WEBVTT

00:14.019 --> 00:22.180
hello and welcome to the day three of week
three this lecture series so i hope you

00:22.180 --> 00:30.270
you tried practising finding out conjugate
elements for was different point groups

00:30.270 --> 00:37.740
if you have not please try some more so today
will start there were we left the other day

00:37.740 --> 00:45.820
so we are talking about how to write
this you know classes for one particular

00:45.820 --> 00:59.680
point group so we showed that for c three
v we write e two c three and c sigma b because

00:59.680 --> 01:04.509
all the sigma v is they belong the same class
and all the proportionate symmetry operations

01:04.509 --> 01:10.009
they belong to another class and e thats identity
from class by itself

01:10.009 --> 01:18.249
so the reason we club them together and
that the reason is that their properties

01:18.249 --> 01:30.310
their characters are same likes the characters
of c three and c three squares they are same

01:30.310 --> 01:35.829
the characters of sigma v prime sigma v double
prime sigma v triple prime they are all same

01:35.829 --> 01:43.659
for c three point group ok but the character
for you know this c three class and character

01:43.659 --> 01:50.599
for sigma v class they may not be same when
i say about character that has a deep meaning

01:50.599 --> 01:56.130
so we will come to that the following
two or three classes and so lets

01:56.130 --> 02:02.490
come back to the you know classification
of symmetry operations so on your screen

02:02.490 --> 02:10.780
you can see that there is d for h point
group and the the classes that are possible

02:10.780 --> 02:14.940
for this particular point group are
written here

02:14.940 --> 02:25.480
so you have ten different classes so you have
class by identity inverse and the sigma h

02:25.480 --> 02:33.310
ok they form class by itself you are already
told that here you can see that and then you

02:33.310 --> 02:40.740
know two different sigma v s they form
you know class two sigma d is they form

02:40.740 --> 02:48.950
a class then c two there is a c two which
is along the z axis in case of d four h

02:48.950 --> 02:59.580
d four h is z axis means is the you know
same axis through which you proper axis

02:59.580 --> 03:04.190
of symmetry c four runs so c two forms
class by itself and then perpendicular c twos

03:04.190 --> 03:11.530
they formed you know class by themselves
and there are you know improper axis of

03:11.530 --> 03:19.980
symmetry so this s four and s four inverse
they form also class by they form

03:19.980 --> 03:32.710
a class ok so you have to try a lot of
point groups and you find out the you

03:32.710 --> 03:38.630
know symmetry operations and then you perform
the simulative transformation on them and

03:38.630 --> 03:40.850
you try to find out classes ok

03:40.850 --> 03:49.870
so suppose i give you this you know
d four h or some other say that d four d then

03:49.870 --> 03:56.370
there are certain you know travels in
front of you what are they like just just

03:56.370 --> 04:02.880
like that if i give you d four d how will
you know what point group or what thats symmetry

04:02.880 --> 04:09.560
operations that this point group has because
if i give you the molecule may be you can

04:09.560 --> 04:23.440
try to look at and find out you know what
are the symmetry operations all you have to

04:23.440 --> 04:30.979
you know think about a molecule which may
have a point group d four d but this is little

04:30.979 --> 04:44.169
bit problematic so i have easy option will
be to just go and find out the book of

04:44.169 --> 04:51.090
book of group theory and at the end of this
book you will always find a set of tables

04:51.090 --> 04:53.281
which are known as character table and this
character table gives you all the symmetry

04:53.281 --> 04:56.879
operations but they will give you in terms
of you know this classes but that will you

04:56.879 --> 05:04.950
give you wait hint but otherwise how will
you find out about the the symmetry operations

05:04.950 --> 05:10.340
that will a point group has so like we are
going in a reverse relation now earlier we

05:10.340 --> 05:13.210
figured out all the symmetry operations for
a particular molecule and then figured out

05:13.210 --> 05:14.520
which point group this molecule belongs to

05:14.520 --> 05:21.310
now what we will do that in general if someone
gives me a particular point group can i find

05:21.310 --> 05:32.490
out what are the symmetry operations that
you know point group will have and thats what

05:32.490 --> 05:38.270
we will look at now so how will you do that
we will do that using something calls stereographic

