`CuSO_(4)` solution reacts with excess of KCN solution to form:-

Options

A. `K_(2)[Cu(CN)_(4)]`
B. `K_(3)[Cu(CN)_(4)]`
C. `CuCN_(2)`
D. `Cu[K Cu(CN)_(4)]`

Answer

To solve the question, we need to analyze the reaction between copper sulfate (CuSO₄) and potassium cyanide (KCN) and determine the product formed when excess KCN is present.

### Step-by-Step Solution:

1. **Identify the Reactants:**
- The reactants are copper sulfate (CuSO₄) and potassium cyanide (KCN).

2. **Initial Reaction:**
- When CuSO₄ reacts with KCN, it forms cupric cyanide (Cu(CN)₂) and potassium sulfate (K₂SO₄).
- The balanced reaction can be represented as:
\[ CuSO₄ + 4KCN \rightarrow Cu(CN)₂ + K₂SO₄ \]

3. **Decomposition of Cupric Cyanide:**
- The cupric cyanide (Cu(CN)₂) can decompose into cuprous cyanide (Cu₂(CN)₂) and cyanide ions (CN⁻):
\[ 2Cu(CN)₂ \rightarrow Cu₂(CN)₂ + 2CN⁻ \]

4. **Formation of Stable Complex:**
- The cuprous cyanide (Cu₂(CN)₂) formed can react with excess KCN to form a stable complex, which is potassium tetracyanocuprate(II), represented as K₃Cu(CN)₄.
- The reaction can be summarized as:
\[ Cu₂(CN)₂ + 4KCN \rightarrow K₃Cu(CN)₄ + 2K^+ \]

5. **Final Product:**
- Therefore, the final product when CuSO₄ reacts with excess KCN is K₃Cu(CN)₄.

### Conclusion:
The correct answer to the question is **K₃Cu(CN)₄**.

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