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What is Lanthanoid contraction? Write in increasing order of basic nature of oxides \( (Ln_2O_3) \) and hydroxides \( [Ln(OH)_3] \) of Lanthanoids.

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Poor shielding by 4f electrons shrinks the ions from La to Lu; smaller ion means weaker base, so La(OH)3 is most basic.
Updated On: Jul 10, 2026
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Solution and Explanation

Step 1: State the phenomenon simply.
Going along the 4f series from La to Lu, every element is a little smaller than the one before it. This continuous shrinking of atomic and ionic radii is the lanthanoid contraction.

Step 2: Why it happens.
Each new proton in the nucleus is balanced by a new electron placed in the deep-lying \(4f\) orbital. Because f-orbitals are diffuse and shield rather badly, the extra nuclear pull is not cancelled for the outer shell. The net (effective) nuclear charge grows, so the electron cloud is drawn inward and the radius falls step by step.

Step 3: Consequence for basicity.
Basic strength of a metal hydroxide depends on how easily it releases \(OH^-\), which is greater for larger, more electropositive ions. Since \(La^{3+}\) is the largest and \(Lu^{3+}\) the smallest, \(La(OH)_3\) is most basic and \(Lu(OH)_3\) least basic; the same trend holds for the oxides.

Step 4: Write the increasing orders.
\[\boxed{Lu(OH)_3 < \ldots < La(OH)_3 \ \text{and}\ Lu_2O_3 < \ldots < La_2O_3}\]
Basic nature rises as the ionic size increases, i.e. from Lu towards La.
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