(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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 The 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:

NEET 2026 All NEET

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 $61$.