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Is there any molecular vibrational mode (or activity) that is neither Raman nor IR active?

In other words. Is there any activity in the molecules that can't be captured by either Raman spectrometer (all wavelenght) or IR absorption spectrometer (all wavelenght)? What are those?

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    $\begingroup$ Why, there are many such modes (silent, as they are called). $\endgroup$
    – Ivan Neretin
    Oct 7, 2019 at 11:14
  • $\begingroup$ Please give some examples of what they could be. Thanks. $\endgroup$
    – Jtl
    Oct 7, 2019 at 11:23
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    $\begingroup$ A molecule's normal modes are determined by its symmetry, the selection rules for transitions by change in dipole (IR) or change in polarisability (raman) which can only be satisfied by certain normal modes. If you look at point groups tables there are many irreps (symmetry species) with neither ($x,y,z$) for IR, ,or product operators ($xy, x^2$, etc), Raman. The silent vibrational frequency can sometimes be worked out by comparing the measured heat capacity with that produced via the partition function using the measured frequencies, the discrepancy is the unknown vibrational energy. $\endgroup$
    – porphyrin
    Oct 7, 2019 at 21:11

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Yes it is possible. While no normal modes can be both IR and Raman active in molecules with a centre of symmetry, as mentioned in a comment molecules with silent modes exist.

The easiest example is probably the HCH out-of-plane twisting of ethylene: enter image description here

Source: http://www.shodor.org/succeed-1.0/compchem/labs/vibrations/ethylene.jpg

Due to the symmetry of the molecule, when the HCH twist out of plane the dipole and polarizability of molecule is unchanged.

Other examples can be found here

https://www.chem.purdue.edu/jmol/vibs/c6h6.html

where the modes of benzene are sketched. The silent modes easily seen are kind of distorted breathing of the ring, that due to the C6 axis of symmetry again let the molecule unchanged both for the electric dipole moment and its polarisability.

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  • $\begingroup$ Does it mean that if a molecule has a change in dipole moment or has electron cloud polarizability, they are always IR and raman active? Are there change in dipole moment or cloud polarization that can't be detected by any spectrometer? $\endgroup$
    – Jtl
    Oct 8, 2019 at 0:18
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    $\begingroup$ @Jtl active mode doesn't implies the intensity of the abs. or scatt. to be high. The latter is always small and just lasers made raman possible. The effect is always there, just not detected using a standard light source. More close to your Q: for IR smaller is the el. dipole change weaker is the vibr. band intensity. For Raman the same though the change involves polarisability.In Raman, is even more complicated as coupling of vibrations with the el. transitions can lead to enhancement of bands. Vibrations within a chromophore can be greatly enhanced, so that remaining vibrations " vanishes". $\endgroup$
    – Alchimista
    Oct 8, 2019 at 9:34
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    $\begingroup$ "Is there a way to influence the molecules without it affecting either the dipole moment or polarization?".Not sure what you mean. If you mean if there is a normal mode that is silent, no. H2O has 3 normal modes which are both IR and Raman active. Its symmetry isn't high enough. $\endgroup$
    – Alchimista
    Oct 10, 2019 at 9:44
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    $\begingroup$ Tentative: That region might sense impurities that are hydrogen bound and vial walls might be involved. Assuming T to be reasonably constant you could see if the band change with time. Moreover I am not sure that the band really vanished if normalized to the main vibrational bands. You are at the detection boundary subtraction of noise and background might be tricky and you have basically two different setups (distance 1 and distance 2). $\endgroup$
    – Alchimista
    Oct 11, 2019 at 9:31
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    $\begingroup$ I mean the time water stayed in the vial. If you are near a vial signal then it just matter of focusing / reflection and I would treat it as cut-off, meaning that region not very informative about water anymore. Note that 1) things are now totally unrelated to the Q 2) people can't guess details $\endgroup$
    – Alchimista
    Oct 11, 2019 at 11:21

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