Problem 56 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Partial roasting of chalcopyrite produces (a) \(\mathrm{Cu}_{2} \mathrm{~S}\) and \(\mathrm{FeO}\) (b) \(\mathrm{Cu}_{2} \mathrm{O}\) and \(\mathrm{FeO}\) (c) \(\mathrm{CuS}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) (d) \(\mathrm{Cu}_{2} \mathrm{O}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\)

Step-by-Step Solution

Answer

Option (a): \( \text{Cu}_2 \text{S} \) and \( \text{FeO} \).

Step 1: Understand Chalcopyrite Composition

Chalcopyrite is a mineral with the formula \( \text{CuFeS}_2 \). It is composed of copper (\( \text{Cu} \)), iron (\( \text{Fe} \)), and sulfur (\( \text{S} \)). Upon heating, it undergoes oxidation-reduction reactions.

Step 2: Determine Partial Roasting Byproducts

In partial roasting, the goal is to convert some sulfur in chalcopyrite to form oxides and sulfides. Primary partial roasting reactions may include: \[ \text{2CuFeS}_2 + 3O_2 \rightarrow \text{Cu}_2\text{S} + 2FeO + 3SO_2 \]. Here, copper sulfide (\( \text{Cu}_2\text{S} \)) and iron(II) oxide (\( \text{FeO} \)) are produced.

Step 3: Choose The Correct Byproducts

The key products of partial roasting are copper(I) sulfide (\( \text{Cu}_2\text{S} \)) and iron(II) oxide (\( \text{FeO} \)). This matches with option (a) \( \text{Cu}_2 \text{S} \) and \( \text{FeO} \).

Step 4: Verification

Review reactions and byproducts formed during partial roasting to ensure correct selection. The process involves oxidizing only some components, which aligns with the production of \( \text{Cu}_2 \text{S} \) and \( \text{FeO} \), validating option (a) as correct.

Key Concepts

Partial Roasting

Partial roasting is a crucial process in the extraction of metals from ore, especially in mineral compounds like chalcopyrite. Chalcopyrite is a complex mixture of copper, iron, and sulfur, represented by the chemical formula \( \text{CuFeS}_2 \). The partial roasting process involves heating the mineral, but carefully controlling the amount of oxygen. This means that not all the sulfur in the compound is actually oxidized.

This process aims at transforming a part of the sulfur present into sulfur dioxide \( \text{SO}_2 \) while leaving enough sulfur to form metal sulfides. This makes it distinct from complete roasting, where all sulfur would be removed as \( \text{SO}_2 \). The reaction for partial roasting of chalcopyrite can be seen as:

• \( 2\text{CuFeS}_2 + 3\text{O}_2 \rightarrow \text{Cu}_2\text{S} + 2\text{FeO} + 3\text{SO}_2 \)

This shows the formation of partial sulfides like copper sulfide \( \text{Cu}_2\text{S} \) and oxides like iron(II) oxide \( \text{FeO} \). Understanding this transformation is key to grasp how different compounds form through controlled roasting methods.

Copper Sulfide Formation

Copper sulfide forms an essential part of the products when chalcopyrite undergoes partial roasting. As the oxygen molecules interact with \( \text{CuFeS}_2 \), not all of the copper bonds directly with oxygen. Instead, some bonds with remaining sulfur.

This leads to the formation of copper(I) sulfide, \( \text{Cu}_2\text{S} \). Copper sulfide is advantageous because it is easier to further process for extracting pure copper. The formation of \( \text{Cu}_2\text{S} \) rather than \( \text{Cu}_2\text{O} \) during partial roasting indicates controlled oxidation allowing certain sulfur presence.

The presence of copper sulfide in the partial roasting byproducts reveals a selective oxidation, which prevents excessive loss of copper to oxides by controlling sulfur's role in the formulation. Future steps in copper extraction can eliminate the remaining sulfur to obtain metallic copper efficiently. This method is both economically beneficial and practical for mineral processing.

Iron(II) Oxide Production

In the partial roasting process of chalcopyrite, an important product is iron(II) oxide, \( \text{FeO} \). This redox reaction in roasting separates out the iron from the chalcopyrite compound. The balanced chemical reaction also involves sulfur converting into sulfur dioxide.

Iron(II) oxide forms as part of selective oxidation. Some oxygens target the iron component, leading to the liberation of \( \text{FeO} \). While some iron would remain bound to other sulfur components in the mineral, releasing iron in the \( \text{FeO} \) state prepares it for additional processing steps.

• The correct roasting conditions, involving sufficient oxygen but not complete oxidation, ensure that \( \text{FeO} \) is produced.

Iron(II) oxide's production is crucial in efficiently separating iron from copper and sulfur in the later stages, making the refinement process more efficient and naturally cost-effective as it needs fewer purification stages.