Why Classify Redox Reactions
By the time a reaction is labelled "redox," two facts are already settled: one element has been oxidised (its oxidation number has risen) and another has been reduced (its oxidation number has fallen), with the loss and gain of electrons exactly balanced. Classification does not change that underlying electron bookkeeping — it organises redox reactions by the structural pattern through which the oxidation-number changes occur, which is precisely what NEET tests when it asks you to name or sort a reaction.
The NCERT scheme recognises four types. The first three — combination, decomposition and displacement — involve two distinct elements changing in opposite directions. The fourth, disproportionation, is structurally unique: a single element in an intermediate oxidation state is oxidised and reduced at the same time. Keeping this distinction in view is the fastest route through the identification questions that the four types generate.
| Type | Structural skeleton | Elements changing ON |
|---|---|---|
| Combination | $\ce{A + B -> C}$ | Two (one rises, one falls) |
| Decomposition | $\ce{C -> A + B}$ | Two (one rises, one falls) |
| Displacement | $\ce{X + YZ -> XZ + Y}$ | Two (X and Y) |
| Disproportionation | one element → higher + lower state | One element, both ways |
Combination Reactions
A combination (synthesis) reaction follows the skeleton $\ce{A + B -> C}$, in which two or more reactants unite into a single product. NCERT adds a sharp condition for it to count as redox: either A or B, or both, must be in the elemental (free) form. Only an element in its free state carries an oxidation number of zero, so only then is there room for the number to change as the bond forms.
All combustion reactions that consume elemental dioxygen belong here, as do reactions between two elements. The burning of carbon and the oxidation of magnesium are textbook cases; the combustion of methane is a third, where carbon is oxidised even though hydrogen's oxidation number is unchanged.
Combustion of carbon (carbon $0 \to +4$, oxygen $0 \to -2$):
$\ce{\overset{0}{C}(s) + \overset{0}{O_2}(g) -> \overset{+4}{C}\overset{-2}{O_2}(g)}$
Magnesium burning in nitrogen (Mg $0 \to +2$, N $0 \to -3$):
$\ce{3\overset{0}{Mg}(s) + \overset{0}{N_2}(g) -> \overset{+2}{Mg_3}\overset{-3}{N_2}(s)}$
Combustion of methane (C $-4 \to +4$; H stays $+1$, O $0 \to -2$):
$\ce{\overset{-4}{C}H_4(g) + 2\overset{0}{O_2}(g) -> \overset{+4}{C}\overset{-2}{O_2}(g) + 2H_2\overset{-2}{O}(l)}$
Decomposition Reactions
Decomposition is the exact opposite of combination: a single compound breaks down into two or more components, following $\ce{C -> A + B}$. The NCERT redox condition mirrors the combination rule — for the decomposition to be a redox reaction, at least one of the products must be in the elemental state, so that some element drops to or rises from oxidation number zero.
The thermal decomposition of water, of sodium hydride, and of potassium chlorate are all redox. In potassium chlorate, note carefully that potassium's oxidation number does not change — it is chlorine and oxygen that move — exactly as hydrogen's number was untouched in the methane combustion above.
Electrolysis of water (O $-2 \to 0$, H $+1 \to 0$):
$\ce{2H_2\overset{-2}{O}(l) -> 2\overset{0}{H_2}(g) + \overset{0}{O_2}(g)}$
Sodium hydride (H $-1 \to 0$, Na $+1 \to 0$):
$\ce{2\overset{+1}{Na}\overset{-1}{H}(s) -> 2\overset{0}{Na}(s) + \overset{0}{H_2}(g)}$
Potassium chlorate (Cl $+5 \to -1$, O $-2 \to 0$; K unchanged at $+1$):
$\ce{2K\overset{+5}{Cl}\overset{-2}{O_3}(s) -> 2K\overset{-1}{Cl}(s) + 3\overset{0}{O_2}(g)}$
The decisive counter-example is calcium carbonate. It decomposes neatly into calcium oxide and carbon dioxide, but no element changes its oxidation number — calcium stays $+2$, carbon stays $+4$, oxygen stays $-2$ — so this decomposition is not a redox reaction. NCERT highlights it precisely to puncture the assumption that every decomposition is automatically redox.
"All decompositions are redox" — false
A decomposition (or a combination) is redox only when an element appears in the free, elemental state on one side. $\ce{CaCO_3(s) -> CaO(s) + CO_2(g)}$ produces no element in its zero state, so nothing is oxidised or reduced. Similarly, $\ce{BaCl_2 + Na_2SO_4 -> BaSO_4 + 2NaCl}$ is a double-displacement (precipitation) with no oxidation-number change — the trap in the 2024 paper.
Test: assign oxidation numbers on both sides. No change anywhere → not redox, whatever the structural pattern looks like.
Displacement Reactions
In a displacement reaction an ion or atom in a compound is replaced by an ion or atom of another element, following $\ce{X + YZ -> XZ + Y}$. The incoming element X ends up combined and the displaced element Y is set free, so an oxidation-number change is unavoidable. NCERT splits displacement into two families based on the nature of the displaced species: metal displacement and non-metal displacement.
