Problem 101 - A Complete Resource Book in Chemistry for JEE Main

Question

Identify the compound (X) and the condition (Y) here \(\mathrm{PbS} \stackrel{\text { Heat }}{\longrightarrow}(\mathrm{X})+\mathrm{PbS} \stackrel{(\mathrm{Y})}{\longrightarrow} \mathrm{Pb}+\mathrm{SO}_{2}\) (a) \(\mathrm{X}=\mathrm{PbSO}_{3}, \mathrm{Y}=\) low temperature (b) \(\mathrm{X}=\mathrm{PbO}\) or \(\mathrm{PbSO}_{4}, \mathrm{Y}=\) high temperature (air supply is cut off) (c) \(\mathrm{X}=\mathrm{Pb}_{3} \mathrm{O}_{4}, \mathrm{Y}=\) high temperature (d) \(X=P b S O_{4}, Y=N_{2}\)

Step-by-Step Solution

Answer

Compound X is PbO; condition Y is high temperature with air supply cut off (option b).

Step 1: Analyze the First Reaction

The first reaction is given as \( \mathrm{PbS} \stackrel{\text{ Heat }}{\longrightarrow}(\mathrm{X}) \). Here, the heat causes the conversion of \( \mathrm{PbS} \) into a compound \( \mathrm{X} \). Upon heating, lead sulfide (\( \mathrm{PbS} \)) can form \( \mathrm{PbO} \) or lead sulfate \( \mathrm{PbSO}_4 \) depending on the conditions of the reaction, as both are typical products of oxidation.

Step 2: Analyze the Second Reaction

The second reaction is \( \mathrm{X} + \mathrm{PbS} \stackrel{(\mathrm{Y})}{\longrightarrow} \mathrm{Pb} + \mathrm{SO}_2 \). This implies that \( \mathrm{X} \) reacts with \( \mathrm{PbS} \) to produce \( \mathrm{Pb} \) and \( \mathrm{SO}_2 \). Such a reaction suggests a reducing condition that removes oxygen or combines with sulfur.

Step 3: Consider the Reaction Conditions

High temperature and lack of oxygen supply favor the formation of \( \mathrm{PbO} \), particularly in the presence of sulfur, as this condition typically leads to direct conversion. Cutting off the air supply minimizes the formation of sulfates (as in \( \mathrm{PbSO}_4 \)) encourages the formation of oxides instead (like \( \mathrm{PbO} \)).

Step 4: Conclusion Based on Options

Based on the above deductions, option (b), where \( X = \mathrm{PbO} \) or \( \mathrm{PbSO}_{4}, \ Y = \) high temperature (air supply is cut off), fits well. \( \mathrm{PbO} \) is likely since low air supply conditions are conducive for oxides rather than sulfates, and high temperature corresponds to such synthesis conditions.

Key Concepts

Lead Sulfide Reactions

Lead sulfide, commonly represented as \( \text{PbS} \), is an important compound in chemistry, especially in the study of oxidation-reduction reactions. These reactions involve the transfer of electrons between substances. Here, \( \text{PbS} \) can transform into various products depending on the specific conditions of the reaction. This is typical in oxidation reactions where lead sulfide can convert into metal oxides such as \( \text{PbO} \) or sometimes into lead sulfate \( \text{PbSO}_4 \), provided the right conditions are met. During decomposition, \( \text{PbS} \) undergoes a chemical change when heat is applied, making it a stepping stone in understanding the broader topic of sulfur compounds and their reaction pathways.

Thermal Decomposition

Thermal decomposition refers to the chemical breakdown of a substance when heat is introduced. For \( \text{PbS} \), heating leads to its transformation into other compounds such as \( \text{PbO} \) or \( \text{PbSO}_4 \).
This process is critical in studying chemical reactions because it showcases how temperature impacts molecular structures:

• The application of heat provides energy that overcomes the chemical bonds holding the \( \text{PbS} \) molecule together, resulting in its breakdown.
• Depending on whether oxygen is plentiful or limited, \( \text{PbS} \) will yield different products due to varying oxidation levels.

This exemplifies how heat acts as a catalyst, initiating reactions that wouldn't occur under normal conditions, thus highlighting the importance of controlling reaction temperatures in experiments.

Reaction Conditions

The conditions under which a reaction takes place are just as important as the reactants themselves. Here, we focus on the high temperature and restricted air supply conditions detailed in the reaction involving \( \text{PbS} \).
When the air supply is cut off, oxygen becomes scarce, favoring the formation of oxides rather than sulfates.

• High temperatures facilitate the transformation of \( \text{PbS} \) to \( \text{PbO} \) as less oxygen leads to fewer opportunities for sulfur to bond with it.
• In such conditions, oxidation involves lesser oxygen leading to the creation of metal oxides, aligning with option \( b \) in the exercise.

These considerations underline the role of environmental variables in synthetic chemistry, demonstrating how minute changes in conditions can lead to vastly different outcomes.

Metal Oxides and Sulfates

Understanding the distinction between metal oxides and sulfates is central to grasping oxidation-reduction reactions. Metal oxides like \( \text{PbO} \) are often the result when a metal combines with oxygen under certain conditions, more likely in limited oxygen scenarios.
In contrast, metal sulfates such as \( \text{PbSO}_4 \) form when a metal combines with sulfate ions, typically in an excess of oxygen scenario.
This chemistry is crucial because:

• Oxides often form in reducing environments where less oxygen is present.
• Sulfates tend to form under conditions where more oxygen is available, allowing sulfate ions to stabilize in the presence of metals.

Recognizing these patterns helps us anticipate and control product formations in industrial and laboratory settings, thus enhancing our proficiency in managing chemical reactions.