Consider the following statements regarding the oxidation states of lanthanoids: I. Europium (Eu) exhibits a +2 oxidation state due to the extra stability of the half-filled 4f⁷ configuration. II. Ytterbium (Yb) in its +2 oxidation state acts as a strong oxidizing agent in aqueous solutions. III. Cerium (Ce) exhibits a +4 oxidation state and acts as an oxidizing agent, reverting to the common +3 state. Which of the s

Step-by-step solution

The electronic configuration of Europium (Eu, Z=63 ) is [ Xe ] 4f⁷ 6s² . By losing two electrons, it forms Eu ²⁺ with a stable half-filled 4f⁷ configuration. Thus, Statement I is correct. The electronic configuration of Ytterbium (Yb, Z=70 ) is [ Xe ] 4f¹⁴ 6s² . It forms Yb ²⁺ with a fully filled 4f¹⁴ configuration. However, since the +3 oxidation state is the most stable state for all lanthanoids, Yb ²⁺ tends to lose an electron to form Yb ³⁺ , thereby acting as a reducing agent, not an oxidizing agent. Thus, Stat