JEE Main · 2020 · 06 Sep 2020 · Shift 1 · d and f Block Elements

A lanthanoid reaches the $+4$ state only if losing a fourth electron leaves a specially stable $f^0$, $f^7$ or $f^{14}$ configuration. Cerium ($f^1$ as $\mathrm{Ce}{}^{3+}$) drops to $f^0$ and terbium ($f^8$ as $\mathrm{Tb}{}^{3+}$) drops to $f^7$, so both show $+4$, as does dysprosium. Europium, however, is $f^7$ in its $+3$ state and strongly prefers to go the other way to the very stable $f^7$ $\mathrm{Eu}{}^{2+}$ rather than to $+4$, so europium does not show the $+4$ oxidation state.