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GENESIS TUTORIALS
Institute for CSIR-UGC-NET/JRF, GATE & IIT-JAM
Group Theory
Introduction:-
The systematic discussion or mathematical study of symmetry is called Group theory.
This theory gives a simple and direct method for arriving at useful conclusions with minimum
calculations about the geometry and electronic structure of molecules. Pericyclic reactions are
mainly based on group theory. The ligand field theory is also supported by group theory.
To understand the various symmetry elements present in a molecule it is necessary to have same
basic knowledge about the geometry of various molecules, obtained from Valence Shell Electron
Pair Repulsion Theory (VSEPR)
Symmetry Elements.
Symmetry elements are defined as imaginary as imaginary geometrical entities such as points, lines
and planes that are present in a molecule, about which when symmetry operations are performed,
the molecules present an indistinguishable configuration.
Symmetry Operations
These are simple geometric operations such as reflection, rotation or inversion which when
performed on the molecule, give rise to an indistinguishable configuration of the same molecule.
Symmetry element Carres ponding symmetry operation
1.
2.
3.
4.
5.
Identity Element (E) or (I)
Plane of Symmetry ( )
Centre of symmetry (i)
Proper axis of symmetry (Cn)
Improper axis of Symmetry
Doing nothing (3600 rotation)
Reflection
Inversion of all coordinates.
Rotation through an angle of 0360
n
about an axis. (n = order)
Rotation through an angle of 0360
n
about an axis followed by reflection about
a plane perpendicular to the original axis.
Improper axis of Symmetry(Sn)
When a molecule is rotated about an axis through an angle of 0360
nfollowed by reflection about a
plane perpendicular to the originally chosen axis, if it presents an indistinguishable configuration,
it is said to possess an improper axis of symmetry, represented by Sn.
Molecule having an improper axis of symmetry can be classified in to two types.
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1 Molecular having both, on axis of
symmetry (Cn) and an perpendicular
plane of symmetry ( )
Eg:- BCl3
2 Molecular having neither an axis of
symmetry nor a plane of symmetry
but having only an improper axis of
symmetry (Sn)
Eg:- Staggered conformation of Ethane.
Associated Operations of Sn
The number of times the process has to be repeated to get the original configuration is raftered to
as the associated operations of the Sn axis.
1) Associated operations of the S2 axis.
1 1
2 2
2 2 1
2 2
S C
S C E
2) Associated operations of S3 axis
1 1
3 3
1 2 2 2
3 3 3
3 3 3
3 3
4 4 4 1 1
3 3 3 3
5 5 5 1
3 3 3
6 6 6
3 3
S C
S C C
S C E
S C C C
S C C
S C EE E
3) Associated operations of the S4 axis.
1 1
4 3
2 2 2
4 4 2 2
4 3 3 3
3 4 4
4 4 4
4 4
S C
S C EC C
S C C
S C E E E
Therefore they can be listed as 1 3
4 2 4S ,C ,S and E
4) Associated symmetry operations of the S5 axis.
1 1
5 5
2 2 2 2 2
5 5 5 5
3 3 3 3
5 5 5
4 4 4 2 4
5 5 5 5
S C
S C E C C
S C C
S C E C C
5 5 5
5 5
6 6 6 1 1
5 5 5 5
7 7 7 2
5 5 5
S C E
S C E E C C
S C C
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8 8 8 3 3
5 5 5 5
9 9 9 2
5 5 5
10 10 10
5 5
S C E C C
S C C
S C E E E E
Sn
n Performed operation
Even n
odd 2n
1. Group and Point Group and its Properties
Group:- Any set or collection of elements which obey a certain set of mathematical rules and are
related to each other by certain rules.
The elements could be number, symmetry elements.
Mathematical Rules
1) Closure Property:- The combination of any two members of a group and square of every member
of a group should be a member of that group.
2) I dentity: The group contain one and only on identity element (E) which has the property.
A E = E A =A for all elements of the group
3) Associative Law: That is A (B C) = (A B) C
4) Inverse: The inverse of every element should also be a member of the group.
The inverse of any element x = x-1
And 1 1x x x E
Point Group Inverse
Cn n 1
nC
Sn (n = even) n 1
nS
Sn (n = odd) 2n 1
nS
Point Group:- A point group is a collection of the symmetry elements present in a molecule that obeys
the mathematical rules for the formation of a group.
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Point Group Symmetry
Elements
C1
CS
Ci
Cn
Cnv
Cnh
Dn
Dnh
Dnd
Sn
Td
On
Ih
2 (E,C1)
2 (E, )
2 (E, i)
n
2n
2n
2n
4n
4n
n (n-even)
2n (n = odd)
29
48
120
Point group is represented by Schoenflies Symbols
Order of Group: The total number of symmetry elements (non redundant) present in a point group
is called the order of a group. Represented by ‘h’
CLASS OF THE GROUP:- A complete set of elements in a group which are conjugated to each
other by similarity transformation. Are raid to be in the some does
“The element which are in the same class always show the same matrix.
Group Multiplication Table
A group multiplication table is a table which gives the binary product of any two members of the
group and the square of every member of a group.
Group multiplication table for C2v
C2v E C2 1V
2V
E E C2 1V
2V
C2 C2 E 2V
1V
1V 1V
2V E C2
2V 2V
1V C2 E
Point Group
3. Cnv Point Group
Schoen flies Symbols
Low symmetry point group
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n vC n E
Eg:-
1 2
2 2 v v
1 2 3
3 3 v v v
V
C v E, C , , ,
C v E, C , , , ,
C v E, C ,
4. Cnh Point group
n nC 1 E
Eg:-
2h 2 n
3h 3 n
C E, C , , i
C E C
2. Dnh point Group.
n 2 n vC nC , ,n
Eg:- 2h 2 2 n vD ,EC 2 C , , 2
3. Dnd point Group
n 2 vC nc ,n
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Points Groups and Dipole Moment.
Ci, Cnh, In, On (possessing i)
Td, Dn, Dnh, Dnd (possessing more than one non –
coincident axis)
Zero
Dipole moment
C1, CS, Cn, Cnv Measurable
dipole moment
Symmetry Number
Symmetry Number of a molecule indictes the number of equivalent configurations that can be obtained
by rotating the molecule around one or more axes.
Point Group Symmetry No.
1 s iC C ,C ,C 1
n nh nvC ,C ,C n (order of axis)
n nh ndD ,D ,D 2n
hD 2
Td 12
On 24
Matrix Representation of Symmetry operations-
The most accurate method is to study that effect of operations on the coordinates and express the
transformations in the form of a matrix.
Trans Formation Matrix For Identity Element:-
Consider a vector Op
having the coordinate (x, y, z,). When the identity operation E is performed
on this vector. Let the coordinate be changed to (x2, y2, z2)
OP
(x1, y1, z1) E
2 2 2Op x , y ,z
Since by definition the identity operation should not charge any coordinates x2 = x1, y2 = y1, z2 =
z1
The now position of the vector can be defined
2 1 1 1
2 1 1 1
2 1 1 1
x 1 x o y 0 z
y 0 x 1 y 0 z
z 0 x o y 1 z
These can be combined to give the following equation
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2 1
2 1
2 1
IdentityMotrix
x x1 0 0
y 0 1 0 y
0 0 1z z
The character of the transformation matrix E.
E sum of diagonal elements (from left top to right bottom)
E
1 0 0
0 1 0
0 0 1
1 1 1 3
Transformation Matrix for Plane of Symmetry
xy
1 1 1 2 2 2OP x , y ,z OP x , y ,z
2 1 1 1
2 1 1 1
2 1 1 1
x 1 x o y 0 z
y 0 x 1 y 0 z
z 0 x o y 1 z
2 1
2 1
2 1
x x1 0 0
y 0 1 0 y
0 0 1z z
Matrix for
xy z yz
z xz
xy 1 y 1
x 1
1 0 0 1 0 0
0 1 0 0 1 0
0 0 1 0 0 1
1 0 0
0 1 0
0 0 1
B. Trans formation Matrix for centre of Symmetry
i
1 1 1 1 1 1 2 2 2
2 1
2 1
2 1
OP x , y , z OP x , y , z OP x , y , z
x x
y y
z z
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The relation ship between 1 1 1 2 2 2x , y , z and x , y , z
2 1 1 1
2 1 1 1
2 1 1 1
x 1 x o y 0 z
y 0 x 1 y 0 z
z 0 x o y 1 z
2 1
2 1
2 1
x x1 0 0
y 0 1 0 y
0 0 1z z
Transformation Matrix for i =
1 0 0
0 1 0
0 0 1
i 3
4. Trans formation Matrix for Axis of Symmetry
Let vector OP
have a magnitude ‘r’ and derive an angle o with respect to the z axis
Then Cn(z)
1 1 1 2 2 2OP x y z OP x , y ,z
2 1
2 1
2 1
x xCos Sin 0
y Sin Cos 0 y
0 0 1z z
Trans formation matrix for Cn(z) =
Cos Sin 0
Sin Cos 0
0 0 1
Character nC (z) 2Cos 1
Transformation Matrix for improper Axis of Symmetry.
n xy n
n
S C (z)
1 0 0 Cos Sin 0
S 0 1 0 Sin Cos 0
0 0 1 0 0 1
Cos Sin 0
Sin Cos 0
0 0 1
Sn 2Cos 1
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Element Character (Contribution per alom)
C2 -1
C3 0
C4 +1
S2, i -3
S3 -2
S4 -1
h d v/ / 1
E 3
Character corresponding to any Symmetry operation R in a representation
R
Number of atoms that do not shift
when a symmetry operation (R) is
performed.
×
Character Corresponding to the
symmetry operation (R) in a
general (3×3) matrix
representation
2
xz
y z
E
C (E)
3 3 9For Water molecule
1 1 1Thecharacter of identity it known
3 1 3as dimension .
1 1 1
Representation of Croups
Representation is set of matrices for a group each corresponding to a single operation in the group,
that can be combine among themselves in a manner parallel to the way in which the group elements.
Reducible Representation:- Representation of higher dimension which can reduce to a lower
dimension representation it known as reducible Representation.
To drive a reducible representation first of all we have to choose the basis set according to our
need.
1) Basis Set:
3-Catesion cordinate
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C2v E C2(z) zx
zy
X 1 -1 1 -1
Y 1 -1 -1 1
Z 1 1 1 1
Reducible
Representation
3 -1 1 1
Character of identity
2) Base sel: 3N Cartesian Coordinate
Method to use unshifted atom
C2v E C2 zx
zy
Number of
unshifted atoms
3 1 3 1
Conlribution per
atoms
3 -1 1 1
R. R, 9 -1 3 1
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Eg:- BF3 i.e. D3h point group
D2h E 2C3 3C2 h
32S v3
Number of unshifted atoms 4 1
1
2 4 1 2
Conlribution per atoms 3 0 -1 1 -2 1
R.R. 12 0 -2 4 -2 2
Eg:- CH4 i.e. Td point group
Td E 8C3 3C2 6S4 d6
Nomber of unshifted
atoms
5 2
1
1 1 3
Conlribution per atoms 3 0 -1 -1 1
R.R. 12 0 -1 -1 3
d n
2
Paralled toC principalAxis
It bisect angle between twoc axis
It bisec t min imum number of atoms.
3) Basis Set: Bond vector
Eg:- CH4 i.e Td point group
E 8C3 3C2 6S4 d6
Nomber of unshifted
atoms
4 1 0 0 2
Conlribution per atoms 1 1 1 1 1
R.R. 4 1 0 0 2
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Eg:- PCl5 i.e. D3h point group, Basis set: Bond vector
E 2C3 3C2 n 2S3
V3
Nomber of unshifted
atoms
5 2 1 3 0 3
Conlribution per atoms 1 1 1 1 1 1
R.R. 5 2 1 3 0 3
General Rule for class.
(i) Inversion will be always in the separate class, if it is present
(ii) h Will be always in the seperate class.
(iii)In the case of proper axis m
nC and n m
nC will be in the same class.
Eg:-
1 2
3 3
1 3
4 4
c and c
c and c
(iv) There will be separate different class for d vand
(v) In the case of improper axis m n m
n nS andS will be in the same class.
The elements which are reciprocal of each other then both lying in the some class.
Irreducible Representation
A representation of lower dimention which can not be further reduce is known as irreducible
representation.
The irreducible representation for a group can be derive using great orthogonality theorem (G.O.T)
Postulates of G.O.T. and Derivation of Irreducible representation of C2V. Point group.
i) Number of irredcible representation in a group are always equal to the no. of classes of the
group.
Here 4 irreducible representation because C2v has 4 class
ii) The sum of square of dimension of all the irreducible represenatation in the group will be
equal to the order of the group.
C2v E C2 nz
zy
Ir R1 k1 1 m1 n1
Ir R2 k2 2 m2 n2
Ir R3 k3 3 m3 n3
Ir R4 k4 4 m4 n4
2v
2v
C Class 4
C order 4
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The character related to identity known as dimension according to rule (ii)
Dimension
iii) The sum of squares fo the character of an irreducible representation will be equal to the
order of group.
2 2 2 2
1 1 1
2 2 2 2
2 2 2
2 2 2 2
3 3 3
2 2 2 2
4 4 4
1 m n 4
1 m n 4
1 m n 4
1 n n 4
2 2 21 1 1 4
iv) The characters of any 2 irreducible representation in the same group will be always
orthogonal to each other ie on multplying any 2 irreducible represent ation the result
should be o
C2v E C2 nz
zy
Ir R1 1 1 m1 n1
Ir R2 1 2 m2 n2
Ir R3 1 3 m3 n3
Ir R4 1 4 m4 n4
C2v E C2 nz
zy
Ir R1 1 1 m1 n1
Ir R2 1 2 m2 n2
Ir R3 1 3 m3 n3
Ir R4 1 4 m4 n4
C2v E C2 nz
zy
Ir R1 1 1 1 1
Ir R2 1 1 -1 -1
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r 2 r 3i.e.I R I R 0
Eg:- (I)
Order = 6
Class = 3 = IR
2 2 21 1 2 6
(II)
Ir R3 1 3 m3 n3
Ir R4 1 4 m4 n4
C2v E C2 nz
zy
Ir R1 1 1 1 1
Ir R2 1 1 1 -1
Ir R3 1 -1 -1 -1
Ir R4 1 -1 -1 1
C3v C3v E 2C3 v3
IR1 1 1 m1
IR2 1 2 m2
IR3 2 3 m3
C3v E 2C3 v3
2 2 2
2 2 2
1 1 1 1 m 1 n 0
1m 1 n 1
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2 3IR IR 0
2. Character Table and Its Derivation.
A character table is a way which can be utilized for application of group theory for the molecular
symmetry.
Each character table has 4 columns and different symbols in them which can be explain by taking
example of 2vC point group.
Coloumn I:- Its Coloumn of character table shows mullein symbol for irreducible representation of
the group.
IR1 1 1 1
IR2 1 1 -1
IR3 2 3 m3
C3v E 2C3 v3
IR1 1 1 1
IR2 1 1 -1
IR3 2 -1 0
i.e. 1 2IR IR 0
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C2V E 2C3 v3
A1 1 1 1
A2 1 1 -1
E 2 -1 0
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Coloumn II : II nd coloumn of character table shown irreducible representation of the group can
be drive using G.O.T.
Coloumn III: III rd coloumn explain transformation propertier of 3-cartesion Co-ordinate, x, y,
z and 3 rotational axis ie Rx, Ry, Rz.
Coloumn-iv ivth Coloumn explain
transformation properties of quadratic
functions of coordinates like
transformation properties of x2, y2, z2.
C2v E C2 xz yz
x2
y2
z2
xy
yz
zx
1
1
1
1
1
1
1
1
1
-1
-1
1
1
1
1
1
-1
-1
1
1
1
-1
1
-1
Relation between Reducible and Zrreducible Representation
C2v E C2 nz
zy
x 1 -1 1 -1 B1
y 1 -1 1 1 B2
z 1 1 -1 1 A1
RX 1 -1 -1 1 B2
Ry 1 1 -1 -1 B1
RZ 1 1 -1 -1 A2
C2V E C2 2x
2y
A1 1 1 1 1
A2 1 1 -1 -1
B1 2 -1 1 -1
B2 1 -1 -1 1
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1
2
1
2
A
A
B
B
1 2 2 2
1n 1.1.9 1.1. 1 1.1. 3 1.1.1
4
19 1 3 1 3
4
1n 1.1.9 1.1 1 3.1. 1 1.1. 1
4
1 49 1 3 1 1
4 4
1n 1.1.9 1. 1 . 1 3.1.1. 1. 1 1
4
112 3
4
1 8n 9 1 3 1 2
4 4
3A A 3B 2B 3 1 3 2 9
No of one dimentional in
Application of Group Theory:-
Hybridization:- The concept is invoked to explain the strong nature of the bonds and their
directional nature. In such a molecules, the hybrid arbiters are used as the base for the
representation to determine the symmetry of the hybridized orbitals.
