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.

Explanation

`Ce(58),Pr(59),Nd(60),Tb(65),Dy(66)=+4O.S`.
`Nd(60),Sm(62),Eu(63),Tm(69),Yb(70)=+2O.S`.
`+2` oxidation state is exhibited when the lanthanoid has the configuration `5d^(0)6s^(2)` so that 2 electrons are easily lost `+4` oxidation state is exhibited when the configuration left is close to `4f^(0)(e.g.4f^(0),4f^(1),4f^(2))` or close to `4f^(7)(e.g.4f^(7) or 4f^(8))`.