To solve this problem, we need to analyze the reaction and the hydrolysis of the given compound to determine the number of moles of HF produced.
1. Analyzing the Reaction:
From the reaction between \( \text{Xe} \) and \( \text{O}_2\text{F}_2 \), a xenon compound \( P \) is formed. The xenon compound formed is likely to be xenon tetrafluoride (XeF₄), as this is a common product in such reactions involving xenon and oxygen difluoride. The reaction can be represented as follows:
\[ \text{Xe} + \text{O}_2\text{F}_2 \rightarrow \text{XeF}_4 + \text{O}_2 \]
2. Hydrolysis of Xenon Tetrafluoride (XeF₄):
When xenon tetrafluoride (\( \text{XeF}_4 \)) undergoes complete hydrolysis, it reacts with water to produce xenon dioxide (XeO₂) and hydrofluoric acid (HF). The reaction for the hydrolysis of 1 mole of xenon tetrafluoride is:
\[ \text{XeF}_4 + 4\text{H}_2\text{O} \rightarrow \text{XeO}_2 + 4\text{HF} \]
3. Conclusion:
For each mole of xenon tetrafluoride (\( \text{XeF}_4 \)) hydrolyzed, 4 moles of hydrofluoric acid (HF) are produced.
Final Answer:
The number of moles of HF produced by the complete hydrolysis of 1 mole of \( P \) (which is \( \text{XeF}_4 \)) is 4.
\(\text{Xe} + 2\text{O}_2 + \text{F}_2 \rightarrow \text{XeF}_4 + 2\text{O}_2\)
\(3\text{XeF}_4 + 6\text{H}_2\text{O} \rightarrow 2\text{Xe} + \text{XeO}_3 + \frac{23}{2}\text{O}_2 + 12\text{HF}\)
∴ One mole of \(\text{XeF}_4\) gives 4 moles of HF on hydrolysis.
Given below are two statements. 
In the light of the above statements, choose the correct answer from the options given below:
Given below are two statements:
Statement I: Nitrogen forms oxides with +1 to +5 oxidation states due to the formation of $\mathrm{p} \pi-\mathrm{p} \pi$ bond with oxygen.
Statement II: Nitrogen does not form halides with +5 oxidation state due to the absence of d-orbital in it.
In the light of the above statements, choose the correct answer from the options given below:
Given below are the pairs of group 13 elements showing their relation in terms of atomic radius. $(\mathrm{B}<\mathrm{Al}),(\mathrm{Al}<\mathrm{Ga}),(\mathrm{Ga}<\mathrm{In})$ and $(\mathrm{In}<\mathrm{Tl})$ Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius $\left(\mathrm{M}^{3+}\right)$ than the other one. The atomic number of the element (X) is
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 