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>> Good morning.
Today the plan is to finish up talking
about electronic spectroscopy,
and to start moving toward talking about [inaudible]
so we're going to have a little detour where we look
at Fourier transforms and talk
about crystallography really briefly, just because it's neat
and it ties into a lot of other things
that we've-that we've been doing.
The use of Fourier transforms involves interactions
between protons and matter; it's not spectroscopy.
We're going to talk about it not in any great depth,
but it's neat and involves symmetry.
Does anybody have any questions about electronic spectroscopy
or anything from last time?
Yes.
>> It's a little hard to hear you back here.
>> It's hard to hear me, OK.
Let me see if I can fix that.
Is that better?
Alright, good.
Any other questions about electronic spectroscopy
or [inaudible] factors or anything like that?
OK, good. Everybody is ready to take the quiz.
Get out a piece of paper- [all talking at once] [laughter]
>> Come one, you walked into that.
I would be happy to stand here asking-answering questions
as much as you want, but you said you're ready
to take the quiz, so-so here it is.
Your lowest quiz is going to be dropped anyway, no matter what,
and in the way the-the way the seminars work is you get points
if you answer the questions.
I've been giving five points for answering the questions.
You have to answer every question
and write a reasonable amount
for each thing to-to get all the points.
So the seminars are in general worth a little more
than the average quiz and how it works is they just get averaged
in with your quiz grade.
So you can go to as many as you want as long
as they're the actual P Chem seminars or things
that I've accrued as being related to P Chem,
and there's no limit; so, you know,
I'm sure this would never happen, but if you mess
up the quiz every time and you go to a lot of seminars,
you can pretty much make it go away.
Not that anyone is worried about that today.
Uh, so let's talk about this a little bit,
so it's-it's not actually that difficult,
but I think it's a little tricky because it's maybe worded
in a different way than you're used to seeing it or you have
to pull together a lot of stuff that you've learned
from different places in the class, and that is kind of hard.
So that's one of the things that I want you to get out of this.
So one of the problems with P Chem, at least starting
out is a lot of the things that people actually do
in our research labs is so involved computationally
that we can't really do a realistic example in the class,
and so what I'd like you to get out of this is an understanding
of how we sort of work through these problems, you know,
what the concepts are, the cases where symmetry helps us,
and to really understand the fundamentals
of some basic problems.
And then, you know, if you become a physical chemist
and you do electronic spectroscopy
in more complicated molecules, then you can learn all the tools
that you need to understand these things computationally.
OK, so I'm not going to spend a huge amount of time on this,
but I just wanted to briefly talk about how to do it.
OK, so for the first one is an electronic transition
from a sigma plus state to a sigma minus state induced
by Z polarized radiation allowed in HCL?
OK, so what do you need to know here?
You need to remember that HCL belongs
to the C infinity B point group, which hopefully everyone figured
out from having the point group table there, if nothing else.
The other important thing about this question is
that if you just remember Laporte's rule,
you've got the wrong answer because that only works
for environments where there's an inversion center,
and of course, HCL doesn't have an inversion center.
So how you actually do this I you take your [inaudible] table
and look at the symmetries for the relevant species,
so sigma plus is A one.
We have a Z for the dipole moment operator
in the Z direction; that's also A one.
And then Sigma minus is A two
and you multiply the coefficients
for those things together and of course what you end
up with belongs to the A two symmetry species,
which is not A one, and so you can say the transition is
not allowed.
So that's how you do that; so, again, it's really easy
if you remember how to do it, and if you don't,
it's confusing, So, you know,
what I-what I want the take home message-what I want the take
home message to be is just, you know, think about the problem
and the information that you're given, and, you know,
figure out how to do it.
Yes.
>> I don't know if this makes sense, but can we also say
that it's because there is no X component in the sigma minus,
uh, even though we have [inaudible] we only care
about the X and Y.
>> No, so Z polarized radiation is telling us that, you know,
it's only along Z, so that's-the idea is kind of there,
but you got confused about the details.
