Arrange the following cyano complexes in decreasing order of their magnetic moment.

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${[Mn (CN)_{6}]}^{3 -} > {[Cr (CN)_{6}]}^{3 -} > {[Fe (CN)_{6}]}^{3 -} > {[Co (CN)_{6}]}^{3 -}$

${[Fe (CN)_{6}]}^{3 -} > {[Cr (CN)_{6}]}^{3 -} > {[Mn (CN)_{6}]}^{3 -} > {[Co (CN)_{6}]}^{3 -}$

${[Cr (CN)_{6}]}^{3 -} > {[Mn (CN)_{6}]}^{3 -} > {[Fe (CN)_{6}]}^{3 -} > {[Co (CN)_{6}]}^{3 -}$

${[Co (CN)_{6}]}^{3 -} > {[Cr (CN)_{6}]}^{3 -} > {[Mn (CN)_{6}]}^{3 -} > {[Fe (CN)_{6}]}^{3 -}$

answer is A.

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Detailed Solution

$\begin{array}{ll} {[Cr (CN)_{6}]}^{3 -} {Cr}^{3 +} in strong ligand field = t_{2 g}^{3} e_{g}^{_{0}} = 3 unpaired electron \\ {[Mn (CN)_{6}]}^{3 -} {Mn}^{3 +} in strong ligand field = t_{2 g}^{4} e_{g}^{0} = 2 unpaired electron \\ {[Fe (CN)_{6}]}^{3 -} {Fe}^{3 +} in strong ligand field = t_{2 g}^{5} e_{g}^{0} = 1 unpaired electron \\ {[Co (CN)_{6}]}^{3 -} {Co}^{3 +} in strong ligand field = t_{2 g}^{6} e_{g}^{0} = Zero unpaired electron \end{array}$