NEET 2026 (Phase 1) · ChemistryLanthanoidsPrevious Year Question
Although the +3 oxidation state is most common in lanthanoids, cerium still shows the +4 oxidation state because:
- A.(a) After losing one more electron, it acquires \ce{4f^14} electronic configuration.
- B.(b) Its nearest inert gas is Radon.
- C.(c) Its atomic number is 61.
- D.(d) After losing one more electron, it acquires \ce{4f^0} electronic configuration.✓
Correct Answer
(D) (d) After losing one more electron, it acquires \ce{4f^0} electronic configuration.
Solution & Explanation
\textbf{Answer:} (d) After losing one more electron, it acquires \ce{4f^0} electronic configuration. \textbf{Solution:} Cerium in the +3 state is \ce{Ce^3+} (\ce{[Xe] 4f^1}). On losing one more electron it forms \ce{Ce^4+} with the configuration \ce{[Xe] 4f^0}, i.e. the stable, empty 4f (noble-gas \ce{[Xe]}) configuration. This extra stability of the \ce{4f^0} configuration is why cerium readily shows the +4 oxidation state, in addition to the usual +3 common to lanthanoids.
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