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a bit of quantitative consideration i will
not derive it but let me explain what it means
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recall that that a polyatomic molecule has
n atoms three n coordinates and during the
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00:00:51,330 --> 00:00:57,340
oscillation or during the vibration each one
of the atoms is expected to go away from its
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00:00:57,340 --> 00:01:04,390
equilibrium position so if you think atom
one as having a coordinate x one equilibrium
5
00:01:04,390 --> 00:01:15,970
and its instantaneous atom one
x one equilibrium and at any instant of time
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it has a position an amplitude x one the difference
between the two x one minus x one equilibrium
7
00:01:25,210 --> 00:01:31,160
is the amplitude of the oscillation x one
is a function of time x one equilibrium is
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not x one is x one of t it's a function of
time therefore during the oscillation the
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00:01:37,080 --> 00:01:42,230
x one changes and as a changes the amplitude
there is a vibrational amplitude which is
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increased and it goes to zero and it increases
and so on
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so this is for the x coordinate of the first
atom and likewise for the y coordinate of
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00:01:50,590 --> 00:01:58,440
the atom y one y one e and z coordinate is
z one minus z one e now when you ask me what
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00:01:58,440 --> 00:02:02,920
these coordinates are if you have to ask me
how do we define the axes system let us assume
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that these axes system the x y z axes remains
fixed at the center of mass of the molecule
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00:02:10,160 --> 00:02:14,570
this is a classical picture please remember
that ok in quantum mechanics there is a problem
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in fixing the center of mass to a particular
point if the total momentum of the molecule
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is conserved there is a problem lets not worry
about that ok
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00:02:23,500 --> 00:02:27,390
lets worry about the fact that we have the
axes system fixed at the center of mass of
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00:02:27,390 --> 00:02:36,030
the molecule and we talk about the three coordinates
of atom one then with respect to the polyatomic
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00:02:36,030 --> 00:02:45,030
molecule if i may draw the picture so lets
assume that some three dimensional blob of
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00:02:45,030 --> 00:02:49,360
a polyatomic molecule with various atoms in
positions in bonded to each other and lets
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assume that the center of mass is somewhere
here we will draw three axes system there
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are reasons there are specific methods for
assigning these axes system not very about
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00:02:59,920 --> 00:03:02,940
it
lets assume as a first drawn picture that
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we can assign an axes system and we can find
out that how this atoms goes away with respect
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to its equilibrium position during the vibration
that vibrational vector amplitude is projected
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on to this axes system into three components
x component y component and z component and
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lets call those as a three coordinates for
atom one and likewise if we write for atom
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two similarly x two minus x two of e as the
x coordinate y two minus y two of e z two
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minus z two of e and likewise we can do this
for all the n atoms n atoms so we have three
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00:03:47,980 --> 00:03:53,960
n coordinates which are likely to change during
vibrational motion of these molecule three
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00:03:53,960 --> 00:04:04,350
n coordinates
therefore the potential energy of the molecule
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during the vibration is a function of all
these coordinates x one y one z one x two
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00:04:10,590 --> 00:04:20,840
y two z two x n y n z n remember this was
not a problem for the diatomic molecule because
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00:04:20,840 --> 00:04:24,960
in the case of diatomic molecule we consider
only the relative displacement of the atoms
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00:04:24,960 --> 00:04:30,340
with respect to each other one coordinate
now we have got three n coordinates the potential
37
00:04:30,340 --> 00:04:35,410
energy in the case of a diatomic molecule
depended on that one coordinate in the form
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00:04:35,410 --> 00:04:42,000
of k x square where x is the amplitude of
the vibration now when you have a polyatomic
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00:04:42,000 --> 00:04:45,430
molecule with three n such coordinates you
see that the potential energy is going to
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00:04:45,430 --> 00:04:51,500
be a function of all these three n coordinates
therefore it's a much more difficult function
41
00:04:51,500 --> 00:05:00,500
to write and this were the approximation of
