E = Dₑq { 1−exp [a (rₑq−r)] }¹/² ...........(1)
Where 'a' is a constant for a particular molecule and Dₑq is dissociation energy, when equation (1) is used in the Schrodinger equation, the vibrational energy expression is obtained as ;
Eᵥ = (v+1/2) hcɯ̄ₑxₑ Joules................... (2)
∈ = (v+1/2)²ɯ̄ₑxₑ cm⁻¹ .....................(3)
Where ω̄ₑ is equilibrium vibrational frequency of the anharmonic oscillator and anharmonicity Xₑ is small and positive. Hence vibrational levels will become closer with increasing 'v'. Some of the vibrational energy levels shown figure. Equation (3) becomes,
∈ᵥ = ω̄ₑ { 1−xₑ (v+1/2)} (v+1/2) ............ (4)
Compare with equation, we can write ;
ω̂ₒₛ꜀ = ω̄ₑ { 1− xₑ (v+1/2)} ....................(5)
An anhormonic oscillator behaving like a harmonic oscillator but with an oscillation frequency which decreasing with increasing 'v'. According theselection rule for anharmonic oscillator is,
∆v = ±1, ±2, ±3,...............etc.
But in practice the lines corresponding to
∆v = ±1, ±2, ±3, are observed and the other lines and sharply decrease in their intensities. In addition, we take the population of the energy levels into account which is governed by Boltzmann equation, given by
Nᵤ = exp ⁻ʰᶜω̄/kT
Nl
If v= 1 for upper state and v= 0 lower state, then N ᵤ ≈ 0.008.
Nl
Thus the transition except from v=0 state. Consider three following transitions.
(1) v = 0 →v = 1, ∆v = +1, with considerable intensity.
∆∈ = ∈ᵥ₌₁ ₋ ∈ᵥ₌₀
= (1+/2) ω̄ₑ - xₑ(1+1/2)² ɯ̄− {1/2ɯₑ−(1/2)²xₑ ω̄ₑ }
= ω̄ₑ (1−2xₑ) cm⁻¹
(2) v = 0 →v =2, ∆v = +2, with small intensity.
∆∈ = (2+1/2) ω̄ₑ −xₑ (2+1/2)² ω̄ₑ − {1/2ω̄ₑ− (1/2)²xₑ ω̄ₑ }
= 2ω̄ₑ (1−3xₑ) cm⁻¹
(3) v = 0 →v = 3, ∆v = ±3, with normally negligible intensity.
∆∈ = (3+1/2) ω̄ₑ − xₑ (3+1/2)² ω̄ₑ − { 1/2 ω̄ₑ − (1/2)² ω̄ₑ xₑ
= 3 ω̄ₑ (1-4xₑ) cm⁻¹
These three transition are shown figure. To a good approximately since xₑ = 0.01. The three spectral lines lie very close to ω̄, 2ω̄ and 3ω̄. The line near ωₑ is called fundamental absorption, while those near 2ω̄ and 3ω̄ are called the first and second overtones respectively. The spectrum of Hydrochloric acid shows a very intense absorption at 2886 cm⁻¹, a weaker one at 5668 cm⁻¹ and a very weak one at 8347 cm⁻¹. From these data we can find equilibrium frequency and anharmonicity constant by solving the following equations.
ω̄ₑ = (1-2xₑ) = 2886 cm⁻¹
2ω̄ₑ = (1-3xₑ) = 5668 cm⁻¹
3ω̄ₑ = (1-4xₑ) = 8347 cm⁻¹
Solving these equation, we obtain ω̄ₑ = 2990 cm⁻¹ and xₑ = 0.0174. The force constant the bond HCl can be calculated from the following equation by the substituting the values reduced mass other constants.
k = 4π²c²ω̄ₑ² μ Nm⁻¹ = 516 Nm⁻¹
If the temperature is raised or if the vibration has particularly low frequency the population of the v = 1 may become appreciable. The transition from v = 1 to v = 2, ∆v = 1, is observation is normally weak. ∆∈ = ω̄ₑ (1- 4xₑ) cm⁻¹. Such weak bonds are called hot bonds are temperature is the one condition for their occurrence.
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