Past exam of the chemistry A course of the University of Cambridge / 2025 / Paper 1 / A3 / d / i / Solution

This is a rovibrational spectrum: the vibrational excitation is accompanied by quantized molecular rotation. The high-wavenumber R branch has $\Delta J=+1$, and the low-wavenumber P branch has $\Delta J=-1$. In the simplest common-$B$ approximation,Consecutive lines within either branch are therefore separated by roughly $2B$, while the gap between P(1) and R(0) is roughly $4B$. There is no Q branch: for this diatomic parallel $\Sigma \to \Sigma$ transition, the electric-dipole rotational selection rule allows $\Delta J=\pm 1$, and the $\Delta J=0$ line strength vanishes.

The intensity envelope comes from rotational populations, approximately $(2J+1)e^{-hcBJ(J+1)/(k_{B}T)}$, together with the rotational line strengths. The paired peaks arise from the two chlorine isotopologues, ${\mathrm{H}}^{35}Cl$ and ${\mathrm{H}}^{37}Cl$: their different reduced masses shift both rotational and vibrational spacings. The more abundant ${}^{35}\mathrm{Cl}$ gives the stronger member. The changing spacing away from the band centre reflects $B_1\ne B_0$ and, at higher $J$, centrifugal distortion; it is not evidence for additional vibrational fundamentals.