you've added past,
and figure out the number of

moles that there are, and
divide by the total volume.

So this was like one of the
problems on the exam, and one

thing that I thought was
interesting on the exam is that

more people seemed to get the
hard problem right than this,

which was the easy problem.

So we'll see on the final,
there will be an acid based

titration problem on the
final, at least one.

So let's see if we can
get, then, the easy and

the hard ones right.

So you've mastered the hard
ones and let's see if you can

learn how to do the easy ones
as well for the final exam.

OK, so we're going to continue
with transition metals.

We were talking about crystal
field theory and magnetism, and

you should have a handout for
today, and you should also have

some equipment to make models
of orbitals and coordination

complexes -- these
are not snacks.

They can be snacks later, right
now they're a model kit.

All right, so I'm going to
introduce you to some terms

that we're going to come back
you at the end of today's

lecture, and then we're going
to talk about the shapes of

coordination complexes.

So, magnetism.

So we talked last time, before
the exam, if you remember,

about high spin and low spin,
unpaired electrons and

paired electrons.

Well, compounds that have
unpaired electrons are

paramagnetic, they're attracted
by a magnetic field, and those

where the electrons are paired
are diamagnetic are repelled

by a magnetic field.

So you can tell whether a
coordination complex is

paramagnetic or diamagnetic,
you can test the magnetism,

and that'll give you some
information about the electron

configuration of the d orbitals
in that coordination complex.

And that can tell you
about the geometry.

And so you'll see that by the
end we're going to talk about

different types of energy
orbitals when you have

different geometries.

So why might you care about the
geometry of a metal center.

Well, people who study proteins
that have metal centers care a

lot about the geometry of them.

So let me just give
you one example.

We talked a lot about energy in
the course this semester, so we

need catalysts for removing
carbon monoxide and carbon

dioxide from the environment.

And nature has some of these --
they have metal cofactors and

proteins that can do this, and
people have been interested in

mimicking that chemistry
to remove these gases

from the environment.

So let me tell you these
enzymes are organisms.

And this is pretty amazing,
some of these microorganisms.

So, over here there's one
-- it basically lives

on carbon monoxide.

I mean that's -- you know
alternative sources of energy

are one thing, but that's
really quite a crazy thing

that this guy does.

So, you can grow it up in these
big vats and pump in carbon

monoxide and it's like oh,
food, and they grow and

multiply, and they're very,
very happy in this carbon

monoxide environment.

There are also microorganisms
that live on carbon dioxide as

their energy and
a carbon source.

And so these organisms have
enzymes in them that have metal

centers, and those metal
centers are responsible for the

ability of these organisms to
live on these kind of bizarre

greenhouse gases
and pollutants.

So people would like to
understand how this works.

So microbes have been estimated
to remove hundred, a million

tons of carbon monoxide from
the environment every year,

producing about one trillion
kilograms of acetate from

these greenhouse gases.

And so, what do these catalysts
look like and these enzymes,

what do these metal clusters
look like that do

this chemistry.

And this was sort of a rough
model of what they look like,

and they thought it had iron
and sulfur and then a nickel in

some geometry, but they had no
idea sort of where the nickel

was and how it was coordinated.

And so before there was any
kind of three dimensional

information, they used
spectroscopy, and they

considered whether it was
paramagnetic or diamagnetic to

get a sense of what the
geometry around the metal was.

So we're going to talk about
different coordination

geometries and how many
unpaired or paired electrons

you would expect, depending
on those geometries today.

And so, crystal field theory,
again, can help you help

explain/rationalize the
properties of these transition

metal complexes or
coordination complexes.

So, to help us think about
geometry, I always find

for myself that it's
helpful to have models.

So not everyone can have such
large models as these, but you

can all have your own little
models of these geometries.

So, what we have available to
you are some mini marshmallows,

which, of course, as we all
know, are representative of d

orbitals, and jelly beans,
which we all know are useful

for making coordination
complexes.

So, what you can do with your
mini marshmallows is you can

put together to make
your different sets.

And so, over here we have --
oh, actually it says gum drops

-- you don't have gum drops
this year, I changed up here, I

forgot to change it down here.

We have mini marshmallows.

Dr. Taylor went out and tried
to purchase enough gum drops to

do this experiment, and
discovered that Cambridge only

had 300 gum drops, so we have
mini marshmallows

instead today.

But this gives you the idea.

