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Which of the xenon fluorides ndergoes disproportionation during the hydrolysisA. `XeF_(2)`B. `XeF_(4)`C. `XeF_(6)`D. All of these |
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Answer» Correct Answer - 2 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` |
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