← Chemical Atlas

Reaction families

How chemists sort reactions by their pattern. Every example below is balanced and classified by the engine — the producer refuses to file a reaction under a family it does not actually belong to — and the net-ionic view strips spectators to the essential change. Theredox tag is machine-derived: an element that appears free on one side and combined on the other changed oxidation state, so electrons moved.

Combustion

rule-sourced (openstax-chemistry-2e)redox
fuel+O2→CO2+H2O\mathrm{fuel} + \mathrm{O_{2}} \rightarrow \mathrm{CO_{2}} + \mathrm{H_{2}O}

A fuel made of carbon and hydrogen burns in oxygen. Complete combustion sends every carbon to carbon dioxide and every hydrogen to water. Because a free element (O₂) becomes combined, combustion is always a redox reaction.

  • Needs O₂ as a reactant and enough of it — incomplete combustion (too little O₂) also makes carbon monoxide or soot, which this complete-combustion form does not cover.
  • The fuel here is a hydrocarbon (only C and H, optionally O); other fuels burn to other oxides.

Verified examples

CH4 (g)+2 O2 (g)→CO2 (g)+2 H2O (g)\mathrm{CH_{4}}\,\text{(g)} + 2\,\mathrm{O_{2}}\,\text{(g)} \rightarrow \mathrm{CO_{2}}\,\text{(g)} + 2\,\mathrm{H_{2}O}\,\text{(g)}

CH₄ burns in O₂, giving only CO₂ and H₂O — complete combustion.

redoxO appears as a free element on one side and combined on the other, so its oxidation state changed — electrons were transferred (a redox reaction). Assigning full oxidation numbers is a later topic; here the free-element signature is enough.

C3H8 (g)+5 O2 (g)→3 CO2 (g)+4 H2O (g)\mathrm{C_{3}H_{8}}\,\text{(g)} + 5\,\mathrm{O_{2}}\,\text{(g)} \rightarrow 3\,\mathrm{CO_{2}}\,\text{(g)} + 4\,\mathrm{H_{2}O}\,\text{(g)}

C₃H₈ burns in O₂, giving only CO₂ and H₂O — complete combustion.

redoxO appears as a free element on one side and combined on the other, so its oxidation state changed — electrons were transferred (a redox reaction). Assigning full oxidation numbers is a later topic; here the free-element signature is enough.

2 C2H6 (g)+7 O2 (g)→4 CO2 (g)+6 H2O (g)2\,\mathrm{C_{2}H_{6}}\,\text{(g)} + 7\,\mathrm{O_{2}}\,\text{(g)} \rightarrow 4\,\mathrm{CO_{2}}\,\text{(g)} + 6\,\mathrm{H_{2}O}\,\text{(g)}

C₂H₆ burns in O₂, giving only CO₂ and H₂O — complete combustion.

redoxO appears as a free element on one side and combined on the other, so its oxidation state changed — electrons were transferred (a redox reaction). Assigning full oxidation numbers is a later topic; here the free-element signature is enough.

Watch out for

  • Combustion just makes a compound disappear into smoke. Nothing disappears — every atom is conserved. Count them: the carbon leaves as CO₂ and the hydrogen as H₂O, both weighed in the balanced equation.
  • You can balance the oxygen by writing CO instead of CO₂. CO (carbon monoxide) is a different substance from CO₂. Balance with coefficients out front, never by changing a formula's subscripts.

Synthesis (combination)

rule-sourced (openstax-chemistry-2e)redox
A+B→AB\mathrm{A} + \mathrm{B} \rightarrow \mathrm{AB}

Two or more reactants combine into a single product. When the reactants are free elements, their atoms go from oxidation state 0 to combined, so element-combining synthesis is a redox reaction.

  • The signature is one product from two or more reactants.
  • Combining two elements is always redox; combining two compounds (e.g. an oxide with water) need not be.

Verified examples

N2 (g)+3 H2 (g)→2 NH3 (g)\mathrm{N_{2}}\,\text{(g)} + 3\,\mathrm{H_{2}}\,\text{(g)} \rightarrow 2\,\mathrm{NH_{3}}\,\text{(g)}

N₂ + H₂ combine into the single product NH₃.

redoxH, N appears as a free element on one side and combined on the other, so its oxidation state changed — electrons were transferred (a redox reaction). Assigning full oxidation numbers is a later topic; here the free-element signature is enough.

