Molecular Vibration Model 1: The Classical Harmonic Oscillator 1. The vibrational frequency is inversely proportional to 𝜇. Replacing an atom by a heavier mass acts to

decrease the vibrational frequency. Bonds involving heavy atoms have lower vibrational frequencies. Bonds involving light atoms – particularly hydrogen – have higher vibrational frequencies.

2. The vibrational frequency is proportional to 𝑘. If the force constant is replaced by a larger value, the vibrational frequency will increase. Stronger bonds have higher vibrational frequencies. Weaker bonds have lower vibrational frequencies.

3. The reduced mass, µ, is given by !!  ×  !!!!!  !!

. If m1 = m2 = m:

µ = !  ×  !!!  !

= !!

!! =  !

!

4. Using the result from question 5, µ(H2) = ½ g mol-1 and µ(D2) = 2/2 g mol-1 = 1 g mol-1. Hence:

!!!!!!

= !!/!

= 2 and !!!!!!

= !/!!

= 1/2

5. The reduced mass of HX becomes increasingly close to 1 g mol-1 as the mass of X increases. For a molecule AB where the mass of A is much smaller than the mass of B, the reduced mass tends towards the smaller mass:

µ = !!  ×  !!!!!  !!

= !!  ×  !!!!

if m2 >> m1. Hence, µ →  𝑚1  ×  𝑚2𝑚2

 =  m1

Model 2: The Quantum Harmonic Oscillator 1. (a) The Born interpretation is that ψ2(x) is proportional to the probability of finding the system at x.

The wavefunction for zero energy would be zero and so ψ2(x) would be zero for all x. This solution is not useful as it the system must be somewhere.

(b) The minimum energy corresponds to ν = 0. This has energy:

ε0 = (0 + ½) ħ!! = ½ ħ

!!

2. The energy separation between levels v and v + 1 is:

Δε = εν + 1 − εν

= [(ν + 1 + ½) ħ!!

] - [(ν + ½) ħ!!

] = ħ!!

The separation of the energy levels is constant and is equal to the classical vibrational frequency. 3. The probability is symmetrical about x = 0 for all harmonic wavefunctions. The average value of x is

zero meaning that the average value of the bond length is the same in all vibrational levels and is equal to the equilibrium bond length.

Molecular Vibration

4. The zero point energy is given by ε0 ½ ħ!!

. The force constant for H2

and D2 are very similar but the zero point energies differ because of the different reduced masses. The ratio of their zero point energies is given by:

!�!!!�!!

= !!!!!!

 !!!!!!

    = !!!!!!!!!!!!

  = !!!!!!

 

= !!   or 𝜀�!! = 0.707 𝜀�!!

where the result from question 6 that 𝜇!! = 2 𝜇!!has been used.

The separation of the energy levels, from question 9, similarly depends on the reduced masses:

!!�!!!!�!!

= !!!!!!

 !!!!!!

    = !!!!!!!!!!!!

  = !!!!!!

  = !!   or Δ𝜀�!! = 0.707 Δ𝜀�!!

13. (a)     At high positive values of x, corresponding to increasing elongation of the bond, a real bond breaks so the real potential energy curve slopes. The harmonic function incorrectly suggests the bond can carry on elongating forever.

At high negative values of x, corresponding to increasing compression, the real potential grows more rapidly than suggested by the harmonic potential. As the bond becomes very compressed, the nuclei get very close and the energy rises extremely quickly.

(b)     The real potential is no longer symmetrical due to the changes outlined in (a). The average bond length gets larger and larger as the energy rises, increasing to infinity for the dissociated molecule. The red line above shows, approximately, the result.

Model 3 The Anharmonic Oscillator 1. (a) G(0) = (0 + ½)ωe - (0 + ½)2ωexe = 1/2ωe – 1/4ωexe

(b) G(1) = (1 + ½)ωe - (1 + ½)2ωexe = 3/2ωe – 9/4ωexe

(c) G(2) = (2 + ½)ωe - (2 + ½)2ωexe = 5/2ωe – 25/4ωexe (d) G(3) = (3 + ½)ωe - (3 + ½)2ωexe = 7/2ωe – 49/4ωexe

2. (a) Fundamental: G(1) - G(0) = [3/2ωe – 9/4ωexe] – [1/2ωe – 1/4ωexe] = ωe – 2ωexe

(b) First overtone: G(2) - G(0) = [5/2ωe – 25/4ωexe] – [1/2ωe – 1/4ωexe] = 2ωe – 6ωexe (c) Second overtone: G(3) - G(0) = [7/2ωe – 49/4ωexe] – [1/2ωe – 1/4ωexe] = 3ωe – 12ωexe

Molecular Vibration

3. The two equations are:

ωe – 2ωexe = 4028.3 cm-1 (i) 2ωe – 6ωexe = 7880.0 cm-1 (ii) Taking (ii) – 2 × (i) gives: [2ωe – 6ωexe] – 2 × [ωe – 2ωexe] = (7880.0 - 2 × 4028.3) cm-1 -2ωexe = -176.6 cm-1 ωexe = 88.3 cm-1 Substituting this into (i) gives: ωe – 2ωexe = ωe – 2 × 88.3 cm-1 = 4028.3 cm-1 ωe = 4024.9 cm-1 4. The dissociation energy from the bottom of the potential well, De, is given by:

De = !!!

!!!!! = (!"#!.!)

!

!(!!.!) = 50060.0 cm-1

From Q1(a), The zero point energy, G(0), is: G(0) = 1/2ωe – 1/4ωexe = (1/2 × 4024.9) – (1/4 × 88.3) = 2080.4 cm-1

The dissociation energy, D0, is therefore

D0 = De – G(0) = !!!

!!!!! – G(0) = (50060.0) – (2080.4) cm-1 = 47980 cm-1