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 regarding the color change when aqueous NaOH is added to an aqueous solution of chromium (III) ions, we can follow these steps:

### Step 1: Understand the Initial Reaction
When sodium hydroxide (NaOH) is added to a solution containing chromium (III) ions (Cr³⁺), a reaction occurs. Chromium (III) ions in aqueous solution are typically present as [Cr(H₂O)₆]³⁺.

**Hint:** Remember that chromium (III) ions are hydrated in solution, meaning they are surrounded by water molecules.

### Step 2: Formation of Precipitate
The hydroxide ions (OH⁻) from NaOH react with the chromium (III) ions to form a precipitate. The initial precipitate formed is chromium (III) hydroxide, which can be represented as:

\[ \text{[Cr(H}_2\text{O)}_6]^{3+} + 3 \text{OH}^- \rightarrow \text{Cr(OH)}_3 (s) + 3 \text{H}_2\text{O} \]

This chromium (III) hydroxide precipitate is typically a green-blue color.

**Hint:** Identify the precipitate formed in the reaction and its color.

### Step 3: Re-dissolution of Precipitate
Upon further addition of NaOH, the precipitate of chromium (III) hydroxide can dissolve in excess hydroxide ions to form a complex ion. This complex ion is known as chromate ion:

\[ \text{Cr(OH)}_3 + \text{OH}^- \rightarrow \text{[Cr(OH)}_4]^{-} \]

This results in a green solution due to the formation of the [Cr(OH)₄]⁻ complex.

**Hint:** Consider how the precipitate can dissolve in excess hydroxide and what complex ion is formed.

### Step 4: Conclusion about the Color
The green color of the resulting solution is due to the presence of the [Cr(OH)₄]⁻ complex ion.

**Hint:** Recall that the color of a solution can often be attributed to the specific ions or complexes present in it.

### Final Answer
The green color is due to the formation of the [Cr(OH)₄]⁻ complex ion.

**Correct Option:** B