3
$\begingroup$

N$_2$ is a canonical example of this aspect of the failure of HF method, for instance, see table 4.9 in "Modern Quantum Chemistry" by Szabo and Ostlund. Here it is shown that in HF, the HOMO of N$_2$ is of $\pi$ symmetry, whereas correlated method and experiments show that its HOMO should have $\sigma$ symmetry, which becomes HOMO-1 in HF.

$\endgroup$
10
  • $\begingroup$ Maybe this answer on a related question helps: chemistry.stackexchange.com/questions/154162/… $\endgroup$
    – Libavius
    Dec 22, 2021 at 12:59
  • $\begingroup$ The linked question discusses N$_2$ only. $\endgroup$
    – nougako
    Dec 23, 2021 at 2:16
  • 1
    $\begingroup$ Yes, but the linked answer discusses the notion of "wrong" orbital ordering and the limits of the interpretability of orbitals in general. $\endgroup$
    – Libavius
    Dec 23, 2021 at 21:51
  • $\begingroup$ The idea of the question is to address the symmetry of the highest occupied orbital in N$_2$ in HF method. The fact that there is a "experiment" column in the cited table above indicates that experiments can actually identify the symmetry of the lowest ionization potential of N$_2$, which is associated with the HOMO of the neutral. The next ionization potential is associated with ionization that brings the cation to its first excited state, and in HF this is approximated by the removal of HOMO-1 of the neutral. $\endgroup$
    – nougako
    Dec 24, 2021 at 1:00
  • $\begingroup$ Therefore, despite the idea of orbital being unphysical in the exact sense, it still bear certain degree of information in the approximate single determinant sense. $\endgroup$
    – nougako
    Dec 24, 2021 at 1:01

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.