$Xe + F_2$ (ratio $1:20$) $\xrightarrow{573 \ K, 60-70 \ bar} A$
What is $A$?

  • A
    $XeF_6$
  • B
    $XeF_5$
  • C
    $XeF_2$
  • D
    $XeF_4$

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Reason : Inert gases have stable configuration.

$Xe_{(g)}$ and $F_{2(g)}$ react in the ratio $1:20$ at $573 \ K$ and $60-70 \ bar$ to form $A$. When $A$ is completely hydrolyzed,$B$ and $HF$ are formed. $A$ and $B$ are respectively:

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