The transition elements exhibit variable oxidation states because (n−1)d electrons can also participate in bonding, as the energy difference between (n−1)d and ns orbitals is very small. The maximum oxidation state that can be exhibited by a transition element is equal to the total number of electrons present in both ns and (n−1)d orbitals. The elements in the middle exhibit more oxidation states, e.g., manganese exhibits all oxidation states from +2 to +7. In the second series, a stable maximum oxidation state is exhibited by technetium, and an unstable maximum oxidation state of +8 is exhibited by ruthenium. In the third series, a stable maximum oxidation state of +8 is exhibited by osmium. Transition metal ions having a completely filled or an exactly half-filled d subshell are stable. The Fe3+ ion is more stable than the Fe2+ ion because of the stable half-filled 3d5 electronic configuration of Fe3+.
The extra stability of the ferric ion (Fe3+) is due to its
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Half-filled d orbitals
The passage states that ions with a completely filled or a half-filled d subshell are stable, and that \( \ce{Fe^3+} \) is more stable than \( \ce{Fe^2+} \) because of its half-filled 3d5 configuration.
Configuration of Fe: Fe (Z = 26) is [Ar]3d64s2.
Electrons removed: the two 4s electrons and one 3d electron, giving \( \ce{Fe^3+} \): 3d5.
Result: each of the five 3d orbitals holds one electron, a half-filled d subshell, which is extra stable.
Why the other options are incorrect
Option (1) Completely filled d orbitals — Incorrect. A completely filled d subshell is 3d10 (as in \( \ce{Zn^2+} \)); \( \ce{Fe^3+} \) has only five d electrons.
Option (3) Empty d orbitals — Incorrect. Empty d orbitals (3d0) belong to ions such as \( \ce{Sc^3+} \); \( \ce{Fe^3+} \) still has five d electrons.
Option (4) Completely filled 4s orbital — Incorrect. The 4s electrons are the first to be lost, so \( \ce{Fe^3+} \) has no 4s electron at all.