The spin only magnetic moment of [MBr_4]^2- is 5.9 BM. M is:

The correct answer is Option (3) → Mn

The spin-only magnetic moment depends only on the number of unpaired electrons, n:

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

With μ = 5.9 BM:

    \( n(n+2) = 5.9^2 \approx 35 \Rightarrow n = 5 \qquad (\sqrt{5 \times 7} = \sqrt{35} = 5.92) \)

The oxidation state x of M in [MBr₄]²⁻:

    \( x + 4(-1) = -2 \Rightarrow x = +2 \)

So M²⁺ must have five unpaired electrons, a 3d⁵ ion (\( \ce{Br^{-}} \) is a weak field ligand and does not pair electrons). Mn (Z = 25) is [Ar]3d⁵4s²; losing the two 4s electrons gives Mn²⁺: 3d⁵, five unpaired electrons. So M is Mn.

Why the other options are incorrect

• Option (1) Fe — Incorrect. Fe (Z = 26) [Ar]3d⁶4s² gives \( \ce{Fe^{2+}} \) 3d⁶, four unpaired electrons: μ = √24 = 4.90 BM.

• Option (2) Cr — Incorrect. Cr (Z = 24) [Ar]3d⁵4s¹ gives \( \ce{Cr^{2+}} \) 3d⁴, four unpaired electrons: 4.90 BM.

• Option (4) Co — Incorrect. Co (Z = 27) [Ar]3d⁷4s² gives \( \ce{Co^{2+}} \) 3d⁷, three unpaired electrons: μ = √15 = 3.87 BM.

NCERT (2026-27 edition):
Magnetic Properties – “… A single unpaired electron has a magnetic moment of 1.73 Bohr magnetons (BM). The magnetic moment increases with the increasing number of unpaired electrons. Thus, the observed magnetic moment gives a useful indication about the number of unpaired electrons present in the atom, molecule or ion.”