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Write the hydrolysis products of XeF2 XeF4, XeF6.

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Write the hydrolysis products of XeF2 XeF4, XeF6.

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(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

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