I've been studying a particular reaction with DFT, where a transition state and the following intermediate are very close to each other. The change in geometry is rather small and therefore the electronic energies differ only by about ~ 1-2 millihartrees. Still, the TS has the higher electronic energy (everything else would be surprising?), but adding thermal corrections to both structures leaves the intermediate about 2 kJ/mol higher in energy.

Does that make sense? Or is there something I did not consider in my calculations?

  • 2
    $\begingroup$ It happens relatively often. The calculations of the electronic energy are the most accurate, vibrational & rotational correction calculations are rather crude. And besides, the typical accuracy of energetics with hybrid functionals (you didn't mention, but I assume you use some hybrid functional) is $\pm 10-15 \, \mathrm{kJ/mol}$ (MAD), so that a difference of few kJ/mol is not a difference at all. $\endgroup$ – Wildcat Jul 14 '16 at 8:39
  • $\begingroup$ Which reaction is it? I realize it is secret and stuff, but at least mention the class. Is it organic, organometallic, other? $\endgroup$ – sixtytrees Jul 14 '16 at 22:44
  • $\begingroup$ It is organometallic, transition metal / ligand / substrate. The transition state is a hydride transfer from the metal center onto the ligand. $\endgroup$ – snurden Jul 18 '16 at 21:22

1 Kcal/mol is indeed quite low. This is around 1 hydrogen bond energy. In general - compute frequencies using freq command in Gaussian or whatever you are using. You should have a single negative frequency. It should be relatively low (abs value compared to lowest positive frequencies). Visualize it using Moleken or a similar tool. If movement is along the reaction coordinate then you are good.

Also, compare all bond lengths and angles to see what changed. Did bonds along the reaction coordinate change the most?

Please use Kcal/mol instead of Hartrees. I realize that Gaussian prints out Hartrees, but it is easy enough to convert to Kcal/mol.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.