Past exam of the chemistry A course of the University of Cambridge / 2025 / Paper 1 / A3 / a

For one normal mode, the fundamental is $v=0\to 1$, with other modes initially in their ground levels. In the linear-dipole harmonic approximation, the transition moment is nonzero for $\Delta v=\pm 1$, so a vibrational fundamental is normally much stronger than its overtones, provided the mode has vibrational infrared activity.

The first vibrational overtone is $v=0\to 2$, with $\Delta v=2$. It lies near twice the fundamental wavenumber, generally slightly below it because of anharmonicity. The linear-dipole harmonic model forbids it; anharmonic wavefunctions or higher terms in the dipole expansion give a weak intensity. Thus it is generally weaker than the corresponding fundamental.

The second vibrational overtone is $v=0\to 3$, with $\Delta v=3$, near three times the fundamental wavenumber. It requires higher-order departures from the linear-dipole harmonic model and is ordinarily weaker again. These are qualitative intensity trends; resonances can redistribute intensity between nearby vibrational states.

A hot band starts from a thermally populated vibrationally excited state, for example $v=1\to 2$. It can satisfy $\Delta v=1$ and need not have a small transition moment, but its lower-state population is reduced by the Boltzmann distribution:It is therefore usually weak at low temperature and grows relative to the fundamental on heating. Anharmonicity shifts its frequency away from the $0\to 1$ transition; in a polyatomic molecule, excitation in another mode can also produce a hot-band sequence.