a harmonic oscillator like approximation comes
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out very handy comes in very handy right the
normal modes as you will see in a few minutes
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00:05:08,030 --> 00:05:14,750
are nothing other than the extension of the
harmonic oscillator diatomic molecule to harmonic
44
00:05:14,750 --> 00:05:21,090
motions of all the three n minus six degrees
of freedom of a polyatomic molecules restricted
45
00:05:21,090 --> 00:05:30,910
to quadratic in the amplitudes k x square
is the quadratic in the amplitude x and if
46
00:05:30,910 --> 00:05:36,820
you do that in the case of a polyatomic molecule
as a quadratic to every such atomic amplitude
47
00:05:36,820 --> 00:05:41,139
then what you get out these the solution called
normal mode solution this is all known in
48
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classical mechanics
and in quantum mechanics again you take this
49
00:05:44,090 --> 00:05:49,199
picture directly and then you construct the
vibrational hamiltonian and solve for the
50
00:05:49,199 --> 00:05:54,460
quantum mechanical normal modes of vibration
for a polyatomic molecule so if we have to
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write the potential energy as a function of
all these coordinates and let me write them
52
00:06:02,910 --> 00:06:12,830
as simply r one vector r two r n if we write
this as n vectors you see that for amplitudes
53
00:06:12,830 --> 00:06:18,740
of vibration the potential energy is nothing
but a constant value corresponding to this
54
00:06:18,740 --> 00:06:28,669
amplitudes all be in zero what is meant by
that every difference x two minus x two e
55
00:06:28,669 --> 00:06:36,150
y two minus y two e z two minus z two e like
wise all the amplitudes are zero ok v zero
56
00:06:36,150 --> 00:06:46,060
corresponding to equilibrium concentration
equilibrium ah geometry then we have sum over
57
00:06:46,060 --> 00:07:00,330
let me not specify what they are but sum over
every coordinate dou v by dou x i x i minus
58
00:07:00,330 --> 00:07:19,759
x i e sum over i equal to one to n plus again
i equal to one to n dou v by dou y i y i minus
59
00:07:19,759 --> 00:07:33,840
y i e this is a taylor series plus some over
i equal to one to n dou v by dou z i times
60
00:07:33,840 --> 00:07:41,660
z i minus z i e and what you have is nothing
but the first derivative in the taylor series
61
00:07:41,660 --> 00:07:49,820
of a multidimensional function dou v dou v
and if you extent this to the quadratic term
62
00:07:49,820 --> 00:08:02,960
which you have is one by two sum over i not
equal to j dou square v by dou x i dou x j
63
00:08:02,960 --> 00:08:13,880
x i minus x i e x j minus x j e and you can
write five more such products corresponding
64
00:08:13,880 --> 00:08:26,520
to y i y j z i z j then x i y j x i z j y
i z j there are six such terms and this is
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00:08:26,520 --> 00:08:35,570
called the quadratic term it is quadratic
because you see this is a constant evaluated
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at equilibrium
this is evaluated at equilibrium therefore
67
00:08:47,769 --> 00:08:52,089
these are all constants this is evaluated
at equilibrium
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now if the potential is a minimum is an absolute
minimum we have a way of saying let us ignore
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the potential these equal to zero this is
again goes back to the harmonic oscillator
70
00:09:05,370 --> 00:09:12,189
model we remember the force as minus k x we
integrated the force by writing them as a
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negative derivative of the potential and then
we set the potential at x is equal to zero
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to be zero as a scale that's our minimum and
in the same way if we assume that the equilibrium
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00:09:24,070 --> 00:09:29,800
geometry has the smallest potential energy
lest assume that to be the minimum therefore
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with respect to the minimum you remember the
minimum is defined by the requirement of all
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the derivatives of the potential with respect
to the first coordinate as you have here this
76
00:09:40,310 --> 00:09:46,929
one all the derivatives will also be zero
radiant is zero at the minimum with respect
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to that coordinate therefore if we write the
potential energy using the small amplitude
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approximation as a taylor series this goes
to zero by our scale this goes to zero by
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the requirement that the potential is minimum
this goes to zero by the requirement that
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00:10:04,110 --> 00:10:10,600
the potential is minimum and also this and
therefore the very first term that is non
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00:10:10,600 --> 00:10:16,800
zero is the quadratic term ok
the quadratic term unfortunately contains
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the x coordinate of one atom and the x coordinate
of the other atom it contains the x coordinate
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00:10:22,760 --> 00:10:28,389
of one atom and z coordinate of the other
atom therefore quadratic is the mixed quadratic
84
00:10:28,389 --> 00:10:35,439
the normal approximation or the procedure
is a procedure by which we are able to rewrite
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this in the form of new coordinates which
will contain only the square of the individual
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00:10:41,259 --> 00:10:53,550
atoms when you write this v in the form of
new coordinates q one q two q three n such
