When aqueous NaOH is added to an aqueous solution of chromium (III ) ions, a green blue precipitate is first formed which re - dissolves to give a green solution. The green colour is due to

Options

A. `[Cr(H_2O)_6]^(3+)`
B. `[Cr(OH)_4]^(-)`
C. `CrO_4^(2-)`
D. `[Cr(OH)_3(H_2O)_3]`

Answer

To solve the question, we need to analyze the reaction that occurs when aqueous NaOH is added to an aqueous solution of chromium (III) ions. Let's break it down step by step.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have chromium (III) ions in aqueous solution, represented as \( \text{Cr}^{3+} \).
- We are adding sodium hydroxide (NaOH), which dissociates into \( \text{Na}^+ \) and \( \text{OH}^- \) ions in solution.

2. **Write the Initial Reaction**:
- When NaOH is added to the chromium (III) ion solution, the hydroxide ions (\( \text{OH}^- \)) react with the chromium ions.
- The initial precipitate formed is chromium (III) hydroxide, represented as \( \text{Cr(OH)}_3 \), which is a green-blue precipitate.
- The reaction can be written as:
\[ \text{Cr}^{3+} + 3 \text{OH}^- \rightarrow \text{Cr(OH)}_3 \downarrow \]

3. **Formation of the Precipitate**:
- The precipitate \( \text{Cr(OH)}_3 \) is initially formed and appears as a green-blue solid.

4. **Re-dissolution of the Precipitate**:
- Upon further addition of NaOH, the precipitate can dissolve in excess hydroxide ions to form a complex ion.
- This reaction can be represented as:
\[ \text{Cr(OH)}_3 + \text{OH}^- \rightarrow \text{Cr(OH)}_4^{-} \]
- The complex ion \( \text{Cr(OH)}_4^{-} \) is soluble and gives a green solution.

5. **Conclusion**:
- The green color of the resulting solution is due to the formation of the complex ion \( \text{Cr(OH)}_4^{-} \).

### Final Answer:
The green color is due to the formation of the complex ion \( \text{Cr(OH)}_4^{-} \).