Question

Among the following, which is insoluble in water?
A) Sodium fluoride
B) Barium fluoride
C) Calcium fluoride
D) Magnesium fluoride
E) All are correct.

Answer 1

Consider the lattice energy between the ions and the size of the ions in the compound. Hydration energy decreases down the group

Complete step by step solution:
In the question four options are given and we have to select the compound which is completely insoluble in water.
We can see that the given options are all ionic compounds, the cations belong to the alkaline earth metal and to alkali metal and the anion common in all options is the fluoride ion.
So now we should compare the answer with respect to the cation, but before that, we should know the relation between solubility, hydration energy and lattice energy. And the basic idea of these terms.
-Solubility defines the ability of the substance or the ionic compound to dissolve in the solvent.
-Hydration energy or hydration enthalpy is the energy released when any one of the substances is undergoing hydration.
-Lattice energy- it is the energy required to break the bonds between the ionic compounds and convert one mole of the solid ionic compound into gaseous form. The lattice energy gives the strength of the bonds between the ions.
Now we have to see the relation between these terms,
Hydration energy is size dependent-it depends on the size of the ions and hydration energy decreases down the group.
Lattice energy is dependent on the charge and size of the ions, as the charge increases lattice energy increases as the size increases lattice energy decreases.
So we can conclude that, if the hydration energy released exceeds the lattice energy then the substance is soluble in water.
$\text{Hydration enthalpy }\propto \,\text{Solubility}\,\propto \dfrac{1}{Lattice\,energy}$
In the option NaF is given which is a fluoride of alkali metals and is soluble in water.
Now we have to see the other options, Barium fluoride, Calcium fluoride and Magnesium fluoride
All the cations here are alkaline earth metals, so let’s compare these options with the aid of above explained terms.
Here the size of cations in the increasing order of the ionic size is, $\operatorname{Mg}\,<\,Ca\,<\,Ba$
As the size increases lattice enthalpy will be less and the hydration energy will be more and will be soluble in water.
Here, $Mg{{F}_{2}}$ is having the smallest size and will have greater lattice energy so the solubility is less. $Mg{{F}_{2}}$ is an inorganic compound which is not soluble in water.

So the correct option for the question is option (D)

Note:
We should know the positions of each element in the periodic table to comment about the charge and the size of the ions, which helps in explaining various properties of the compounds formed by the elements.
Calcium fluoride is also almost insoluble in water as it possesses a greater lattice energy.$Ca{{F}_{2}}$ is sparingly soluble in water.
$Mg{{F}_{2}}$ is also insoluble in ethanol, it is soluble in nitric acid.