Chemistry · Redox Equilibrium · NEET
It is a redox reaction where ONE element, sitting in ONE oxidation state, is both oxidised and reduced in the same reaction. Some atoms of that element move to a HIGHER oxidation state (oxidised) and other atoms move to a LOWER oxidation state (reduced). Example: 2Cu+ -> Cu2+ + Cu. Here copper starts as +1. One copper goes up to +2 and the other goes down to 0. The starting element MUST be in a middle oxidation state, because it needs room to go both up and down. This matters for NEET because 'which of these is disproportionation' is asked almost every year.
Step 1: Pick one element. Step 2: Find its oxidation number in the reactant (it must be a single value in ONE species). Step 3: Find its oxidation number in the products. If the SAME element ends up in TWO different products — one higher and one lower than the start — it is disproportionation. Quick test on 3MnO4^2- + 4H+ -> 2MnO4^- + MnO2 + 2H2O: Mn starts at +6 (in MnO4^2-). In products Mn is +7 (MnO4^-) and +4 (MnO2). +6 splits into +7 and +4, so YES, Mn disproportionates. If more than one element changes oxidation state, it is usually NOT a pure disproportionation.
They are exact reverses. Disproportionation: one middle oxidation state SPLITS into a higher and a lower state (1 state -> 2 states). Comproportionation: two different oxidation states of the same element MERGE into one middle state (2 states -> 1 state). Example of comproportionation: 2MnO4^- + 3Mn^2+ + 2H2O -> 5MnO2 + 4H+. Here Mn(+7) and Mn(+2) come together and both become Mn(+4). NEET 2019 used exactly this reaction as a trap answer — students wrongly ticked it as disproportionation. Remember: if two states become one, it is comproportionation, NOT disproportionation.
No. It looks like one because Mn(+7) goes to Mn(+6) in K2MnO4 and Mn(+4) in MnO2. But there is a second element also changing: oxygen goes from -2 to 0 in O2. In a true disproportionation ONLY ONE element changes oxidation state, and that single element must go both up and down. Because oxygen is also oxidised here, this is a thermal DECOMPOSITION (redox), not a disproportionation. NEET 2019 put this as option (c) exactly to catch students who forget to check the oxygen.
Because copper starts in ONE state (+1 in Cu+) and ends in TWO states. One Cu+ loses an electron to become Cu2+ (oxidised, +1 -> +2). The other Cu+ gains an electron to become Cu metal, Cu0 (reduced, +1 -> 0). Same element, one starting state, split into higher and lower. NCERT states Cu+ is unstable in water and disproportionates because the E° value for this is favourable. This is the most common textbook example, so memorise it.
An element can disproportionate ONLY if it is in an INTERMEDIATE (middle) oxidation state, because it needs room to go both up and down. If an element is already in its HIGHEST oxidation state, it cannot be oxidised further, so it cannot disproportionate. Example: in ClO4^-, chlorine is +7 (its highest), so ClO4^- does NOT disproportionate. But ClO^- (Cl is +1) can: 3ClO^- -> 2Cl^- + ClO3^- (Cl goes to -1 and +5). This 'highest-state cannot disproportionate' idea is the next concept — study it right after this.
Which of the following reactions are disproportionation reactions? (a) 2Cu+ -> Cu2+ + Cu (b) 3MnO4^2- + 4H+ -> 2MnO4^- + MnO2 + 2H2O (c) 2KMnO4 -> K2MnO4 + MnO2 + O2 (d) 2MnO4^- + 3Mn^2+ + 2H2O -> 5MnO2 + 4H+ Select the correct option.
For bromine, the standard reduction potentials for each step are shown: BrO4^- --(1.82 V)--> BrO3^- --(1.5 V)--> HBrO --(1.595 V)--> Br2 --(1.0652 V)--> Br^- Then the species undergoing disproportionation is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. Oxidation and reduction both happen, so it is a true redox reaction. The special point is that the SAME element does both jobs at once, instead of one element being oxidised and a different element being reduced.
2Cu+ -> Cu2+ + Cu (copper +1 splits into +2 and 0). Another common one: Cl2 + 2OH^- -> ClO^- + Cl^- + H2O, where chlorine goes from 0 to +1 and -1. This chlorine reaction is used to make household bleach.
Comproportionation is the reverse of disproportionation: two different oxidation states of one element combine into one middle state. Example: 2MnO4^- + 3Mn^2+ + 2H2O -> 5MnO2 + 4H+, where Mn(+7) and Mn(+2) both become Mn(+4).
No. To disproportionate, an element must go both up and down. If it is already at its highest oxidation state (like Cl in ClO4^-, +7), it cannot go higher, so it cannot disproportionate. This is covered in the next concept, 'why some species cannot disproportionate'.
In a normal redox reaction, ONE element is oxidised and a DIFFERENT element is reduced. In disproportionation, the SAME element in the SAME starting oxidation state is both oxidised and reduced.
