WEBVTT

00:13.780 --> 00:20.539
hello welcome back so where we were looking
at the selection rules of fundamental transitions

00:20.539 --> 00:34.739
right so we will start from there today again
so

00:34.739 --> 00:46.559
for fundamental transition so we explained
in the last class what is a fundamental

00:46.559 --> 01:01.989
transition will again recap that so if
i have a n number of normal modes for a

01:01.989 --> 01:08.930
given particular molecule and the out of
you know initially the ground state means

01:08.930 --> 01:16.430
that all of the normal modes will be staying
at vibrational level zero ok so which is return

01:16.430 --> 01:38.549
as say the product of all the wave functions
ok now if i say this is my quantum number

01:38.549 --> 01:46.759
vibrational quantum number then if n equals
to zero then if all of them all of wave functions

01:46.759 --> 01:53.250
are at zero vibrational level then this is
my molecular ground state and if one of the

01:53.250 --> 02:03.270
normal modes you know is excited to the you
know level one then i have a situation

02:03.270 --> 02:12.000
where i have suppose this is the jth normal
mode so kth normal mode is was stain or residing

02:12.000 --> 02:20.800
at vibrational level one and rest of the
normal modes which is i that number will

02:20.800 --> 02:33.400
be now i minus one we still stay in the
level zero so this transition is a fundamental

02:33.400 --> 02:49.710
transition ok
so so we we we want to study fundamental transition

02:49.710 --> 02:54.230
because that is the most intense one compared
to any other possible transition that we mentioned

02:54.230 --> 03:03.790
like an overtone or combination banks combination
transition now we we got ultimately

03:03.790 --> 03:11.920
the rules in terms of the transition moment
integral so when we were looking at the infrared

03:11.920 --> 03:25.980
spectroscopy transition we got you know the
transition moment integral as one as integration

03:25.980 --> 03:33.629
psi v j we use this particular terminology
ok instead of using this terminology we use

03:33.629 --> 03:41.480
this one ok v is vibration ok and j is the
particular normal mode jth normal mode which

03:41.480 --> 03:58.849
is being excited and you have either x
or y or z so this was one and second was a

03:58.849 --> 04:20.900
singular integral with y as operator and
third one was integration for z and we say

04:20.900 --> 04:33.639
that since this one this psi v sorry this
psi v i is my ground state and we already

04:33.639 --> 04:42.720
said in the last class that the you know
ground state of you know any in normal

04:42.720 --> 04:47.820
mode in the ground state will always form
the basis for totally symmetry irreducible

04:47.820 --> 04:53.009
representations so this one i dont have to
care about so i have to only care about this

04:53.009 --> 05:02.840
part ok so the normal mode that is being excited
at the operator ok so if i know the you know

05:02.840 --> 05:10.259
irreducible representation for which this
x forms the basis of then all i need to know

05:10.259 --> 05:17.319
what is the irreducible representation for
which this normal mode that is being excited

05:17.319 --> 05:23.830
from the bases off and as per the rule of
transition moment integral whether that will

05:23.830 --> 05:33.060
be you know zero or non zero will depend
on whether this overall integral is you

05:33.060 --> 05:39.870
know forms the basis of totally is symmetric
irreducible representation or contain the

05:39.870 --> 05:46.530
total symmetric irreducible representations
so when we represent this particular product

05:46.530 --> 05:54.009
by a direct product of the respective irreducible
representations so that will mean that i need

05:54.009 --> 06:00.860
to know the particular irreducible representations
of this normal mode ok so the moment this

06:00.860 --> 06:08.130
and this belong to the same irreducible representations
i have a nonzero value for this integral so

06:08.130 --> 06:16.050
this three all together will give me that
selection rule for a fundamental transition

06:16.050 --> 06:26.569
in infrared spectroscopy so a normal mode
will be inferred active if this belongs to

06:26.569 --> 06:34.259
the irreducible representations for which
either one or more of the cartesian coordinates

