Normal modes of vibration of
a XY3 pyramidal molecule
by
Dr.D.UTHRA
Head
Department of Physics
DG Vaishnav College
Chennai-106
For XY3 pyramidal molecule,
Calculate normal modes of
vibration by all three methods
Using 3N-6 formula
For an XY3 pyramidal molecule
N=4 ;
3N-6 = 6 modes of vibration
Using internal coordinates
In XY3 pyramidal molecule,
b = 3 (no.of bonds)
 a = 4 (no.of atoms)
 a1= 3 (no.of atoms with multiplicity one)
Hence,

nr = 3 ,i.e three stretching vibrations
 nΦ = 4*3-3*4+3 =3, i.e three bending

vibrations

nτ = 3-3 = 0
3N-6 = nr+nΦ+nτ = 3+3= 6
Using Character table
XY3 pyramidal molecule belongs to
C3v point group
From character table,
h - order of the point group = 6
ℓ - dimension of species A1 and A2 = 1
dimension of species E = 2
C3v
E
2C3
3σv
A1
1
1
1
A2
1
1
-1
E
2
-1
0
To calculate χj' (R)
C3v
E
2C3
3σv
A1
1
1
1
A2
E
1
1
-1
2
-1
0
NR
4
1
2
Θ(in degrees)
0
120
0
2cosθ
2
-1
2
(1+2cosθ)
3
0
-
(-1+2cosθ)
-
-
1
(NR-2)*(1+2cosθ)
6
0
-
NR*(-1+2cosθ)
-
-
2
ni'= (1/h)∑R ℓΨ (R)χ' (R)
h - order of the C3v point group = 6
ℓ - dimension of species A
and A2 =1
dimension of species E = 2
Ψj(R) - from the character table
1
χ' (E) = 6; χ' (C3) = 0; χ' (σv) = 2
To find ni‘(A1)
Ψ (E) = 1; Ψ (C3) = 1; Ψ (σv) = 1
To find ni‘(A2)
Ψ (E) = 1; Ψ (C3) = 1; Ψ (σv) = -1
To find ni‘(E)
Ψ (E) = 2; Ψ (C3) = -1; Ψ (σv) = 0
ni‘(A1)= (1/6) [ 1*1*6 + 1*1*0*2 + 1*1*2*3] =(1/6)*12=2
ni‘(A2)= (1/6) [ 1*1*6 + 1*1*0 *2+ 1*-1*2*3 ]=(1/6)*0=0
ni‘(E)= (1/6) [ 2*2*6 + 2*-1*0*2 + 2*0*2*3]=(1/6)*24=2
ni‘ = 2 A1+ 2 E = 2+ (2*2) = 6
Note : Remember you have two C3 operations and three σv operations
E is a doubly degenerate species
Enjoy learning!
-uthra mam