Problem 136 - A Complete Resource Book in Chemistry for JEE Main

Question

Two ores of the same metal \((\mathrm{M})\) are \(\left(\mathrm{A}_{1}\right)\) and \(\left(\mathrm{A}_{2}\right)\) (1) \(\mathrm{A}_{1} \stackrel{\text { calcination }}{\longrightarrow}\) Black residue \(\mathrm{C}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (2) \(\mathrm{A}_{1} \stackrel{\mathrm{HCl}, \mathrm{Kl}}{\longrightarrow} \mathrm{I}_{2}+\mathrm{D}\) precipitate (3) \(\mathrm{A}_{2} \stackrel{\text { roasting }}{\longrightarrow} \operatorname{Gas}(\mathrm{G})+\operatorname{Metal}(\mathrm{M})\) (4) \(\mathrm{G} \stackrel{+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}, \text { acidified }}{\longrightarrow}\) Green solution Identify \(\mathrm{A}_{1}\) and \(\mathrm{A}\), here? (a) \(\mathrm{A}_{1}=\mathrm{CuCO}_{3} . \mathrm{Cu}(\mathrm{OH})_{2}, \mathrm{~A}_{2}=\mathrm{Cu}_{2} \mathrm{~S}\) (b) \(\mathrm{A}_{1}=\mathrm{Cu}_{2} \mathrm{~S}, \mathrm{~A}_{2}=\mathrm{CuCO}_{3} \cdot \mathrm{Cu}(\mathrm{OH})_{2}\) (c) \(\mathrm{A}_{1}=\mathrm{CuFeS}_{2}, \mathrm{~A}_{2}=\mathrm{Cu}_{2} \mathrm{~S}\) (d) none of these

Step-by-Step Solution

Answer

(a) \(A_1 = CuCO_3 \cdot Cu(OH)_2, A_2 = Cu_2S\)

Step 1: Understand Process (Calcination)

The calcination of \(A_1\) leads to the formation of a black residue \(C\), \(CO_2\), and \(H_2O\). This suggests that \(A_1\) contains carbonates and hydroxides, as \(CO_2\) is released from carbonates and \(H_2O\) from hydroxides.

Step 2: Reaction with HCl and KI

\(A_1\) reacts with HCl and KI to release \(I_2\) and form a precipitate \(D\). This indicates that \(A_1\) contains a halide ion that can react with KI to release \(I_2\). A common reactant for this reaction is Cuprous Chloride, CuCl.

Step 3: Understand Process (Roasting)

Roasting of \(A_2\) results in the formation of a gas \(G\) and metal \(M\). This suggests that \(A_2\) contains a sulfide. During roasting, metal sulfides typically release sulfur dioxide (\(SO_2\)) and leave behind the metal.

Step 4: Reaction with Acidified K2Cr2O7

The gas \(G\) reacts with acidified \(K_2Cr_2O_7\) to give a green solution, indicating that \(G\) is \(SO_2\), as it forms a green chromate solution after reacting with \(K_2Cr_2O_7\). This confirms \(A_2\) as a copper sulfide ore.

Step 5: Infer the Identity of A1 and A2

Putting all steps together, \(A_1\) appears to be \(CuCO_3 \cdot Cu(OH)_2\) that releases \(CO_2\) and \(H_2O\) upon calcination, and reacts to produce \(I_2\). \(A_2\) is \(Cu_2S\), which during roasting, releases \(SO_2\) (\(G\)) and leaves behind the metal \(Cu\).

Key Concepts

Ore Calcination

Calcination is a thermal treatment process applied to ores and other solid materials. Its main purpose is to drive off volatile substances such as water, carbon dioxide, or other impurities. The primary goal is to transform the ore into a more thermally stable form. During calcination, certain chemical changes take place; for instance, carbonates decompose into oxides, releasing carbon dioxide.
For example, when cupric carbonate \( \ ext{CuCO}_3 \cdot \ ext{Cu(OH)}_2 \) undergoes calcination, it breaks down into copper oxide, water vapor, and carbon dioxide:\[\text{CuCO}_3 \cdot \ ext{Cu(OH)}_2 \rightarrow \ ext{CuO} + \ ext{H}_2\text{O} + \ ext{CO}_2\]
This process is crucial in preparing ores for further processing, such as metal extraction via reduction or other methods.

Metal Roasting

Roasting is another significant step in metal extraction, especially when dealing with sulfide ores. It involves heating the ore in the presence of oxygen. The aim is to convert sulfides into oxides, which are easier to reduce. However, it also frequently results in the release of gaseous by-products, such as sulfur dioxide.
For example, when copper(I) sulfide (\( \ ext{Cu}_2 \ ext{S} \)) is roasted, sulfur dioxide is released, and copper oxide is formed:\[ \text{Cu}_2 \ ext{S} + \ ext{O}_2 \rightarrow \ ext{Cu}_2 \ ext{O} + \ ext{SO}_2\]
This method is especially advantageous in metallurgy because it not only clarifies the metal content but also concurrently removes sulfur, an undesired impurity, from the ore.

Copper Ores

Copper is commonly extracted from its ores, such as chalcopyrite (\( \ ext{CuFeS}_2 \)) and cuprite (\( \ ext{Cu}_2 \ ext{S} \)). These ores require different processing approaches based on their chemical compositions to make copper extraction feasible.

• Chalcopyrite: Often undergoes roasting and then further conversion into copper metal through smelting and electrolysis.
• Cuprite: Ideal for direct roasting to yield metallic copper due to its high copper content and simpler composition.

Each ore's structure dictates the specific extraction methods used, influencing economics, energy requirements, and environmental impact. Understanding the unique properties of these ores is vital for efficient copper production.

Chemical Reactions in Metallurgy

Metallurgy relies heavily on chemical reactions to transform raw ores into pure metals. The reactions typically involve oxidation-reduction processes, where the metal ions lose or gain electrons to form their metallic form.

• Oxidation: The metal ion or compound loses electrons. For example, in roasting, sulfides are oxidized to oxides.
• Reduction: Involves gaining electrons, where oxidized ores are reduced back to their metal forms, such as during smelting.

These reactions are usually facilitated by heat, catalysts, or specific chemical conditions. An example includes the reaction of sulfur dioxide \( \ ext{SO}_2 \) with potassium dichromate in an acidic medium, resulting in a color change to green, indicating the formation of chromium sulfate:\[ \ ext{SO}_2 + \ ext{K}_2 \ ext{Cr}_2 \ ext{O}_7 \rightarrow \ ext{Cr}_2 \ ext{(SO}_4 \ ext{)}_3\]
Understanding these chemical reactions helps control processes, optimize outcomes, and minimize unwanted by-products.