Solution of alkali metal cyanide containing freshly prepared ion `(II)` sulphate solution and dilute `H_(2)SO_(4)` on exposure to air produces prussian blue precipitate

Explanation

To solve the problem, we will analyze the reaction step by step, focusing on the formation of the Prussian blue precipitate from the given components.

### Step-by-Step Solution:

1. **Identify the Components**:
- We have an alkali metal cyanide (let's take potassium cyanide, KCN).
- Freshly prepared iron(II) sulfate (FeSO4).
- Dilute sulfuric acid (H2SO4).
- The reaction is exposed to air.

2. **Initial Reaction**:
- When KCN reacts with FeSO4, the following reaction occurs:
\[ 2 \text{KCN} + \text{FeSO}_4 \rightarrow \text{Fe(CN)}_2 + \text{K}_2\text{SO}_4 \]
- This produces iron(II) cyanide, which can be represented as Fe(CN)2.

3. **Formation of Iron(III) Ions**:
- On exposure to air, the iron(II) ions (Fe²⁺) get oxidized to iron(III) ions (Fe³⁺):
\[ \text{Fe}^{2+} + \text{O}_2 + 4 \text{H}^+ \rightarrow \text{Fe}^{3+} + 2 \text{H}_2\text{O} \]

4. **Formation of Prussian Blue**:
- The iron(III) ions then react with the cyanide ions to form the complex known as Prussian blue:
\[ \text{Fe}^{3+} + \text{Fe(CN)}_6^{4-} \rightarrow \text{Fe}_3[\text{Fe(CN)}_6] \text{ (Prussian blue)} \]
- The complex formed is often represented as Fe3[Fe(CN)6]2, which is the Prussian blue precipitate.

5. **Final Product**:
- The final product is a blue precipitate of Prussian blue, which is indicative of the presence of Fe³⁺ ions and cyanide ions.

### Summary of the Reaction:
- The alkali metal cyanide reacts with iron(II) sulfate to form iron(II) cyanide.
- Upon exposure to air, iron(II) is oxidized to iron(III).
- The iron(III) ions then react with cyanide to form the blue precipitate of Prussian blue.