Displacement Reactions

Drop a shiny iron nail into a beaker of blue copper sulphate solution and wait a few minutes. The blue slowly fades to pale green, and a reddish-brown coating creeps over the nail. Nothing was added, nothing was heated — yet a completely new set of substances has appeared. This is a displacement reaction: a more reactive element shoulders its way into a compound and shoves a less reactive element out, taking its place. It is one of the four basic reaction types in the CBSE Class 10 chapter "Chemical Reactions and Equations," and it is also the first place students meet the idea of reactivity — the notion that some elements are chemically "stronger" than others and can dislodge them from their compounds. This article builds the general pattern $A + BC \rightarrow AC + B$ , works through the classic iron-copper-sulphate and zinc-lead-nitrate examples with full state symbols, and sharpens the boundary between displacement and its easily-confused cousin, double displacement.

The core insight

A more reactive element displaces a less reactive element from its compound, following the pattern $A + BC \rightarrow AC + B$ , with reactivity — not any calculation — deciding whether the swap happens at all.

The core idea

A displacement reaction always has the same skeleton: one free element walks in, one free element walks out, and exactly one compound is rearranged.

A + BC \rightarrow AC + B

where: $A$ is a free element (almost always a metal in the CBSE syllabus), $BC$ is a compound — typically a salt dissolved in water — containing element $B$ , $AC$ is the new compound formed once $A$ takes $B$ 's place, and $B$ is the element that has been set free. The reaction proceeds in the direction written only if $A$ is more reactive than $B$ ; if $A$ is less reactive, nothing happens, no matter how long you wait or how concentrated the solution is. Two features make this pattern easy to spot on paper. First, there is exactly one compound among the reactants and exactly one among the products — never two compounds trading partners. Second, exactly one element appears free before the reaction and exactly one element appears free after it, but they are different elements. If you scan an equation and see two compounds reacting to give two new compounds, with no lone element anywhere, that is not displacement — it is double displacement, and the two reaction types are graded very differently in board answers. The everyday, memorable version of this rule is: the more reactive metal always wins. Iron is more reactive than copper, so iron displaces copper from copper sulphate. Zinc is more reactive than lead, so zinc displaces lead from lead nitrate. Zinc is even more reactive than the hydrogen "locked" inside an acid, so zinc displaces hydrogen gas out of dilute sulphuric acid. In every case, the identity of $A$ and $B$ is fixed by the question; your job is to judge who is more reactive, write the swap, and balance the equation.

Figure

A displacement reaction diagram showing an iron nail sitting in blue copper sulphate solution on the left turning into a copper coated nail in pale green iron sulphate solution on the right with an arrow of time between them

Why it works

There is no formula to compute here — the outcome of a displacement reaction is decided entirely by the relative reactivity of the two elements involved, not by any numerical rule. Reactivity measures how strongly an element "wants" to exist in a chemically combined, compound form rather than as a free element; a highly reactive metal like iron would rather be locked into a compound than sit around as bare metal, so it readily takes copper's place in copper sulphate. Copper, being less reactive, is comfortable being displaced and ends up as free metal. If the situation were reversed — a copper strip dipped into iron sulphate solution — nothing would happen, because copper is not reactive enough to force iron out of its compound. Three checks confirm a displacement reaction has actually occurred, and CBSE examiners routinely ask for them. First, a visible sign: a colour change in the solution (blue CuSO₄ fading as it becomes pale green FeSO₄) and/or a new solid deposit forming on the metal that was dipped in (the reddish-brown copper coating). Second, a compositional check: the free element on the reactant side (Fe) must reappear combined in a compound on the product side, while the element that was combined (Cu) must reappear free. Third, a balance check: every atom, including the atoms inside a polyatomic group like SO₄ or NO₃, must appear in equal numbers on both sides — polyatomic groups should be balanced as a single unit wherever possible rather than atom by atom, since they usually pass through unchanged. Because the reasoning rests entirely on "who is more reactive," the full ordered list of metals — the reactivity series — is not required at this stage; only enough of it to compare two named metals at a time. The complete, ordered series (potassium down to gold) and the deeper explanation of why some metals lose electrons more easily than others is developed formally later, in the "Metals and Non-metals" chapter.

Worked example

A strip of zinc metal is dipped into a solution of lead nitrate, Pb(NO₃)₂(aq). Zinc is more reactive than lead. Write the balanced equation with state symbols, name the reaction type, and state what you would observe.

