`XeF_(6)` (s) can be prepared by:

Options

A. Xe:`F_(2)` in 1:20 mole ratio and high pressure (60-70 bar) and temperature 573 K
B. `Xe:F_(2)` in 1:5 mole ratio and pressure (7 bar) and temperature 873 K
C. From product of (b), followed by an interaction with `O_(2)F_(2)` at 143 K
D. Taking Xe in excess at atmospheric pressure (1 bar ) and temperature (673 K)

Answer

A, C

Explanation

`Xe(g)+F_(2)(g) overset(673K,1 " bar") rarr XeF_(2)(s)`
(xenon in excess)
`Xe(g)+2Fe_(2)(g) overset(873K,1 " bar ")rarr XeF_(4)(s)`
(1:5 ratio)
` Xe(g)+3F_(2)(g) overset(573K,60-70" bar") rarr XeF_(6)(s)`
(1:20 ratio)
`XeF_(4)+O_(2)F_(2) rarr XeF_(6)+O_(2)`