2 Na (s)+Cl2 (g)→2 NaCl (s)2\,\mathrm{Na}\,\text{(s)} + \mathrm{Cl_{2}}\,\text{(g)} \rightarrow 2\,\mathrm{NaCl}\,\text{(s)}

Na + Cl₂ combine into the single product NaCl.

redoxCl, Na appears as a free element on one side and combined on the other, so its oxidation state changed — electrons were transferred (a redox reaction). Assigning full oxidation numbers is a later topic; here the free-element signature is enough.

2 Mg (s)+O2 (g)→2 MgO (s)2\,\mathrm{Mg}\,\text{(s)} + \mathrm{O_{2}}\,\text{(g)} \rightarrow 2\,\mathrm{MgO}\,\text{(s)}

Mg + O₂ combine into the single product MgO.

redoxMg, O appears as a free element on one side and combined on the other, so its oxidation state changed — electrons were transferred (a redox reaction). Assigning full oxidation numbers is a later topic; here the free-element signature is enough.

Watch out for

  • Any reaction with two reactants is a synthesis. Synthesis is defined by the products, not the reactants: it must end in a single product. Two reactants giving two products is a replacement, not a synthesis.

Decomposition

rule-sourced (openstax-chemistry-2e)redox varies
AB→A+B\mathrm{AB} \rightarrow \mathrm{A} + \mathrm{B}

A single compound breaks apart into two or more products — the reverse pattern of synthesis. Whether it is a redox reaction depends on the reaction: releasing a free element (like O₂) is redox; splitting into two compounds (a carbonate into an oxide and CO₂) is not.

  • The signature is one reactant giving two or more products.
  • An energy input (heat, light, or electricity) usually drives it — but this Atlas describes the change, not how to carry it out.

Verified examples

2 KClO3 (s)→2 KCl (s)+3 O2 (g)2\,\mathrm{KClO_{3}}\,\text{(s)} \rightarrow 2\,\mathrm{KCl}\,\text{(s)} + 3\,\mathrm{O_{2}}\,\text{(g)}

The single compound KClO₃ breaks apart into 2 products.

redoxO appears as a free element on one side and combined on the other, so its oxidation state changed — electrons were transferred (a redox reaction). Assigning full oxidation numbers is a later topic; here the free-element signature is enough.

CaCO3 (s)→CaO (s)+CO2 (g)\mathrm{CaCO_{3}}\,\text{(s)} \rightarrow \mathrm{CaO}\,\text{(s)} + \mathrm{CO_{2}}\,\text{(g)}

The single compound CaCO₃ breaks apart into 2 products.

2 H2O (l)→2 H2 (g)+O2 (g)2\,\mathrm{H_{2}O}\,\text{(l)} \rightarrow 2\,\mathrm{H_{2}}\,\text{(g)} + \mathrm{O_{2}}\,\text{(g)}

The single compound H₂O breaks apart into 2 products.

redoxH, O appears as a free element on one side and combined on the other, so its oxidation state changed — electrons were transferred (a redox reaction). Assigning full oxidation numbers is a later topic; here the free-element signature is enough.

Watch out for

  • Decomposition always means a redox reaction. Not always. When a carbonate splits into a metal oxide and carbon dioxide, no element goes free — every atom keeps its oxidation state. Releasing a free element (O₂, H₂) is what makes the other examples redox.

Single replacement

rule-sourced (openstax-chemistry-2e)redox
A+BC→B+AC\mathrm{A} + \mathrm{BC} \rightarrow \mathrm{B} + \mathrm{AC}

A free element trades places with an element inside a compound: the newcomer combines, the displaced element goes free. One element is oxidized and the other reduced, so single replacement is always a redox reaction — its net-ionic view shows exactly which atoms swap electrons.

  • One free element and one compound on each side (A + BC -> B + AC).
  • Whether the swap actually happens depends on an activity ordering of the elements — a more reactive metal displaces a less reactive one. That ordering is its own sourced table.

Verified examples

Zn (s)+2 HCl (aq)→ZnCl2 (aq)+H2 (g)\mathrm{Zn}\,\text{(s)} + 2\,\mathrm{HCl}\,\text{(aq)} \rightarrow \mathrm{ZnCl_{2}}\,\text{(aq)} + \mathrm{H_{2}}\,\text{(g)}
net ionicZn (s)+2 H+ (aq)→Zn2+ (aq)+H2 (g)\mathrm{Zn}\,\text{(s)} + 2\,\mathrm{H}^{+}\,\text{(aq)} \rightarrow \mathrm{Zn}^{2+}\,\text{(aq)} + \mathrm{H_{2}}\,\text{(g)}

The free element Zn displaces an element from a compound, releasing H₂.

redoxH, Zn appears as a free element on one side and combined on the other, so its oxidation state changed — electrons were transferred (a redox reaction). Assigning full oxidation numbers is a later topic; here the free-element signature is enough.

