Which one of the lanthanoids given below is the most stable in divalent form?

We need to determine which lanthanoid is most stable in the divalent ($$Ln^{2+}$$) form. Lanthanoids are most stable in the +3 oxidation state. However, some lanthanoids can exhibit the +2 oxidation state, and the stability of the +2 state depends on the electronic configuration of the resulting ion. Ions with empty ($$4f^0$$), half-filled ($$4f^7$$), or completely filled ($$4f^{14}$$) configurations are exceptionally stable.

Option A: Yb (Z = 70) Yb atom: $$[Xe] 4f^{14} 6s^2$$, so $$Yb^{2+}$$ is $$[Xe] 4f^{14}$$ — completely filled 4f shell.

Option B: Sm (Z = 62) Sm atom: $$[Xe] 4f^6 6s^2$$, so $$Sm^{2+}$$ is $$[Xe] 4f^6$$ — neither half-filled nor completely filled.

Option C: Eu (Z = 63) Eu atom: $$[Xe] 4f^7 6s^2$$, thus $$Eu^{2+}$$ has $$[Xe] 4f^7$$ — exactly half-filled 4f shell.

Option D: Ce (Z = 58) Ce atom: $$[Xe] 4f^1 5d^1 6s^2$$, giving $$Ce^{2+}$$ as $$[Xe] 4f^1 5d^1$$ — neither half-filled nor completely filled.

Both $$Eu^{2+}$$ ($$4f^7$$, half-filled) and $$Yb^{2+}$$ ($$4f^{14}$$, completely filled) have stable configurations. However, $$Eu^{2+}$$ is considered the most stable divalent lanthanoid because the half-filled $$4f^7$$ configuration provides exceptional stability due to maximum exchange energy, and europium has the strongest tendency among all lanthanoids to form the +2 state, as evidenced by its highly negative reduction potential for the $$Eu^{3+}/Eu^{2+}$$ couple ($$-0.36$$ V), meaning $$Eu^{2+}$$ strongly resists oxidation to $$Eu^{3+}$$.

The correct answer is Option C: Eu (Atomic Number 63).