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[ Silence ]
[ Inaudible Discussion ]
>> Good morning everybody.
Let's go ahead and get started.
I get the feeling my microphone is not working
so well, is that correct?
All right.
[ Noise ]
OK. How about now?
Is that better?
>> It's still quiet.
>> It's still quiet.
OK. Maybe I can move it.
How about that?
>> Good. That's good.
>> OK, good.
All right.
So, last time, we talked about transformation matrices.
And I warned you to be really, really careful about making sure
that you know what basis you're in and being consistent
and then I made a mistake involving not doing that.
So let's put up the slide again where that happened.
This is the corrected version.
So, basically, what I did is I had half of it
in the different basis than the other half which is not good,
and so thanks to the person who pointed out.
Here's the corrected version of how we set
up these transformation matrices.
The matrices were right,
it's just that the answer was not consistent
with the vector that I applied it to.
So, sorry about that, mistakes do happen
but I always want to correct them.
So I do want to point out like if things like that happen
in class, I really do want you to tell me
and so I appreciate that when people do.
You know, sometimes we'll have time to stop them and clear it
up right away, other times we'll just have to move on and deal
with it later but, you know, we always want to make sure to get
that cleared up, so stuff is clear.
Does anybody have any questions about these or other stuff
that has gone on so far?
OK, that's good.
Yes?
>> Sorry. For the projection on the X axis,
I was wondering why it was [inaudible] right there
which is the 000?
>> Well, so you're just sending up with your--
you just want the projection on the X axis.
So, that's the only-- so 1, 1 is the only cell that you want
to have anything in it.
[ Inaudible Remark ]
The first element on the diagonal.
>> OK.
>> OK, cool.
All right.
So, we're going to go and talk about Group Theory.
And today is going to be cool because we're going to learn how
to take the two scales that we've learned so far
and put them together to do something useful.
So, so far we've learned how to put molecules
and actually any kind of different shapes
into point groups, and we also learned how to set
up transformation matrices.
And now, we're going to see how this actually has an application
in chemistry.
The first application we're going
to do is the chemical bonding.
And I like to do these examples
because we learned the general formalism for going through
and using the group theory, but the advantage is
that the answer you get is something
that you all already know from general chemistry.
So it's really easy to check your work.
If you get the wrong answer you're going
to know right away that it's wrong.
This is going to be helpful because later on we're going
to be doing these applications, just spectroscopy
where it won't be so obvious what's happening.
So it's good to make sure that you get what you expect
when we talk about this in relation to bonding.
Another sort of general strategy, the thing that I want
to point out before I forget is with respect to putting things
in the right basis and making sure you're consistent,
when you go to do these things on the exam, we give lots
of partial credit for different things
and also there are just different ways
to get the right answer.
Please make sure that whatever you're doing,
you write down what basis you're using, draw pictures,
draw your coordinate system, just make sure that the TAs
and I understand what you're doing whatever it may be.
OK. So let's go through talking about point groups.
So we're going to start out talking about water which is
in the C2v point groups.
So, of course, we could be talking about any object
that fits in that group.
And if we look at our character table,
we see that we have these operations.
And last time we went through all the various things that are
in the character table.
And we talked about the irreducible representations
which are these things that are called A, B, C, E et cetera.
And I said that they described how things transform
under these operations.
We're going to talk about what that means.
So, we need to introduce the idea of being able to look
at something and say, "OK.
When I perform this operation,
it's a valence symmetry operation but we have to be able
to distinguish between whether it leaves the object completely
unchanged or not.
And so, in order to do this,
I've labeled these hydrogens red and blue.
In reality of course, with the real water molecule,
we can't tell the difference
between those hydrogens are the same.
But in our minds, we can, you know, designate them as red
and blue and keep track of where they go.
OK. So if we perform the identity operation,
it doesn't change.
That's what the identity means.
But if we do a C2 rotation, it's a fine symmetry operator.
It leaves the molecule unchanged as far
as the hydrogens being indistinguishable
but it does swap them.
Now, if we look at our reflections in the XZ plane
which notice my little coordinate system here,
that's one, that's the plane of the screen.
