Question:
The most stable oxidation state of lanthanides is
The most stable oxidation state of lanthanides is
Updated On: Jun 8, 2024
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The Correct Option is D
Solution and Explanation
The general electronic configuration of lanthanides is
$[X e](n-2) f^{1-14}(n-1) d^{1} n s^{2}$
$\because$ After the loss of both of the 65 electrons and also the solitary $d$ electrons, the lanthanoids gain stable configurations.
$\therefore+I I I(+3)$ oxidation state is most common among lanthanides.
$[X e](n-2) f^{1-14}(n-1) d^{1} n s^{2}$
$\because$ After the loss of both of the 65 electrons and also the solitary $d$ electrons, the lanthanoids gain stable configurations.
$\therefore+I I I(+3)$ oxidation state is most common among lanthanides.
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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.