`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[KCu(CN)_4]`

Answer

To solve the problem of what happens when a solution of CuSO4 reacts with excess KCN, we can break it down into several steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this reaction are Copper(II) sulfate (CuSO4) and Potassium cyanide (KCN).

2. **Initial Reaction**:
When CuSO4 reacts with KCN, it forms Potassium sulfate (K2SO4) and Copper(II) cyanide (Cu(CN)2). The balanced equation for this step is:
\[ CuSO4 + 2KCN \rightarrow K2SO4 + Cu(CN)2 \]

3. **Stability of Cu(CN)2**:
Copper(II) cyanide (Cu(CN)2) is unstable and can undergo reduction. In this reaction, Cu(CN)2 is reduced to Cu(I) cyanide (CuCN) and releases cyanogen gas (C2N2). The reduction can be represented as:
\[ Cu(CN)2 \rightarrow 2CuCN + C2N2 \]

4. **Final Reaction with Excess KCN**:
Since the question states that there is excess KCN, the CuCN formed will react with additional KCN to form a complex ion. The reaction is:
\[ CuCN + 3KCN \rightarrow K3Cu(CN)4 \]

5. **Final Product**:
The final product of the reaction is Potassium tetracyanocuprate(III), which is represented as K3Cu(CN)4.

### Conclusion:
The overall reaction can be summarized as:
\[ CuSO4 + 4KCN \rightarrow K2SO4 + K3Cu(CN)4 + C2N2 \]
Thus, the product formed when CuSO4 reacts with excess KCN is **K3Cu(CN)4**.