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

The members of lanthanide series along with their electronic configuration are given in the below table:

The typical oxidation state of the lanthanides is +3. The oxidation state of +2 and +4 are exhibited by some of the elements. These are shown by those elements which by losing 2 or 4 elements acquire a stable configuration.

+2 oxidation state is exhibited when the lanthanide has the configuration 5d⁰6s² so that 2 electrons are lost easily. Hence the members which will show +2 oxidation state are:

+2 = ₆₀Nd, ₆₂Sm, ₆₃Eu, ₆₉Tm, ₇₀Yb

+4 oxidation state is exhibited when the configuration left (by losing 2 electrons) is close to 4f⁰ (example 4f⁰ , 4f¹, 4f², 4f³) or close to 4f⁷ (example 4f⁷or 4f⁸) or close to 4f¹⁴ ( 4f¹³ or 4f¹⁴)

Hence, the members which will show +4 oxidation state are:

+4 = ₅₈Ce, ₅₉Pr, ₆₀Nd, ₆₅Tb, ₆₆Dy

Note: Each case tends to revert to the more stable oxidation state of +3 by loss or gain of an electron. That is why Sm²⁺, Eu²⁺ and Yb²⁺ ions in solutions are good reducing agents and aqueous solution of Ce⁴⁺ and Tb⁴⁺ are good oxidizing agents.