`CuSO_(4)` reacts with excess `KCN` to form

Options

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

Answer

To solve the question of how `CuSO4` reacts with excess `KCN`, we can break down the reaction into several steps:

### Step 1: Initial Reaction
When copper(II) sulfate (`CuSO4`) reacts with potassium cyanide (`KCN`), the first reaction that occurs is the formation of copper(I) cyanide (`CuCN`) and potassium sulfate (`K2SO4`). The balanced equation for this reaction is:

\[ CuSO_4 + 2 KCN \rightarrow K_2SO_4 + 2 CuCN \]

### Step 2: Formation of Insoluble Copper(I) Cyanide
In this step, the `CuCN` formed is insoluble in water. Therefore, we have:

\[ 2 CuCN \text{ (s)} \]

This is a solid precipitate that forms during the reaction.

### Step 3: Formation of Cyanogen
The `CuCN` can further react with excess `KCN` to form a soluble complex. The reaction can be represented as:

\[ 2 CuCN + 2 KCN \rightarrow 2 K_3Cu(CN)_4 \]

This reaction indicates that the insoluble `CuCN` dissolves in excess `KCN` to form the soluble complex `K3Cu(CN)4`.

### Final Product
Thus, the final product of the reaction between `CuSO4` and excess `KCN` is:

\[ K_3Cu(CN)_4 \]

### Conclusion
The answer is that `CuSO4` reacts with excess `KCN` to form `K3Cu(CN)4`.

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