0620/23 May/June 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

40 questions · 40 marks · 45 minutes

Question papers and mark schemes are copyright Cambridge International. We do not reproduce them: the worked answers here are written by The Practice Book. Have the paper open alongside. Get the official paper from Cambridge

1. Question 1

   1 marksEffect of pressure and temperature on gas volume
   Step 1: For a fixed amount of gas, the volume increases when the pressure is decreased and when the temperature is increased. Step 2: The biggest increase therefore comes from the largest fall in pressure combined with a rise in temperature. Step 3: Decreasing the pressure by 20% (the largest decrease) while increasing the temperature by 10% gives the greatest increase in volume.
   Method:

   Recall how pressure and temperature each change gas volume, then choose the pair that maximises the increase.
   Examiner tips

   • Lower pressure and higher temperature both expand a gas; combine the largest of each.

2. Question 2

   1 marksParticle behaviour on a cooling curve
   Step 1: As G cools, its temperature falls because the molecules lose energy and move more slowly, so statement 1 is correct. Step 2: Cooling means thermal energy is given out by G to the surroundings, so statement 3 is correct. Step 3: On cooling the molecules become more closely packed (not less), and the substance is solidifying rather than melting, so statements 2 and 4 are wrong, leaving 1 and 3.
   Method:

   Decide what cooling does to molecular speed, packing, energy flow and state, then pick the two true statements.
   Examiner tips

   • Cooling: molecules slow, pack closer, and release thermal energy to the surroundings.

3. Question 3

   1 marksNeutron to proton ratio in atoms
   Step 1: "Twice as many neutrons as protons" means the number of neutrons must be exactly two times the number of protons. Step 2: The atom with 1 proton and 2 neutrons has 2 = 2 × 1 neutrons, which fits the rule. Step 3: The others have 0, 1 or an equal number of neutrons to protons, so only the atom with 1 proton and 2 neutrons is correct.
   Method:

   Work out the protons and neutrons in each atom, then find where neutrons are double the protons.
   Examiner tips

   • Neutrons = mass number − proton number; check whether that is twice the proton number.

4. Question 4

   1 marksDefinition of nucleon number
   Step 1: The nucleus contains protons and neutrons, together called nucleons. Step 2: The nucleon number counts all of these nucleons, that is the protons plus the neutrons. Step 3: Electrons are outside the nucleus and are not counted, so the nucleon number is the number of neutrons and protons in the nucleus.
   Method:

   Recall which particles are in the nucleus and that the nucleon number counts all of them.
   Examiner tips

   • Nucleon number = protons + neutrons; proton number = protons only.

5. Question 5

   1 marksPredicting a formula from Periodic Table positions
   Step 1: A Group II element forms an ion with a charge of 2+, and a Group VII element forms an ion with a charge of 1−. Step 2: To balance the charges, one $\text{P}^{2+}$ ion needs two $\text{Q}^-$ ions. Step 3: This gives the formula $\text{PQ}_2$.
   Method:

   Read the group of each element to find its ion charge, then combine the ions so the charges cancel.
   Examiner tips

   • Use the group number to find the ion charge, then cross over to balance.

6. Question 6

   1 marksProperties of ionic compounds
   Step 1: Ionic compounds are built from oppositely charged ions arranged in a regular giant lattice, which is formed when they crystallise from solution, so that statement is correct. Step 2: In a lattice, ions of opposite charge attract and sit closest together, while ions of the same charge repel, so they are not the closest. Step 3: Melting an ionic solid breaks the strong ionic bonds (electrostatic forces) between ions, not intermolecular forces, and the electron counts of the ions vary, so only the first statement is correct.
   Method:

   Recall the giant ionic lattice model and test each statement against it.
   Examiner tips

   • Ionic = giant lattice of oppositely charged ions held by strong ionic bonds.