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that the potential energy is to first approximation
nothing other than sum over i a constant lambda
88
00:11:04,389 --> 00:11:18,010
q i square i equal to one to three n as opposed
to what you have here what you have here is
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00:11:18,010 --> 00:11:24,179
a constant which is the function of the two
labels atom i and atom j this is a lambda
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00:11:24,179 --> 00:11:29,259
i j equivalent but now if you are able to
do an analytical procedure so that this is
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00:11:29,259 --> 00:11:36,319
changed into this new coordinate so that the
potential energy is nothing other than lambda
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00:11:36,319 --> 00:11:43,360
i q i square where now q i are defined as
a new amplitudes this procedure is something
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00:11:43,360 --> 00:11:49,029
that one can do analytically there is no problem
with that and if you cannot do analytically
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00:11:49,029 --> 00:11:54,579
for large molecules we can subject the procedure
to a computer program and we can get the results
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00:11:54,579 --> 00:11:58,679
numerically you will see all the animations
that i will show you in the next ah few minutes
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are based on a procedure of converting them
into this form and so what you have is h is
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00:12:03,430 --> 00:12:09,170
nothing but some over all the atoms of the
individual harmonic oscillator i equal to
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one to three n except that now we have three
n coordinates and therefore what we have is
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half these three n coordinates if we remove
the six degrees of freedom for a linear for
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a non linear system we get three n minus six
normal coordinates corresponding to the q
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00:12:23,860 --> 00:12:29,049
i this is a very very grass way of explaining
things but mathematically if we have to do
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it it gets some what complicated the procedures
are well known
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now let us see for ah some real molecular
cases what this will lead to for vibrational
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00:12:41,749 --> 00:12:48,939
motions of the polyatomic molecule let us
see it one by one and with a few examples
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ok so what i have here in this program i am
i am using a program known as gaussian in
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chemistry its one of the most important programs
in computational chemistry for the discovery
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00:13:07,119 --> 00:13:12,959
of this program ah professor john porter from
carning university when he did most of this
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work ah got the noble prize he shared it with
walter corn for computational chemistry walter
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00:13:19,899 --> 00:13:24,910
corn devised the procedure what is called
density functional theory for computing p
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00:13:24,910 --> 00:13:29,709
energies and the the quantum mechanics of
atoms and molecules
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00:13:29,709 --> 00:13:37,399
this is a program which provides you the animations
and calculations that are shown ah that use
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00:13:37,399 --> 00:13:42,459
to provide you the animation the calculations
are all analy they are all numerically very
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00:13:42,459 --> 00:13:49,050
accurate calculations so here what i have
is the tetrahedral molecule of ah methane
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ok again you see that methane as its does
not have a dipole moment but there are going
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00:13:54,439 --> 00:13:59,380
to be some vibrational modes in which the
atoms change unsymmetrically and therefore
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00:13:59,380 --> 00:14:05,749
dipole moments will be introduced let us see
that by way of simple definitions that this
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00:14:05,749 --> 00:14:10,850
a five atom molecule therefore there are fifteen
degrees of freedom six degrees these are non
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00:14:10,850 --> 00:14:16,439
linear molecule tetrahedral therefore six
degrees of freedom go towards rotation and
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00:14:16,439 --> 00:14:19,510
translation there are remaining nine degrees
of freedom
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00:14:19,510 --> 00:14:25,160
now molecule is highly symmetric and therefore
some of these vibrational degrees of freedom
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00:14:25,160 --> 00:14:30,829
have the same energies and same frequencies
let us ah look at this table which is not
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00:14:30,829 --> 00:14:36,040
very clear to you but in this table there
are nine vibrational degrees of freedom each
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00:14:36,040 --> 00:14:41,980
with the vibrational frequency given here
and what is here is the first three vibrational
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00:14:41,980 --> 00:14:48,100
frequency or the same the next two vibrational
frequencies of the same this is different
