A metal ion (M) of group-4 is showing magnetic moment of 1.73 BM, its…

The correct answer is Option (3) → \(M^{3+}\)

The magnetic moment of a transition metal ion comes mainly from its unpaired electrons and is given by the spin-only formula:

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

For μ = 1.73 BM:

    \( \displaystyle \sqrt{n(n+2)} = 1.73 \;\Rightarrow\; n(n+2) = 3 \;\Rightarrow\; n = 1 \)

The first group-4 metal is titanium: Ti (Z = 22) is [Ar]3d²4s². To leave one unpaired electron (3d¹), the two 4s electrons and one 3d electron must be removed, which gives Ti³⁺. So the oxidation state is +3.

Why the other options are incorrect

• Option (1) M⁺ — Incorrect. \( \ce{Ti^{+}} \) is [Ar]3d²4s¹, with 3 unpaired electrons: μ = √15 ≈ 3.87 BM.

• Option (2) M²⁺ — Incorrect. \( \ce{Ti^{2+}} \) is 3d², with 2 unpaired electrons: μ = √8 ≈ 2.83 BM.

• Option (4) M⁴⁺ — Incorrect. \( \ce{Ti^{4+}} \) is 3d⁰, with no unpaired electron: μ = 0 (diamagnetic).

NCERT (2026-27 edition):
Magnetic Properties – “… 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. 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.,
\( \mu = \sqrt{n(n+2)} \)
where n is the number of unpaired electrons and μ is the magnetic moment in units of Bohr magneton (BM). A single unpaired electron has a magnetic moment of 1.73 Bohr magnetons (BM).”

Note: The same/similar question has been asked in different years by NTA as per the details given below: