WEBVTT

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hello everyone today is the say five of the
sixth week of this lecture series so let us

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quickly recap whatever we learnt in last class
so we took an example of naphthalene to illustrate

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all the knowledge about salc formation using
the projection operator and ultimately you

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know utilizing those salcs to form the you
know energy state of the molecule so essentially

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the molecular orbital of naphthalene molecule
so what we did let us go through that quickly

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before we move on to our next lesson

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so first we figured out point group of the
naphthalene which is d two eight and then

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we took the ten (Ref Time: 01:00) p by orbital
sitting on ten carbon atoms so using those

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ten p pie orbital we formed an irreducible
representation which is gamma ok so we also

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said that we can actually you know separate
this p pie orbital into three different sets

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because this you know different carbon atoms
forms different sets of equivalent carbon

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atoms

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so we therefore met set one set two set three
comprising four and two p pie orbital and

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individually we form the representations of
using the you know basic set one two and three

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and ultimately we reduce them and we got the
irreducible representation to which this

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we found the irreducible representation that
occured in the irreducible representations

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that is found out of one two and three and
we also you know said that either way that

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is you you know form the separate three
representation and then ultimately reduce

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it to the irreducible representation or you
know take all the you know p pie orbitals

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form at you know representation which is ten
dimensional and then reduce it the result

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will be ultimately same

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so after you do that you use the projection
operator for each individual irs that are

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you know obtained and you find the salcs so
we saw that there are certain number of

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irreducible representation occurring for example
like you know au irreducible representations

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occurs two times b one u occurs three times
b three g occurs three times while b two g

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occurs two times so that means that i will
have total ten molecular orbital which are

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you know i mean given by the salcs that we
have formed that these ten salcs will have

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their own energies

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so in the next step we are supposed to find
out these energies so how did we do that we

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form the you know the secular equations in
terms of the secular determines and here in

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this case you know because we form this
you know individual says and from there we

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found that essentially four individual irreducible
representations are there and correspondingly

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we had like two or three number of salcs which
transform according to the particular irreducible

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representation that we ultimately found so
we get you know this secular determinant to

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be a you know block factored determinant and
(Ref Time: 04:00) we took out each block and

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then form four individual secular equations
in terms of four individual you know this

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blocks and solving that utilizing huckels
approximation which says that you know interaction

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energy between two orbitals are zero until
or less these two orbital are adjacent to

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each other ok and that you know energy of
the interaction is expressed in terms of beta

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and beta is the unit of energy here and the
interaction with itself that is pie i h pie

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i this quantity is written as alpha and alpha
is taken to be the zero of energy according

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to huckel approximation and the interaction
the overlap integral that is you know five

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and pie two (Ref Time: 05:00) integration
over all the space is taken to be zero under

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huckel approximation

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so utilizing that we could find out all the
energies of the you know ten different salcs

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that we can have so ultimately we found you
know energies of the possible mos to be like

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this so two salcs which you know has the
same symmetry au has two different energies

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alright similarly for b one u and b two g
and b three g we found you know number of

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energy levels now our job was to adding this
energy levels so you know beta is negative

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so therefore the ordering of energies is done
in such a way that is shown on your screen

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so these are (Ref Time: 06:00) expressed in
terms of beta and beta itself is negative

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and zero as it mentioned corresponds to the
energies of the atomic orbital which are not

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interacting that means before the formation
of the molecular bonds whatever the energies

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the atoms have they are taken to be the zero
of energies and whenever there is an interaction

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favourable interaction we have you know lowering
of energy which is given here so these are

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expressed in terms of beta so beta is negative
so overall this energies are all negative

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while this terms here that is minus point
six one eight that is also expressed in terms

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of beta which is negative so overall this
is a positive energy correct so these are

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higher energy in the next step we are supposed
to fill out this energy levels with the electrons

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so (Ref Time: 07:00) there we utilize the
you know hunds rule and the exclusion principle

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so what does that say that you know in case
of the order of filling of molecular orbitals

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for the ground state of the molecule it it
follows the same rule as does the filling

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of atomic orbitals in the ground state of
an atom

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so the an electron will go into the lowest
and filled level subject to the certain restriction

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what are those restriction the only two electrons
may occupy a single level and their spins

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must be opposite that is the exclusion principle
right pauls exclusion principle when electron

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are to be placed in a pair of degenerate orbitals
suppose you have two orbitals having same

