Question

The most soluble fluoride among those of ${2^{nd}}$ group is:
$(A)Ca{F_2}$
$(B)Sr{F_2}$
$(C)Ba{F_2}$
$(D)Be{F_2}$

Answer 1

Hint: Solubility of any compound is generally considered in water. The compounds can be soluble (fully soluble), partially soluble and insoluble in water. We can determine the solubility of any compound by observing its hydration energy, lattice energy and size of its cation and anion. We can also determine the solubility of any compound by determining the difference in the sizes of its cationic and anionic part.

Complete answer:
We know that solubility of any compound is determined by the amount of solute (compound) which dissolves in the solvent (generally water).
Solubility of any compound is generally determined by its hydration energy and lattice energy. If the lattice energy of the compound is greater than its hydration energy, then the compound will be insoluble in water. If the lattice energy of the compound is less than its hydration energy, then the compound will be soluble in water.
In the above given compounds, the anion fluoride is common. And we know that smaller the cation, greater will be its hydration energy.
Therefore, $B{e^{2 + }}$ is the smallest second group cation and it will have high hydration energy. This means that the hydration energy of $Be{F_2}$ is more than its lattice energy and it will be the most soluble fluoride among the given compounds.
Hence, the correct option is $(D)Be{F_2}$ .

Note:
We now know that if the lattice energy of a compound is greater than its hydration energy, then the compound is insoluble in water. This happens because due to high lattice energy, the compound will not be able to dissociate when mixed with water and hence, it will not get dissolved in the water.