Question

Although +3 oxidation state is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why?

Answer 1

The characteristic oxidation state of lanthanoids is +3, as they generally lose 5d and 6s electrons.

But in case of Cerium $(Ce=[Xe\left]4f^{1}5d^{1}6s^2\right)$, it tends to attain stable configuration $\left[Xe\right]4f^0$ which is a noble gas configuration after losing four electrons.

Therefore, Ce exhibits +4 oxidation state as well.