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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.
-