The correct order of magnetic moment (spin values in B.M.) is(Atomic no. Mn = 25, Fe = 26, Co= 27)

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Fe(CN)64>CoCl42−>MnCl42

MnCl42−>Fe(CN)64−>CoCl42−

Fe(CN)64>MnCl42−>CoCl42−

MnCl42−>CoCl42−>Fe(CN)64

answer is D.

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Detailed Solution

Magnetic moment of [MnC14]2− is 5.9 BM.Mn is in +2 oxidation state having configuration of [Ar]3d5. It has 5 unpaired electrons.μ=n(n+2) =(5(5+2) =5.9BMMagnetic moment of [CoC14]2− is 3.9 BM.Co is in +2 oxidation state having configuration of [Ar]3d7. It has 3 unpaired electrons.μ=n(n+2) =(3(3+2) =3.9BMMagnetic moment of [FeCN4]2− is 0.0 BM.Fe is in +2 oxidation state having configuration of [Ar]3d6. In this complex CN− ion is a strong ligand and all electrons are paired and there are no unpaired electrons.μ=n(n+2) =0(0+2) =0BMThe decreasing order of the magnetic moments is:[MnCl4]−2>[CoCl4]2−>Fe(CN)6]4−

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