AlF_3 is soluble in HF only in the presence o | vedclass

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(C) $AlF_3$ is insoluble in anhydrous $HF$ because it lacks free $F^-$ ions to interact with the $Al^{3+}$ center. In the presence of $KF$,$KF$ dissoc

Vedclass · Accepted Answer

$[AlF_6]^{3-}$

Vedclass · Answer

(C) $AlF_3$ is insoluble in anhydrous $HF$ because it lacks free $F^-$ ions to interact with the $Al^{3+}$ center. In the presence of $KF$,$KF$ dissociates to provide $F^-$ ions,which react with $AlF_3$ to form a soluble complex. The reaction is: $AlF_3 + 3KF \rightarrow K_3[AlF_6]$ This complex dissociates into $3K^+$ and $[AlF_6]^{3-}$. Therefore,the correct option is $[AlF_6]^{3-}$.

$AlF_3$ is soluble in $HF$ only in the presence of $KF$ due to the formation of:

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