Q.When an oxide of manganese (A) is fused with KOH in the presence of air and dissolved in water, dark green compound (B) is obtained. Compound (B) disproportionates in neutral or acidic medium to give purple compound (C). Compound (C), in alkaline medium, oxidises potassium iodide solution to compound (D) and compound (A) is also formed. Identify compounds A to D and explain the reactions involved.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Disproportionation Reaction
Disproportionation Reactions: The Self-Oxidation-Reduction
The Intuition
Imagine you have a group of friends who are all equally wealthy — each has exactly ₹100. Now suppose one friend decides to give ₹50 to another. After this transaction, one friend has ₹50 (lost money), another has ₹150 (gained money), and the rest are unchanged. Notice something: the same action — transferring money — made one person poorer and another richer.
A disproportionation reaction works on a similar principle, but with electrons instead of money. One atom of an element simultaneously gets oxidised (loses electrons) and reduced (gains electrons). The same element ends up in two different oxidation states — one higher, one lower — starting from a single intermediate oxidation state.
The word "disproportionation" literally means "breaking apart into unequal parts." The original state splits into two different states.
The Precise Definition
A disproportionation reaction is a redox reaction in which a single substance (element or compound) in an intermediate oxidation state is simultaneously oxidised and reduced, producing two different products — one with a higher oxidation state and one with a lower oxidation state.
The general form looks like this:
Element in intermediate state⟶Higher oxidation state+Lower oxidation state
The Key Condition
For disproportionation to occur, the element must be in an intermediate oxidation state — meaning it can both increase and decrease its oxidation number. If the element is already in its highest possible oxidation state, it can only be reduced. If it's in its lowest, it can only be oxidised. No disproportionation possible.
Disproportionation requires the element to have at least three accessible oxidation states: one lower, one intermediate (the starting point), and one higher.
Classic Example: Hydrogen Peroxide
Hydrogen peroxide (H2O2) is the textbook example. Oxygen in H2O2 has an oxidation state of -1. This is intermediate — oxygen can go to 0 (in O2) or to -2 (in H2O).
When H2O2 decomposes:
2H2O2⟶2H2O+O2
Let's track the oxygen:
- In H2O2: oxidation state = -1
- In H2O: oxidation state = -2 (reduction — gained an electron)
- In O2: oxidation state = 0 (oxidation — lost an electron)
The same oxygen atoms (from the same molecule) undergo both oxidation and reduction. That's disproportionation.
Another Common Example: Copper(I) in Solution
Copper(I) ion (Cu+) is unstable in aqueous solution and disproportionates:
2Cu+⟶Cu+Cu2+
- Cu+ (oxidation state +1) is the intermediate
- Cu (oxidation state 0) is the reduced product
- Cu2+ (oxidation state +2) is the oxidised product
A common mistake is to think that a single atom does both oxidation and reduction. In reality, two atoms of the same element are involved — one gets oxidised, the other gets reduced. The reaction requires at least two formula units of the starting substance.
How to Identify a Disproportionation Reaction
- Look for a single reactant that contains an element in an intermediate oxidation state.
- Check the products — the same element must appear in two different oxidation states (one higher, one lower than the starting state). …
Why this formula?
Disproportionation Reaction — Understanding the Why
A disproportionation reaction is a redox reaction where the same element in one oxidation state simultaneously undergoes oxidation (increase in oxidation number) and reduction (decrease in oxidation number).
The key formula that governs whether such a reaction is spontaneous is based on the standard electrode potentials (E∘) of the two half-reactions.
The Core Idea: Why Does Disproportionation Happen?
For an element in an intermediate oxidation state, it can be both oxidised and reduced.
Whether this happens spontaneously depends on the relative ease of these two processes.
Consider an element X in oxidation state +n:
-
Oxidation half-reaction:
X+n→X+(n+1)+e−
(loss of electron, oxidation number increases)
-
Reduction half-reaction:
X+n+e−→X+(n−1)
(gain of electron, oxidation number decreases)
The overall disproportionation reaction is:
2X+n→X+(n+1)+X+(n−1)
The Key Formula: Spontaneity Condition
For a disproportionation reaction to be spontaneous (under standard conditions), the overall cell potential Ecell∘ must be positive.
Derivation:
-
Identify the two half-reactions and their standard reduction potentials (E∘):
-
Reduction half-reaction (the one that gains electrons):
X+n+e−→X+(n−1)
Let its standard reduction potential be Ered∘.
-
Oxidation half-reaction (the one that loses electrons):
X+n→X+(n+1)+e−
This is the reverse of a reduction. So its standard oxidation potential is −Eox∘, where Eox∘ is the standard reduction potential for:
X+(n+1)+e−→X+n
-
-
Overall cell potential is:
Ecell∘=Ereduction half-cell∘−Eoxidation half-cell∘
But careful: The oxidation half-cell is the reverse of a reduction. So we write:
Ecell∘=Ered∘−Eox∘
where:
- Ered∘ = standard reduction potential for X+n→X+(n−1)
- Eox∘ = standard reduction potential for X+(n+1)→X+n
- Spontaneity condition:
Ecell∘>0⇒Ered∘>Eox∘
In words: Disproportionation is spontaneous if the reduction potential for the lower oxidation state is greater than that for the higher oxidation state.
Why This Makes Sense — A Conceptual Explanation
- Ered∘ tells you how easily X+n gets reduced to X+(n−1). …
The manganese story is the classic MnO₂ → K₂MnO₄ → KMnO₄ sequence: pyrolusite fused with KOH in air gives green manganate, which disproportionates to purple permanganate, which then oxidises iodide to iodate in alkali (reforming MnO₂). …
A = MnO2, B = K2MnO4 (dark green potassium manganate), C = KMnO4 (purple potassium permanganate), D = KIO3 (potassium iodate). This is the standard laboratory preparation and chemistry of KMnO4.
