Question

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

• A. \( \text{Ce (Atomic Number 58)} \)
• B. \( \text{Sm (Atomic Number 62)} \)
• C. \( \text{Eu (Atomic Number 63)} \)
• D. \( \text{Yb (Atomic Number 70)} \)

Hint:

The stability of lanthanide ions depends on their electronic configuration and reduction potential. Fully filled (\( 4f^{14} \)) and half-filled (\( 4f^7 \)) configurations are especially stable.

Answer 1

The Correct Option is C

The stability of the divalent state in lanthanide ions is influenced by their electronic configuration. Fully filled and half-filled configurations are particularly stable: \[ \text{Ce}^{2+} \to 4f^1, \quad \text{Sm}^{2+} \to 4f^6, \quad \text{Eu}^{2+} \to 4f^7, \quad \text{Yb}^{2+} \to 4f^{14}. \]
Reduction potential comparison: \[ E^\circ_{M^{3+}/M^{2+}}: \quad \text{Eu} = -0.35 \, \text{V}, \quad \text{Yb} = -1.05 \, \text{V}. \]
Europium (\( \text{Eu}^{2+} \)) is more stable than \( \text{Yb}^{2+} \), as it has a lower reduction potential and a more stable half-filled \( 4f^7 \) configuration. Final Answer: \[ \boxed{\text{Europium} \, (\text{Eu}^{2+})} \]