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energy we call them as degenerate orbitals
so if you have to fill them so you have to

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fill both of them (Ref Time: 08:00)having
the same spin so that overall the total spin

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becomes one so which is the hunds rule right
so using this exclusion principle and hunds

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rule we fill out this energy level

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now after getting this this overall molecular
orbital picture and their energies next we

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will talk about the you know this interaction
between different energy levels meaning the

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you know transition from one electronic level
to another electronic level and the corresponding

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selection rules so we will stop the discussing
right here about this you know molecular orbitals

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and the corresponding electronic levels and
the transitions involves between the electronic

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levels we will come back to this after a while
so in the meantime what we will try to

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look at is (Ref Time: 09:00) to look at certain
internal motions of the molecule the bonds

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and the specifically we will deal with the
molecular vibrations the normal modes and

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we will try to see how symmetry properties
of the molecule can be applied to molecular

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vibration to you know find out exact normal
modes their symmetries and their you know

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you know many vibrational transitions or
raman transitions and their probabilities

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and the corresponding selection of rules

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so this is what we are going to you know study
for another five or six lectures so to start

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with if i take any given molecule they will
have three different ways (Ref Time: 10:00)

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to store the internal energy so one is translation
you know rotation vibration so if if i

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have a molecule having certain you know atoms
and then certain bonds there so this bonds

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can move in any particular direction or by
changing an angle with respect to another

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bond in such a way that there this motion
will not displays the molecule in a

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particular direction that means it will not
lead to any translational motion that is there

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will be no movement of the centre of mass
of the molecule and also this type of motion

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that i am talking about that you know may
not change the net angular momentum of this

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molecule meaning that there will be no rotation
and such motions of the you know bonds in

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a molecule is known as the vibration ok

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so if if if i if i take you know any
particular molecule and consider the internal

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motions of this bonds or atoms we can see
that you know there are bonds are moving in

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different directions or you know changing
the angles like for examples if i take just

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water molecule so i can see there is you know
edge bonds can move in different way or you

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know it can changes angles (Ref Time: 12:00)
apparently it may seem that all this motions

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are very much random but if it take a close
look at that observe it you will see that

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nothing is random but they are well you know
this motion is very much regular the moment

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i say that motion is regular that means that
there must be some symmetry into it right

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because in the very first class if you remember
we said like you know what are the meaning

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of this word symmetry one of the thing is
there must be some regularity in whatever

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the phenomena or you know process or in structure

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so sometime it may be that you know thats
you know regular you know regularity in this

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motion can be bit time taking so that you
know apparently it may seem like that there

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is no regularity but actually there it is
and you know how do we classify this you know

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particular vibration because there may be
several bond several bonds are moving you

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know executing vibration and all of this can
be superimposed is actually superimposed and

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then we get an overall you know internal motion
of this bonds in a molecule and these are

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actually quantized by the so called normal
modes ok

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so many of us are familiar with terms like
you know for a given molecule that there is

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a stretching or there is a bending there is
you know symmetry stretching or asymmetric

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stretching or you know in plane bending out
of plane bending so these are actually the

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individual normal mode and we will be dealing
with this normal mode to find out about the

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molecular vibrations and their you know associated
properties of this molecular vibrations now

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first and for most thing is that how do we
find out what is the number of normal mode

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in a given molecule and probably many of you
already know that because many times you get

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a you know simple formula and then you just
use that formula and get the number of normal

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mode that is possible but many occasion you
do that without actually knowing the reason

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behind this formula

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so if you if you take any given molecule say
for example i will i will take a very simple

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example say i have ammonia ok so
so you know each of this bond can you know

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execute certain motion ok such that this doesnt
you know lead to any translation or notation

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now what i can think of that for each and
every atom will have their coordinates correct

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so x y and z this is true for each and every
atom on the molecule now each coordinate is

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independent of the other coordinates

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so therefore this n can move in the x direction
by certain amount without worrying about what

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is going to happen in y and z direction and
h also can do the same thing and it can do

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it in any direction it can do it in the z
direction or you know y direction or in x

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direction which are all independent now suppose
a situation when all the atoms starts moving

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in x direction by definite amount right so
what happens in that case you are moving the

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centre of mass in x direction by that definite
amount meaning the whole molecule is translating

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in x direction by a definite amount

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so i can do the same thing in y direction
as well as in z direction so each of this

