# Disproportionation Reactions > Disproportionation reactions for JEE Chemistry: intermediate oxidation states, chlorine in hot and cold alkali, peroxide, manganate, P4 in NaOH, balancing. - Canonical URL: https://prepelephant.com/topics/jee/chemistry/disproportionation-reactions - Exam / course: JEE · Subject: Chemistry - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Disproportionation Reactions", PrepElephant, https://prepelephant.com/topics/jee/chemistry/disproportionation-reactions ## 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.