How would you explain the following observations?

(i) BeO is almost insoluble but BeSO4 is insoluble in water.

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

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

Asked by Pragya Singh | 1 year ago |  121

1 Answer

Solution :-

(i) The sizes of Be2+ and O2- are small and are highly compatible with each other. Due to this, a high amount of lattice energy is released during its formation. The hydration energy, when it is made to dissolve in water, is not enough to overcome the lattice energy. Thus, BeO is almost insoluble in water.

Whereas the size of an SO42- is large compared to Be2+ and there is lesser compatibility and lattice energy which can be easily overcome by the hydration energy. Thus, BeSO4 is easily soluble in water.

(ii) The sizes of Ba2+ and SO42- are large and are highly compatible with each other. Due to this, a high amount of lattice energy is released during its formation. The hydration energy, when it is made to dissolve in water, is not enough to overcome the lattice energy. Thus, BaSO4 is insoluble in water.

Whereas the size of an O2- is small compared to Be2+ and there is lesser compatibility and lattice energy which can be easily overcome by the hydration energy. Thus, BaO is easily soluble in water.

(iii) The lithium-ion has a smaller size and as a result of that, it has a higher polarizing capability. This enables it to polarize the electron cloud around an iodide ion thus resulting in a greater covalent character in LiI than KI. Thus, LiI is easily soluble in ethanol.

Answered by Abhisek | 1 year ago

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