An alkaline earth metal 'M' readily forms water soluble sulphate and water insoluble hydroxide. Its oxide MO is very stable to heat and does not have rock-salt structure. M is

We recall that the elements of the alkaline-earth (Group 2) family are $$\text{Be},\ \text{Mg},\ \text{Ca},\ \text{Sr},\ \text{Ba}.$$ The statement tells us three experimental facts about the unknown metal $$M:$$

(i) Its sulphate $$\text{MSO}_4$$ is readily soluble in water.

(ii) Its hydroxide $$\text{M(OH)}_2$$ is almost insoluble in water.

(iii) Its oxide $$\text{MO}$$ is highly refractory (very stable to heat) and does not crystallise in the rock-salt (NaCl) lattice.

Let us examine each fact in the light of periodic trends.

First trend - solubility of sulphates. For Group 2 the solubility order is known to decrease smoothly down the group:

$$\text{BeSO}_4 \; > \; \text{MgSO}_4 \; > \; \text{CaSO}_4 \; ( \text{sparingly soluble} ) \; > \; \text{SrSO}_4 \; ( \text{almost insoluble} ) \; > \; \text{BaSO}_4 \; ( \text{insoluble} ).$$

Thus a “readily water-soluble sulphate” restricts $$M$$ essentially to the top two members, $$\text{Be}$$ or $$\text{Mg}.$$

Second trend - solubility of hydroxides. For hydroxides the order is the reverse of sulphates; solubility increases down the group:

$$\text{Be(OH)}_2 \; ( \text{almost insoluble} ) \; < \; \text{Mg(OH)}_2 \; ( \text{sparingly soluble} ) \; < \; \text{Ca(OH)}_2 \; ( \text{moderately soluble} ) \; \cdots$$

An “insoluble hydroxide” again points only to $$\text{Be}$$ or, at most, $$\text{Mg}.$$

Third fact - nature of the oxide lattice. Most alkaline-earth oxides crystallise in the rock-salt (NaCl-type) structure. The single exception is beryllium oxide $$\text{BeO},$$ which adopts the hexagonal wurtzite lattice. Moreover, $$\text{BeO}$$ has a melting point above $$2500^{\circ}\text{C},$$ making it one of the most heat-resistant simple oxides. Magnesium oxide $$\text{MgO}$$, although also refractory, does possess the rock-salt structure.

Therefore the requirement “oxide does not have rock-salt structure” can be satisfied only by $$\text{BeO}.$$ Combining this with the previous two solubility observations, we conclude unequivocally that

$$M = \text{Be}.$$

Hence, the correct answer is Option D.