06:34.259 --> 06:46.190
that is x y z said ok from the bases of
so this is for ir activity ok so the particular

06:46.190 --> 06:52.580
normal modes which we lead two nonzero value
we will say they are ir active normal modes

06:52.580 --> 07:07.330
ok on the other side for raman spectroscopy
we got an integral which is psi v j p per

07:07.330 --> 07:19.180
p is that polarizability things are component
right so just like here we can look at the

07:19.180 --> 07:23.410
you know symmetry properties of the polarization
[comp/component] polarizability component

07:23.410 --> 07:35.250
which are like this binary products x y
or x z or y z or the square like x square

07:35.250 --> 07:42.660
or y square z square or any of the combinations
like you know x square minus y square and

07:42.660 --> 07:51.520
so on so this kind of functions we form
the polarizability tensor they they represent

07:51.520 --> 07:58.930
the polarizeability tensor component
so i need to know the symmetry of those

07:58.930 --> 08:08.849
x y or y z x square or y square so on and
the normal mode that is being excited ok that

08:08.849 --> 08:15.599
should from the basis of the same irreducible
representation for which the particular polarizability

08:15.599 --> 08:23.729
component from the basis of so suppose i have
a component x y suppose which forms the basis

08:23.729 --> 08:34.479
for say au then my selection rules says that
this transition will be allowed only when

08:34.479 --> 08:46.470
psi v j also from the basis for au right because
the direct product of au two au s will give

08:46.470 --> 08:53.360
me the totally symmetric irreducible representation
and then my integral will be will survived

08:53.360 --> 09:00.700
so here what is the selection rule for raman
spectroscopy sectional for fundamental

09:00.700 --> 09:08.150
transition that will be raman active that
is that a particular normal mode of vibration

09:08.150 --> 09:16.650
will be raman active when the normal mode
belongs to the same irreducible representation

09:16.650 --> 09:25.030
to which this binary combinations that
is xy xz etcetera which form the basis form

09:25.030 --> 09:30.530
alright
so after getting this thing we are in a position

09:30.530 --> 09:41.590
to talk about the you know that which
particular modes which particular normal

09:41.590 --> 09:53.690
modes that we deduced say for co three two
minus or n two f two so which one of

09:53.690 --> 10:02.200
them will be ir active which one of them will
be infrared active right so say for co three

10:02.200 --> 10:11.220
two minus we got the you know normal modes
which are genuine with represent genuine vibration

10:11.220 --> 10:27.220
they were a one prime a two prime i think
a two prime a two double prime and two

10:27.220 --> 10:41.070
e primes now if you consult the cut the
character table you will see that among these

10:41.070 --> 10:52.870
this a one prime that you know that
has this cartesian coordinate component ok

10:52.870 --> 11:05.330
as its basis ok so this one will be infrared
active while this one has this binary function

11:05.330 --> 11:16.440
as its basis so this one will be raman active
right and this e prime it has both the no

11:16.440 --> 11:19.990
only cartesian coordinate component coordinate
at basis as well as the combination of cartesian

11:19.990 --> 11:27.860
in binary combination of this curtesian coordinate
so this will be ir active as well as raman

11:27.860 --> 11:48.810
active alright ok so we also had this
n two f two molecule for which we had this

11:48.810 --> 12:06.000
vibrational modes like three ag plus au plus
two bu right now if i look at the character

12:06.000 --> 12:22.370
table of c two eight
so what i see that ag has a this only the

12:22.370 --> 12:33.120
binary functions like x square or y square
z square xy ok so this binary functions

12:33.120 --> 12:46.450
tells me that ag will be a raman active mode
while this au it has z as its basis and bu

12:46.450 --> 13:03.430
has x or y as its basis so x y and z get that
transformed as either au or bu so this au

13:03.430 --> 13:12.770
and bu you know the most which transform
as au and bu will be infrared active correct