OK, yes. [inaudible question] Uh, like transition from plus
to minus is for-but what case, what?
So we talked about various specific cases.
You know, as far as getting credit,
if you get the right answer,
and you wrote some reasonable rationality,
you get the right answer; but, I guess, what I worry
about on exams is that if you get the wrong answer,
but you had a reasonable thought process, please make sure
that you write down enough so that we can make sure
to give you partial credit.
OK, so that's the first one.
The second one, basically I just wanted you
to draw a potential diagram for these electronic states.
I said it only has two to make it relatively simple,
and the business about the upper state being shifted
in the X direction by, you know,
one point five times the equilibrium bond distance is
just to show that the upper state is, you know,
shifted over in the X direction as far as where its minimum is,
and so if you drew something that looks kind of like
that and, you know, drawing some vibration line energy levels,
then that's good.
It's useful to be able to visualize these things.
And then for the expression for the amplitude of the transition,
basically what you need to do is recognize there
that the electronic part of it doesn't really enter,
We're talking about the Frank Connor [phonetic] factor
between these vibration line states;
so mu double prime equals zero means
that this first hermite [phonetic] polynomial represents
your initial state, and then, uh, new prime equals three,
so your final state is represented
by this other hermite polynomial, and then you have
to stick the X operator in between them
because that's your-that's your dipole moment operator
and integrate that with respect to the X
and that's your-that's your transition dipole,
and then the Frank Connon [phonetic] factor is related
to that squared, so that's basically what you do.
Yes. [inaudible questions] It's, uh, it's the whole way function,
but for this example it's-it's related
to these hermite polynomials.
So, anyway, you didn't need to evaluate it;
that wasn't part of the issue.
I just wanted to write it down.
Yes. [inaudible question] If you put the disproportion overlaps
over the two states, that is also fine, yeah.
OK, so-well, that's annoying.
OK, technical difficulties.
Alright, let's finish up our discussion
of electronic spectroscopy.
Alright, so term symbols are necessary
for describing the states of these molecules.
And I just want to talk about this in a little more detail
for diatomic molecules because I think some people are confused
about it.
I think everyone gets the atomic part from last quarter.
The people that I've talked to seem
to have a really good handle on that,
but I think for-where this comes
from for diatomic molecules is a little bit confusing,
and at this point like for-I don't want to spend a lot
of time learning how to generate these for complex molecules.
Let's just worry about the diatomic case
and mostly I want you to understand what they mean.
So if we have our diatomic molecule, we have values of L
and S for the whole thing, and our term symbol looks like this,
so we've got the super script is the spin multiplicity,
which is two X plus one.
And here S is the total spin quantum number for the molecule;
so to get-you know, to get this, we have to sum
over all the electrons in the molecule, and then this thing,
which is going to be Sigma pi delta, etcetera,
just like it's SPD for the atomic case, that just tells you
about the value of Lambda for the molecule; again,
summed over all the electrons.
And then this thing, the subscript,
which was J in the atomic case, here it's Omega, and same thing,
you're adding up the Z projections of LNS,
and here's a little diagram of that for the molecule.
I also posted a PDF of a tutorial on this stuff
that I found online that I think might be helpful,
so you can check that out if you-if you want to
or if you still feel you need to review all of this stuff.
OK, so-and, again, just terminology.
In this particular thing, Sigma is the projections of S
on the inter nuclear access, so same things
as when we were talking about in the atomic case;
we had like a total angular of momentum and the Z component
of the angular momentum.
Here we are projecting everything
in the inter nuclear access, but the idea is the same.
So that's where these things are coming from
and what they're about.
OK, so let's look at some specific examples
of how to build this up.
So if we have our general chemistry level
on molecular orbital diagram,
we start with some P orbitals [phonetic] and here we're going
to define Z as the inter nuclear access and say,
this is where we get
when our-when our PZ orbitals overlap.
So we know that we get two molecular orbitals;
we get a bonding and an anti bonding orbital, and now we know
that we can describe these as having G and U symmetry based
on whether they're even or odd with respect to inversion;
and since we started out with the total value
of emzybel [phonetic] equals zero and added that up,
that's going to give us-that's going to give us sigma terms,
so we're going to get sigma orbitals out of this.
