JEE Main 2022 · 25 June · morning · Single correct · +4 / −1
Cerium (IV) has a noble gas configuration. Which of the following is correct statement about it?
- (a) It will not prefer to undergo redox reactions.
- (b) It will prefer to gain electron and act as an oxidizing agent.
- (c) It will prefer to give away an electron and behave as reducing agent.
- (d) It acts as both, oxidizing and reducing agent.
Answer and solution
Answer (b) It will prefer to gain electron and act as an oxidizing agent.
Solution
Chemistry ma’am · 4 steps
Ce(IV) has no electrons left to give, so it can only take one and drop back to the comfortable +3 state.
- Write configurations. Cerium is [Xe]4f¹ 5d¹ 6s². In Ce⁴⁺ all four outer electrons are gone, leaving the bare [Xe] core. Ce³⁺ is [Xe]4f¹.
- Cannot lose more. Removing another electron from Ce⁴⁺ would mean breaking into the noble gas core, which costs far too much energy. So Ce⁴⁺ cannot act as a reducing agent.
- Pull towards +3. For lanthanoids, +3 is the most stable oxidation state. Ce⁴⁺ readily accepts one electron to become Ce³⁺, and the large positive reduction potential confirms this.
Ce⁴⁺ + e⁻ → Ce³⁺ E° = +1.74 V
- So. Ce⁴⁺ prefers to gain an electron and acts as a strong oxidising agent, which is option (b).
The idea to keep. A lanthanoid ion in any state other than +3 wants to return to +3. Above +3 it is an oxidising agent, like Ce⁴⁺. Below +3 it is a reducing agent, like Eu²⁺ or Yb²⁺.
Answer from the NTA final answer key. Solution by PrepWiser.