Although the +3 oxidation state is most common in lanthanoids, cerium still shows the +4 oxidation state because:

Solution & Explanation

\textbf{Answer:} (d) After losing one more electron, it acquires $4\mathrm{f}{}^0$ electronic configuration. \textbf{Solution:} Cerium in the +3 state is $\mathrm{Ce}{}^{3+}$ ($[\mathrm{Xe}]4\mathrm{f}{}^1$). On losing one more electron it forms $\mathrm{Ce}{}^{4+}$ with the configuration $[\mathrm{Xe}]4\mathrm{f}{}^0$, i.e. the stable, empty 4f (noble-gas $[\mathrm{Xe}]$) configuration. This extra stability of the $4\mathrm{f}{}^0$ configuration is why cerium readily shows the +4 oxidation state, in addition to the usual +3 common to lanthanoids.