WEBVTT

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hello all and welcome to the first day of
the seventh week of the lecture series

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so last week we started looking at the
molecular vibration and his symmetry properties

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so just i quickly recap of whatever we
discussed in the last class so we mentioned

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that there you know there are different
degrees of freedom available for given

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molecule so out of this three different
degrees of freedom that we know about one

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is translation one is another is vibration
and this period of the lecture this week

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or will part of the next week also we will
be dealing with this molecular vibration and

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their symmetry properties

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so we mention that you know molecule
can you know vibrate its bond can stange

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its angles can bend and it can go you know
in a plane or you know out of a plane so

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different bonds will make different kinds
of would mean different angles will make different

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you know movement so apparently it may
look like that it's a random motion but if

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one takes a choose look at this motion then
one all will find that you know in this randomness

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there is actually a symmetry involve there
is a regularity there and the you know

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this you know this seemingly you know aperiodic
or random internal motions of vibrating

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molecule that are the result of superposition
of number of relatively simple vibratory motions

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which are known as the normal mode of vibration
now each of these normal mode of vibrations

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they have their fixed frequencies ok and we
need to know you know for a given molecule

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what are the normal mode of vibrations and
what are their symmetries so that we can develop

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something further in you know understanding
this molecular vibration emotion and you

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know the energic transaction form the one
vibration state to another vibrational state

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how it is governed by the symmetry rules all
right

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so we started talking about the you
know properties of these normal mode of vibration

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and in that case that we said is that you
know each of the you know fluid we actually

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actually defined the the normal mode of vibrations
for a particular molecular ion and which

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is carbonate ion ok so you looked at and this
carbon c o three two minus and we depicted

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all the possible normal modes so first we
did used you know what is the number of

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normal mode of vibration that is possible
for this so we also looked at the you know

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way to find out the number of normal modes
for a given molecule and we found that

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a two molecule is non linear then ten minus
six is the number of normal modes by n is

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the number of atom in that molecule and
for a for a linear molecule the number

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will be three and minus five so using that
formula we could say that for carbonate ion

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if you are taking as a planar molecule
the number of normal modes are six and

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we do all six normal modes of vibration and
we mention that ok this you know normal mode

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are depicted by arrows which tells us about
the particular duration and a little magnitude

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of the motion of the atom involved ok in a
particular direction or in a change of angle

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because of this motion both in plane and
out plane

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so we said that you know this six motions
that are possible that is three six normal

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modes of vibration you know we given them
a this discrete frequencies say that's

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a property of normal mode that each and individual
normal modes will have a distinct frequency

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so we mentioned this one as nu one this is
nu two and this is nu three a and this is

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as nu three b while this is nu four b and
nu sorry this is nu four a and this is nu

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four b now it will be clear in the movement
why we are calling this as three or three

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v not like three four five six ok there is
a reason and that is also given by the symmetric

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properties of this motions

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now there are two distinct properties
of this you know this type of motion that

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is normal mode of vibration that is you
know here this vectors that we are represent

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you know using to represent their motion
of a particular atom or you know a particular

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bond in a particular direction they can
be expressed as you know a linear combination

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of the basis vector that will use ok and second
that each one of this normal modes so nu one

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nu two nu three and nu three b and so on all
of them each one of them will transform

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as one as the universal representation to
which this particular molecule belong to so

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here this particular structure if we look
at we easily can say that this molecule as

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a point know for d three h if you you know
remember you have basic perform this course

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so it has as a c three axis and it as three
perpendicular c two primes and it as sigma

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h this is the spay because we i will considering
a planar molecules so it as a point group

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of d three h so what you says there is that
each of this normal mode will belong to the

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highers of this particular point of d three
h right

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now what to first you know ask that
how to assign the symmetries of this

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these normal modes ok so what what we need
to do for that so we need to look at the

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you know effect of all the symmetry operations
of d three h point two on this particular

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normal modes so wil look at this one two
and three a three b four a and four b so if

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we look at say first this structures for the
which the frequencies nu one will see that

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whatever be the symmetry operation so what
are the symmetry operation for d three h so

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for that if you want a quick answer for
that you what you have to do you have to look

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at the character table right you will know
by now that if you look at so this is our

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form of character table you can get it anywhere
so you look at the character table for say

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any the particular point group

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so let's look at the point group of d three
h right and so here we go so if you look at

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the point group of d three h he will find
all the symmetry operations there ok and the

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corresponding illusive representations the
terminologies you will also get the you know

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linear or non linear functions which also
transform as the you know you know illusive

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representation of this particular point group
so if i look at this and we can easily

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figure out what are the symmetric operations
that is have so i will just quickly mention

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so it as identity it as c three c three square
it has three c two primes it has sigma h it

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has s three and s three five and it as three
sigma vs right all right

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so now a what you have to do you have to
deal with the you know transformation of how

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this you know vectors that are that
are representing the motion of this atom within

