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Question

# The correct order of magnetic moment (spin only values in B.M.) among is: (Atomic no.s : Mn=25; Fe=26; Co=27)

A
[Fe(CN)6]4>[MnCl4]2>[CoCl4]2
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B
[MnCl4]4>[Fe(CN)6]4>[CoCl4]2
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C
[MnCl4]2>[CoCl4]2>[Fe(CN)6]4
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D
[Fe(CN)6]4>[CoCl4]2>[MnCl4]2
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Solution

## The correct option is C [MnCl4]2−>[CoCl4]2−>[Fe(CN)6]4−[Fe(CN)6]4−→Fe2+→[Ar]3d6 (Due to strong field ligand of (CN−) all unpaired electrons get paired.) [MnCl4]2−→Mn2+→[Ar]3d5 (due to weak field ligand of (Cl−) no change in unpaired electrons [CoCl4]2−→Co2+→[Ar]3d7 (Due to weak field ligand of (Cl−) no change in unpaired electrons) Magnetic moment =√n(n+2) B.M. Where, n = number of unpaired electrons. In [MnCl4]2−, Mn contains maximum no. of electrons So, order of magnetic moment [MnCl4]2−>[CoCl4]2−>[Fe(CN)6]4−

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