Question
Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because :
NEET
Chemistry
d and f Block Elements
2026
NEET 2026 (Cancelled)
NEET Chemistry Question (2026) — Solution
Options
- A. After losing one more electron, it acquires 4f^ 14 electronic configuration.
- B. Its nearest inert gas is Radon.
- C. Its atomic number is 61.
- D. After losing one more electron, it acquires 4f^ 0 electronic configuration.
Answer
D. After losing one more electron, it acquires 4f^ 0 electronic configuration.
Step-by-step 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.
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