question_answer

The correct order of magnetic moments (spin only values in B.M.) among is [AIEEE 2004]

A) \[{{[Fe{{(CN)}_{6}}]}^{4-}}>{{[MnC{{l}_{4}}]}^{2-}}>{{[CoC{{l}_{4}}]}^{2-}}\]

B) \[{{[MnC{{l}_{4}}]}^{2-}}>{{[Fe{{(CN)}_{6}}]}^{4-}}>{{[CoC{{l}_{4}}]}^{2-}}\]

C) \[{{[MnC{{l}_{4}}]}^{2-}}>{{[CoC{{l}_{4}}]}^{2-}}>{{[Fe{{(CN)}_{6}}]}^{4-}}\]

D) \[{{[Fe{{(CN)}_{6}}]}^{4-}}>{{[CoC{{l}_{4}}]}^{2-}}>{{[MnC{{l}_{4}}]}^{2-}}\] (Atomic nos. \[Mn=25,\,Fe=26,\,Co=27\])

Correct Answer: C

Solution :

Magnetic moment \[=\sqrt{n+2}\] Where, n = number of unpaired electron i.e., greater the number of unpaired electron greater will be the paramagnetic character.