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Let's get started. We have a lot more chemistry to
learn today. And, as you are probably
becoming aware, the octahedron plays a very
important role in the coordination chemistry of the
d-block elements. So we are going to begin today
also with the octahedron. And I just want to remind you
of the approach that we were taking on Monday to understand
how d-orbitals split under the influence of the presence of a
set of ligands. And this splitting I am
referring to is an energy splitting.
We are talking once again about how we can use energy level
diagrams to understand and interpret the properties of
molecules. And today, in particular,
we are going to be thinking about how we can go from the
angular properties of the d orbitals all the way to some of
the magnetic and spectroscopic or color properties of ions that
contain these d-block transition elements.
And so, you will remember last time on Monday,
we looked at placing ligands at positions one through six on the
octahedron in reference to our coordinate system that we had
chosen, x, y, and z here, as specified.
And what we were doing was, at each of these positions that
corresponds to a ligand atom, evaluating each of the
d-orbitals for the value of theta and phi that is found at
that ligand position. And so, I am going to put up
part of the table that we generated last time.
And that part will correspond to ligand position,
where we have ligands one, two, three, four,
five, and six. Those six ligand positions.
And two of the d orbitals here, d z squared and d x
squared minus y squared, were found to give
non-zero values at these six ligand positions that correspond
to an octahedral complex. The other three d-orbitals gave
zero at those positions because all six of these ligands lie on
nodal planes of xz, yz, and xy.
And what we are doing is just writing down,
in tabular form, the relative magnitude of what
you get if you evaluate the orbital at those ligand
positions. We had one in positions one and
six, which are on the big lobes of d z squared along
the z-axis. And we set that to one because
that is the largest value that we get when we evaluate any of
the orbitals. And then, relative to that,
these ligands two through five that are in the x,y-plane and
which interact with the torus of the d z squared
orbital in the x,y-plane, have a value of one-quarter of
what it is along the z-axis. And then, d x squared minus y
squared evaluated at positions one and
six gave us a zero. And that is because positions
one and six fall on the z-axis, which is actually an
intersection of two nodal planes for the d x squared minus y
squared orbital. And then, here we have at
positions two through five, three-quarters.
And if you are having trouble understanding what these numbers
mean, remember that we started out with a sphere,
so we are interested in seeing essentially, what the
probability is in finding that electron in that particular d
orbital at that point on the surface of the sphere.
And so what this is saying is that, if you will,
the radial probability of finding an electron in d z
squared along the torus at some distance from the
center is one-quarter what it is along the z-axis at the same
distance from the center. And then, at that same
distance, if we are in the x,y-plane along x or along y,
because that is where the lobes of d x squared minus y squared
extend, that probability drops to
three-quarters relative to the two big lobes of d z squared
along z. You can think of it as telling
you about the size of the lobes. And, accordingly,
the way that these lobes of the d-orbitals can interact with the
ligands at these positions on the surface of the sphere coming
right out of evaluating the angular parts of the wave
function. And then, there are other
important coordination geometries that we have to
consider. Because coordination number
six, which corresponds to the octahedron, is relatively
common. But coordination number four is
also quite common. And one of the important
coordination geometries for coordination number four is a
tetrahedral coordination geometry.
The tetrahedron is not limited to being important in organic
chemistry. It is also important for
coordination chemistry. And you will recall that when
we look at a tetrahedron we talk about ligands at alternating
corners of a cube. And I will call these ligand
positions seven, eight, nine,
and ten. And, if we have a tetrahedral
metal complex with four ligands, be they water ligands or
chloride ligands and be the central metal ions,
something like cobalt two or nickel two,
a tetrahedral array like this is produced.
And we can go ahead and evaluate the d orbital wave
functions at theta and phi values that correspond to these
positions, seven, eight, nine,
and ten, all at the same distance from the center.
And then we can generate a similar table,
where we have our ligand position and our set of d
orbitals. And I will start here with
d(xz), d(yz) and d(xy), and over here,
d z squared and d x squared minus y squared.
And we need our ligands
positioned at seven, eight, nine,
and ten, which are these alternating corners of the cube,
which is the tetrahedral geometry.
