The reaction of HClO3 with HCl gives a paramagnetic gas, which upon reaction with O3 produces
The reaction of HClO3 with HCl gives a paramagnetic gas, which upon reaction with O3 produces
- $Cl _2 O$
- $ClO _2$
- $Cl _2 O _6$
- $Cl _2 O _7$
The Correct Option is C
Solution and Explanation
Step 1: Reaction of \( \text{HClO}_3 \) and \( \text{HCl} \)
The reaction of chloric acid (\( \text{HClO}_3 \)) with hydrogen chloride (\( \text{HCl} \)) results in the formation of chlorine dioxide (\( \text{ClO}_2 \)), which is a paramagnetic gas. This is because chlorine dioxide has an unpaired electron in its structure, making it paramagnetic. The reaction can be written as: \[ \text{HClO}_3 + \text{HCl} \rightarrow \text{ClO}_2 + \text{H}_2\text{O} \]
Step 2: Reaction of \( \text{ClO}_2 \) with \( \text{O}_3 \)
When chlorine dioxide (\( \text{ClO}_2 \)) reacts with ozone (\( \text{O}_3 \)), it forms dichlorine hexaoxide (\( \text{Cl}_2\text{O}_6 \)). The reaction can be written as: \[ 2 \text{ClO}_2 + \text{O}_3 \rightarrow \text{Cl}_2\text{O}_6 \] This reaction shows that \( \text{ClO}_2 \) reacts with ozone to form the compound \( \text{Cl}_2\text{O}_6 \).
Step 3: Conclusion
Based on the reactions described above, the correct compound formed after the reaction of the paramagnetic gas with \( \text{O}_3 \) is \( \text{Cl}_2\text{O}_6 \), which corresponds to option C.
The correct option is C: \( \text{Cl}_2\text{O}_6 \).
Top Questions on p -Block Elements
- For \( P \), the incorrect statement is:
- JEE Main - 2026
- Chemistry
- p -Block Elements
- Two elements of p-block can form following halides XF$_3$ and YF$_3$. XF$_3$ can act as Lewis acid while YF$_3$ can act as Lewis base. Then hybridization of ‘X’ and ‘Y’ in XF$_3$ and YF$_3$ is respectively.
- JEE Main - 2026
- Chemistry
- p -Block Elements
- An element of p-block forms a species of type EH$_4^{+}$, which when passed through a basic solution of K$_2$[HgI$_4$], forms a brown ppt. Select the correct option:
- JEE Main - 2026
- Chemistry
- p -Block Elements
- Statement-I: An element ‘X’ of P-block forms a hydride H–X, which has the longest bond length, then element ‘X’ will have the shortest covalent radius.
Statement-II: An element ‘E’ of Group 15 forms hydride EH$_3$, that has least B.P. The maximum covalency of E is 4.
- JEE Main - 2026
- Chemistry
- 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
Questions Asked in JEE Advanced exam
- Let $ x_0 $ be the real number such that $ e^{x_0} + x_0 = 0 $. For a given real number $ \alpha $, define $$ g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} $$ for all real numbers $ x $. Then which one of the following statements is TRUE?
- JEE Advanced - 2025
- Fundamental Theorem of Calculus
- A linear octasaccharide (molar mass = 1024 g mol$^{-1}$) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is 58.26 % (w/w) of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is _____.
Use: Molar mass (in g mol$^{-1}$): ribose = 150, 2-deoxyribose = 134, glucose = 180; Atomic mass (in amu): H = 1, O = 16- JEE Advanced - 2025
- Biomolecules
Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?
- JEE Advanced - 2025
- Conic sections
- Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from 10 mg g$^{-1}$ and 16 mg g$^{-1}$ aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be 4 mg g$^{-1}$ and 10 mg g$^{-1}$, respectively. At this temperature, the concentration (in mg g$^{-1}$) of adsorbed phenol from 20 mg g$^{-1}$ aqueous solution of phenol will be ____. Use: $\log_{10} 2 = 0.3$
- JEE Advanced - 2025
- Adsorption
- At 300 K, an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height (h) of the solution (density = 1.00 g cm$^{-3}$) where h is equal to 2.00 cm. If the concentration of the dilute solution of the macromolecule is 2.00 g dm$^{-3}$, the molar mass of the macromolecule is calculated to be $X \times 10^{4}$ g mol$^{-1}$. The value of $X$ is ____. Use: Universal gas constant (R) = 8.3 J K$^{-1}$ mol$^{-1}$ and acceleration due to gravity (g) = 10 m s$^{-2}\}$
- JEE Advanced - 2025
- Colligative Properties
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