7. Question 7

   1 marksWriting an ionic equation for a precipitation
   Step 1: The sodium ions and chloride ions appear unchanged on both sides, so they are spectator ions and are left out of the ionic equation. Step 2: The actual change is the iron(III) ions joining with hydroxide ions to make the insoluble solid iron(III) hydroxide. Step 3: So the ionic equation is $\text{Fe}^{3+}(aq) + 3\text{OH}^-(aq) \rightarrow \text{Fe(OH)}_3(s)$.
   Method:

   Identify the spectator ions, remove them, and write the equation for the precipitate that forms.
   Examiner tips

   • An ionic equation shows only the ions that actually change, here forming the precipitate.

8. Question 8

   1 marksEmpirical formula from percentage composition
   Step 1: Divide each percentage by the relative atomic mass: carbon is 85.7 ÷ 12 = 7.14, and hydrogen is 14.3 ÷ 1 = 14.3. Step 2: Divide both by the smaller value (7.14): carbon becomes 1 and hydrogen becomes about 2. Step 3: So the simplest whole-number ratio of C to H is 1 : 2, giving the empirical formula $\text{CH}_2$.
   Method:

   Convert each percentage to moles by dividing by atomic mass, then find the simplest ratio.
   Examiner tips

   • Empirical formula = simplest whole-number ratio of atoms; always reduce to lowest terms.

9. Question 9

   1 marksConcentration from a titration
   Step 1: Moles of barium hydroxide = 0.75 × (15.2 ÷ 1000) = 0.0114 mol. Step 2: From the equation, 1 mol of $\text{Ba(OH)}_2$ reacts with 2 mol of $\text{HNO}_3$, so moles of nitric acid = 2 × 0.0114 = 0.0228 mol. Step 3: Concentration of nitric acid = 0.0228 ÷ (20.0 ÷ 1000) = 0.0228 ÷ 0.0200 = 1.14 mol/dm³.
   Method:

   Calculate moles of base, apply the 1:2 stoichiometry, then divide the acid moles by its volume in dm³.
   Examiner tips

   • Always use the balanced ratio: Ba(OH)2 needs twice as many moles of HNO3.

10. Question 10

    1 marksCharge carriers in the external circuit
    Step 1: In the external metal wires, charge is carried by moving electrons, as in any metallic conductor. Step 2: Ions (anions and cations) carry the charge only through the molten electrolyte, not through the wires. Step 3: So in the external circuit it is the electrons that transfer the charge.
    Method:

    Separate the wires (metallic conduction by electrons) from the electrolyte (ionic conduction).
    Examiner tips

    • External circuit (wires) = electrons; electrolyte = ions.

11. Question 11

    1 marksHydrogen as a fuel and exothermic combustion
    Step 1: Burning hydrogen in air is an exothermic reaction that produces water and releases thermal energy, so statement 2 is correct. Step 2: Because burning hydrogen releases energy, hydrogen is useful as a fuel, so statement 1 is also correct. Step 3: The energy released when hydrogen burns is the reason it is used as a fuel, so statement 2 explains statement 1.
    Method:

    Check each statement is true, then decide whether the energy release explains why hydrogen is a fuel.
    Examiner tips

    • Fuels release energy when they burn; that released energy is what makes them useful.

12. Question 12

    1 marksProducts of electrolysis of concentrated copper(II) chloride
    Step 1: At the negative electrode (cathode), copper ions are discharged in preference to hydrogen because copper is below hydrogen in reactivity, so copper is deposited. Step 2: At the positive electrode (anode), because the solution is concentrated in chloride ions, chlorine is discharged in preference to oxygen. Step 3: So copper forms at the negative electrode and chlorine at the positive electrode.
    Method:

    Apply the discharge rules at each electrode, noting the concentrated chloride favours chlorine.
    Examiner tips

    • Concentrated chloride solutions give the halogen at the anode; unreactive metal ions give the metal at the cathode.

13. Question 13

    1 marksEnthalpy sign and surroundings for an exothermic reaction
    Step 1: An exothermic reaction gives out energy to the surroundings, so the enthalpy change ΔH is negative. Step 2: Because energy is released to the surroundings, the temperature of the surroundings increases. Step 3: So an exothermic reaction has a negative ΔH and the surroundings warm up.
    Method:

    Recall that exothermic means energy out, giving negative ΔH and warmer surroundings.
    Examiner tips

    • Exothermic = negative ΔH + warmer surroundings; endothermic = positive ΔH + cooler surroundings.

