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

Cerium (Ce) has atomic number 58. Its ground‑state electron configuration is:

${\mathrm{Xe}}_{} 4f1 5d1 6s2$

When cerium forms the common +3 oxidation state, it loses the two 6s electrons and the 5d electron, leaving a single 4f electron:

$Ce^\{3+\}\; :\; \backslash ;\; [Xe] 4f1$

To reach the +4 oxidation state, one more electron is removed. That electron comes from the 4f subshell, which then becomes empty:

$Ce^\{4+\}\; :\; \backslash ;\; [Xe] 4f0$

The empty 4f subshell (4f⁰) is relatively low in energy because there is no electron‑electron repulsion within that shell. This extra stability makes the +4 state accessible for cerium, even though +3 is the more typical oxidation state for lanthanides.

Therefore, the reason cerium can exhibit a +4 oxidation state is that after losing one more electron, it acquires a 4f⁰ electronic configuration.