For a d^{4} metal ion in an octahedral field, | vedclass

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Question

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

Vedclass · Accepted Answer

$t_{2g}^{3} e_{g}^{1}$ when $\Delta_{o} < P$

Vedclass · Answer

(C) In an octahedral field,the $d$-orbitals split into $t_{2g}$ and $e_{g}$ sets.For a $d^{4}$ ion,if the crystal field splitting energy $\Delta_{o}$ is less than the pairing energy $P$ (weak field ligand),the electrons will occupy the orbitals singly before pairing occurs.Thus,the configuration is $t_{2g}^{3} e_{g}^{1}$.If $\Delta_{o} > P$ (strong field ligand),the fourth electron will pair in the $t_{2g}$ orbital,resulting in $t_{2g}^{4} e_{g}^{0}$.

For a $d^{4}$ metal ion in an octahedral field,the correct electronic configuration is:

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