The reaction of $Xe$ and $O _2 F _2$ gives a Xe compound $P$. The number of moles of HF produced by the complete hydrolysis of $1 \,mol$ of $P$ is ______
Correct Answer: 4
Approach Solution - 1
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.
Approach Solution -2
\(\text{Xe} + 2\text{O}_2 + \text{F}_2 \rightarrow \text{XeF}_4 + 2\text{O}_2\)
\(3\text{XeF}_4 + 6\text{H}_2\text{O} \rightarrow 2\text{Xe} + \text{XeO}_3 + \frac{23}{2}\text{O}_2 + 12\text{HF}\)
∴ One mole of \(\text{XeF}_4\) gives 4 moles of HF on hydrolysis.
Top Questions on p -Block Elements
Given below are two statements.
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Chemistry
- p -Block Elements
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
- JEE Main - 2025
- Chemistry
- p -Block Elements
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:- JEE Main - 2025
- Chemistry
- p -Block Elements
- Thionyl chloride on reaction with white phosphorus gives a compound of phosphorus 'C' which on hydrolysis gives an oxo acid 'O'. The correct statements about C and O are
I. Shape of 'C' is pyramidal
II. 'O' is a dibasic acid
III. 'O' is a monobasic acid
IV. 'C' on reaction with acetic acid gives 'O'- TS EAMCET - 2025
- Chemistry
- p -Block Elements
- Consider the following
Assertion (A): phosphorus can form both phosphorus (III) and phosphorus (V) chlorides but nitrogen cannot form nitrogen(V) chloride.
Reason (R): The electronegativity of nitrogen is more than that of phosphorus
The correct answer is- TS EAMCET - 2025
- Chemistry
- p -Block Elements
Questions Asked in JEE Advanced exam
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
- If $$ \alpha = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} x}{2x^2 - 3x + 2} \, dx, $$ then the value of $ \sqrt{7} \tan \left( \frac{2\alpha \sqrt{7}}{\pi} \right) $ is.
(Here, the inverse trigonometric function $ \tan^{-1} x $ assumes values in $ \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) $.)- JEE Advanced - 2025
- Integral Calculus
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 ________.
- JEE Advanced - 2025
- binomial expansion formula
- The total number of real solutions of the equation $$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $$ is
(Here, the inverse trigonometric functions $ \sin^{-1} x $ and $ \tan^{-1} x $ assume values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)- JEE Advanced - 2025
- Inverse Trigonometric Functions
A temperature difference can generate e.m.f. in some materials. Let $ S $ be the e.m.f. produced per unit temperature difference between the ends of a wire, $ \sigma $ the electrical conductivity and $ \kappa $ the thermal conductivity of the material of the wire. Taking $ M, L, T, I $ and $ K $ as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity $ Z = \frac{S^2 \sigma}{\kappa} $ is:
- JEE Advanced - 2025
- Dimensional Analysis
Concepts Used:
P-Block Elements
- P block elements are those in which the last electron enters any of the three p-orbitals of their respective shells. Since a p-subshell has three degenerate p-orbitals each of which can accommodate two electrons, therefore in all there are six groups of p-block elements.
- P block elements are shiny and usually a good conductor of electricity and heat as they have a tendency to lose an electron. You will find some amazing properties of elements in a P-block element like gallium. It’s a metal that can melt in the palm of your hand. Silicon is also one of the most important metalloids of the p-block group as it is an important component of glass.
P block elements consist of:
- Group 13 Elements: Boron family
- Group 14 Elements: Carbon family
- Group 15 Elements: Nitrogen family
- Group 16 Elements: Oxygen family
- Group 17 Elements: Fluorine family
- Group 18 Elements: Neon family