You can take one toothpick and
you can make d z squared,

putting on your orbitals, you
have your donut in the middle,

and then your two lobes,
which run along the z-axis.

And then for your other sets of
orbitals, you can take these

two toothpicks and put on these
sets of mini marshmallows, and

handily, you can just have one
for all of the other d

orbitals, because depending on
how you hold it, it can

represent all of the other d
orbitals just very well.

So, you can just have one of
these for all the others

and then your d z squared.

So what we're going to do when
we have our orbitals set up,

then we can think about how
ligands in particular

positions, in particular
geometries would clash with our

orbitals -- where there'd be
big repulsions or

small repulsions.

So, any other people missing
their jelly beans or

their marshmallows?

Please, raise your
hand, we have extras.

So, those of you who have
them, go ahead and start

making your d orbitals.

All right, so if you're
finished with your two d

orbitals, you can start making
an octahedral complex.

So in your geometries set,
you'll have a big gum which can

be your center metal -- you'll
have a big jelly bean -- sorry,

big jelly beans and small jelly
beans are our ligands, or our

negative point charges, and
you can set up and make an

octahedral geometry here.

OK, so as you're finishing this
up, I'm going to review what we

talked about before the exam --
so this isn't in today's

lecture handouts, it was in
last time, which we

already went over.

But sometimes I've discovered
that when there's an exam in

the middle, there needs to be a
bit of a refresher, it's hard

to remember what happened
before the exam, and you

have your models to
think about this.

So, before the exam, we had
talked about the octahedral

case, and how compared to a
spherical situation where the

ligands are everywhere
distributed around the metals

where all d orbitals would be
affected/repulsed by the

ligands in a symmetric fashion
equally, when you have them put

as particular positions in
geometry, then they're going to

affect the different d
orbitals differently.

And so, if you have your d z
squared made, and you have your

octahedral made, you can sort
of hold these up and realize

that you would have repulsion
from your ligands along the

z-axis directly toward your
orbitals from d z squared.

So that would be
highly repulsive.

The ligands are along the
z-axis, the d orbitals are

along the z-axis, so the
ligands, the negative point

charge ligands are going
to be pointing right

toward your orbitals.

And if you hold up this as a d
x squared y squared orbital

where the orbitals are right
along the x-axis and right

along the y-axis and you hold
that up, remember, your ligands

are right along the x-axis
and right along the y-axis.

So, you should also have
significant repulsion for d x

squared minus y squared, and
octahedrally oriented ligands.

In contrast, the ligands set
that are 45 degrees off-axis,

so d y z, d x z, and d x y,
they're all 45 degrees off.

Your ligands are along the
axis, but your orbitals

are 45 degrees off-axis.

So if you look at that
together, you'll see that

whichever one you look at, the
ligands are not going to be

pointing directly toward
those d orbitals.

The orbitals are off-axis,
ligands are on-axis.

So there will be much
smaller repulsions there.

And that we talked about the
fact that for d x squared minus

y squared and d z squared,
they're both have experienced

large repulsions, they're both
degenerate in energy, they go

up in energy, whereas these
three d orbitals, smaller

repulsion, and they're also
degenerate with respect to each

other, and they're stabilized
compared to these guys up here.

So you can try to hold those up
and convince yourself that

that's true for the
octahedral case.

So, that's what we talked about
last time, and now we want to

-- oh, and I'll just remind you
we looked at these splitting

diagrams as well.

We looked at the average energy
of the d orbitals -- d z

squared and d x squared minus
y squared go up in energy,

and then the other three d
orbitals go down in energy.

So now we want to consider
what happens with

different geometries.

So now you can turn your
octahedral case into a

square planar case, and
how am I going to do that?

Yeah, so we can just take off
the top and the bottom and we

have our nice square planar
case, and try to make a

tetrahedral complex as well.

And here's an example
of a tetrahedral one.

Again, you can take a jelly
bean in the middle, and big

jelly bean, and then the
smaller ones on the outside.

So what angles am I going for
here in the tetrahedral case?

109 .

5.

So you can go ahead and make
your tetrahedral complex,

and don't worry so
much about the 0 .

5, but we'll see if people can
do a good job with the 109.

OK, how are your tetrahedral
complexes coming?

Do they look like this sort of?

So let me define for you how
we're going to consider

the tetrahedral case.

So, in the tetrahedral case,
we're going to have the x-axis

comes out of the plane, the
y-axis is this way, z-axis

again, up and down.

