Which of the following ions is the strongest oxidizing agent? (Atomic Number of Ce=58, Eu=63, Tb=65, Lu=71)

Solution

Chemistry ma’am · 3 steps

For lanthanides, +3 is home: an ion sitting above +3 grabs an electron to get back, so it oxidises.

1. The rule. The most stable oxidation state for every lanthanide is +3. An ion in a higher state takes an electron to return to +3, so it acts as an oxidising agent. An ion in a lower state gives an electron away, so it acts as a reducing agent.

   Ion	4f configuration	Compared with +3	Behaviour
   Lu³⁺	4f¹⁴	already +3	stable, little redox tendency
   Eu²⁺	4f⁷	below +3	loses an electron, reducing agent
   Tb⁴⁺	4f⁷	above +3	gains an electron, oxidising agent
   Ce³⁺	4f¹	already +3	can only lose an electron (to Ce⁴⁺), not an oxidant

2. Why Tb4+. Terbium is [Xe] 4f⁹6s², so Tb⁴⁺ is 4f⁷. Even with this half-filled shell, the strong pull back to the +3 state makes Tb⁴⁺ readily accept an electron, Tb⁴⁺ + e⁻ → Tb³⁺.
3. So. Tb⁴⁺ is the only ion here above +3, so it is the strongest oxidising agent: option (c).

The idea to keep. For lanthanide ions, compare the charge with +3. Above +3 (like Ce⁴⁺, Tb⁴⁺) means oxidant. Below +3 (like Eu²⁺, Sm²⁺, Yb²⁺) means reductant.

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