Skip to content
NCERT Exemplar · Q15

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.)

(i) Phosphorus is undergoing reduction only.
(ii) Phosphorus is undergoing oxidation only.
(iii) Phosphorus is undergoing oxidation as well as reduction.
(iv) Hydrogen is undergoing neither oxidation nor reduction.
Punjab PsebMCQ· 1mImportance★★★★★est
71% · 55/78 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

In this reaction phosphorus simultaneously undergoes both oxidation (to HX2POX2X−\ce{H2PO2^-}) and reduction (to PHX3\ce{PH3}) — 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 PX4\ce{P4} sits at oxidation state 00. The products show phosphorus in two entirely different forms: PHX3\ce{PH3} (phosphine) and HX2POX2X−\ce{H2PO2^-} (hypophosphite ion). To decide what has happened, we track the oxidation number of phosphorus in each species.


Step-by-step oxidation-state analysis

  1. Phosphorus in PX4\ce{P4}:

    Elemental form, so oxidation state = 00.

  2. Phosphorus in PHX3\ce{PH3}:

    Hydrogen is +1 (as usual in covalent hydrides with non-metals).

    Let P be xx:

x+3(+1)=0  ⟹  x=−3.x + 3(+1) = 0 \implies x = -3.

Phosphorus has been reduced from 00 to −3-3.

  1. Phosphorus in HX2POX2X−\ce{H2PO2^-}: Hydrogen is +1, oxygen is −2-2, overall charge is −1-1. Let P be yy:

2(+1)+y+2(−2)=−1  ⟹  2+y−4=−1  ⟹  y=+1.2(+1) + y + 2(-2) = -1 \implies 2 + y - 4 = -1 \implies y = +1.

Phosphorus has been oxidised from 00 to +1+1.

  1. Hydrogen throughout: In OHX−\ce{OH^-}: oxygen is −2-2, so H is +1+1. In HX2O\ce{H2O}: H is +1+1. In PHX3\ce{PH3}: H is +1+1. In HX2POX2X−\ce{H2PO2^-}: H is +1+1. Hydrogen's oxidation state never changes; it is a spectator in redox terms.

What is happening?

Phosphorus in PX4\ce{P4} is simultaneously:

  • losing electrons (oxidation) to form HX2POX2X−\ce{H2PO2^-} with P at +1, and
  • gaining electrons (reduction) to form PHX3\ce{PH3} with P at −3-3.

This is the hallmark of disproportionation: one element undergoing both oxidation and reduction in the same reaction. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.