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00:14:48,100 --> 00:14:52,980
and then there is another vibrational mode
for which all the three frequencies are degenerate
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00:14:52,980 --> 00:14:58,610
but the vibrational motions will be different
lets take the first one thirteen seventy three
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00:14:58,610 --> 00:15:05,119
point five centimeter inverse ok will start
the vibrational motion and see what it looks
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00:15:05,119 --> 00:15:16,149
like so this is a clearly the way you look
at it this is the bending motion look at it
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00:15:16,149 --> 00:15:26,970
from angle you see that what you have here
is the bending of two atoms these two atoms
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00:15:26,970 --> 00:15:33,369
in one plane and those two atoms in the other
plane just look around ok this is clear in
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00:15:33,369 --> 00:15:40,299
this view the pair of atoms in a plain perpendicular
to the pair of atoms in the other plain undergoing
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00:15:40,299 --> 00:15:49,749
the bending motion this is one of the normal
vibrational modes for methane lets stop this
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00:15:49,749 --> 00:15:56,749
and there is another vibrational mode with
the same energy now its easy for you to remember
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00:15:56,749 --> 00:16:09,419
that there should be at least two other modes
let us look at why it should be
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00:16:09,419 --> 00:16:20,669
see what kind of bending that takes place
which modes are bent here see that these two
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00:16:20,669 --> 00:16:29,179
atoms are meant and this is the other vibrational
normal mode lets look at whats the third one
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00:16:29,179 --> 00:16:40,089
is this is the third vibrational normal mode
so the bending about various axes there are
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00:16:40,089 --> 00:16:51,139
three such equivalent bending modes in this
case what about ah the next one ok the frequency
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00:16:51,139 --> 00:16:57,171
is now fifteen ninety three centimeter inverse
and again this is a bending vibrational mode
140
00:16:57,171 --> 00:17:03,620
you see that these two bonds are meant and
there are relative variations positional angle
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00:17:03,620 --> 00:17:10,270
variations and all of this the bending motion
is generally of a lower energy than the stretching
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00:17:10,270 --> 00:17:16,059
motion the stretching means you are stretching
a given bond like a c h which is a fairly
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00:17:16,059 --> 00:17:20,980
strong in energy therefore let us look at
one of the stretching motions and see what
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00:17:20,980 --> 00:17:33,390
it is with a very high energy
you can see that this bond is being stretched
145
00:17:33,390 --> 00:17:37,850
and likewise the other bond is being stretched
and there is a certain amount of synchronization
146
00:17:37,850 --> 00:17:44,240
look at it very carefully all the four atoms
are displaced from the equilibrium by the
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00:17:44,240 --> 00:17:51,800
same during the same time by the same amount
because all four are hydrogen same mass and
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00:17:51,800 --> 00:17:59,310
the the the positions of the atom instantaneously
at any point is such that you see that there
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00:17:59,310 --> 00:18:07,510
is a dipole moment created during the motion
there is a net dipole moment which gets created
150
00:18:07,510 --> 00:18:12,960
increases and then decreases to zero and so
on therefore this is an infrared active more
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00:18:12,960 --> 00:18:21,450
that it can be detected using infrared spectroscopy
now let me look at another such bending mode
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00:18:21,450 --> 00:18:27,800
sorry another such ah the stretching mode
and what you see here is again the stretching
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00:18:27,800 --> 00:18:34,510
of the atoms in a different way there are
very specific rules for determining these
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00:18:34,510 --> 00:18:40,320
oscillations these amplitudes and these were
given elaborately by professor hertbert in
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00:18:40,320 --> 00:18:49,670
his most famous ah the monograph on molecular
spectroscopy professor hertbert ah did most
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00:18:49,670 --> 00:18:55,360
of his research on this in canada let us look
at one vibrational mode which is very interesting
157
00:18:55,360 --> 00:19:05,220
namely the symmetric stretch lets looks at
that 19 :00)all the atoms vibrate about they
158
00:19:05,220 --> 00:19:10,340
breath in and breath out with respect to the
equilibrium position and in this motion of
159
00:19:10,340 --> 00:19:15,840
course there is not a single instance where
there is a dipole moment therefore this there
160
00:19:15,840 --> 00:19:20,650
is no dipole moment during the motion and
there is no dipole moment for equilibrium
161
00:19:20,650 --> 00:19:26,860
ah structure of methane and therefore this
is not infrared active this cannot be detected
162
00:19:26,860 --> 00:19:33,360
this is for benzene this is for methane
lets look at some interesting case like the
163
00:19:33,360 --> 00:19:40,501
slightly ah more numerous atom molecule ah
