Which of the following ions is paramagnetic in nature?

The correct answer is Option (4) → \(V^{2+}\)

A substance is paramagnetic when it has unpaired electrons; an ion with no d electrons or with all d electrons paired is diamagnetic. The 4s electrons are lost first when the ions form:

• \( \ce{Sc^{3+}} \): Sc (Z = 21) [Ar] 3d¹ 4s² → [Ar] 3d⁰ — no unpaired electron.

• \( \ce{Ti^{4+}} \): Ti (Z = 22) [Ar] 3d² 4s² → [Ar] 3d⁰ — no unpaired electron.

• \( \ce{Zn^{2+}} \): Zn (Z = 30) [Ar] 3d¹⁰ 4s² → [Ar] 3d¹⁰ — all paired.

• \( \ce{V^{2+}} \): V (Z = 23) [Ar] 3d³ 4s² → [Ar] 3d³ — three unpaired electrons.

So V²⁺ is paramagnetic (μ = √15 ≈ 3.87 BM).

Why the other options are incorrect

• Option (1) Sc³⁺ — Incorrect. 3d⁰, no unpaired electrons, so it is diamagnetic (and colourless).

• Option (2) Ti⁴⁺ — Incorrect. 3d⁰, no unpaired electrons, so it is diamagnetic.

• Option (3) Zn²⁺ — Incorrect. 3d¹⁰, a completely filled d subshell, so it is diamagnetic.

NCERT (2026-27 edition):
Magnetic Properties – “… Diamagnetic substances are repelled by the applied field while the paramagnetic substances are attracted. Substances which are attracted very strongly are said to be ferromagnetic. In fact, ferromagnetism is an extreme form of paramagnetism. Many of the transition metal ions are paramagnetic. Paramagnetism arises from the presence of unpaired electrons, each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum.”