Before You Begin

Question #44891

Difficulty: 8/10 | 🟢 Free

The Question

Consider the thermodynamic and equilibrium properties of lanthanoid hydroxides, $\ce{Ln(OH)3}$ , and substituted phenols at $298\text{ K}$ . The solubility product ( $K_{sp}$ ) of $\ce{La(OH)3}$ is approximately $2 \times 10^{-20}$ , and the $pK_a$ of phenol is $10.0$ . Which of the following statements correctly distinguishes between their intrinsic electronic properties and effective aqueous behavior?

Options

A.
While $\ce{La(OH)3}$ is the most intrinsically basic lanthanoid hydroxide due to the lowest covalent character in its $\ce{Ln-OH}$ bond, its saturated aqueous solution has a lower $\ce{[OH^-]}$ than a $0.1\text{ M}$ solution of sodium phenoxide, because the effective alkalinity of $\ce{La(OH)3}$ is severely restricted by its high lattice energy.
B.
The saturated aqueous solution of $\ce{Lu(OH)3}$ is more alkaline than that of $\ce{La(OH)3}$ because the highly exothermic hydration enthalpy of the smaller $\ce{Lu^{3+}}$ ion overcomes its lattice energy, leading to a higher $K_{sp}$ .
C.
The addition of phenol to a suspension of $\ce{La(OH)3}$ results in complete dissolution to form lanthanum phenoxide, because the intrinsic basicity of $\ce{La(OH)3}$ is thermodynamically sufficient to deprotonate phenol ( $K_{eq} > 1$ ).
D.
The intrinsic basicity of $\ce{Ln(OH)3}$ increases from $\ce{La}$ to $\ce{Lu}$ because the increasing effective nuclear charge ( $Z_{eff}$ ) enhances the ionic character of the $\ce{Ln-OH}$ bond, analogous to how electron-withdrawing groups increase the acidity of phenols.

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

F-block Basic Character|Phenol Acidity vs. Substituents|Carboxylic Acid Strength

Why This Question Is Tricky

Common Mistakes

This question tests the critical distinction between 'intrinsic basicity' (an electronic parameter governed by Fajans' rules) and 'effective alkalinity' (a thermodynamic parameter governed by the solubility product, $K_{sp}$ ).

Students often fall into the trap of treating the 'most basic' lanthanoid hydroxide, $\ce{La(OH)3}$ , as a strong Arrhenius base capable of neutralizing weak organic acids like phenol.

By forcing a quantitative comparison between the hydrolysis of sodium phenoxide and the dissolution of $\ce{La(OH)3}$ , the problem highlights how the high lattice energy of f-block hydroxides severely restricts their practical basicity in aqueous solutions, mimicking how electronic effects dictate organic acid-base equilibria.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	D and F Block Elements - Class 12
Concept	F-block Basic Character
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free