That one leaves it unchanged whereas the other one which goes
through the middle of the molecule swaps them.
So-- OK, simple enough right?
We want to be able to look at how we can use these.
[ Pause ]
So, now we need to set up matrices
to perform our transformations.
And in this case, the basis
that we're looking at is the hydrogens.
So the basis is that's the thing that we're talking about.
And what we're going to see is that the transformation matrix
for a particular operation depends
on what basis you're using.
So I can't say make a-- I can't make a transformation matrix
for a C2 rotation completely in general.
I have to specify what I'm talking about.
So last time we talked
about little unit vectors pointing along the axis either
in a plane or in free space now our basis is the hydrogens
on this water molecule.
And so, for the identity matrix we need to look at the thing
that keeps our hydrogens unchanged.
So now, notice I'm calling them H and H prime.
And so, our identity matrix is always going to be--
we're always going to get something that has 1s
on the diagonal and zeros everywhere else
but how big it is depends on what it is.
So here we have two things.
We got a 2 by 2 matrix for that.
And notice that its character is 2, so that's the character chi.
So remember, that's what you get when you add the elements
on the diagonal and it's-- we're working with character tables
so we know that that has something to do
with those elements that go into the irreducible representations
and we're going to see how in a little bit.
So, in many point groups, there's a shortcut
that we can use or we can say, "All right, I don't need to set
up this whole matrix."
I can just say, you know, we have two things and neither one
of them changed and so we can add them up
and get the character 2.
So, this is a shortcut that often works.
I want to point out that it doesn't always.
So if we have things that have trigonal kinds of symmetry,
so like if we're in the D3h or C3v point group.
This becomes a lot harder
and we'll see some examples of that later.
But in many cases, we can use the shortcut
where we can just count the things that changed
and find the character right away.
And I'll point out when we're able to do that.
OK. So now, we can look at our C2 operation.
And we know-- we already know what that does,
it swaps the hydrogens.
So if that's our basis, if one changes places with the other,
then that character is going to be zero.
And we can use that shortcut
and we can also set up the actual matrix.
So that's the matrix that makes these things flip position.
So, I'm not going to do the other two
for this thing right this minute.
I will leave it as an exercise question.
>> Why do you change their primes,
because before blue have the prime and now red has the prime?
You said [inaudible].
>> That's just 'cause my picture is long.
There has to be at least one every lecture.
Thanks. So, the colors are right and the vector is right
but yeah, the prime shouldn't be on the red one.
That is correct.
OK. So now, let's looks at what happens if we're talking
about objects that can change their sign.
So, for hydrogen atoms our only option here is they can stay
in the same place, they can move to a different place
and so we get 1s or zeros for the character.
Question?
>> Sorry. Why is X equals to zero again?
>> It's not X. It's chi.
>> Or chi?
>> And that's the trace of the matrix or the character
of the matrix and it's what you get when you add
up all the elements on the diagonal.
>> How did you find it [inaudible]?
>> So that is the shortcut that we can sometimes use
and sometimes not and I'm going
to keep giving examples of these.
So in the first example, we have two hydrogen atoms
and neither one of them change place or change at all.
And so that-- so we can say, OK, there are two things
and they didn't change, so that means the character
of this thing is 2.
In this case, they swap places
and so the character of that is zero.
So we're saying-- you know, in the cases
where the shortcut works,
I'm going to show you some examples where it doesn't.
You can say, if something says the same it gets 1,
if it changes places with another element it get zero,
and if it changes sign it gets minus 1.
>> OK. So [inaudible] definitions?
>> Yeah.
>> OK.
>> So, of course, and it's--
you know, again, this is a heuristic.
It's a shortcut that you can use sometimes.
It doesn't always work.
And it doesn't work when you have things where you can't--
where you can't map one element onto another very easily
with the symmetry operations we have
and I'll show you cases where that's true.
Yes?
>> So, for clarification, that's suggest
where the transformation--
that's the character of the transformation?
>> That's the character that goes under that transformation.
So what we're doing here is we're making a reducible
representation for how a particular thing
which is our basis transforms
under all those symmetry elements.
