The  electronic configurations of  bivalent europium and trivalent cerium are :

(atomic number : $\mathrm{Xe}=54,\mathrm{Ce}=58,\mathrm{Eu}=63$ )

The atomic numbers of the given elements, Europium and Cerium are 63 and 58 respectively and their electronic configuration are:

For europium (Eu) =$$[Xe]4f^{7}6s^2$$
For cerium (Ce)=$\left[Xe\right]4f^{1}5d^{1}6s^2$

Europium atom is made bivalent after removing 2 electrons and the cerium atom is made trivalent by removing three electrons from the atom. Their electronic configuration becomes:

$$\begin{gathered} {{\text{Eu}}^2}^{\text{+}}\to \left[\text{Xe}\right]{\text{4f\hspace{0.17em}}}^{\text{7}} \\ {\text{Ce}}^{\text{3+}}\to \left[\text{Xe}\right]{\text{4f\hspace{0.17em}}}^{\text{1}} \end{gathered}$$