lets look at the normal modes of vibration
164
00:19:40,501 --> 00:19:47,170
of benzene it's a very classic example one
of the most ah important aromatic molecules
165
00:19:47,170 --> 00:19:56,910
of course c six h six twelve atoms thirty
six degrees of freedom three n three n minus
166
00:19:56,910 --> 00:20:04,420
six benzene is is a non linear molecule planer
in its equilibrium ah geometry and what you
167
00:20:04,420 --> 00:20:11,250
have is ah thirty vibrational degrees of freedom
also benzene has a very symmetric structure
168
00:20:11,250 --> 00:20:19,060
and therefore by the rule of nature by the
law of nature benzene also has several vibrational
169
00:20:19,060 --> 00:20:25,510
modes which are degenerate the important thing
to note is that benzene does not have any
170
00:20:25,510 --> 00:20:32,470
vibrational mode which is more than w degenerate
the previous molecule that i talked about
171
00:20:32,470 --> 00:20:38,740
the ben the methane molecule has two degrees
of freedom which are triply degenerate because
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00:20:38,740 --> 00:20:43,550
of certain types of symmetry the molecular
symmetry and molecular vibrations are intimately
173
00:20:43,550 --> 00:20:47,990
connected to each other is not about this
lecture is not about it but let us look at
174
00:20:47,990 --> 00:20:52,990
the benzene vibrational modes to see a little
bit more about what normal modes are
175
00:20:52,990 --> 00:21:00,400
so let me first locate a very low energy vibration
four one four three not four point three not
176
00:21:00,400 --> 00:21:06,100
four that is four hundred and fourteen centimeter
inverse this is theoretically calculated numbers
177
00:21:06,100 --> 00:21:11,610
there are thirty vibrational degrees of freedom
lets look at the first one and start the animation
178
00:21:11,610 --> 00:21:16,880
to see how they look like nothing seems to
be happening on the screen but if i just move
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00:21:16,880 --> 00:21:21,820
the plane of the molecule to a slightly better
perspective you you can see that the atoms
180
00:21:21,820 --> 00:21:30,370
are vibrating this is nothing but the ring
the benzene ring getting the started ok see
181
00:21:30,370 --> 00:21:36,230
that two of the atoms come out of the plane
and other four atoms remain and this motion
182
00:21:36,230 --> 00:21:43,740
is such that a there is no net dipole moment
created because whatever is the dipole that
183
00:21:43,740 --> 00:21:49,930
is created by these two groups is annulled
by the other four groups there is no dipole
184
00:21:49,930 --> 00:21:54,070
moment therefore this cannot be detected by
infrared spectroscopy
185
00:21:54,070 --> 00:21:59,540
now let me look at another benzene vibrational
degrees of freedom degree of freedom and here
186
00:21:59,540 --> 00:22:07,580
is another one this is also such that the
you can see that the rings are getting distorted
187
00:22:07,580 --> 00:22:13,280
ok almost like the benzene is dancing into
the tune of a molecular motion again if you
188
00:22:13,280 --> 00:22:17,630
are look at carefully there is no dipole moment
at any poi any instant of time because the
189
00:22:17,630 --> 00:22:23,670
atoms are exactly oppositely plays as the
molecule vibrates for this atoms position
190
00:22:23,670 --> 00:22:29,480
this is compensating for this atoms position
this is compensating and for this this is
191
00:22:29,480 --> 00:22:34,650
compensating there is no net dipole moment
during the vibration therefore this is also
192
00:22:34,650 --> 00:22:42,150
infrared inactive lets look at another motion
where which has a large ben dipole moment
193
00:22:42,150 --> 00:22:51,770
the all of the hydrogen atoms come down on
the plane to keep center of mass of the molecule
194
00:22:51,770 --> 00:22:58,820
constant you see the benzene atom the carbon
atoms go up the plane look at it very carefully
195
00:22:58,820 --> 00:23:03,390
this way you can see it let me just rotate
the molecule in such a way that you can actually
196
00:23:03,390 --> 00:23:09,020
see it yes ok you can see that that with respect
to this that there is a very slight displacement
197
00:23:09,020 --> 00:23:13,880
of the oxy ah the carbon atoms are above the
plane while there is a large displacement
198
00:23:13,880 --> 00:23:20,300
of the hydrogen atoms below the plane above
and below the if you look at it that is simultaneously
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00:23:20,300 --> 00:23:24,710
a large dipole moment is created in this motion
because there is no compensation between the
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dipole moment is created and destroyed it
oscillates between zero and some value therefore
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this is infrared active
lets look at another infrared inactive motion
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here this is also a sort of a benzene ah ring
which in which some of the atoms you see it
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clearly that one this one two three the three
carbon atoms move while the other carbon atoms
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move down and as this carbon atoms moves up
the hydrogen atom moves down these normal
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mode there is a very clear mathematical procedure
for drawing these normal mode arrows and how
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they should oscillate and the fact that this
is a three n minus six or three n e is not
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magical you know there is mathematical structure
for doing that therefore vibrational motions
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at the level of very elementary what is called
harmonic approximation give you a little bit
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about the molecular motion benzene in a highly
excited state of course throws away all of
