I saw this image from the Wikipedia article on the Franck-Condon principle:
enter image description here

But couldn't find an explanation for the shift $q_{01}$ to the right (i.e. increasing the inter-nuclear distance) as the electron energy level goes from $E_0$ to $E_1$. As far as I understand it, the absorbed energy in an electron will push it to an outer-more electron shell, but is this the sole reason for the shift towards right in this graph? Is it simply, the further binding electrons are to the nucleus, the further apart the nuclei binds? And if so, does this trend continue for each subsequent energy level, $E_2$, $E_3$, etc. or are there other factors at play here?


1 Answer 1


There is no reason why equipotential curves should have their minima lined up at the same internuclear distance or why the shapes would be similar.

As an example, when in a single-electron picture an electron is excited from a bonding orbital to an antibonding orbital, if there still is a minimum, it would be at a larger internuclear distance. Or when an electron is excited from a $\sigma$ orbital to a $\pi$ orbital.

I see no obvious reason why excited states could not have a shorter equilibrium distance, for example when an electron is excited from an antibonding orbital to a bonding orbital.

In general, it is all more complicated because of electron correlation.

  • $\begingroup$ Thanks! So what would be the conditions for the similarities in shape and shift to the right in the picture above? $\endgroup$
    – Void
    Jun 1, 2020 at 21:03
  • $\begingroup$ All bonding potential surfaces will look more or less similar but with very different vibrational frequencies and dissociation energies. The dissociative excited states look totally different. $\endgroup$
    – user137289
    Jun 1, 2020 at 21:49

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.