Which of the following acts as a strong reducing agent? (Atomic number: Ce=58, Eu=63, Gd=64, Lu=71)

Solution

Chemistry ma’am · 3 steps

For lanthanides, +3 is home: an ion sitting below +3 is eager to give up an electron and get there.

1. The rule. The most stable oxidation state for every lanthanide is +3. An ion in +2 tends to lose an electron and rise to +3, so it is a reducing agent. An ion in +4 tends to gain an electron and fall to +3, so it is an oxidising agent.

   Ion	Configuration	Tendency	Acts as
   Eu²⁺	[Xe] 4f⁷	loses e⁻ to reach +3	strong reducing agent
   Gd³⁺	[Xe] 4f⁷	already +3, half-filled	stable, neither
   Lu³⁺	[Xe] 4f¹⁴	already +3, fully filled	stable, neither
   Ce⁴⁺	[Xe] 4f⁰	gains e⁻ to reach +3	strong oxidising agent

2. Why Eu2+ exists. Europium ([Xe] 4f⁷ 6s²) loses only its two 6s electrons, leaving a half-filled 4f⁷ shell. That gives Eu²⁺ some stability, but the pull towards the common +3 state wins. That is why E°(Eu³⁺/Eu²⁺) is about -0.35 V.
3. So. Eu²⁺ readily changes to Eu³⁺ and is the strong reducing agent. The answer is (a).

The idea to keep. In the lanthanides, compare every ion with +3. Below +3 (like Eu²⁺, Sm²⁺) means reducing agent. Above +3 (like Ce⁴⁺, Tb⁴⁺) means oxidising agent.

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