Xenon forms different binary compounds with fluorine.

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Solution

Xenon forms three fluorides:
$X e (g) + F_{2} (g) 400^{\circ} C 1 a t m - ------ \to X e F_{2} (s)$ {Xe in excess}
$X e (g) + 2 F_{2} (g) 600^{\circ} C 6 a t m - ------ \to X e F_{4} (s) {X e : F_{2} = 1 : 1.5}$
$X e (g) + 3 F_{2} (g) 300^{\circ} C 50 a t m - ------- \to X e F_{6} (s) {X e : F_{2} = 1 : 20}$

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