C2v E C2 xz yz
A1
A2
B1
B2
1
1
1
b) 1
1
1
-1
c) -1
1
-1
1
d) -1
1
-1
-1
1
P.R e) 9 -1 3 1
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Using the orthogonatity theorem and the character table, this can be reduced to give the
IR& of the group as given below.
order 24
Class 5
Number of A1
2
1
1
11.4.1 8.1.1 3.0.1 6.0.1 6.2.1
24
241
24
1 1A 1.4.1 8.1.1 0 0 6.2. 1 12 12 0
24 24
1 1E 1.4.2 8.1 1 0 0 6.2.0 8 8 0
24 24
1 1T 1.4.3 0 0 0 6.2. 1 12 12 0
24 24
1 1T 1.4.3 0 0 0 6 2 1 12 1
24 24
1 2
2 1
R.R A T
Td E 8C3 3C2 6S4 d6
Nomber of unshifted atoms 4 1 0 0 2
Conlribution per bond veclor 1 1 1 1 1
R.R. 4 1 0 0 2
Td
R
E
4
8C3
1
3C2
0
6S4
0
d6
2
A1 1 1 1 1 1
A2 1 1 1 -1 -1
E 2 -1 2 0 0
T1 3 0 -1 1 -1
T2 3 0 -1 -1 1
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A1 is most symmetric representation and it must there fore correspond. to the highly symmetric &
orbital. Area III of character table indicates x, y, z vectors corresponding to the T2 symmetry which
represent the Px, Py and Pz orbital. Thus we con say that the hybrid orbitals are formed by the mixing
of one s orbital and three p orbitals resulting in sp3 hybridization.
Eg: NH3 The ammonia molecule belongs to the C3v point group. In ammonia, there are there bonded
and one lone pair of electrons. Let the four hybrid oerbital 1 2 3 4, , and be chosen as then base
for arriving at the character representation.
C3v E 2C3 V3
Nomber of unshifted
atoms
4 1 2
Conlribution per atoms 1 1 1
R.R. 4 1 2
Bu using the orthogonality theorem and the character table, it can be reduced to give the IRs of the
group as follows.
C3v E 2C3 V3
A1 4 1 2
A2 1 1 -1
E 2 -1 0
R.R 4 1 2
Number of A1 = 1
4 2 6 26
Number of A2 = 1
4 2 6 06
Number of E = 1
8 2 0 16
R.R = 2A1 + E
The orbitals corresponding to symmetry A1 and E combine to give hybrid orbitals. From the
character table, it can be seen that these symmetries represent the s, px,
P4, P2 orbitals. Thus the hybridization in NH3 is sp3.
N
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2) Spectroscopy:-
STEP I Reducible Representation using 3N Cartesian coordinate.
Example C2v (H2O)
STEP II Develop Relation between Reducible Representation and Irreducible representation.
1 2 1 23A A 3B 2B
Total degree of freedom (Total Number of mode) = 3N
For water = 3×3 = 9
1 2 1 23A A 3B 2B
STEP III Determine Number of translational, Rotational and Vibrational degree of freedom.
1) Translational degree of Freedom;- Transforms according to x, y, z axis (Column II)
X B1 y = B2 Z = A1
i.e Trnslational = B1 + B2 + A1
2) Rotational degree of freedom:-transform according to RX, Ry, Rz (coloum III)
Rx B2 Ry = B1 R2 = A2
B1 + B2 + A2
3) Vibrational = Total – (Translational + Rotational)
1 2 1 2 1 2 1 1 2 2
1 1
3A A 3B 2B B B A B B A
2A B
1 1FundamentalVibration 2A B
3N 6 9 6 3
C2V E C2(z) xz yz
R R 9 -1 3 1
C2v
E C2(z) xz vz
A1 1 1 1 1 Z x2, y2, z2
A2 1 1 -1 -1 R2 Xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 Y, Ry yz
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STEP IV: IR Active vibration Transforns according to x. y and z axis. (coloum III)
1 1
IRActive
2 A B IR Active
Number of Normal IR mode = 3
STEP-V RAMAN Active Vib. Transform according to quandratic functions (square) of x, y,z
(column IV)
2 2 2
1
1
1 1
RamanActive
RamanActive
x , y , z A
xz B
2A B
STEP-VI Determine parallel (Rocking) and perbendicular Vibration.
Vibration is symmetric with respect to principal axis then it is defined as paraller vibration
and if is Anti symmelric with respect to principal axis it is perpendicular vibration.
A1 C1 Symmelric Vibration
B1 -1 Antisymmetric Perperdicualr.
Q. Refer character table of D3h point group and Answer the following question.
a) Derive the reducible representation using 3N cartesion coordinates.
b) Calculate all possible symmetry type of fundamental Vibration
c) Determine translational, rotational & Vibrational degree of freedom.
d) Determine IR active and Raman activties of each vibration
e) Classify the vibration as in plane & out of plane.
Sol D3h example BF3
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' ' "
1 2 2IR A A 3E ' 2A E" Totaldeg reeof freedom
Translational x E’ Z "
2A
ie Translational = E’ + "
2A
Rotational = E” + "
2A
Vibrational = Total – (Translational + Rotational)
' ' 1 " " 1
1 2 2 2 2A A 3E 2A E" E ' A E" A
Vibrational = ' "
1 2A 2E ' A
Check F3 (Non linear)
' 1
1 c3N 6 A 2E ' A
3 4 6 1 2 2 1 6
6
STEP-IV
IR Active
' 1 "
1 2
IR IR ActiveActive
A 2 E A
'
2i.e IR ACTIVE 2E ' A
RAMAN Active:-
D3h E 2C2
3C3
n 33S
v3
'
1A 1 1 1 1 1 1
'
2A 1 1 -1 1 1 -1
E’ 2 -1 0 2 -1 0
"
1A 1 1 1 -1 -1 -1
"
2A 1 1 -1 -1 -1 1
E” 2 -1 0 -2 1 0
R.R. 12 0 -2 4 -2 2
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1 "
1 2
RamanRamanActiveActive
A 2E ' A
1 1
2RAMAN ACTIVE 2E A
STEP-V Classification of in plane or out of plane Vibration.
