WEBVTT

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hello and welcome so we were discussing about
how to determine the symmetry of you know

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the normal symmetric type of the normal modes
so what we are going to do we are going

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to take one particular example of this carbonate
ion and that we discussed in the previous

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class itself so what we will do to start with
we will take all the x y z coordinates

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on sitting on each and every atom ok so the
atoms of the nuclear to be in the origin

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and then the corresponding x one y one
z one or x two y two z two and all this cartesian

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coordinates will be there to act as a basis
so we have total twelve basis functions in

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our case
so now what we have to do we have to form

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early presentation based on this so we have
already learned that so now we are going to

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see what are the you know the characters of
the representation technical form using these

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basis functions now when we will operate the
by the way so here we have differentiated

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each one of these atoms by in a given number
this is one two three and four ok so the carbon

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atom is given the number four now when we
operate the symmetry operations on this basis

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function what you will do we will operate
on the x y z this you know this is vectors

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but not on this atoms alright
so lets try to figure out what will be

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the you know characters of this representation
so identity will give me matrix which is twelve

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by twelve and it is an unique matrix of twelve
by twelve so ultimately on the you know

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diagonal element will give me one each
and total twelve diagonal element so i will

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get a character of twelve now when i operate
c three on this particular structure so what

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do i get so c three this is a c three symmetry
right so c three is right here so when it

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operates one two and three they shift their
position by one right so one will go here

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two will go here three will come here so as
we discussed earlier this x one y one z one

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or x two y two z two x three y three z three
are not going to contribute anything to the

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diagonal
so this fourth atom that is the carbon atom

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and its this is functions that is basis vectors
here all of our use ok and since the c three

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is not changing this atom or it is not shifting
this atom so this basis functions will contribute

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so let us target you know one by one so this
say i have the axis c three axis along z axis

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right so what will happen this z will remain
as it is correct so it will contribute one

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to the diagonal now x and x four and y four
upon c three operation what what will be the

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total you know diagonal contribution from
there so if you you know look at that geometrically

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you can easily solve for that so we are not
going to do that but i can tell you that ultimately

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if you consider only this three basis you
know vectors because all the other atoms are

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moving completely shifting so we have to consider
only this part so therefore this whole thing

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falls down to a three by three matrix right
so z will remain unchanged so that will contribute

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one that is without any doubt i can immediately
write down

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now what will be the form of x and y so if
you solve it geometrically you will see this

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will be something like this ok because if
you apply c three on x four it will give you

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minus half of x four minus half of y four
so the other part will be root three

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by two and here it will be minus half ok so
this side you can figure out yourself ok so

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therefore if you look at the trace of this
you will have zero so the rest of the three

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oxygens are contributing zero and this one
also though they contribute to the diagonal

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but overall sum will be zero so you will get
zero character for c three

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now if you look at the c two so say you choose
one of the c twos ok so if you can solve it

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for one c three c two then you automatically
have the characters for other two c two types

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so if you consider this x c two axis then
what you have this o one and c four they remaining

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at their place but o three and o two are interchanging
so we do not consider this so we had now six

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by six matrix so to start with this c two
is along x axis correct so the x will remain

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and unchanged so i get a contribution one
from here and one from here right so i get

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already a contribution of plus two
now this y one will become minus y one if

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i rotate along this x-axis so so will be path
for y four therefore this and this will contribute

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totally minus two so i have minus two contribution
coming from the y side and then the z side

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z will also give me and you know minus
z one so i have minus two contribution from

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the z so out of this six basis vectors
i get total character to be equals to minus

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two right so i can simply write minus two
for all the c two primes ok now let us consider

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the next symmetric element that is sigma n
so what is sigma n is going to do so sigma

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n which will which will not change any of
the atoms so everything is remaining their

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case now x and y constitute the plane of sigma
n so there is nothing going to happen to x

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and y so there are total four such x y so
total eight you know this is vectors will

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be contributing one each to the diagonal so
i have a contribution coming from the x s

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and y s i have a contribution of total plus
eight correct because nothing is changing

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while this eight if you look at it will be
completely inverted it will be reflected onto

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the negative z side so also we contribute
one minus one each

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so therefore i have a contribution from z
s to be minus four so the resultant of this

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is going to be plus four correct so you have
this one to be plus four next is s three so

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see when you talk about s three here s three
has only one difference from that of c three

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so c three gave me this matrix by which
is actually this is the block in the total

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twelve by twelve matrix so this three by three
block you know gave some contribution to the

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diagonal but ultimately overall this is going
to be zero for c three

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now what happened here that is this z remain
unchanged therefore it contributed plus one

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so this plus one and then minus of minus of
it all together becomes zero now in case of

