(C) Cerium $(Ce)$ has an atomic number of $58$. Its electronic configuration is $[Xe] 4f^1 5d^1 6s^2$. By losing four electrons, it reaches the stable

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

(C) Cerium $(Ce)$ has an atomic number of $58$. Its electronic configuration is $[Xe] 4f^1 5d^1 6s^2$. By losing four electrons, it reaches the stable noble gas configuration of Xenon $(4f^0)$, which is why it readily exhibits the $+4$ oxidation state.

Class 12 Chemistry d-and f-Block Elements Lanthanoids and Actinoids

Medium

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

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A
After losing one more electron, it acquires $4f^{14}$ electronic configuration.
B
Its nearest inert gas is Radon.
C
After losing one more electron, it acquires $4f^0$ electronic configuration.
D
Its atomic number is $61$.