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Which change in ${[Mn (CN)_{6}]}^{4 -}$ results in symmetrical distribution of electrons in non-degenerated-orbitals of metal ion?
(P) ${[Mn (CN)_{6}]}^{4 -} \frac{oxidation}{by O_{2}} {[Mn (CN)_{6}]}^{3 -}$
(Q) ${[Mn (CN)_{6}]}^{4 -} \underset{by Zn}{\overset{reduction}{⟶}} {[Mn (CN)_{6}]}^{5 -}$

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P Only

Q Only

P and Q both

None of these

answer is B.

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Detailed Solution

Due to low oxidation state of $Mn (+ 1) Δ_{0} < P$ , hence, pairing does not occur.

$\begin{aligned} {[{Mn}^{+ 2} (CN)_{6}]}^{5 -} \Rightarrow d^{5}, t_{2 g}^{2, 2, 1} e_{g}^{0, 0} \\ {[{Mn}^{+ 1} (CN)_{6}]}^{5 -} \Rightarrow d^{6}, t_{2 g}^{2, 1, 1} e_{g}^{1, 1} \end{aligned}$

Hence, option B is correct.

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