If vibration which is symmetric (+1) with respect to h are called in plane vibration and
which is antisymmelric with resped toh is called out of plane vibration.
1 "
1 2
Inplane Outof
PlaneInPlane
A 2E ' A
3. Direct Product
Q. The EXE product in D3 point group contain the irreducible representation:
D3 E 2C3 23c
A1 1 1 1
A2 1 1 -1
E 2 -1 0
R-E×E 4 -1 0
A1
A2
A3
1n 4 2 0 1
6
1n 4 2 0 1
6
1n 4 2 1
6
Ans. A1 + A2 + E
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GENESIS TUTORIALS
Institute for CSIR-UGC-NET/JRF, GATE & IIT-JAM
Assignment-I: Group Theory
1. The symmetry point group of ethane in its staggered conformation is
(a) C3v (b) D3d (c) D3h (d) S6 [NET Dec. 2011]
2. The molecule that has an S6 symmetry element is [NET Dec 2012]
(a) B2H6 (b) CH4 (c) PH5 (d) SF6
3. Identify the point group symmetry of the following molecule (all C—C bond lengts are
equal) [NETJUNE-2013]
(a) C2v (b) S4 (c) D2d (d) D4d
4. The result of the product C2(x) C2(y) is [NET June 2014]
(a) E (b) σxy (c) C2(z) (d) i
5. The low temperature (-98°C) 19F NMR spectrum of SF4 shows doublet of triplets. It is
consistent with the point group symmetry. [NET June 2014]
(a) C3v (b) C4v (c) Td (d) C2v
6. The point group symmetries for trans-[Cr(en)2F2]+ and [TiCl6]3−, respectively, are
(a) D4d and D3d (b) D3d and D4d
(c) D4h and D3h (d) D3h and D4h [NET Dec 2014]
7. The symmetry point group of propyne is [NET-June-2014]
(a) C3 (b) C3v (c) D3 (d) D3d
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8. The product σxy. S4z (S4
z is the four fold improper axis of rotation around the z-axis, and σxy
is the reflection in the xy plane) is [NET-Dec-2014]
(a) C4z (b) C4
z. i (c) C4y (d) C2
z
9. The product C2xσxy (C2
x is the two–fold rotation axis around the x–axis and σxy is the xy
mirror plane) is [NET-June-2015]
(a) σxz (b) σyz (c) C2y (d) C2
z
10. The molecule diborane belongs to the symmetry point group [NET-June-2015]
(a) C2v (b) C2h (c) D2d (d) D2h
11. The symmetry point group of the most stable geometry of the following molecule Cl(H) C
= C = C(H) Cl is [NET-June-2016]
(a) C2 (b) C1 (c) C2v (d) C2h
12. The ponit group obtained by additing symmetry operation σh to the point C4 is
(a) S4 (b) C4h (c) D2h (d) D4 [NET-June-2016]
13. The molecule that possesses S4 symmetry element is [NET-Dec-2016]
(a) ethylne (b) allene (c) benzene (d) 1, 3–butadiene
14. The pair of symmetry points groups that are associated with only polar molecules is
(a) C2v, D∞h (b) C3v, C2h
(c) D2h, Td (d) C2v, C∞v [NET-Dec-2016]
15. The compound N2F2 has two isomers. Choose the correct option from the following:
(a) both isomers possess σv plane
(b) both isomers possess σh plane
(c) one isomer has σh plane while the other has a σv plane
(d) none of them have a σh plane [NET-June-2017]
16. The correct realtion involving symmetry operations [NET-June-2017]
(a) S42 = S2 (b) σ(xz)σ(yz) = C2(x)
(c) S43 = C4
3 (d) S63 = S2
17. The molecule with a C2 axis of symmetry [NET-Dec-2017]
(a) BH2Cl (b) CH3Cl (c) NH2Cl (d) HOCl
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GENESIS TUTORIALS
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Assignment-2: Group Theory
1. Refer to the character table of the point group C3V given above. Find which of the following
transition is forbidden- (CSIR JUNE 2011, PART B)
(a) a1 ↔ a1 (b) a1 ↔ e (c) a2 ↔ e (d) a1 ↔ a2
2. Given the character table of the point group C3V,
E 2C3 v
A1
A2
E
1
1
2
1
1
-1
1
-1
0
Z
(x,y)
E 2C3 v
6 3 0
Its reducible components are (CSIR JUNE 2011, PART C)
(a) E + 2A1 + 2A2
(b) 2E + A1 + A2
(c) 3A1 + 3A2
(d) E2 + 2A1
3. The character table of the C2V point group is given below:
C2V E C2 v v
A1
A2
B1
B2
1
1
1
1
1
1
-1
-1
1
-1
1
-1
1
-1
-1
1
1 = p1 + 2p2 + 2p3 + p4 2 = 2p1 - p2 - p3 + 2p4 (where pk is the p-
orbital on the kth atom of cis- v is the molecular plane) belong to
(a) A1 and A2 respectively (b) Both A2 (CSIR DECEMBER 2011, PART C)
(c) Both B2 (d) B1 and B2 respectively
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4. If the displacement vectors of all atoms in cis-butadiene are taken as the basis vectors, the
characters of the reducible representation of E, C2 v v are
(a) 30, 10, 30, 0 (b) 30, 0, 10, 0 (c) 30, 20, 0, 0 (d) 30, 0, 20, 0
5. A square pyramidal, MX4, molecule belongs to C4V point group. The symmetry operations
are; E, 2C4 v d. The trace for the reducible representation, when symmetry
operations of C4V applied to MX4, is (CSIR JUNE 2012, PART C)
(a) 51113 (b) 11111 (c) 51111 (d) 41113
6. The character table of the C2V point group is (CSIR JUNE 2012, PART C)
C2V E C2 v v
A1
A2
B1
B2
1
1
1
1
1
1
-1
-1
1
-1
1
-1
1
-1
-1
1
z
-
x
y
If the initial and final states belong to A1 and B1 irreducible representations respectively,
the allowed electronic transition from A1 to B1 is
(a) z-polarised (b) y-polarised (c) x-polarised (d) x, z-polarised
7. Among the following, the CORRECT statement is- (CSIR JUNE 2012, PART B)
(a) The number of irreducible representations is equal to classes of symmetry operations.
