Question

Aluminium trifluoride is insoluble in anhydrous $HF$ but dissolves on addition of
$NaF$. Aluminium trifluoride precipitates out of the resulting solution when gaseous $B{F_3}$ is bubbled through. Give reasons.

Answer 1

Solubility of any compound depends on different factors one of them is the type of bond they have. As we know that, Aluminium trifluoride is an ionic compound. On the other hand, HF is a covalent compound.

Step by step answer: Aluminium trifluoride is ionic in nature because we know that the bond between an electronegative element and electropositive element is ionic due to complete transfer of electron from aluminium to fluorine.
$Al{F_3}$ is insoluble in $HF$ but dissolves in $NaF$. It is because $HF$ has a covalent bond. As fluorine has a strong affinity for hydrogen atom therefore free ${F^ - }$ ions are not formed and hence are not available for the formation of complexes with$Al{F_3}$. However, $NaF$ is an ionic compound and hence ions are easily available for the formation of complexes of type${\left[ {Al{F_3}} \right]^{3 - }}$.
$Al{F_3} + 3NaF \to N{a_3}\left[ {Al{F_6}} \right]$
On bubbling gaseous $B{F_3}$, aluminium trifluoride is precipitated, boron due to its small size and Higher electronegativity tends to more susceptible to form complexes than aluminum. Hence, precipitation of tetrafluoroaluminate takes place when $B{F_3}$ is passed through a sodium hexafluoroaluminate solution.
$N{a_3}\left[ {Al{F_6}} \right] + 3B{F_3} \to Al{F_3} \downarrow + 3Na\left[ {B{F_4}} \right]$
Aluminium trifluoride precipitated because $B{F_3}$ is stronger Lewis acid than $Al{F_3}$ and hence shall have more tendency to form complex.

Note: Aluminium trifluoride, $Al{F_3}$, is actually used in production of aluminum metal, but it is not reacted in the procedure. It is more or less a catalyst that reduces the melting point of its mixture with bauxite so it’s cheaper to pump enough electricity into the mix to melt and deoxygenate the bauxite.