Question:
Out of the following redox reactions,
\[
\text{I. NH}_4\text{NO}_3 \xrightarrow{\Delta} \text{N}_2\text{O} + 2\text{H}_2\text{O}
\]
\[
\text{II. NH}_4\text{NO}_2 \xrightarrow{\Delta} \text{N}_2 + 2\text{H}_2\text{O}
\]
\[
\text{III. PCl}_5 \xrightarrow{\Delta} \text{PCl}_3 + \text{Cl}_2
\]
Disproportionation is not shown in
Out of the following redox reactions,
\[
\text{I. NH}_4\text{NO}_3 \xrightarrow{\Delta} \text{N}_2\text{O} + 2\text{H}_2\text{O}
\]
\[
\text{II. NH}_4\text{NO}_2 \xrightarrow{\Delta} \text{N}_2 + 2\text{H}_2\text{O}
\]
\[
\text{III. PCl}_5 \xrightarrow{\Delta} \text{PCl}_3 + \text{Cl}_2
\]
Disproportionation is not shown in
Show Hint
Disproportionation needs same element both oxidized and reduced.
Updated On: Apr 23, 2026
- I and II
- II and III
- I and III
- I, II and III
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
Disproportionation = same element oxidized and reduced
Step 1: In reaction I, N has different oxidation states initially.
Step 2: In reaction II, N again starts with different oxidation states.
Step 3: In reaction III, P changes but no simultaneous oxidation and reduction of same species.
Step 4: Hence none are true disproportionation reactions.
Conclusion:
Disproportionation not shown in all (I, II, III)
Step 1: In reaction I, N has different oxidation states initially.
Step 2: In reaction II, N again starts with different oxidation states.
Step 3: In reaction III, P changes but no simultaneous oxidation and reduction of same species.
Step 4: Hence none are true disproportionation reactions.
Conclusion:
Disproportionation not shown in all (I, II, III)
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