Step 1. Identify the free element and the compound. Zinc, Zn(s), is the free element; lead nitrate, Pb(NO₃)₂(aq), is the compound, with lead, Pb, as the element currently combined inside it.

Step 2. Compare reactivity and predict the swap. Since zinc is more reactive than lead, zinc takes lead's place in the compound, freeing lead as the metal and forming zinc nitrate as the new compound. This matches the general pattern $A + BC \rightarrow AC + B$ with $A = \text{Zn}$ , $B = \text{Pb}$ , $C = (\text{NO}_3)_2$ .

Step 3. Write the skeletal equation. \text{Zn} + \text{Pb(NO}_3)_2 \rightarrow \text{Zn(NO}_3)_2 + \text{Pb}

Step 4. Check the balance, treating NO₃ as one unit. Left side: 1 Zn, 1 Pb, 2 NO₃ groups. Right side: 1 Zn (in Zn(NO₃)₂), 2 NO₃ groups, 1 Pb. Both sides match with every coefficient equal to 1, so no further balancing is needed.

Step 5. Add state symbols and classify. \text{Zn}(s) + \text{Pb(NO}_3)_2(aq) \rightarrow \text{Zn(NO}_3)_2(aq) + \text{Pb}(s) A single free element (Zn) has replaced a single free element (Pb) inside one compound, so this is a displacement reaction, driven by zinc being more reactive than lead.

Step 6. State the observation. A grey deposit of lead metal gradually forms on the surface of the zinc strip as the reaction proceeds, and the zinc strip itself slowly thins as it dissolves into solution as zinc nitrate — the visible evidence that displacement has taken place is the appearance of a new metal deposit, Pb(s), on the zinc strip.

Competitive stretch

The same reasoning extends to non-metal displacement inside acids: dilute sulphuric acid contains "combined" hydrogen, and a metal more reactive than hydrogen can displace it as hydrogen gas, $\text{Zn}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{H}_2(g)$ . Here $B$ in the general pattern is hydrogen rather than a metal — a reminder that "element $B$ " need not be a metal, only whatever is currently combined and less reactive than $A$ .

Common traps

Mixing up displacement with double displacement — displacement always has exactly one free element among reactants and exactly one among products; double displacement has two compounds exchanging partners and no free element anywhere.
Assuming a reaction will happen simply because a metal is dipped into some salt solution — check reactivity first; if the free metal is less reactive than the metal in the compound, there is no reaction at all, not even a slow one.
Balancing SO₄ or NO₃ atom by atom instead of as a single polyatomic unit, which leads to unnecessary errors when the group appears on both sides unchanged.
Describing the observation vaguely ("something changed colour") instead of naming the specific colour change and the specific new deposit, which is what CBSE answer keys reward.

Ask the tutor: want to check whether a particular metal will displace another from its salt or acid — try naming the two metals and ask the AI tutor to reason through the reactivity comparison with you.

How do I know which metal is more reactive without memorising the whole reactivity series?

At this stage, CBSE questions almost always tell you directly or make it obvious from context — for example, stating that "zinc is more reactive than lead," or giving you the classic iron-versus-copper case, where iron displacing copper is a standard fact you are expected to know from the textbook. You only need to compare the two named metals in the question, not rank the entire series. The full ordered reactivity series, and the reasoning behind why one metal is more reactive than another, is built up properly in the "Metals and Non-metals" chapter — use this chapter's pairwise comparisons as a stepping stone toward that fuller picture.

Why does zinc displacing copper from copper sulphate look the same as zinc displacing hydrogen from an acid?

Because both are governed by the exact same rule: a more reactive element (zinc) forces a less reactive element (copper, or the hydrogen locked in the acid) out of its compound and takes its place. The only difference is what gets freed — copper comes out as a solid metal deposit, while hydrogen comes out as a gas that you would see bubbling off. Both fit the pattern $A + BC \rightarrow AC + B$ ; only the identity and physical state of $B$ changes.

If I dip a copper strip into iron sulphate solution, will anything happen?

No, and that "nothing happens" is itself an important, testable fact. Copper is less reactive than iron, so copper cannot displace iron from iron sulphate — the solution stays the same colour, and no deposit forms on the copper strip. Displacement reactions are one-directional: they proceed only when the free element is more reactive than the element already combined in the compound. If a question describes a metal sitting in a salt solution with no visible change, that is usually the examiner testing whether you recognise the metal is less reactive, not a sign you made an error.