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x three everything will be same except for
the fact that there be reflection in the x

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y plane which will cause this plus one to
be actually minus one so in case of if i consider

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this you know c three then say in if i
right in white color this will be c three

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then this is plus one and if i consider s
three in this color then this will be minus

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one so when you consider s three it will have
total character to be minus two so i have

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minus two for this s three now we consider
sigma v here also we can choose from one of

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them so lets take this particular sigma v
alright so this contains x z plane correct

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so what will happen to o two and o three they
will be shifted completely because they are

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interchanging so again i am left with this
o one and c four correct and their this is

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vectors
so since we are considering x z plane then

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i have no contribution coming from x s and
z s to the plus one each because nothing is

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going to be change so then i have two x s
and two z so total four while the contribution

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from y s will be minus one because this x
s plane will reflect this plus y one two minus

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y one or plus y four to minus y four so there
will be contribution of minus two from the

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side of y s to the overall diagonal so therefore
i have four minus two because two plus two

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so sigma v s we contribute plus two to the
overall character of the reducible presentation

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so this is my reducible representation that
i got using three n cartesian coordinates

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as my basis z all right so what is my next
job so my next job is to deduce this correct

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so in order to deduce that what we have to
do you have to use the formula that we have

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already deduced earlier so if you deduce that
what you are going to get you are going to

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get say if i
so essentially i have gamma three n equals

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to a one prime plus a two prime plus three
e prime and you have two a two prime double

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prime and you have e double prime so this
is what you will get ultimately so you solve

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this meaning that you deduce it and then you
are supposed to get this particular expression

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so now i could i could find out what are
the irreducible representations which

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to which all this my basis vectors x y and
z transform as so we have use three in basis

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vectors meaning that we have considered all
the you know possible motions translations

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rotations as well as vibrations and they all
to be transfer as this many number of you

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know this many number of this thing number
of reducible presentation so if you can

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cross that here you will see this two are
one dimensional and this e s are two dimensional

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so one plus one two then six eight and then
again two ten and then again two twelve so

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total dimensionality is twelve which is matches
with this as well as this matches with overall

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number of this is vector that we have used
so everything is in ordered

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now what is our job we have to find out from
this list of irreducible representation we

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have to find out what are the irreducible
representation that actually are you

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know for which the molecular vibration or
the normal modes are actually the basis for

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so how do we do that that is very easy so
again you have to take help from the character

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table of the particular point view so again
we will resort to the character table of

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this particular table point group so we have
got our this list of irreducible presentation

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correct now what we have to do essentially
we have to strike out all

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the options that are for translation and rotational
motion then the residual you know irreducible

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presentation that we get will be for molecular
violation so if you concentrate on this particular

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character table here the area three and area
four for you will see that there are different

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linear functions that you have already discussed
so now what we have to do so for translation

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we have to look at the irreducible presentation
for which the linear functions that is cartesian

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coordinates x y and z they from the basis
for so here if we look at what we see is this

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representation e prime for representation
e prime x and y from the basis together while

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z from the basis for a two prime correct so
this x y and z this forms the basis of translational

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motion so therefore what we can write that
out of this if i go for say for example gamma

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for translation so gamma t r if it i then
what we have we have e prime plus we have

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a two double prime ok so i will i will again
till this part so in order to find out which

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irreducible presentation you know will corresponded
to the translational motion if you have to

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find that all you have to do you have to look
at the character table and particularly area

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three because area three gives you the you
know linear basis functions x y and z

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now in area three this x or y or z they will
be written you know besides one particular

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irreversible presentation or another particular
irreducible presentation and you have to find

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out which representations they are and what
we saw that this x and y to get that from

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the basis for e prime irreducible presentation
while the z function transform as a two double

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prime so if i take out this two part i know
that ok this to irreducible representation

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they they actually belongs to this there is
possible for the translational motion ok in

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other word actually this translational motion
for this particular molecule it you know transforms

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as a two prime and e prime in irreducible
presentation

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now similarly i have to find out what are
the irreducible presentation for rotational

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motion so again if i look at the character
table and we look for the functions this r

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x r y and r z so we have to find out which
irreducible presentation you have this r x

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or r y or r z ok find out those irreducible
presentation and you have got the irreducible

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presentations for rotational motion so if
i look at again we will see that r z that

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is rotation about the z axis it form basis
for e two prime all right so if i write gamma

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rotation to be a two prime and r x and r y
they transform together at the basis of e

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double prime so therefore so how many rotational
degrees of freedom are there three how many

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transmission degrees of freedom there three
so if you check here that you have total dimension

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of this is three and here also three so that
is again you know tells see that everything

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is in order so i have found out the irreducible
presentation which corresponds to translational

