Question

How are xenon fluorides XeF2 ,XeF4, XeF6 obtained?

Answer 1

Answer:

They are formed by the direct reaction of elements under appropriate experimental conditions.

XeF2

xe (g) + F2(g) ............... XeF2 (s)

(xe excess) 673k, 1 bar

XeF4

xe (g) + 2F2(g) .................. XeF4(s)

(1:5 ratio) 873k,7bar

XeF6

xe (g) + 3F2(g) .................... XeF6(s)

(1:20 ratio) 573k, 60-70 bar

XeF6 can also be prepared by the interaction of XeF4 and O2F2 at 143k.

XeF4 + O2F2 ............... XeF6 + O2

Explanation:

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Answer 2

Answer:

Explanation:

The preparation of xenon fluorides are given below:

Xe(g,excess)+F2​(g)673 K1 bar​XeF2​(s)

Xe(g,1mole)+2F2​(g,5 moles)873 K7 bar​XeF4​(s)

Xe(g,1 mole)+3F2​(g,20 moles)60−70 bar​XeF6​(s)