`CuSO_(4)` decolourises on addition `KCN` , the product is

Options

A. `[Cu(CN)_(4)]^(2-)`
B. `Cu^(2+)` get reduced to form `[Cu(CN)_(4)]^(3-)`
C. `Cu(CN)_(2)`
D. `CuCN`

Answer

To solve the question regarding the reaction of copper(II) sulfate (CuSO₄) with potassium cyanide (KCN), we will follow these steps:

### Step 1: Identify the Reactants
The reactants involved in the reaction are:
- Copper(II) sulfate (CuSO₄)
- Potassium cyanide (KCN)

### Step 2: Write the Initial Reaction
When KCN is added to CuSO₄, a complex reaction occurs. The initial reaction can be written as:
\[ \text{CuSO}_4 + 2 \text{KCN} \rightarrow \text{Cu(CN)}_2 + \text{K}_2\text{SO}_4 \]

### Step 3: Identify the Product
The product formed from the reaction is cuprous cyanide (CuCN) and potassium sulfate (K₂SO₄). However, Cu(CN)₂ is unstable and decomposes:
\[ \text{Cu(CN)}_2 \rightarrow \text{CuCN} + \text{CN}_2 \]

### Step 4: Formation of White Precipitate
Cuprous cyanide (CuCN) is formed as a white precipitate. The reaction can be summarized as:
\[ \text{CuSO}_4 + 2 \text{KCN} \rightarrow \text{CuCN (s)} + \text{K}_2\text{SO}_4 + \text{CN}_2 \]

### Step 5: Final Product
The final product in this reaction is cuprous cyanide (CuCN), which appears as a white precipitate.

### Conclusion
Thus, the product formed when CuSO₄ is treated with KCN is **CuCN**.

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