Q.Identify the correct statements with reference to the given reaction
P4 + 3OH^- + 3H2O → PH3 + 3H2PO2^- (Note: more than one of the given options may be correct.)
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Start your 14-day free trial to unlock the full solution →In this reaction phosphorus simultaneously undergoes both oxidation (to ) and reduction (to ) — a classic disproportionation. Hydrogen remains at +1 throughout, so it experiences neither. Correct: (iii) and (iv).
The heart of this problem is recognizing a disproportionation reaction: a single element in one oxidation state splits into two different oxidation states, acting as both its own oxidising agent and reducing agent.
Phosphorus in elemental sits at oxidation state . The products show phosphorus in two entirely different forms: (phosphine) and (hypophosphite ion). To decide what has happened, we track the oxidation number of phosphorus in each species.
Step-by-step oxidation-state analysis
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Phosphorus in :
Elemental form, so oxidation state = .
-
Phosphorus in :
Hydrogen is +1 (as usual in covalent hydrides with non-metals).
Let P be :
Phosphorus has been reduced from to .
- Phosphorus in : Hydrogen is +1, oxygen is , overall charge is . Let P be :
Phosphorus has been oxidised from to .
- Hydrogen throughout: In : oxygen is , so H is . In : H is . In : H is . In : H is . Hydrogen's oxidation state never changes; it is a spectator in redox terms.
What is happening?
Phosphorus in is simultaneously:
- losing electrons (oxidation) to form with P at +1, and
- gaining electrons (reduction) to form with P at .
This is the hallmark of disproportionation: one element undergoing both oxidation and reduction in the same reaction. …
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