Chemistry - f-block Question with Solution | TestHub

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

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

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

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

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

${\mathrm{E}}_{{\mathrm{M}}^{3+}/{\mathrm{M}}^{2+}}^{°}\Rightarrow \begin{array}{cc}\mathrm{Eu}& \mathrm{Yb}\\ -0.35& -1.05\end{array}$

Hence, due to more reduction potential in $\mathrm{Eu}$ as compared to $\mathrm{Yb}$, it can concluded that ${\mathrm{Eu}}^{2+}$ is more stable than ${\mathrm{Yb}}^{2+}$.