Question:
The reaction of chlorine with hot and concentrated NaOH solution gave two chlorine containing products X, Y and \(H_2O\). Oxidation number of chlorine in X and Y is
The reaction of chlorine with hot and concentrated NaOH solution gave two chlorine containing products X, Y and \(H_2O\). Oxidation number of chlorine in X and Y is
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Cold dilute NaOH gives \(Cl^-\) and \(ClO^-\), whereas hot concentrated NaOH gives \(Cl^-\) and \(ClO_3^-\).
Updated On: Jun 18, 2026
- +1, +5
- -1, +1
- -1, +5
- +1, +3
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The Correct Option is C
Solution and Explanation
Concept:
Chlorine undergoes disproportionation with hot concentrated alkali.
Step 1: Write the reaction.
\[ 3Cl_2+6NaOH \rightarrow 5NaCl+NaClO_3+3H_2O \] Products are \[ NaCl \] and \[ NaClO_3 \]
Step 2: Determine oxidation numbers.
In NaCl, \[ Cl=-1 \] In NaClO_3, \[ Cl=+5 \] Therefore, \[ \boxed{(-1,\,+5)} \]
Step 1: Write the reaction.
\[ 3Cl_2+6NaOH \rightarrow 5NaCl+NaClO_3+3H_2O \] Products are \[ NaCl \] and \[ NaClO_3 \]
Step 2: Determine oxidation numbers.
In NaCl, \[ Cl=-1 \] In NaClO_3, \[ Cl=+5 \] Therefore, \[ \boxed{(-1,\,+5)} \]
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