There is a "famous" rule in spectroscopy,1 that goes this way:
If a compound is centrosymmetric, then its normal vibrational modes cannot be simultaneously IR and Raman active.
and this is simple enough to prove: for a vibrational mode to be IR active, it has to transform under the same irrep as (one of) the cartesian axes $x$, $y$ or $z$, which are all ungerade. For a vibrational mode to be Raman active, it has to transform under the same irrep as one of $(x^2, y^2, z^2, xy, yz, xz)$, and since $u \otimes u = g$, these irreps cannot be the same as the irreps of $(x,y,z)$. Hence, a vibrational mode, which transforms under a single irrep, cannot simultaneously satisfy both conditions at once.
However, I am curious about the converse statement.
If a compound does not have any vibrational modes which are simultaneously IR and Raman active, is it necessarily centrosymmetric?
If the answer is yes, I would like a proof; if it is no, a counterexample, please.
(Just in case anybody is confused, because I am: A counterexample would be a molecule without a centre of inversion, which has no vibrational mode which is simultaneously IR and Raman active.)
1 Famous enough for me to know, at least.