Problem 48 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A mixture consists of \(A\) (yellow solid) and \(B\) (colourless solid) which gives lilac colour in flame. (a) Mixture gives black precipitate \(C\) on passing \(\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) through its aqueous solution. (b) \(C\) is soluble in aqua-regia and on evaporation of aqua-regia and adding \(\mathrm{SnCl}_{2}\), gives greyish black precipitate \(D\). The salt solution with \(\mathrm{NH}_{4} \mathrm{OH}\) gives a brown precipitate. (i) The sodium carbonate extract of the salt withCCl \(_{4} / \mathrm{FeCl}_{3}\) gives a violet layer. (ii) The sodium carbonate extract gives yellow precipitate with \(\mathrm{AgNO}_{3}\) solution which is insoluble in \(\mathrm{NH}_{3}\). Identify \(A\) and \(B\), and the precipitates \(C\) and \(D\)

Step-by-Step Solution

Answer

A is lead(II) chloride, B is potassium chloride; C is lead(II) sulfide, and D is elemental lead.

Step 1: Identify the flame test indication

The lilac color observed in the flame test indicates the presence of potassium. Therefore, substance \( B \) is likely potassium chloride (\( KCl \)), as it is a common potassium salt that is colorless and provides a lilac flame test.

Step 2: Analyzing the black precipitate

The black precipitate \( C \) formed upon passing \( \text{H}_2\text{S} \) gas through the solution suggests the presence of lead ions. This makes \( A \), the yellow solid, likely lead(II) chloride (\( \text{PbCl}_2 \)), which forms a black precipitate of \( \text{PbS} \) with hydrogen sulfide gas.

Step 3: Process with aqua regia

The fact that precipitate \( C \), which is black, dissolves in aqua regia and upon evaporation, adding \( \text{SnCl}_2 \) results in a greyish-black precipitate \( D \) further confirms \( C \) is \( \text{PbS} \). The greyish-black precipitate \( D \) is likely elemental lead, resulting from a reduction reaction with \( \text{SnCl}_2 \).

Step 4: Ammonium hydroxide test

When the salt solution reacts with \( \text{NH}_4\text{OH} \) and forms a brown precipitate, this indicates the presence of ferric ions \( \text{Fe}^{3+} \), which form brown ferric hydroxide \( \text{Fe(OH)}_3 \).

Step 5: Sodium carbonate extract with \(\text{CCl}_4\) and \(\text{FeCl}_3\)

The extract's formation of a violet layer with \(\text{CCl}_4/\text{FeCl}_3\) indicates the presence of iodine ions \( \text{I}^- \). This suggests the possible presence of iodide as a minor contaminant or secondary component in the mixture.

Step 6: Sodium carbonate extract with \(\text{AgNO}_3\) solution

A yellow precipitate that is insoluble in \( \text{NH}_3 \) suggests lead iodide \( \text{PbI}_2 \). This confirms the presence of lead (from \( A \)) as a major component in the mixture and another verification for halide ions.

Key Concepts

Flame Test

The flame test is a simple yet powerful method in chemistry to identify the presence of certain metal ions in compounds. When a substance is heated in a flame, the electrons in the metal ions absorb energy and get excited to higher energy levels. As they return to their ground state, they release energy in the form of light. Each metal ion emits light of a characteristic color. This unique color serves as a "fingerprint" to identify the elements.
In the context of our exercise, the lilac color observed in the flame test signifies the presence of potassium ions. Potassium ions emit a lilac flame, which helps to identify one of the substances in our mixture as potassium chloride (\( \text{KCl} \)).
The simplicity of the flame test makes it an accessible tool in the laboratory. However, it is important to note that the method can sometimes give ambiguous results if multiple ions are present, as the dominant color might overshadow fainter emissions from other ions.

Precipitate Formation

The formation of a precipitate is a clear indication of a chemical reaction where insoluble compounds form and separate out of a solution. Precipitates are often formed when two solutions containing soluble salts are mixed, and an insoluble salt is produced.
In our exercise, when hydrogen sulfide (\( \text{H}_2\text{S} \)) gas is passed through the aqueous solution of the mixture, a black precipitate forms. This black precipitate, identified as lead sulfide (\( \text{PbS} \)), indicates the reaction between hydrogen sulfide and lead ions.

• This confirms the presence of lead(II) chloride (\( \text{PbCl}_2 \)) in the original mixture.

Further, the precipitation process can involve a series of transformations indicative of the compounds in question. Upon treatment with aqua regia, the precipitate dissolves and, upon reduction with \( \text{SnCl}_2 \), forms another precipitate, supporting the presence of lead through this systematic process.

Aqua Regia

Aqua regia, a highly corrosive mixture of nitric acid (\( \text{HNO}_3 \)) and hydrochloric acid (\( \text{HCl} \)), is renowned for its ability to dissolve gold and platinum - metals that are typically refractory to most acids. This unique property makes it an essential reagent in chemistry for dissolving metals for qualitative and quantitative analysis.
In the exercise, the black precipitate (\( \text{PbS} \)) dissolves in aqua regia. This leads to the formation of a solution that, upon evaporation and treatment with stannous chloride (\( \text{SnCl}_2 \)), produces a greyish-black precipitate of elemental lead. This tells us two things:

• Aqua regia's role in preparing complex mixtures for analytical processes is indispensable due to its ability to dissolve metallic compounds efficiently.
• The behavior of the precipitate in aqua regia helps verify the chemical transformations involving lead, corroborating identification tasks in the solution.

Understanding aqua regia is not only crucial for theoretical chemistry but also for practical applications, particularly in fields like metal recovery and purification.