Discussion
It is given that the electronic ground state of a diatomic molecule \(X_{2}\) has even parity and the nuclear spin of \(X\) is 0. Which one of the following is the CORRECT statement with regard to the rotational Raman spectrum (\(J\) is the rotational quantum number) of this molecule?
Answer: C
Only lines of even \(J\) values are present.
The nuclei have spin 0, so they are bosons and the total molecular wavefunction must be symmetric when the two nuclei are exchanged. With nuclear spin 0 the nuclear-spin part is symmetric, and the electronic ground state is even, so the rotational part must be symmetric too. The rotational wavefunction has exchange symmetry \((-1)^{J}\), so only even \(J\) levels exist.
Rotational Raman lines obey \(\Delta J = 0, \pm 2\), which links even levels only to even levels, so every line comes from an even \(J\) and the odd-\(J\) lines are missing altogether, as in the Raman spectrum of CO2, whose oxygen nuclei also have spin 0.
Lines of all \(J\) would need a heteronuclear molecule. The 3 : 1 intensity alternation belongs to spin-1/2 nuclei such as in H2. Only odd \(J\) would require the opposite electronic or nuclear symmetry.
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