Group Theory - Genesis Tutorials

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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




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.




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




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




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




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




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




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




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




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




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




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




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




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




C2V E 2C3 v3

A1 1 1 1

A2 1 1 -1

E 2 -1 0




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




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




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




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




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




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




' ' "

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




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


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





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





GENESIS TUTORIALS


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





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





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





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:





(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





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





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)





GENESIS TUTORIALS



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





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





(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





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





GENESIS TUTORIALS



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





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


Group Theory - Genesis Tutorials

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