And what we find, when we go through and evaluate
for those theta and phi values that correspond to these ligand
positions using our polar coordinate system and then
scaling it the same way that we do over here for the octahedron,
is that all of these values come out to be one-third.
Notice that, in the case of the octahedron,
we are interacting with some of the lobes, like the lobes of d z
squared, the large ones,
directly along the axis that makes it a pure sigma
interaction, a cylindrically symmetric interaction.
A big lobe of d z squared and a ligand coming
in along the z-axis. But if you think of the
positions of the lobes of the xz, yz, and xy orbitals relative
to these ligand positions, seven, eight,
nine, and ten, the interaction is kind of
oblique. It is not zero.
We are not on a node, but it is pretty small.
The overlap is not going to be as good.
And it evaluates to one-third for all of those positions.
What we find is that while xz, yz, and xy were all zero over
here for the octahedral case, over here, for the tetrahedral
case, they are all the same again, but they are not at zero.
They are at one-third. And then, for x squared minus y
squared, remember x squared minus y
squared has its lobes along x and y.
What that means is that ligands seven, eight,
nine, and ten actually lie on a nodal surface of the d x squared
minus y squared orbital. And that means that this is
zero, this is zero, this is zero,
and this is zero. And the amazing thing about the
math of the tetrahedral geometry is that these ligands,
seven, eight, nine, and ten also lie on the
nodal surface of d z squared.
Remember that d z squared looks like this and has our opposite
phase torus in the middle and then has a conical nodal
surface. We actually went ahead and --
If you set the equation for d z squared equal to zero
and solved for it, we found this angle here.
We found that angle. And you will remember that that
was the arc cosine of root three over three.
It turns out that that value is exactly half of the tetrahedral
angle, which is the angel between seven,
the center and eight. And, because of this,
all ligands, seven, eight,
nine, and ten lie on the conical nodal surface of d z
squared, such that we get zeros here,
too. One thing that relates very
nicely, the orbital picture for the tetrahedral case for the
octahedral case, is that z squared has the same
energy as x squared minus y squared.
When you sum over all the ligand positions,
we are going to get three here, and we are going to get three
here. And xz, yz, and xy all have
zero. And over here xz,
yz, and xy all have four-thirds.
And z squared and x squared minus y squared
both have zero. And so I will redraw that over
here.
What we are doing is drawing up d orbital splitting diagrams for
two different possible coordination geometries for a
coordination complex. Let me just make that very
clear.
And I have my energy units here, zero, one,
two, three. And this energy level diagram
looks like this for the octahedral.
And we symbolize the octahedral geometry with this Oh for
octahedral. And we have here an energy
splitting. We have a triply degenerate set
of orbitals which we recognize as xz, yz, and xy.
And the name that we give to that triply degenerate set is
t(2g). This t(2g) triply degenerate
set, for this whole set, is the t(2g) set.
If I use that label t(2g), you should remember the
specific d orbitals that contribute to the t(2g) set and
make up the t2g set. Up here we have the d z squared
and the d x squared minus y squared
at the same energy, which are three relative energy
units here. And those two together
constitute the e(g) set. And I will often write that
with a star because we are going to find out later,
when we do the molecular orbital theory of coordination
complexes, that these are antibonding.
And the reason they are antibonding is because,
if you go over here and look at it, if you have a ligand that
wants to act as a Lewis base with respect to d z squared,
then putting an electron in d z squared
leads to repulsion of that ligand, generating
antibonding characters. You want your Lewis acid to
have empty orbitals to receive lone pairs of electrons.
And, if instead they are populated, that corresponds to
antibonding character, much like we have talked about
for other types of molecular orbital diagrams.
This d-orbital splitting diagram for Oh is really a
simplified molecular orbital diagram for the molecule in
which we just look at those orbitals that derive from the
set of five d orbitals on that metal ion.
Because these are the orbitals, --
-- and, together with the electrons that occupy them,
they control the properties of color and magnetism that we will
be talking about for the rest of today.
And let me remind you here that we have a name for the magnitude
of this splitting, which is called delta O for
octahedral. And then, we also have the
tetrahedral case. And how do we get the diagram
from this? We just sum over the
contributions from each of the ligand positions.
