XeF$$_6$$ on partial hydrolysis with water, produces a compound X. The same compound X is formed when XeF$$_6$$ reacts with silica. The compound X is:

We start with the given xenon compound $$XeF_6$$. Whenever a fluoroxenon compound meets water, step-wise or “partial” hydrolysis can occur. The general idea is that every mole of water replaces two fluorine atoms by one oxygen atom, liberating two moles of $$HF$$ each time.

For the very first, most limited, stage of hydrolysis the reaction is

$$XeF_6 + H_2O \;\rightarrow\; XeOF_4 + 2\,HF$$

Thus, after the loss of only two fluorine atoms, the product obtained is $$XeOF_4$$. This compound still contains both xenon-oxygen and xenon-fluorine bonds, so we call it a mixed oxide-fluoride.

If more water were supplied, subsequent steps would be

$$XeF_6 + 2\,H_2O \;\rightarrow\; XeO_2F_2 + 4\,HF$$

$$XeF_6 + 3\,H_2O \;\rightarrow\; XeO_3 + 6\,HF$$

However, the question explicitly says “partial hydrolysis,” which stops after the very first step, giving $$XeOF_4$$.

Next, we examine the reaction with silica ($$SiO_2$$). Fluoride ions have a strong affinity for silicon; therefore $$XeF_6$$ reacts with silica according to

$$XeF_6 + SiO_2 \;\rightarrow\; XeOF_4 + SiF_4$$

We see that the very same product, $$XeOF_4$$, appears here as well. Because both routes—partial hydrolysis with water and reaction with silica—furnish the identical species, that species must be the compound X mentioned in the problem.

Looking at the options:

A. $$XeF_4$$     B. $$XeF_2$$     C. $$XeO_3$$     D. $$XeOF_4$$

The only choice matching our deduction is $$XeOF_4$$.

Hence, the correct answer is Option D.