What is the electronic arrangement of a metal | vedclass

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(C) In an octahedral complex,the $d$-orbitals split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy).For a $d^4$ configuration,if the

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$t^3_{2g}e^1_g$

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(C) In an octahedral complex,the $d$-orbitals split into two sets: $t_{2g}$ (lower energy) and $e_g$ (higher energy).For a $d^4$ configuration,if the crystal field splitting energy $\Delta_0$ is less than the pairing energy $P$ (i.e.,$\Delta_0 In this case,the fourth electron will occupy the higher energy $e_g$ orbital rather than pairing up in the $t_{2g}$ orbital because the energy required to pair is greater than the energy required to promote the electron to the $e_g$ level.Thus,the first three electrons occupy the $t_{2g}$ orbitals singly,and the fourth electron occupies one of the $e_g$ orbitals.The electronic configuration is $t^3_{2g}e^1_g$.

$I$. $NCS^{-}$	$II$. $S^{2-}$
$III$. $en$	$IV$. $SCN^{-}$

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

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