Metal Displacement
A metal in a compound is pushed out into the uncombined state by a more reactive metal. The driving force is the activity (reactivity) series: the displacing metal is a better reducing agent — it loses electrons more readily — so it reduces the ion of the less reactive metal. These reactions underpin metallurgy, where pure metals are won from their ores.
Zinc displacing copper (Zn $0 \to +2$, Cu $+2 \to 0$):
$\ce{\overset{+2}{Cu}SO_4(aq) + \overset{0}{Zn}(s) -> \overset{0}{Cu}(s) + \overset{+2}{Zn}SO_4(aq)}$
Calcium reducing vanadium pentoxide (Ca $0 \to +2$, V $+5 \to 0$):
$\ce{\overset{+5}{V_2}O_5(s) + 5\overset{0}{Ca}(s) -> 2\overset{0}{V}(s) + 5\overset{+2}{Ca}O(s)}$
The thermite (aluminothermic) reaction — aluminium displacing chromium (Al $0 \to +3$, Cr $+3 \to 0$):
$\ce{\overset{+3}{Cr_2}O_3(s) + 2\overset{0}{Al}(s) -> \overset{+3}{Al_2}O_3(s) + 2\overset{0}{Cr}(s)}$
From the comparisons in the chapter, zinc releases electrons to copper and copper to silver, fixing the reducing order $\ce{Zn > Cu > Ag}$. This same competition for electrons is later harnessed in galvanic cells, the bridge to the Electrochemistry chapter.
Non-Metal Displacement
Here the displaced species is a non-metal — most often hydrogen, occasionally oxygen. Very reactive metals such as the alkali metals and the heavier alkaline-earth metals (Ca, Sr, Ba) displace hydrogen even from cold water; less active metals like magnesium and iron need steam; and many metals that resist water still liberate hydrogen from acids.
Sodium with cold water (Na $0 \to +1$, H $+1 \to 0$):
$\ce{2\overset{0}{Na}(s) + 2\overset{+1}{H_2}O(l) -> 2\overset{+1}{Na}OH(aq) + \overset{0}{H_2}(g)}$
Zinc with hydrochloric acid (Zn $0 \to +2$, H $+1 \to 0$):
$\ce{\overset{0}{Zn}(s) + 2H\overset{+1}{Cl}(aq) -> \overset{+2}{Zn}Cl_2(aq) + \overset{0}{H_2}(g)}$
Hydrogen evolution from acid is slowest for the least active metal (Fe) and fastest for the most reactive (Mg); silver and gold do not react even with hydrochloric acid. An activity series for the halogens exists too: oxidising power falls down group 17, $\ce{F2 > Cl2 > Br2 > I2}$, so a higher halogen displaces the halide ion of a lower one.
Chlorine displacing bromide (Cl $0 \to -1$, Br $-1 \to 0$):
$\ce{\overset{0}{Cl_2}(g) + 2\overset{-1}{Br}^{-}(aq) -> 2\overset{-1}{Cl}^{-}(aq) + \overset{0}{Br_2}(l)}$
Chlorine displacing iodide (basis of the laboratory layer test):
$\ce{\overset{0}{Cl_2}(g) + 2\overset{-1}{I}^{-}(aq) -> 2\overset{-1}{Cl}^{-}(aq) + \overset{0}{I_2}(s)}$
Fluorine is so reactive it cannot be used for halogen displacement in water — it attacks water itself: $\ce{2H_2O(l) + 2\overset{0}{F_2}(g) -> 4H\overset{-1}{F}(aq) + \overset{0}{O_2}(g)}$.
Every label above rests on assigning oxidation numbers correctly. If the rules feel shaky, revisit Oxidation Number before tackling identification questions.
Disproportionation Reactions
Disproportionation is a special type of redox reaction in which a single element in one oxidation state is simultaneously oxidised and reduced. The reacting substance contains an element that can exist in at least three oxidation states; it begins in the intermediate state, and both a higher and a lower oxidation state of that same element are formed in the products.
The decomposition of hydrogen peroxide is the familiar entry point. Oxygen sits in its unusual $-1$ state in $\ce{H2O2}$; in the products it both rises to $0$ in $\ce{O2}$ and falls to $-2$ in $\ce{H2O}$ — a single element going both ways.