And we're going to build up to how to actually use that to find
out something about the molecule
but for right now we're just practicing.
We're just seeing how different things transform
under the symmetry elements.
Yes?
>> Are those the characters the numbers in these tables?
>> So that's a good question.
So, the characters in those tables are the characters
of the irreducible representations.
So those are-- they're the very basic representations
of how objects transform under the symmetry operations.
And I don't think that's clear right now but it will be
as we go through more examples.
What we're making now is called the reducible representation.
And so, we can add up different combinations
of those irreducible representations in the table
and make a reducible representation,
which is what we're making up now.
And we're going to see how that works and we're going
to learn a formula for reducing it and that's going
to enables us to find how-- you know, for instance, how many
and which kind of orbitals can be involved
in making a particular kind of bond.
And again, that's why I like this kind of examples
because you already know the answer to that.
You've all learned it in general chemistry.
So it's very-- the procedure that we're going
to learn is a little bit involved but it's very easy
to check your work because the answer that you get is something
that you already know for right now.
Later, we're going to do harder things, but-- OK.
So here's another example.
We're going to look at the same kind of thing.
So now, we're still in the C2v point group, right?
But now, we're looking at a different molecule.
So we have OCl2.
And our basis is two particular P orbitals
on these chlorine atoms.
And the way I draw the picture,
you can see that they have opposite phase from each other.
And one thing that people get confused
about is why did I pick this as my basis?
What about all the rest of the P orbitals on the chlorine atoms?
What about the P orbitals on the oxygen atom?
And the answer is, I don't care.
I'm not looking at those right now.
So, you get to set up the basis.
You're making up the, you know,
the problem that we're looking at, you know,
in this case, I'm making it up.
So this is what we're happening to look at.
The choice of the basis here is arbitrary
because we're just practicing seeing how things transform.
In a real scenario where you're actually trying
of figure something out, you're going to be looking
at which direction bonds are pointing because you might want
to learn which orbitals going to making up those bonds.
When we start talking about molecular motions,
our basis is going to be a little X, Y,
and Z axis on every atom in the molecules
so that we can learn about its motions.
Those are examples of realistic things that we might come
up with to solve a problem.
This is just for practice.
We just want to see how it works.
OK. So, again, if we look
at the identity we can use the shortcut here and say,
"All right, we have these two P orbitals and I know
that they're not going to change when we do the identity."
So, our character equals 2.
We can also make up the transformation matrix.
And we always know what it's going to be for the identity.
It's just 1s on the diagonal and zeros everywhere else.
And now, look what happens when we do a C2 rotation.
So, A and B swapped places but their signs don't change.
And this is important to realize,
the signs of the orbitals
or I'm saying are going with the positions.
So the sign of the thing that on the left is the same
as it was before, however, the identity
of these orbitals swapped places.
So as soon as they change places, the character
of that matrix is zero and it looks just like it did
in the hydrogen atom place-- in the hydrogen atom case
and we can see how the transformation matrix works.
OK. So now, let's look at what happens with a reflection.
So again, my XZ plane is in the screen--
the plane of the screen.
So if I do that reflection, these orbitals are going
to reflect through the screen and they stay in the same place.
The identities of A and B don't change
but they both changed sign.
So that tells me that the character now is minus 2 using
our little heuristic that we have two things
and they both changed sign.
And then we can also set
up the appropriate transformation matrix.
So, what I want you to get
out of this it is kind of a toy example.
There is not necessarily anything that useful
about knowing about these particular P orbitals.
What I want you to see is, you know, just how we can--
you know, first of all, telling the difference
between the elements changing places, things swapping around
and changing sign and the fact
that those are separate from each other.
I also want to point out that exactly this problem was
on the exam last year.
So this is-- you know, it's a good example of things
that you might need to be able to do.
Yes?
>> For the last one, did you kind of like demonstrate where,
you know, how you rotated?
>> I didn't rotate it, I reflected it.
So the XZ plane is the plane of the screen
and I reflected it about through that--
[ Inaudible Remark ]
Yeah. Yes?
>> You chose both the-- you chose one of the shaded to be
up in one and shaded to be down, down on A up on B. What
if you chose both the shaded to be up on A and up on B?
