WEBVTT

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hello and welcome today is the third day of
the seventh week of the series so we have

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been discussing about the molecular vibration
and its symmetries so we looked at a one example

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of co three two minus which had d three
h symmetry so what we did we use all the

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curtizen coordinates seeking on each atom
and using those coordinates we found a

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representation and reducing those representations
we could ultimately find out what are the

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irreducible representation that any kind of
molecular motion can have and from there

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with help of the character table we eliminated
that you know that are you know that

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represent translation and rotational motion
so there by we got the irreducible representations

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that can be assigned to the genuine modes

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so till that point we did and we all says
that if

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we do not have any idea about the particular
normal mode then use in something call

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internal coordinates as my basis function
one can actually find out that what are

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the different motions that this particular
molecule can execute in terms of vibrational

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motion so today we will take you
know another example here and we will

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find out not only the irreducible representations
that its vibrational motions can transform

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as but also we will try to use the internal
coordinates to find out about the normal

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modes themselves so we would like to find
out the exact you know directional motion

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this which this the atoms of this
molecule will execute about it about its equilibrium

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opposition

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and also we will try to find out that out
of these vibrational modes which which

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terms some as certain irreducible representations
which one of them is infrared active which

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one of them is roman active because
this two are you know you know this

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two gives as a information about the vibrational
transitions in a molecule so lets get started

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with one particular molecule so the one
that will first consider is n two f two

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and one particular form that is trans planer
consideration will be consider today so

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as usual so see this particular example will
take you through a many various case that

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you have learned so far so each every example
that we discus here will give you you know

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a recap of whatever we have learned and every
time something added to it so that you get

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an overall idea what we can do with the symmetry
and group theory so first thing first

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so we have to know the structure this is the
trans planer so usual they have their lone

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pairs of electrons which will not consider
here even if you want to take it you can take

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it as per your wish so if you take it in this
way this is fine but we will ignore the presence

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of the lone pairs of electrons here

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so first we have to find out what is the point
group of this particular molecule so this

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has c two which is toward you so along
z access if i ensure and then this is the

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this is the sigma h plane so i have a i
have c two i have sigma h and also i have

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an inversion center right because if i start
from here it will go here and it will be coming

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to the place of one and this two and three
they will also interchange as a opposition

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so i have also i and never less i have
e so my point group is c two h all right fine

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so we have found the point view next what
we suppose to do so next we have to find

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out the representation for this particular
point group and the we have choose up basis

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sets for doing that so in this cases when
we want to deal with the molecular motion

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we choose generally the coordinate system
on each and every atom so we will consider

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the x y z coordinates on each one of this
atoms ok

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so total i have four into three equals
to twelve cut then coordinate you know

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coordinates as my as my basis function
now you know when ever we talk about the

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molecular motions so we talk about the degrees
of freedoms so in whatever direction i can

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have the motion right translation rotation
or vibration so now if i talk specifically

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about vibration so vibration is nothing but
the moment of any particular atom you

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know with in a molecule in a particular
direction by a particular amount ok so this

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vectorial motion can be you know be a result
of three unit vectors in that directions in

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in on that atom ok so i can take this orthogonal
cartesian coordinate system x y and z coordinates

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through find that particular you know
you know motion of atom right so it can be

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vibration it can be bending of an angle
ok it can be out of plane or in plane motions

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so all this things you know say like translation
i can i can express all this motions by combining

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this you know x y z cartesian coordinate
system on each and every atom rather i said

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each and every nuclei so that why we used
this you know cartesian coordinate system

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as my basis set whenever i deal with
this kind of problem particularly involving

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molecular vibration so i have to form a representation
for for this particular molecule using

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cartesian coordinate so i have e i have
c two i will specified this as c two z then

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i have inversion symmetry and then i have
sigma h all right so i have to find out what

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are the characters of my representation so
identity operation we not you know change

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any of the you know atomic positions so there
by all the this is functions also we contribute

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to the diagonal ok because all the basis functions
we also remain identical

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so therefore total twelve basis functions
will contribute twelve to the overall diagonal

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and thereby to the character now what will
be the of c two z c two z will which is

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right here right toward you so that will shift
all the atoms so shifting all the atoms from

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his place means the basis functions are also
will change their place thereby we will not

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be able to contribute anything to the diagonal
and then ultimately you get a character zero

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now if you know about the c two then i does
the pretty same thing so it will inter change

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all atoms like one and four will be interchange
their place two and three will interchange

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their places therefore the overall contribution
coming from any un shifted atom in this molecule

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by the operation i is zero so to character
this zero and sigma h what it will do sigma

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h will not disturb any of atom in the molecule
therefore all the basis functions will contribute

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to the diagonal now by what and which amount
right so this sigma h playing is nothing but

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x y plain so x and y coordinates will contribute
you know one one each so over all i have total

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four z and four y so total contribution coming
to the diagonal is plus eight while z will

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be reflected to the minus z so four z coordinates
will give me minus four contribution

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so over all contribution toward the diagonal
of the representation for sigma h is eight

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minus four equals to four so this is what
i get as my representation and we have got

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this representation based on cartesian coordinate
so i will name it as gamma three n so in the

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next step what i have to do i have to see
whether it is reducible or irreducible so

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how do i know if it is reducible so if
my sum over square of this characters sum

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over all the symmetry operations is greater
than the order of group now what is this amount

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for our presentations so it is twelve square
plus four square total is hundred and sixty

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right so this is hundred and sixty which is
very greater than the order of the group which

