There is a sample that we know is all made up by only one of these two molecule: enter image description here

how you can see they have the same formula but two different geometries. The question is if it is possible to know by which of these two molecules the sample is made with an experiment involving vibrational spectroscopy (IR or Raman). The following is a quite wide answer I gave to myself but I'm not sure it is right, so please correct me or confirm what I have written.
We know that if we are considering small oscillations, the vibrational part of the wave function of a molecule with $N$ atoms is the product of $3N-6$ quantum armonic oscillators eigenfunctions, and we also know that in a classic way to proceed the nuclei can move in $3N-6$ differents normal modes. Fixed one of these normal modes, if it is for example IR active, we know that all the $3N-6$ oscillators moving in that normal mode can absorb photons from the external radiation. So if the external radiation has a frequency band enough large to contain all the proper frequencies of the oscillators, we'll notice $3N-6$ absorbtion lines. While if we are considering a not IR active normal mode of course we won't see any absorbtion lines. In general the nuclei of a molecule move on a linear combination of all the normal modes. So we can think that each IR active normal mode of the combination gives the same $3N-6$ lines while each not IR active mode gives zero lines. So I would expect to observe for my real sample 12 absorbtion lines (I am not considering overtones and the noise of the media) which correspond to the proper frequencies of the oscillators. So since the two molecule have exactly the same oscillators (it only changes their position), I should observe the same 12 lines on both cases, so I'd say that it is not possible to distinguish between the two. The same argument can be used for the Raman. However I'm almost sure that I'm wrong and a way there must exists.

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    $\begingroup$ see ideals.illinois.edu/handle/2142/79485 $\endgroup$ – MaxW Jan 24 '19 at 20:58
  • $\begingroup$ Basically your question is how to identify wich modes of a molecules are IR or Raman active. It seems you are aware. The point is that you wrote "have the same armonic oscillators" which is not true even at a first glance as you have already realised if you have followed the link given above. After that you see how to discern between the two in finer details. $\endgroup$ – Alchimista Jan 25 '19 at 9:18

The starting point is to look at the potential point groups of the two molecules and work out if there are common vibrational mode symmetries between IR and Raman spectra, assuming that you are able to obtain both. In the 3rd and 4th columns of the point group are the $x, \,y, \,z$ and $xy, \,xz$ etc operators. The $x, \,y, \,z$ operators transform as does a dipole (so IR transitions) and act as an alias for the symmetry species.

For example in the $C_{2v}$ point group each IR transition must be one of $A_1,\,B_1 ,\, B_2$ species and as there are 12 transitions there must be several of the same symmetry but different frequency.

The Raman transitions correspond to product terms , $x^2,\, xy$ etc. (because Raman spectroscopy depends of the polarisability which is proportional volume and so area in cross-section ) and these have also $A_1,\,B_1 ,\, B_2$ symmetry species as well as $A_2$ species (in $C_{2v}$) so we do expect to see some different transition frequencies between IR and Raman spectra. We do not necessarily know what the symmetry species are for these transitions, just that there is a difference.

In the case of $D_{2h}$ point group as there is a centre of inversion, there are no common transitions between IR and Raman spectra. Look at the point group and you will see this.

In more detailed analysis the polarisation of Raman signals can be used to determine which transition is $A_1$ etc. Spectroscopy books give details of this.

look at Understanding group theory easily and quickly for an explanation of point groups.


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