The molecule as an anharmonic Oscillator: A comparison of the observed infrared spectrum with that expected from a diatomic molecule is treated as a harmonic oscillator shows an important disagreement.
The molecule as an anharmonic Oscillator
The harmonic oscilator would give a single band at ω(cm^-1) which is the classical frequency of vibration of the molecule.
The actual infrared spectrum is however found to consists of an intense band (fundamental band) at ω plus a number of the weak band (overtones) at slightly lesser & lesser than 2ω, 3ω, …..
The observed overtones indicate that the selection rule Δv = ±1 is not strictly obeyed.
The dipole moment of the molecule is not strictly linear with respect to internuclear displacement.
This is expressed as electrical and ‘anharmonicity of the molecule’.
The observations that the overtone appears not exactly at 2ω, 3ω, ….. but a lesser than it, shows that the vibrational energy levels are not equally spaced but converse slowly.
This is due to the fact that, “the potential energy of the molecule is not strictly linear” but it is given by :
V(x) = gx^2 – fx^3
where g << f
On Substituting this value of V(x) in Schrodinger equation & solving by Perturbation method, we got the vibrational energy of anharmonic oscillator, is given by :
The quantity ωe is the spacing of energy levels, Xe is the anharmonicity constant, which is much smaller than ωe & is always positive.
This equation indicates that energy levels in anharmonic oscillator are not equidistant but the separation decreases slowly with increasing v.
When the molecule receives energy more than that corresponding to uppermost vibrational level, it dissociates into atoms and the excess energy appears as kinetic energy of these atoms. Hence a continuum joins the uppermost level.
The zero point energy of the anharmonic oscillator is obtained by putting v = 0 in eq. (ii) Thus,
The observed wave number for the transition :
Thus, observed absorption frequency for ground state is equal to ωo.