Solution:
Hydrogen fluoride is a covalent compound with an exceptionally solid intermolecular clinging to hydrogen. Along these lines, it doesn’t give particles and doesn’t break up aluminum fluoride in it. Sodium fluoride is an ionic compound, and Alf breaks down when added to the blend. This is a direct result of the Free $F^{-}$available. The response associated with the cycle is:
\[3NaF\text{ }+\text{ }Al{{F}_{3}}~\to {{a}_{3}}[Al{{F}_{6}}]\]
At the point when boron trifluoride is added to the arrangement, aluminum fluoride becomes accelerated out of arrangement. This is on the grounds that the boron’s inclination to shape edifices is a lot more noteworthy than that of aluminum. In this way, when boron trifluoride is added to the arrangement, B replaces Al of the edifices by the accompanying response
$N a_{3}\left[A l F_{6}\right]+3 B F_{3} \rightarrow 3 N a\left[B F_{4}\right]+A l F_{3}$