Question:medium

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

Show 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.
Updated On: Jan 13, 2026
  • \( \text{Ce (Atomic Number 58)} \)
  • \( \text{Sm (Atomic Number 62)} \)
  • \( \text{Eu (Atomic Number 63)} \)
  • \( \text{Yb (Atomic Number 70)} \)
Show Solution

The Correct Option is C

Solution and Explanation

The stability of divalent lanthanide ions depends on their electronic configuration. Fully filled and half-filled configurations exhibit enhanced stability, as seen in: \[\text{Ce}^{2+} \to 4f^1, \quad \text{Sm}^{2+} \to 4f^6, \quad \text{Eu}^{2+} \to 4f^7, \quad \text{Yb}^{2+} \to 4f^{14}.\]
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 \( \text{Yb}^{2+} \) due to its lower reduction potential and the greater stability of its half-filled \( 4f^7 \) configuration.

Final Answer: \[\boxed{\text{Europium} \, (\text{Eu}^{2+})}\]
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