The d-electron configuration of $[Ru {(en)}_{3}] {Cl}_{2}$ and $[Fe {(H_{2} O)}_{6}] {Cl}_{2}$ respectively are:

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$t_{2 g}^{6} e_{g}^{0} and t_{2 g}^{4} e_{g}^{2}$

$t_{2 g}^{4} e_{g}^{2} and t_{2 g}^{4} e_{g}^{2}$

$t_{2 g}^{4} e_{g}^{2} and t_{2 g}^{6} e_{g}^{0}$

$t_{2 g}^{6} e_{g}^{0} and t_{2 g}^{6} e_{g}^{0}$

answer is B.

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Detailed Solution

Fe and Ru belong to the same group and their oxidation states in the given complexes are ${Fe}^{2 +}$ and ${Ru}^{2 +}$ respectively.

Both ${Fe}^{2 +}$ and ${Ru}^{2 +}$ have d⁶configuration. In $[Ru {(en)}_{3}] {Cl}_{2},$ the electronic arrangement becomes $t_{2 g}^{6} {eg}^{0}$ .

In ${[Fe {(H_{2} O)}_{6}]}^{2 +},$ the electronic arrangement becomes $t_{2 g}^{4} {eg}^{2}$ because $H_{2} O$ is weak ligand.

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