${\mathrm{AlF}}_3$ is soluble in $\mathrm{HF}$ only in presence of $\mathrm{KF}$. It is due to the formation of

• A

  ${\mathrm{K}}_3\left[{\mathrm{AlF}}_3{\mathrm{H}}_3\right]$
• Correct answer
  B

  ${\mathrm{K}}_3\left[{\mathrm{AlF}}_6\right]$
• C

  ${\mathrm{AlH}}_3$
• D

  $\mathrm{K}\left[{\mathrm{AlF}}_3\mathrm{H}\right]$

Hence, the correct answer is option: 'B'.