>> I'm so glad you ask
that 'cause that's what we're going to do.
[ Inaudible Remark ]
OK. So-- OK.
So, as we-- you know, as we anticipate,
this really shows you that how these matrices end up coming
out depends on both what's the operation
and what are you doing it to.
The basis is really important.
And so this is why I'm saying, you know, when you go to do this
on the exam, draw lots of pictures,
tell us what you're using for the basis
because there might be equivalent ways
of doing the same thing and also you want to be able
to get partial credit for you work,
so make sure you make it clear what you're doing.
OK. So for this case, we are looking at, you know, first,
we have the identity matrix, that's always going
to be the same so we don't need talk to about it again.
But now if we do a C2 rotation,
not only do these things change sign
or not only do they change places
but they also change sign, because now when we turn it
around the darker lobes
that were pointing forward are now pointing back.
So in this case, we get negative 1s in the off-diagonal elements.
So, that transformation depends not only on the operation
that we're doing but also on the object that we're doing it too.
And so again, just to bring it back to what's
in the character table,
all of those irreducible representations are objects
that transform in different ways under all
of the symmetry operations in the point group.
Again, the things that we're making here are reducible
representations that can be generated by adding
up different irreducible representations
and we're going to see how.
OK. So again, let's reflect it in the XZ plane.
Same thing, nothing changes places but they do changed sign.
So that one is the same as it was in the other example.
So-- OK. So now, we've illustrated how we can look
at these things and make up transformation matrices
for operations in a particular basis.
We have learned that the basis is really important to figure
out what these matrices are.
And, you know, we're starting to get a sense
of how this goes together with the things that are
in the character table.
OK. So now, let's look at some objects that transform in ways
that give us reducible representations.
So again, here's a point that people often get confused
about when they're first learning this concept.
So here we have a PX orbital.
And you might say, if we just have a PX orbital off by itself
in space, it doesn't belong to the C2v point group, you know,
yet I'm doing these examples in the C2v point group.
That's because we decided that that's the basis
that we're working in because of something about the problem.
So we were looking at OCl2, that molecule is
in the C2v point group, that's what we're doing.
So now we can consider various orbitals and objects
that are associated with that molecule and look
at how it transforms under operations in that point group
but we're not starting over
and assigning the orbital to a point group.
It makes sense to everybody?
I bring that up because I did something that people tend
to get confused about.
Yes?
>> All right.
Are the different signs in this P orbital different color is
just to show the different orientations on it?
>> The two sides of the P orbital are different colors
because P orbitals have a node
in the middle so it changes sign.
So that's the phase of the orbital.
>> OK.
>> And, you know, of course that's going
to important 'cause things might change sign
under some of the operations.
OK. So now, we have PX orbital.
We're in the C2v point group because I said so.
I know that a P orbital alone in space does not belong
to that point group but we're using this to set up a problem
that involves a molecule that belongs to that point group.
And we're going to consider how that orbital transforms
under each symmetry operation.
And it helps if you have your character table out and open
to the right point group when we do this.
OK. So, if I do the identity nothing changes
so it gets a character of 1.
If I do a C2 rotation it moves around and so it changes sign,
and so that means it gets a character of minus 1.
If I do a reflection in the XZ plane which again is the plane
of the screen, nothing happens, doesn't change.
So it gets a character of 1.
If we do the same thing in the YZ plane,
now it does change sign and so it gets a character of minus 1.
And so that's how we set up our representation in this group.
And so, if we go to the character table and look
at the C2v table, what you'll notice is you have this pattern
of characters 1 minus 1, 1 minus 1 and that belongs to
or it corresponds to the B1 symmetry species,
so that's an irreducible representation.
And then if you notice in the first column on the right there,
it says X in that column.
So that means that the character table is telling you how
something that has the symmetry so you can thing of it
as a Px orbital or as a little unit vector pointing
in the X direction.
We're describing how that transforms
under all the operations.
And if you look below it under B2 here, that one is--
that one has a Y, and so that tells you
that that's how a PY orbital is going to transform
under these representations, all right,
under these symmetry operations.