05:38.270 --> 05:45.960
positions ok let me be cleared this point
that we will not go into the deep you know

05:45.960 --> 05:51.330
details of the stereographic positions
or deviation of the stereographic positions

05:51.330 --> 05:58.530
we will we will tell you the basic
things of the stereographic positions and

05:58.530 --> 06:05.599
we will show you how to use this stereographic
positions to find out the symmetry operations

06:05.599 --> 06:15.509
that are possible and also we will try to
show you that by using stereographic positions

06:15.509 --> 06:21.240
you can find out the products of symmetry
operations thats an added advantage of this

06:21.240 --> 06:22.240
stereographic positions

06:22.240 --> 06:23.819
so what is this stereographic positions so
any molecule will be dispose in a space ok

06:23.819 --> 06:28.800
so its a three dimension of space so anything
you know if you take it like three dimension

06:28.800 --> 06:37.060
sphere and you know you project everything
on to this you know plane at the middle of

06:37.060 --> 06:45.999
the sphere you can take it like you know
mapping plane right so the stereographic

06:45.999 --> 06:49.770
positions is a particular mapping that
projects a sphere on to a plane so what

06:49.770 --> 06:57.080
we do here we we defines our working
space ok or working area and we perform

06:57.080 --> 07:06.249
this you know positions of any point and
utilises symmetry operations to generate several

07:06.249 --> 07:13.349
you know points that are also positions
of any point the three dimensional space so

07:13.349 --> 07:22.129
we generate different points and thereby we
figure out what are the symmetry operations

07:22.129 --> 07:28.600
other than which are very clearly defines
by the point group rotation that represent

07:28.600 --> 07:29.600
ok

07:29.600 --> 07:35.619
so lets its started so that it becomes bit
more clear so ok so there there are

07:35.619 --> 07:40.259
some issue in this particular try here
so this circle actually is meant to be here

07:40.259 --> 07:43.039
so what i will do i just draw circle here
ok that will solve the problem alright so

07:43.039 --> 07:46.460
this circle big circle here shown
by a dash line is our working space and everything

07:46.460 --> 07:52.219
that we do here will be in this working space
and then what we do we take any general

07:52.219 --> 07:58.849
point on that particular working area or
working space and then we perform the symmetry

07:58.849 --> 08:03.809
operations now which symmetry operations that
we will be performing see when will be

08:03.809 --> 08:13.639
look at the symbols that is the rotation
for any point group certain symmetry operations

08:13.639 --> 08:24.680
which having that particular point group pretty
obvious lets say for examples if i take

08:24.680 --> 08:37.681
the case of c two v ok so what what we
know about that you know c two that means

08:37.681 --> 08:45.440
there is definitely there is c two axis and
v is and v is transfer of vertical plane of

08:45.440 --> 08:52.340
symmetry that is there is a sigma v so this
two at least i know and e is always implied

08:52.340 --> 09:02.400
so i will start from first one that is a c
two then after i will operate all the possible

09:02.400 --> 09:11.000
you know operations that is generate to
c twos so here it is only one c two because

09:11.000 --> 09:19.630
c to two or c two square is identity and then
we will move to the next symmetry operations

09:19.630 --> 09:26.590
that is obvious from that point group
rotation that is sigma v s you will apply

09:26.590 --> 09:31.740
sigma v so similarly if we have say like you
know d six eight what will you do so d stands

09:31.740 --> 09:36.910
for like c n so d six means a c n then
there are perpendicular c twos and h means

09:36.910 --> 09:42.040
there is a sigma h square so this three at
least i know now i do not know whether other

09:42.040 --> 09:45.110
than this there are you know centre of
inversion whether there are other sigma v

09:45.110 --> 09:48.510
is the other sigma d is whether there are
improper axis of symmetry all those things

09:48.510 --> 09:53.430
we can actually find out by using this stereographic
positions so lets get started now before we

09:53.430 --> 09:57.100
actually use if we before we start stereographic
positions we will first you know

09:57.100 --> 10:03.880
look at certain symbols that are used here
so any any proper axis of rotation is defined

10:03.880 --> 10:09.470
by some solid symbols so like done here this
is for c two c three c four c five c six and

10:09.470 --> 10:15.560
you can keep going so you know and triangle
is for c three you know a square is for c

