Given below are two statements :
Statement I : CeO₂ can be used for oxidation of aldehydes and ketones.
Statement II : Aqueous solution of EuSO₄ is a strong reducing agent.
In the light of the above statements, choose the correct answer from the options given below :

Solution

Chemistry ma’am · 4 steps

For lanthanoids, +3 is home: Ce(IV) wants to drop to +3, Eu(II) wants to climb to +3.

1. The usual state. The most stable oxidation state for every lanthanoid is +3. An ion in any other state tends to change back to +3, and that tendency decides whether it oxidises or reduces.
2. Statement I. In CeO₂ cerium is +4. Ce⁴⁺ readily gains one electron to become Ce³⁺ (E° = +1.74 V), so it is a strong oxidising agent. It can oxidise organic compounds such as aldehydes and ketones. True.

   Ce⁴⁺ + e⁻ → Ce³⁺ Eu²⁺ → Eu³⁺ + e⁻

3. Statement II. In EuSO₄ europium is +2. Eu²⁺ readily loses one electron to become Eu³⁺, so in water it acts as a strong reducing agent. True.
4. So. Both statements are true, so the answer is (d).

The idea to keep. For any lanthanoid not in the +3 state, ask which way it must move to reach +3. Moving down (like Ce⁴⁺) makes it an oxidant. Moving up (like Eu²⁺, Yb²⁺) makes it a reductant.

Answer from the NTA final answer key. Solution by PrepWiser.