A metal `M` readily forms its sulphate `MSO_(4)` which is water soluble. It forms its oxide `MO` which becomes inert on heating. It forms its insoluble hydroxide `M(OH)_(2)` which is soluble in `NaOH` solution. Then `M` is

Options

A. `Be`
B. Ba
C. Ca
D. Mg

Answer

To determine the identity of the metal \( M \) based on the given properties, we can analyze each characteristic step by step.

### Step 1: Analyze the properties of the metal \( M \)
1. **Forms a water-soluble sulfate \( MSO_4 \)**: This indicates that the metal \( M \) is likely from the group of metals that form soluble sulfates. Common soluble metal sulfates include those of alkali metals and some alkaline earth metals.

### Step 2: Examine the oxide \( MO \)
2. **Forms an oxide \( MO \) that becomes inert on heating**: This suggests that the oxide does not react further when heated, which is characteristic of certain metal oxides, particularly those of alkaline earth metals like beryllium.

### Step 3: Analyze the hydroxide \( M(OH)_2 \)
3. **Forms an insoluble hydroxide \( M(OH)_2 \)**: This indicates that the hydroxide does not dissolve in water. For alkaline earth metals, beryllium hydroxide is known to be insoluble in water.

### Step 4: Solubility in NaOH
4. **The hydroxide \( M(OH)_2 \) is soluble in \( NaOH \)**: This property is indicative of amphoteric behavior, which is characteristic of beryllium hydroxide. Amphoteric hydroxides can react with both acids and bases, leading to solubility in strong bases like sodium hydroxide.

### Conclusion
Based on the analysis:
- The metal \( M \) forms a water-soluble sulfate, has an oxide that becomes inert upon heating, and its hydroxide is insoluble in water but soluble in sodium hydroxide. The only metal that fits all these criteria is **Beryllium (Be)**.

Thus, the metal \( M \) is **Beryllium**.

### Final Answer:
**M is Beryllium (Be)**.

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