Which of the xenon fluorides undergoes disproportionation during the hydrolysis

Options

A. `XeF_(2)`
B. `XeF_(4)`
C. `XeF_(6)`
D. All of these

Answer

B

Explanation

Hydrolysis of `XeF_(4) (O.N." of "Xe" is "(+4))` gives `Xe (O.N" is "O)` and `XeO_(3) (O.N." is "+6)`:
`6XeF_(4)+12H_(2)Orarr4Xe+2XeO_(3)+24HF+3O_(2)`
`XeF_(2)` is hydrolysed to give `Xe, He`, and `O_(2)`
` 2XeF_(2)(s)+2H_(2)O(l)rarr2Xe(g)+4HF(aq.)+O_(2)(g)`
Complete hydrolysis of `XeF_(6)` with water gives `XeO_(3)`:
`6XeF_(4)+12H_(2)O rarr 4He+2XeO_(3) +24HF+3O_(2)`
Partial hydrolysis of `XeF_(6)` gives oxyfluroides `XeOF_(4)` and `XeO_(2)F_(2)`:
`XeF_(6)+H_(2)OrarrXeOF_(4)+2HF`
`XeF_(6)+2H_(2)Orarr XeO_(2)F_(2)+4HF`