We're going to have one ligand
coming out here, another going

back, and then these two
are pretty much in the

plane of the screen.

So this is sort of how I'm
holding the tetrahedral complex

with respect to the x, z,
and y coordinate system.

So, there is a splitting,
energy splitting, associated

with tetrahedral, and it's
going to be smaller than

octahedral because none of
these ligands will be pointing

directly toward the orbitals.

But let's consider which
orbitals are going to be most

affected by a tetrahedral case.

So, let's consider d z squared.

What do you think?

Is that going to be
particularly -- are the ligands

pointing toward d z squared?

No.

And d x squared minus y
squared, we can think of,

what about that one?

No, not really.

What about d x y,
d y z, and d x y?

Moreso.

So, if you try holding up your
tetrahedral in our coordinate

system, and then hold your d
orbitals 45 degrees off-axis,

it's not perfect, they're not
pointing directly toward them,

but it's a little closer than
for the d orbitals that

are directly on-axis.

So, if we look at this, we see
that the orbitals are going to

be split in the exact opposite
way of the octahedral system.

In the octahedral system, the
ligands are on-axis, so the

orbitals that are on-axis, d x
squared minus y squared and d

z squared are going to
be the most affected.

But with tetrahedral, the
ligands are off-axis, so the

d orbitals that are also
off-axis are going to

be the most affected.

But they're not going to be as
dramatically affected, so the

splitting is actually
smaller in this case.

So here, with tetrahedral,
you have the opposite of

the octahedral system.

And you can keep these and
try to convince yourself

of that later if you have
trouble visualizing it.

So, you'll have more repulsion
between the ligands as negative

point charges, and the d
orbitals that are 45 degrees

off-axis than you do with
the two d orbitals

that are on-axis.

So here, d x squared minus y
squared and d z squared have

the same energy with respect to
each other, they're degenerate.

And we have our d y z, x z,
and x y have the same energy

with respect to each other,
they are also degenerate.

So it's the same sets that
are degenerate as with

octahedral, but they're
all affected differently.

So now let's look at the energy
diagrams and compare the

octahedral system with
the tetrahedral system.

Remember an octahedral, we
had the two orbitals going

up and three going down.

The splitting, the energy
difference between

them was abbreviated.

The octahedral crystal field
splitting energy, with a

little o for octahedral.

We now have a t for
tetrahedral, so we have

a different name.

And so here is now
our tetrahedral set.

You notice it's the opposite of
octahedral, so the orbitals

that were most destabilized in
the octahedral case are now

more stabilized down here, so
we've moved down in energy.

And the orbitals that are
off-axis, 45 degrees off-axis,

which were stabilized in the
octahedral system, because none

of ligands were pointing right
toward them, now those ligands

are a bit closer so they jump
up in energy, and so we have

this swap between the two.

So, we have some new
labels as well.

So, we had e g up here as an
abbreviation for these sets

of orbitals, and now that's
just referred to as e.

Notice the book in one place
has an e 2, but uses e in all

the other places, so just
use e, the e 2 was a

mistake in the book.

And then we have t 2 g
becomes t 2 up here.

So we have this slightly
different nomenclature and we

have this flip in direction.

So, the other thing that is
important to emphasize is that

the tetrahedral splitting
energy is smaller, because none

of those ligands are pointing
directly toward any

of the d orbitals.

So here there is a much larger
difference, here there is a

smaller difference, so that's
why it's written much closer

together, so that's smaller.

And because of that, many
tetrahedral complexes are high

spin, and in this course, you
can assume that they're

all high spin.

So that means there's a weak
field, there's not a big

energy difference between
those orbital sets.

And again, we're going to --
since we're going to consider

how much they go up and down
in energy, the overall

energy is maintained.

So here we had two orbitals
going up by 3/5, three

orbitals going down by 2/5.

So here, we have three orbitals
going up, so they'll go up in

energy by 2/5, two orbitals go
down, so they'll be going

down in energy by 3/5.

So again, it's the opposite
of the octahedral system.

It's opposite pretty much in
every way except that the

splitting energy is much
smaller, it's not as large

for the tetrahedral complex.

All right, so let's look at an
example, and we're going to

consider a chromium, and like
we did before, we have to first

figure out the d count, so
we have chromium plus 3.

So what is our d count here?

You know where chromium is,
what its group number --

here is a periodic table.

So what is the d count?

3.

So we have 6 minus 3,
3 -- a d 3 system.