Spectators: Cl⁻

Fe (s)+CuSO4 (aq)→FeSO4 (aq)+Cu (s)\mathrm{Fe}\,\text{(s)} + \mathrm{CuSO_{4}}\,\text{(aq)} \rightarrow \mathrm{FeSO_{4}}\,\text{(aq)} + \mathrm{Cu}\,\text{(s)}
net ionicFe (s)+Cu2+ (aq)→Fe2+ (aq)+Cu (s)\mathrm{Fe}\,\text{(s)} + \mathrm{Cu}^{2+}\,\text{(aq)} \rightarrow \mathrm{Fe}^{2+}\,\text{(aq)} + \mathrm{Cu}\,\text{(s)}

The free element Fe displaces an element from a compound, releasing Cu.

redoxCu, Fe appears as a free element on one side and combined on the other, so its oxidation state changed — electrons were transferred (a redox reaction). Assigning full oxidation numbers is a later topic; here the free-element signature is enough.

Spectators: SO₄²⁻

Mg (s)+2 HCl (aq)→MgCl2 (aq)+H2 (g)\mathrm{Mg}\,\text{(s)} + 2\,\mathrm{HCl}\,\text{(aq)} \rightarrow \mathrm{MgCl_{2}}\,\text{(aq)} + \mathrm{H_{2}}\,\text{(g)}
net ionicMg (s)+2 H+ (aq)→Mg2+ (aq)+H2 (g)\mathrm{Mg}\,\text{(s)} + 2\,\mathrm{H}^{+}\,\text{(aq)} \rightarrow \mathrm{Mg}^{2+}\,\text{(aq)} + \mathrm{H_{2}}\,\text{(g)}

The free element Mg displaces an element from a compound, releasing H₂.

redoxH, Mg appears as a free element on one side and combined on the other, so its oxidation state changed — electrons were transferred (a redox reaction). Assigning full oxidation numbers is a later topic; here the free-element signature is enough.

Spectators: Cl⁻

Watch out for

  • A single-replacement reaction is a plain double replacement. Look for a free element. A single replacement always has an uncombined element on each side; a double replacement has none — both partners stay combined and only swap.

Precipitation

rule-sourced (openstax-chemistry-2e)not redox
AB (aq)+CD (aq)→AD (s)+CB (aq)\mathrm{AB}\,\text{(aq)} + \mathrm{CD}\,\text{(aq)} \rightarrow \mathrm{AD}\,\text{(s)} + \mathrm{CB}\,\text{(aq)}

Two soluble salts swap partners, and one new pairing is insoluble — it drops out as a solid precipitate. The solubility rules decide which pairing is the solid, and the net-ionic equation strips away the spectator ions to show only the ions that actually combine. No element goes free, so precipitation is not a redox reaction.

  • Both reactants dissolved (aqueous); the driving force is that one product is insoluble.
  • Which product is insoluble comes from the sourced solubility rules — the engine classifies it, it is not guessed.

Verified examples

CaCl2 (aq)+Na2CO3 (aq)→CaCO3 (s)+2 NaCl (aq)\mathrm{CaCl_{2}}\,\text{(aq)} + \mathrm{Na_{2}CO_{3}}\,\text{(aq)} \rightarrow \mathrm{CaCO_{3}}\,\text{(s)} + 2\,\mathrm{NaCl}\,\text{(aq)}
net ionicCa2+ (aq)+CO32− (aq)→CaCO3 (s)\mathrm{Ca}^{2+}\,\text{(aq)} + \mathrm{CO_{3}}^{2-}\,\text{(aq)} \rightarrow \mathrm{CaCO_{3}}\,\text{(s)}

CaCO₃ is insoluble (insol-carbonate) and drops out as a solid.

Spectators: Cl⁻, Na⁺

MgCl2 (aq)+2 NaOH (aq)→Mg(OH)2 (s)+2 NaCl (aq)\mathrm{MgCl_{2}}\,\text{(aq)} + 2\,\mathrm{NaOH}\,\text{(aq)} \rightarrow \mathrm{Mg(OH)_{2}}\,\text{(s)} + 2\,\mathrm{NaCl}\,\text{(aq)}
net ionicMg2+ (aq)+2 OH− (aq)→Mg(OH)2 (s)\mathrm{Mg}^{2+}\,\text{(aq)} + 2\,\mathrm{OH}^{-}\,\text{(aq)} \rightarrow \mathrm{Mg(OH)_{2}}\,\text{(s)}

Mg(OH)₂ is insoluble (insol-hydroxide) and drops out as a solid.

