Which of the following metal ions will display maximum value of spin-…

The correct answer is Option (1) → Fe³⁺

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

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

The more unpaired electrons, the larger the moment. By Hund’s rule, the five 3d orbitals are first filled singly, so the number of unpaired electrons rises up to d⁵ (five unpaired) and then falls. To form an ion of a 3d metal, the 4s electrons are removed first, then 3d electrons.

• Fe³⁺: Fe (Z = 26) is [Ar]3d⁶4s²; removing two 4s and one 3d electron gives 3d⁵, n = 5, μ = √35 = 5.92 BM.

• Co³⁺: Co (Z = 27) is [Ar]3d⁷4s²; removing the two 4s electrons and one 3d electron gives 3d⁶, n = 4, μ = √24 = 4.90 BM.

• Mn³⁺: Mn (Z = 25) is [Ar]3d⁵4s²; removing the two 4s electrons and one 3d electron gives 3d⁴, n = 4, μ = 4.90 BM.

• Cr²⁺: Cr (Z = 24) is [Ar]3d⁵4s¹; removing the 4s electron and one 3d electron gives 3d⁴, n = 4, μ = 4.90 BM.

So Fe³⁺, with five unpaired electrons, has the maximum spin-only magnetic moment.

Shortcut: A d⁵ ion always has the largest number of unpaired electrons (five) among free 3d ions, so just look for the d⁵ ion.

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