A) \[{{[MnC{{l}_{4}}]}^{2-}}>{{[CoC{{l}_{4}}]}^{2-}}>{{[Fe{{(CN)}_{6}}]}^{4-}}\]
B) \[{{[MnC{{l}_{4}}]}^{2-}}>{{[Fe{{(CN)}_{6}}]}^{4-}}>{{[CoC{{l}_{4}}]}^{2-}}\]
C) \[{{[Fe{{(CN)}_{6}}]}^{4-}}>{{[MnC{{l}_{4}}]}^{2-}}>{{[CoC{{l}_{4}}]}^{2-}}\]
D) \[{{[Fe{{(CN)}_{6}}]}^{4-}}>{{[CoC{{l}_{4}}]}^{2-}}>{{[MnC{{l}_{4}}]}^{2-}}\]
Correct Answer: A
Solution :
\[{{[MnC{{l}_{4}}]}^{2-}}\] Number of unpaired \[{{e}^{-}}=5\] \[[CoC{{l}_{4}}){{]}^{2-}}\]: Number of unpaired \[{{e}^{-}}=3\] \[{{[Fe{{(CN)}_{6}}]}^{4-}}\]: Number of unpaired \[{{e}^{-}}=0\] Magnetic moment \[=n\sqrt{n+2}\] where n = Number of unpaired \[{{e}^{-}}\] Greater the number of unpaired electrons, greater the magnetic moment. Hence the order of magnetic moment is as \[{{[MnC{{l}_{4}}]}^{2-}}>{{[CoC{{l}_{4}}]}^{2-}}>{{[Fe{{(CN)}_{6}}]}^{4-}}\]
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