NEETChemistryd and f Block Elements
The +3 oxidation state is the most characteristic state for lanthanoids. Considering this, what is the expected chemical behavior of Ce⁴⁺ and Eu²⁺ ions when present in an aqueous solution?
Options
- ACe⁴⁺ acts as a strong oxidizing agent, while Eu²⁺ acts as a strong reducing agent.
- BCe⁴⁺ acts as a strong reducing agent, while Eu²⁺ acts as a strong oxidizing agent.
- CBoth Ce⁴⁺ and Eu²⁺ act as strong oxidizing agents to achieve the +3 state.
- DBoth ions are chemically inert in aqueous solution due to their stable f⁰ and f⁷ configurations respectively.
Correct answer
A. Ce⁴⁺ acts as a strong oxidizing agent, while Eu²⁺ acts as a strong reducing agent.
Step-by-step solution
The most stable and characteristic oxidation state for all lanthanoids in aqueous solution is +3 . Cerium in the +4 state has a stable 4f⁰ configuration, but it strongly tends to revert to the more thermodynamically stable +3 state. By gaining an electron ( Ce⁴⁺ + e⁻ Ce³⁺ ), it acts as a strong oxidizing agent. Europium in the +2 state has a stable half-filled 4f⁷ configuration. However, it also tends to achieve the common +3 state by losing an electron ( Eu²⁺ Eu³⁺ + e⁻ ). Thus, it acts as a strong reducing agent.