Which transition metal ion M^2+ gives magnetic moment 5.92 BM?

The correct answer is Option (1) → Mn²⁺

For first-row transition metal ions the magnetic moment is given by the spin-only formula, μ = √[n(n + 2)] BM, where n is the number of unpaired electrons.

    √[n(n + 2)] = 5.92 ⇒ n(n + 2) = 35 ⇒ n = 5

Five unpaired electrons means a d⁵ ion with one electron in each d orbital. Mn (Z = 25), [Ar] 3d⁵4s², loses its two 4s electrons to give \( \ce{Mn^{2+}} \) = [Ar] 3d⁵, so μ = √35 = 5.92 BM.

Why the other options are incorrect

• Option (2) Cr²⁺ — Incorrect. Cr (Z = 24), [Ar] 3d⁵4s¹, loses the 4s electron and one 3d electron: \( \ce{Cr^{2+}} \) = 3d⁴, n = 4, μ = √24 = 4.90 BM.

• Option (3) Fe²⁺ — Incorrect. Fe (Z = 26), [Ar] 3d⁶4s², loses both 4s electrons: \( \ce{Fe^{2+}} \) = 3d⁶. Five d orbitals take the first five electrons singly and the sixth pairs up, leaving n = 4, μ = 4.90 BM.

• Option (4) Ni²⁺ — Incorrect. Ni (Z = 28), [Ar] 3d⁸4s², gives \( \ce{Ni^{2+}} \) = 3d⁸: three orbitals are full and two hold one electron each, so n = 2, μ = √8 = 2.84 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).”