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and rotational motion so now if the altogether
translation to rotational i have a two prime

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a two double prime e prime and e double prime
if i you know take out this four irreducible

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presentation then from this list i have got
the irreducible presentation which you know

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which gives me the symmetries of the molecular
vibration right so in order to do that i have

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to get gamma so i can write gamma vibration
equals to comma three n minus gamma translation

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minus gamma rotation ok so what do i have
here a two prime i have one a two prime

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here correct so if i remove a two prime from
this list so let me strikeout this and then

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a two double prime i have two a two double
prime so i will strike one of them ok i still

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have one a two double prime and then one e
prime and one e two e double prime from this

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list so e double prime is gone and then i
have now left i am now left with two e primes

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so if i write that ultimately what do i have
i have a one prime plus two e prime all right

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and i have one a two prime how many number
of vibrations means normal modes we have i

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have total six so the dimensionality wise
i should have a match so one plus one plus

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four total six everything is in order so these
are the normal modes and these are the symmetries

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of the possible normal mode of vibration or
c o three two minus molecule

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now if you just go back to those structures
that i you know showed and you know i showed

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you that symmetries also there i just wrote
down but how did i you know get this one is

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this way now if we have to suppose i i
i i i do not know what are the normal modes

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right no one has drawn those you know structures
modes with atoms particularly then how do

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i know which modes which particular you know
which are correspond to so now what we need

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to do there so if we look at the characters
here again so i have to look at the character

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corresponding to a one prime so a one prime
has all the characters to be plus one so this

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is totally symmetric higher so if i have to
assumes something completely symmetric about

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this molecular vibration what can i think
of i can think of just a structure where this

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all this atoms are straights (Refer Time:
23 :00) in equal amount in the same direction

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as the direct bonds ok so that means is
that symmetric stretching so you know i can

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accordingly draw the normal mode of that and
then i have the characters for you know e

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prime so e prime what it has it has two for
identity and minus one for c three right

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you have minus one for s three also and sigma
v s are zero so in this case you know

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i cannot comment on you know the extract stretching
bond stretching or movement of a particular

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angle but it is not so you know it cannot
be expressed in superior form but i can say

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some combination of you know bond stretch
are you know combination of one you know angle

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movement will give rise to this e type of
symmetry alright

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and this you know a two prime that has that
has you know character of negative one

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with respect to sigma each which means that
that is not symmetry this motion is not symmetric

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with respect to the molecular plane that initially
tells you that this must be something related

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to the out of plane motion ok then only you
will have a negative character for this sigma

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s so now having said all these things next
question is like could we come to this you

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know normal modes and their symmetries
without dealing with all this coordinate systems

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x y z and all these things so the answer is
yes so how can you do that so you can do it

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by taking the internal coordinates into account
so how to do that so you have to consider

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either certain bonds or the bond angles to
be your internal coordinates so here we

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can consider this c o bonds as my basis function
or this o c o bond angle as my basis function

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or any diagonal angle as my basis function
now which one you consider first or you

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know which one you consider late and there
is no last and first rule so you can start

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with any one of them so for example if i start
with this co bonds then what we have we can

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form a formal representation where by number
of basis functions are three so so i con

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consider as set one which has three c o bonds
they act as my basis function ok so then on

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this three c o bonds i operate all the
symmetry operations of the point group and

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i form the representation all right and then
i deduce that so if you if you you know formal

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representation where you are considering this
only this right so this three bonds can either

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straights in the same amount at the same time
which will very much symmetric so definitely

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that will be giving me totally symmetric universal
presentation a one so here it is a one prime

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and the rest two it can be like you know one
is stretching in this direction one is contracting

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so there will be a you know combination of
you know two motions so i can intuitively

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we can say that that will belong to one of
the e representations ok so call it e prime

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so i can say that intuitively
and i can similarly form another state of

23:50.020 --> 23:51.020
basic functions so the lets call it as set
two while i consider this c three o

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c o bond angles ok so now unlike this you
know c o bonds i cannot have all the three

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angles being expanded in a you know same way
that is not possible so therefore you know

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two angles will you know changes in a same
way but other one will be in a different way

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so again in that way i can you know
intuitively say that ok i will have a three

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dimensional representation that i can deduce
it to say one one dimensional and two have

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two dimensional presentation and thats how
we can formed you know we we can we can

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say or comment about the normal mode of
vibration from the symmetry angles so this

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particular part we will elaborate in our next
class where will show that how using the internal

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coordinate we can exactly deduce the symmetries
of the normal modes ok so thats what we

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will deal in the following class and then
you know next after that we will go to

24:18.410 --> 24:19.410
the you know go and find out how to
get the selection rule of vibrational transition

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and thank you for your attention see you in
the next class