$\text{The product of complete hydrolysis of}{\mathrm{ XeF}}_6\text{in the following reaction is}$
$${\mathrm{XeF}}_6+H_{2}O\to ?+\mathrm{HF}$$

The correct answer is XeO₃.

Let us understand the step-by-step chemical behavior of xenon hexafluoride (${\mathrm{XeF}}_6$) during hydrolysis to see why this is the case.

Xenon hexafluoride reacts with water in stages depending on the amount of water available. This can lead to either partial hydrolysis or complete hydrolysis.

1. Partial Hydrolysis:
When a limited amount of water is present, the reaction proceeds partially, replacing only some of the fluorine atoms with oxygen:
$${\mathrm{XeF}}_6+H_{2}O\to {\mathrm{XeOF}}_4+2\mathrm{HF}$$
With a bit more water, it can hydrolyze further to form xenon dioxide difluoride:
$${\mathrm{XeF}}_6+2H_{2}O\to {\mathrm{XeO}}_2{F}_2+4\mathrm{HF}$$

2. Complete Hydrolysis:
When excess water is available, all six fluorine atoms in ${\mathrm{XeF}}_6$ are replaced by oxygen atoms. Because oxygen is divalent and fluorine is monovalent, three oxygen atoms replace the six fluorine atoms, yielding xenon trioxide (XeO₃) as the final xenon-containing product:
$${\mathrm{XeF}}_6+3H_{2}O\to {\mathrm{XeO}}_3+6\mathrm{HF}$$
Therefore, the compound represented by the "?" in the complete hydrolysis equation is XeO₃.