(b) The number of irreducible representations is equal to the order of symmetry point group
(c) The irreducible representations contained in any point group are always of one dimension
(d) A symmetry point group may not contain a totally symmetric irreducible representation
8. The number of IR active Vibrational modes of pyridine is-
C2V E C2 v v
A1
A2
B1
B2
1
1
1
1
1
1
-1
-1
1
-1
1
-1
1
-1
-1
1
Z
Rz
x, Ry
y, Rx
(a) 12 (b) 20 (c) 24 (d) 33 (CSIR December 2012, PART C)
9. A molecule contains the following symmetry operations: E, 2C6, 2C3, C2 d v. The
number of classes and order of the symmetry point group is: (CSIR June 2013, PART B)
(a) 3, 12 (b ) 5, 12 (c) 6, 12 (d) 6, 6
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10. Identify the Mulliken notation for the following irreducible representation
E Cn nC2 i h
1 1 -1 -1 -1
(CSIR June 2013, PART C)
(a) A′1u (b) A′′2u (c) B′2u (d) A′2u
11. Identify the point group symmetry of the following molecule (all C-C bond length are equal)
x
x
xx
(a) C2V (b) S4 (c) D2d (d) D4d (CSIR June 2013, PART C)
12. The transition that is allowed by x-polarised light in trans-butadiene is
(The character table for C2h is given below)
C2h E C2 i h
Ag
Bg
Au
Bu
1
1
1
1
1
-1
1
-1
1
1
-1
-1
1
-1
-1
1
Rx, x2,y2,z2,xy
Rx, Ry, xz,yz
z
x, y
(a) 1Au → 1Au (b) 1Au → 1Bg
(c) 1Bu → 1Bg (d) 3Bg → 1Ag (CSIR DECEMBER 2013, PART C)
13. The E E direct product in D point group contains the irreducible representationsX3
D3 E 2C3 3C2
A1
A2
E2
1 1 1
1 1 -1
1 - 1 0
(a) A1 + A2 + E (b) 2A1 + E (CSIR JUNE 2014, PART C)
(c) 2A2 + E (d) 2A1 + 2A2
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14. The character table of C2V point group is given below. In cis-butadiene molecule the
Vibrational modes belonging to A2 irreducible representation are IR inactive. The remaining
IR active modes are- (CSIR DECEMBER 2014, PART C)
C2V E C2 v v
A1
A2
B1
B2
1
1
1
1
1
1
-1
-1
1
-1
1
-1
1
-1
-1
1
z, x2, y2, z
Rz, xy
x, Ry, xz
y, Rx, yz
(a) 7A1 + 5B1 + 8B2 (b) 9A1 + 4B1 + 7B2 (c) 7A1 + 3B1 + 7B2 (d) 9A1 + 3B1 + 8B2
15. IR active normal modes of methane belong to the irreducible representation
Td E 8C3 3C2 6S4 d
A1 1 1 1 1 1 x2+y2+z2
A2 1 1 1 -1 -1
E 2 -1 2 0 0 2z2-x2-y2, x2-y2
T1 3 0 -1 1 -1 Rx, Ry, Rz
T2 3 0 -1 -1 1 x, y, z, xy, yz,zx
(a) E + A1 (b) E + A2 (CSIR JUNE 2015, PART C)
(c) T1 (d) T2
16. The irreducible representations of C2h are Ag, Bg, Au and Bu. The Raman active modes of
trans-1,3-butadiene belong to the irreducible representations
(a) Ag and Bg (b) Ag and Au (CSIR DECEMBER 2015, PART C)
(c) Au and Bg (d) Bg and Bu
17. The molecule diborane belongs to the symmetry point group
(a) C2V (b) C2h (CSIR DECEMBER 2015, PART C)
(c) D2d (d) D2h
18. The character table for the D3 point group is provided below:
D3 E 2C3 3C
A1 1 1 1 x2 + y2, z2
A2 1 1 -1 z, Rz
E 2 -2 0 x, y, (Rx, Ry) (x2 + y2, xy), xz, yz
For this point group, the correct statement among the following is:
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(a) it is possible to have a totally symmetric normal mode of vibration which is IR-active
(b) All IR-active normal modes are necessarily Raman-inactive
(c) All IR-active normal modes are necessarily IR-active
(d) It is possible to have a pair of IR-active normal modes that are degenerate
(CSIR JUNE 2016, PART C)
19. The character table for C3V point group is provided below, along with an additional
E 2C3 v
A1 1 1 1
A2 1 1 -1
E 2 -1 0
6 0 2
(CSIR JUNE 2016, PART C)
(a) A1 + A2 + 2E (b) 2A1 + 2E (c) 2A2 + 2E (d) 2A1 + 2A2 + E
20. For H2O molecule, the electronic transition from the ground state to an excited state of
B1 symmetry is (CSIR DECEMBER 2016, PART C)
C2V E C2 v v
A1 1 1 1 1 z, z2, x2, y2
A2 1 1 -1 -1 xy
B1 1 -1 1 -1 x, xz
B2
1 -1 -1 1 y, yz
(a) not allowed (b) allowed with x-polarisation
(c) allowed with y-polarisation (d) allowed with z-polarisation
