The correct order of spin only magnetic moments for the following complex ions is

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${[{MnBr}_{4}]}^{2 -} < {[{CoF}_{6}]}^{3 -} < {[Fe {(CN)}_{6}]}^{3 -} < {[Mn {(CN)}_{6}]}^{3 -}$

${[{CoF}_{6}]}^{3 -} < {[{MnBr}_{4}]}^{2 -} < {[Fe {(CN)}_{6}]}^{3 -} < {[Mn {(CN)}_{6}]}^{3 -}$

${[Fe {(CN)}_{6}]}^{3 -} < {[Mn {(CN)}_{6}]}^{3 -} < {[{CoF}_{6}]}^{3 -} < {[{MnBr}_{4}]}^{2 -}$

${[Fe {(CN)}_{6}]}^{3 -} < {[{CoF}_{6}]}^{3 -} < {[{MnBr}_{4}]}^{2 -} < {[Mn {(CN)}_{6}]}^{3 -}$

answer is A.

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Detailed Solution

${[Fe (CN)_{6}]}^{3 -} {Fe}^{+ 3} \to Electronic configuration is [Ar] 4 s^{0} 3 d^{5} 4 p^{0} {CN}^{(-)} act as strong ligand and electronic arrangement is t_{2 g}^{5} {eg}^{0} and contains 1 unpaired electron$
${[Mn (CN)_{6}]}^{3 -} {Mn}^{3 +} \to E . C is [Ar] 4 s^{0} 3 d^{4} 4 p^{0} {CN}^{(-)} act as strong ligand and electronic arrangement is t_{2 g}^{4} {e_{g}}^{0} and contains 2 unpaired electrons$
${[{CoF}_{6}]}^{3 -} {Co}^{3 +} \to E . C is [Ar] 4 s^{0} 3 d^{6} 4 p^{0} F^{(-)} act as weak ligand and electronic arrangement is t_{2 g}^{4} e_{g}^{2} and contains 4 unpaired electrons$
${[{MnBr}_{4}]}^{2 -} {Mn}^{2 +} \to {Mn}^{2 +} \to E.C is [Ar] 4 s^{0} 3 d^{5} 4 p^{0}$
${Br}^{-} act as weak ligand and shape is tetrahedral and contains 5 unpaired electrons$
$The correct order is {[Fe (CN)_{6}]}^{3 -} < {[Mn (CN)_{6}]}^{3 -} < {[{CoF}_{6}]}^{3 -} < {[{MnBr}_{4}]}^{2 -}$