The electronic configuration of gadolinium (Z = 64) is [Xe] 4f7 5d1 6…

The correct answer is Option (3) → 7.94 BM

The spin-only magnetic moment depends only on the number of unpaired electrons (n): \( \mu = \sqrt{n(n+2)} \) BM. So the first task is to find how many unpaired electrons Gd³⁺ has.

• Gd (Z = 64): [Xe] 4f⁷ 5d¹ 6s².

• Gd³⁺: the two 6s electrons and the one 5d electron are removed, leaving [Xe] 4f⁷.

• The seven 4f orbitals each hold one electron (Hund’s rule), so n = 7.

• \( \mu = \sqrt{7(7+2)} = \sqrt{63} = 7.94 \) BM.

The half-filled 4f⁷ set is the reason Gd³⁺ has the largest possible number of unpaired f electrons.

NCERT (2026-27 edition):
4.5.1 Electronic configuration with 6s Configurations – “…the electronic configurations of all the tripositive ions (the most stable oxidation state of all the lanthanoids) are of the form 4fⁿ (n = 1 to 14 with increasing atomic number).”

Table 4.9: Electronic Configurations and Radii of Lanthanum and Lanthanoids (row for gadolinium)

4.3.9 Magnetic Properties – “…the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula, i.e., μ = \( \sqrt{n(n+2)} \) where n is the number of unpaired electrons and μ is the magnetic moment in units of Bohr magneton (BM).”