WEBVTT

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hello welcome to the day four of a second
week of a this lecture series i hope all of

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you are doing well we have been trying
to figure out whether we can find out the

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point groups of different molecular structure
so in the last class we looked at

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the structures of hydrogen peroxides
and we saw that for hydrogen peroxide i can

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have two different structure one is the
plane axis when we consider the planar configuration

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so we got c two v point group for cis planar
while we got c two h point group for trans

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planar right so i guess you still
remember what were the symmetry operations

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in case you have not then lets again look
at those so for c two v that we got for

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planar cis planar configuration of h two o
two we got the following symmetry operations

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so we had e we got a c two we got a two
sigma v which are differentiated as sigma

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v sigma v prime while so this is a group
and for trans planar configuration we got

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identity we got c two we got sigma h and we
got the inversion symmetry

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now if i ask you can you tell me what are
the orders of this groups both of them have

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four elements ok in a group there are
four elements and each element is a symmetry

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operation so the answer is four for both for
them ok so order we have mention it earlier

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of a group this equals to the number of elements
in it ok so in case of symmetry point groups

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also all the symmetry operations when you
have found all the symmetry operations

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belonging to that particular group so you
form a close group and the number of such

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symmetry operations including identity will
give you the order of the group

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and in many cases you will be using
notation called h h stands for order

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of the group in some groups you will find
out that point group for order but it doesnt

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matter we mostly will use this term h to
define the order of the group now one more

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thing have you noticed about this particular
groups some kind of relations that you

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can figure out which exists you know
among the symmetry operations so we should

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think a little bit about it ok later on
we will come back to this topic again so

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what we will do today we will continue a looking
at some more molecules molecular structures

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and we find out there point group
symmetries ok systematically you remember

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we had a you know shown nice future
which is a systematic procedure which describes

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systematic procedure to find out the point
group of any giving molecular structure when

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you know all the symmetry operations for
that particular molecular structure so

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we will use the same scheme and try to find
out point group such certain molecules

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we will go from you know simpler to little
harder problems

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so index molecule that we will look at is
on your screen so this is one three five seven

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tetramethylcyclooctatetraene ok the structure
is giving here so for this molecule you

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what are the symmetry elements you can find
out ok the symmetry elements or if you say

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here directly symmetry operations identity
is into a by default its present there

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what are the other element that you can think
of does it have any proper axis of symmetry

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proper axis of rotation we don't a you know
really see a proper axis of you know axis

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of rotation which is not consequence of
something else by something else what i mean

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is an improper axis of symmetry which is indicated
here so rotation about this axis by ninety

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degree followed by a reflection perpendicular
to this axis we give you an you know indistinguishable

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structure we can just verify that one

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so if i put a star on this particular
ethyl group then a ninety degree will

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take it to somewhere here right so from here
from here to somewhere there and then when

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you invert it so the one the carbon which
was a you know protruding outside it will

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be going in below the plane of the paper so
that will actually you know come here while

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the guy sitting here will now move up here
so ultimately you will get back and in

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this indistinguishable structure so there
is a valid s four now if there is an s four

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we have learned in the you know one of
the previous classes that for even order improper

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axis of symmetry when n is greater than
two ok in that case one always will have

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a proper axis of symmetry which is like
c n by two ok so here we have s four so s

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four will automatically give you c two
so c two must be present so if you again

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think about this particular axis is s four
if you imagine c two axis there you will

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see that really that exists an c two about
the same axis

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so you know you can figure it out like
this portion which i have made this given

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this star symbol so this will come over to
this place and this guy will go here so ultimately

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you are getting the indistinguishable structure
so there existence c two but this c two is

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not natural not you know proper
axis of symmetry a principle axis of symmetry

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by this own merit but this is consequence
of having even odder in proper axis of

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symmetry so by rule you know we had
the step one two and three so if you remember

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that step three that will be looked at so
found that ok is it like s four is on only

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s n axis of even odder is the only symmetry
axis is possible yes so i can generate c two

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from you know s four itself so this
you know point group you can easily now

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figure out for this particular molecule

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so lets move on to another molecule which
is cycleoctatetraene now you see in the

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previous example as well as this current example
it is very vital to know the structure

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and that's why i in the you know one of this
previous classes i mentioned that in order

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to find out the point group first you have
to do you have to find out the structure