Spectators: Cl⁻, Na⁺

CuSO4 (aq)+Na2CO3 (aq)→CuCO3 (s)+Na2SO4 (aq)\mathrm{CuSO_{4}}\,\text{(aq)} + \mathrm{Na_{2}CO_{3}}\,\text{(aq)} \rightarrow \mathrm{CuCO_{3}}\,\text{(s)} + \mathrm{Na_{2}SO_{4}}\,\text{(aq)}
net ionicCu2+ (aq)+CO32− (aq)→CuCO3 (s)\mathrm{Cu}^{2+}\,\text{(aq)} + \mathrm{CO_{3}}^{2-}\,\text{(aq)} \rightarrow \mathrm{CuCO_{3}}\,\text{(s)}

CuCO₃ is insoluble (insol-carbonate) and drops out as a solid.

Spectators: Na⁺, SO₄²⁻

Watch out for

  • The spectator ions disappear once the precipitate forms. Spectator ions stay dissolved and are still there — the complete-ionic equation shows them on both sides, and the net-ionic equation cancels them because they are unchanged, not gone.
  • The reactant present in the smaller amount is what limits the precipitate. Amount alone does not decide it — the mole ratio does. Run the ledger: the reactant that runs out first at the balanced ratio is limiting, which can be the one you have more of.

Acid-base neutralization

rule-sourced (openstax-chemistry-2e)not redox
acid+base→salt+H2O\mathrm{acid} + \mathrm{base} \rightarrow \mathrm{salt} + \mathrm{H_{2}O}

An acid supplies H+ and a base supplies OH-; the two combine into water, leaving the acid's anion and the base's cation together as a dissolved salt. Strip the spectators and every strong-acid + strong-base neutralization has the same net-ionic core: H+ + OH- -> water. No element goes free, so neutralization is not a redox reaction.

  • Needs a species that donates H+ (an acid) and one that donates OH- (a base).
  • The strong-acid + strong-base examples here go essentially to completion; a weak acid or weak base only partly ionizes and the story is subtler.

Verified examples

HCl (aq)+NaOH (aq)→NaCl (aq)+H2O (l)\mathrm{HCl}\,\text{(aq)} + \mathrm{NaOH}\,\text{(aq)} \rightarrow \mathrm{NaCl}\,\text{(aq)} + \mathrm{H_{2}O}\,\text{(l)}
net ionicH+ (aq)+OH− (aq)→H2O (l)\mathrm{H}^{+}\,\text{(aq)} + \mathrm{OH}^{-}\,\text{(aq)} \rightarrow \mathrm{H_{2}O}\,\text{(l)}

hydrochloric acid neutralizes sodium hydroxide: H+ and OH- combine to water, leaving a dissolved salt.

Spectators: Cl⁻, Na⁺

H2SO4 (aq)+2 NaOH (aq)→Na2SO4 (aq)+2 H2O (l)\mathrm{H_{2}SO_{4}}\,\text{(aq)} + 2\,\mathrm{NaOH}\,\text{(aq)} \rightarrow \mathrm{Na_{2}SO_{4}}\,\text{(aq)} + 2\,\mathrm{H_{2}O}\,\text{(l)}
net ionicH+ (aq)+OH− (aq)→H2O (l)\mathrm{H}^{+}\,\text{(aq)} + \mathrm{OH}^{-}\,\text{(aq)} \rightarrow \mathrm{H_{2}O}\,\text{(l)}

sulfuric acid neutralizes sodium hydroxide: H+ and OH- combine to water, leaving a dissolved salt.

Spectators: Na⁺, SO₄²⁻

HNO3 (aq)+KOH (aq)→KNO3 (aq)+H2O (l)\mathrm{HNO_{3}}\,\text{(aq)} + \mathrm{KOH}\,\text{(aq)} \rightarrow \mathrm{KNO_{3}}\,\text{(aq)} + \mathrm{H_{2}O}\,\text{(l)}
net ionicH+ (aq)+OH− (aq)→H2O (l)\mathrm{H}^{+}\,\text{(aq)} + \mathrm{OH}^{-}\,\text{(aq)} \rightarrow \mathrm{H_{2}O}\,\text{(l)}

nitric acid neutralizes potassium hydroxide: H+ and OH- combine to water, leaving a dissolved salt.

