• # question_answer 1) How would you explain? (i) $BeO$ is insoluble but $BeS{{O}_{4}}$ is soluble in water. (ii) $BaO$ is soluble but$BaS{{O}_{4}}$ is insoluble in water. (iii) $LiI$ is more soluble than $KI$ in ethanol.

(i) Due to small size of Beryllium ion,$BeO$is covalent and polymeric in nature and hence, is insoluble in water. The hydration energy of$BeS{{O}_{4}}$ is higher than its lattice energy and hence it is soluble in water. (ii) Both $BaO$ and $BaS{{O}_{4}}$ are ionic compounds. The size of ${{O}^{2-}}$ion is very small in size in comparison to size of $SO_{4}^{2-}$ion while $B{{a}^{2+}}$ ion is comparatively a large ion. Since, bigger anion stabilizes a bigger cation,$BaS{{O}_{4}}$ is more stable than$BaO$. Hence,$BaO$ is soluble due to small size of ${{O}^{2-}}$ion and high hydration energy while$BaS{{O}_{4}}$ is insoluble due to low hydration energy. (iii) Covalent character of $LiI$ is greater than that of $KI$ hence $LiI$ is more soluble in ethanol (non-polar solvent) than $KI$.