Types of reactions — combination, decomposition, displacement, double displacement, redox: Worked examples

The scenario matrix

Before solving, let us list every distinct kind of case these reactions can throw at you. Think of each row as a "cell" you must be able to handle. Each worked example below is tagged with the cell it fills.

Cell	Case class	The tricky thing it tests
C1	Combination that IS redox	element + element, oxidation numbers change
C2	Combination that is NOT redox	oxide + water, charges stay put
C3	Decomposition — thermal	heat splits one into many
C4	Decomposition — electrolytic / photo	electricity or light as the energy source
C5	Displacement that works	more reactive kicks out less reactive
C6	Displacement that fails (degenerate)	reactant below in series ⇒ no reaction
C7	Double displacement with a driving force	precipitate / gas / water leaves solution
C8	Double displacement with no driving force (degenerate)	all products soluble ⇒ no reaction
C9	Real-world word problem	translate a story into an equation
C10	Exam twist — pattern overlap	one reaction fits two labels at once

The word "degenerate" here just means the boundary case where nothing happens — the reaction that doesn't go. Those cells (C6, C8) are where most marks are lost, so we give them full examples.

Tools you will reuse: Balancing Chemical Equations, the Reactivity Series of Metals, and Oxidation Number Rules. Keep them open.

C1 — Combination that IS redox

Worked example Hydrogen burns in chlorine

Statement: $H_{2} + C l_{2} \to ?$

Forecast: (guess before reading on) Two elements, nothing else present — what is the only pattern where two things become one? … Combination. And since both start as free elements, expect electrons to move ⇒ redox too.

Identify the pattern. Two separate substances, no third reactant ⇒ this must be $A + B \to A B$ . Why this step? The skeleton is chosen by counting reactants and products — one product means combination.
Write the product. H (wants to share/lose 1 e⁻) with Cl (wants 1 e⁻) ⇒ $H C l$ .
Balance (see Balancing Chemical Equations). $H_{2}$ gives 2 H, $C l_{2}$ gives 2 Cl ⇒ we need 2 molecules of $H C l$ : $H_{2} + C l_{2} \to 2 H C l$ Why this step? One $H C l$ has only 1 H and 1 Cl; the diatomic reactants force the coefficient 2.
Redox check (using Oxidation Number Rules): free elements are $0$ . In $H C l$ , H is $+ 1$ , Cl is $- 1$ . So H: $0 \to + 1$ (oxidised), Cl: $0 \to - 1$ (reduced). Electrons moved ⇒ redox: yes.

Verify: Left = 2 H, 2 Cl. Right = 2 H, 2 Cl. ✔ Atom-balanced. Total oxidation number on right $= (+ 1) + (- 1) = 0$ , matching the neutral molecule. ✔

C2 — Combination that is NOT redox

Worked example Sulfur trioxide meets water

Statement: $S O_{3} + H_{2} O \to ?$ (this is how acid rain forms)

Forecast: An acidic oxide plus water. Two reactants, likely one product ⇒ combination. But will oxidation numbers change? Watch closely.

Pattern: $A + B \to A B$ , combination.
Product: acidic oxide + water → the corresponding acid ⇒ $H_{2} S O_{4}$ . $S O_{3} + H_{2} O \to H_{2} S O_{4}$ Why this step? No atoms are left over: 1 S, 4 O, 2 H on each side.
Redox check: In $S O_{3}$ : O is $- 2$ (×3 = $- 6$ ), so S $= + 6$ . In $H_{2} S O_{4}$ : H $= + 1$ (×2), O $= - 2$ (×4 $= - 8$ ), so S $= + 6$ again. Sulfur stays +6. No element changes ⇒ redox: no.

Verify: Atoms: S 1=1, O 3+1=4, H 2=2. ✔ For $H_{2} S O_{4}$ : $2 (+ 1) + (+ 6) + 4 (- 2) = 2 + 6 - 8 = 0$ ✔ (neutral molecule). Cross-check with the parent's lime example (CaO + H₂O) — same story: combination, not redox.

C3 — Thermal decomposition

C4 — Electrolytic / photo decomposition

Worked example Two energy sources, two decompositions

Statement (a): electrolysis of molten sodium chloride, $N a C l electricity ?$ (see Electrolysis). Statement (b): silver chloride left in sunlight, $A g C l light ?$

Forecast: Both are one compound broken apart, so decomposition — but the sub-type is named by the energy source: electricity vs light.

