A metal M reacts with sodium hydroxide to give a white precipitate X which is soluble in excess of NaOH to give Y. Compound X is soluble in HCI to form a compound Z. Identify M.X,Y and Z
M , X , Y, Z

Options

A. `Si, SiO_(2), Na_(2)siO_(3), SiCI_(4)`
B. `AI, AiI(OH)_(3), NaMgO_(3), AICI_(3)`
C. `Mg, Mg(Oh)_(3), NaMgO_(3), MgCI_(2)`
D. `Ca, Ca(OH)_(2) , Na_(2) CO_(3), NaHCO_(3)`

Explanation

To solve the problem, we need to identify the metal (M), the white precipitate (X), the compound formed in excess NaOH (Y), and the compound formed when X is dissolved in HCl (Z).

### Step-by-Step Solution:

1. **Identify the Metal (M)**:
The metal M must be amphoteric, meaning it can react with both acids and bases. A common metal that fits this description is aluminum (Al).

2. **Reaction with Sodium Hydroxide (NaOH)**:
When aluminum reacts with sodium hydroxide, it forms aluminum hydroxide, which is a white precipitate. The reaction can be represented as:
\[ 2 \text{Al} + 2 \text{NaOH} + 6 \text{H}_2\text{O} \rightarrow 2 \text{Al(OH)}_3 (s) + 2 \text{Na}^+ \]
Here, the white precipitate (X) is aluminum hydroxide, Al(OH)₃.

3. **Solubility in Excess NaOH**:
Aluminum hydroxide (X) is soluble in excess sodium hydroxide, forming sodium aluminate (Y). The reaction is:
\[ \text{Al(OH)}_3 (s) + \text{OH}^- \rightarrow \text{NaAlO}_2 (aq) + 2 \text{H}_2\text{O} \]
Thus, Y is sodium aluminate, NaAlO₂.

4. **Solubility in HCl**:
When aluminum hydroxide (X) is dissolved in hydrochloric acid (HCl), it forms aluminum chloride (Z). The reaction can be represented as:
\[ \text{Al(OH)}_3 (s) + 3 \text{HCl} \rightarrow \text{AlCl}_3 (aq) + 3 \text{H}_2\text{O} \]
Therefore, Z is aluminum chloride, AlCl₃.

### Summary of Identifications:
- **M**: Aluminum (Al)
- **X**: Aluminum hydroxide (Al(OH)₃)
- **Y**: Sodium aluminate (NaAlO₂)
- **Z**: Aluminum chloride (AlCl₃)