In tetrahedral, there are only four ligands.
For xz, yz, and zy we have to sum up four times one-third,
and that gives us, for that set of three orbitals
using the same energy scale as over there, three orbitals at a
value of four-thirds and two orbitals down here at our zero
of energy. And these will be d z squared,
located on that node, and x squared minus y
squared. Whereas, up here we have xz,
yz, and zy. And these get the names e and
t(2) in the tetrahedral geometry.
And the splitting between the levels, the doubly degenerate
and the triply degenerate pair of energy levels,
is here referred to as delta t. And from the math of these two
tables and the magnitude of this splitting being three relative
to this one being four-thirds, you can easily derive the
relation that delta t is equal to four-ninths of delta O.
One of the neat things about
this result qualitatively is that if you have more ligands,
six versus four, you have a larger splitting of
the energy levels. And that corresponds to putting
more Lewis bases, more negative charges,
around that sphere that we talked about,
in the coordination sphere. And also, if you have
interactions between ligands and metal d orbitals that are end-on
sigma symmetry, directed at each other,
you get bigger splitting because that leads to better
overlap than the type of oblique overlap that you get if you have
a ligand at a position such as seven relative to an orbital
like d(xz). You can now see that to fully
appreciate this, you really have to have a very
good grasp of the nodal properties of the d orbitals.
This is why I am going to re-emphasize that you should
spend some time on the computer visualizing these things and
rotating them around and seeing where these ligand positions for
these different coordination geometries are with respect to
each of the five d orbitals. That is quite important.
Also, you are going to see that you could consider other
possible coordination geometries that involve ligands at these
positions or at other positions. For coordination number five,
you could have a trigonal bipyramid coordination geometry.
And you could go ahead and evaluate the d orbitals at the
relevant positions. You would already have three of
them here because in a trigonal bipyramid, you would have
ligands at positions one and six, and two.
And then two more back here at 120 degrees to two,
but in the x,y-plane. So you can do a trigonal
bipyramid the same way and generate the d-orbital splitting
diagram for it. You can also go ahead and do a
d-orbital splitting diagram for a different geometry
corresponding to coordination number four, which would be the
square planar coordination geometry.
If you just took ligand positions two,
three, four, and five you could generate a
diagram. And that diagram would be
relevant for systems that do have square planar coordination
around a metal center, and these are actually fairly
common. Those are the four coordination
geometries, octahedral, tetrahedral,
trigonal bipyramid, and square planar that are all
pretty commonly encountered and that you can actually interpret
pretty nicely using the d-orbital splitting diagrams.
Now, one of the properties that we are going to want to
interpret will be the magnetism.
And, in talking about the magnetism of coordination
complexes, we are going to start by mentioning two terms that you
have probably heard, but I would like to define
explicitly today. One is paramagnetic.
That term is given to substances that are attracted
into a magnetic field. It is somehow a bulk phenomenon
that we are going to relate to properties of d orbital
splitting diagrams.
And systems can also be described as diamagnetic.
And there are many other types of magnetic behavior that we
won't be discussing. I'm a little too close to the
board, here. Diamagnetic.
And that means that the substance is repelled out of or
away from a magnetic field. And one thing that you should
keep in mind is that the magnitude of paramagnetism is
usually orders of magnitude greater than the magnitude of
diamagnetism. So, this is much larger.
And what this means is if a substance behaves as a
paramagnet, you can often neglect its diamagnetism that
also may be present, also must be present.
The substances that are paramagnetic are also
diamagnetic, but the paramagnetism is much larger in
terms of its order of magnitude and wins out.
If you want to measure one of these quantities for a system,
that which is far more difficult to measure is the
diamagnetism because it is a much smaller number.
And then, I want to bring up this quantity S.
This is the spin quantum number.
And you can readily calculate S from looking at a populated d
orbital spitting diagram because it will be the total number of
electrons multiplied by one-half, which is the spin per
electron.
If you are confronted with a particular coordination complex,
the kind of thought process you will need to go through is,
what is the coordination number, what is the coordination
geometry, what does the d orbital splitting diagram look
like, how many electrons do I have with which to populate the
d orbital splitting diagram? After I have figured out how to
populate the d orbital splitting diagram, how many of those
electrons are unpaired? That gives me this.