10:15.560 --> 10:16.714
four and pentagon is for c five and so and
so its not very difficult to remember also

10:16.714 --> 10:20.230
and this this this particular geometry have
been two points on the two sides is

10:20.230 --> 10:26.510
is symbol of c two for improper axis what
will we use is the same symbol but not

10:26.510 --> 10:34.000
solid one but the hallow now any sigma plane
is given by a solid line so so let me

10:34.000 --> 10:38.440
clarify this thing little bit here so suppose
this is my working area and then i find out

10:38.440 --> 10:40.760
that there is a c two which is a proper axis
symmetry which is here then what i do we

10:40.760 --> 10:48.770
mark it here so that it means there is a c
two so for any given object here will you

10:48.770 --> 10:57.660
know come to this point by a c two operation
ok

10:57.660 --> 11:04.880
so if there is a c three then i mark it here
now there if there perpendicular c twos that

11:04.880 --> 11:09.920
that is very interesting thing like so this
region is for peaceful axis of rotation or

11:09.920 --> 11:12.910
even improper axis of rotation that is given
in centre while any perpendicular c twos

11:12.910 --> 11:15.320
are you know drawn at the periphery ok suppose
there is a perpendicular c twos here so say

11:15.320 --> 11:20.830
suppose this is c two perpendicular c two
axis then you draw a that sign here ok this

11:20.830 --> 11:24.690
this means perpendicular c twos or if i write
c two prime ok so all the axis you know

11:24.690 --> 11:28.310
proper axis or improper axis they are written
here or any perpendicular c two are written

11:28.310 --> 11:33.700
on the periphery fine now you know this
whenever i draw a line particularly when

11:33.700 --> 11:40.540
it is dash line that means nothing so that
has no physical significance here whenever

11:40.540 --> 11:47.160
you see the solid line that means something
so this dash lines are just for you know guidance

11:47.160 --> 11:48.160
thats all

11:48.160 --> 11:53.100
now suppose suppose this line is by
sigma plane then i will specify this one

11:53.100 --> 12:00.500
by a solid line ok across this is sigma plane
if there is an sigma e what will you do this

12:00.500 --> 12:05.500
is sigma e means this whole plane molecular
plane you can imagine is a plane of symmetry

12:05.500 --> 12:12.750
so then this whole thing will become solid
i will you will make it solid two signify

12:12.750 --> 12:16.130
that is sigma e that is shown here ok with
this basic knowledge we will start looking

12:16.130 --> 12:18.240
at stereographic positions of second particular
point group

12:18.240 --> 12:23.620
so the easiest one will be look at something
like c two v ok i know by it law you know

12:23.620 --> 12:30.570
what is the symmetry operations that c
two v point group has but it would be a good

12:30.570 --> 12:38.290
idea also to this c two v it will verify immediately
whether you can use stereographic positions

12:38.290 --> 12:45.890
to to find out what are the symmetry
operations that are possible so will start

12:45.890 --> 12:53.380
here so you look at you know the figure
here so we have the working space here as

12:53.380 --> 13:01.300
usual and what you know from c two v point
group that we have a c two ok so from c two

13:01.300 --> 13:07.720
v what i know we have a c two definitely and
then i have a sigma v sigma v you know what

13:07.720 --> 13:09.850
is the definition of sigma v

13:09.850 --> 13:20.740
so now we start with c two so we take a general
point here and then perform c two so if you

13:20.740 --> 13:25.860
perform c two if you do a clock wise rotation
then this point will come on the way and go

13:25.860 --> 13:30.880
right here and that what it shown on this
figure so ok so from here so you know originally

13:30.880 --> 13:35.600
it was here and now i have here ok this of
course from reason this writing function

13:35.600 --> 13:39.620
but thats ok so so thats what we generate
by using c two now we know that we have sigma

13:39.620 --> 13:46.910
v so next i will use a sigma v prime so where
we will use sigma v so lets try sigma v right

13:46.910 --> 13:50.950
here ok in this direction ok now will you
generate something new and we can see that

13:50.950 --> 13:55.590
s this will have a replica here and this line
have a replica over here so the result of

13:55.590 --> 13:57.480
applying of sigma after applying this c two
is this ok

13:57.480 --> 14:06.784
so now you see iam marking here also this
point was one and this point was two now