Identification and reasoning.
- A = MnO2 (pyrolusite): the oxide of manganese that is fused with KOH in air.
- Fusion with KOH in presence of air (oxidising) converts Mn(+4) to Mn(+6), giving the dark green manganate ion — so B = K2MnO4: 2MnO2+4KOH+O2→2K2MnO4+2H2O
- Manganate disproportionates in neutral/acidic medium (Mn +6 → Mn +7 and Mn +4), giving the purple permanganate — so C = KMnO4: 3MnO42−+4H+→2MnO4−+MnO2+2H2O
- In alkaline medium KMnO4 oxidises iodide I− to iodate IO3−, while Mn(+7) is reduced back to MnO2 (= A) — so D = KIO3: 2KMnO4+KI+H2O→2MnO2+KIO3+2KOH (ionic form: 2MnO4−+I−+H2O→2MnO2+IO3−+2OH−).
[!NOTE]
OR alternative asked:
(a) Permanganate–oxalate in acid: 2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O …
- CBSE 2026Set ANNUAL1 markMCQQ.In the following reaction 4P + 3KOH + 3H2O -> 3KH2PO2 + PH3, which statement is correct?(a) 'P' is oxidised only(b) 'P' is reduced only(c) 'P' is oxidised as well as reduced(d) 'P' is neither oxidised nor reduced
›Reveal solutionSolution
When the same element, starting from a single oxidation state, ends up in both a higher and a lower oxidation state among the products, that is a disproportionation reaction.
Reaction: 4P + 3KOH + 3H2O -> 3KH2PO2 + PH3
Oxidation state of P in elemental phosphorus (P4, written here as P): 0 (element in its standard state).
Oxidation state of P in KH2PO2 (potassium hypophosphite): Using K = +1, H = +1 (bonded to O, standard H), O = -2:
(+1) + 2(+1 for the two H bonded to O) ... more directly: for the hypophosphite ion H2PO2-, charge = -1. With 2 H at +1 and 2 O at -2: 2(+1) + x + 2(-2) = -1 => 2 + x - 4 = -1 => x = +1.
So P is +1 in KH2PO2 -- this is an INCREASE from 0, i.e. P is OXIDISED here.
Oxidation state of P in PH3: H bonded to P (a less electronegative element than H here, since P and H have similar/P slightly higher electronegativity by the modified scale used for this compound) is taken as -1 in this hydride convention: x + 3(-1) = 0 => x = +3?
…
- CBSE 2026Set ANNUAL1 markMCQQ.In the disproportionation reaction 2 Cu⁺ (aq) ⇌ Cu (s) + Cu²⁺ (aq), the cuprous ion, Cu⁺(a) undergoes reduction only(b) undergoes both reduction and oxidation(c) undergoes oxidation only(d) does not undergo redox
›Reveal solutionSolution
Disproportionation means the same species is simultaneously oxidised and reduced; here half the Cu⁺ goes to Cu (reduction) and half to Cu²⁺ (oxidation) — option (B).
Disproportionation is a redox reaction in which an element in one intermediate oxidation state is simultaneously oxidised and reduced.
Track the oxidation number of copper in 2Cu+→Cu+Cu2+:
- One Cu+ (oxidation state +1) → Cu (oxidation state 0): gain of electron = reduction. …
- CBSE 2022Set ANNUAL1 markMCQQ.Which of the following is strongest acid?(a) HClO(b) HClO3(c) HClO2(d) HClO4
›Reveal solutionSolution
For oxoacids of the same halogen, acid strength rises with the number of oxygen atoms / the oxidation state of the central atom. HClO4 (Cl = +7) is the strongest. Option (D).
The chlorine oxoacids and the oxidation state of Cl are:
- HClO (hypochlorous), Cl = +1
- HClO2 (chlorous), Cl = +3
- HClO3 (chloric), Cl = +5
- HClO4 (perchloric), Cl = +7 …
- CBSE 2019Set ANNUAL1 markQ.MnO4⁻ and ________ are formed by the disproportionation of MnO4²⁻ in acidic medium.
›Reveal solutionSolution
In acidic medium, manganate (MnO4^2-) disproportionates into permanganate (MnO4-) and manganese dioxide (MnO2).
Manganate ion, MnO4^2- (Mn in +6 oxidation state), is unstable in acidic solution and undergoes disproportionation - simultaneous oxidation and reduction of the same species. Mn(VI) is oxidised to Mn(VII) (as MnO4-) while another portion is reduced to Mn(IV) (as MnO2).
Balanced equation: …
- CBSE 2019Set ANNUAL1 markQ.Write the reaction of Cl₂ with water.
›Reveal solutionSolution
Chlorine hydrolyses in water to give hydrochloric acid and hypochlorous acid; chlorine's oxidation state changes from 0 to both −1 and +1 (disproportionation).
When chlorine gas is passed into water, it undergoes hydrolysis:
Cl2+H2O→HCl+HOCl
(hydrochloric acid + hypochlorous acid)
…
- CBSE 2018Set ANNUAL1 markQ.Complete the following equation : 2NaOH + Cl₂ → NaCl + .... + H₂O
›Reveal solutionSolution
Cold dilute NaOH with Cl2 gives NaCl and sodium hypochlorite, NaOCl.
Concept. Chlorine undergoes disproportionation with alkalis. With cold and dilute NaOH, chlorine (oxidation state 0) is simultaneously reduced to Cl− (−1) and oxidised to OCl− (+1).
Reaction.
2NaOH+Cl2→NaCl+NaOCl+H2O …
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