Question

Which of the following is an uncommon hydrolysis product of XeF2 and XeF4?​

Answer 1

Explanation:

(i) 2XeF2 (s) + 2H2O(l) → 2Xe (g) + 4 HF(aq) + O2(g)

(ii) 6XeF4 + 12 H2O → 4Xe + 2XeO3 + 24 HF + 3O2

(iii) XeF6 + 3 H2O → XeO3 + 6 HF Partial hydrolysis of XeF6 gives oxyfluorides,

XeF6 + H2O → XeOF4 + 2HF XeF6 + 2 H2O → XeO2F2 + 4HF

Answer 2

Answer:

The answer is $XeO_{3}$.

Explanation:

Hydrolysis of $XeF_{2}$ :

$2XeF_{2} +2H_{2} O$ → $2Xe + O_{2} +4HF$

Hydrolysis of $XeF_{4}$ :

$2XeF_{4} +12H_{2} O$ → $4Xe + 2XeO_{3} +24HF+3O_{2}$

The product which is uncommon in the above mentioned hydrolysis is $XeO_{3}$.

Hydrolysis : Hydrolysis is a typical type of chemical reaction in which water is used to break down chemical bonds between two or more substances.

The word hydrolysis comes from the Greek words 'hydro', which means water, and 'lysis', which means to break or unbind.