Disproportionation Reactions
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Direct answer
Chlorine gas in cold dilute sodium hydroxide becomes bleach; in hot concentrated alkali it becomes chlorate. The same element is oxidised and reduced in one stroke — a disproportionation reaction, possible only for an element in an intermediate oxidation state with reachable states on both sides. In the cold: Cl2 + 2NaOH → NaCl + NaClO + H2O, chlorine splitting into −1 and +1. In the hot: 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O, splitting into −1 and +5. The JEE roster continues with hydrogen peroxide (oxygen −1 → −2 and 0), manganate (+6 → +7 and +4), phosphorus in alkali (0 → −3 and +1) and, in organic chemistry, the Cannizzaro reaction of formaldehyde.
What you must remember
- Eligibility test: the element must start intermediate — an element at its highest or lowest state cannot disproportionate; fluorine never does (no positive state).
- Chlorine duo: cold alkali gives Cl^- + ClO^- (hypochlorite, bleaching action); hot alkali gives Cl^- + ClO3^- (chlorate) — temperature decides, a permanent MCQ pair.
- Hydrogen peroxide self-split: 2H2O2 → 2H2O + O2, oxygen at −1 dividing into −2 and 0 — why peroxide bottles fizz on standing.
- Manganate disproportionation: 3MnO4^2- + 4H+ → 2MnO4^- + MnO2 + 2H2O, dark green manganate (+6) yielding purple permanganate (+7) and brown dioxide (+4) except in strongly alkaline solution.
- Phosphorus in alkali: P4 + 3OH^- + 3H2O → PH3 + 3H2PO2^- (0 → −3 and +1), the laboratory source of phosphine.
- Organic case: 2HCHO + concentrated NaOH → CH3OH + HCOONa, carbon 0 → −2 and +2 — Cannizzaro as redox.
- Reverse gear: comproportionation (synproportionation) collapses two states into one, as in BrO3^- + 5Br^- + 6H+ → 3Br2 + 3H2O.
Balancing the classic honestly
Set up chlorine in hot concentrated alkali by electrons. Reduction half: Cl2 + 2e^- → 2Cl^-. Oxidation half (alkaline): Cl2 + 12OH^- → 2ClO3^- + 6H2O + 10e^-. Balance electrons: multiply the reduction by 5, add, cancel to get 3Cl2 + 6OH^- → 5Cl^- + ClO3^- + 3H2O, or with sodium: 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O. The ratio 5:1 is not decoration — it is the electron ledger: each of the two oxidised chlorines (from one Cl2) surrenders five electrons, exactly enough to reduce five chlorines from the other two Cl2 molecules.
Now check by oxidation numbers alone: six chlorine atoms at 0 become five at −1 (gain 5e total) and one at +5 (lose 5e) — the books balance before any hydrogen or oxygen is counted. Two methods, one answer, and either alone is fast enough for the exam.
Where students slip
The vocabulary trip first: every disproportionation is a redox reaction, but not every decomposition is one — CaCO3 → CaO + CO2 involves no change in oxidation state, and options bank on the confusion. Second, students misidentify the element being split: in H2O2 it is oxygen (−1), not hydrogen (already +1). Third, the cold-versus-hot alkali distinction decides product identity, and half-remembered equations produce hypochlorite in a hot-alkali question. JEE Advanced adds the electrochemical view — species with the right potentials (on a Frost diagram, a point above the line joining its neighbours) disproportionate spontaneously, which is why Cu^+ in water barely exists: 2Cu^+ → Cu^2+ + Cu. When a question hands you an unfamiliar species, run the oxidation-number scan first; if the element sits intermediate with both neighbours stable, disproportionation is on the table.
Frequently asked questions
What condition makes disproportionation possible?
The element must begin in an intermediate oxidation state with both a higher and a lower state accessible — terminal states cannot split.
Why does chlorine give different products in cold and hot alkali?
Cold dilute alkali stops the split at hypochlorite (+1); hot concentrated alkali drives further oxidation to chlorate (+5) — the same disproportionation, different depth.
Why can fluorine never disproportionate?
It is the most electronegative element and shows no positive oxidation state, so it has nowhere "up" to go; it can only be reduced from 0 to −1.
Is the Cannizzaro reaction a disproportionation?
Yes — formaldehyde's carbon at 0 divides into methanol carbon (−2) and formate carbon (+2) in concentrated alkali, organic chemistry's cleanest self-redox.
What is comproportionation, with an example?
The reverse process: two different oxidation states of one element converge to a single state, as bromate and bromide give bromine in acid.