Question

The outer electronic configuration of two members of lanthanoids series are as follows $4f^{1}5d^{1}6s^2$ and $4f^{7}5d^{0}6s^2$ what are their atomic numbers? Predict heir oxidation states.

Answer 1

$\left[Xe\right]4f^{1}5d^{1}6s^2=58⟶$ Cerium ($Ce$)

Oxidation states of $Ce=+3$ and $+4$ are stable oxidation states of $Ce$.

$\left[Xe\right]4f^{7}5d^{0}6s^2=63⟶$ Europium ($Eu$)

Stable oxidation state of $Eu$ is $+2$. Also has an oxidation state of $+3$.