Which one of the lanthanoids given below is the most stable in divalent form?
Select the correct option:
A
Ce (Atomic Number 58)
B
Sm (Atomic Number 62)
C
Eu (Atomic Number 63)
D
Yb (Atomic Number 70)
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$
\mathrm{E}_{\mathrm{M}^3+\mathrm{M}^2}^{\circ} \Rightarrow \begin{array}{cc}
\mathrm{Eu} & \mathrm{Yb} \\
-0.35 & -1.05
\end{array}
$$
Hence, due to more reduction potential in Eu as compared to Yb , it can concluded that $\mathrm{Eu}^{2+}$ is more stable than $\mathrm{Yb}^{2+}$.
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