MCQ
Which one of the lanthanoids given below is the most stable in divalent form?
- A$Ce$ (Atomic Number $58$)
- B$Sm$ (Atomic Number $62$)
- ✓$Eu$ (Atomic Number $63$)
- D$Yb$ (Atomic Number $70$)
✓
Answer
Correct option: C.
$Eu$ (Atomic Number $63$)
c
$E _{ M ^{3+} / M ^{2+}}^{0} \Rightarrow Eu(-0.35) Yb(-1.05$
$E _{ M ^{3+} / M ^{2+}}^{0} \Rightarrow Eu(-0.35) Yb(-1.05$
Hence, due to more reduction potential in $Eu$ as compared to $Yb$, it can concluded that $Eu ^{2+}$ is more stable than $Yb ^{2+}$.
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