How are xenon fluorides $X e F_{2}, X e F_{4}$ and $X e F_{6}$ obtained?

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Solution

Xenon flurides are prepared by direct reaction of xenon and fluorine under different reaction conditions.

(i) $X e F_{2} : X e (g) + F_{2} (g) 673 K; 1 b a r - ----- \to X e F_{2} (s) X e n o n i n e x c e s s F l u o r i n e X e n o n d i f l u o r i d e$

(ii) $X e F_{4} : X e (g) + 2 F_{2} (g) 873 K; 7 b a r - ----- \to X e F_{4} (s) (1 p a r t) (5 p a r t) X e n o n t e t r a f l u o r i d e$

(iii) $X e F_{6} : X e (g) + 3 F_{2} (g) 573 K; 60 - 70 b a r - -------- \to X e F_{6} (s) (1 p a r t) (20 p a r t) X e n o n h e x a f l u o r i d e$

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