In acidic solution $\mathrm{Ce}{}^{4+}$ is a strong oxidising agent even though it has the stable $4f^0$ configuration. The best explanation is that

1. A$\mathrm{Ce}{}^{4+}$ has $4f^{14}$ configuration
2. B$\mathrm{Ce}{}^{4+}$ has a half-filled $4f$ subshell that it wishes to complete
3. C$\mathrm{Ce}{}^{4+}$ has a very small ionisation enthalpy
4. D$\mathrm{Ce}{}^{3+}$ is the characteristic stable oxidation state of lanthanoids, so $\mathrm{Ce}{}^{4+}$ tends to gain an electron