Transition metal ions with zero unpaired electrons are:
A. \(Sc^{3+}\)
B. \(Ti^{3+}\)
C. \(Cu^{2+}\)
D. \(Zn^{2+}\)
E. \(Mn^{2+}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A and D only
To count unpaired electrons in a transition metal ion, write the configuration of the neutral atom, remove the 4s electrons first and then 3d electrons, and fill the five 3d orbitals singly before pairing (Hund’s rule). An ion with d0 or d10 has no unpaired electron, so it is diamagnetic and colourless.
A. Sc3+ — Correct. Sc (Z = 21) is [Ar]3d14s2; removing three electrons gives Sc3+: [Ar]3d0, a noble gas configuration with 0 unpaired electrons.
B. Ti3+ — Incorrect. Ti (Z = 22) is [Ar]3d24s2; Ti3+ is 3d1, with 1 unpaired electron.
C. Cu2+ — Incorrect. Cu (Z = 29) is [Ar]3d104s1; losing the 4s electron and one 3d electron gives Cu2+: 3d9, with 1 unpaired electron.
D. Zn2+ — Correct. Zn (Z = 30) is [Ar]3d104s2; losing the two 4s electrons gives Zn2+: 3d10, completely filled, with 0 unpaired electrons.
E. Mn2+ — Incorrect. Mn (Z = 25) is [Ar]3d54s2; Mn2+ is 3d5, one electron in each d orbital: 5 unpaired electrons.
NCERT (2026-27 edition):
Trends in the M3+/M2+ Standard Electrode Potentials – “… The low value for Sc reflects the stability of Sc3+ which has a noble gas configuration. The highest value for Zn is due to the removal of an electron from the stable d10 configuration of Zn2+.”