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.When $\Delta_{0} < P$ (where $\Delta_{0}$ is the crystal field splitt

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

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(C) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.When $\Delta_{0} For a $d^4$ configuration,the first three electrons occupy the lower energy $t_{2g}$ orbitals singly according to Hund's rule.The fourth electron will occupy the higher energy $e_{g}$ orbital because the energy required to pair it in the $t_{2g}$ orbital is greater than the energy required to promote it to the $e_{g}$ orbital.Therefore,the electronic arrangement is $t_{2g}^3 e_g^1$.

$I$. $H_2O$	$II$. $CO$
$III$. $NH_3$	$IV$. $I^{-}$
$V$. $F^{-}$

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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