Question

Answer the following

Europium and ytterbium behave as good reducing agents in the +2 oxidation state explain.

Answer 1

i. Europium (Eu) and Ytterbium (Yb) show +2 oxidation states. Their electronic configurations are given below:

Eu = [Xe] 4f⁷ 6s² ; Eu²⁺ = [Xe] 4f⁷

Yb = [Xe] 4f¹⁴ 6s² ; Yb²⁺ = [Xe] 4f¹⁴

ii. It is clear from the configuration of Eu that Eu²⁺ is favoured by its half-filled f-subshell. But it can be easily converted into stable Eu³⁺ by the loss of an electron. Due to this reason, Eu²⁺ is a good reducing agent.

iii. Similarly, Yb²⁺ ion is stabilized due to completely filled f-subshell. It can be easily converted into Yb³⁺ by the loss of an electron. Due to this reason, Yb²⁺ is a good reducing agent.