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of the molecule or the shape of the molecule
for that you can use this you know the structure

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and apply find out the structure and the
shape then only you can actually visualize

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the you know symmetry operations that may
exist in those molecular structure so here

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like when i look at this cyclooctatetraene
you know you need to know that this is

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a three dimensional structure like what is
shown on your screen so its its like a its

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like like a boat right so the cyclooctatetraene
also its has an alternative double bond

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now if you look at this structure what
you have here just like the you know previous

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case you have an s four but here other
than s four there are other symmetry

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elements so you can see that we have to
pointed out here here is one axis which joints

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this point and this point this is c two
axis so if you rotate hundred and twenty degree

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you will get back the indistinguishable structure
and so is this axis so you have two c twos

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and this c two is perpendicular to this axis
which is s four and obviously it will have

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c two axis as well coinciding ok on the
same axis so you essentially have now c

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two and to perpendicular c two primes
so this molecule definitely belongs to

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t type of point group right so without about
anything i can write its d and here the

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you know principle axis of symmetry i have
because this not the case which is only present

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here so other than s is also are there

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so i have a c two and two perpendicular
c twos to this original c two axis so it

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will it belongs to d and since the principle
axis has an order two because c two means

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the order of rotation is two so immediately
you can write up to d two now then you have

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to ask the next question that whether
the there is a sigma h ok so now looking at

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this molecule do i do a c any plane which
is perpendicular to this c two or s four

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axis that can give you an indistinguishable
structure and answer an obvious yes no

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right so i cannot really you know have this
reflection symmetry which you know is

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based on the plane perpendicular to c two
axis so there is no sigma h but there are

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sigma ds ok so the moment i know that there
are sigma ds i have to find out how many sigma

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ds so it demands that i should have at you
know two sigma ds then i can call this point

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group as d two d

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so lets find out so i have this two c two
prime right here and here and the definition

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of a you know of sigma d that is a for
a sigma plane to be called sigma d that

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plane has to bisect the and dehydral angle
formed by the principle axis and the two

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perpendicular c twos ok or you you can say
that ok it can also you know bisect the

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this c two prime the angle angle created by
two c two prime axis so i can have an plane

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so if i can draw the plane symbol like this
so one plane will be here and another plane

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will be on to this plane ok so if i can draw
the plane like this so i have two sigma ds

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so i like sigma d and sigma d prime ok so
i have two sigma ds so now my point group

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is complete so this molecule belongs to point
t to d so you have everything

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explicitly written here also so you can go
through this one so that you can understand

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it better
so lets move on to some other molecules

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benzene we have been showing you the benzene
models ok and this is probably one of the

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simplest thing that you know so the benzene
we have we have already seen that it has

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principle axis of symmetry which is
you know perpendicular to the plane of

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this scream ok so there are you know hydrogen
atoms

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i am not drawing the i am not drawing the
pi cloud here so assume that pi cloud is there

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that is in symmetry so it should be you
know implied so now you have along you know

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if i can imagine an axis which is a you know
passing through this point and perpendicular

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to this plane of the molecule then that is
my c six axis right and i do not have any

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other c six axis fine so c six is my principle
axis of symmetry so proper rotation is there

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and obviously this molecule for this molecule
i don't have to care about possibility

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of any special groups because its not that
on this you know sigma s there are only

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i is there is not linear molecule and since
i can see that you know c six is there

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so it can generate lot of other symmetry
elements i can see lot of planes there lot

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of perpendicular c two is there so they have
many other symmetry operations

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so we directly go to state four and
five so in order to go to state four and

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five what we have to do we have to ask
a question ok are there six c twos perpendicular

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to the c six so the question we ask is six
c two primes lets see so i can you know

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imagine an axis here so this is very very
obvious to ask right so if i give an hundred

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and eighty degree i get the equivalent structure
so this is one c two prime because this is

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perpendicular my c six similarly i can add
this two opposite hydrogen atoms so already

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i got three c two primes now similar to this
this c twos there are other c twos right so

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like this like this and like this so i got
my six c twos so the answer is yes so automatically

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my point group will be d six i had not
completed it but it will be d six followed

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by something ok either h or d or nothing lets
figure that out

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so now the following question will be is
there a sigma h and answer is pretty obvious

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because this molecule is planar so you know
and your principle axis of symmetry is

