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Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because:
A
Its nearest inert gas is Radon.
B
After losing one more electron, it acquires $4f^{14}$ electronic configuration.
C
Its atomic number is 61
D
After losing one more electron, it acquires $4f^0$ electronic configuration.
Detailed Solution
$Ce \rightarrow [_{54}Xe]\,6s^2\,5d^1\,4f^1$
$Ce^{4+} \rightarrow [_{54}Xe]\,4f^0$
After losing one more electron (from $Ce^{3+}$), cerium acquires the stable $4f^0$ (noble gas) configuration, hence it shows +4 oxidation state.