The most common oxidation state of Lanthanoid elements is +3. Which of the following is likely to deviate easily from +3 oxidation state?
- (A)Ce (At. No. 58)✓ Correct
- (B)La (At. No. 57)
- (C)Lu (At. No. 71)
- (D)Gd (At. No. 64)
Answer: (A)
Solution
Cerium (Ce) can easily deviate from the +3 oxidation state to +4.
Electronic configuration:
The +4 oxidation state of cerium is particularly stable because it achieves a noble gas configuration with an empty 4f subshell. This makes Ce the lanthanoid most likely to deviate from the +3 oxidation state.
The other lanthanoids listed do not achieve such stable configurations in the +4 state.
Electronic configuration:
- Ce: [Xe]
- : [Xe]
- : [Xe] (noble gas configuration)
The +4 oxidation state of cerium is particularly stable because it achieves a noble gas configuration with an empty 4f subshell. This makes Ce the lanthanoid most likely to deviate from the +3 oxidation state.
The other lanthanoids listed do not achieve such stable configurations in the +4 state.
JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Single correct option