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perpendicular to that molecular plane so the
molecular plane what a planar molecule always

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host up a between a plane of symmetry so this
kind of symmetry is perpendicular to the principle

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axis of symmetry c six hence i have sigma
h so if that question is ask sigma h so the

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answer is again yes so ultimately my point
group is d six h so that is very easy now

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you can see that there are lot of symmetry
elements present there and then there will

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be lot of symmetry operations generated from
that we can now find out what is a point group

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for a given molecular structure and in
many cases we don't have to even go i find

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out all the symmetry operations because now
we have learned a little bit that ok what

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are the main things that we should looked
at in order to find out what is the point

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group so that's why i dint look at any other
possibilities but you can have a quick look

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at this structure again for benzene and you
can see there is an inversion centre right

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you start from any hydrogen passed through
the you know origin the centre point of the

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benzene molecule go to the other direction
the same same distance you will find identical

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hydrogen there so there is an i so there will
be improper axis of symmetries there will

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be a many other sigma planes which are not
in the molecular plane but perpendicular of

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the molecular plane so there will be six
sigma planes there ok so there are so many

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other symmetry operations

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now there will be a question that if someone
tells me a point group can i you know find

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out what are the symmetry operations that
this particular group can have answer is

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yes obviously yes so there there are you know
techniques by which you can find out so which

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are you know i will talk about in the following
week ok so you you will remember that we

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will be you know discussing about the possibility
of finding out the symmetry operations from

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the point group symmetry so one thing i
i forgot to tell you i should have this

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this rotations that we are using to do
you know to symbolized a particular point

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group like c two v c two s d two d two s d
two e whatever these rotations are called

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rotation ok so that's for your information

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so is move on to other some other molecules
again p f five that must be now by now very

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very easy for you right p f five what will
be structure you know first lecture you

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can think of is triangle by parameter so
let me erase the rest of the part so p

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f five will have a structure something like
this oops so if we call a there will be one

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florine and in axial direction there will
be one florine while here there will be one

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more florine and prosperous will be sitting
here so now what are the symmetry operations

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first like we will look at the principle axis
of symmetry what is the principle axis of

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symmetry we can see for very twenty one twenty
degree rotation i can get an equivalent structure

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so there is c three fine

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now next question to be asked is is there
any perpendicular c two so what will be

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the answer answer is easy right so i can i
can see one c two here because this three

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x in the equatorial plane will form as if
an equilateral triangle so you start from

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one tip of the equilateral triangle and
go till the mid of the base you get an

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axis which is a nothing but a c two right
so this is one c two prime time because it

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is perpendicular to the principle axis of
symmetry so principle axis of symmetry is

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along this as we wrote already c three so
since i have started in one tip of the equilateral

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triangle and got you know extend it to
the base and got my c two prime so i can similarly

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find other c two primes if i start from other
two tips and and rest to the respective basis

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right so similarly i get one c two here another
c two here so i got three perpendicular c

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twos that is c two primes so my point groups
belongs d type that must you know and it will

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be also d three type

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now is there is sigma h yes this molecule
this equatorial plane act as a mirror plane

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right so the top four florine will be reflected
here and this will go here so i have a mirror

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plane which is in the molecular plane and
perpendicular to the principle axis of symmetry

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so i have the point group right now which
is d three h that was easy right ok now

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you will see how important is to know
the structure in this and even in some

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of the following examples ok so we are going
to talk about ferrocene ferrocene is he have

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an iron and then two cycle cyclopentyl
groups groups so how it will look like

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so it depends right so do cyclopentane rings
will the iron in a center now this two

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cyclopentane they can be you know they can
the freely can you know they can rotate freely

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ok so they can have an eclipse configuration
and they can have a staggered confirmation

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so if i take this ferrocene molecule and look
from the top ok and if the molecule is in

23:48.150 --> 23:55.860
eclipse confirmation what do i have i have
something like this right because i was looking

23:55.860 --> 24:02.510
for the top in reality it will have a structure
which is like you know almost like a you know

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kind of shape right a with structure on the
other hand the eclipse the on the other

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hand the staggered configuration will have
something like this

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so

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ok fine so now we have to find out the you
know point groups for eclipse and staggered

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ok confirmations lets starts with the eclipse
which is the easiest i tell you at the way

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beginning this will have a t five h point
group why its a pentagon cyclopentane group