And now, why don't you tell me
how you would fill in those

three electrons in a
tetrahedral case.

Have a clicker question there.

So, notice that in addition to
having electron configurations

that are different, the d
orbitals are labelled

differently.

OK, 10 more seconds.

OK, very good, 80%.

So, let's take a look at that.

So down here, we're going to
have then our d x squared minus

y squared, d z squared orbitals
up in the top, we have

x y and x z and y z.

Again, the orbitals that are
on-axis are repelled a little

less than the orbitals that are
off-axis in a tetrahedral case.

And then we put in our
electrons, we start down here.

And then one of the questions
is do we keep down here and

pair up or go up here, and the
answer is that you

would go up here.

Does someone want to tell me
why they think that's true?

Yeah.

STUDENT: [INAUDIBLE]

PROFESSOR: Right, because it
has a smaller splitting energy.

So, the way that we were
deciding before with the weak

field and the strong field, if
it's a weak field, it doesn't

take much energy to
put it up there.

So you go they don't want to
be paired, there's energy

associated with pairing.

But if there's a really huge
splitting energy, then it takes

less energy to pair them up
before you go that big

distance up there.

But in tetrahedral cases, the
splitting energy's always

small, so you're just going to
always fill them up singly

to the fullest extent
possible before you pair.

So this is like a weak field
case for the octahedral system,

and all tetrahedral complexes
are sort of the equivalent of

the weak field, because the
splitting energy is always

small in an octahedral case,
because none of the ligands'

negative point charges are
really pointing toward any of

those orbitals that much, so
it's not that big a difference.

So, here we have this and now
we can practice writing our d

to the n electron
configuration.

So what do I put here?

What do I put first?

So we put the e and then what?

Yup.

There are two electrons in the
e set of orbitals, and in the

t 2 orbitals, there's one.

So that is our d n
electron configuration.

And then we're also asked how
many unpaired electrons.

Unpaired electrons
and that is three.

All right.

So that's not too bad, that's
the tetrahedral case.

The hardest part is
probably making your

tetrahedral complex.

Now square planar.

So again, with the square
planar set you have your square

planar model -- we have
a bigger one down here.

And the axes is defined such
that we have ligands right

along x -- one coming out at
you and one going back, and

also ligands right
along the y-axis.

So as defined then, we've
gotten rid of our ligands

along the z-axis.

So, what do you predict?

Which two of these will be
the most destabilized now?

What would be the most
destabilized, what

do you guess?

You can hold up your
little sets here.

What's the most destabilized,
what's going to go up

the most in energy here?

Yeah, d z squared
minus y squared.

What do you predict might
be next, in terms of

most unfavorable?

Yeah, the x y one.

So these two now are going to
be the most destabilized, with

d x squared minus y squared
being a lot more destabilized

than just the x y, because
again, those d orbitals

are on-axis and these
ligands are on-axis.

So, let's take a look
at all of these again.

So in the octahedral case,
these were degenerate.

That's no longer true,
because there are no ligands

along the z-axis anymore.

So we took those off in going
from the octahedral to the

square planar, so you have much
less repulsion, but with the d

x squared minus y squared, you
still have a lot repulsion.

so then if we start building up
our case, and this diagram is,

I think, on the next page of
your handout, but I'm going to

start building it
all up together.

So now d x squared, y squared
is really high up, it's very

much more destabilized
than anybody else.

D z squared, on the
other hand, is down.

It's not -- it would be
stabilized compared -- it's

not nearly as destabilized
as the other system.

So then we go back
and look at these.

You told me that d x y would
probably be next, and

that's a very good guess.

You see you have more repulsion
than in the other two, because

the other orbitals have
some z component in them.

So you have less repulsion than
d x squared minus y squared,

because it's 45 degrees off,
but still that one is probably

going to be up a little bit
more in energy than

the other set.

These two here are stabilized
compared to the others, so

they're somewhere down here.

Now the exact sort of
arrangement can vary a little

bit, but the important points
are that the d x squared minus

y squared is the most
destabilized, d x y would be

next, and the other are
much lower in energy.

And we're not going to do this
how much up and down thing,

like the 3/5 and the
2/5 because it's more

complicated in this case.

So just the basic rationale you
need to know here, not the

exact energy differences
in this particular case.

OK, so now we've thought about
three different kinds of

geometries -- octahedral,
tetrahedral, and

the square planar.

You should be able to
rationalize, for any

geometry that I give
you, what would be true.

