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Question #44101

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The Question

An aqueous solution contains $1.0 \times 10^{-3}\text{ M } AgNO_3$ at $298\text{ K}$ . To this, an equal volume of a buffer solution containing $0.2\text{ M}$ of an aromatic amine ( $L$ ) and $0.2\text{ M}$ of its corresponding hydrochloride salt ( $L \cdot HCl$ ) is added. This experiment is performed separately with three different amines: Ligand I: 4-Methylaniline Ligand II: Aniline Ligand III: 2-Methylaniline

The formation of the linear $[Ag(L)_2]^+$ complex competes with the precipitation of $AgCl_{(s)}$ ( $K_{sp} = 1.8 \times 10^{-10}$ ). A student hypothesizes that since methyl groups exert a +I effect, Ligand III should be a stronger electron donor than Ligand II, thereby forming a more stable complex and reducing the likelihood of precipitation.

Which of the following correctly identifies the actual outcome regarding the precipitation of $AgCl_{(s)}$ and the primary chemical principle responsible?

Options

A.
Precipitation of $AgCl_{(s)}$ is most extensive with Ligand III; the ortho-effect induces severe steric hindrance that drastically lowers the complex formation constant ( $\beta_2$ ), maximizing the free $[Ag^+]$ available to exceed $K_{sp}$ .
B.
Precipitation of $AgCl_{(s)}$ is least extensive with Ligand III; the proximity of the ortho-methyl group provides additional inductive stabilization to the $[Ag(L)_2]^+$ complex, keeping free $[Ag^+]$ below the precipitation threshold.
C.
Precipitation of $AgCl_{(s)}$ is most extensive with Ligand I; the strong +I and hyperconjugation effects of the para-methyl group increase the basicity of the amine, leading to the preferential precipitation of $Ag_2O_{(s)}$ .
D.
Precipitation of $AgCl_{(s)}$ is identical across all three solutions; the common ion effect of the $0.1\text{ M } Cl^-$ overwhelmingly dominates, as aromatic amines are too weakly basic to form stable coordination complexes with $Ag^+$ .

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Solubility Product (Ksp) vs. Ionic Product (Qsp)|Ortho-Effect in Anilines

Why This Question Is Tricky

Common Mistakes

The question tests the 'Ortho-Effect' as a critical exception to standard electronic substituent effects in coordination chemistry.

Students often mechanically apply inductive (+I) and mesomeric (+M) effects to predict ligand strength, assuming that any electron-donating group will increase basicity and complex stability.

The trap lies in assuming 2-methylaniline is a better ligand than aniline due to the methyl group's +I effect.

However, the hidden variable is steric hindrance (F-strain).

The ortho-methyl group physically blocks the approach of the bulky silver ion, drastically lowering the formation constant of the complex.

This failure to complex the silver ion allows the traditional common ion effect (from the chloride in the buffer) to dominate, leading to unexpected precipitation.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	Solubility Product (Ksp) vs. Ionic Product (Qsp)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free