21. The pair of symmetry point groups that are associated with only polar molecules is
(a) C2v, D∞h (b) C3v, C2h
(c) D2h, Td (d) C2v, C∞v (CSIR DECEMBER 2016, PART C)
22. For a point group, an incomplete character table is given below with one irreducible
representation missing (CSIR JUNE 2017, Part C)
E 2C3 3σv
A1 1 1 1
- - - -
E 2 -1 0
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The mulliken symbol and characters of the missing representation are
1. A1΄ 1 -1 1
2. B1 1 -1 -1
3. A2 1 1 -1
4. B2 1 -1 1
23. Given below is a specific vibrational mode of BCl3 with (+) and (-) denoting movements of
the respective atoms above and below the plane of the molecule respective atoms above
and below the plane of the molecule respectively. The irreducible representation of the
vibrational mode and its IR/ Raman activity are (CSIR JUNE 2017, Part C)
D3h E 2C3 3C2 σh 2S3 3σv
A1 1 1 1 1 1 1 x2+y2, z2
A2 1 1 -1 1 1 -1 Rz
E΄ 2 -1 0 2 -1 0 (x,y) (x2-y2, xy)
A1˝ 1 1 1 -1 -1 -1
A2˝ 1 1 -1 -1 -1 1 Z
E˝ 2 -1 0 -2 1 0 (Rx, Ry) (xz,yz)
(a) 𝐀𝟐′ ; neither IR nor Raman active (b) E΄ ; both IR and Raman active
(c) 𝐀𝟏′
; Raman active (d) 𝐀𝟐′′; IR active
24. A part of the character table of a point group (of order 4) is given below
E X1 X2 X3
Г𝟏 1 1 1 1
Г𝟐 1 -1 1 -1
Г𝟑 1 -1 -1 1
Г𝟒 ? ? ? ?
The four character of Г4 are, respectively (CSIR DEC 2017, Part C)
(a) 1, 1, -1, -1 (b) 2, 0, 0, 1 (c) 1, i, i, 1 (d) 1, -i, i, -1
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25. N2O molecule belongs to the point group (CSIR DEC 2018, Part B)
(a) D∞h (b) C∞v (c) C2v (d) S2
Considering the table 1 answer the following Q. 26. Q.27
26. The πu orbital of ethylene, when placed in the xy-plane with the C = C bond aligned to
the x-axis, transforms according to the irreducible representation (use table I)
(a) au (b) b1u (c) b2u (d) b3u (CSIR DEC 2018, Part C)
27. The b1u→ b2g transition in ethylene is (CSIR DEC 2018, Part C)
(a) not allowed. (b) allowed by x-polarized light.
(c) allowed by y-polarized light. (d) allowed by z-polarized light
28. The total number of symmetry elements in diborane molecule is
(a) 2 (b) 4 (c) 6 (d) 8 (CSIR JUNE 2019, Part B)
29. The 𝜋 orbital p1+ p2- p3- p4 of cis butadiene belongs to the irreducible representation:
𝐂𝟐𝐯 E 𝐂𝟐 𝛔𝐯 𝝈𝒗𝟏
A1 1 1 1 1
A2 1 1 -1 -1
B1 1 -1 1 -1
B2 1 -1 -1 1
(a) A1 (b) A2 (c) B1 (d) B2 (CSIR JUNE 2019, Part C)
30. The number of times the A1 representation appears in the representation Г of the C2v
point group given below is:
𝐂𝟐𝐯 E 𝐂𝟐 𝛔𝐯 𝝈𝒗𝟏
Г 3 1 1 3
(a) 1 (b) 2 (c) 3 (d) 4 (CSIR JUNE 2019, Part C)
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GENESIS TUTORIALS
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Assignment-3: Group Theory
1. The symmetry point group of the BF3 molecule is: (GATE-2001)
(a) C3v (b) D3h (c) C2v (d) D2h
2. The point group symmetry of the free nirate ion is: (GATE-2002)
(a) D3h (b) C3v (c) C3h (d) D3
3. List-I List-II (GATE-2003)
P. [Cr(H2O)6]3+ 1. C3v
Q. Fe2(CO)9 2. D3h
R. Eclipsed Ferrocene 3. Oh
4. D3d
5. D5h
6. D4d
(a) P—3, Q—2, R—5 (b) P—2, Q—4, R—1
(c) P—6, Q—2, R—5 (d) P—3, Q—6, R—4
4. The number and symmetry type of normal modes of vibration of H2O are (GATE-2004)
(a) 3 and 2A1 + B2 (b) 3 and 2A1 + A1
(c) 3 and 2A1 + B1 (d) 4 and 3A1 + B2
5. The orbital Ψ = lSHA− lSHB
of water belongs to the irreducibile representation
(a) A1 (b) B1 (c) A2 (d) B2 (GATE-2005)
6. A molecule has a 2–fold axis and a mirror plane pependicular to that. The point group
must have a (GATE-2006)
(a) C2 axis (b) centre of inversion (c) σh plane (d) σv plane
7. The symmetry elements that are present in BF3 are (GATE-2006)
(a) C3, σv, σh, 3C2 (b) C3, 3C2, S2, σv (c) C3, 3C2, σh, S2 (d) C3, σh, σv, i
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Common Data for Q.8, Q.9, Q.10:
Trans 1,2—difluoroethylene molecule has a 2–fold rotational axis, a symmetry plane
perpendicular to the rotational axis and an inversion centre.