13:12.770 --> 13:25.870
so
so this two will be infrared active infrared

13:25.870 --> 13:40.330
active normal mode why this three ag will
constitute raman active normal mode now one

13:40.330 --> 13:49.520
thing you notice here in this particular molecule
co three two minus i have one more which is

13:49.520 --> 13:57.190
ir active but raman inactive another mode
i have raman active but ir inactive and we

13:57.190 --> 14:04.710
have certain modes which are both ir active
as well as raman active ok so when you look

14:04.710 --> 14:15.410
at n two f two i see that none of the modes
are raman as well as ir active so those modes

14:15.410 --> 14:20.580
which are ir active are not raman active and
vice versa so there is a difference between

14:20.580 --> 14:25.950
this two we will talk about this one will
take another example another particular

14:25.950 --> 14:32.760
molecule for example like ammonia and look
at the practical example and see what happens

14:32.760 --> 14:38.540
there ok then we will come to some you know
conclusion about you know this are the reason

14:38.540 --> 14:49.220
behind this kind of observation ok
so now after we we could separate out the

14:49.220 --> 14:59.610
raman active our infrared active modes using
this symmetry now we will go back to the

14:59.610 --> 15:09.980
original problem of this n two f two molecule
and its normal modes ok so we got this you

15:09.980 --> 15:18.580
know this six normal modes and there is symmetry
it now we do not have any idea what are this

15:18.580 --> 15:25.930
normal mode so how do they look like ok
so how do we do that so here at this point

15:25.930 --> 15:31.010
we will take help from so called internal
coordinates ok so what are internal coordinates

15:31.010 --> 15:35.930
we mentioned it earlier let me tell you again
so this internal coordinate will be either

15:35.930 --> 15:50.170
the bond distance or a bond angle ok
so in case of n two f two i have ok this is

15:50.170 --> 15:56.440
the structure so how many bond distances i
can take i can have n f to n f bond distance

15:56.440 --> 16:09.640
and one n f bond distance ok so i have two
n f i have one n n then i have this angle

16:09.640 --> 16:23.110
as well as this angle so i have both of them
are same so i have n n f angle right

16:23.110 --> 16:32.500
so i you know there is no hard and fast rule
that ok which one i will use as my coordinate

16:32.500 --> 16:37.430
and which one i will use first which one will
i was later so it depends on what you want

16:37.430 --> 16:42.630
to do now since we are dealing with the vibration
which means either the bond distance will

16:42.630 --> 16:49.270
alter or the bond angle you know temperature
doing altar in plain or out of plain ok

16:49.270 --> 16:57.450
so lets start with this now before even starting
starting with this internal coordinate what

16:57.450 --> 17:04.520
we will do we will have a look at the character
table once again so if i look at the character

17:04.520 --> 17:19.120
table here for sit wait what we have so
if we look at the we have three different

17:19.120 --> 17:26.040
types of irreducible representation i know
according to which the normal modes get transformed

17:26.040 --> 17:37.760
so out of this three ag au and bu i have this
au particularly this one this has a character

17:37.760 --> 17:52.850
of minus one for so sigma h equals to minus
one ok why for ag and bu the character for

17:52.850 --> 18:02.180
sigma h is plus one so what does that mean
that in this au mode the you know bond displacements

18:02.180 --> 18:12.180
or when a bond angle change such that the
structure is you know not symmetric with respect

18:12.180 --> 18:20.100
to the sigma is plane so whether it is antisymmetric
so what does that mean that there is some

18:20.100 --> 18:27.140
you know part of the structures you know
you know about the plane or below the plane

18:27.140 --> 18:35.590
ok so that means the particular normal mode
which transforms according to au has some

18:35.590 --> 18:41.850
motion which goes out of this plane now we
started with a planar molecule right and now

18:41.850 --> 18:50.990
i have a situation where i see that sigma
eight produces you know i minus one as the