14:06.784 --> 14:13.250
this one prime and two prime have been generated
by reflecting the points one and two through

14:13.250 --> 14:20.620
this plane ok which is our sigma v now you
see that you know one prime can be formed

14:20.620 --> 14:23.700
through reflection on a plane which is here
ok of the point two so two and one prime can

14:23.700 --> 14:26.870
be generated from each other if there is a
(Refer Time :15:00) plane here right so i

14:26.870 --> 14:38.880
dont need to you know so i get another
actually sigma v here so we call it as sigma

14:38.880 --> 14:46.290
v prime so now from this next part we should
look at is that i have applied both the symmetry

14:46.290 --> 14:51.070
operations that i could find out from the
rotation itself that is c two sigma v and

14:51.070 --> 15:01.570
operates both of them and generates all the
point that are possible ok so that those

15:01.570 --> 15:09.160
are the i know actual number of points
maximum that you can generate and after generating

15:09.160 --> 15:16.900
those you see if you know one of the points
could be generated from another point through

15:16.900 --> 15:20.780
any other symmetry operations which you have
already you have not looked at already ok

15:20.780 --> 15:25.055
and we could figure out that less there are
stood exits plane here which we are calling

15:25.055 --> 15:28.910
sigma v prime by which you can generate (Refer
Time :16:00) one prime starting from two or

15:28.910 --> 15:33.210
vice versa similarly one to two prime or two
prime to one ok

15:33.210 --> 15:40.770
so we have now figured out that ok they are
all c two sigma v sigma v prime and no matter

15:40.770 --> 15:46.470
what identity so these are the four symmetry
operations that are possible for the point

15:46.470 --> 15:50.670
group c two v and you already know that so
you can see how useful this stereographic

15:50.670 --> 15:56.850
position can be in order to find out the si
symmetry operations for a given point group

15:56.850 --> 16:05.150
alright so this is on your screen this
what i just now said that is you know this

16:05.150 --> 16:11.740
this one prime and two prime could be generated
from a reflection of the point you know two

16:11.740 --> 16:14.430
and one through a reflection plane here ok
so from this example also you find out that

16:14.430 --> 16:22.110
the prells of c two axis and ones sigma v
plane will definitely give rise to other sigma

16:22.110 --> 16:28.450
v plane here one sigma v prime ok so this
is no matter what this is true if you have

16:28.450 --> 16:36.050
one c two axis sigma v prime there will be
another sigma v plane and similarly

16:36.050 --> 16:42.550
presence of a c n axis and sigma v plane guarantee
that they are there must be a total n sigma

16:42.550 --> 16:47.780
v planes we have already you know learned
that while forming the point group ok will

16:47.780 --> 16:52.070
had gone through the various steps right if
you remember you know step one two three four

16:52.070 --> 16:53.070
five

16:53.070 --> 16:59.150
so in those steps we mention that ok if
you have c n do you have a n c two primes

16:59.150 --> 17:06.539
do you have a n sigma v do you have a n sigma
d is so we know that if you have you know

17:06.539 --> 17:18.480
principle axis of symmetry that c n n if you
have one sigma v if you pick out the doubt

17:18.480 --> 17:24.010
from that so implies rotation point group
then you are sure they all will be such to

17:24.010 --> 17:30.530
tell a n sigma vs ok so thats why this
cn axis and sigma v plane they are called

17:30.530 --> 17:35.150
the generating elements of c n v point group
ok for example if you would have taken

17:35.150 --> 17:42.630
c c three v you should be able to that very
quickly right so so let me let me help

17:42.630 --> 17:47.010
you about in doing that very quick so what
you have for c three v you first find out

17:47.010 --> 17:52.340
the working space so you took trying to
figure out c three v ok so c three its principle

17:52.340 --> 17:58.240
axis of what i know from this is that i have
a c three axis and i have sigma v plane and

17:58.240 --> 18:05.950
already we have learned that for that for
c three there must be three sigma v plane

18:05.950 --> 18:14.160
so i already know that now i will start from
here so for a c three lets put the c three

18:14.160 --> 18:22.680
symbol here ok alright this is solid triangle
symbolising c three and then lets take any

18:22.680 --> 18:29.740
any general point here ok this is my generate
point and for my own understanding i

