Eu2+ (Eu = 63) is stable because

The correct answer is Option (2) → It has half-filled electronic configuration of f-orbitals

Subshells that are half-filled or completely filled have extra stability, because of their symmetrical electron distribution and larger exchange energy. In the lanthanoids, the common oxidation state is +3, but an ion with an empty (4f⁰), half-filled (4f⁷) or completely filled (4f¹⁴) 4f subshell can also be stable in a +2 or +4 state.

For europium:

    \( \displaystyle \mathrm{Eu}\ (Z=63):\ [\mathrm{Xe}]\,4f^{7}\,6s^{2} \)

    \( \displaystyle \mathrm{Eu^{2+}}:\ [\mathrm{Xe}]\,4f^{7}\quad(\text{the two 6s electrons are lost}) \)

\( \ce{Eu^{2+}} \) has seven electrons in the seven 4f orbitals, one in each: a half-filled f subshell. That is why Eu shows a stable +2 state.

Why the other options are incorrect

• Option (1) It has half-filled electronic configuration of d-orbitals — Incorrect. \( \ce{Eu^{2+}} \) has no 5d electrons (5d⁰); its stability comes from the 4f subshell.

• Option (3) It has fully-filled electronic configuration of f-orbitals — Incorrect. A full f subshell is 4f¹⁴, as in \( \ce{Yb^{2+}} \): Yb (Z = 70) is [Xe]4f¹⁴6s², and losing the two 6s electrons leaves 4f¹⁴. \( \ce{Eu^{2+}} \) has only 4f⁷.

• Option (4) It has noble gas configuration — Incorrect. A noble gas configuration ([Xe], 4f⁰) is found in \( \ce{Ce^{4+}} \): Ce (Z = 58) is [Xe]4f¹5d¹6s², and losing all four outer electrons leaves [Xe]. \( \ce{Eu^{2+}} \) still has seven 4f electrons.

NCERT (2026-27 edition):
Oxidation States