Q.Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3 is bubbled through. Give reasons.

Excess F$^-$ from NaF turns insoluble AlF$_3$ into soluble $\text{Na}_3[\text{AlF}_6]$. Boron is a stronger Lewis acid than aluminium towards fluoride (small size, no d-orbitals to fill but very high charge density), so BF$_3$ scavenges F$^-$ ions out of the aluminofluoride complex to form the very stable $[\text{BF}_4]^-$ ion, and …

This is a competition between two Lewis acids — Al$^{3+}$ (in AlF$_3$) and B (in BF$_3$) — for the Lewis base F$^-$.

1. Why AlF$_3$ is insoluble in pure anhydrous HF. AlF$_3$ is a giant ionic lattice with very high lattice energy; anhydrous HF alone cannot supply enough F$^-$ to break it down into a complex ion.
2. Why it dissolves on adding NaF. NaF is a ready source of F$^-$ ions. Al$^{3+}$, being a small, highly-charged Lewis acid, readily accepts three more F$^-$ ions to complete an octahedral coordination sphere:

This complex anion is soluble, so the solid dissolves. …

Identify this as competing Lewis-acid–Lewis-base (complex formation) equilibria: compare the relative affinity of Al$^{3+}$ and B (in BF$_3$) for F$^-$. Write the complexation step ($\text{AlF}_3 + 3\text{F}^- \to [\text{AlF}_6]^{3-}$), then the fluoride-abstraction step by the stronger acid BF$_3$ ($[\te …

• Saying AlF$_3$ is simply "insoluble" without explaining the complex-ion mechanism by which NaF dissolves it.
• Forgetting that the reprecipitation is due to BF$_3$'s stronger affinity for F$^-$, not some unrelated reaction. …