I can then calculate S. And, if I have S,
I can make a prediction about the value of something called
the spin-only magnetic moment.
Let me mention one more thing back here, and that would be the
units.
When we talk about magnetism, we are going to be using units.
This is called the Bohr magneton.
It is sometimes written as mu sub b.
It is also sometimes written as capital B., capital M.
That is the Bohr magneton.
And it is equal to 9.2741x10^-24 joules per tesla.
This is the value of the Bohr magneton.
And the Bohr magneton is the unit, also, of our magnetic
moment.
And in particular, today, I am going to be talking
about the spin-only value of the magnetic moment.
So we are going to be making predictions using spin-only
considerations. And for those considerations,
we need to be able to calculate S.
And we can do that pretty quickly after we have it
correctly populated, correctly chosen d orbital
splitting diagram for the system.
And so the spin-only value for the magnetic moment is given by
this formula, which is mu is equal to 2.00
times the square root of S times S plus one.
And, in this formula,
this part, this square root of S times S plus one,
from the quantum mechanics, is the value of the angular
momentum, the electron spin angular momentum.
And this value 2.00, I am calling it 2.00.
It is actually a number that differs only very slightly from
two. So we can, for most purposes,
use 2.00. This is called the gyromagnetic
ratio of the electron.
And, as its name implies, it is the ratio of the angular
momentum to the magnetic moment for the electron.
This is our conversion factor to go back and forth between
angular momentum and magnetic moment.
And that factor is the gyromagnetic ratio for the
electron. You will see this as gamma,
sometimes. And, for the free electron
value, it is a number very close to two.
And so, if you think about different ions that could be at
the center of a coordination complex, you wonder,
well, what are my possible values for the spin-only
magnetic moment, as given by this formula?
Well, we can make a table for that.
We have the number of electrons that are unpaired,
remember. And then, we will be able to
calculate S. And then we will want to get
mu, which is this spin-only value calculated according to
that formula. And ions from the part of the
Periodic Table that we may consider working with can have
one, two, three, four, five, six,
up to seven unpaired electrons. These numbers,
like six and seven, are more important for the
lanthanide ions in systems with f-electrons than for systems
with d electrons. But, nonetheless,
we can easily make these predictions the same way.
And so S would be one-half for one electron,
one, then three-halves, then two for four unpaired
electrons, then five-halves for five unpaired electrons,
then three, and then seven-halves finally for seven
unpaired electrons. And if you put this into this
formula, you would see that for one unpaired electron,
it would be one-half times one-half plus one square root
times two. That would be square root of
three, so this would be a spin-only magnetic moment of
And then for S equals one,
we actually have a value of about 2.83.
And then for three unpaired electrons, we have a value of,
let's say approximately 3.87. And then for S equals two,
or four unpaired electrons, we have a value of 4.90.
For the spin-only magnetic moment, if you go through with
these values of S, these numbers of unpaired
electrons. Let's just make sure we are
very clear on that.
And then 5.92, 6.93, 7.94 approximately.
You can check those numbers, but they are approximately
right. And in each case what you are
seeing is the spin-only prediction for the magnetic
moment is always, in units of Bohr magnetons,
a little less than one plus the number of unpaired electrons.
It is not like you have four unpaired electrons,
so your spin-only magnetic moment is around four.
It is actually a little less than five.
And that is because of the quantum mechanics of the
electron spin angular momentum. And this kind of consideration,
here, leads you into thinking about what happens with
different d^n counts for a particular type of metal
complex.
This means that we are going to attack and address the high
spin/low spin problem.
These systems are not so terribly difficult,
but let's face it; --
-- once you have been presented with a particular coordination
complex, and you think you know the geometry,
and you think you know the d orbital splitting diagram that
corresponds to that geometry, and you think you know how many
d electrons go into that diagram, there is one more
thing. And that is sometimes you have
to figure out if it is high spin or if it is low spin.
And what that means is for a particular d^n count,
there might be two possible choices over here,
depending on how the electrons occupy the orbitals.
And so let's look at an example of this.
