Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because :
- After losing one more electron, it acquires 4f14 electronic configuration.
- Its nearest inert gas is Radon.
- After losing one more electron, it acquires 4f0 electronic configuration.
- Its atomic number is 61.
Answer
(3) After losing one more electron, it acquires 4f0 electronic configuration.
Solution
Ce (Z = 58): [Xe] 4f1 5d1 6s2 ⇒ Ce3+: [Xe] 4f1
Losing one more electron ⇒ Ce4+: [Xe] 4f0
As per NCERT, the occurrence of CeIV is favoured by its noble gas configuration (empty 4f subshell).
Answer: (3)