Question

8. Calculate the frequency of transitions using the following informations: Asymmetric, bending and symmetric modes vi, v2, and va for hydrogen eyanide, HCN molecule are 3311 cm1, 713 em' and 2097 cm respectively. a) Wavenumber of the transition 000-040 b) Wavenumber of the transition: 111222 c) wavenumber of the transition 011 → 022 d) Wavenumber of the transition 211 132 e) Wavenumber of the transition 000001

Answer 1

Answer:

$\nu$₁ = 3311cm^-1$\nu$₂ = 713 cm^-1$\nu$₃ = 2097 cm^-1

energy required for going from a,b,c to x,y,z

= ( E(x,y,z) - E(a,b,c) )

= h$\nu$₁(x + 1/2) + h$\nu$₂(y + 1/2) + h$\nu$₃(z + 1/2) - ( h$\nu$₁(a + 1/2) + h$\nu$₂ (b + 1/2) + h$\nu$₃(z + 1/2) )

= h ($\nu$₁(x - a) + $\nu$₂(y - b) + $\nu$₃(z - c))

So the wavenumbers corrsponding to above energy = E/h = $\nu$₁(x - a) + $\nu$₂(y - b) + $\nu$₃(z - c)

So the answer for the above questions is

a) 000 --> 040

E( in cm^-1) = ($\nu$₁(0 - 0) + $\nu$₂(4 - 0) + $\nu$₃(0 - 0)) = 4$\nu$₂ = 4 x 713 cm^-1 = 2852 cm^-1

b) 111 --> 222

E( in cm^-1) = ($\nu$₁(2 - 1) + $\nu$₂(2 - 1) + $\nu$₃(2 - 1)) = $\nu$₁ + $\nu$₂ + $\nu$₃ = (3311 + 713 + 2097) cm^-1 = 6121 cm^-1

c) 011 --> 022

E( in cm^-1) = ($\nu$₁(0 - 0) + $\nu$₂(0 - 0) + $\nu$₃(2 - 1)) = $\nu$₂ + $\nu$₃ = (713 + 2097) cm^-1 = 2810 cm^-1

d) 211 --> 132

E( in cm^-1) = ($\nu$₁(1 - 2) + $\nu$₂(3 - 1) + $\nu$₃(2 - 1)) = -$\nu$₁ + 2$\nu$₂ + $\nu$₃ = (-3311 + (2 x 713) + 2097) cm^-1 = 212 cm^-1

d) 000 --> 001

E( in cm^-1) = ($\nu$₁(0 - 0) + $\nu$₂(0 - 0) + $\nu$₃(1 - 0)) = 0$\nu$₁ + 0$\nu$₂ + $\nu$₃ = $\nu$₃ = 2097 cm^-1