AlF 3 is soluble in H F only in the presence of K F due to formation of

${\mathrm{AlF}}_3$ is not soluble in the anhydrous $\mathrm{HF}$ because of non availability of the ${\mathrm{F}}^{-}$ ions to the intermolecular hydrogen which is bonded to $\mathrm{HF}$ but it possesses solubility in the presence of $\mathrm{KF}$ that is due to the formation of soluble complex. The full reaction is shown below:

.${\mathrm{AlF}}_3 + 3\mathrm{KF} \to {\mathrm{K}}_3\left[{\mathrm{AlF}}_6\right]$

Thus, the correct option is ${\left[{\mathrm{AlF}}_6\right]}^{3-}$