Question

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

Options

• Ce (Atomic Number 58) 

• Sm (Atomic Number 62) 

• Eu (Atomic Number 63) 

• Yb (Atomic Number 70)

Answer 1

Eu (Atomic Number 63)

Explanation:

\[\ce{Ce^{2+} -> 4f^1}\]

\[\ce{Sm^{2+} -> 4f^6 -> half filled configuration}\]

\[\ce{Eu^{2+} -> 4f^7}\]

\[\ce{Yb^{2+} -> 4f^14 -> fully filled configuration}\]

\[\ce{E^\circ_{M^{3+}/M^{2+}} => \underset{- 0.35}{Eu}\phantom{..} \underset{- 1.05}{Yb}}\]

Hence, due to more reduction potential in Eu than Yb, it can be concluded that Eu²⁺ is more stable than Yb²⁺.