Which one of the lanthanoids given below is the most stable in...

${Ce}^{+ 2}: \left(xe\right)4{f}^{2}$

${Eu}^{+ 2}: \left(xe\right)4{f}^{7}$

${Sm}^{+ 2}: \left(xe\right)4{f}^{6}$

${Yb}^{+ 2}: \left(xe\right)4{f}^{14}$

As ${Yb}^{+ 2}and{Eu}^{+ 2}$ has completely filled and half filled $4f$ orbital, so they would be the most stable. But among the these two ${Eu}^{+ 2}$ is more stable based on reduction potential value.

${E}_{{M}^{3 +} / {M}^{2 +}}^{°} ⇒$\begin{cases}Eu & Yb \\ - 0 . 35 & - 1 . 05\end{cases}$$</p><p>Hence, due to more reduction potential in$Eu$as compared to$Yb$, it can concluded that${Eu}^{2 +}$is more stable than${Yb}^{2 +}$.