(a) Products: electricity pumps electrons in, reducing $N a^{+}$ to $N a$ metal and oxidising $C l^{-}$ to $C l_{2}$ gas. $2 N a C l electricity 2 N a + C l_{2} ↑$ Why this step? $C l_{2}$ is diatomic, so we need an even number of Cl ⇒ coefficient 2 on $N a C l$ and $N a$ .
Redox check (a): Na $+ 1 \to 0$ (reduced), Cl $- 1 \to 0$ (oxidised). Redox: yes (electrolytic decompositions always are).
(b) Products: light supplies energy; $A g^{+}$ is reduced to grey metallic $A g$ (this darkening is photographic film's "image"): $2 A g C l light 2 A g + C l_{2} ↑$ Redox check (b): Ag $+ 1 \to 0$ , Cl $- 1 \to 0$ . Redox: yes.

Verify: (a) Na 2=2, Cl 2=2 ✔. (b) Ag 2=2, Cl 2=2 ✔. Both are decomposition (one → two) AND redox — an overlap, previewing C10.

C5 — Displacement that works

The reactivity series is our referee. Look at it as a ladder: the higher metal always wins.

Figure — Types of reactions — combination, decomposition, displacement, double displacement, redox

Worked example Zinc dropped into silver nitrate

Statement: $Z n + A g N O_{3} \to ?$

Forecast: One free element (Zn) + a compound of another metal (Ag). Pattern $A + B C \to A C + B$ . It only "goes" if Zn is above Ag on the ladder.

Check the ladder (Reactivity Series of Metals): Zn sits above Ag ⇒ Zn is more reactive ⇒ it can displace Ag. Why this step? Direction of displacement is decided entirely by who is higher; skip this and you may write an impossible reaction.
Swap partners: Zn takes the nitrate, Ag is set free.
Balance. Zn is $+ 2$ , so it needs two $N O_{3}^{-}$ ; each $A g N O_{3}$ carries one ⇒ two of them: $Z n + 2 A g N O_{3} \to Z n (N O_{3})_{2} + 2 A g$ Why this step? The charge on Zn ( $+ 2$ ) dictates it needs 2 nitrate groups, forcing coefficient 2 on silver.
Redox check: Zn $0 \to + 2$ (oxidised), Ag $+ 1 \to 0$ (reduced). Redox: yes — displacements of a metal from its salt are always redox.

Verify: Zn 1=1, Ag 2=2, N 2=2, O 6=6. ✔ Electrons lost by Zn ( $2$ ) = electrons gained by 2 Ag ( $2 \times 1 = 2$ ). ✔

C6 — Displacement that FAILS (the degenerate case)

Worked example Copper in zinc sulfate — the reaction that isn't

Statement: $C u + Z n S O_{4} \to ?$

Forecast: It looks like displacement (element + salt). But look at the ladder first before you write anything.

Check the ladder: Cu is below Zn ⇒ Cu is less reactive than Zn.
Apply the rule: a less reactive metal cannot displace a more reactive one. Why this step? Cu cannot "want" the sulfate more than Zn does, so no swap occurs. $C u + Z n S O_{4} \to No reaction$
Sanity check by reversal: the reverse, $Z n + C u S O_{4}$ , does go (parent note, iron/copper logic). A reaction that runs one way only tells you the forward direction here is blocked.

[!mistake] Writing products anyway Under pressure students "balance" $C u + Z n S O_{4} \to C u S O_{4} + Z n$ because it looks symmetric. Fix: Always consult the ladder first. If the free metal is lower, the answer is literally the words "no reaction." That is the correct, full-marks answer.

Verify: No products form, so there is nothing to balance — the absence is the answer. Contrast: the working direction $Z n + C u S O_{4} \to Z n S O_{4} + C u$ balances (Zn 1=1, Cu 1=1, S 1=1, O 4=4). ✔

C7 — Double displacement with a driving force

A double displacement only "goes" if a product leaves the solution — as a solid (precipitate), a gas, or water. Picture the ions swapping partners and one pair falling out:

Worked example Lead nitrate meets potassium iodide (the "golden rain")

Statement: $P b (N O_{3})_{2} + K I \to ?$

Forecast: Two dissolved salts ⇒ pattern $A B + C D \to A D + C B$ . It goes only if one new pairing is insoluble. $P b I_{2}$ is a famous bright-yellow precipitate.

Pattern: double displacement (ions exchange).
Swap partners: $P b^{2 +}$ pairs with $I^{-}$ ; $K^{+}$ pairs with $N O_{3}^{-}$ .
Driving force: $P b I_{2}$ is insoluble ⇒ it precipitates (falls out) ⇒ the swap is permanent. Why this step? Without something leaving solution, ions just mingle (that is C8).
Balance. $P b^{2 +}$ needs 2 iodides ⇒ 2 $K I$ ; that gives 2 $K^{+}$ ⇒ 2 $K N O_{3}$ : $P b (N O_{3})_{2} + 2 K I \to P b I_{2} ↓ + 2 K N O_{3}$
Redox check: Pb stays $+ 2$ , I stays $- 1$ , K stays $+ 1$ , $N O_{3}$ intact ⇒ redox: no (double displacements almost never are).

