Paramagnetic substances are attracted by an external magnetic field, that is, they move from a weaker to a stronger part of the field. Paramagnetic substances contain unpaired electrons. Diamagnetic substances are repelled by an external magnetic field, that is, they tend to move away from the stronger part to the weaker part of the field. In diamagnetic substances, all the electrons are paired.
The magnetic moment of a substance arises from the spin angular momentum and the orbital angular momentum of its electrons. In many compounds of the 3d series metals, the orbital contribution is neglected, because the orbital motion of these electrons is quenched by the surrounding species in the compound or solution. The magnetic moment is then calculated by the spin-only formula, in Bohr magnetons (BM):
\[\mu_s = \sqrt{n(n + 2)}\ \text{BM}\]
where n is the number of unpaired electrons.
Which of the following ions shows the highest spin-only magnetic moment?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Fe3+
The spin-only magnetic moment \( \mu=\sqrt{n(n+2)}\ \mathrm{BM} \) rises with the number of unpaired electrons n, so the ion with the most unpaired d electrons has the highest moment. Each ion below loses its 4s electrons first and then 3d electrons.
V3+ — V (Z = 23) is [Ar]3d34s2; V3+ is 3d2, n = 2, \( \mu=\sqrt{8}=2.83\ \mathrm{BM} \).
Cr3+ — Cr (Z = 24) is [Ar]3d54s1; Cr3+ is 3d3, n = 3, \( \mu=\sqrt{15}=3.87\ \mathrm{BM} \).
Fe3+ — Fe (Z = 26) is [Ar]3d64s2; Fe3+ is 3d5, one electron in each 3d orbital, n = 5, \( \mu=\sqrt{35}=5.92\ \mathrm{BM} \).
Co3+ — Co (Z = 27) is [Ar]3d74s2; Co3+ is 3d6, n = 4, \( \mu=\sqrt{24}=4.90\ \mathrm{BM} \).
Fe3+, with the half-filled 3d5 set, has the most unpaired electrons and the highest spin-only moment.
NCERT (2026-27 edition):
4.3.9 Magnetic Properties – “The magnetic moment increases with the increasing number of unpaired electrons.”