10.29 How would you explain the following observations? (i) BeO is almost insoluble but BeSO4 is soluble in water, (ii) BaO is soluble but BaSO4 is insoluble in water, (iii) LiI is more soluble than KI in ethanol.

Answers (1)
R Rakesh

(i) Be^{2+} is small in size so it has high polarising power and O^{2-} is also small in size. Compatibility of both the cation and anion are very high. So their lattice energy is also very high. When BeO is dissolved in water it's hydration energy is not sufficient to overcome its lattice energy. So, therefore, it is insoluble in water.

On the other hand, SO_{4}^{2-} ions are large in size. Hence Be^{+} ion can easily polarise SO_{4}^{2-} ions and making it unstable and  because of that lattice energy of BeSO_{4} is not very high and so it is soluble in water.

(ii) BaO is soluble because Ba^{2+} cation is large in size as compare to O^{2-} anion. Size compatibility between them is not good. Therefore BaO is unstable. Hence lattice energy during the formation of their lattice is not high So it can be easily overcome by hydration energy. Therefore BaO is soluble in water. In case of BaSO_{4}, we know that down the group hydration enthalpy decreases and both the anion and cation have very good size compatibility. So, lattice energy cannot be overcome by hydration energy. As a result, BaSO_{4}is not soluble in water.

(iii)  The Li^{+} ion has high polarising power. It is very small in size as compare to K^{+} ion. So, it has a high tendency to distort the electron cloud around the negative iodide (I^{-}) ion. As a result of high polarizability, it has high covalent character than.KI Hence it is more soluble in methanol.