And for sigma terms we have an additional symmetry descriptor
that we need to worry about.
So G and U refer to what happens
when you go through an inversion.
Does it change sign or stay the same?
And then we also have plus and minus, and plus and minus refers
to what happens when you reflect
through a plane containing the inter nuclear access,
so here's a picture of that.
So that is going to be a sigma minus term
because when you reflect through that plane, it changes sign;
whereas something that looks
like this has a bonding molecule orbital, that's going
to be sigma plus because it stays the same
when you reflect it through that plane.
So those are the symmetry descriptors for sigma terms.
When we get into things that have larger values of Lambda,
then some of these things disappear,
so we don't have the plus and minus descriptor any more,
but we can still write term symbols for these things.
So now let's say we have PX or PY orbitals, and they're going
to be the same, so we can just look at either PX or PY.
Same thing, these can overlap constructively or destructively.
We started with two atomic orbitals so we need
to get two molecular orbitals at the end, and we get a pi
and a pi star molecular orbital,
but now we have emzybel being plus or minus one
for the PX and PY orbitals.
And again, we can describe our pi or pi star molecule orbitals
as having G or U symmetry with respect to inversion,
and these things give us pi terms, and we need to sum
over all electrons to-to get that,
and it's plus or minus one.
So hopefully that helps, seeing some concrete examples
as to what these things mean.
Let's talk about Frank Connon factors a little bit more.
So we have looked at this mathematically
and we've seen how to write down expressions for them.
Let's just look at some pictures and see what
that looks like graphically.
So basically if we have the bonding character
of two states being pretty similar-so in this case both
of these way functions looks like there's a lot
of electron density between the atoms;
they're- they have a lot of bonding character.
In that case, there's going to be a lot of overlap right
at the point where the inter nuclear separation is
at the equilibrium distance, and we're not going to see a lot
of different vibration lines going on there
because there's no reason for the nuclei
to change position very much as a result of the electron popping
up to that excited state.
So remember, we said the mechanism for that is
that the electrons change state
and then suddenly the nuclei are feeling all kinds
of electronic potentials than they were before
because the electron cloud has changed shape,
and then they start to move around
and then we see these vibration line progressions.
If the states were pretty similar in bonding character
to begin with, then there's not very much change,
and we don't see a whole bunch of lines in the spectrum;
whereas if the bonding character
of the two state is really different-so
in this case we've got the electronic ground state,
you know, doesn't have a node in the middle of it
and then it hops up to this excited state
where there's a lot of nodes,
it does not have-does not have very much bonding character
in the middle of the molecule, that causes a big change
in the shape of the electron cloud,
and so we see this progression that has a lot of peaks in it,
and also the potential is shifted in the X direction
in relation to the ground state.
OK, so we can also look at these things and learn something
about disassociation interviews.
So in some cases we can estimate this really directly.
So again, G of mu is the energy
of this electronic transition expressed in wave numbers,
and mu is the quantum number here, so we can write this
down in terms of the frequency of the transition and, you know,
X is the correction to the potential
for a morse [phonetic] oscillator,
which I know we've done some-some practice problems
like that in the homework.
And so we can look at what happens at mu max,
so when mu is maximized here, that means we're
at the disassociation level.
And so if we maximize this and look at what happens
when it equals zero that gives us an expression for mu max.
And the value for the energy of the transition
when that condition is satisfied tells us
about the disassociation energy,
and so sometimes you can do that.
You can estimate it directly.
Another thing you can do is use something like the [inaudible]
which we talked about a little bit;
there's some practice problems on that.
That's where you plot that frequency versus the separation
and take the, you know, formally you should take the area,
but a lot of times you have to extract it
because you don't see lines going all the way
up to the-the disassociation limit here.
So these are the kinds of things that we can get
out of the electronic spectrum,
but a lot of what they're actually used
for in practical applications are more like things
that we saw earlier on when I talked
about just some applications.
