Symmetry and Group Theory (CHMS3202)
BSc, Chemistry - CHMS
Semester: Second Semester
Level: 300
Year: 2018
UNIVERSITY OF BAMENDA FACULTY OF SCIENCE
SECOND SEMESTER EXAMINATIONS 2017-2018
Department: CHEMISTRY Course instructors: Paul Chongwain
Month: July, 2018 Course code: CHMS 3202 CREDIT VALUE: 04
TIME ALLOWED: 3 HOURS Course title: Symmetry and Group theory in Chemistry
Instructions: Attempt all Questions.
Q1) Define (i) Symmetry element, (ii) Symmetry operation
(b) Identify clearly the symmetry elements and their operations in the molecule 1,3,5-trichlorobenzene. The
molecule is modified such that the chlorine atom in position five (-5) is replaced by a hydrogen atom. What
becomes of the symmetry of this modified molecule compared to the parent compound?
Deduce the point groups for these molecules. 18 mks
Q2. What do you understand by (i) representations (ii) the character of a representation? Show that the
following set of matrices F, G, H and L form a group under the C2V point group .(Use the symbols as given).
1 0 0 -1 0 0 1 0 0 -1 0 0
F =
0 1 0
G=
0
-1
0
H=
0
-I
0
L =
0 1 0
0 0 1
0 0 1
0 0 1
0 0 1
Obtain the multiplication table for this group. 17 mks
Q3. The ammonia molecule belongs to the point group C
3V
(character table provided). Using a Cartesian basis
for representation or any suitable basis, determine: (i) all the vibrational modes associated with this molecule,
(ii) Identify the Infra-red and (iii) Raman active modes of vibration. 15 mks
C
3V
Character table
C
3V
E 2C
3
36
v
A
1
1 1 1 z
x
2
+y
2
, x
2
A
2
1 1 -1
Rz
E
2 -1 0
(x, y)(Rx, R
Y
) (x
2
- y
2
, xy)(xz, yz)
Q4) (a) Give the answers to the following matrix transformations
(i)
6 7 8
9
10
11
(ii)
4
1 7
6 −3
3 2
4 7
(iii)
6
7
(iv)
√
√
0
0 0 1
√
−
√
0
∗
1 2 3
2 1 3
3 3 9
∗
√
0
√
√
0 −
√
0 1 0
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(b)The secular determinant for cyclopropene is given below:
1 1
1 1
1 1
=0 X= (α-E)/(β-SE.
(i) Define the symbols α, β, S.
(ii) Determine the value of E for the roots of this determinantal equation,
(iii) Arrange the energies in increasing order.
(iv) Calculate the total energy of the occupied pie molecular orbitals. 20mks
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