18:50.990 --> 18:59.000
character that means that this bonds when
one of this one is upward one is downward

18:59.000 --> 19:05.260
meaning that this motion of this you know
a bonds they are not in plane rather they

19:05.260 --> 19:11.810
are out of plane so they are doing some kind
of out of plane motion right so i can safely

19:11.810 --> 19:21.240
to start with i can describe this one as out
of plane bend why bending because this bond

19:21.240 --> 19:27.740
stretching if it goes on in the same direction
it will still remain planner but in this particular

19:27.740 --> 19:32.590
case of au type of irreducible representations
i can clearly see that will be out of the

19:32.590 --> 19:39.260
plane so therefore the angle say if i say
this you know this angle here which was in

19:39.260 --> 19:47.270
the plane so after this you know movement
of this f out of the plane that is a creation

19:47.270 --> 19:55.930
of an angle here ok
so without worrying about anything i can see

19:55.930 --> 20:09.920
that this is a bending motion ok because from
planarity i have au i say this is out of plane

20:09.920 --> 20:25.630
bending so it is like you know this one is
plus and this one is minus ok so in that way

20:25.630 --> 20:33.750
or it can be in this direction
now we are left with five other so ag three

20:33.750 --> 20:40.880
ag and two bu and if we use this as my internal
coordinate then i find because i am using

20:40.880 --> 20:46.410
five internal coordinates and i am going to
find out about the properties of the normal

20:46.410 --> 20:57.060
modes of five five six normal mode belonging
to three ag and two bu ir well so lets do

20:57.060 --> 21:14.840
that now here we use this to enough bonds
together right and the so lets look at the

21:14.840 --> 21:26.410
representation so i have this and then
i can have three different presentation based

21:26.410 --> 21:39.040
on n f n n our lets say n n f which is an
angle and then we have the representation

21:39.040 --> 21:51.430
based on n double bond n distance alright
so he will not do anything so where two basis

21:51.430 --> 22:05.340
set its basis functions will give me a character
of two so if i do c two hear this e f together

22:05.340 --> 22:13.200
they are getting you know interchange so i
get nothing here same is true for i and sigma

22:13.200 --> 22:21.190
x i get two now if i look at the angle it
has the same fact took it is that the two

22:21.190 --> 22:30.550
angles that i am getting so now when i have
this n n bond distance then identity of course

22:30.550 --> 22:38.120
it gives one and c two and i sigma eight all
of them will contribute one to the diagonal

22:38.120 --> 22:44.460
so i get this is ready presentation so this
one i dont have to do anything i immediately

22:44.460 --> 22:50.480
see this is all symmetric with respect to
all the symmetric operations there for the

22:50.480 --> 22:55.320
representation that i am having this belongs
to totally symmetric irreducible representations

22:55.320 --> 23:07.330
so this was was very easy alright now i have
to reduce this two ok and what about i you

23:07.330 --> 23:13.570
know that is a like get for gamma n f will
be same as gamma n n f correct because there

23:13.570 --> 23:21.520
you know sorry this one is two they are identical
and if you reduce this one then what you will

23:21.520 --> 23:35.230
see that this one is equal to ag plus bu and
obviously this one will also be ag plus bu

23:35.230 --> 23:44.860
alright ok
so i got back my original result that we got

23:44.860 --> 23:51.120
earlier using that three in cartesian coordinate
system that's good that says that i have chosen

23:51.120 --> 23:56.510
the right you know internal coordinates know
why did i do all these things i have already

23:56.510 --> 24:00.740
got this one so what was the need for getting
this one this is because i need to have some

24:00.740 --> 24:08.370
idea about that particular motion of the normal
modes right so now i get this ag and bu based

24:08.370 --> 24:19.340
on this nf bond that means the normal mode
you know which transforms as ag and bu they

24:19.340 --> 24:29.640
we have some motion of this bond distance
ok nf bond distance now if you look at

