JEE Main 2023 · 6 April · morning · Single correct · +4 / −1
Strong reducing and oxidizing agents among the following, respectively, are :
- (a) Ce⁴⁺ and Eu²⁺
- (b) Eu²⁺ and Ce⁴⁺
- (c) Ce⁴⁺ and Tb⁴⁺
- (d) Ce³⁺ and Ce⁴⁺
Answer and solution
Answer (b) Eu²⁺ and Ce⁴⁺
Solution
Chemistry ma’am · 4 steps
For lanthanides, +3 is home: an ion in +2 wants to lose an electron, one in +4 wants to gain one.
- Most stable state. For every lanthanide, the +3 oxidation state is the most stable one. Ions in +2 or +4 states tend to return to +3.
- Europium(II). Eu²⁺ readily gives up one electron to become Eu³⁺, even though Eu²⁺ has a half-filled 4f⁷ shell. Losing an electron means it reduces something else, so it is a strong reducing agent (E°_(Eu³⁺/Eu²⁺) ≈ -0.35 V).
- Cerium(IV). Ce⁴⁺ readily takes one electron to become Ce³⁺, even though Ce⁴⁺ has the stable 4f⁰ configuration. Gaining an electron means it oxidises something else, so it is a strong oxidising agent (E°_(Ce⁴⁺/Ce³⁺) = +1.74 V).
- So. The strong reducing agent is Eu²⁺ and the strong oxidising agent is Ce⁴⁺, in that order. The answer is (b).
The idea to keep. For lanthanides, a +2 ion goes back to +3 by losing an electron, so it is a reducing agent. A +4 ion goes back to +3 by gaining one, so it is an oxidising agent.
Answer from the NTA final answer key. Solution by PrepWiser.