So, all of these things are collected for you here
in the character table.
And this is-- this contains a lot of useful information.
OK. So what else is in here?
We have RZ, RY, and RX.
Anybody know what those are?
[ Inaudible Remarks ]
Yeah. Those are rotations.
So those are rotations about those axes.
So, it takes a little bit of work to visualize how to sign
up rotation to a symmetry group but, you know, if you think
about it for a while you can probably visualize it.
We'll talk about it later when we get
into to talking about spectroscopy.
And then these other things, the XY, XZ, YZ et cetera,
these are quadratic terms, you know,
in terms of the Cartesian coordinates.
You can think of them as D orbitals for now or we'll see
that they belong to, you know, vibrational motions
that change the polarizability of a molecule.
So, hopefully you get an idea
of just how much information is already tabulated for you
in the point group table.
Question?
>> Could you clarify what A1 and A2 [inaudible].
>> Those are just the names of the irreducible representations.
And so, all you really need to know about them is A
and B are non-degenerate,
so there's only one symmetry species of that energy there.
If it's called E then it's going to be doubly degenerate
and if it's called T then it's triple E degenerate
and we'll see more examples of that.
The other thing that you should know is
that every point group has something--
sometimes it's called A, sometimes it's called A1,
sometimes it's called A1 prime et cetera.
It's always called A something,
where the characters are 1s for everything.
So there's always a species
where it remains invariant under any operation.
And that's a sphere, right?
A sphere always has that property.
And so, if we're talking about chemical bonding, you can think
about that as an S orbital and every point group will have
that representation, otherwise, you don't really need to worry
about what those things are.
They're just the names of the irreducible representations
in that particular group.
OK. So now, let's look at Y and Z orbitals.
So I'm going to go through this a little bit fast because it's--
you don't need to write it down, it's already
on the character table which is the joy of this whole thing.
There's all this information already tabulated for you
and all you have to do is figure out how to read it.
You don't have to generate it yourself every time.
OK. So we already went through X. We know how that one behaves.
So if we look at the PZ orbital,
we can see that that one doesn't change under the identity,
C2 doesn't do anything to it,
and neither do these reflections.
So, in this particular case, the Z orbital--
the PZ orbital belongs to that symmetry group that, you know,
that doesn't change under any transformation,
which in this case is A1.
OK. So now, I know some people might be thinking but what
about in that case where we were looking at the OCl2
when we had these orbitals, you know,
pointing in different directions
and we got different answers for these?
That's because in that case, my basis was the set
of the two orbitals together.
And so, that means that, you know, when we have--
so when we have an orbital and isolation like this,
in this case they belong to the irreducible representations.
But if you start building up sets of them,
then that might transform differently and you can make
up reducible representations for it.
We're going to learn how to deal with that.
OK. So if we look at a Y orbital,
we can do the same thing.
You know, again, I'm just going to go through it quickly.
You can do it yourself for extra practice if you need to.
And then now, the next thing is let's look
at this thing that's called A2 here.
So, that doesn't have anything that belongs to the X, Y,
and Z unit vectors but there are still some characters for it.
If we think about it as a DXY orbital,
we can see that that's going to behave like this.
So E and C2 aren't going to do anything to it
but it will change sign under those reflections and so we get
that under the A2 operation.
And so, sometimes if you look at these different point groups,
there might be irreducible representations
that are included for completeness,
but you might notice that there's nothing listed
under that, you know, in terms of linear quadratic terms
of the Cartesian coordinates or rotations, you know,
there might just be nothing in those categories.
That's OK.
It's just means that there's nothing
of chemical interest belonging to that symmetry species there.
Yes?
>> So the XY, XZ, YZ on the right,
can be basically visualizes the D orbital as well?
>> You can.
Yeah.
>> And they'll all work [inaudible] functions
like the 1, negative 1, negative 1, 1, they're all be the same?
>> They will be transformed according
to that symmetry species.
And, you know, again, how this is going to work
and where are these things go?
Totally depends on the point group.
So if you look around the different 1s, they'll belong
to different representations, and so that's why it
so important when you start doing these problems
to assign your molecule to the correct point group.
