A) \[{{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}>{{\left[ MnC{{l}_{4}} \right]}^{2}}^{-}>{{\left[ CoC{{l}_{4}} \right]}^{2-}}\]
B) \[{{[MnC{{l}_{4}}]}^{2-}}>{{[Fe{{(CN)}_{6}}]}^{4-}}>{{[CoC{{l}_{4}}]}^{2-}}\]
C) \[{{[MnC{{l}_{4}}]}^{2-}}>{{[Co{{(Cl)}_{4}}]}^{2-}}>{{[Fe{{(CN)}_{6}}]}^{4-}}\]
D) \[{{[Fe{{(CN)}_{6}}]}^{4-}}>{{[Co{{(Cl)}_{4}}]}^{2-}}>{{[MnC{{l}_{4}}]}^{2-}}\]
Correct Answer: C
Solution :
[c] \[{{[Fe{{(CN)}_{6}}]}^{4-}}\]
no of unpaired electron = 0 
no of unpaired electron = 5 
no of unpaired electron = 3 The greater the number of unpaired electrons, greater the magnitude of magnetic moment. Hence the correct order will be \[{{[MnC{{l}_{4}}]}^{2-}}>{{[CoC{{l}_{4}}]}^{2+}}>{{[Fe{{(CN)}_{6}}]}^{4-}}\]
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