Which of the following will show the highest value of spin only magne…

The correct answer is Option (3) → Mn²⁺

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

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

To find n, write the configuration of the ion. A 3d-series metal loses its 4s electrons first, and the d electrons occupy the five d orbitals singly before pairing (Hund’s rule).

Mn (Z = 25) is [Ar]3d⁵4s²; losing the two 4s electrons gives \( \ce{Mn^{2+}} \): 3d⁵, one electron in each d orbital, so n = 5 and μ = \( \sqrt{35} \) = 5.92 BM.

So Mn²⁺, with five unpaired electrons, has the highest magnetic moment.

Shortcut: The moment rises with n, so there is no need to calculate it. A d⁵ ion has the maximum of five unpaired electrons.

Why the other options are incorrect

• Option (1) Zn²⁺ — Incorrect. Zn (Z = 30) is [Ar]3d¹⁰4s²; losing the two 4s electrons gives 3d¹⁰, all electrons paired, so it is diamagnetic (0 BM).

• Option (2) Ni²⁺ — Incorrect. Ni (Z = 28) is [Ar]3d⁸4s²; losing the two 4s electrons gives 3d⁸, two unpaired electrons, μ = \( \sqrt{8} \) = 2.83 BM.

• Option (4) V²⁺ — Incorrect. V (Z = 23) is [Ar]3d³4s²; losing the two 4s electrons gives 3d³, three unpaired electrons, μ = \( \sqrt{15} \) = 3.87 BM.

NCERT (2026-27 edition):
Magnetic Properties – …