Question:

The most basic hydroxide from following is

Updated On: Jan 23, 2024
  • $\ce{Pr (OH)_3 \, (Z = 59)}$
  • $\ce{Sm (OH)_3 \, (Z = 62)}$
  • $\ce{Ho (OH)_3 \, (Z = 67)}$
  • $\ce{La (OH)_3 \, (Z = 57)}$
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The Correct Option is D

Solution and Explanation

Due to lanthanoid contraction, the basic character of hydroxides decreases from $La ^{3+}$ to $Lu ^{3+}$

As a result, covalent nature of $Ln� OH$ bond also increases. Thus, the most basic hydroxide among the given option is $La( OH )_{3}$
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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.