Question

Consider the following complex ions:
P = $[FeF_6]^{3-}$ 
Q = $[V(H_2O)_6]^{2+}$ 
R = $[Fe(H_2O)_6]^{2+}$ 
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 1

The Correct Option is C

• [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.