18:29.740 --> 18:37.300
am writing this this this dash line ok
so this is us to help myself ok this does

18:37.300 --> 18:45.140
not have any other meaning alright so three
means hundred and twenty degree rotation so

18:45.140 --> 18:51.710
let me i have taken this generalise point
so this is a generalised point any general

18:51.710 --> 18:55.870
point also and then i will operates c three
ok

18:55.870 --> 19:04.560
so c three will take this one from here to
somewhere here ok now my job is to complete

19:04.560 --> 19:10.380
the operations that can be generated from
a c three element right so then other operations

19:10.380 --> 19:13.800
that can be generated is c three square that
means c three and another c three so this

19:13.800 --> 19:19.030
element will reach somewhere here after c
three operations correct c three square operations

19:19.030 --> 19:26.670
sorry now i have sigma vs so next i have
to operate the sigma v so what i will do i

19:26.670 --> 19:33.470
will say first take this one as my sigma v
one of the sigma v so i will get one that

19:33.470 --> 19:38.710
here now this point here will have a point
one here right if i reflect and then this

19:38.710 --> 19:40.450
will have all similar points here

19:40.450 --> 19:44.690
so now i have generated all the points
that i could using the information that i

19:44.690 --> 19:49.800
have from the implies rotation ok so i have
use the c three performed all the operation

19:49.800 --> 19:57.500
c three c three square and then i have applied
one sigma v and i see the effect here now

19:57.500 --> 20:04.685
so i have got c three c three square i
have got one sigma v now lets see so now lets

20:04.685 --> 20:10.400
this point so like you know so say this
is my point one two and three now this is

20:10.400 --> 20:17.179
say three prime and this is two prime and
this is one prime now we just ok so

20:17.179 --> 20:19.770
from here you can see that this this x
three and x two prime they could be generated

20:19.770 --> 20:24.310
from a reflection through this plane ok so
if you drive properly it will be much more

20:24.310 --> 20:30.660
clear and then this x two and x three prime
they could be generated that is x two

20:30.660 --> 20:37.180
x three prime could be generated from x two
through a reflection on this particular plane

20:37.180 --> 20:43.220
so what we get we got total three sigma
vs fine ok so you can you can try doing

20:43.220 --> 20:50.320
many other point groups and you can
you can you know from the point group

20:50.320 --> 20:53.350
symbols you can find out all the symmetry
operations that you can generate now

20:53.350 --> 20:57.630
we we we also showed that you know with
stereographic position we can find out what

20:57.630 --> 21:02.710
is the result of combining to symmetry operations
that is you know successive you

21:02.710 --> 21:06.770
know successively if you operate two symmetric
operations on any particular structure

21:06.770 --> 21:10.040
what will happen can be figured out by a stereographic
position that will you tell me that what will

21:10.040 --> 21:15.440
be the result at symmetry operations ok now
while looking at the group multiplication

21:15.440 --> 21:21.870
table we also we have figured out that like
you know here in front of you have the

21:21.870 --> 21:26.190
you know group multiplication for c two v
prime groups and there you can see that all

21:26.190 --> 21:28.550
the binary products are written here now
stereographic position also does the same

21:28.550 --> 21:31.540
thing so using stereographic position you
can actually find the group multiplication

21:31.540 --> 21:32.540
table ok

21:32.540 --> 21:40.510
so there there all inter link you can see
that so now you will try to verify that you

21:40.510 --> 21:47.130
know what will be the effect of you know
doing this successive of operations ok

21:47.130 --> 21:54.340
so now suppose i try to find out that
what happens when you operate c two and sigma

21:54.340 --> 21:59.559
v successfully ok so that means i have to
get sigma v c two which means first you operates

21:59.559 --> 22:03.050
c two and then operate sigma v so first we
operates c two so this point it goes over

22:03.050 --> 22:06.679
here ok so this this particular place is
the original you know point that i took

22:06.679 --> 22:13.470
now after i got this structure then if i operates
sigma v here what we get is the reflection

22:13.470 --> 22:20.201
and in that case sigma vs in this plane ok
so reflection will take a back to the original

22:20.201 --> 22:29.100
point ok so we get it we get it we get here
ok now so so sorry there was a mistake

