A sulphide ore is first converted into its oxide before reduction. This is done because :

Options

A. a sulphide ore cannot be reduced to metal at all
B. no reducing agent is found suitable for reducing a sulphide ore
C. the enthalpy of formation of `CO_(2)` is more than that of `CS_(2)`
D. a metal oxide is generally less stable than the metal sulphide

Answer

To solve the question "A sulphide ore is first converted into its oxide before reduction. This is done because:", we will follow these steps:

### Step 1: Understanding the Process
The conversion of a sulphide ore to its oxide form is a crucial step in metallurgy. This process is known as roasting, where the ore is heated in the presence of oxygen.

**Hint:** Think about the chemical reactions involved in roasting and the products formed.

### Step 2: Chemical Reaction
During roasting, the sulphide ore reacts with oxygen to form a metal oxide and sulfur dioxide (SO₂). The general reaction can be represented as:
\[ \text{Metal Sulphide} + \text{Oxygen} \rightarrow \text{Metal Oxide} + \text{SO}_2 \]

**Hint:** Recall the products of roasting and how they relate to the reduction process.

### Step 3: Stability Considerations
The metal oxide formed is generally less stable than the corresponding metal sulphide. This means that it requires less energy to reduce the metal oxide back to the metal compared to reducing the sulphide directly.

**Hint:** Consider the thermodynamic stability of metal oxides versus metal sulphides.

### Step 4: Energy Efficiency
If we were to reduce the metal sulphide directly, it would require very high temperatures and energy. By converting it to the oxide first, we can perform the reduction at lower temperatures, making the process more energy-efficient.

**Hint:** Think about the implications of energy requirements in metallurgical processes.

### Step 5: Conclusion
Thus, the reason for converting a sulphide ore into its oxide before reduction is that the stability of metal oxides is generally lower than that of metal sulphides, allowing for a more efficient reduction process.

**Final Answer:** The correct option is D: A metal oxide is generally less stable than the metal sulphide.