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is equals to four

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therefore this is a reducible representation
so in the next step i have to reduce it how

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do i reduce i know that if i want to find
out if ith irreducible representation of this

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particular point group point exist in the
representation that i have just formed then

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the number of such you know occurrence given
as a i is equals to this is for i gamma three

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n multiplied by the character of the same
operation belonging to the ithe irreducible

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representation remembered that we have ignored
the complex conduit here ok and because

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the character table that we have [consi/considering]
considering you know considering right now

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it does not contain any complex quantity
but remember that it is there in general

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so if i use this formula then i can find out
what will be the what will be the numbers

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of time any particular irreducible presentation
will occur in this reducible presentation

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so hm what i can write here is that this gamma
three n is equals to so we will now solve

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this explicitly rather what we will do we
just review the final result ok so if i write

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this it will be four a g plus two b subscript
g plus twice a u and b u will occur four times

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so if i just check if this one is correct
or not then all the dimensions will match

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up to the dimension of this particularly representation

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so all of them are one dimensional representation
and the i have total twelve as my dimension

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which is same as this so my you know
reduction of this representation is correct

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now next step what we have to do we have to
find out which of these irreducible representation

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will will give me the genuine vibrational
normal modes or in other words which one of

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these all know possible modes of degree
of freedom how many will be the vibrational

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normal mode that answer we know that will
be six so i have to find out so which six

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ideas actually form the set of genuine vibrations
so in order to do that what i have to do i

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have to eliminate all the other contributions
those are coming from translation and rotational

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motion so translation it transforms as translational
motion so that transforms as one of this [transfo/transform]

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transform translation motion transforms as
either x or y or z cartesian coordinates and

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rotational motion take transform as r x or
r y or r z fine so what i can do i can take

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this x y z or r x or r y or r z individually
as my basis function form a representation

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and then find out the which particular
irreducible representation that is

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and assigned it to a particular translational
mode or rotational mode on the other hand

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easily what i can do without taking their
trouble of formining the representation we

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can just take the help of the character table
of this particular point group c two h so

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we will now check with the character table
and concentrate on the area three right so

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area three will give me that you know
functions which transform as one of this ideas

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correct so if i look at the area three what
i get so i gave that x and y transform

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together now both x and y they transform as
the b u representation for c two h ok and

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(Refer Time: 17 :00) z transforms as a u representation
ok so x transforms as b u y also transforms

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as b u so you can immediately figure out
that x y do not transform as b u together

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but individually because this is on one dimensional
representation right so one function can

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which is select its like x is not like
you know x y or x z so x alone can give you

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only one dimensional representation so x also
transform as b u y also transform as b u and

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z transforms as a u while r x transform as
b g irreducible representation r y also transforms

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as b g so r x as well as r y both transforms
as b g representation and a g transform as

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z also i can write r z transform as
a g irreducible representation

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so these are the information that i require
correct and we have taken them from the character

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table so so this is my over all irreducible
representation and then i have to get rid

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of the translational rotational contribution
so for translational contribution so translational

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contribution comes from this x y and z so
both x and y transforms as b u so and z transforms

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as a u so i have ultimately contribution coming
from x y and z as two b u plus a u right why

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two because both x and y transform as b u
so each contribution i have to take care of

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and then i get this two and rotational i have
i have two b g similarly plus a g

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now i have to remove this contributions from
this total list so if i do that so b u and

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b g occurs twice in that list so b g is gone
and b u becomes two right so the genuine vibration

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refers to a g occurs one time less so from
four i remove one so i have three a g i have

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two b g here so b g does not occur in the
final vibrational modes and then i have two

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a u so i have one a u here so i remove one
contribution so i have one a u and then i

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have two [b a/ b u] b u here so if i subtract
two from this four b us i am left with two

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b u correct so lets cross check i have
got three a g and one a u and two b u right

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so total number of irreducible representation
that form you know to which the

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normal modes of vibration form the basis of
is this ok so we have we have got the

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irreducible representations that is that we
wanted now what we need to do we need to assign

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particular

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you know um you know normal mode to their
symmetries

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so right now do not know what are the particular
normal modes

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that exists here right so i have to do that
and second i think that this vibrational

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motions ok how can i talk about this vibrational
how can i you know find out about

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this vibrational motion experimentally so
i can find that out by using spectroscopy

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so what are the vibrational spectroscopic
techniques that will you know give me

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the transition energies or any particular
vibrational mode so one is in infra red

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spectroscopy and another is roman spectroscopy
spectroscopy so i have infra red spectroscopy

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and roman spectroscopy as my tool to find
out about the vibrational transition frequencies

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for any given particular normal mode

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now i have got number of vibrational
mode total six vibrational mode so each

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one of them will make some kind of transitions
now how will i know that which particular

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mode will be infra red active infra red spectroscopy
active and which one will be roman active

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or in other word what i
can say is that particular vibrational mode

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can be seen using infra red spectroscopy you
know and some particular vibrational modes

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can be seen

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by roman spectroscopy it
is possible that i can see one or more normal

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modes transition involving one or more
normal modes using both i r or roman spectroscopy

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or it can be you know you know either of
them

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so
we need to know about such selection rule

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then only we

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can comment on that out of this six vibrational
modes which one

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will be the roman active which one would be
infra red active and also we should be

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able to tell that ok if some mode is i r active
or some mode is you know roman active then

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what particular co realization this

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transition

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will follow so

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with those thing will come back in the
following be ok

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so

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till
thank you

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for your attention

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and

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see you tomorrow