24:29.640 --> 24:41.610
the characters of ag and bu let's have a look
at the respective characters

24:41.610 --> 25:01.840
so ag is totally symmetric correct so then
i can safely say that ag is my symmetric stretching

25:01.840 --> 25:15.350
mode why because it is concerned with the
nf bond distance so nf bond will increase

25:15.350 --> 25:22.840
in both the direction in equal manner if it
has to contract both of them will contact

25:22.840 --> 25:32.470
the same manner and then only it will belong
to ag representation ok so this particular

25:32.470 --> 25:38.370
mode is obviously the symmetric stretching
mode so this is symmetry stretching and if

25:38.370 --> 25:49.640
of course belongs to ag alright while if i
look at the beauty presentation then i see

25:49.640 --> 25:57.620
that with respect to
i always expect to see to you know this is

25:57.620 --> 26:05.660
a symmetric so what does that mean here since
this is best on nf bond distance again it

26:05.660 --> 26:12.640
has to do something with the change in the
bond distance and it says that if i operate

26:12.640 --> 26:17.610
you know c two c two what it will do it will
take this guy over here and this guy will

26:17.610 --> 26:27.400
here so this nf bonds they will get interchange
and that will give me minus one as my you

26:27.400 --> 26:36.070
know character for the overall molecules
factor that means that change in the bond

26:36.070 --> 26:45.081
distance due to the normal mode of vibration
which transforms as this bu will change to

26:45.081 --> 26:59.730
nf bonds in different ways meaning it will
be something like this

26:59.730 --> 27:19.190
so one will you know increase while other
one will decrease so it is like i have you

27:19.190 --> 27:27.450
know say n two so like this ok so this
is my n n bond and this is f two my this bond

27:27.450 --> 27:32.420
is increasing while this is shrinking and
then when this one is shrinking and this one

27:32.420 --> 27:45.320
is increasing so this is asymmetric stretching
so this bu gives me and asymmetry stretching

27:45.320 --> 27:52.559
which is soon shown here alright
so that was obtained from the representation

27:52.559 --> 28:00.500
that we got using nth bond distance now if
i look at the representation that i got out

28:00.500 --> 28:09.570
of n n f alright so n n f is an angle so whatever
the normal modes that i i am talking about

28:09.570 --> 28:17.600
has to do something with the change of this
angles ok so ag is a total symmetric ir so

28:17.600 --> 28:25.260
normal mode which transforms according to
this ag must change this bond angles in a

28:25.260 --> 28:33.760
free symmetric way that means if this bond
goes here making this angle shorter and this

28:33.760 --> 28:41.400
will also go in this direction making this
one you know similarly shorter so this will

28:41.400 --> 29:04.440
be a symmetry bending mode so if i can write
here in what it will happen so ok

29:04.440 --> 29:12.090
so both of them will move in opposite direction
by equal amount so that this angle is changed

29:12.090 --> 29:23.280
by the same amount ok so this is the symmetric
bending ok so this is symmetric bending and

29:23.280 --> 29:32.780
it corresponds to bu i am sorry this will
be corresponding to ag i am talking about

29:32.780 --> 29:38.980
this particular ag ok now what will bu the
bu will also be a bending mode because it

29:38.980 --> 29:46.970
deals with this angle and in this case if
i look at the character table of the bu

29:46.970 --> 29:54.270
like before it has a it is a symmetry with
respect to see two or i so there it should

29:54.270 --> 30:05.390
be a asymmetric kind of movement right so
therefore it will be a symmetric bending so

30:05.390 --> 30:14.830
if i have f here f here then if this angle
becomes shorter then this will become larger

30:14.830 --> 30:21.530
so if this goes in this direction this also
goes in this direction or vice versa ok

30:21.530 --> 30:33.890
so this is asymmetric in bending motion and
has a symmetry of bu alright so we are left

30:33.890 --> 30:42.011
with this last normal mode having ag symmetry
and this was born out of this in an bond distance

