In the presence of KF, AlF3 is soluble in HF. Find the complex formed

The correct option/answer is K₃[AlF₆].

Aluminum fluoride (${\mathrm{AlF}}_3$) is an ionic compound with a high lattice energy, which makes it insoluble in anhydrous hydrogen fluoride (HF) on its own. HF is a polar covalent solvent but cannot overcome the strong lattice enthalpy of ${\mathrm{AlF}}_3$ directly.

However, when potassium fluoride (KF) is added to the mixture, KF acts as a source of fluoride ions ($F^{-}$) in the solution. These available fluoride ions readily coordinate with aluminum.

Aluminum ($\mathrm{Al}$), being in the third period of the periodic table, has vacant 3d orbitals. This allows it to expand its coordination number from 3 to 6. The ${\mathrm{AlF}}_3$ reacts with the fluoride ions supplied by KF to form a soluble coordination complex.

The balanced chemical equation for this reaction is:
${\mathrm{AlF}}_3+3\mathrm{KF}\to K_3\left[{\mathrm{AlF}}_6\right]$

In this reaction, the complex compound formed is potassium hexafluoroaluminate, which is represented by the formula K₃[AlF₆]. This complex is highly soluble in HF, resulting in the dissolution of aluminum fluoride.