`Co^(2+) + KCN` (not in excess) `to` precipitate.
Select the correct statement(s) with respect to the precipitate.

Options

A. It is yellow in colour
B. It is reddish-brown in colour
C. It dissolves in excess of the reagent forming a brown solution
D. It is obtained when brown solution (option C) is acidified with dilute `HCl` in the cold.

Explanation

To solve the question regarding the reaction of Co²⁺ with KCN (not in excess) and the properties of the resulting precipitate, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in this case are Co²⁺ ions (cobalt(II) ions) and KCN (potassium cyanide).

2. **Reaction in Limited Amount**: When Co²⁺ reacts with KCN in limited amounts, it forms a complex compound. The reaction can be represented as:
\[ \text{Co}^{2+} + 2 \text{KCN} \rightarrow \text{Co(CN)}_2 \text{ (Cobalt(II) cyanide)} \]
This compound is known to precipitate out of the solution.

3. **Color of the Precipitate**: The precipitate formed, Co(CN)₂, is reddish-brown in color.

4. **Excess KCN Reaction**: If KCN is added in excess, it forms a different complex:
\[ \text{Co(CN)}_6^{4-} \]
This complex is brown in color.

5. **Acidification of the Complex**: When the brown solution (which contains the complex Co(CN)₆⁴⁻) is treated with dilute HCl, it leads to the formation of CoCl₂, which is also a precipitate.

6. **Correct Statements**: Based on the above information, we can analyze the options provided in the question:
- **Option 1**: Incorrect, because the precipitate is reddish-brown, not just brown.
- **Option 2**: Correct, as it describes the precipitate accurately.
- **Option 3**: Correct, since the precipitate dissolves in excess KCN forming a brown solution.
- **Option 4**: Correct, as it states that the precipitate is obtained when the brown solution is acidified with dilute HCl.

### Summary of Correct Statements:
- Option 2 is correct.
- Option 3 is correct.
- Option 4 is correct.