Problem 107 - A Complete Resource Book in Chemistry for JEE Main

Question

In this reaction (C) and (D) are respectively, \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} \stackrel{100^{\circ} \mathrm{C}}{\longrightarrow}(\mathrm{A}) \stackrel{230 \mathrm{C}}{\longrightarrow}\) (B) \(\stackrel{800^{\circ} \mathrm{C}}{\longrightarrow}(\mathrm{C})+(\mathrm{D})\) (a) \(\mathrm{Cu}_{2} \mathrm{O}, \mathrm{SO}_{2}\) (b) \(\mathrm{CuO}, \mathrm{SO}_{3}\) (c) \(\mathrm{CuO}, \mathrm{SO}_{2}\) (d) \(\mathrm{Cu}_{2} \mathrm{O}, \mathrm{SO}_{3}\)

Step-by-Step Solution

Answer

(C) and (D) are Cu_2O and SO_2, so the correct option is (c).

Step 1: First Decomposition

Upon heating CuSO_4 · 5H_2O at 100°C, it loses its water molecules to form anhydrous CuSO_4. This is Chemical A.

Step 2: Second Decomposition

Heating CuSO_4 further at 230°C causes it to decompose into CuO and SO_3. Here, CuO is Chemical B.

Step 3: Final Decomposition

When CuO is heated further at 800°C with SO_3, it undergoes a reduction reaction to form Cu_2O and SO_2. Therefore, Chemicals C and D are Cu_2O and SO_2, respectively.

Key Concepts

Copper(II) sulfate

Copper(II) sulfate, commonly known by its chemical formula \( \mathrm{CuSO}_{4} \), is an inorganic compound that frequently appears in chemistry laboratories. It's particularly recognizable in its pentahydrate form, \( \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} \), which is a bright blue crystalline solid. This compound is often used to demonstrate changes in chemical properties when exposed to heat, making it a popular example for studying thermal decomposition reactions.

When \( \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} \) is heated, it undergoes dehydration, losing water molecules and resulting in an anhydrous form of copper sulfate, which is a white powder. This specific transformation demonstrates how water molecules are integral to the crystalline structure, and without them, the color changes significantly from blue to white. Notably, the compound in its anhydrous form is often used as a dehydrating agent and in other industrial applications.

Copper(II) sulfate is a versatile compound:

• It serves as a fungicide, algaecide, and root killer in agriculture.
• It’s used in the preparation of other copper compounds.
• It can act as an analytical reagent to detect reducing sugars or proteins.

Dehydration Reaction

A dehydration reaction is a type of chemical reaction where a water molecule is removed from a compound. This reaction plays a key role in transforming copper(II) sulfate pentahydrate into its anhydrous form. Let's delve deeper into this process.

At around \( 100^{\circ} \mathrm{C} \), \( \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} \) loses its water content, resulting in an anhydrous compound, \( \mathrm{CuSO}_{4} \). The transition from the hydrated to the anhydrous form is a classic example of a dehydration reaction, highlighting how water can be structurally integral to chemical compounds.

Here are key characteristics of a dehydration reaction:

• Involves removal of water from a substance.
• Commonly initiated by heating.
• It is often reversible, meaning that absorption of water can return the material to its hydrated state.

This type of reaction is broader than just laboratory experiments. You'll find comparable processes in natural settings, such as the drying of clay or cement, where water loss leads to hardening and structural changes.

Reduction Reaction

Reduction reactions involve the gain of electrons by a molecule, atom, or ion. These reactions are central to the final stage of the copper(II) sulfate decomposition sequence, where \( \mathrm{CuO} \) reacts to form \( \mathrm{Cu}_{2} \mathrm{O} \).

During the reduction of \( \mathrm{CuO} \) at \( 800^{\circ} \mathrm{C} \), it gains electrons to form \( \mathrm{Cu}_{2} \mathrm{O} \), while releasing \( \mathrm{SO}_{2} \). Here, \( \mathrm{SO}_{3} \) initially involved in the compound decomposes, facilitating the reduction process.

In the reaction:

• Ironically \( \mathrm{CuO} \), a copper oxide, acts as an oxidizing agent because it accepts electrons.
• This process leads to the formation of \( \mathrm{Cu}_{2} \mathrm{O} \), a different copper oxide with a lower oxidation state of copper.
• The reaction is significant as it changes the chemical properties of the copper oxide, allowing it to behave differently in further chemical processes or applications.

Reduction reactions like these are fundamental to many industrial applications, including metal refining and processing. Understanding these principles can provide insights into broader concepts of chemistry, like redox reactions and electron transfer.