How would you explain the following observations?

(i) $\mathrm{BeO}$ is almost insoluble in water and $\mathrm{BeSO}_{4}$ is soluble in water. $\mathrm{Be}^{2+}$ is a small cation with a high polarising power and $\mathrm{O}^{2-}$ is a small anion. The size compatibility of $\mathrm{Be}^{2+}$ and $\mathrm{O}^{2-}$ is high. Therefore, the lattice energy released during their formation is also very high. When $\mathrm{BeO}$ is dissolved in water, the hydration energy of its ions is not sufficient to overcome the high lattice energy. Therefore, $\mathrm{BeO}$ is insoluble in water. On the other hand, $\mathrm{SO}_{4}^{2-}$ ion is a large anion. $\mathrm{Hence}, \mathrm{Be}^{2+}$ can easily polarise $\mathrm{SO}_{4}^{2-}$ ions, making $\mathrm{BeSO}_{4}$ unstable. Thus, the lattice energy of $\mathrm{BeSO}_{4}$ is not very high and so it is soluble in water.

(ii) $\mathrm{BaO}$ is soluble in water, but $\mathrm{BaSO}_{4}$ is not. $\mathrm{Ba}^{2+}$ is a large cation and $\mathrm{O}^{2-}$ is a small anion. The size compatibility of $\mathrm{Ba}^{2+}$ and $\mathrm{O}^{2-}$ is not high. As a result, BaO is unstable. The lattice energy released during its formation is also not very large. It can easily be overcome by the hydration energy of the ions. Therefore, BaO is soluble in water. In $\mathrm{BaSO}_{4}, \mathrm{Ba}^{2+}$ and $\mathrm{SO}_{4}^{2-}$ are both large-sized. The lattice energy released is high. Hence, it is not soluble in water.

(iii) Lil is more soluble than $\mathrm{KI}$ in ethanol. As a result of its small size, the lithium ion has a higher polarising power than the potassium ion. It polarises the electron cloud of the iodide ion to a much greater extent than the potassium ion. This causes a greater covalent character in Lil than in KI. Hence, Lil is more soluble in ethanol.