24:52.450 --> 25:03.500
and all the c c bond distance are same right
so for every it will be what will be

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the angle it should be three sixty degrees
by five so it should be seventy two degrees

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so every seventy two degree rotation will
give me an indistinguishable position about

25:20.410 --> 25:27.520
this axis ok so this is my ferrocene molecule
so axis you know passing through the center

25:27.520 --> 25:34.780
of one cyclopentyl group iron and then the
center another cyclopentyl group that will

25:34.780 --> 25:44.620
my c five so this two are eclipse so a
plane which is perpendicular to this c five

25:44.620 --> 25:52.580
and containing the iron will reflect the top
top part that top cyclopentyl group to the

25:52.580 --> 26:00.980
bottom and visa versa so i have a sigma h
right and this is a you know cyclopentyl

26:00.980 --> 26:11.930
group so just like the p f three i can have
c twos which are perpendicular right

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so i I can find out all the possible c
twos so there will be total five c twos which

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are perpendicular ok so i have c five as a
principle axis of symmetry i have five c two

26:27.060 --> 26:38.070
primes and i have sigma h so this becomes
d five h point group for confirmation now

26:38.070 --> 26:47.980
what about the staggered confirmation ok here
what do i have do i still have c five answer

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is yes i have c five so you can figure this
one out because this one will move here similarly

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this one will move here right so i will not
find any you know distinguishable change

27:02.320 --> 27:14.490
there so c five still exist in this staggered
confirmation now what about perpendicular

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c twos so now you know you know let me use
another colour so that i can i can separate

27:25.750 --> 27:41.440
this so let me use something like green
so now say this is my top cyclopentyl group

27:41.440 --> 27:46.840
and the other one that is in red colour that
is at the bottom so i am discussing the

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possibility of having a c two prime here now
we had c two prime for this eclipse confirmation

27:53.660 --> 28:01.679
right so along this there was a c two correct
now do i have a same kind of possibility here

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suppose i have i draw axis through this
now what will happen after performing a c

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two type of rotation ok along an axis which
is similar to this ok here also this top green

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part will now go to the bottom ok and this
red part which was in the bottom will come

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up here so this green part it will go to the
bottom not only go to the bottom so like you

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know this this tip of my this finger is the
is this point i am talking about so this will

28:45.490 --> 28:52.210
now flip completely and this part which was
you know projecting toward you will now come

28:52.210 --> 28:58.679
in this side that is projecting toward me
ok and its in the bottom hemisphere just like

28:58.679 --> 29:08.600
this similarly the red part which is like
fluct you know the the red part which was

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like you know tip here so that will go above
here so ultimately i have a c two which is

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perpendicular

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so very easily i can find out other five other
four c two primes so i have essentially five

29:25.720 --> 29:36.309
c two primes now the question is sigma h here
i do not see a plane which will give me

29:36.309 --> 29:45.350
you know an indistinguishable structure
upon reflection on that plane and particularly

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the plane which is perpendicular to the
c five axis here so i do not have any sigma

29:52.980 --> 30:04.510
h do i have a sigma d yes i do so any of this
you know sigma ds will be very easily

30:04.510 --> 30:11.970
found because i have the c twos now so for
you know taking if i find out any two adjacent

30:11.970 --> 30:17.660
c twos if you have a plane which contains
the principle axis of symmetry and bisects

30:17.660 --> 30:23.100
that you know angle formed by two c two primes
you have got your sigma d and you should be

30:23.100 --> 30:27.270
able to do that ok using plane and paper
very easily

30:27.270 --> 30:34.610
so you will see if you perform that you will
there are five sigma ds so that is sufficient

30:34.610 --> 30:44.580
to get the point group so i have d five
d point group for the staggered confirmation

30:44.580 --> 30:55.080
of ferrocene molecule ok so we will
stop here today and in the following class

30:55.080 --> 31:01.929
i will look at few more a quite complicated
molecules and then we will also learn about

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some properties of the symmetry operations
and we also will try to find out

31:09.770 --> 31:17.100
if the you know certain physical properties
like polarity or you know chirality and

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hence optical isomerism with the symmetries
of the molecules and therefore with the

31:25.020 --> 31:28.350
point group of the molecule so see you tomorrow
again

31:28.350 --> 31:29.080
thank you very much