Question

How would you explain the following observation?

A $BeO$ is almost insoluble but $BeS{O}_4$ is soluble in water.

B $BaO$ is soluble but $BaS{O}_4$ is insoluble in water.

C $Lil$ is more soluble than $Kl$ in ethanol.

Answer 1

$\text{Lattice energy}$

The amount of energy required to convert one mole of an ionic solid into gaseous ionic constituents
Or
The amount of energy released when one mole of ionic solid is formed from its gaseous ionic constituents.

$\text{Hydration energy}$

Hydration energy is an amount of energy which is released on hydration of one mole number of ions. When the solvent is not water is known as a solvation energy.

$\text{Condition for a substance to become soluble}$

The solubility of a compound in water depends on the balance between lattice energy and hydration energy. When hydration energy of any ion overcomes the amount of lattice energy the bond dissociates, and solubility occurs.

$\text{Explanation}$

$BeO$ is almost insoluble in water and $BeS{O}_4$ is soluble in water. Because $B{e}^{2+}$ is a small cation with a high polarising power and $O^{2-}$ is a small anion. The size compatibility of $B{e}^{2+}$ and $O^{2-}$ 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.
Hence, $BeO$ is almost insoluble but $BeS{O}_4$ is soluble in water


$\text{Lattice energy}$

The amount of energy required to convert one mole of an ionic solid into gaseous ionic constituents
Or
The amount of energy released when one mole of ionic solid is formed from its gaseous ionic constituents.

$\text{Hydration energy}$

Hydration energy is an amount of energy which is released on hydration of one mole number of ions. When the solvent is not water is known as a solvation energy.

$\text{Condition for a substance to become soluble}$

The solubility of a compound in water depends on the balance between lattice energy and hydration energy. When hydration energy of any ion overcomes the amount of lattice energy the bond dissociates, and solubility occurs.

$\text{Explanation}$

$BaO$ is soluble in water, but $BaS{O}_4$ is not. $B{a}^{2+}$ is a large cation and $O^{2-}$ is a small anion. The size compatibility of $B{a}^{2+}$ and $O^{2-}$ 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 overcomed by the hydration energy of the ions. Therefore, $BaO$ is soluble in water.
In case of $BaS{O}_4$, $B{a}^{2+}$ and $S{O}_4^{2-}$ are both large-sized, the hydration energy is lower than its lattice energy.

When $BaS{O}_4$ is dissolved in water, the hydration energy of its ions is not sufficient to overcome the high lattice energy. So, it is not soluble in water.
Hence, $BaO$ is soluble but $BaS{O}_4$ is insoluble in water.


$\text{Explanation}$

$Lil$ is more soluble than $Kl$ in ethanol. Because lithium ion small in size compared to potassium ion, due to this 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 $Lil$ than in $Kl$. And as we know ethanol also is covalent in nature. So $Lil$ become more soluble in ethanol.

Hence, $Lil$ is more soluble in ethanol compared to $Kl$