Question 10.29:

How would you explain the following observations?

(i) BeO is almost insoluble but BeSO4 in soluble in water,

(ii) BaO is soluble but BaSO4 is insoluble in water,

(iii) LiI is more soluble than KI in ethanol.

(i) BeO is almost insoluble in water and BeSO4 is soluble in water. Be2+ is a small cation with a high polarising power and O2- is a small anion. The size compatibility of Be2+ and O2- is high. Therefore, the lattice energy released during their formation is also very high. When BeO is dissolved in water, the hydration energy of its ions is not sufficient to overcome the high lattice energy. Therefore, BeO is insoluble in water. On the other hand, ${\mathrm{SO}}_{4}^{2-}$ ion is a large anion. Hence, Be2+ can easily polarize ${\mathrm{SO}}_{4}^{2-}$ ions, making BeSO4 unstable. Thus, the lattice energy of BeSO4 is not very high and so it is soluble in water.
(ii) BaO is soluble in water, but BaSO4 is not. Ba2+ is a large cation and O2- is a small anion. The size compatibility of Ba2+ and O2- 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 BaSO4, Ba2+ and ${\mathrm{SO}}_{4}^{2-}$ are both large-sized. The lattice energy released is high. Hence, it is not soluble in water.
(iii) LiI is more soluble than 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 LiI than in KI. Hence, LiI is more soluble in ethanol.