OK. So now, we have made up-- we have looked at--
we know what irreducible representations are.
We have at least seen some practice examples of how to make
up reducible representations.
Let's look at how we get from one to the other.
And there's a formula that we can use.
That's, you know, you have to keep track of what you're doing
but it's conceptually not really hard.
OK. So, what we want to know is we have some reducible
representation that we're going to make up because it has to do
with bonding or molecular motion or some property of the molecule
that we're interested in.
And we want to see how can we add
up the various irreducible representations and--
sorry, and get that reducible representation.
So, what we're looking for is, for example, the number of A1s
that appears in your representation.
So, sometimes we can do this by inspection
but often it's not really that easy particularly when we get
into harder problems, we need to use this formula.
OK. So, what we're going to do is add up the following things.
So first of all, we have a 1 over H. So what's H?
Remember, that's what you get when you add
up the total number of operations.
It's also called the order of the group.
So, the character table I gave you unfortunately doesn't give
you H so you have to add it up yourself but that's OK.
And then, what's in the sum here?
Chi R is the character in the reducible representation.
Chi I is the character in the irreducible representation.
So, remember, the reducible representation is whatever you
generated that's going to help you solve the molecular problem.
The character in the irreducible representation is the one
from the table under that particular operation.
And N is the number
of equivalent operations in the class.
So that means the coefficient in front of the operation
at the top of the table.
So let's look at this.
[ Multiple Speaker ]
Yeah, sure.
Do you have questions about it or it's just hard
to absorb everything getting it written down?
>> I have question [inaudible].
>> OK.
>> You said you have to add something yourself [inaudible]?
>> OK. So H is the number--
it's the total number of operations in the class.
So like for C2v we have-- so there's the identity,
there is C2, and then there's sigma V and sigma V prime.
>> OK.
>> So H is 4 for that group.
Some character tables give you the value of H
but this one doesn't say if to add them.
OK. So let's go through an example of how to use this.
So here's a reducible representation.
Again, if we're doing this in a realistic scenario,
we're going to have made this reducible representation
up ourselves because it's telling us something interesting
that we want to know.
In this case, for the sake of not taking forever,
I'm just giving you an example of one that works.
OK. So now we're in the C3v point group.
You know, Y again, it's just an example.
OK. So for this one, H equal 6 because we have the identity,
we have two C3s and we have three sigma Vs so we add
up all of those and get 6.
And so now, we have to look at the character table
and see how many irreducible representations we have.
There are three.
There's A1, A2, and E. And we have to figure out how many
of each of kind are added
up to give us this reducible representation.
OK. So we have 1 over 6 times, now we're going
to start adding this up.
So for each operation in the table, we take the character
in the reducible representation, that's the problem
that we're trying to solve.
I see your question and I'll get to it
when I finish explaining this really quick.
OK. So here, the character
in the reducible representation is 4 under E, and again,
that's what we were given.
1 is the character under E in the irreducible representation
that we looked up from the table and then 1 is the number
of symmetry elements in that class.
There's only one identity.
And now, we have to go through and do this
for every operation that's possible in the C3v point group.
So now for C3, we have 1 in our reducible representation,
that's what we got from the problem, 1 from the character
that we looked up, and there are two C3 operations.
And then we go through the same thing for the sigma V planes.
And question back there?
[ Inaudible Remark ]
OK, cool. OK.
So again, this has picky things that you have to keep track of
but it's not that hard.
So we just have the character that we're given
in the reducible representation.
We have the character in the irreducible representation.
In this case, A1 because that's what we're trying to figure
out which we looked up from the table and the number
of operations and then we just have to go through
and add these things for each element in the table.
Question?
>> So for the character in the reducible,
where did you get that [inaudible]?
>> That's the problem.
That's what I gave you.
So that thing here that I'm calling gamma 1--
>> Oh, that was given [inaudible].
>> That's given.
That's the question.
In real life, you're going to make that up yourself
because that's going to help you answer our question
that you want to know the answer to.
I'm just giving you some to practice.
We'll do a real example later on.
If we don't get to it this time, we'll do it next lecture.