If I tell you the geometry and
how it compares with our frame,

with our axis frame of where
the z-axis is, you should be

able to tell me which
orbital sets would be

the most destabilized.

And to give you practice,
why don't you try

this one right here.

So we have a square pyramidal
case as drawn here with the

axes labeled z, y and x,
coming in and coming out.

Tell me which of the following
statements are true.

And if you want, you can take
your square planar and turn it

into the geometry
to help you out.

Let's just take
10 more seconds.

All right.

That was good.

People did well on
that question.

So, if we consider that we
had the top two are correct.

So, if we consider the d z
squared, now we've put a ligand

along z, so that is going to
cause that to be more

destabilized for this geometry
rather than square planar,

which doesn't have anything in
the z direction. ah And then in

terms, also, other orbitals
that have a component along z

are going to be affected a
little bit by that, but our

other one here is not going to
be true, so we just have all of

the above is not correct,
so we have this one.

So if we had up those, that's
actually a pretty good score.

And so you could think about,
say, what would be true of a

complex that was linear along
z, what would be the most

stabilized, for example.

So these are the kinds of
questions you can get, and

I think there are a few
on the problem-set.

All right, so let's come
back together now and talk

about magnetism again.

So, we said in the beginning
that magnetism can be used to

figure out geometry in, say, a
metal cluster in an enzyme, and

let's give an example of
how that could be true.

So, suppose you have a nickel
plus 2 system, so that would be

a d 8 system, so we have group
10 minus 2 or d 8, and it was

found to be diamagnetic.

And from that, we may be able
to guess, using these kinds of

diagrams, whether it has
square planar geometry,

tetrahedral geometry,
or octahedral geometry.

We can predict the geometry
based on that information.

Let's think about
how that's true.

We have a d 8 system.

Think about octahedral
for a minute.

Are there two options for how
this might look in this case?

Is there going to be a
difference in electron

configurations if it's a weak
field or a strong field?

So, write it out on your
handout and tell me whether

it would be true, think
about it both ways.

Is there a difference?

So, you would end up
getting the same thing

in this particular case.

So if it's a weak field and
you put in 1, 2, 3, then jump

up here, 4, 5, and then you
have to come back, 6, 7, 8.

Or you could pair up all the
ones on the bottom first and

then go up there, but you
actually get the same result no

matter which way you put them
in, the diagram looks the same.

So it doesn't matter in this
case if it is a weak or strong

field, you end up with those
number of electrons with the

exact same configuration.

So, we know what
that looks like.

Well, what about square planar.

So let's put our
electrons in there.

We'll start at the bottom,
we'll just put them in.

I'm not going to worry too much
about whether we can jump up or

not, we'll just go and pair
them up as we go down here, and

then go up here, and now we've
put in our eight electrons.

So, how close these are, we're
just going to put them all in.

We're just going to be very
careful not to bump up any

electrons there unless we
absolutely have to, because d x

squared minus y squared is very
much more destabilized in the

square planar system, so we're
going to want to pair all

our electrons up in those
lower energy orbitals.

So even if we sort of
did it a different way,

that's what we would get.

So we're going to want to pair
everything up before we go

up to that top one there.

So there's our square planar.

Well, what about tetrahedral.

How are we going
to fill these up?

Do we want to pair first, or
we do want to put them to the

full extent possible singly?

Single, right, it's going to be
a weak field, there's not a big

splitting here between these,
so we'll put them in, there's

1, 2, 3, 4, 5, 6, 7, 8.

All right, so now we can
consider which of these will

be paramagnetic and which
will be diamagnetic.

What's octahedral?

It's paramagnetic, we
have unpaired electrons.

What about square planar?

Square planar's diamagnetic.

And what about tetrahedral?

Paramagnetic.

So, if the experimental data
told us that a nickel center in

an enzyme was diamagnetic, and
we were trying to decide

between those three geometries,
it really seems like square

planar is going to
be our best guess.

And so, let me show
you an example of a

square planar system.

And so this particular nickel
is in a square planar system.

It has four ligands that are
all in the same plane, and it

is a square planar center for a
nickel, so that's one example.

And this is a cluster
that's involved in life

on carbon dioxide.

All right, so that's
different geometries,

you're set with that.

Monday we're going to talk
about colors of coordination

complexes, which all have to do
with the different geometries,

paired and unpaired electrons,
high field, low spin,

strong field, weak field.

Have a nice weekend.