So a big thing that is done
with electronic spectroscopy is just Beer's law just looking at,
OK, I have some substance that absorbs light and I want
to know the concentration of it, and you just use Beer's law
to figure out how much of it you have.
That's very common application of electronic spectroscopy.
Of course, there are a lot
of other uses involving learning something about the molecule
as we've been talking about here, and an important branch
of physical chemistry research is taking these kinds
of electronic spectrum and using that to find
out about the bonding energy of molecules,
what kind of bonding is being formed, what do some
of these excited states look like, and of course,
that feeds into a lot of things
like in synthetic chemistry learning about how the symmetry
of different excited states affects what kinds
of molecules you can make.
And before we finish this up I want to talk
about one more application.
So, so far we've mostly been talking
about electronic spectroscopy in the UV and visible range,
so that has to do
with [inaudible] electrons being promoted their relatively low
energy transition as the things go.
Of course, their higher energy
than the vibration line rotational transitions,
but we're mostly talking
about [inaudible] electrons jumping up and down.
If we want to learn more about the-the bonding structure
of the molecule or atom, we can do photoelectron spectroscopy,
and so what we're doing here is a brute force approach,
so instead of sweeping the frequency and looking
at where things absorb or meet,
we are just bombarding the sample with high energy photons
at a fixed wave blank.
This is often in the X ray, so the idea is we have plenty
of energy available to ionize all kinds of electrons,
even deep within the core of the molecule.
And then we can measure the kinetic energy of the electrons
that are detected, and here's a schematic of how that works.
So we have a beam of-here's it's shown as atoms;
could be molecules; depends on what you're trying to measure.
And that is being blasted with high energy photons.
So, again, a lot of times these X rays are done at synchrotrons,
pretty often; and then the electrons get ejected out
and they are placed in an electric field and so they bend,
so the faster ones are over here
and the slower ones are over there.
And so you can measure the kinetic energy
of the electrons that-that come out of the sample
and that tells us something
about the bonding energy because, of course,
the amount of energy that it takes
to disassociate those electrons is related to the energy
of that bond, and so that can teach us
about the bonding structure.
So here are some typical values, so this is our energy
in mega Jules per mole, so it takes a lot of energy
to knock these things off.
And, you know, as we get, you know, as we get into, you know,
first, if we look at Boron here,
the balanced electrons are relatively easy to knock off,
and then it starts to get a little bit harder,
but then when we get down to that one S shell, it takes a lot
of energy to ionize these electrons.
So is as you'd expect, the electrons that are closest
to the nucleus are most tightly bound, but we do have plenty
of energy to ionize all of them, and so when you do this,
and look at all the peaks that you get, it tells you something
about the bonding in the-it tells you something
about the electronic structure of the atom or molecule.
OK, so in this case where we're just looking at atoms,
you might think it's really boring
because surely somebody has measured all of the, you know,
the ionization energies for different electronic states
in common atoms, and that's true, but you can use
that to your advantage.
So one of the primary uses
for X ray photo electronic spectroscopy is looking
at a surface and figuring out what kinds
of atoms are on that surface.
So since these things are well known, there are tables
of what these energies look like,
you can find really low levels of different kinds of things
on the surface and learn about, you know, what that looks like.
You can also do this for molecules, so here's one
for N two; so again, here's our general chemistry molecular
orbital level dia-molecular orbital diagram for N two,
which is a reasonable description of the boding.
OK, so as we saw before when we were kind of talking about this
in a theoretical sense-you know,
so here there are three bands in-in the spectrum,
so A is what we get when we remove a weekly bonding
electron, so that's the two P sigma G orbital
and that transition has relatively few lines,
so ionizing that to-you know, it tells us that hoping
up from the ground state to that state doesn't change your
nuclear separation very much;
whereas B here is removing a strongly bonding electron-
that's the pi U-it's form the Pi U molecular orbital,
so that's-that's down here.
