Which one of these tripositive ions is the most stable in aqueous solution?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(Cr^{3+}\)
An ion is stable in aqueous solution when it shows no tendency to be oxidised or reduced by water or dissolved air. In the 3d series this depends largely on the d-electron configuration: a half-filled \( t_{2g} \) level (d3) in an octahedral aqua complex is particularly stable. First derive the configuration of each ion:
Ti (Z = 22) is \( \mathrm{[Ar]\,3d^{2}\,4s^{2}} \). Losing the two 4s electrons and one 3d electron gives \( \ce{Ti^3+} \): \( \mathrm{3d^{1}} \).
V (Z = 23) is \( \mathrm{[Ar]\,3d^{3}\,4s^{2}} \); \( \ce{V^3+} \) is \( \mathrm{3d^{2}} \).
Cr (Z = 24) is \( \mathrm{[Ar]\,3d^{5}\,4s^{1}} \). Losing the 4s electron and two 3d electrons gives \( \ce{Cr^3+} \): \( \mathrm{3d^{3}} \).
Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); \( \ce{Mn^3+} \) is \( \mathrm{3d^{4}} \).
\( \ce{Cr^3+} \) (d3) has the three \( t_{2g} \) orbitals singly filled. That is why \( \ce{Cr^2+} \) (d4) is a reducing agent that readily becomes \( \ce{Cr^3+} \), while \( \ce{Cr^3+} \) itself neither gains nor loses an electron easily. So the most stable tripositive ion in water is \( \ce{Cr^3+} \), chromium(III).
Why the other options are incorrect
Option (2) \(Ti^{3+}\) — Incorrect. With a single 3d electron, \( \ce{Ti^3+} \) is a reducing agent; air oxidises it to the d0 state of titanium(IV).
Option (3) \(V^{3+}\) — Incorrect. \( \ce{V^3+} \) (d2) is also slowly oxidised in air to vanadium(IV), \( \ce{VO^2+} \).
Option (4) \(Mn^{3+}\) — Incorrect. \( \ce{Mn^3+} \) (d4) is a strong oxidising agent: gaining one electron gives the extra-stable half-filled d5 configuration of \( \ce{Mn^2+} \), so it oxidises water and disproportionates.
NCERT (2026-27 edition):
Trends in the M2+/M Standard Electrode Potentials – “… Cr2+ is reducing as its configuration changes from d4 to d3, the latter having a half-filled t2g level (see Unit 5). On the other hand, the change from Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra stability.”