14. Question 14

    1 marksLabelling a reaction pathway diagram
    Step 1: The energy from the reactants up to the top of the curve is the activation energy, so X is the activation energy. Step 2: The energy difference between the reactants and the products is the enthalpy change, so Y is the enthalpy change. Step 3: The products are higher in energy than the reactants, which means energy is taken in, so the reaction is endothermic.
    Method:

    Match each arrow to the energy it spans on the diagram, then use the product energy to classify the reaction.
    Examiner tips

    • Activation energy = reactants to the peak; enthalpy change = reactants to products; products higher = endothermic.

15. Question 15

    1 marksEffect of concentration on reacting particles
    Step 1: A higher concentration means more acid particles in the same volume, so collisions between reacting particles happen more often and the collision rate increases. Step 2: The proportion of particles with enough energy to react depends on the temperature, which has not changed, so this proportion stays the same. Step 3: So increasing the concentration increases the collision rate but leaves the proportion of energetic particles unchanged.
    Method:

    Separate the effect on collision frequency from the effect on particle energy, which needs temperature.
    Examiner tips

    • Higher concentration = more frequent collisions; only higher temperature raises the proportion with enough energy.

16. Question 16

    1 marksPosition of equilibrium and colour change
    Step 1: Adding more thiocyanate ions increases the concentration of a reactant, so the equilibrium shifts to the right to oppose the change. Step 2: Shifting right makes more of the red FeSCN²⁺ ions. Step 3: So the mixture becomes more red.
    Method:

    Use Le Chatelier reasoning: adding a reactant shifts equilibrium to products and deepens the product colour.
    Examiner tips

    • Add a reactant and the equilibrium moves to make more product; follow the colour of that product.

17. Question 17

    1 marksIdentifying oxidising and reducing agents
    Step 1: The bromine is reduced (it gains electrons, 0 to −1), so it is the oxidising agent. Step 2: The iron(II) in $\text{FeSO}_4$ is oxidised (it loses electrons, +2 to +3), so $\text{FeSO}_4$ is the reducing agent. Step 3: So the oxidising agent is $\text{Br}_2$ and the reducing agent is $\text{FeSO}_4$.
    Method:

    Track which reactant gains electrons (oxidising agent) and which loses electrons (reducing agent).
    Examiner tips

    • Oxidising agent is itself reduced; reducing agent is itself oxidised.

18. Question 18

    1 marksChange in oxidation number
    Step 1: In $\text{Fe}_2\text{O}_3$ the oxygen is −2, so the iron must be +3 to balance the charges. Step 2: In the product the iron is the free metal Fe, which has an oxidation number of 0. Step 3: So the oxidation number of iron changes from +3 to 0 (the iron is reduced).
    Method:

    Assign the oxidation number of iron in the oxide and in the metal, then state the change.
    Examiner tips

    • A free, uncombined element always has an oxidation number of 0.

19. Question 19

    1 marksIdentifying acid and base by proton transfer
    Step 1: In the forward reaction, the water molecule gives away a hydrogen ion (proton) to become the hydroxide ion, so water is the proton donor, the acid. Step 2: The ammonia molecule accepts that hydrogen ion to become the ammonium ion, so ammonia is the proton acceptor, the base. Step 3: So water acts as the acid and ammonia acts as the base.
    Method:

    Follow the hydrogen ion: the donor is the acid and the acceptor is the base.
    Examiner tips

    • Acid donates H+, base accepts H+; track the proton from one species to the other.

