The reaction of $HClO _3$ with $HCl$ gives a paramagnetic gas, which upon reaction with $O _3$ produces
- $Cl _2 O$
- $Cl _2 O$
- $Cl _2 O _6$
- $Cl _2 O _7$
The Correct Option is C
Approach Solution - 1
The correct option is (C): \(Cl _2 O _6\)
Approach Solution -2
When chloric acid (\(HClO_3\)) reacts with hydrochloric acid (\(HCl\)), it forms chlorine dioxide (\(ClO_2\)), which is a paramagnetic gas. The balanced reaction is:
\[ HClO_3 + HCl \rightarrow ClO_2 + Cl_2 + H_2O \]
Chlorine dioxide (\(ClO_2\)) can react further with ozone (\(O_3\)). The reaction of \(ClO_2\) with \(O_3\) can produce dichlorine hexoxide (\(Cl_2O_6\)).
The balanced chemical equation for the reaction of chlorine dioxide with ozone is:
\[ 2ClO_2 + 2O_3 \rightarrow Cl_2O_6 + 2O_2 \]
So, the paramagnetic gas \(ClO_2\) reacts with ozone to produce dichlorine hexoxide (\(Cl_2O_6\)).
The correct Answer is option is (C):\(Cl_2O_6\)
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
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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
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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
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- $\textbf{Choose the correct statements about the hydrides of group 15 elements.}$
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
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- $\textbf{Reduction potential of ions are given below:}$
\[ \begin{array}{ccc} \text{ClO}_4^- & \text{IO}_4^- & \text{BrO}_4^- \\ E^\circ = 1.19 \, \text{V} & E^\circ = 1.65 \, \text{V} & E^\circ = 1.74 \, \text{V} \\ \end{array} \]
The correct order of their oxidising power is:- JEE Main - 2024
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- p -Block Elements
Questions Asked in JEE Advanced exam
- A region in the form of an equilateral triangle (in x-y plane) of height L has a uniform magnetic field 𝐵⃗ pointing in the +z-direction. A conducting loop PQR, in the form of an equilateral triangle of the same height 𝐿, is placed in the x-y plane with its vertex P at x = 0 in the orientation shown in the figure. At 𝑡 = 0, the loop starts entering the region of the magnetic field with a uniform velocity 𝑣 along the +x-direction. The plane of the loop and its orientation remain unchanged throughout its motion.
Which of the following graph best depicts the variation of the induced emf (E) in the loop as a function of the distance (𝑥) starting from 𝑥 = 0? - A positive, singly ionized atom of mass number \(A_M\) is accelerated from rest by the voltage \(192 V\). Thereafter, it enters a rectangular region of width 𝑤 with magnetic field \(\bar{B_0} = 0.1𝑘̂\)Tesla, as shown in the figure. The ion finally hits a detector at the distance 𝑥 below its starting trajectory. [Given: Mass of neutron/proton =\((5/3) × 10^{−27}\) kg, charge of the electron \(= 1.6 × 10^{−19} C.\)]
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- An organic compound \(P\) with molecular formula \(C_6H_6O_3\) gives a ferric chloride test and does not have an intramolecular hydrogen bond. The compound \(P\) reacts with 3 equivalents of \(NH_2OH\) to produce oxime \(Q\). Treatment of \(P\) with excess methyl iodide in the presence of \(KOH\) produces compound \(R\) as the major product. Reaction of \(R\) with excess iso-butylmagnesium bromide followed by treatment with \(H_3O^+\) gives compound \(S\) as the major product. The total number of methyl (\(-CH_3\)) groups in compound \(S\) is \_\_\_\_\_.
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- An organic compound P having molecular formula C6H6O3 gives ferric chloride test and does not have intramolecular hydrogen bond. The compound P reacts with 3 equivalents of NH2OH to produce oxime Q. Treatment of P with excess methyl iodide in the presence of KOH produces compound R as the major product. Reaction of R with excess iso-butylmagnesium bromide followed by treatment with H3O+ gives compound S as the major product.
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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