JEE Main · 2022 · 28 Jun 2022 · Shift 1 · d and f Block Elements

Among lanthanoids, the +2 oxidation state is most stable when the resulting ${\mathrm{Ln}}^{2+}$ ion has a stable electronic configuration (half-filled $f^7$ or fully-filled $f^{14}$).

Europium (Eu): ${\mathrm{Eu}}^{3+}$ is $[\mathrm{Xe}]4f^6$, so ${\mathrm{Eu}}^{2+}$ is $[\mathrm{Xe}]4f^7$ (half-filled, very stable).

Ytterbium (Yb): ${\mathrm{Yb}}^{3+}$ is $[\mathrm{Xe}]4f^{13}$, so ${\mathrm{Yb}}^{2+}$ is $[\mathrm{Xe}]4f^{14}$ (fully-filled, stable).

The standard reduction potential $E_{{\mathrm{M}}^{3+}/{\mathrm{M}}^{2+}}^{\circ }$ for Eu is $-0.35\mathrm{V}$ and for Yb is $-1.05\mathrm{V}$. A less negative (more positive) reduction potential means the +2 state is more stable. Therefore, ${\mathrm{Eu}}^{2+}$ is more stable than ${\mathrm{Yb}}^{2+}$, making Eu the most stable in the divalent form among the options.