20. Question 20

    1 marksClassifying acidic, basic and amphoteric oxides
    Step 1: Sulfur dioxide is a non-metal oxide, so it is acidic. Step 2: Sodium oxide is a Group I metal oxide, so it is basic, while aluminium oxide reacts with both acids and bases, so it is amphoteric. Step 3: So the acidic oxide is $\text{SO}_2$, the basic oxide is $\text{Na}_2\text{O}$ and the amphoteric oxide is $\text{Al}_2\text{O}_3$.
    Method:

    Classify each oxide by its element type and reactivity with acids and bases, then match the row.
    Examiner tips

    • Sodium oxide is basic; sulfur dioxide is acidic; aluminium oxide and zinc oxide are amphoteric.

21. Question 21

    1 marksDensity trends in Groups I and VII
    Step 1: Density increases down a group, so of lithium and potassium, the lower element potassium has the higher density. Step 2: Of chlorine and iodine, the higher element chlorine has the lower density. Step 3: So the higher-density metal (potassium) reacts with the lower-density halogen (chlorine) to form potassium chloride.
    Method:

    Apply the down-group density trend to each group, then combine the chosen elements.
    Examiner tips

    • Going down a group, density increases; pick the bottom metal and the top halogen.

22. Question 22

    1 marksUsing groups and proton number to identify an element
    Step 1: Silicon is in Group IV, so element X must also be in Group IV. Step 2: Having more protons than silicon means X is lower down the group than silicon. Step 3: Germanium is the Group IV element directly below silicon, so it has more protons and could be X.
    Method:

    Identify silicon's group, keep X in the same group, and pick the element with more protons.
    Examiner tips

    • Same group = same column; more protons = lower in that column.

23. Question 23

    1 marksProperties of the noble gases
    Step 1: The noble gases all exist as single, separate atoms in the gas state at room temperature, so they are all gases. Step 2: They are monatomic (not diatomic), and helium has only two outer electrons (not eight), so those statements are wrong. Step 3: They are very unreactive and do not react with oxygen, so the only correct statement is that they are all gases at room temperature.
    Method:

    Recall the key noble-gas properties and rule out the statements that are not always true.
    Examiner tips

    • Noble gases: monatomic, unreactive, full outer shell (helium has 2, the rest have 8).

24. Question 24

    1 marksIdentifying a transition element from its properties
    Step 1: Transition elements have high melting points, high densities and form coloured compounds. Step 2: The element with melting point 1907 °C and density 7.2 g/cm³ has both a high melting point and a high density, and its green oxide is a coloured compound. Step 3: So this element fits all the properties of a transition element.
    Method:

    Compare each element against the transition-element properties and pick the one that meets them all.
    Examiner tips

    • Transition metals: high melting point, high density, coloured compounds.

25. Question 25

    1 marksComparing reactivity of metals with acid
    Step 1: The reactivity order of these metals is magnesium > zinc > iron > copper, and copper does not react with dilute acid. Step 2: Magnesium reacts most vigorously, zinc fairly rapidly, and copper not at all, so iron sits between zinc and copper. Step 3: Iron therefore gives a slow but steady stream of bubbles, less vigorous than magnesium or zinc but still reacting, unlike copper.
    Method:

    Order the four metals by reactivity, then match iron to the moderate bubbling rate.
    Examiner tips

    • Reactivity order Mg > Zn > Fe > Cu; copper gives no reaction with dilute acid.

26. Question 26

    1 marksExplaining the strength of an alloy
    Step 1: In pure copper the atoms are the same size and are arranged in regular layers that can slide over each other, making it softer. Step 2: Adding zinc atoms of a different size distorts the layers, so the layers can no longer slide easily over each other. Step 3: This makes brass harder and stronger than pure copper, so the correct statement is that the layers cannot easily slide.
    Method:

    Recall how alloying distorts the metal layers and decide how that affects strength.
    Examiner tips

    • Alloys are stronger because different-sized atoms prevent the layers from sliding.

27. Question 27

    1 marksProducts of thermal decomposition of calcium carbonate
    Step 1: Heating calcium carbonate, $\text{CaCO}_3$, strongly causes thermal decomposition. Step 2: It breaks down into calcium oxide, CaO (a white solid), and carbon dioxide, $\text{CO}_2$ (a colourless gas). Step 3: So substance U is calcium oxide and substance V is carbon dioxide.
    Method:

    Recall the thermal decomposition of a carbonate to give the metal oxide and carbon dioxide.
    Examiner tips

    • Metal carbonate → metal oxide + carbon dioxide on strong heating.

