Question:
Ozonolysis of \(ClO_2\) produces oxide of chlorine. The average oxidation state of chlorine in this oxide is ____.
Ozonolysis of \(ClO_2\) produces oxide of chlorine. The average oxidation state of chlorine in this oxide is ____.
Updated On: June 02, 2025
Hide Solution
Verified By Collegedunia
Correct Answer: 6
Solution and Explanation
Step 1: Understanding the Given Reaction
We are given the following reaction:
\( 2 \, \text{ClO}_2 + 2 \, \text{O}_3 \rightarrow \text{Cl}_2\text{O}_6 + 2 \, \text{O}_2 \)
This reaction shows chlorine dioxide (ClO2) reacting with ozone (O3) to give dichlorine hexoxide (Cl2O6) and oxygen gas (O2).
Step 2: Paramagnetism of ClO2
ClO2 contains an odd electron, which makes it paramagnetic. This is because it has an unpaired electron in its molecular orbitals, which causes it to be attracted to magnetic fields.
Step 3: Oxidation States of Chlorine in ClO2 and Cl2O6
To determine the oxidation state of chlorine in the given compounds, we need to consider the oxidation state of oxygen (which is typically -2 in most compounds) and apply the rules for calculating oxidation states.
- In ClO2, oxygen has an oxidation state of -2. Since the compound is neutral overall, we can use the equation to determine the oxidation state of chlorine: \[ x + 2(-2) = 0 \quad \Rightarrow \quad x - 4 = 0 \quad \Rightarrow \quad x = +4 \] So, the oxidation state of chlorine in ClO2 is +4.
- In Cl2O6, there are six oxygens with an oxidation state of -2 each. The total oxidation state for oxygen is: \[ 6(-2) = -12 \] Since the compound is neutral, we can use the following equation to find the oxidation state of chlorine: \[ 2x + (-12) = 0 \quad \Rightarrow \quad 2x = 12 \quad \Rightarrow \quad x = +6 \] Therefore, the oxidation state of chlorine in Cl2O6 is +6.
Step 4: Conclusion
From the above calculations, we can conclude that in Cl2O6, the average oxidation state of chlorine is +6.
We are given the following reaction:
\( 2 \, \text{ClO}_2 + 2 \, \text{O}_3 \rightarrow \text{Cl}_2\text{O}_6 + 2 \, \text{O}_2 \)
This reaction shows chlorine dioxide (ClO2) reacting with ozone (O3) to give dichlorine hexoxide (Cl2O6) and oxygen gas (O2).
Step 2: Paramagnetism of ClO2
ClO2 contains an odd electron, which makes it paramagnetic. This is because it has an unpaired electron in its molecular orbitals, which causes it to be attracted to magnetic fields.
Step 3: Oxidation States of Chlorine in ClO2 and Cl2O6
To determine the oxidation state of chlorine in the given compounds, we need to consider the oxidation state of oxygen (which is typically -2 in most compounds) and apply the rules for calculating oxidation states.
- In ClO2, oxygen has an oxidation state of -2. Since the compound is neutral overall, we can use the equation to determine the oxidation state of chlorine: \[ x + 2(-2) = 0 \quad \Rightarrow \quad x - 4 = 0 \quad \Rightarrow \quad x = +4 \] So, the oxidation state of chlorine in ClO2 is +4.
- In Cl2O6, there are six oxygens with an oxidation state of -2 each. The total oxidation state for oxygen is: \[ 6(-2) = -12 \] Since the compound is neutral, we can use the following equation to find the oxidation state of chlorine: \[ 2x + (-12) = 0 \quad \Rightarrow \quad 2x = 12 \quad \Rightarrow \quad x = +6 \] Therefore, the oxidation state of chlorine in Cl2O6 is +6.
Step 4: Conclusion
From the above calculations, we can conclude that in Cl2O6, the average oxidation state of chlorine is +6.
Was this answer helpful?
0
0
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 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
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
- In which of the following species, the ratio of s-electrons to p-electrons is the same?
- TS EAMCET - 2024
- Chemistry
- p -Block Elements
- $\textbf{Choose the correct statements about the hydrides of group 15 elements.}$
A. The stability of the hydrides decreases in the order \(\text{NH}_3 > \text{PH}_3 > \text{AsH}_3 > \text{SbH}_3 > \text{BiH}_3\)
B. The reducing ability of the hydrides increases in the order \(\text{NH}_3 < \text{PH}_3 < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\)
C. Among the hydrides, \(\text{NH}_3\) is a strong reducing agent while \(\text{BiH}_3\) is a mild reducing agent.
D. The basicity of the hydrides increases in the order \(\text{NH}_3 < \text{PH}_3 < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\)
Choose the most appropriate from the option given below:- JEE Main - 2024
- Chemistry
- p -Block Elements
View More Questions
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 $. Then which one of the following statements is TRUE?
- JEE Advanced - 2025
- Fundamental Theorem of Calculus
- At 25$^\circ$C, the concentration of H$^+$ ions in $ 1.00 \times 10^{-3} \, \text{M} $ aqueous solution of a weak monobasic acid having acid dissociation constant $ K_a = 4.00 \times 10^{-11} $ is $ X \times 10^{-7} \, \text{M} $. The value of $ X $ is ____. {Use: Ionic product of water $ K_w = 1.00 \times 10^{-14} $ at 25$^\circ$C
- JEE Advanced - 2025
- Ionic Equilibrium
- 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 ________.
Two identical concave mirrors each of focal length $ f $ are facing each other as shown. A glass slab of thickness $ t $ and refractive index $ n_0 $ is placed equidistant from both mirrors on the principal axis. A monochromatic point source $ S $ is placed at the center of the slab. For the image to be formed on $ S $ itself, which of the following distances between the two mirrors is/are correct:
- JEE Advanced - 2025
- Ray optics and optical instruments
- Let $$ \alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}. $$ Then the value of $$ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 $$ is \rule{1cm}{0.15mm}.
- JEE Advanced - 2025
- Some Applications of Trigonometry
View More Questions
JEE Advanced Notification
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