Question

How would you explain the following observations: (i) BeO is almost insoluble but $BeSO_{4}$ is soluble in water. (ii) BaO is soluble but $BaSO_{4}$ is insoluble in water. (iii) Lil is more soluble than KI in ethanol.

Answer 1

"1. Lattice energy of BeO is comparatively higher than the "hydration energy". Therefore, it is almost "insoluble" in water whereas $Be{ SO }_{ 4 }$ is "ionic in nature" and its "hydration energy dominates" the "lattice energy".

2. Both BaO and$Ba{ SO }_{ 4 }$are "ionic compounds" but the "hydration energy" of BaO is "higher" than the "lattice energy" therefore it is "soluble in water".

3. Since the "size" of${ Li }^{ + } ion$ is "very small" in comparison to ${ K }^{ + } ion$, it "polarises the electron cloud" of ${ I }^{ - } ion$ to a "great extent". Thus, "Lil" dissolves in "ethanol more easily" than the "KI"."

Answer 2

Answer:

(i) Due to high lattice enthalpy and covalent nature, BeO is insoluble in water. In case of ionic BeSO

4

, the hydration enthalpy is much more than its lattice enthalpy. Hence, it is water soluble.

(ii) Oxide ion has smaller size than sulphate ion. BaO has smaller lattice energy than BaSO

4

as bigger cation stablizes bigger anion to greater extent than smaller cation stabilizes bigger anion. Hence, BaO is water soluble and BaSO

4

is water insoluble.

(iii) As per Fajan rules, smaller Li

+

ion polarizes bigger I

−

to a greater extent than K

+

ion. Hence, LiI is more covalent than KI and hence, more soluble in organic solvents such as ethanol.