164. Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit + 2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.

Cerium (Ce) and Terbium (Tb) show +4 oxidation state. Their electronic configurations are given below:
Ce = [Xn] 4f¹ 5d¹ 6s²
Tb = [Xn] 4f⁰ 6s²
It is clear from the configuration of Ce that Ce⁺⁴ is favoured by its noble gas configuration i.e., [Xn] 4f⁰ 5d⁰ 5s⁰, but can be easily converted into Ce³⁺ ([Xn] 4f¹ 5d⁰ 6s⁰). Due to this reason Ce⁺⁴ is an oxidising agent.

Tb⁴⁺ ion is stabilized due to half filled f-subshell i.e., [Xn] 4f⁷. It also acts as an oxidant.
Europium (63) and ytterbium (70) show +2 oxidation state, this acts as reducing agents because they can be converted into common oxidation state +3. The electronic configuration of Eu and Y are as follows:

Eu = [Xn] 4f 6s² Y = [Xn] 4f¹⁴ 6s² Formation of Eu²⁺ ion leaves 4f⁷ configuration and Y²⁺ ion leaves 4f¹⁴ configuration. These configurations can stable due to half filled and full filled f-subshell. Samarium, Sm (62) 4f⁶ 6s² also shows both +2 and +3 oxidation states like europium.