Question

The force constant of 14^N^16O is 1550 N m^-1: (i) Predict its bond order. (ii) Calculate the reduced mass. (iii) What is the fundamental frequency in cm^-1? (iv) What will be the fundamental frequency if the O atom is isotopically labeled as 18^O?

Answer 1

Nitrogen forms : N ≡ N ,   Oxygen forms:  O = O
N: 2s2 2p3     O: 2s2 2p4

Bond order   for  ¹⁴N - ¹⁶O  bond:
    = (number of bonding electrons - number of antibonding electrons)/2
    = [ (2 from N +4 from O) -  1 from N ] / 2 = 2.5

m1 = 14u        m2 = 16u
reduced mass = μ = m1 * m2 / (m1 + m2)
                              = 112/15 = 7.47 u
   μ = 7.47 * 0.001 kg / 6.023 * 10²³  = 1.239 * 10⁻²⁶ kg

k = force constant = 1150 N/m
c = speed of light = 3 * 10¹⁰ cm/s

The fundamental frequency of oscillation of N & O atoms about the center of mass is given by:   f = 1/2π * √(k/μ)      in  Hz
 
                 f = 1/(2π c) * √(k/μ)    in  cm⁻¹    = wavenumber of the oscillation
                   = 1 876 cm⁻¹

If O¹⁸ isotope is used in NO bond, then:
         μ = 63/8 amu = 1.307 * 10⁻²⁶ kg
         f = 1.826  cm⁻¹   (wave number of vibration)