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IIT-JEE advanced 2013 has a question:

Give the correct order of the complex ions given below according to their spin-only magnetic moment: A. $\ce{[FeF6]^{3-}}$ B. $\ce{[V(H2O)6]^{2+}}$ C. $\ce{[Fe(H2O)6]^{2+}}$

So, my reasoning is more unpaired electron results in greater spin-only magnetic moment. So, the order should be B>A>C

But the answer is given A>C>B.

I don't know where I'm wrong.

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The first choice lists $\ce{Fe^3+}$ as the central metal atom (CMA), so you have the configuration: $\mathrm{3d^5 4s^0}$. Remember that fluoride is a weak field ligand and would not cause pairing of electrons. This gives you 5 unpaired electrons.

The second choice has $\ce{V^2+}$ as the CMA, the configuration being $\mathrm{3d^3 4s^0}$. This gives you 3 unpaired electrons.

Finally, the third choice has $\ce{Fe^2+}$ as the CMA. Yet again water is a weak field ligand and does not cause pairing of electrons. The isolated configuration although is $\mathrm{3d^6 4s^0}$, but now you'd have 4 electrons in the $\mathrm{t_2g}$ set and 2 electrons in the $\mathrm{e_g}$ set. So now you've 4 unpaired electrons.

more unpaired electron results in greater spin-only magnetic moment.

This is true and I can see it being followed as $\mathrm{A>C>B}$.


Also see my comment for the strength of ligands.

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The answer A>C>B is correct. I think you might have considered Fluoride to be a strong field ligand but it's actually not that strong so there won't be any pairing in the first compound as a result it will have five unpaired electrons. The remaining compounds have lesser unpaired electrons so yeah the order is justified.

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