Which one of the lanthanoids given below is the most stable in divalent form?
- Ce (Atomic Number 58)
- Sm (Atomic Number 62 )
- Eu (Atomic Number 63)
- Yb (Atomic Number 70)
The Correct Option is C
Solution and Explanation
The correct answer is (C) : Eu (Atomic Number 63)
Yb+2 is 4f14
Eu+2 is 4f7
but Eu+2 is more stable than Yb+2 because
\(E°_{Eu|Eu^{2+}} > E°_{Yb|Yb^{2+}}\)
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Concepts Used:
Lanthanoids
Lanthanoids are at the top of these two-row, while actinoids are at the bottom row.
Properties of Lanthanoids
Lanthanoids are inclusive of 14 elements, with atomic numbers 58-71:
- Cerium - Xe 4f1 5d1 6s2
- Praseodymium - Xe 4f3 6s2
- Neodymium - Xe 4f4 6s2
- Promethium - Xe 4f5 6s2
- Samarium - Xe 4f6 6s2
- Europium - Xe 4f7 6s2
- Gadolinium - Xe 4f7 5d1 6s2
- Terbium - Xe 4f9 6s2
- Dysprosium - Xe 4f10 6s2
- Holmium - Xe 4f11 6s2
- Erbium - Xe 4f12 6s2
- Thulium - Xe 4f13 6s2
- Ytterbium - Xe 4f14 6s2
- Lutetium - Xe 4f14 5d1 6s2
These elements are also called rare earth elements. They are found naturally on the earth, and they're all radioactively stable except promethium, which is radioactive. A trend is one of the interesting properties of the lanthanoid elements, called lanthanide contraction.