Examine Statement I

Cerium has atomic number $58$.

Its electronic configuration is approximately: $\mathrm{Ce}: [\mathrm{Xe}]\,4f^1 5d^1 6s^2$

For $\mathrm{Ce}^{3+}$, removal of three electrons gives: $\mathrm{Ce}^{3+} : [\mathrm{Xe}]\,4f^1$

For $\mathrm{Ce}^{4+}$, removal of one more electron gives: $\mathrm{Ce}^{4+} : [\mathrm{Xe}]$

This is the noble gas configuration of xenon. Hence, formation of $\mathrm{Ce}^{4+}$ is especially favoured.

So, Statement I is true.

Examine Statement II

In lanthanoids, the most common oxidation state is $+3$.

Since $\mathrm{Ce}^{4+}$ tends to revert to the more common and stable $\mathrm{Ce}^{3+}$ state, it readily gains an electron: $\mathrm{Ce}^{4+} + e^- \rightarrow \mathrm{Ce}^{3+}$

Therefore, $\mathrm{Ce}^{4+}$ acts as a strong oxidising agent.

So, Statement II is true.

Choose the correct option

Since both statements are true, the correct option is: $\boxed{\text{C}}$

Comparison with stored answer

Stored correct answer: $\text{C}$

My derived answer: $\text{C}$

They match.