Copper sulphate dissolves in excess of KCN to give :

Options

A. CuCN
B. `[Cu(CN)_(4)]^(3-)`
C. `[Cu(CN)_(4)]^(2-)`
D. `Cu(CN)_(2)`

Explanation

To solve the question "Copper sulphate dissolves in excess of KCN to give:", we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are copper sulfate (CuSO4) and potassium cyanide (KCN).

### Step 2: Write the Initial Reaction
When copper sulfate reacts with potassium cyanide, the following reaction occurs:
\[ \text{CuSO}_4 + 2 \text{KCN} \rightarrow \text{K}_2\text{SO}_4 + \text{Cu(CN)}_2 \]

### Step 3: Analyze the Product
The product formed, cupric cyanide (Cu(CN)2), is unstable and will decompose.

### Step 4: Decomposition of Cupric Cyanide
The cupric cyanide decomposes to form cuprous cyanide (CuCN) and cyanogen gas (C2N2):
\[ \text{Cu(CN)}_2 \rightarrow 2 \text{CuCN} + \text{C}_2\text{N}_2 \]

### Step 5: Formation of the Complex
When excess potassium cyanide is present, the cuprous cyanide (CuCN) will react with additional KCN to form a soluble complex:
\[ \text{CuCN} + 3 \text{KCN} \rightarrow \text{K}_3[\text{Cu(CN)}_4] \]

### Step 6: Conclusion
Thus, the final product when copper sulfate dissolves in excess KCN is the complex ion:
\[ \text{K}_3[\text{Cu(CN)}_4] \]

### Final Answer
The answer to the question is that copper sulfate dissolves in excess KCN to give the complex ion \(\text{K}_3[\text{Cu(CN)}_4]\).

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