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the normal mode of vibration how they behave
right so if you apply identity on this

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particular structure what you will get
everything will remain unchange ok so if you

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take this structure as a whole right so
if i can just redraw the same structure here

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what i have three equal arrows which are
along the three bonds ok in the same direction

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of this see your bonds right so now if
you apply identity it will give back that

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the same structure all right now if you apply
c three on this motion so if you rotate c

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three and if you are not numbering them actually
just looking at their overall you know this

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position of this arrows then c three also
will give you the exact you know indistinguishable

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structure so that will also give you a same
structure back so if this gives me one and

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this gives me one and next you look at the
c two so c two also because after rotating

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above say i am considering this as my c two
axis so if i do c two around this one so this

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will rotate and since this are not numbers
so i have an indistinguishable structure ok

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so ultimately applying c after applying c
two you will get just an indistinguishable

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structure so over all it will give me a character
one and while am writing this characters i

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am recreating that here i am not taking this
three as separate three basis vectors but

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as a whole this is my structure and this is
as if acting as one basis function right

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so similarly sigma which also will give
you one because there will be no change s

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three is also going to be mean one so will
sigma v so looking at this character we can

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find out that this is a totally symmetric
illusive representation so what is there totally

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illusable representation in d three h say
that it we gives a term a one prime all right

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so in the same way if you look at the structure
nu two so if you look at the structure nu

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two what you see that three oxygen atoms are
out of the plane that is above the plane while

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the carbon atoms in central carbon that is
bellow the plane that is you know given by

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the plus and minus sign so plus means its
you know the about the plane and minus is

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below the plane so what happens here so
if you apply identity so identity will give

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you say if i write down for this particular
structure say here so identity will give me

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a character of one while say c three that
will also give me one because after rotation

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by two phi by three i will getting indistinguishable
structure so if i go for c three that is also

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going to give me one

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now for let's these c two prime c two prime
is also going to give me the same thing so

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c two will give me character of one and then
we have sigma h so sigma h is going to

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inward the science right so whichever was
above the plane they will now go below the

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plane and whatever was the below the plane
now will come up so that will inward the structure

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so it is going to give me minus one right
and then we have s three so s three is

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what it is the effect of c three which doesnt
do anything but then there is a perpendicular

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sigma plane so that is going to reward you
know invert everything so then again s three

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is going to give me minus one and then we
are left with sigma v so sigma v is not going

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to change anything right so sigma v will give
me plus one so now if i compare this particular

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illusive representation because here the representation
i given me only you know you know one as

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the matrix element only one matrix elements
is there so it is the one dimensional representation

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and therefore i can just quickly check in
the you know i correct it in the and find

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out which illusive representation this one
is ok

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so only the character of sigma h and the
s three they are negative right so if i

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compare that then i have to so for d three
h if i have so here sorry there was a mistake

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this c two actually if look at so if i take
this as c two so whats happening this two

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are you know going bellow the plane and
so is base one so every atom and you know

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and their position know that position is getting
inverted for c two as well all right so

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the c two primes so therefore what we have
is c two is also giving me a minus one so

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that was a mistake so if i correct everything
so now i have c two sigma h and s three as

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negative one so now let's compare so that
gives me a two double prime so this belongs

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a two double prime right so now a what you
can see that you know both of this normal

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mode but we consider when we operate this
symmetry operator we are finding that they

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transform according to one particular illusive
representation the first one we found it to

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be a one prime and this one now we are finding
to be a two prime so in this way we can keep

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going on so here like this nu three and the
structure corresponds to nu three b they actually

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form the basis for the illusive representation
e prime together so that can be shown and

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we will show that will quick shortly

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so what we can think of is the following
so when you apply the symmetry operation

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say for example c three on this one of the
structures nu three a so what it will do it

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will you know it will transform to
a nu you know configuration which will

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be nu three a prime say for example so
what will be the you know the form

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of nu three a prime so what you are saying
is this nu three a prime will be a linear

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will be a linear combination of nu three a
and nu three b and similarly nu three b when

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you operate say any symmetry operation that
will also be a linear combination of nu three

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a and nu three b so let's show how it is going
to work so here here see an example how that

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you know c three aspects this structure
with nu three a so this is so after you know

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operating c three what will happen so this
arrow here on this atom one o here when we

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are doing this operations so we are only dealing
with this this vectors all right so this

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guy will now go over here right so accordingly
its orientation will also change now you see

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this one which was on atom a is now coming
over here with this orientation and this guy

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on the carbon atom this big arrow will now
go to a particular hundred eighty degree

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duration so it will come over here and let's
have a look at this this is the nu position

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of this arrow so in this way a you can find
out what will be the final structure now this

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final structure which we are calling as nu
prime three a how we can express this in terms

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of nu three a and nu three b so if i say that
this nu structure that is three a prime is

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expressing this passion and if you can show
that and prove that then you know you are

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proved our point so you you take you know
a particular mode of normal mode of

