Hydrolysis of XeF₂ gives

The balanced chemical equation for the hydrolysis of 2 moles of ${\mathrm{XeF}}_2$ (xenon difluoride) in the presence of water is:

$2 {\mathrm{XeF}}_2 + 4 {\mathrm{H}}_2\mathrm{O} \to 2 \mathrm{Xe} + 4\mathrm{HF} + {\mathrm{O}}_2$

In this reaction, two molecules of ${\mathrm{XeF}}_2$ react with four molecules of water to produce two atoms of xenon gas (Xe), four molecules of hydrofluoric acid ($\mathrm{HF}$), and one molecule of oxygen gas (${\mathrm{O}}_2$).

The reaction proceeds via a series of steps where ${\mathrm{XeF}}_2$ is first hydrolyzed to form ${\mathrm{XeOF}}_4$ and $\mathrm{HF}$, and${\mathrm{XeOF}}_4$ is further hydrolyzed to produce ${\mathrm{XeO}}_2{\mathrm{F}}_2$ and HF. The ${\mathrm{XeO}}_2{\mathrm{F}}_2$ is then decomposed to form Xe, ${\mathrm{O}}_2$, and $\mathrm{HF}$

Overall, this reaction is a redox reaction where xenon in ${\mathrm{XeF}}_2$ is reduced from +2 to 0, while oxygen in water is oxidized from -2 to 0.