Hydrogen peroxide (O: $-1 \to 0$ and $-1 \to -2$):
$\ce{2H_2\overset{-1}{O_2}(aq) -> 2H_2\overset{-2}{O}(l) + \overset{0}{O_2}(g)}$
Phosphorus in hot alkali (P: $0 \to -3$ and $0 \to +1$):
$\ce{\overset{0}{P_4}(s) + 3OH^{-}(aq) + 3H_2O(l) -> \overset{-3}{P}H_3(g) + 3H_2\overset{+1}{P}O_2^{-}(aq)}$
Chlorine in cold dilute alkali — household bleach (Cl: $0 \to +1$ and $0 \to -1$):
$\ce{\overset{0}{Cl_2}(g) + 2OH^{-}(aq) -> \overset{+1}{Cl}O^{-}(aq) + \overset{-1}{Cl}^{-}(aq) + H_2O(l)}$
The chlorine reaction produces the hypochlorite ion $\ce{ClO^-}$, the active species in household bleaching agents. Bromine and iodine follow the same trend, but fluorine does not disproportionate: as the most electronegative element it has no positive oxidation state to climb to, so with alkali it gives $\ce{F^-}$ and $\ce{OF2}$ (forcing oxygen, not fluorine, positive): $\ce{2F_2(g) + 2OH^{-}(aq) -> 2F^{-}(aq) + OF_2(g) + H_2O(l)}$.
Disproportionation = same element, both oxidised AND reduced
The single most-tested idea here: in disproportionation, one element in an intermediate state is the oxidising agent and the reducing agent at once. A species can disproportionate only if its element is not already in its highest (or lowest) possible state. That is why $\ce{ClO4^-}$, with Cl at its maximum $+7$, cannot disproportionate, while $\ce{ClO^-}$, $\ce{ClO2^-}$ and $\ce{ClO3^-}$ all can. Likewise fluorine ($-1$, no positive state) never disproportionates.
Quick screen: an element already at its highest or lowest oxidation state CANNOT disproportionate — it can only be reduced or oxidised respectively.
$\ce{ClO4^-}$ (Cl at $+7$, highest) does not disproportionate. The other three do:
$\ce{3\overset{+1}{Cl}O^{-} -> 2\overset{-1}{Cl}^{-} + \overset{+5}{Cl}O_3^{-}}$
$\ce{6\overset{+3}{Cl}O_2^{-} -> 4\overset{+5}{Cl}O_3^{-} + 2\overset{-1}{Cl}^{-}}$
$\ce{4\overset{+5}{Cl}O_3^{-} -> \overset{-1}{Cl}^{-} + 3\overset{+7}{Cl}O_4^{-}}$
Comproportionation — The Reverse
Comproportionation is the exact reverse of disproportionation: two species containing the same element in different oxidation states — one higher, one lower — combine to give a single product in an intermediate state. It is the mirror image of the splitting shown in Figure 1, with the arrows reversed and converging. NCERT does not develop comproportionation as a named fifth category; it is best understood simply as disproportionation run backwards, useful for recognising why some redox reactions merge two oxidation states into one.
Iodate and iodide converging to iodine ($+5$ and $-1$ → intermediate $0$):
$\ce{\overset{+5}{I}O_3^{-} + 5\overset{-1}{I}^{-} + 6H^{+} -> 3\overset{0}{I_2} + 3H_2O}$
Master Table of the Four Types
The table below consolidates the four NCERT categories — definition, a representative reaction with oxidation-number labels, and the oxidation-number changes that make each one redox. This is the highest-yield summary for revision.
| Type | Definition | Example reaction | ON changes |
|---|---|---|---|
| Combination | $\ce{A + B -> C}$; redox if A and/or B is elemental. | $\ce{C + O_2 -> CO_2}$ | C: $0 \to +4$; O: $0 \to -2$ |
| Decomposition | $\ce{C -> A + B}$; redox if a product is elemental. | $\ce{2KClO_3 -> 2KCl + 3O_2}$ | Cl: $+5 \to -1$; O: $-2 \to 0$ (K unchanged) |
| Displacement (metal) | More reactive metal displaces a metal ion. | $\ce{CuSO_4 + Zn -> Cu + ZnSO_4}$ | Zn: $0 \to +2$; Cu: $+2 \to 0$ |
| Displacement (non-metal) | Metal displaces H; higher halogen displaces a halide. | $\ce{Cl_2 + 2Br^- -> 2Cl^- + Br_2}$ | Cl: $0 \to -1$; Br: $-1 \to 0$ |
| Disproportionation | One element in an intermediate state is oxidised and reduced together. | $\ce{2H_2O_2 -> 2H_2O + O_2}$ | O: $-1 \to -2$ and $-1 \to 0$ |
Lock these in before the mock
- Four types: combination, decomposition, displacement (metal + non-metal), disproportionation.
- Combination / decomposition are redox only when an element appears in the free state on one side; $\ce{CaCO3 -> CaO + CO2}$ is the standard non-redox decomposition.
- Metal displacement follows the activity series (e.g. thermite, $\ce{Zn}$ + $\ce{CuSO4}$); non-metal displacement covers H from water/acid and halogen-by-halogen.
- Disproportionation: one element, intermediate state, oxidised and reduced together — H₂O₂, $\ce{P4}$/$\ce{Cl2}$ in alkali. An element at its highest or lowest state cannot disproportionate (so $\ce{ClO4^-}$ and $\ce{F2}$ cannot).
- Comproportionation is disproportionation reversed: two states converge to one intermediate state.