So, yeah, this capital gamma is just--
that's a reducible representation
that we're looking at.
OK. And so, after all of these we go through and find
that there are no A1s in our particular representation.
That's fine.
I want to point out here that if you do this
and you don't get an integer number,
something went badly wrong and you need to check your work.
OK. So now, let's look at A2.
We're just going down the table.
We have to check out all
of the possible irreducible representations.
And so we do the same thing.
So again, the identity, the character
in the reducible representation is 4, that's given.
The character under A2 is 1 that we looked up on the table
and there's only one of it.
And then, we just go through and follow the characters under A2
and do the same thing for that symmetry species and we get
that there two of them in this case.
And then, we have to do the same thing for E. So,
this is one of the things that's a little bit challenging.
Don't get confused about E, the symmetry species
and the identity operator E. They have the same name
which is not so ideal but you can tell it from context.
OK. So we're looking for the number
of the symmetry species E. And again, we go through
and use our reduction formula for each operation
in the table, and we get 1.
And so, we went through all
of the possible irreducible representations
in this point group and we used this reduction formula,
and so we are done.
And here's what the form of the answer should look like.
We should say gamma 1 equals 2A2 plus E. So again,
it's not hard it's just--
you just have a lot of stuff to keep track of.
And, you know, it's fine to get zero for these and you'll find
that some reducible representation don't contain a
particular irreducible representation.
What you don't want to get is fractions.
You really do need to get whole numbers for these things.
So if you don't, go check your work.
Yes?
[ Inaudible Remark ]
So-- I'm sorry, can you say it again?
>> Yeah.
[ Inaudible Remark ]
>> Negative 2 below the three sigma V. That's the problem.
That's what I gave you.
[ Inaudible Remark ]
The problem is, here is this reducible representation gamma
1, reduce it.
And, you know, again, I want to point out when we do real ones,
that's going to come from something that we do as part
of setting up the problem.
Another question?
>> So basically instead of writing [inaudible] group,
we're going to write it in terms of the already [inaudible]?
>> That's right.
And the reason that we want to do that is
because we have all sorts of information about that
on the character table and we can use it to learn stuff.
OK. Let's try to bring this back to one discussion instead
of many 'cause it's really hard to hear what's going on.
Another question?
>> Oh, yeah.
The third term-- OK,
so the first term was the reducible second term was
irreducible for the third term?
>> Yes. The third term is the number of operations
in that class that are equivalent.
So that's the coefficient in front of that operation.
[ Inaudible Remark ]
It's on the character table.
So if you look at C3 there's a 2 in front of it.
>> OK.
>> If you look at sigma V there's a 3 in front of it.
OK. So, everybody clear on our terminology and what we're going
to do here 'cause let's move on.
OK. Let's see another example.
So I'm going to give you another example.
We're in the same point group and we're going
to go through this again.
And we're going to do it fast because we've already gone
through it however, if you get confused I'm happy
to answer questions about it.
OK. So we need to go through the number of A1 again.
And so now we have a different reducible representation
which we're calling gamma 2.
And so in this case, when we go through this,
we find that the number of A1s equals 1.
And I guess what I want to point out here is
that in this reducible representation, the same two--
the first two characters are the same
and only the third one is different.
We can get quite different answers from that.
So, it's not always easy to do them by inspection.
Sometimes you can if you can just see what adds up.
That's a fine way to do it but it's not always easy
and the formula always work, so we definitely want
to learn how to do that.
OK. So, if we go through and get the number
of A2s now again using this formula, we get 1
and the same thing for the E symmetry species.
[ Pause ]
And now, we get 1 for that one.
And so, we can write gamma 2 as A1 plus A2 plus E.
[ Pause ]
OK. So now, we know how to use this reduction formula
to reducible representations
and get some arbitrary representation in terms
of the irreducible representations.
As you can imagine, you could make up 1s
that don't work, right?
So if you make up a reducible representation
without corresponding to something real in terms
of chemical bonding or something like that,
you can imagine coming up with sets of numbers
that don't give you integers in terms of the point group.
