Although +3 oxidation state is most common in lanthanoids, cerium...

The atomic number of Cerium (Ce) is $58$.

The electronic configuration of Ce is $[Xe] 4f^{1} 5d^{1} 6s^{2}$.

In $+3$ oxidation state, the electronic configuration of $Ce^{3+}$ is $[Xe] 4f^{1}$.

By losing one more electron, it forms $Ce^{4+}$ ion.

The electronic configuration of $Ce^{4+}$ is $[Xe] 4f^{0}$, which is a highly stable noble gas configuration.

Thus, Cerium shows $+4$ oxidation state because after losing one more electron from $+3$ state, it acquires $4f^{0}$ electronic configuration.

Answer: After losing one more electron, it acquires $4f^{0}$ electronic configuration.