Given below are two statements:
Statement I: \( \text{S}_8 \) solid undergoes disproportionation reaction under alkaline conditions to form \( \text{S}^{2-} \) and \( \text{S}_2\text{O}_3^{2-} \).
Statement II: \( \text{ClO}_4^- \) can undergo disproportionation reaction under acidic condition.
In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I: \( \text{S}_8 \) solid undergoes disproportionation reaction under alkaline conditions to form \( \text{S}^{2-} \) and \( \text{S}_2\text{O}_3^{2-} \).
Statement II: \( \text{ClO}_4^- \) can undergo disproportionation reaction under acidic condition.
In the light of the above statements, choose the most appropriate answer from the options given below:
- Statement I is correct but statement II is incorrect.
- Statement I is incorrect but statement II is correct
- Both statement I and statement II are incorrect
- Both statement I and statement II are correct
The Correct Option is A
Approach Solution - 1
To evaluate the given statements and determine which is correct, let's analyze each statement based on known chemical reactions and principles.
- Statement I: \(S_8\) solid undergoes disproportionation reaction under alkaline conditions to form \(S^{2-}\) and \(S_2O_3^{2-}\).
- Disproportionation is a specific type of redox (reduction-oxidation) reaction in which a single substance is simultaneously reduced and oxidized to form two different products.
- In alkaline conditions, elemental sulfur \((S_8)\) can actually disproportionate. The reaction is:
\(3S_8 + 6OH^- \rightarrow 2S_2O_3^{2-} + 4S^{2-} + 3H_2O\)
- This confirms that under alkaline conditions, sulfur can indeed form \(S^{2-}\) (sulfide) and \(S_2O_3^{2-}\) (thiosulfate).
- Statement II: \(ClO_4^-\\) can undergo disproportionation reaction under acidic conditions.
- Perchlorate ion \((ClO_4^-)\) is known for being very stable, especially under acidic conditions. It is already in its highest oxidation state, where chlorine is +7.
- Generally, in a disproportionation reaction, a substance must be capable of existing in a higher as well as a lower oxidation state. However, chlorine in \(ClO_4^-\) cannot oxidize further (higher than +7 is unfeasible under normal conditions).
- Therefore, no disproportionation reaction of \(ClO_4^-\) occurs under acidic conditions, confirming that Statement II is incorrect.
Based on the analysis above:
- Statement I is correct as sulfur in \(S_8\) can undergo disproportionation to form \(S^{2-}\) and \(S_2O_3^{2-}\) under alkaline conditions.
- Statement II is incorrect as \(ClO_4^-\\) does not undergo disproportionation under acidic conditions.
Therefore, the correct answer is: Statement I is correct but statement II is incorrect.
Approach Solution -2
To determine which of the given statements are correct, let's analyze each statement individually and understand the chemical processes involved.
Statement I: \( \text{S}_8 \) solid undergoes disproportionation reaction under alkaline conditions to form \( \text{S}^{2-} \) and \( \text{S}_2\text{O}_3^{2-} \).
Disproportionation reactions involve a substance being simultaneously oxidized and reduced. The reaction for sulfur in \( \text{S}_8 \) under alkaline conditions can be represented as follows:
\[ \text{3S}_8 + \text{12OH}^- \rightarrow \text{4S}_2\text{O}_3^{2-} + \text{2S}^{2-} + \text{6H}_2\text{O} \]
In this reaction:
- Sulfur is reduced from \( \text{S}_8 \) (0 oxidation state) to \( \text{S}^{2-} \) (-2 oxidation state).
- Sulfur is also oxidized from \( \text{S}_8 \) (0 oxidation state) to \( \text{S}_2\text{O}_3^{2-} \) (+2.5 oxidation state).
This confirms that \( \text{S}_8 \) undergoes a disproportionation reaction under alkaline conditions. Hence, Statement I is correct.
Statement II: \( \text{ClO}_4^- \) can undergo disproportionation reaction under acidic condition.
\( \text{ClO}_4^- \), or perchlorate ion, is already in its highest oxidation state (+7 for chlorine). In a disproportionation reaction, an element must be able to undergo both oxidation and reduction. Since chlorine in \( \text{ClO}_4^- \) is in its highest oxidation state, it cannot be oxidized further; hence, it cannot participate in a disproportionation reaction under normal conditions, regardless of the acidity of the medium.
Therefore, Statement II is incorrect.
Conclusion:
Based on the analysis:
- Statement I is correct because \( \text{S}_8 \) indeed undergoes disproportionation under alkaline conditions to form \( \text{S}^{2-} \) and \( \text{S}_2\text{O}_3^{2-} \).
- Statement II is incorrect as \( \text{ClO}_4^- \) cannot undergo disproportionation under any condition.
Thus, the correct answer is: Statement I is correct but statement II is incorrect.
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Redox reactions Questions
- \[ 2 \text{MnO}_4^{-} + b \text{I}^{-} + c \text{H}_2\text{O} \rightarrow x \text{I}_2 + y \text{MnO}_2 + z \text{OH}^{-} \]
If the above equation is balanced with integer coefficients, the value of \( z \) is \( \_\_\_\_\_\_ \). - 1 mole of PbS is oxidised by “X” moles of O3>to get “Y” moles of O2. X + Y = __________
- Chlorine undergoes disproportionation in alkaline medium as shown below:
\(a \, \text{Cl}_2(g) + b \, \text{OH}^-(\text{aq}) \rightarrow c \, \text{ClO}^-(\text{aq}) + d \, \text{Cl}^-(\text{aq}) + e \, \text{H}_2\text{O}(\text{l})\)
The values of \( a \), \( b \), \( c \), and \( d \) in a balanced redox reaction are respectively: - In alkaline medium, \( \text{MnO}_4^- \) oxidizes \( \text{I}^- \) to:
- In acidic medium, \( K_2Cr_2O_7 \) shows oxidizing action as represented in the half-reaction: \(Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O\)
\( X \), \( Y \), \( Z \), and \( A \) are respectively:
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,