So, you know, if you're going to look for practice examples,
you know, don't just make up arbitrary ones it may not work.
OK. So let's-- I had another one in here
but I don't think we need to do it.
Let's quickly go through how we're going to set this
up in terms of a real problem.
We won't have time to finish this problem but I at least want
to set it up so that you can see it for next time
and have a chance to think about it.
OK. So let's think about bonding in a trigonal planar molecule.
So, to give you a concrete example, let's say it's BCl3.
And what I want to know is which orbitals make
up the sigma bonds in BCl3.
And we all know the answer, right?
What's the answer?
What's the hybridization of the central atom?
>> SP2.
>> SP2. So, at the end we should get something that's consistent
with that answer.
But we're going to go through the process
because we'll see examples later that are not so simple.
OK. So, in order to figure out what--
which orbitals are involved in the bonding in this molecule,
I'm going to take a basis that has to do
with the problem that I want to solve.
So I want to know about those sigma bonds.
And so my basis is three vectors,
I'm just calling them A1, A2, and A3 pointing in the direction
from the central atom to each outer atom.
And that's my basis and I'm going to figure out by symmetry
which orbitals have the appropriate symmetry
to belong to those bonds.
And I'm going to do that by making a reducible
representation that corresponds to this basis and then I'm going
to reduce it and look at the character table and see
which irreducible representations fit
with that symmetry.
So, this is what we're going to do.
So here's our basis, it's these three vectors.
And now, what I need to do is go through and figure
out the character of my reducible representation
under each of the symmetry operations.
And so, I can do that in a number of different ways.
I can use our heuristic if I'm careful.
I can also set up the actual matrix for transforming this
and then take its character, that'll always work.
Let's just look at how we do this, if we set up a vector
that contains, you know, our three representations
of the bonds and then just look at how that gets transformed
under these operations.
So we already know what the identity is going to be.
That doesn't change anything
so we just get 1s along the diagonal
and zeros everywhere else.
And so, the character of the identity matrix is 3.
OK. So now let's do a C3 rotation counterclockwise.
And, you know, those who want to look for mistakes,
this is a prime way to-- this is a prime place to make them,
so make sure I didn't do it clockwise in the matrix.
OK. So-- yeah, it looks like I didn't mess it up, good.
All right.
So we have our vectors.
I did a C3 rotation and we can look at how they change places.
And now I need the matrix that generates that.
And so, this is what we get.
And its character is zero.
And again, we can use the heuristic that all
of these things swapped places,
so we know our character has to be zero.
OK. So now, how about C2?
Question?
>> So for problem like this we would get the rotated
[inaudible] and we'll just have to determine what the matrix is
to achieve that rotation?
>> Yeah. I'm just-- I'm going to through, you know,
what's the process for thinking about this?
And, you know, sometimes when you get really good
at it you can just kind of see it in your head, that's fine.
But, you know, for-- you know, it's kind of hard
when you first start learning how to do it,
so I want to make sure to go through step by step
for all of these things.
Question?
>> So for counterclockwise we see that A3 takes place
for A1 and [inaudible].
>> Right.
>> What would-- clockwise would be the opposite?
>> Clockwise would be--
>> Like A1.
>> -- you know, A2 would take the place of A1.
And so then my matrix would be reflected
about the diagonal if that's--
>> My question is how do you know which one is clockwise
and which one is [inaudible]?
>> You just have to do it.
Set up the matrix and multiply it by your original thing
and see what gives you the result that you want.
>> OK.
[ Inaudible Remark ]
>> So it's the picture going counterclockwise?
>> The picture is going counterclockwise, yeah.
>> OK. So, then in that case the Z axis
or what actually is going outside--
>> OK. This is-- wait.
Everybody listen.
This is an important question.
The question is what's the Z axis?
The Z axis is always the principal axis
that you determine when you put it in a point group.
And then--
[ Inaudible Remark ]
So in this case, it's coming out at you.
OK. We're about out of time and this is a decent place to stop.
We're going to finish this next time.
Also, you know, although our quizzes are unannounced,
this is kind of a good point to maybe think about having one.
So that's something that you should keep
in mind for next week.
[ Noise ]
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