28. Question 28

    1 marksSources of water pollution
    Step 1: Harmful microbes such as bacteria come from untreated human and animal waste. Step 2: Sewage is exactly this kind of waste, so it introduces harmful microbes into the river water. Step 3: Fertilisers, detergents and plastics are chemical or solid pollutants and are not sources of microbes, so the answer is sewage.
    Method:

    Pick the pollutant that contains living organisms rather than dissolved chemicals.
    Examiner tips

    • Sewage carries living microbes; fertilisers and detergents are chemical pollutants.

29. Question 29

    1 marksColour changes in chemical tests for water
    Step 1: Anhydrous copper(II) sulfate is white and turns blue when water is added, so the change is white to blue. Step 2: Anhydrous cobalt(II) chloride is blue and turns pink when water is added, so the change is blue to pink. Step 3: So copper(II) sulfate goes white to blue and cobalt(II) chloride goes blue to pink.
    Method:

    Recall the colours of each anhydrous salt and the colour after water is added.
    Examiner tips

    • These colour changes are the standard chemical tests for the presence of water.

30. Question 30

    1 marksGreenhouse gases
    Step 1: Greenhouse gases absorb and re-emit thermal energy, increasing global warming. Step 2: Carbon dioxide is a well-known greenhouse gas produced by burning fossil fuels. Step 3: Argon, nitrogen and oxygen are not greenhouse gases, so the answer is carbon dioxide.
    Method:

    Identify the gas known to absorb thermal energy and rule out the unreactive air gases.
    Examiner tips

    • Carbon dioxide and methane are the key greenhouse gases at IGCSE.

31. Question 31

    1 marksHow greenhouse gases cause global warming
    Step 1: The Earth's surface, warmed by the Sun, radiates thermal energy back outwards. Step 2: Greenhouse gases absorb this thermal energy and then re-emit it, some of it back towards the Earth. Step 3: This traps energy near the surface and warms the planet, so the correct statement is that greenhouse gases absorb and re-emit thermal energy radiated from the Earth's surface.
    Method:

    Trace the energy: from the warmed Earth, absorbed and re-emitted by the gases to trap heat.
    Examiner tips

    • Greenhouse effect: gases absorb and re-emit the Earth's radiated heat, trapping it near the surface.

32. Question 32

    1 marksReactions in a catalytic converter
    Step 1: Carbon monoxide forms when there is not enough oxygen, that is during incomplete combustion, not complete combustion, so this statement is not correct. Step 2: In the converter the carbon monoxide is oxidised to carbon dioxide, and the nitrogen monoxide is reduced to nitrogen, so those statements are correct. Step 3: Nitrogen monoxide does form from nitrogen and oxygen reacting in the hot engine, so the only incorrect statement is that carbon monoxide comes from complete combustion.
    Method:

    Check each statement for accuracy and identify the one that misstates how carbon monoxide forms.
    Examiner tips

    • Incomplete combustion (limited oxygen) makes carbon monoxide; complete combustion makes carbon dioxide.

33. Question 33

    1 marksIdentifying structural isomers
    Step 1: Structural isomers must have the same molecular formula as but-2-ene, which is $\text{C}_4\text{H}_8$. Step 2: Structure 1 (but-1-ene) and structure 3 (methylpropene) are both $\text{C}_4\text{H}_8$, so they are isomers of but-2-ene. Step 3: Structure 2 (methylpropane) is $\text{C}_4\text{H}_{10}$, a different molecular formula, so it is not an isomer, leaving only structures 1 and 3.
    Method:

    Work out the molecular formula of each structure and keep those that match but-2-ene (C4H8).
    Examiner tips

    • Count the carbons and hydrogens: only structures matching C4H8 are isomers of but-2-ene.

