Normal modes of vibration of a XY 3 pyramidal molecule by Dr.D.UTHRA Head Department of Physics DG Vaishnav College Chennai-106

For XY 3 pyramidal molecule, Calculate normal modes of vibration by all three methods

Using 3N-6 formula For an XY 3 pyramidal molecule N = 4 ; 3N-6 = 6 modes of vibration

Using internal coordinates In XY 3 pyramidal molecule, b = 3 (no.of bonds) a = 4 (no.of atoms) a 1 = 3 (no.of atoms with multiplicity one) Hence,  n r = 3, i.e three stretching vibrations  n Φ = 4*3-3*4+3 =3, i.e three bending vibrations  n τ = 3-3 = 0 3N-6 = n r +n Φ +n τ = 3+3= 6

Using Character table XY 3 pyramidal molecule belongs to C 3v point group From character table, h - order of the point group = 6 ℓ - dimension of species A 1 and A 2 = 1 dimension of species E = 2 C 3v E 2C 3 3σv3σv A1A1 111 A2A2 11 E2 0

To calculate χ j ' (R) C 3v E 2C 3 3σv3σv A1A1 111 A2A2 11 E 2 0 NRNR 412 Θ(in degrees) cosθ 22 (1+2cosθ) 30- (-1+2cosθ) --1 (N R -2)*(1+2cosθ) 60- N R *(-1+2cosθ) --2

n i '= (1/h)∑ R ℓΨ (R)χ ' (R) h - order of the C 3v point group = 6 ℓ - dimension of species A 1 and A 2 =1 dimension of species E = 2 Ψ j (R) - from the character table χ ' (E) = 6; χ ' ( C 3 ) = 0; χ ' ( σ v ) = 2 To find n i ‘( A 1 ) Ψ (E) = 1; Ψ ( C 3 ) = 1; Ψ ( σ v ) = 1 To find n i ‘( A 2 ) Ψ (E) = 1; Ψ ( C 3 ) = 1; Ψ ( σ v ) = -1 To find n i ‘( E) Ψ (E) = 2; Ψ ( C 3 ) = -1; Ψ ( σ v ) = 0 n i ‘(A 1 )= (1/6) [ 1*1*6 + 1*1*0*2 + 1*1*2*3] =(1/6)*12=2 n i ‘(A 2 )= (1/6) [ 1*1*6 + 1*1*0 *2+ 1*-1*2*3 ]=(1/6)*0=0 n i ‘(E)= (1/6) [ 2*2*6 + 2*-1*0*2 + 2*0*2*3]=(1/6)*24=2 Note : Remember you have two C 3 operations and three σ v operations E is a doubly degenerate species n i ‘ = 2 A E = 2+ (2*2) = 6

Enjoy learning! -uthra mam