The magnetic moment of Cu^2+ ion (Z = 29) is:

The correct answer is Option (1) → \(\sqrt{3}\)

Cu (Z = 29): [Ar] 3d¹⁰ 4s¹. To form \( \ce{Cu^{2+}} \) the 4s electron and one 3d electron are removed:

\( \ce{Cu^{2+}} \): [Ar] 3d⁹

Nine electrons in five 3d orbitals: four orbitals hold a pair each (8 electrons) and the fifth holds one. So n = 1 unpaired electron.

Spin-only magnetic moment:
μ = √[n(n + 2)] BM = √[1 × (1 + 2)] = √3 BM ≈ 1.73 BM

NCERT (2026-27 edition):
Magnetic Properties – “… For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula, i.e., μ = √n(n+2) where n is the number of unpaired electrons and μ is the magnetic moment in units of Bohr magneton (BM).”