Double displacement reactions

A double displacement reaction (also called a metathesis reaction) is a chemical reaction in which two compounds react by exchanging their ionic partners, producing two new compounds. Nothing is oxidised or reduced in the usual sense here — the radicals simply rearrange.

where: $A$ and $C$ are the positive parts of the two reactant compounds (typically a metal or the ammonium radical), and $B$ and $D$ are the negative parts (acid radicals such as chloride, nitrate or sulphate). Notice the structural signature that distinguishes this from every other reaction type at Class 8: both reactants are compounds — there is no free element anywhere on the reactant side. That single check is enough to rule out simple displacement (which needs one free element) and combination or decomposition (which involve only one compound changing). The classic ICSE example is sodium chloride reacting with silver nitrate:

Sodium (originally paired with chloride) now pairs with nitrate, while silver (originally paired with nitrate) now pairs with chloride, forming an insoluble white precipitate of silver chloride. The downward arrow and the $(s)$ state symbol together flag that this product separates out of the solution rather than staying dissolved.

There is no formula to derive here — double displacement is a pattern-recognition and bookkeeping skill, so the reasoning that makes it "work" is a short checklist you apply to any pair of solutions.

Step 1. Confirm both reactants are compounds. If either side of the reactant arrow has a free, uncombined element, the reaction is not double displacement — it is simple displacement instead. Both $\text{NaCl}$ and $\text{AgNO}_3$ are compounds, so this check passes.

Step 2. Identify the positive and negative parts of each compound. Every compound splits conceptually into a positive part (a metal or ammonium) and a negative part (an acid radical). In $\text{NaCl}$: $\text{Na}^+$ and $\text{Cl}^-$. In $\text{AgNO}_3$: $\text{Ag}^+$ and $\text{NO}_3^-$.

Step 3. Swap the negative partners across the two positive parts, respecting valency. $\text{Na}^+$ (valency 1) takes $\text{NO}_3^-$ (valency 1), giving $\text{NaNO}_3$; $\text{Ag}^+$ (valency 1) takes $\text{Cl}^-$ (valency 1), giving $\text{AgCl}$. When the valencies of the swapped ions differ — as with $\text{Ba}^{2+}$ needing two $\text{Cl}^-$ ions — the new formula must reflect that ratio, or the equation will not balance.

Step 4. Balance atoms, then decide which product is the precipitate. Count each element on both sides; add coefficients only if needed. Then use solubility knowledge (silver chloride and barium sulphate are insoluble; sodium and potassium salts are almost always soluble) to mark the insoluble product with $(s)\downarrow$ and the soluble one with $(aq)$.

Question: A solution of barium chloride is mixed with a solution of sodium sulphate, and a white precipitate forms. Write the balanced equation with state symbols, name the reaction type, and identify the precipitate.

Step 1. Check both reactants are compounds. $\text{BaCl}_2$ and $\text{Na}_2\text{SO}_4$ are both compounds — no free element is present — so this is a candidate for double displacement, not simple displacement.

Step 2. Split each compound into its positive and negative parts. $\text{BaCl}_2$ gives $\text{Ba}^{2+}$ and $\text{Cl}^-$; $\text{Na}_2\text{SO}_4$ gives $\text{Na}^+$ and $\text{SO}_4^{2-}$.

Step 3. Swap partners, respecting valency. Barium (valency 2) pairs with sulphate (valency 2) to give $\text{BaSO}_4$ directly. Sodium (valency 1) pairs with chloride (valency 1), and since there are two chlorides to place, the product is $2\text{NaCl}$.

Step 4. Balance and write the full equation. Left side: 1 Ba, 2 Cl, 2 Na, 1 S, 4 O. Right side: 1 Ba, 1 S, 4 O (in $\text{BaSO}_4$) plus 2 Na, 2 Cl (in $2\text{NaCl}$) — matched exactly, so no further coefficients are needed.

Step 5. Identify the precipitate and name the reaction. Barium sulphate is insoluble in water and separates out as the white precipitate, while sodium chloride remains dissolved. This is a double displacement reaction. Final answer: $\text{BaCl}_2(aq) + \text{Na}_2\text{SO}_4(aq) \rightarrow \text{BaSO}_4(s)\downarrow + 2\text{NaCl}(aq)$, with barium sulphate as the white precipitate.

• Confusing double displacement with simple displacement. If both reactants are compounds, it is double displacement; if one reactant is a free element and the other a compound, it is simple displacement — always check the reactant side first, not the products.
• Writing net-ionic equations. At Class 8, always write the complete formula equation for every reactant and product, including spectator ions such as $\text{Na}^+$ and $\text{NO}_3^-$; net-ionic equations belong to higher classes.
• Forgetting the precipitate marker. When an insoluble solid forms, ICSE convention shows this with the state symbol $(s)$ and usually a downward arrow $\downarrow$.
• Swapping labels without checking valency. $\text{Ba}^{2+}$ needs two $\text{Cl}^-$ ions, not one — blindly exchanging radicals without matching charges gives a wrongly written, unbalanced formula.

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