Question

Which one of the following is the least soluble?
i) $Be{{F}_{2}}$
ii) $LiF$
iii) $Mg{{F}_{2}}$
iv) $Ca{{F}_{2}}$

Answer 1

The compounds given in the question are known as fluorides as these all compounds contain fluoride ion represented by the symbol ${{F}^{-}}$ it is an inorganic monatomic anion of fluorine whereas anion are those ions which are negatively charged.

Complete answer:
The salts of fluorides are colorless or white in color. Fluoride salts have different types of tastes, some bitter in taste and it is odorless in nature. It is classified as a weak base as it partially dissociates in solution.
Out of all the compounds given Li i.e. lithium ions have the highest hydration energy due to its size i.e. it is small as compared to other alkali metal ions so we can say that it should have high solubility. However the small size of $L{{i}^{+}}$ and ${{F}^{-}}$ ions interact very strongly with each other and resulting in the high lattice energy of $LiF$ and this is responsible for its insolubility. Hence we can say that $LiF$ is the least soluble among the fluorides of alkali metals. Lithium fluoride is also of covalent nature because of the high polarizing power of $L{{i}^{+}}$ ion due to its very small size and high effective nuclear charge.

Hence we can conclude that $LiF$is least stable out of given compounds.

Note:
There are many applications of fluorides like the salts and minerals of fluorides are important chemical reagents and can also be used as industrial chemicals mainly in the production of hydrogen fluoride for fluorocarbons.