Given below are two statements :
Statement (I) : In the Lanthanoids, the formation Ce⁺⁴ is favoured by its noble gas configuration.
Statement (II) : Ce⁺⁴ is a strong oxidant reverting to the common +3 state.
In the light of the above statements, choose the most appropriate answer from the options given below :

Solution

Chemistry ma’am · 4 steps

Write cerium's configuration first: Ce⁴⁺ is [Xe], yet +3 is still the more stable state.

1. Configuration of Ce. Cerium (Z = 58) is [Xe] 4f¹ 5d¹ 6s². Removing all four outer electrons leaves Ce⁴⁺ as [Xe], a noble gas configuration.
2. Statement I. That empty 4f⁰ arrangement is the reason cerium can reach +4, which most lanthanoids cannot. So Statement I is true.

   Ce⁴⁺ + e⁻ → Ce³⁺ E° = +1.74 V

3. Statement II. For the lanthanoids as a family, +3 is the most stable state. Ce⁴⁺ readily takes an electron to go back to Ce³⁺, and the large positive E° makes it a strong oxidant. So Statement II is true.
4. So. Both statements are true, which is option (c).

The idea to keep. In lanthanoids, a noble gas, f⁰, f⁷ or f¹⁴ configuration explains why +2 or +4 can form. +3 still wins, so Ce⁴⁺ oxidises and Eu²⁺ reduces to get back there.

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