Here is energy. Again, we are really focusing a
lot on energy level diagrams. Sometimes we focus on molecular
orbital diagrams. Right now, we are using these d
orbital splitting diagrams. And I will give you an example
that is octahedral. And this is a species that we
looked at in a recent demonstration in class,
which is hexaaquo meaning six waters around an iron two plus
in an octahedral array.
This will be our metal complex for which we will draw the d
orbital splitting diagram. It is an octahedral complex,
so we draw our diagram like this.
And of course, we have t(2g) and e(g)* levels
that we can populate with our d electrons.
And we are going to need to figure out how many d electrons
we have to put into this diagram for the octahedral system.
The oxidation state of the iron here is +2.
And you can tell that because the way you tell oxidation state
is to take into account both the charge on the ligands.
Chlorides, for example, are minus one.
Because water is a neutral molecule, it is simply neutral.
And then, you have to also take into account the charge on the
whole ion. So this is iron two plus.
The iron in the plus two
oxidation state with six neutral ligands.
If I had six NH three ligands in the same charge on
the ion, then it would still be iron plus two.
And you are going to get some practice figuring out oxidation
state because if you cannot do that correctly you cannot
identify the number of d electrons to put into the
diagram correctly. And this is iron plus two.
You need to know that iron is
in Group 8 of the periodic table.
The equation for number of d electrons here is eight minus
two equals six. If it is in Group 8,
that means that an iron atom has eight valance electrons,
the same way that carbon has four valance electrons or oxygen
has six valance electrons. And in Group 8 of the periodic
table, we are subtracting two from the eight valance electrons
because there is a two plus charge on the system.
That leaves six valance electrons on the iron.
And that means this is what we call a d six system.
By convention, we are considering all the
valance electrons that remain after the metal has been
oxidized to plus two as d electrons, meaning they are
going to go into our d orbital splitting diagram.
Now, the reason we do this, in fact, is because in ions
like this, the d orbitals are the orbitals on the metal,
on the iron, that are the lowest in energy.
After, in the case of iron, iron has principle quantum
number three. There is also a 4s and a 4p set
of orbitals available to iron, but they are much higher in
energy. We are simplifying the system
and just looking at what happens with the five d orbitals,
which are the lowest valance energy orbitals available to the
iron. Just like a carbon has a 2s and
a set of three 2p orbitals, an iron has a set of five 3d
orbitals, a 4s, and three 4p orbitals.
But those s and p ones are higher.
We are just focusing on our d orbitals, here,
using our d orbital only splitting diagrams with which we
have six electrons for population of a diagram.
And so what you see immediately is you will identify with the
notion that there are two ways to populate this diagram with
six electrons.
Because we can do this.
Here, we are following Hund's rule of maximum multiplicity,
trying not to pair up any electrons until we run out of
spots for them, and so we have to.
There is one way to populate the diagram.
Two electrons in e(g)*. Four electrons in t(2g).
One of them, spin inverted relative to the
other five because you cannot put two electrons in the same
orbital unless they have different spins.
And, finally, we could populate the diagram
this way. The case on the left is what we
call high spin.
The case on the right is what we call low spin.
And, as you think about it, you will realize that the high
spin/low spin problem only arises for certain d^n counts.
It doesn't arise, for example,
for d one because we would just put one electron down
here in t(2g) and that is not a problem.
And you will see that, as you run through the d one
all the way up to the d ten.
d^n counts can go from d^0 to d^10, and that is the limit of
the possibilities here. Zero, one, two,
three, four, five, six, seven,
eight, nine, and ten.
Here I have chosen a d^6 case to illustrate the high spin
versus low spin dichotomy. And you might wonder when do
you -- Well, let me get back to that
in a moment. And let me just say here that
we have four unpaired electrons in d^6 high spin.
We have one, two, three, four.
That makes this an s equals two case.
And, over here, we have s equals zero.
And so this one on the right, this low spin case is s equals
zero. We would say that this is not
paramagnetic. It is going to be only
diamagnetic. It has no unpaired electrons to
give it a magnetic moment according to the spin-only
formula that we looked at over here.
