When `CuSO_4` solution is added to `K_4[Fe(CN)_6]`, the formula of the product formed is

Options

A. `Cu_2Fe(CN)_6`
B. `KCN`
C. `Cu(CN)_3`
D. `Cu(CN)_2`

Explanation

To solve the question regarding the reaction between `CuSO4` and `K4[Fe(CN)6]`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this case are copper(II) sulfate (`CuSO4`) and potassium ferrocyanide (`K4[Fe(CN)6]`).

2. **Understand the Reaction**:
When `CuSO4` is added to `K4[Fe(CN)6]`, a double displacement reaction occurs. In this reaction, the copper ions from `CuSO4` will react with the ferrocyanide ions from `K4[Fe(CN)6]`.

3. **Write the Reaction**:
The reaction can be represented as:
\[ CuSO_4 + K_4[Fe(CN)_6] \rightarrow Cu_2[Fe(CN)_6] + K_2SO_4 \]
Here, `Cu2[Fe(CN)6]` is the product formed, which is known as copper(II) ferrocyanide.

4. **Determine the Product's Formula**:
The formula of the product formed is `Cu2[Fe(CN)6]`.

5. **Final Answer**:
Therefore, the formula of the product formed when `CuSO4` solution is added to `K4[Fe(CN)6]` is `Cu2[Fe(CN)6]`.