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 the crystal field splitting energy $\Delta_0$ is less than the p

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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 the crystal field splitting energy $\Delta_0$ is less than the pairing energy $P$ (weak field ligand),the electrons prefer to occupy higher energy orbitals rather than pairing up.For a $d^4$ configuration,the first three electrons occupy the $t_{2g}$ orbitals singly.The fourth electron enters the higher energy $e_g$ orbital because $\Delta_0 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 $d^4$ configuration,if $\Delta_0 < P$ (pairing energy)?

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