Verify: Pb 1=1, N 2=2, O 6=6, K 2=2, I 2=2. ✔ Charges unchanged ⇒ not redox, consistent with the parent's rule.

C8 — Double displacement with NO driving force (degenerate)

Worked example Sodium chloride meets potassium nitrate

Statement: $N a C l + K N O_{3} \to ?$

Forecast: Two soluble salts. Swap gives $N a N O_{3}$ and $K C l$ . Ask: is any product insoluble, a gas, or water?

Swap on paper: $N a^{+}$ with $N O_{3}^{-}$ → $N a N O_{3}$ ; $K^{+}$ with $C l^{-}$ → $K C l$ .
Test each product: $N a N O_{3}$ soluble, $K C l$ soluble. Nothing leaves the solution. Why this step? With no precipitate, gas, or water, there is no force to pull the swap forward. $N a C l + K N O_{3} \to No reaction$

[!recall]- Why C8 is the mirror of C6 C6 fails because the reactivity ladder blocks it; C8 fails because no product escapes solution. Both are the degenerate "no reaction" cells — different tests, same conclusion. The two questions to ask ::: (1) displacement: is my free element higher on the ladder? (2) double displacement: does any product leave solution?

Verify: All four ions stay dissolved and unchanged ⇒ no net change ⇒ nothing to balance. The correct answer is "no reaction."

C9 — Real-world word problem

Worked example Rust-proofing a boiler with an antacid twist

Statement: A factory washes a boiler with dilute hydrochloric acid; the acid then has to be safely neutralised before disposal using calcium hydroxide (slaked lime). Write the neutralisation and name the pattern.

Forecast: "Neutralise an acid with a base" is a phrase you should translate to acid + base → salt + water, a double displacement (specifically neutralisation, links to Acids, Bases and Salts).

Translate the story: acid $= H C l$ , base $= C a (O H)_{2}$ .
Swap partners: $H^{+}$ joins $O H^{-}$ → water (the driving force — water is very stable, un-ionised); $C a^{2 +}$ joins $C l^{-}$ → the salt $C a C l_{2}$ .
Balance. $C a (O H)_{2}$ has 2 $O H^{-}$ , so we need 2 $H C l$ to supply 2 $H^{+}$ : $2 H C l + C a (O H)_{2} \to C a C l_{2} + 2 H_{2} O$ Why this step? Two $O H^{-}$ demand two $H^{+}$ to make two waters, which fixes the acid coefficient at 2.
Redox check: H $+ 1$ , O $- 2$ , Ca $+ 2$ , Cl $- 1$ throughout ⇒ not redox.

Verify: H: left $2 + 2 = 4$ , right $4$ (in 2 H₂O). ✔ Cl 2=2, Ca 1=1, O 2=2. ✔ Forming water is the driving force ⇒ a genuine double displacement.

C10 — Exam twist: one reaction, two labels

Worked example Rusting-in-fast-forward: iron wool burning in oxygen

Statement: $F e + O_{2} \to ?$ (as in a sparkler). Name every label that applies.

Forecast: Two elements → one product smells like combination. But both start at oxidation number 0, so electrons must move ⇒ also redox. Two labels at once — that is the twist.

Pattern: $A + B \to A B$ , combination.
Product: iron(III) oxide, $F e_{2} O_{3}$ .
Balance (this one needs care — see Balancing Chemical Equations). Try 4 Fe and 3 $O_{2}$ : $4 F e + 3 O_{2} \to 2 F e_{2} O_{3}$ Why this step? 2 $F e_{2} O_{3}$ has 4 Fe and 6 O; to get 6 O from diatomic $O_{2}$ we need 3 molecules ⇒ coefficient 3.
Redox check: Fe $0 \to + 3$ (oxidised, loses 3 e⁻ each), O $0 \to - 2$ (reduced). Redox: yes.
Both labels: combination AND redox. This is exactly the parent note's warning that redox is an overlay, not a sixth shape.

Verify: Fe 4=4, O 6=6. ✔ Electrons: 4 Fe lose $4 \times 3 = 12$ e⁻; 6 O gain $6 \times 2 = 12$ e⁻. Balanced electron transfer ✔ — the deep reason the coefficients come out to 4:3:2.

Every cell filled — the map

Recall Self-test before the exam

Forecast, name the pattern, say redox or not, then balance: $2 K + 2 H_{2} O \to ?$ ::: $2 K O H + H_{2} ↑$ — displacement + redox (K above H in the ladder) $C a C O_{3} Δ ?$ ::: $C a O + C O_{2} ↑$ — thermal decomposition, not redox $B a C l_{2} + N a_{2} S O_{4} \to ?$ ::: $B a S O_{4} ↓ + 2 N a C l$ — double displacement (precipitate), not redox $A g + C u S O_{4} \to ?$ ::: No reaction — Ag is below Cu on the ladder (degenerate C6)