Spectators: K⁺, NO₃⁻

Watch out for

  • Neutralization always leaves a neutral (pH 7) solution. Only when a strong acid exactly cancels a strong base. A leftover of either, or a weak partner, leaves the resulting salt solution slightly acidic or basic — 'neutralize' names the reaction, not a guaranteed pH 7.
  • The salt and water are brand-new atoms made in the reaction. Every atom is conserved. The water's H and O come from the acid's H+ and the base's OH-; the salt's ions are the acid's anion and the base's cation, just re-paired.

Gas evolution

rule-sourced (openstax-chemistry-2e)not redox
acid+carbonate→salt+H2O+CO2 (g)\mathrm{acid} + \mathrm{carbonate} \rightarrow \mathrm{salt} + \mathrm{H_{2}O} + \mathrm{CO_{2}}\,\text{(g)}

A double replacement would form an unstable product that immediately breaks down and releases a gas — that gas escaping is the driving force. A carbonate plus an acid makes carbonic acid, which falls apart into water and carbon dioxide; an ammonium salt plus a base makes aqueous ammonia, which releases ammonia gas. No element goes free, so it is not a redox reaction.

  • The trigger is a would-be product that is unstable: carbonic acid (from a carbonate) or aqueous ammonia (from an ammonium salt + base).
  • The gas leaving the solution is what pulls the reaction forward.

Verified examples

2 HCl (aq)+Na2CO3 (aq)→2 NaCl (aq)+H2O (l)+CO2 (g)2\,\mathrm{HCl}\,\text{(aq)} + \mathrm{Na_{2}CO_{3}}\,\text{(aq)} \rightarrow 2\,\mathrm{NaCl}\,\text{(aq)} + \mathrm{H_{2}O}\,\text{(l)} + \mathrm{CO_{2}}\,\text{(g)}
net ionic2 H+ (aq)+CO32− (aq)→H2O (l)+CO2 (g)2\,\mathrm{H}^{+}\,\text{(aq)} + \mathrm{CO_{3}}^{2-}\,\text{(aq)} \rightarrow \mathrm{H_{2}O}\,\text{(l)} + \mathrm{CO_{2}}\,\text{(g)}

carbonic acid would form but is unstable — it decomposes to CO₂ gas and water. Carbonic acid is unstable and breaks down into water and carbon dioxide gas.

Spectators: Cl⁻, Na⁺

2 HCl (aq)+CaCO3 (s)→CaCl2 (aq)+H2O (l)+CO2 (g)2\,\mathrm{HCl}\,\text{(aq)} + \mathrm{CaCO_{3}}\,\text{(s)} \rightarrow \mathrm{CaCl_{2}}\,\text{(aq)} + \mathrm{H_{2}O}\,\text{(l)} + \mathrm{CO_{2}}\,\text{(g)}
net ionic2 H+ (aq)+CaCO3 (s)→Ca2+ (aq)+H2O (l)+CO2 (g)2\,\mathrm{H}^{+}\,\text{(aq)} + \mathrm{CaCO_{3}}\,\text{(s)} \rightarrow \mathrm{Ca}^{2+}\,\text{(aq)} + \mathrm{H_{2}O}\,\text{(l)} + \mathrm{CO_{2}}\,\text{(g)}

carbonic acid would form but is unstable — it decomposes to CO₂ gas and water. Carbonic acid is unstable and breaks down into water and carbon dioxide gas.

Spectators: Cl⁻

NH4Cl (aq)+NaOH (aq)→NaCl (aq)+NH3 (g)+H2O (l)\mathrm{NH_{4}Cl}\,\text{(aq)} + \mathrm{NaOH}\,\text{(aq)} \rightarrow \mathrm{NaCl}\,\text{(aq)} + \mathrm{NH_{3}}\,\text{(g)} + \mathrm{H_{2}O}\,\text{(l)}
net ionicNH4+ (aq)+OH− (aq)→NH3 (g)+H2O (l)\mathrm{NH_{4}}^{+}\,\text{(aq)} + \mathrm{OH}^{-}\,\text{(aq)} \rightarrow \mathrm{NH_{3}}\,\text{(g)} + \mathrm{H_{2}O}\,\text{(l)}

aqueous ammonia would form but is unstable — it decomposes to NH₃ gas and water. Aqueous ammonia (ammonium hydroxide) breaks down into ammonia gas and water.

Spectators: Cl⁻, Na⁺

Watch out for

  • Carbonic acid (H₂CO₃) is one of the final products. It forms for an instant, then decomposes — the products you actually collect are water and carbon dioxide gas. The Atlas shows the two-step logic behind the single written equation.
  • The bubbles are just the solvent boiling. The gas is a real product of the reaction (CO₂ or NH₃), formed atom-for-atom from the reactants — count them in the balanced equation. It is chemistry, not heat.