Which one of the following is a diamagnetic ion? (Atomic numbers: Sc = 21, Mn = 25, Co = 27, Cu = 29)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Sc3+
An ion is diamagnetic only when it has no unpaired electrons; any unpaired electron makes it paramagnetic. Each ion loses its 4s electrons first, then 3d electrons:
\( \ce{Cu^2+} \): Cu (Z = 29) is [Ar]3d104s1; removing 4s1 and one 3d electron gives 3d9 → 1 unpaired electron, paramagnetic.
\( \ce{Mn^2+} \): Mn (Z = 25) is [Ar]3d54s2; removing 4s2 gives 3d5 → 5 unpaired electrons, paramagnetic.
\( \ce{Sc^3+} \): Sc (Z = 21) is [Ar]3d14s2; removing 4s2 and 3d1 gives 3d0 → 0 unpaired electrons, diamagnetic.
\( \ce{Co^2+} \): Co (Z = 27) is [Ar]3d74s2; removing 4s2 gives 3d7 → 3 unpaired electrons, paramagnetic.
NCERT (2026-27 edition):
4.3.9 Magnetic Properties – “When a magnetic field is applied to substances, mainly two types of magnetic behaviour are observed: diamagnetism and paramagnetism. Diamagnetic substances are repelled by the applied field while the paramagnetic substances are attracted. … 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.”