Oxides from thermal decomposition

A cement kiln roars at over a thousand degrees, and out of it comes quicklime — calcium oxide — driven off from ordinary limestone by heat alone. No other reactant is added; the heat itself is enough to shake calcium carbonate apart into a solid oxide and a gas that escapes as bubbles or fumes. This single idea, that heating a metal carbonate, nitrate, or sulphate breaks it into a metal oxide plus a gas, is one of the most exam-tested decomposition patterns in ICSE Class 8 Chemistry. It sits inside the Chemical Reactions chapter, and it hides a trap: an oxide can also appear from a completely different reaction — a metal simply combining with oxygen in air — so telling the two routes apart, and balancing each equation correctly with the heat symbol and state symbols, is exactly what this article drills.

The core insight

Heating a metal carbonate, nitrate, or sulphate is a decomposition reaction that breaks one compound into a metal oxide plus a gas, whereas heating a metal element in air is a separate combination reaction that also happens to give an oxide.

The core idea

Thermal decomposition is the breaking down of a single compound into two or more simpler substances by the action of heat alone — no second reactant is needed. When the starting compound is a metal carbonate, nitrate, or sulphate, one of the products is almost always the metal oxide, with a gas escaping alongside it. Three patterns to know cold for Class 8:

Carbonates release carbon dioxide:

\text{CaCO}_3(s) \xrightarrow{\Delta} \text{CaO}(s) + \text{CO}_2(g)

where: $\Delta$ over the arrow means "on heating," $(s)$ marks a solid and $(g)$ a gas.

Nitrates release nitrogen dioxide together with oxygen — two gases, not one:

2\text{Pb(NO}_3)_2(s) \xrightarrow{\Delta} 2\text{PbO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)

Sulphates release sulphur trioxide:

\text{ZnSO}_4(s) \xrightarrow{\Delta} \text{ZnO}(s) + \text{SO}_3(g)

There is also a second, entirely distinct way to end up with an oxide that involves no decomposition at all: direct combination of a metal (or non-metal) element with oxygen in the air on heating, where several substances join into one instead of one substance splitting into several:

2\text{Mg}(s) + \text{O}_2(g) \xrightarrow{\Delta} 2\text{MgO}(s)

So before writing any equation, the first question to ask is: does the reactant list start with one compound (carbonate, nitrate, or sulphate) that is about to split apart, or with an element plus oxygen that is about to join together? The figure below lines up the two routes side by side.

Figure

Two routes to a metal oxide, decomposition of a carbonate nitrate or sulphate splitting into oxide and gas versus combination of a metal and oxygen joining into oxide

Why it works

There is no numeric formula behind this topic — the reasoning is entirely chemical, resting on how heat energy interacts with the bonds inside a compound, and on the bookkeeping rule that atoms are conserved. Heat is a form of energy, and enough of it, delivered as continuous strong heating, can overcome the bonds holding a large ionic compound like a carbonate together. The carbonate ion, $\text{CO}_3^{2-}$ , is not perfectly stable at high temperature: past a certain temperature it splits into an oxide ion, which stays bonded to the metal as the solid oxide, and a molecule of carbon dioxide gas, which has enough kinetic energy to escape as a gas and leave the solid behind. The same idea explains nitrates splitting into an oxide plus $\text{NO}_2$ and $\text{O}_2$ , and sulphates splitting into an oxide plus $\text{SO}_3$ : in every case the metal-oxygen bond in the oxide is the strongest bond in the original compound, so it is the part that survives the heating while the rest breaks off as a gas. Combination is the mirror-image process. Here the reactants are already an element and oxygen gas, sitting apart with no bonds between them; heating supplies the activation energy needed to start new metal-oxygen bonds forming, and once that starts, the metal atoms and oxygen atoms lock together into a single new compound, the oxide, releasing energy as they do. Nothing is left over to escape as a gas because there was only one kind of product being built, not a compound being taken apart. In both routes, the number of atoms of each element on the reactant side must equal the number on the product side — this is why every equation above is balanced with whole-number coefficients before it is accepted, using the hit-and-trial method rather than any calculation.

Worked example

Question: Limestone (calcium carbonate) is strongly heated in a lime kiln, and separately, copper powder is heated in an open dish in air until it turns black. For each case, write the balanced equation, name the type of reaction, and identify the oxide formed.

Step 1. Classify the limestone case. Calcium carbonate is a single compound, a metal carbonate, so heating it follows the carbonate-decomposition pattern: one compound splitting into two simpler substances.

