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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.

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    $\begingroup$ I just went through a number of common molecule point groups and in all of them that were note inversion-symmetric there was at least one irreducible representation that included a simple translational and a product mode. Looks like if there is a counterexmple it will be an extremely rare case. $\endgroup$ – Jan Sep 30 '16 at 16:10
  • $\begingroup$ Have a look at the point groups O (not $\ce{O_h}$) and $\ce{D_{5h}, C_{5h}}$. I will try and find a pic of O point group molecules. $\endgroup$ – porphyrin Sep 30 '16 at 20:38
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Have a look at the point groups O (not $\ce{O_h}$) I, $\ce{D_{5h} and C_{5h}}$. A pic of an O point group molecule is shown below. The $\ce{c_2}$ axes are not shown. Taken from molecule-viewer.com web site. The molecule is 'vanadium hexaoxo phosphonato' or more properly $\ce{C8H24O30P8V^+Cl^-}$, octakis(μ 3-methylphosphonato)-hexaoxo-penta- vanadium(v)-vanadium(iv) chloride clathrate !!!

vanadium O point group

Edit: I checked a bit more and found ; $\ce{ S8 }$

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  • $\begingroup$ One reference says "Number of Vibrational Active IR Bands: Only Rx, Ry, Rz, x, y, and z can be ir active" which would seem to rule out D5h and C5h. chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/… another reference oppositely says "is Raman active if the symmetry is the same as x2, y2, z2, or one of the rotational functions Rx, Ry, Rz" lasalle.edu/~prushan/IC-articles/lecture%204.pdf which would exclude O and I. $\endgroup$ – DavePhD Sep 30 '16 at 22:03
  • $\begingroup$ D5h, C5h, O and I are all fair counter examples if OP's statement "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 (x2 ,y2 ,z 2 ,xy,yz,xz)" is completely right, but not if one of those references is correct. $\endgroup$ – DavePhD Sep 30 '16 at 22:08
  • $\begingroup$ hi, Rx, Ry Rz do not refer to vibrations but to 'rotational' operators such as spin-orbit coupling and whole body rotations $\endgroup$ – porphyrin Oct 1 '16 at 7:49
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If there is a molecule of $O$ (as opposed to $O_h$) or $I$ (as opposed to $I_h$) symmetry, it would be such a counter example (no mode that is both IR and raman active, and not centrosymmetric).

http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=903&option=4

http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=905&option=4

Apparently a snub cube is an example of the $O$ point group and in Chemistry for the 21st Century it is stated:

9.5.2.5 Snub Cube (6) We have recently demonstrated the ability of six resorcin[4]arenes and eight water molecules to assemble in apolar media to form a spherical molecular assembly which conforms to a snub cube (Fig. 9.3)

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