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Using simple arguments from crystal field theory, how do I rationalize the quartic state of octahedral Fe(III)? I am not able to in a logical way distribute the 5 d electrons in such a a way as to end up with three unpaired electrons in the normal splitting diagram for octahedral complexes.

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  • $\begingroup$ Fe(III) has $5 d$ electrons. If the $\Delta_O$ value is small, the five $d$ levels are all occupied by $1$ electron. This is a stable situation. $\endgroup$ – Maurice Sep 3 '20 at 17:17
  • $\begingroup$ Do you mean the quartic excited state? start with low spin $t_{2g}^5$ and excite one of the paired $t_{2g}$ electrons to $e_g$. $\endgroup$ – Andrew Sep 3 '20 at 20:56

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