Xenon hexafluoride on partial hydrolysis produces compounds 'X' and 'Y'. Compounds 'X' and 'Y' and the oxidation state of Xe are respectively:

We are asked what happens when xenon hexafluoride, $$\mathrm{XeF_6}$$, undergoes partial hydrolysis. “Partial” means that we add water step-by-step, not enough to carry the reaction all the way to the fully hydrolysed oxide $$\mathrm{XeO_3}$$.

First we let one mole of water react with $$\mathrm{XeF_6}$$. The experimentally known equation is

$$\mathrm{XeF_6 + H_2O \;\longrightarrow\; XeOF_4 + 2\,HF}$$

Thus the first product that appears is $$\mathrm{XeOF_4}$$. We call this compound ‘X’.

Now we add a second mole of water. The product $$\mathrm{XeOF_4}$$ reacts further but is still not hydrolysed completely. The next step is

$$\mathrm{XeOF_4 + H_2O \;\longrightarrow\; XeO_2F_2 + 2\,HF}$$

The new product that appears is $$\mathrm{XeO_2F_2}$$. We call this compound ‘Y’.

So, after two successive (but still partial) hydrolysis steps we obtain the pair

$$X = \mathrm{XeOF_4}, \qquad Y = \mathrm{XeO_2F_2}.$$

Next we calculate the oxidation state of xenon in each compound.

For $$\mathrm{XeOF_4}$$ we let the oxidation state of Xe be $$x$$. We use the rule “sum of oxidation numbers = overall charge (which is 0 here)”. Fluorine is always $$-1$$ and oxygen is $$-2$$. Hence

$$x + (-2) + 4(-1) = 0.$$

That is

$$x - 2 - 4 = 0 \;\Longrightarrow\; x - 6 = 0 \;\Longrightarrow\; x = +6.$$

For $$\mathrm{XeO_2F_2}$$ we again put xenon’s oxidation state as $$x$$. Now there are two oxygens and two fluorines, so

$$x + 2(-2) + 2(-1) = 0.$$

Simplifying we get

$$x - 4 - 2 = 0 \;\Longrightarrow\; x - 6 = 0 \;\Longrightarrow\; x = +6.$$

Therefore xenon is present in the $$+6$$ oxidation state in both ‘X’ and ‘Y’.

The pair of compounds and the common oxidation state correspond to Option C: $$\mathrm{XeOF_4(+6)}$$ and $$\mathrm{XeO_2F_2(+6)}.$$

Hence, the correct answer is Option C.