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directional motion will me one translational
degree of freedom so in x and y and z direction

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i get translational
motion right so this three directional three

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individually you know individual direction
of motion will give you total three translational

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degrees of freedom now if i think about motion
of this you know you know atoms in a angular

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path about either x or y or z axis now if
all the atoms start you know executing this

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angular motion about one particular axis say
x axis then that will amount to the rotation

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of a whole molecule by a definite amount right

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so it can be about x axis separately it can
be about y axis and z axis so this three different

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you know angular motion about three different
axis or three rotations so i have three (Refer

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Time:18:00) rotational degrees of freedoms
now overall how many i have so i say that

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the each and every atom can execute certain
motion in either x or y or z direction ok

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or altogether it can do and all three motions
will be independent of each other therefore

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if i have n number of atoms so total number
of motion directional motions that can be

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executed by this whole molecule will be three
n right three n is a total number of motion

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possible that is total number of degrees of
freedom that it can execute

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so therefore i have three translational and
three rotational so the rest should be the

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vibration so i have total three n minus six
number of degrees of freedom left for the

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molecule to execute vibrational motion so
thats how we get this number three n minus

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six now this is in general for any molecule
but when we consider a linear molecule so

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for example suppose i have a case of carbon
di oxide this is a linear molecule so here

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each and every molecule can execute translational
in x or y in z direction so the molecule can

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have three translational motion like this
when i talk about the rotation then it can

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have you know about this axis it can have
a perpendicular axis to this but there is

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no meaning of having a rotation about this
axis because all the nuclei are contained

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by this axis ok so i have essentially two
rotational motions so instead of three here

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i have two for linear molecule

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so therefore for linear molecule we have three
n minus five ok for linear alright so let

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us take some you know definite example so
this suppose i have a non linear molecule

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and i have three n minus six number of you
know vibrational degrees of freedom so each

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of this vibrational degrees of freedom is
expressed by the you know normal vibration

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or you know normal modes so we will take an
example of a planar molecule that is carbonate

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ion so we will be talking about co three two
minus ok so if i represent this you know carbonate

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oxygen by this circle then
i can take this to demine carbonate ion ok

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so the central part is the carbon and then
three oxygen are the end

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so i will have total how many there are
four atom so four into two twelve minus six

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so i will have six number of degrees of freedom
let us see how this normal mode actually look

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like so i will first have six structures of
carbonate ion (Refer Time:22:00) and then

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we will look at the normal modes ok so take
that all the bonds are actually identical

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ok so so i have got total six number of
structure to show six conditional normal modes

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ok so say first mode is i will use the different
colour to identify this so

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is this colour visible ok so this arrows one
two three arrows they say signify the direction

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of the movement of the atom right so also
the you know the link that we you know provide

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to this arrows they signifying the amount
of the movement though it is quite a bit of

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exaggerated ok now let us draw all the other
modes then it will be lot more clear

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so here ok so these are given as sign so
plus and minus i will explain what the sign

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means (Refer Time:24:00) ok and

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so we need to understand the exact meanings
of this arrows their links you know from where

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they are originating and all this things you
need to understand this motions properly and

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then the last one
ok and we will also give this motion because

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these are some vibrations so they will have
their associated frequencies so we will have

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nu one and we will call write it here this
si nu one this is nu two this is nu three

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and we will call this nu three a there is
a reason for doing that we will talk about

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that and this is as nu three b and then we
have nu four a and nu four b alright so now

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here we have mostly this motion showed by
arrow and in one case it is shown by the plus

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and minus sign so now you can see that in
manys many of this normal mode picture the

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arrows are having different link

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so what does that mean suppose here you have
longer arrow compared to the arrow which is

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sitting on the centre carbon this meaning
that this atom over here this oxygen is moving

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in this direction without worrying about what
happening to the rest of the molecules while

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this carbon is also moving in this direction
but the amount of movement in that particular

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direction by this atom or this atom are not
equal and the length you know the length the

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ratio of the length is actually the ratio
of the amount of movement that two individual

26:51.830 --> 26:59.630
atoms are making in a particular direction
ok now as i said this links are quite exaggeration

26:59.630 --> 27:05.330
because this this atoms not really move this
much which are comparable of to the bond length

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ok this is just to make things be prominent
to you and here this plus and minus sign this

27:12.910 --> 27:22.040
mean that you know this particular atom is
coming up above the plane of the board while