34. Question 34

    1 marksCharacteristics of a homologous series
    Step 1: Members of a homologous series all contain the same functional group, which gives them similar chemical properties. Step 2: As the chain gets longer, their physical properties (such as boiling point) change gradually, showing a trend. Step 3: So the correct pair is the same functional group together with a trend in physical properties.
    Method:

    Recall the defining features of a homologous series and pick the pair that is both true.
    Examiner tips

    • Homologous series: same functional group, same general formula, similar chemistry, trend in physical properties, differ by CH2.

35. Question 35

    1 marksIdentifying a saturated hydrocarbon
    Step 1: A hydrocarbon contains only carbon and hydrogen, which rules out alcohols and carboxylic acids because they also contain oxygen. Step 2: Containing single bonds only means it is saturated, which rules out alkenes because they have a C=C double bond. Step 3: A saturated hydrocarbon with single bonds only is an alkane, so W is an alkane.
    Method:

    Use the two clues (hydrocarbon, single bonds only) to eliminate the other series.
    Examiner tips

    • Alcohols and carboxylic acids contain oxygen, so they are not hydrocarbons; alkenes have a double bond.

36. Question 36

    1 marksPhotochemical substitution of methane
    Step 1: To make chloromethane, methane reacts with chlorine in a substitution reaction, so it is mixed with chlorine gas (not hydrogen chloride). Step 2: Ultraviolet light supplies the energy needed to start the reaction, that is it provides the activation energy. Step 3: So methane is mixed with chlorine gas and the ultraviolet light provides the activation energy.
    Method:

    Identify the correct reactant (chlorine) and the role of UV light (providing activation energy).
    Examiner tips

    • Substitution of an alkane uses the halogen itself (chlorine) with UV light to start it.

37. Question 37

    1 marksLinkages and types of polymer
    Step 1: Nylon is a polyamide, so it contains the amide linkage where a C=O carbon is bonded to the nitrogen of an N–H group, making the first statement correct. Step 2: Nylon is a polyamide, not a polyester, and PET contains an ester linkage (not a C=C double bond). Step 3: Propane is a saturated alkane with no C=C double bond, so it cannot undergo addition polymerisation, leaving the amide-linkage statement as the correct one.
    Method:

    Recall the linkage in each named polymer and the requirement for addition polymerisation.
    Examiner tips

    • Nylon = polyamide (amide linkage); PET = polyester (ester linkage); addition polymers need a C=C bond.

38. Question 38

    1 marksFiltration terminology
    Step 1: To dissolve the soluble solid you add the mixture to a solvent, so gap 1 is solvent. Step 2: When the mixture is filtered, the insoluble solid that stays on the filter paper is called the residue, so gap 2 is residue. Step 3: So gap 1 is solvent and gap 2 is residue (the liquid that passes through would be the filtrate).
    Method:

    Match each gap to the correct filtration term: solvent for dissolving, residue for the solid kept.
    Examiner tips

    • Residue stays on the filter paper; filtrate passes through; solvent dissolves the soluble solid.

39. Question 39

    1 marksApparatus for separating liquids by boiling point
    Step 1: Liquids with different boiling points are separated by (fractional) distillation, which needs a heat source, a fractionating column, a thermometer and a condenser. Step 2: As the mixture is heated, the liquid with the lower boiling point evaporates first, is cooled in the condenser and collected separately. Step 3: So the distillation apparatus with a heated flask, column, thermometer and condenser is correct; the filter and the electrolysis cell do not separate liquids by boiling point.
    Method:

    Match the separation method (distillation) to the apparatus that includes a condenser and thermometer.
    Examiner tips

    • Distillation (with a condenser) separates liquids by their different boiling points.

40. Question 40

    1 marksThe Rf value in paper chromatography
    Step 1: The Rf value compares how far the substance moves with how far the solvent moves. Step 2: It is defined as the distance travelled by the substance divided by the distance travelled by the solvent. Step 3: Because the substance never moves further than the solvent, this gives a value between 0 and 1.
    Method:

    Recall the Rf definition with the substance distance on top and the solvent distance underneath.
    Examiner tips

    • Rf is always substance ÷ solvent, giving a value between 0 and 1.