Which of the following lanthanoid elements exhibits maximum spin-only…

The correct answer is Option (2) → Gadolinium

The spin-only magnetic moment depends only on the number of unpaired electrons (n):

    \( \displaystyle \mu=\sqrt{n(n+2)}\ \mathrm{BM} \)

Lanthanoid ions in the +3 state have the configuration [Xe]4fⁿ, where n = atomic number − 57 (the three electrons removed are the 6s² and one 5d or 4f electron). The seven 4f orbitals are filled singly first (Hund’s rule), so unpaired electrons rise from f¹ to a maximum of 7 at f⁷ and then fall again.

• Nd (Z = 60) is [Xe]4f⁴6s²; losing the two 6s electrons and one 4f electron gives \( \ce{Nd^{3+}} \): 4f³, 3 unpaired, μ = √15 ≈ 3.87 BM

• Pm (Z = 61) is [Xe]4f⁵6s²; losing the two 6s electrons and one 4f electron gives \( \ce{Pm^{3+}} \): 4f⁴, 4 unpaired, μ = √24 ≈ 4.90 BM

• Gd (Z = 64) is [Xe]4f⁷5d¹6s²; losing the two 6s electrons and the 5d electron gives \( \ce{Gd^{3+}} \): 4f⁷, 7 unpaired, μ = √63 ≈ 7.94 BM

• Dy (Z = 66) is [Xe]4f¹⁰6s²; losing the two 6s electrons and one 4f electron gives \( \ce{Dy^{3+}} \): 4f⁹, 5 unpaired, μ = √35 ≈ 5.92 BM

So gadolinium has the largest spin-only magnetic moment in the +3 state.

Shortcut: The half-filled 4f⁷ ion always has the most unpaired electrons, and \( \ce{Gd^{3+}} \) (Z = 57 + 7 = 64) is that ion.