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vibration so you take part of it so here like
you know you take you look at the structures

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nu three a and nu three b so you see here
right so this one if you operate and

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you you will get an ultimate structure that
we have just shown now if you take some portion

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of this one and some portion of this one and
can generate that results in nu prime three

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a then our job is done and that's what shown
here so if we say that minus of of nu three

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a plus plus of of nu three b is equals to
nu three a prime so this is a linear combination

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of this know this is some sort of combination
so what is minus half of nu three a that is

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this one so if you take the original structure
the arrow and if you you know apply this

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minus half on that so you will get this structure
similarly plus half of this nu three b will

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give a this so you will see that this length
as been reduce to half of this original length

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and here the length as been reduce to half
of its original length as well as the direction

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as been reversed because the minus sign and
now if you add this two you get this one which

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is if you compare here you see this one

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so you could if you know express this this
nu structure as a combination of this nu three

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a and nu three b similarly for any other
you know particular motion on the any particular

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atom say one two three or four you can express
them in this particular passion so this is

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true for this whole normal mode right so
all this things are like you know breaks

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up those a individual arrows that are
you know situated on each and every atom

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and you take this particular combination rule
and you see that ultimately you can generate

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this structure starting from nu three similarly
you can show that for nu three b and there

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will be another type of combination where
here this nu three a prime is a combination

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where is minus of nu three a plus the plus
of nu three b and for say for nu three

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b if i operate c three on that so that i can
write as minus three by two nu three a minus

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half nu three b ok

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so in this combination i can express the new
form of nu three b after operating c three

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all right so in this way if i can you
know show for the nu four a and nu four

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b also now what does it mean mean that nu
three and nu three b upon the symmetry

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operations on them they cannot transform in
as one of the illusive representation of

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d three h point group alone but together that
is nu three a and nu three b together they

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can form the basis for a particular representation
and that must be a two dimensional representation

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because we are taking two basis function as
if right so nu three a and nu three b together

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they will form you know two dimensional
illusive representation and if you you

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know find out all the operations the effect
of this all the operations on of nu three

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a and nu three b you will see that it belongs
to it forms a basis of e prime representation

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of the d three h point group and the ultimately
you you can you can you know see that all

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the normal modes they form the basis for
illusive representation to which for the

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point group to views the molecule belongs
to ok all right

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so now here we started with showing the
normal modes of a particular groups and then

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we showed that they have their symmetric properties
that is they belongs to the form the basis

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for the illusive presentations now how
do i if i do not have this normal modes

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you know present in my in front of me can
we find the normal modes and its symmetry

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properties ok so we can show that by using
the same particular molecule so how do

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we how do we do that so first of all
we need the you know set of basic functions

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right so we can do that in two ways because
we are dealing with the molecular you know

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the the internal motion of the you know
atoms and bonds in a molecule while you

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talked about vibration then we can actually
take the all three cartesian co ordinates

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sitting on each atoms of the molecule because
we stated at the beginning that all these

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arrows right all this motions of this individual
atoms when they execute one mode of particular

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vibration this each arrow can be expressed
as a resultant of three basis of the resultant

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of the basis vectors that we choose if we
take x y and z co ordinates or each atom so

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this arrow which defines of the part of
this normal mode that can be expressed as

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some sort of combination of this x y z so
what i mean is following so if i have say

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let's let's just say tell it like this
so i have right so now in one of the motion

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that i showed like this what you say is i
can take this x y z you know and take this

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atom to be sitting at the origin all right
so then you know if i take the co ordinate

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accordingly then this particular arrow i can
express as some sort of combination of this

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basis vectors so i can take the cartesian
coordinate of all the atoms ok

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so when in each case what i will do i will
take the you know atoms to be siting on the

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origin and i will consider there x y and z
coordinate system to be acting my basis

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so therefore what i can do for this
one i will have total four atoms so four into

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three total twelve basis functions right in
my set and we can do whatever we want to do

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that the other option is that we can take
the internal motion of this molecules as such

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what do i mean so here what i do is like i
have this arrows defining my internal motions

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right so suppose i take this carbon oxygen
bond so by this arrow what i say that this

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oxygen is moving away from carbon so in other
word this c o bond is getting right or while

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it is coming back then it is you know its
its its reducing the bond so i can also take

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the angle between this two c o bonds so i
can take this angle as one of my basis function

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i can also take an dihedral angle meaning
that i have a sigma plane here and i have

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have this angle here ok or i have this plane
here and then i have this bond here

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so the angle between that i can take all this
you know internal co ordinates to be my

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basis function and we will show in the
following class that you know in either of

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the cases i can get the required information
and in certain cases without you know knowing

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or you know with this simple use of symmetry
i can really talk about what are the particular

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normal modes that this molecule have in other
words i can draw exact the normal modes of

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this molecules of any given molecule so that's
what we will start within the following class

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till then

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thank you very much for your attention