Question

$Al{F}_3$ is soluble in $HF$ only in presence of $KF$. It is due to the formation of :

• A. $Al{H}_3$
• B. $K\left[Al{F}_{3}H\right]$
• C. $K_3\left[Al{F}_3{H}_3\right]$
• D. $K_3\left[Al{F}_6\right]$

Answer 1

Answer: (D)

$K_3\left[Al{F}_6\right]$

Aluminium forms fluoride complexes more easily than in the case of silicon. So, $Al{F}_3$ easily dissolve in mixture of $(HF+KF)$.

$Al{F}_3$ is soluble in $HF$ due to the formation of $K_3\left[Al{F}_6\right]$.

$Al{F}_3+3KF\stackrel{HF}{\to }{K}_3\left[Al{F}_6\right]$

Hence, option D is correct.