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these things this is not going to be the picture
but that's a slightly more advanced part of
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spectroscopy since we are concerned with the
basic or the elementary aspects we will restrict
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ourselves to the normal modes so let us look
at this and all of these are some sort of
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bending mode as you see it they are low frequency
modes let us take somewhere in the middle
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go to again slightly different frequency and
what you see is now the bending of groups
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one one two three four groups with respect
to one two three four you see that this slightly
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more energetic note the you can see it using
any perspective angle that you want take this
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is probably better to see how the individual
atoms are its its like nodding its head this
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way and that way
lets look at this motion this is one of the
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normal modes clearly you see the you see these
two these into compensate for any dipole changes
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these two seem to compensate for any dipole
changes and these two seem to compensate for
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any dipole change do they move they don't
seem to move there is very little of that
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but this is clear that this is unlikely to
be infrared active yes these not detected
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by infrared spectroscopy and lets look at
ah a very high energy one of the most beautiful
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oscillation of benzene the symmetric stretch
of benzene about its equilibrium position
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ok this is easier to see this is like all
the oxy the ah carbon atoms shrink while all
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the hydrogen atoms get extended and there
has to be because this is the center of mass
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and the center of mass cannot move in a genuine
vibration
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we have already removed the translational
and the rotational motion of the molecule
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therefore what you see that's why this is
called the genuine vibration there is no other
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contamination by translation and the rotation
in this kind of motion so what you see here
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is obviously molecular motion in which which
is very symmetric through out this motion
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there is no dipole moment therefore this cannot
be detected by infrared spectroscopy how do
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we do this the great sir c v raman found a
procedure for determining all the missing
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vibrational modes of infrared spectroscopy
through his procedure of spectroscopy through
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scattering and you would see that his is a
very complementary spectroscopy to the infrared
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spectroscopy that was known till then and
every mode that is not infrared active for
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a molecule which has a center of symmetry
will be raman active this is the contribution
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this is one of the most pioneering contribution
by the indian physicists ah sir c v raman
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for which he eventually even taught to win
the noble prize in physics ok
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this is the benzene symmetric mode let us
see a molecular motion in which there is a
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large dipole moment for benzene here is one
example this is a high energy stretching mode
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and you see that the dipole moments add because
when this increases this decreases which means
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the dipole moments vectors and and the vectors
and therefore this has to this has a large
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dipole moment and this is infrared active
this is a stretching mode bond stretching
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takes much more energy than bending about
the bond than tartion of the ring and so on
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therefore you see there is a pattern in describing
the molecular motion of the polyatomic motion
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tartional motion has small energies bending
modes have slightly higher energies and the
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stretching modes symmetric and anti symmetric
stretching which involve the bond stretching
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have higher vibrational frequencies this is
for benzene
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let me give the last example of a very complicated
molecule to tell you that this can go to any
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extent as a last example for today let us
look at t and t trinitre row telvin explosive
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hopefully the molecule does not explode on
the screen let us see what it does when it
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vibrates the explosion happens when the vibrations
