The following content is
provided under a Creative
Commons license.
Your support will help MIT
OpenCourseWare continue to
offer high quality educational
resources for free.
To make a donation or view
additional materials from
hundreds of MIT courses, visit
MIT OpenCourseWare at
ocw.mit.edu.
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.
