When filling electrons in atomic or molecular orbitals we fill first the lower energy orbitals and then the higher, according to Pauli and Hund's rule. But this seem not to be the case when considering metal complexes. According to WikipediaWikipedia

For example why the $d_{xy}$ orbital in the $d^4$ high spin is not fully occupied? Is the pairing energy that changes the order of filling the orbitals?


1 Answer 1


This was just an illustrative example. The tendency of a complex to favor low-spin or high-spin configurations is dictated by the relative magnitudes of the pairing energy, P, and the splitting energy, $\Delta$, between $e_g$ and $t_{2g}$.*

If $\Delta$ is greater than the energy required to pair two electrons, then electrons pair in $t_{2g}$ before occupying $e_g$. In the example you've shown of compression, this would be the $d_{xy}$ orbital.

*For octahedral complexes. Jahn-Teller can also happen for tetrahedral complexes.

  • $\begingroup$ Thanks for the answer. So the fact that a complex is low or high spin affect the way of ordering after the Jahn-Teller distortion? For example considering compression for $d^4$ low spin. Should now the $d_{xy}$ be fully occupied? $\endgroup$
    – ado sar
    Apr 17, 2020 at 17:27
  • $\begingroup$ @adosar exactly. In the compression example, $d_{xy}$ would be doubly occupied, defining the $d^4$ low spin configuration. $\endgroup$
    – jezzo
    Apr 17, 2020 at 17:32
  • $\begingroup$ But we can't predict if elongation or compression will happen right? Also when filling atomic orbitals or molecular we neglect pairing energy because it is negligible compared to $\Delta E$ of the orbitals? $\endgroup$
    – ado sar
    Apr 17, 2020 at 17:35
  • $\begingroup$ @adosar I only mentioned compression because I could name the exact orbital. In the case of elongation and $\Delta$ larger than the pairing energy, then you'd have an electron go into either $d_{xz}$ or $d_{xy}$ as they are degenerate. $\endgroup$
    – jezzo
    Apr 17, 2020 at 17:43
  • $\begingroup$ @adosar the pairing energy is exactly what we are comparing to the $\Delta$, so you definitely do not ignore it :) $\endgroup$
    – jezzo
    Apr 17, 2020 at 17:48

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