Which one of the lanthanoids given below is the most stable in divalent form?

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KollegeApply · Accepted Answer

Eu (Z = 63) Explanation: \ce{Ce^{2+} -> 4f^1} \ce{Sm^{2+} -> 4f^6 -> half filled configuration} \ce{Eu^{2+} -> 4f^7} \ce{Yb^{2+} -> 4f^14 -> fully filled configuration} \ce{E^\circ_{M^{3+}/M^{2+}} $\Rightarrow$ \underset{- 0.35}{Eu}\phantom{..} \underset{- 1.05}{Yb}} Hence, due to more reduction potential in Eu than Yb, it can be concluded that Eu2+ is more stable than Yb2+.

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