Which transition metal ion is most stable against oxidation to its +3 state?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(Mn^{2+}\)
How easily a \( \ce{M^2+} \) ion is oxidised to \( \ce{M^3+} \) depends on the stability of its d configuration before and after the change, which shows up in the value of \( E^{\circ}(\ce{M^3+}/\ce{M^2+}) \). A large positive value means \( \ce{M^3+} \) is readily reduced, so \( \ce{M^2+} \) is hard to oxidise. A half-filled \( \mathrm{d^{5}} \) set is especially stable.
\( \ce{Mn^2+} \): Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); losing the two 4s electrons gives \( \mathrm{3d^{5}} \), half-filled. Oxidation to \( \ce{Mn^3+} \) (\( \mathrm{3d^{4}} \)) would break this stable set, and \( E^{\circ}(\ce{Mn^3+}/\ce{Mn^2+}) = +1.57 \) V is high. So \( \ce{Mn^2+} \) resists oxidation most.
Why the other options are incorrect
Option (2) \(Fe^{2+}\) — Incorrect. Oxidation to \( \ce{Fe^3+} \) produces the stable \( \mathrm{3d^{5}} \) configuration, so \( \ce{Fe^2+} \) is easily oxidised, for example by air. Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \); \( \ce{Fe^2+} \) is \( \mathrm{3d^{6}} \), and losing one more electron gives \( \mathrm{3d^{5}} \) (E° = +0.77 V).
Option (3) \(Cr^{2+}\) — Incorrect. It is oxidised so readily to \( \ce{Cr^3+} \) that it is one of the strongest reducing agents among these ions. Cr (Z = 24) is \( \mathrm{[Ar]\,3d^{5}\,4s^{1}} \); \( \ce{Cr^2+} \) is \( \mathrm{3d^{4}} \), and \( \ce{Cr^3+} \), \( \mathrm{3d^{3}} \) (half-filled \( t_{2g} \)), is very stable (E° = −0.41 V).
Option (4) \(Fe\) — Incorrect. This is the metal, not an ion; iron is easily oxidised (it rusts) to the +2 and +3 states.
NCERT (2026-27 edition):
Trends in the M3+/M2+ Standard Electrode Potentials