On the other hand, over here, we have an s equals
two state that predicts a magnetic moment of 4.90 Bohr
magnetons. And experimentally,
mu was found for this system to be 5.1 Bohr magnetons.
So we say this one turned out to be high spin.
And in a little while, we will be talking more about
what are the physical factors that determine whether a system
is low spin or high spin. Remember, we have this delta O
here. And there is not one value for
delta O, and this is something I am going to emphasize again in a
moment. Delta O varies according to a
number of factors, and these factors include the
specific nature of the ligands. It includes the charge on the
metal center. It includes the specific metal
that you are using. And at some point,
if delta O gets bigger and bigger and bigger and bigger and
exceeds what we call the pairing energy --
This pairing energy is something I will talk more about
later, but it is the energy required to put two electrons in
the same d orbital.
If you get to a condition where delta O is greater than this
pairing energy, in other words,
there is a big splitting between t(2g) and e(g)*,
then the electrons will prefer to pair up down here in the
t(2g). Whereas, if you think of it
going to the limit of an infinitely small delta O,
they would have S equals two because the gap between t(2g)
and e(g)* in the limit of vanishing delta O,
for example, when ligands go to an infinite
distance away from the metal center, they would then,
in that limit, have all five the same energy.
And so your S would naturally equal two.
If delta O is much greater than the pairing energy giving a
large value of delta O, then that favors low spin.
Factors that give rise to large values of delta O favor low
spin, and I will talk about a few more factors like that in a
moment.
Here is a brief introduction to the magnetic properties of
coordination complexes. And I want to show that we can
use the same types of diagrams to talk about the colors of
transition metal complexes. We looked at the color change
in that demonstration I showed you.
And one of the really fascinating aspects of
transition metal chemistry is the colors that arise because of
absorption of a photon coincident with promotion of an
electron from t(2g) to e(g)*. And so we will talk briefly now
about electronic / absorption spectroscopy.
And the most important thing I want to impart to you here is a
selection rule wherein delta S is equal to zero.
This is the spin selection rule.
And this means that when a photon comes in and interacts
with d electrons in a system of the type we are discussing,
that photon can be absorbed coincident with promotion of an
electron from t(2g) to e(g)*, as long as it doesn't violate
this spin selection rule. And that means that the number
of unpaired electrons should be the same in the system before
the photon is taken up, as it is after.
And so here is an example of that, a simple example.
Here is an octahedral system, titanium with six waters and a
three plus charge.
Titanium is in Group 4 of the periodic table.
Our water ligands are neutral, with three plus charges to be
subtracted from four. We have a d^1 system.
This is one of the simplest cases for talking about
electronic absorption spectroscopy and is the first
thing that you encounter in your textbook.
By the way, let me just remind you that right now you should be
reading Chapter 16 in your textbook, that deals with
coordination chemistry, electronic structure of d-block
elements, colors and magnetism, and so forth.
And the ground state, here, is one in which t(2g) has
a single electron in it for this d^1 ion and e(g)* is vacant.
And, if that system, in the case of titanium three
aquo in aqueous solution, is a pale blue color,
a pretty pale blue -- If a photon comes in of the
appropriate energy delta O to promote the electron into e(g)*,
then that transforms this system into an excited state,
as follows, with a configuration (t two g) zero and
(e g star) one.
These diagrams, just like the molecular orbital
diagrams we developed for diatomic molecules,
can be associated with configurations.
And this transition can be written as( t two g) one (e g
star) zero, photon in, electron promoted
into e g star, t two g zero and e g star one
now corresponding to the excited
state. And this excited state is
produced in a vibrationally excited manner.
And it and the ground state interact with the environment in
different ways, such that the absorption
spectrum for a system of this type gives rise to not a sharp,
but a broad peak. And, in fact,
the spectrum would look something like this.
Going from 200 to nanometers in wavelength,
you would see a profile like this with one broad peak
somewhere in the red, such that we would be behind
the pale blue color of these solutions of aqueous titanium
three. So this would be an absorption
spectrum, and this is wavelength, down here.
And this is the peak for what we call this d-d transition,
from t(2g) to e(g)*. There are a lot more subtleties
on this, which is where I will begin on Monday.
Have a great weekend.