30:42.011 --> 30:50.410
is totally symmetric so this in nn bond distance
either increasing or decreasing ok its purely

30:50.410 --> 30:58.630
symmetric right you can say so like you know
if you take talk about this one so they will

30:58.630 --> 31:06.720
go apart and they will you know come closer
so in a very symmetric way so overall that

31:06.720 --> 31:12.640
will be a completely symmetric with respect
to all this symmetric operations so this one

31:12.640 --> 31:31.610
will be there nn symmetric stretch ok so that
will be the symmetric stretch ok

31:31.610 --> 31:40.690
so see now if you look at all the symmetry
in all the normal modes and you know when

31:40.690 --> 31:46.280
you look at this presentation of this normal
mode it is absolutely [clea/clear] clear right

31:46.280 --> 31:51.650
so to start with we didnt have any idea like
what are those normal modes how will they

31:51.650 --> 31:56.480
look like because if i will give you the molecule
structure and ask you to drop down almost

31:56.480 --> 32:02.670
structure you may not be able to draw exactly
what is you know what we operated just now

32:02.670 --> 32:10.620
but if you go in this way you exactly no ok
i have use this as my so what about changes

32:10.620 --> 32:15.180
is happening is happening on this particular
bond or on this particular angle for this

32:15.180 --> 32:22.040
particular bond so looking at the symmetry
ok of that particular presentation that we

32:22.040 --> 32:29.700
got using those basis functions i can unequivocally
comment on this particular normal modes so

32:29.700 --> 32:36.390
this is how we get the you know the character
of the normal modes and knowing get symmetric

32:36.390 --> 32:46.000
properties all right now before we stop today
will mention one more thing suppose i take

32:46.000 --> 33:00.500
another molecule ok say ammonia and tick all
three and coordinates and for the representation

33:00.500 --> 33:10.860
so for ammonia belong into to c three point
group i have e c three and sigma v correct

33:10.860 --> 33:21.500
so if i form gamma three n we have twelve
for e c three v e nothing and sigma v will

33:21.500 --> 33:28.120
two two so you can verify yourself this one
and once you reduce it you will get you know

33:28.120 --> 33:34.820
the total number you know all the irreducible
representations containing this representation

33:34.820 --> 33:45.750
which will be three one plus a two last four
e and out of these representations what you

33:45.750 --> 34:03.070
will get is for normal vibration you get to
a one plus two e fine now if you if you try

34:03.070 --> 34:12.130
to find out which one of this you know
normal modes will be infrared active and raman

34:12.130 --> 34:24.840
active you will see both of this both of this
will be ir as well as raman active

34:24.840 --> 34:31.010
now lets compare the three compounds that
we looked at one belonging into d three eight

34:31.010 --> 34:35.679
symmetry another belonging to c two eight
symmetry and this third one that we just looked

34:35.679 --> 34:43.040
at is it c three v symmetry now d three h
had a thing like certain normal modes will

34:43.040 --> 34:49.960
be infrared active only certain normal modes
will be raman active only and sudden normal

34:49.960 --> 34:58.680
mode will be both raman as well as infrared
active in case of c three v i have all the

34:58.680 --> 35:05.080
normal modes are simultaneously raman active
as well as infrared active in case of that

35:05.080 --> 35:12.600
molecule n two f two which belongs to c two
h point group and these sent to symmetry i

35:12.600 --> 35:25.369
get those you know particular normal modes
which are raman active are not infrared active

35:25.369 --> 35:32.200
and vice versa meaning that for that particular
molecule and a particular point group the

35:32.200 --> 35:38.980
raman active modes and infrared active modes
are mutually exclusive ok

35:38.980 --> 35:49.210
so this is a special character of molecules
which are central symmetric in nature ok so

35:49.210 --> 35:55.050
we will stop here today and we will elaborate
this issue in the next class thank you very

35:55.050 --> 35:55.830
much for your attention