`Fe(CN)_(2)darr+4KCN to K_(4)[Fe(CN)_(6)]`

Options

A. For precipitate formation reaction
B. For precipitate dissolution reaction
C. For precipitate exchange reaction
D. For no reaction

Answer

To determine the type of reaction represented by the equation:

\[ \text{Fe(CN)}_2 + 4 \text{KCN} \rightarrow \text{K}_4[\text{Fe(CN)}_6] \]

we can follow these steps:

### Step 1: Identify the Reactants and Products
- The reactants are \(\text{Fe(CN)}_2\) and \(4 \text{KCN}\).
- The product is \(\text{K}_4[\text{Fe(CN)}_6]\), which is a complex ion.

### Step 2: Analyze the Nature of the Reactants
- \(\text{Fe(CN)}_2\) is a compound that can exist as a precipitate in solution. The downward arrow in the equation suggests that it is indeed a solid precipitate.

### Step 3: Analyze the Nature of the Product
- The product \(\text{K}_4[\text{Fe(CN)}_6]\) is a soluble complex. This indicates that it does not form a precipitate.

### Step 4: Determine the Type of Reaction
- Since we start with a precipitate (\(\text{Fe(CN)}_2\)) and end with a soluble complex (\(\text{K}_4[\text{Fe(CN)}_6]\)), we can conclude that the reaction involves the dissolution of the precipitate into a soluble form.

### Conclusion
- Therefore, this reaction is classified as a **precipitate dissolution reaction**.

The answer is **B: Precipitate dissolution reaction**.

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