8. The number of distinct symmetry operations that can be performed on the molecule is
(a) 2 (b) 4 (c) 6 (d) 8 (GATE-2007)
9. The number of irreducible representations of the point group of the molecule is:
(a) 1 (b) 2 (c) 3 (d) 4 (GATE-2007)
10. If two H atoms of the above molecule are also replaced by F atoms, the point group of the
resultant molecule will be- (GATE-2007)
(a) C1 (b) C2h (c) C2v (d) D2h
11. The point group of NSF3 is: (GATE-2008)
(a) D3d (b) C3h (c) D3h (d) C3v
12. The Td point group has 24 elements and 5 classes. Given that it has two 3–dimensional
irreducible representation, the number of one dimensional irreducible representation is:
(a) 1 (b) 2 (c) 0 (d) 3 (GATE-2008)
Common data for Q.13 and Q.14 (GATE-2009)
Character table for the point group C2v is given below:
13. The reducible representation corresponding to the three translational degrees of freedom
Гu, is:
(a) 3, 1, 1, 1 (b) 3, -1, 1, 1 (c) 3, -1, -1, -1 (d) 3, 1, -1, -1
14. The asymmetric stretching mode of the H2O is shown below. The molecular plane is yz
and the symmetry axis of H2O is z.
This vibration transferms as the irreducible representation
C2V E C2 y (xz) y(yz)
A1 1 1 1 1 z x2, y2, z2
A2 1 1 -1 -1 Rz Xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 1 y, Rx yz
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(a) A1 (b) B1 (c) A2 (d) B2
15. [CoCl4]2− is a blue coloured complex. Controlled-treatment of this complex with water
generates two isomeric light pink coloure complexes of composition [Co(H2O)4Cl2].
Identify the correct point groups for [CoCl4]2− and two isomeric complexes [Co(H2O)4Cl2].
(a) D2h and (C2v and C2h) (b) Td and (C2v and D4h)
(c) D4h and (C2v and D4h) (d) Td and (C2v and C4v) (GATE-2010)
16. The point group of ClF3 molecule and its corresponding number of irreducible
representation are respectively. (GATE-2010)
(a) C3v and 4 (b) C2v and 4 (c) C3v and 3 (d) C2v and 3
17. The point group symmetry of the given planar shape is: (GATE-2011)
(a) D3h (b) C3 (c) C3h (d) C3v
18. Symmetry operations of the four C2 axes perpendicular to the principle axis belong to the
same class in the point group (s) (GATE-2012)
(a) D4 (b) D4d (c) D4h (d) D4h and D4d
19. The point group symmetry of CH2 = C = CH2 is: (GATE-2013)
(a) D2h (b) C2h (c) C2v (d) D2d
20. The number of C2 axes in CCl4 is_______ (GATE-2014)
21. The point group of IF7 is (GATE-2015)
(a) D6h (b) D5h (c) C6v (d) C5v
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22. Character table of point group D8 is given below (GATE 2019)
The value of
(a + b + c + d + e + f + g + h + I + j + k) is equal to ___
𝐃𝟖 E 𝟐𝐂𝟖 𝟐𝐂𝟒 𝟐𝐂𝟖𝟑 𝐂𝟐 𝟒𝐂𝟐
′ 𝟒𝐂𝟐′′
𝐀𝟏 a 1 1 1 1 1 1
𝐀𝟐 b 1 1 1 1 h i
𝐁𝟏 c -1 1 -1 1 1 j
𝐁𝟐 d -1 1 -1 1 -1 1
𝐄𝟏 e √2 0 −√2 -2 0 0
𝐄𝟐 f 0 -2 0 k 0 0
𝐄𝟑 g −√2 0 √2 -2 0 0
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GENESIS TUTORIALS
Institute for CSIR-UGC-NET/JRF, GATE & IIT-JAM
Assignment-4: Group Theory
1. Below is a diagram by Maurits Cornelis Escher (1898-1972), a Dutch graphic artist. What
elements of symmetry are present in this Escher diagram generated by tessellation of fish
images? (TIFR 2011)
a) no symmetry
b) 2 fold rotation axis perpendicular to plane of paper
c) mirror plane and 2-fold rotation axis perpendicular to plane of paper
d) mirror plane
2. Point group of 1, 2-propadiene is (TIFR-2013)
(a) C2h (b) C2v (c) D2h (d) D2d
3. Write down the matrix representing a two-step transformation of a general point (x, y, z):
rotation through 180° (about the z-axis) followed by reflection in an yz mirror plane
(a) [1 0 00 −1 00 0 1
] (b) [−1 0 00 1 00 0 1
] (TIFR-2014)
(b) [1 0 00 −1 00 0 −1
] (d) [−1 0 00 1 00 0 −1
]
4. What is the point group of Fe2(CO)9? (TIFR-2015)
(a) C3h (b) D3h (c) C3v (d) D3d
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5. Show below are the front and side views of the structure a molybdenyum-based metal
organic polygon. What is the symmetry of the molecule? [TIFR 2016]
(a) D4h (b) C4h (c) C2v (d) C4
6. Which of the following belong to the same symmetry group as NH3? [TIFR 2018]
(a) BF3 (b) CH4 (c) CH3OH (d) CHCl3
7. Below is a picture of an Origami pinwheel. What elements of symmetry are present in the
pinwheel? [TIFR 2019]
(a) Only one 4-fold rotation axis perpendicular to the plane of the pinwheel
(b) A mirror plane perpendicular to the plane of the paper and a 2-fold axis of rotation
perpendicular to the plane of the pinwheel.
(c) 4 fold rotation axis perpendicular to the plane of the pinwheel and an inversion centre.
(d) 2 mirror planes perpendicular to the plane of the paper