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Question -

Howwould you explain the following observations?

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

(ii) BaO is soluble but BaSO4 is insoluble inwater,

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



Answer -

(i) BeOis almost insoluble in water and BeSO4 issoluble in water. Be2+ isa small cation with a high polarising power and O2– is a small anion. The size compatibilityof Be2+ and O2– ishigh. Therefore, the lattice energy released during their formation is alsovery high. When BeO is dissolved in water, the hydration energy of its ions isnot sufficient to overcome the high lattice energy. Therefore, BeO is insolublein water. On the other hand, ion is a large anion. Hence, Be2+ can easily polarise ions, making BeSO4 unstable. Thus, thelattice energy of BeSO4 is not very high and so it is solublein water.


(ii) BaO is soluble in water, but BaSOis not. Ba2+ isa large cation and O2– isa small anion. The size compatibility of Ba2+ andO2– is not high. As a result, BaO is unstable.The lattice energy released during its formation is also not very large. It caneasily be overcome by the hydration energy of the ions. Therefore, BaO issoluble in water. In BaSO4,Ba2+ and are both large-sized. The lattice energy releasedis high. Hence, it is not soluble in water.

(iii) LiI is more soluble than KI in ethanol. As aresult of its small size, the lithium ion has a higher polarising power thanthe potassium ion. It polarises the electron cloud of the iodide ion to a muchgreater extent than the potassium ion. This causes a greater covalent characterin LiI than in KI. Hence, LiI is more soluble in ethanol.

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