Q83 / 180 · NEET 2026 (Cancelled) · Chemistry
Although $+3$ oxidation state is most common in lanthanoids, cerium still shows $+4$ oxidation state because :
- AAfter losing one more electron, it acquires $4f^{14}$ electronic configuration.
- BIts nearest inert gas is Radon.
- CIts atomic number is $61$.
- Correct answerDAfter losing one more electron, it acquires $4f^{0}$ electronic configuration.
Solution
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.
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.
Hence, the correct answer is option: 'D'.