I have done a geometry optimization of cis and trans-difluoroethene using RHF/STO-3G in Gaussian03. Although this is a simple method and a small basis set, this should ideally put place the molecule in a minimum on its potential energy curve. Within this minimum are vibrational states. I am wondering if it is possible to see, from the calculated IR and Raman spectra whether the optimisation was successful, i.e. whether the molecule was indeed placed in a minimum.

I am thinking like this: If the molecule is in a minimum, then all vibrational IR and Raman active vibrational states should be visible (given a high enough activity, of course) in the theoretical spectra. However, if the molecule is not in the minimum, then only those vibrational states of higher energy should be visible. Hence, fewer IR/Raman active vibrational modes would be observed if the geometry optimization was not complete.

According to this question, all frequencies at a local minimum would be positive, and all negative on a maximum. I assume then that somewhere in-between, you would have both negative and positive frequencies. However, given not the table of IR/Raman active modes and their frequencies, but the actual spectra instead, would I then be able to decide if the molecule is in a minimum or not?

  • $\begingroup$ I'd expect some relative peak intensity changes over large temperature changes. But I don't think that you would be able to get a chemical hot enough to eliminate any vibrational modes entirely. The IR vibrational nodes detected are "resonance vibrations" and a molecule can't be resonating at all frequencies at once. $\endgroup$ – MaxW Nov 10 '15 at 22:36
  • $\begingroup$ My guess is that you're going to get all real (positive) frequencies as opposed to imaginary (negative) frequencies. I say this because, as you point out, you're basically at an energy minimum so you're not gonna get imaginary frequencies which I believe you only get at inflection points and plateaus. $\endgroup$ – jheindel Nov 11 '15 at 7:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.