Why is the basic strength of lanthanoid hydroxides observed to decrease progressively across the series from La(OH)3 to Lu(OH)3?

Answer and explanation

Correct answer: B. Lanthanoid contraction decreases Ln3+ radius to increase charge density and covalent character of Ln-OH

  1. Step 1: Across the lanthanoid series (), the ionic radius decreases continuously from 106 pm to 86 pm due to lanthanoid contraction.
  2. Step 2: According to Fajan's rules, smaller cation size and higher charge density increase the polarizing power of the ion.
  3. Step 3: Consequently, the covalent character of the bond increases from to .
  4. Step 4: Greater covalent character retards the cleavage and release of ions in aqueous solution, making the most basic and the least basic hydroxide in the series.
Answer
Due to lanthanoid contraction, the ionic radius of Ln3+ decreases, increasing charge density and covalent character of the Ln-OH bond
Why not the others: 'Because Lu(OH)3 contains more hydroxide ions...' is false since all trivalent lanthanoid hydroxides have identical stoichiometry; 'Because lanthanum is a non-metal...' is false as all lanthanoids are electropositive metals; 'Because Lu3+ has completely empty 4f orbitals...' is incorrect because has a completely filled configuration.

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