Why does $B{{F}_{6}}^{3-}$ not exist?

Hint: Boron is an element of 13 groups, therefore, it has 3 electrons in its valence shell. Its valence electron is present in the p-block, and the p-block has 3 subshells. Boron atoms have 2 free subshells in p-orbital.

Complete answer:
Boron is an element of p-block. It has atomic number 5 and is in group 13 and period 2. The electronic configuration of boron atoms is $1{{s}^{2}},2{{s}^{2}},2{{p}^{1}}$ . Therefore, it has one electron in p-orbital. It has a valency of 3 and can show a maximum valency of 4 and Beyond this value, boron will become unstable. It has no d-orbital present to expand its valency.
Since boron has maximum covalency of 4 due to its non availability of d orbital, it is unable to expand to its octet configuration. So $B{{F}_{6}}^{3-}$ does not exist.
Aluminium has valence d–orbital therefore, it can expand its octet to form $Al{{F}_{6}}^{3-}$ unlike boron atoms.
In the form of boric acid or borates, traces of boron are necessary for growth of many land plants and thus are indirectly essential for animal life. Pure boron exists in at least four crystalline modifications or allotropes.

Note:
Boron is concentrated on Earth by the water-solubility of its more common naturally occurring compounds, the borate minerals. These are mined industrially as evaporites, such as borax and kernite. Elemental boron is a metalloid that is found in small amounts in meteoroids but chemically uncombined boron is not otherwise found naturally on Earth.