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 $\De

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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 $\Delta_o First,three electrons occupy the $t_{2g}$ orbitals singly.The fourth electron will occupy the $e_g$ orbital because the energy required to pair in $t_{2g}$ $(	ext{pairing energy})$ is greater than the energy required to promote the electron to the $e_g$ level $(\Delta_o)$.Thus,the configuration is $t^3_{2g} e^1_g$.

What is the electronic arrangement of a metal atom/ion in an octahedral complex with a $d^4$ configuration,if $\Delta_o < \text{pairing energy}$?

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