`Cr(OH)_(3) darr+NH_(3)(Excess) to [Cr(NH_(3))_(6)]^(3+)`

Options

A. For coloured ppt./Black ppt
B. For coloured solution.
C. for clear/colourless solution
D. For white ppt.

Answer

To solve the question regarding the reaction between chromium hydroxide \((Cr(OH)_3)\) and ammonia \((NH_3)\) in excess, we will analyze the reaction step by step.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in the reaction are chromium hydroxide \((Cr(OH)_3)\) and ammonia \((NH_3)\).

2. **Understand the Reaction**:
The reaction can be represented as:
\[ Cr(OH)_3 + 6NH_3 \rightarrow [Cr(NH_3)_6]^{3+} + 3OH^- \]
Here, chromium hydroxide reacts with ammonia to form a complex ion \([Cr(NH_3)_6]^{3+}\).

3. **Determine the Nature of the Product**:
The product \([Cr(NH_3)_6]^{3+}\) is a coordination complex. Coordination complexes are typically soluble in solution, especially when formed with ligands like ammonia.

4. **Analyze the Color of the Complex**:
The oxidation state of chromium in the complex \([Cr(NH_3)_6]^{3+}\) is +3. It is known that chromium(III) complexes with ammonia exhibit a characteristic color. Specifically, \([Cr(NH_3)_6]^{3+}\) is known to be purple.

5. **Conclusion**:
Since the reaction produces a soluble complex ion and the complex is colored, we conclude that the reaction results in a colored solution.

### Final Answer:
The reaction between \(Cr(OH)_3\) and excess \(NH_3\) forms a colored solution.

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