Copper (II) ions gives reddish brown precipitate with potassium ferrocyanide. The formula of the precipitate is:

Options

A. `Cu_(4)[fe(CN)_(6]`
B. `Cu_(2)[Fe(CN)_(6)]`
C. `Cu_(3)[Fe(CN)_(6)]`
D. `Cu_(3)[Fe(CN)_(6)]_(2)`

Answer

To determine the formula of the reddish brown precipitate formed when Copper (II) ions react with potassium ferrocyanide, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants in this case are Copper (II) ions (Cu²⁺) and potassium ferrocyanide (K₄[Fe(CN)₆]).

2. **Understand the Reaction**:
- When Copper (II) ions react with potassium ferrocyanide, they form a complex compound. Potassium ferrocyanide is known to react with various metal ions to form precipitates.

3. **Write the Formula for Potassium Ferrocyanide**:
- The formula for potassium ferrocyanide is K₄[Fe(CN)₆]. This indicates that it contains four potassium ions (K⁺), one iron ion (Fe²⁺), and six cyanide ions (CN⁻).

4. **Determine the Product**:
- The product formed from the reaction of Cu²⁺ and K₄[Fe(CN)₆] is known as cupric ferrocyanide. The general formula for the precipitate formed with copper (II) ions is Cu₂[Fe(CN)₆].

5. **Identify the Color of the Precipitate**:
- The precipitate formed is described as reddish brown, which is characteristic of cupric ferrocyanide.

6. **Final Formula**:
- Therefore, the formula of the reddish brown precipitate is Cu₂[Fe(CN)₆], which is also referred to as cupric ferrocyanide.

### Conclusion:
The formula of the reddish brown precipitate formed when Copper (II) ions react with potassium ferrocyanide is **Cu₂[Fe(CN)₆]**.

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