goes on control and what we have here is control
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very small amplitude oscillations for the
molecule lets look at a few small amplitude
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motions and then examine likewise the amplitude
ah vibrations for other higher oscillations
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higher energy vibrational motion what to you
see here is a very low energy vibrational
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motion and it looks like most of the atoms
do not move but they do in the normal mode
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approximation every atoms moves from its equilibrium
position by a small amount but its inversely
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its masses therefore the telvin group this
is the c h three group two four six trinitro
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00:29:30,340 --> 00:29:36,070
telvin that's what you have here the nitrogens
are the blue atoms the oxygen are the red
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atoms the carbons are the gray atoms and the
hydrogens are the white small ones
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so what you have here is a t and t molecule
which in which you have a tartional motion
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of the methyl group relative to rest of thing
ring and to compensate for the ah the effect
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00:29:58,000 --> 00:30:05,000
of this the rest of the rings moves by a very
very small amount you can see this in various
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00:30:05,000 --> 00:30:11,950
angles to see that there is infracted very
small ah motional change in the rest of the
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00:30:11,950 --> 00:30:16,300
molecule lets look at the very large amplitude
motion in fact lets look at the high frequency
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00:30:16,300 --> 00:30:22,200
motion highest frequency lets look at what
it does hm that's not very interesting only
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00:30:22,200 --> 00:30:27,990
two atoms seem to be moving away from each
other for this motion but again as you see
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00:30:27,990 --> 00:30:33,020
that the rest of them are nearly stationery
because the amplitudes of the vibration relative
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00:30:33,020 --> 00:30:39,330
to that stretching of the two hydrogens is
going to be very small it's a stretching motion
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00:30:39,330 --> 00:30:44,430
therefore it has a high frequency motion and
here is the stretching of the methyl group
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00:30:44,430 --> 00:30:52,430
here is a stretching again of the methyl group
symmetrically here this is now the tarsh the
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let us look at a other is a stretching all
over the places this is nitro group for example
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00:30:57,620 --> 00:31:03,310
is stretched this carbon carbon group is stretched
see that this is a very complicated motion
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00:31:03,310 --> 00:31:11,441
how many degrees of freedom does this molecule
have i leave it you as an example and will
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00:31:11,441 --> 00:31:17,720
tell you that the answer is fifty seven find
out why ok here is an another stretching motion
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00:31:17,720 --> 00:31:26,940
lets go up and look at that yes here is ah
the what you see here is again the stretching
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of the ah n o group and so on now the current
day computational chemistry programs can pictorially
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00:31:40,150 --> 00:31:46,480
obtain the normal modes of vibrational of
polyatomic molecules including several tens
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00:31:46,480 --> 00:31:53,150
and twenty thirty s of atoms and even more
with a little bit of approximate theories
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00:31:53,150 --> 00:31:59,050
we can obtain the normal modes of vibrations
of many many atom polyatomic molecules these
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are collective motions i will not be in a
i not had the time to describe the local motion
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but we will continue this in the next lecture
with a description of the local motion and
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vibrational frequencies and so on how we do
this in the laboratory that let us summarize
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00:32:13,570 --> 00:32:19,540
todays lecture by saying that the normal modes
of vibrations of polyatomic molecules are
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the exact analog of the normal vibrational
or the harmonic vibrations of a diatomic molecule
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00:32:27,840 --> 00:32:34,050
and what we did today is a qualitative description
of the vibrational motion we tell little bit
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of about the origin of how they come come
come into effect there are three n minus six
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or three n minus degrees of freedom depending
on the molecular type and its possible for
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us to view under small amplitude motion all
these vibrational motions very very accurately
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we will continue this in the next lecture
to study the local modes of vibration until
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then
thank you very much