${\mathrm{K}}_1$ and ${\mathrm{K}}_2$ are the respective equilibrium constant for the reactions
(i) $\mathrm{X}\mathrm{e}{\mathrm{F}}_6\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{X}\mathrm{e}\mathrm{O}{\mathrm{F}}_4\left(\mathrm{g}\right)+2\mathrm{H}\mathrm{F}\left(\mathrm{g}\right);{\mathrm{K}}_1$
(ii) $\mathrm{X}\mathrm{e}{\mathrm{O}}_4\left(\mathrm{g}\right)+\mathrm{X}\mathrm{e}{\mathrm{F}}_6\left(\mathrm{g}\right)\rightleftharpoons \mathrm{X}\mathrm{e}\mathrm{O}{\mathrm{F}}_4\left(\mathrm{g}\right)+\mathrm{X}\mathrm{e}{\mathrm{O}}_3{\mathrm{F}}_2;{\mathrm{K}}_2$
The equilibrium constant for the reaction
$\mathrm{X}\mathrm{e}{\mathrm{O}}_4\left(\mathrm{g}\right)+2\mathrm{H}\mathrm{F}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{X}\mathrm{e}{\mathrm{O}}_3{\mathrm{F}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$ will be

• A

  $\frac{{\mathrm{K}}_1}{{\mathrm{K}}_2^2}$
• B

  ${\mathrm{K}}_1-{\mathrm{K}}_2$
• C

  $\frac{{\mathrm{K}}_1}{{\mathrm{K}}_2}$
• Correct answer
  D

  $\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}$

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