Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2021 / Paper 2A / Section A / 14

The Stokes vibrational band has O, Q and S branches, with $\Delta N=-2,0,+2$, respectively. Here $N$ denotes end-over-end rotational angular momentum; treating electronic fine structure as unresolved gives the usual elementary rotor calculation. The absolute incident wavenumber isand energy conservation for Stokes Raman scattering givesFor the actual ${}^{16}{\mathrm{O}}_2$ ground state, only odd $N$ have nonzero nuclear-spin statistical weight. Thus the lowest initial levels are $N=3$ for O and $N=1$ for Q and S. Ignoring the small change in $B$ upon vibration, the lines are:

The odd-$N$ restriction follows from the zero nuclear spin and exchange symmetry of ${}^{16}{\mathrm{O}}_2$; the electronic spin splits each such $N$ into several $J$ levels without changing this simplified branch-centre calculation.^[ref]

If the phrase simplest approximation is instead taken to discard nuclear-spin statistics as well, a fictitious unrestricted rotor gives O $2\to 0$ at $7826.9{\mathrm{cm}}^{-1}$ and S $0\to 2$ at $7810.1{\mathrm{cm}}^{-1}$, with Q still at $7818.5{\mathrm{cm}}^{-1}$. Those even-$N$ lines are not physical ground-state ${}^{16}{\mathrm{O}}_2$ lines.