The question asks the following:

'The molecule NO has a low-lying electronic state 0.015 eV above the ground electronic state. Explain why this is the case and state the electronic degeneracies of these two states. [2]'

MO diagram of an NO molecule:

enter image description here

I am a bit confused by the question as I can't find anything on this in my textbook. The energy difference is small, so the electron configuration can't be very different to what's shown in the diagram - i.e. we can't just promote the electron into the σ* orbital as that would surely change the energy by a fair bit. I'm not sure what other actions can be taken to change the energy of an electronic state. Would appreciate it if someone could point me in the right direction.

  • 2
    $\begingroup$ The orbital diagram you show describes only the ground state. There is one unpaired electron and the multiplicity (degeneracy) is $2S+1$ where $S$ is electron spin. If the diagram is correct in its relative energies, the lowest electronic transition would be allowed from the partially filled doubly degenerate antibonding orbital to the next one. $\endgroup$
    – porphyrin
    Commented Apr 11, 2023 at 17:07
  • 2
    $\begingroup$ There's spin–orbit coupling in NO, which isn't reflected in the MO diagram. The low-lying state has the same electron configuration but has a different total angular momentum... (can't remember the symbol for it in a molecule, I remember spin AM is $\Sigma$ and orbital AM is $\Lambda$... is it $\Omega$? It's been a while, sorry.) $\endgroup$ Commented Apr 11, 2023 at 17:33
  • 1
    $\begingroup$ But if you just google for 'spin–orbit coupling in nitric oxide' you'll find plenty of info. $\endgroup$ Commented Apr 11, 2023 at 17:35
  • $\begingroup$ @Orthocresol the ground state is $^2\Pi_{1/2}$ and the first excited state at $\approx 121\,\mathrm{cm^{-1}}$ is $^2\Pi_{3/2}$ which is hardly an excited state at all! (The next one is $^2\Sigma^+$ at $43965\,\mathrm{cm^{-1}}$.) $\endgroup$
    – porphyrin
    Commented Apr 12, 2023 at 9:20
  • $\begingroup$ @porphyrin Yup; that is the number that OP is looking for, though (121 cm-1 is 0.015 eV exactly, unless I miscounted?). $\endgroup$ Commented Apr 12, 2023 at 15:20


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