p-Amino-N,N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with a few drops of aqueous solution of Y to yield blue colouration due to the formation of methylene blue. Treatment of aqueous solution of Y with reagent potassium hexacyanoferrate (II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, the treatment of the solution of Y with the solution of potassium hexacyanoferrate (III) leads to a brown colouration due to the formation of Z.
Q. Compound Y is

Options

A. `MgCl_2`
B. `FeCl_2`
C. `FeCl_3`
D. `ZnCl_2`

Explanation

To determine the compound Y in the given scenario, we will analyze the information step by step.

### Step-by-Step Solution:

1. **Understanding the Initial Compound**:
- The question starts with p-Amino-N,N-dimethylaniline. This compound has an amino group (-NH2) and two methyl groups attached to the nitrogen. Its structure can be represented as:
\[ \text{C}_6\text{H}_4(\text{NH}_2)(\text{N(CH}_3\text{)}_2) \]

2. **Reaction with Strongly Acidic Solution (X)**:
- The compound is added to a strongly acidic solution. A common strong acid used in such contexts is hydrochloric acid (HCl) or sulfuric acid (H2SO4). However, the video suggests that the acidic solution X is sodium sulfide (Na2S), which releases sulfide ions (S^2-). This indicates that the compound is likely protonated in the acidic environment.

3. **Formation of Methylene Blue**:
- The solution is then treated with an aqueous solution of Y, which results in the formation of methylene blue. Methylene blue is formed when p-Amino-N,N-dimethylaniline is oxidized. This oxidation typically requires an oxidizing agent, which is likely to be iron(III) ions (Fe^3+).

4. **Identifying Compound Y**:
- Since the formation of methylene blue is due to the presence of Fe^3+ ions, we can deduce that compound Y must be a source of Fe^3+. A common source of Fe^3+ in aqueous solution is ferric chloride (FeCl3). Therefore, we can conclude that:
\[ \text{Compound Y} = \text{FeCl}_3 \]

5. **Further Reactions with Potassium Hexacyanoferrate(II) and (III)**:
- The video mentions that treating Y with potassium hexacyanoferrate(II) results in an intense blue precipitate, which is likely to be Prussian blue (Fe3[Fe(CN)6]2). This precipitate dissolves upon the addition of excess reagent, which is characteristic of complex formation.
- Additionally, treating Y with potassium hexacyanoferrate(III) leads to a brown coloration due to the formation of a brown compound Z, which is likely to be a different iron complex.

### Conclusion:
- Based on the above analysis, the compound Y is identified as:
\[ \text{Y} = \text{FeCl}_3 \]