Step 2. Write and check the limestone equation. $\text{CaCO}_3(s) \xrightarrow{\Delta} \text{CaO}(s) + \text{CO}_2(g)$ . Counting atoms: the left side has 1 Ca, 1 C, 3 O; the right side has 1 Ca + 1 O in CaO, plus 1 C + 2 O in $\text{CO}_2$ , giving 3 O in total — balanced on both sides.

Step 3. Classify the copper case. Here the reactant is a metal element (copper) heated in air, not a compound, so this is a combination reaction, not a decomposition, even though an oxide is still the result.

Step 4. Write and check the copper equation. $2\text{Cu}(s) + \text{O}_2(g) \xrightarrow{\Delta} 2\text{CuO}(s)$ . Counting atoms: the left side has 2 Cu and 2 O (from one $\text{O}_2$ molecule); the right side has 2 Cu and 2 O in $2\text{CuO}$ — balanced.

Answer: limestone gives calcium oxide (quicklime), CaO, by decomposition; copper gives copper(II) oxide, CuO, by combination. Both are basic oxides, but they were made by opposite kinds of reaction — one compound splitting apart versus two substances joining together — and that distinction, not just the final product, is what ICSE questions usually test first.

Competitive stretch

If asked to predict the products of heating zinc carbonate, $\text{ZnCO}_3$ , apply the same carbonate pattern directly without needing a new fact: $\text{ZnCO}_3(s) \xrightarrow{\Delta} \text{ZnO}(s) + \text{CO}_2(g)$ , giving zinc oxide and carbon dioxide, balanced exactly as calcium carbonate was.

Common traps

Assuming every oxide comes from decomposition. As the copper example shows, an oxide can equally well form by combination of an element with oxygen; always check whether the starting material is one compound (decomposition) or an element plus oxygen (combination).
Forgetting the gas product. When a carbonate, nitrate, or sulphate decomposes, always name and include the escaping gas ( $\text{CO}_2$ , or $\text{NO}_2$ together with $\text{O}_2$ , or $\text{SO}_3$ ) — omitting it breaks the atom balance.
Mixing up nitrate and carbonate products. Carbonates give only $\text{CO}_2$ ; nitrates give $\text{NO}_2$ and $\text{O}_2$ together — do not swap or merge these.
Skipping the heating symbol or state symbols. ICSE marking expects the $\Delta$ over the arrow to show that heat is required, and the (s)/(g) state symbols to be shown wherever the pattern shows them.

Ask the tutor: Want to practise classifying more oxide-forming equations as decomposition or combination, or see how nitrate decomposition differs for different metals? Ask your AI tutor to walk through a few more examples with you.

Why does heating a carbonate give carbon dioxide but heating a sulphate gives sulphur trioxide instead of carbon dioxide?

Because the gas released is always the part of the original ion that breaks away from the metal-oxygen bond, and that depends on which polyatomic ion you started with. A carbonate ion, $\text{CO}_3^{2-}$ , splits into an oxide ion (which stays with the metal) and $\text{CO}_2$ (which leaves). A sulphate ion, $\text{SO}_4^{2-}$ , splits into an oxide ion and $\text{SO}_3$ instead, because sulphate simply contains sulphur and oxygen, not carbon. The metal oxide product is common to both, but the escaping gas always reflects the non-metal atoms present in the original ion.

Is it possible for a metal oxide to form without any heating at all?

For this Class 8 topic, no — every route covered here specifically requires strong heating, shown by the $\Delta$ symbol, and that is what makes both patterns "thermal" reactions. Some metals do react with oxygen in air slowly at room temperature over long periods, such as iron rusting, but that gradual process is treated separately as corrosion, not as the thermal decomposition or thermal combination reactions tested in this chapter. Whenever you see an oxide-forming equation with $\Delta$ over the arrow in this syllabus, it is one of the two patterns covered here.

How can I quickly tell decomposition and combination apart in an exam without writing the full equation first?

Look only at what is written on the reactant side before you balance anything. If there is a single compound listed — a carbonate, nitrate, or sulphate — heating it can only be decomposition, because decomposition by definition breaks one substance into several. If there are two separate reactants listed, typically a metal element and oxygen gas, heating them can only be combination, because combination by definition joins several substances into one. Making this classification first, before writing or balancing the equation, also tells you in advance whether to expect a gas escaping (decomposition) or not (combination).