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

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$e_{g}^{2} t_{2 g}^{2} when Δ_{0} < P$

$t_{2 g}^{3} e_{g}^{1} when Δ_{0} > P$

$t_{2 g}^{3} e_{g}^{1} when Δ_{0} < P$

$t_{2 g}^{4} e_{g}^{0} when Δ_{0} < P$

answer is A.

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Detailed Solution

When $∆_{\circ} < P,$ the electrons will remain unpaired in t_2gand e_g orbitals because energy gap between the orbitals is small. This lead to the formation of a high spin complex. Hence, electronic arrangement becomes $t_{2 g}^{3} e_{g}^{1} .$

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