22:29.100 --> 22:31.350
let me correct that

22:31.350 --> 22:39.760
so you have you know operates c two it
comes here and you take a sigma v which

22:39.760 --> 22:45.320
is around this plane and you generate a similar
point over here that you can see here and

22:45.320 --> 22:51.650
then you know you if you would have taken
a plane which is here perpendicular to

22:51.650 --> 23:04.160
the sigma v plane i would call it sigma
v prime then reflection through that sigma

23:04.160 --> 23:13.640
v prime plane could take the original point
to this point which we are got after operating

23:13.640 --> 23:20.290
c two and sigma v so that means that sigma
v c two it becomes to sigma v prime now lets

23:20.290 --> 23:23.890
go back to the group multiplication table
and see if that is correct ok so if you operates

23:23.890 --> 23:30.500
c two and then sigma v then you get this product
which is sigma v prime so this works

23:30.500 --> 23:37.090
now similarly if we try to figure out what
happens if we operates sigma v prime and then

23:37.090 --> 23:43.040
sigma v what is the result ok so we start
from the general point and the sigma v

23:43.040 --> 23:47.740
is this plane so if you have reflection then
i will go somewhere here right here and then

23:47.740 --> 23:54.810
if i operates sigma v which is along this
then i will generate a point over here thats

23:54.810 --> 24:35.540
what we generated here now if i would have
taken a c two so here it is shown like a inform

24:35.540 --> 24:45.650
here i can go to this point by c two similarly
from here i can would come to this point ok

24:45.650 --> 24:52.980
so if could operate c two from this original
point i could reach over here which is identical

24:52.980 --> 24:56.770
identical to this structure

24:56.770 --> 25:02.440
so that mean the sigma v sigma v prime is
nothing but equal to c two and if we just

25:02.440 --> 25:09.450
have a quick look at the a group multiplication
table then we can see that sigma v sigma v

25:09.450 --> 25:19.420
prime gives me c two and also from here you
can see that you know it doesnot matter

25:19.420 --> 25:34.150
which here come that sigma v is sigma v prime
or sigma v prime sigma v it gives me c two

25:34.150 --> 26:05.440
which also mean set i can find out about the
ok so you can verify the all the you know

26:05.440 --> 26:33.120
other products that you can form ok so
you can take any other point group

26:33.120 --> 26:43.990
and i will i will suggest to you to go steps
like you you go from c two v c two eight

26:43.990 --> 26:56.360
then higher orderly c three v and then you
know d four d four eight d for v d for d

26:56.360 --> 27:11.600
and try to generate all the you know
possible stereographic positions and you know

27:11.600 --> 27:23.890
start verifying thats the group group multiplication
tables that you can have so what it will do

27:23.890 --> 27:36.809
it will you know you know it will give
you enough practise

27:36.809 --> 27:51.070
so that you know
if you have to answer the questions that

27:51.070 --> 27:59.500
what are the symmetry elements present
for this particular point group you will

27:59.500 --> 28:03.809
not take much time and you can do it [imme/immediately]
immediately but the most important thing

28:03.809 --> 28:11.950
is that in the following groups what you
will do we will start moving towards the

28:11.950 --> 28:20.840
actual applications of group theory and symmetry
aspects ok into actual chemistry problems

28:20.840 --> 28:33.460
so essentially for you know you know
so far we have them preparing the you know

28:33.460 --> 28:43.660
basis for this this applications ok
so while we go for this those parts particularly

28:43.660 --> 28:49.520
like you know find out section rules or forming
a particular symmetrical after linier

28:49.520 --> 29:00.830
combination will be using something called
representations so we have to actually prepare

29:00.830 --> 29:07.660
a presentation of this you know symmetric
point groups and then symmetry elements

29:07.660 --> 29:14.190
so there are you need know this symmetry
you know operations elements very

29:14.190 --> 29:38.480
mush ok and in order to you know we able to
that very quick so that you can go to the

29:38.480 --> 29:47.540
actual problems you should you know
practise doing this stereographic positions

29:47.540 --> 29:57.760
more number of point groups ok so i will leave
at that point today and we will come back

29:57.760 --> 30:13.549
in the following class which something more
by will look at the matrix representations

30:13.549 --> 30:41.950
of different symmetry operations till then
good bye