What is the electronic arrangement of a metal | vedclass

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(C) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.For a $d^4$ configuration,if the crystal field splitting energy $\Del

Vedclass · Accepted Answer

$t_{2g}^3 \ e_g^1$

Vedclass · Answer

(C) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.For a $d^4$ configuration,if the crystal field splitting energy $\Delta_0$ is less than the pairing energy $P$ $(\Delta_0 According to Hund's rule,electrons will occupy the higher energy $e_g$ orbitals before pairing in the $t_{2g}$ orbitals.Therefore,the electronic configuration is $t_{2g}^3 \ e_g^1$.

What is the electronic arrangement of a metal atom/ion in an octahedral complex with a $d^4$ configuration,if $\Delta_0 < P$ (pairing energy)?

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