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this oxygen is coming above the board and
this carbon atom is having the minus sign

27:30.150 --> 27:37.270
means that it is moving down the board so
this is an out of plane motion while all these

27:37.270 --> 27:43.540
things all these other structures that nu
barring nu two all the others motions are

27:43.540 --> 27:49.070
in plane that is this molecule is a planet
and all the you know motions of this atoms

27:49.070 --> 27:50.300
are within the plane

27:50.300 --> 27:58.180
so these are in plane vibrational motion while
this is out of plane alright now one thing

27:58.180 --> 28:05.320
you should notice that you know many cases
we have you know misconception about how

28:05.320 --> 28:13.170
to draw this normal mode so here if you look
at carefully the arrows starts from an atom

28:13.170 --> 28:19.550
ok and goes to a another particular direction
so this particular thing means that this atom

28:19.550 --> 28:28.540
is moving in many cases you know you will
find that people write the motion in a way

28:28.540 --> 28:37.380
like they they they write like this ok or
this this this is completely a wrong way of

28:37.380 --> 28:42.620
representation so when you want to present
the normal mode you exactly have to do it

28:42.620 --> 28:47.620
in this way that is start on that particular
atom and then put the arrow in the direction

28:47.620 --> 28:55.250
into which it is executing the motion alright
and you have to take care of the relative

28:55.250 --> 29:01.190
length of the arrows you know within the particular
normal mode right correct say for example

29:01.190 --> 29:06.240
if you look at this one or this one this is
very clear or this one right

29:06.240 --> 29:10.590
so that will tell you that by which amount
one particular atom is moving in a particular

29:10.590 --> 29:21.810
direction relative to other atom within the
molecule so now now here looking at this i

29:21.810 --> 29:28.240
can have two important properties of these
normal modes ok so these are the normal modes

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these are possible now right now i am just
drawing this on the board but you know how

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can i get this one suppose i draw some arbitrary
you know motions with arbitrary arrow direction

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will i able to tell that this is the right
normal mode or not so that must have certain

29:50.720 --> 29:59.260
rules ok and this you know rules are governed
by the symmetry and group theory and you know

29:59.260 --> 30:07.070
in the following lessons we will learn how
to actually use group theory to find out exact

30:07.070 --> 30:12.890
normal modes right now before ending to this
lecture i will mention two important properties

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of these normal modes one is that you know
each of the vectors so all these arrows that

30:18.760 --> 30:23.600
we are you know using i said that they have
a definite you know length and they have of

30:23.600 --> 30:28.120
course the direction so they are the vectors
ok vectors of motion

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so each of this vector representing this instantaneous
movement of this atoms and they you know giving

30:35.690 --> 30:41.760
the displacement they may be regarded as the
resultant of three basis vectors so we can

30:41.760 --> 30:48.500
consider each of this atoms having their own
cartesian coordinate x one y one and z one

30:48.500 --> 30:54.730
and then we can think of this movement in
a particular direction by a particular amount

30:54.730 --> 31:01.390
to be the resultant of this three basis vectors
right and second thing is that which is extremely

31:01.390 --> 31:08.820
important to us that each of this normal mode
they form a basis for an irreducible representation

31:08.820 --> 31:16.880
of the molecule or point group they may the
basis by themselves or they may actually

31:16.880 --> 31:24.640
belong to one of this ir of the molecules
so therefore the structure that i drew they

31:24.640 --> 31:27.890
will be assigned to particular irreducible
representation

31:27.890 --> 31:37.650
so just for your information if i put so this
one will form a representation of a one this

31:37.650 --> 31:49.340
one will form the representation of a two
while the nu three a will form the representation

31:49.340 --> 32:02.580
e prime three b will form the representation
of again e prime while nu four will also form

32:02.580 --> 32:11.770
the representation for e prime so we will
do the normal mode nu prime four b so these

32:11.770 --> 32:16.620
are the two main points about the normal modes
that is their motions can be expressed by

32:16.620 --> 32:22.860
the resultant of three basis vector and second
the each of this normal mode can you know

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form the basis you not only can but they form
the basis for the irreducible representation

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of the molecule molecular point group

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so we will utilize this ideas and then we
built on our you know knowledge about this

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molecular vibration their relation with the
symmetry and ultimately having the applications

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ok so we will commit those things in the following
week and till then have a good day

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thank you very much for your attention