`Cu(OH)_(2)darr+4NH_(3)("soln.")to[Cu(NH_(3))_(4)]^(2+)+2OH^(-)`

Options

A. For precipitate formation 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:

\[ \text{Cu(OH)}_2 + 4\text{NH}_3 \text{(soln.)} \rightarrow [\text{Cu(NH}_3)_4]^{2+} + 2\text{OH}^- \]

we can follow these steps:

### Step 1: Identify the Reactants and Products
- **Reactants**: Copper(II) hydroxide \(\text{Cu(OH)}_2\) and ammonia \(\text{NH}_3\).
- **Products**: A complex ion \([\text{Cu(NH}_3)_4]^{2+}\) and hydroxide ions \(2\text{OH}^-\).

### Step 2: Analyze the Reactants
- Copper(II) hydroxide \(\text{Cu(OH)}_2\) is a solid and typically appears as a precipitate in aqueous solutions.
- Ammonia \(\text{NH}_3\) is in solution and acts as a ligand in this reaction.

### Step 3: Analyze the Products
- The product \([\text{Cu(NH}_3)_4]^{2+}\) is a complex ion that is soluble in water.
- The presence of \(2\text{OH}^-\) indicates that hydroxide ions are also produced.

### Step 4: Determine the Type of Reaction
- Since \(\text{Cu(OH)}_2\) is a precipitate that dissolves to form a soluble complex ion \([\text{Cu(NH}_3)_4]^{2+}\), we can conclude that the reaction involves the dissolution of a precipitate.
- The reaction does not involve a simple exchange of ions but rather the formation of a complex from a precipitate.

### Conclusion
This reaction is classified as a **precipitate dissolution reaction** because a solid precipitate (copper(II) hydroxide) is dissolving to form soluble products.

### Final Answer
The type of reaction is **precipitate dissolution reaction**.

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