That requires a lot more deviation
in the inter nuclear distance,
and so we see a bunch more vibration lines;
and then C comes from removing a weakly anti bonding electron
and, uh, you know, here we have a weaker transition
and we only see one peak,
so it's a relatively short progression.
So just to give you an idea of how these things work
and how that's used-OK, so we don't have a lot of time left,
but I do want to start talking
about X ray crystallography a little bit and we'll finish
up that part next time.
So just to give you an idea where we're going with this,
there's a whole chapter on solids that contains a bunch
of stuff about crystallography; I think it's chapter nine.
We mostly skipped it.
It might be useful to go and skim that and have a review
of things like the difference
between crystal and amorphous solids.
So you know, something isn't a crystal it's
in a really regular repeating lattice;
if it's amorphous it's still solid,
but it's a lot more disordered.
A lot of the stuff that is
in that chapter is pretty descriptive,
and there is not a lot that we could really do with it,
so it's useful to look at it, but we're not going
to spend a lot of time in it.
What I want to talk about is crystallography
as an interaction between a periodic lattice
of your molecules in the crystal and X rays, and, of course,
this happens because we can have a constructive
and destructive interference between the wave function
of the electrons and the incoming X ray photons.
And crystallography is not spectroscopy.
We're just looking at X rays diffracting off the wave
function of electrons, but it's related to a lot
of the other stuff we're doing because it involves these ideas
of symmetry, so instead of point groups, when we start to talk
about crystal lattices, we need to assign things
to space groups, and anybody who is
in a crystallography lab has solved a crystal structure
of an organic molecule has seen some of these things.
It's kind of the next level of symmetry arguments
that we're talking about, and the periodic structure
of the crystal is what enables it to diffract X rays.
So how many of you have been involved
in solving a crystal structure in some-in some way,
either in research or in labs?
OK, so a few, but-how about crystallizing stuff
in an organic chemistry lab to purify it?
Has anybody seen some of that?
OK. Yes, so, we're able
to get-so crystallizing your compound is a good purification
method because you make this regular lattice where molecules
that are the wrong shape don't fit in there, and then we end
up with this periodic function
that can actually diffract X rays.
And so that happens because electrons have some wave
character; they can interact constructively and destructively
with the photons, and so you get something that looks like this.
So we shoot the X ray beam at your crystallized molecule, and,
you know, the crystal in this illustration looks pretty messy,
but, you know, in general you need a very nice crystal
in order to get diffraction.
And then, the X rays that get scattered off the molecule are
then detected and you get a regular pattern of spots
that has something to do with the crystal lattice.
In fact, it gives you the inverse of the dimensions
of the crystal lattice in an indirect kind of way.
And then that enables you to get to an idea
of what the unit cell looks like for a molecule.
So here's one from a crystal structure of redoxon [phonetic]
so we talked about redoxon when we were talking
about how our eyes work and how we have to be careful
about calibrating our instruments.
So we understand redoxon because it's been crystallized;
there have also been a bunch of [inaudible] structures
of all kinds of redoxon.
But so here's what the unit cell of this looks like, and again,
if you want to review what units cells of crystals look like,
go check out-I think it's chapter nine; there are a bunch
of simple examples, and here, this is a complicated one,
but it obeys the same principles.
So here A, B, and C are the dimensions of the unit cell,
so that's a repeating unit, so this is the origin here.
We're looking down the A axis and then B and C are shown here.
Here are just some examples of diffraction patterns
that can be used to solve molecular structures,
so this is the original fiber diffraction pattern
of a DNA double helix.
My DNA picture didn't show up here,
but hopefully everybody knows what it looks like.
So, we see these regular repeating units.
In the case of the DNA we've got patterns here
and then reflections here;
that's reflecting the repeating pattern of the DNA.
Of course, we don't-in that case it's a fiber,
so it's only crystal in one dimension;
whereas as if we have a three dimensional crystal,
we see these really regular patterns of spots,
which we can then analyze and use
to get the molecular structure.
And I'm going to quit there for today because we're out of time,
but next time we're going to talk
about this mysterious process by which that happens.
See you on Friday. ------------------------------d542d96d39a3--