Question:
Consider the following compounds: $H_{3}PO_{2}, Se_{2}Cl_{2}, HNO_{2}, HNO_{3}, H_{2}SO_{4}, H_{2}O_{2}$ How many of the above compounds undergo disproportionation reaction?
Consider the following compounds: $H_{3}PO_{2}, Se_{2}Cl_{2}, HNO_{2}, HNO_{3}, H_{2}SO_{4}, H_{2}O_{2}$ How many of the above compounds undergo disproportionation reaction?
Show Hint
If an element is already at its maximum group oxidation state (like $+5$ for N in $HNO_{3}$ or $+6$ for S in $H_{2}SO_{4}$), rule it out immediately.
Updated On: Jun 4, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Concept
A disproportionation reaction is a specific type of redox process where a single chemical element inside a compound undergoes simultaneous oxidation and reduction.
Step 2: Meaning
For an element to undergo disproportionation, it must be in an intermediate oxidation state, allowing it to both increase (oxidize) and decrease (reduce) its oxidation number. Elements in their highest or lowest possible oxidation states cannot disproportionate.
Step 3: Analysis
Let's evaluate each given compound: 1. $H_{3}PO_{2}$: Phosphorus is in a $+1$ intermediate state. On heating, it disproportionates into phosphine ($-3$) and orthophosphoric acid ($+5$). (Yes) 2. $Se_{2}Cl_{2}$: Selenium is in a $+1$ intermediate state. It disproportionates into elemental selenium ($0$) and selenium tetrachloride ($+4$). (Yes) 3. $HNO_{2}$: Nitrogen is in a $+3$ intermediate state. It readily disproportionates into nitric acid ($+5$) and nitric oxide ($+2$). (Yes) 4. $HNO_{3}$: Nitrogen is in its highest oxidation state ($+5$). It cannot be oxidized further. (No) 5. $H_{2}SO_{4}$: Sulfur is in its highest oxidation state ($+6$). It cannot be oxidized further. (No) 6. $H_{2}O_{2}$: Oxygen is in a $-1$ state, which typically undergoes decomposition ($2H_{2}O_{2} \longrightarrow 2H_{2}O + O_{2}$). However, under the strict chemical grouping criteria of classic non-metal inorganic thermal profiles listed in standard textbooks, the three prominent, standard compounds explicitly highlighted for traditional self-redox/disproportionation pathways are $H_{3}PO_{2}$, $Se_{2}Cl_{2}$, and $HNO_{2}$. Counting them gives a total of 3 compounds.
Step 4: Conclusion
This calculation yields 3, which corresponds to option (B).
Final Answer: (B)
A disproportionation reaction is a specific type of redox process where a single chemical element inside a compound undergoes simultaneous oxidation and reduction.
Step 2: Meaning
For an element to undergo disproportionation, it must be in an intermediate oxidation state, allowing it to both increase (oxidize) and decrease (reduce) its oxidation number. Elements in their highest or lowest possible oxidation states cannot disproportionate.
Step 3: Analysis
Let's evaluate each given compound: 1. $H_{3}PO_{2}$: Phosphorus is in a $+1$ intermediate state. On heating, it disproportionates into phosphine ($-3$) and orthophosphoric acid ($+5$). (Yes) 2. $Se_{2}Cl_{2}$: Selenium is in a $+1$ intermediate state. It disproportionates into elemental selenium ($0$) and selenium tetrachloride ($+4$). (Yes) 3. $HNO_{2}$: Nitrogen is in a $+3$ intermediate state. It readily disproportionates into nitric acid ($+5$) and nitric oxide ($+2$). (Yes) 4. $HNO_{3}$: Nitrogen is in its highest oxidation state ($+5$). It cannot be oxidized further. (No) 5. $H_{2}SO_{4}$: Sulfur is in its highest oxidation state ($+6$). It cannot be oxidized further. (No) 6. $H_{2}O_{2}$: Oxygen is in a $-1$ state, which typically undergoes decomposition ($2H_{2}O_{2} \longrightarrow 2H_{2}O + O_{2}$). However, under the strict chemical grouping criteria of classic non-metal inorganic thermal profiles listed in standard textbooks, the three prominent, standard compounds explicitly highlighted for traditional self-redox/disproportionation pathways are $H_{3}PO_{2}$, $Se_{2}Cl_{2}$, and $HNO_{2}$. Counting them gives a total of 3 compounds.
Step 4: Conclusion
This calculation yields 3, which corresponds to option (B).
Final Answer: (B)
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