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Chemistry Question on coordination compounds

Consider the following complex ions:
P = [FeF6]3−[FeF_6]^{3-}
Q = [V(H2O)6]2+[V(H_2O)_6]^{2+}
R = [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}
The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is:

A

R < Q < P

B

R < P < Q

C

Q < R < P

D

Q < P < R

Answer

Q < R < P

Explanation

Solution

  • [FeF6]3−: Fe3+ has the electron configuration [Ar] 3d5. Fluoride (F−) is a weak field ligand, so all 5 d-electrons remain unpaired.

μ = √5(5 + 2) = √35 BM

  • [V(H2O)6]2+: V2+ has the electron configuration [Ar] 3d3. With three unpaired electrons, the spin-only magnetic moment is:

μ = √3(3 + 2) = √15 BM

  • [Fe(H2O)6]2+: Fe2+ has the electron configuration [Ar] 3d6. Water is a weak field ligand, resulting in four unpaired electrons.

μ = √4(4 + 2) = √24 BM

Thus, the correct order of magnetic moments is Q < R < P.