0620/21 May/June 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

40 questions · 40 marks · 45 minutes

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1. Question 1

   1 marksGas pressure and the kinetic particle model
   Step 1: Heating gives the gas particles more kinetic energy, so they move around faster, making statement 1 correct. Step 2: Faster particles hit the container walls more often, so statement 2 is also correct. Step 3: More frequent and harder collisions in a fixed volume increase the pressure, so statement 3 is wrong, leaving 1 and 2.
   Method:

   Use the kinetic particle model to judge how heating changes particle speed, collisions and pressure.
   Examiner tips

   • In a fixed volume, heating a gas raises both the speed of the particles and the pressure.

2. Question 2

   1 marksHeating curve and changes of state
   Step 1: The plateaux are where the state changes: 2 to 3 is melting and 4 to 5 is boiling, and the longer 4 to 5 plateau means more energy is taken in. Step 2: More energy is needed to separate the particles completely on boiling than to free them on melting, so more energy is absorbed between 4 and 5. Step 3: So the correct statement is that more energy is absorbed between 4 and 5 than between 2 and 3.
   Method:

   Identify the plateaux as melting and boiling, then compare how much energy each absorbs.
   Examiner tips

   • On a heating curve, plateaux are changes of state; boiling takes more energy than melting.

3. Question 3

   1 marksEqual numbers of protons, neutrons and electrons
   Step 1: In a neutral atom the number of protons equals the number of electrons, and the number of neutrons is the nucleon number minus the proton number. Step 2: For nitrogen-14 there are 7 protons, 7 electrons and 14 − 7 = 7 neutrons, so all three are equal. Step 3: The other atoms have more neutrons than protons, so nitrogen-14 is the answer.
   Method:

   Work out protons, neutrons and electrons for each atom and find the one where all three match.
   Examiner tips

   • Neutrons = nucleon number − proton number; equal counts need nucleon number = 2 × proton number.

4. Question 4

   1 marksIsotopic abundance from relative atomic mass
   Step 1: Let the fraction of europium-151 be $x$, so $151x + 153(1-x) = 152$. Step 2: Expanding gives $153 - 2x = 152$, so $2x = 1$ and $x = 0.500$. Step 3: So europium-151 makes up 50.0% of the sample.
   Method:

   Write the weighted-average equation for the abundances and solve for the percentage of the lighter isotope.
   Examiner tips

   • A relative atomic mass exactly between the two isotope masses means a 50:50 mixture.

5. Question 5

   1 marksProperties of ionic compounds
   Step 1: Ionic compounds conduct electricity only when the ions are free to move, which happens when molten or dissolved. Step 2: Molten potassium chloride has mobile ions, so it conducts, making that statement correct. Step 3: Its high melting point is due to strong ionic (not covalent) bonds, it does not conduct as a solid, and it is soluble in water, so the other statements are wrong.
   Method:

   Use the properties of ionic compounds to test each statement about conductivity, bonding and solubility.
   Examiner tips

   • Ionic solids conduct only when molten or in solution, because the ions then move freely.

6. Question 6

   1 marksDot-and-cross diagram of nitrogen
   Step 1: Each nitrogen atom has five outer electrons and needs to reach a full outer shell of eight. Step 2: The two atoms share three pairs of electrons, forming a triple bond, which gives each nitrogen a share of eight electrons. Step 3: That uses three of each atom's five outer electrons, leaving one lone pair on each nitrogen, so the correct diagram has a triple bond and one lone pair per atom.
   Method:

   Give each nitrogen a full outer shell by sharing three pairs, then place the remaining lone pairs.
   Examiner tips

   • Nitrogen forms a triple bond in N2, with one lone pair left on each atom.

7. Question 7

   1 marksStructure and bonding in silicon(IV) oxide
   Step 1: Silicon(IV) oxide is a giant covalent structure in which silicon and oxygen atoms are joined by single covalent bonds. Step 2: Each covalent bond is one shared pair of electrons, so that statement is correct. Step 3: It contains atoms (not ions), it does not conduct electricity, and its tetrahedral arrangement is like diamond rather than graphite, so the other statements are wrong.
   Method:

   Recall that silicon(IV) oxide is giant covalent like diamond and test each statement against that.
   Examiner tips

   • Silicon(IV) oxide is a giant covalent solid like diamond: single bonds, no conduction.

8. Question 8

   1 marksEmpirical formulae of organic molecules
   Step 1: The empirical formula is the simplest whole-number ratio of atoms. Step 2: Lactic acid $\text{C}_3\text{H}_6\text{O}_3$ divides by 3 to give $\text{CH}_2\text{O}$, and glucose $\text{C}_6\text{H}_{12}\text{O}_6$ divides by 6 to give $\text{CH}_2\text{O}$, so they match. Step 3: Propanoic acid $\text{C}_3\text{H}_6\text{O}_2$ does not simplify to $\text{CH}_2\text{O}$, so only lactic acid and glucose share an empirical formula.
   Method:

   Reduce each molecular formula to its simplest ratio and find the molecules that share it.
   Examiner tips

   • Empirical formula = molecular formula divided by the highest common factor of the subscripts.

9. Question 9

   1 marksLimiting reactant and volume of gas
   Step 1: Moles of calcium carbonate = 1.00 ÷ 100 = 0.0100 mol, and moles of HCl = 0.0500 × (50.0 ÷ 1000) = 0.00250 mol. Step 2: The acid is limiting because 0.00250 mol of HCl needs only 0.00125 mol of CaCO3, and 2 mol of HCl give 1 mol of CO2, so moles of CO2 = 0.00250 ÷ 2 = 0.00125 mol. Step 3: Volume of CO2 = 0.00125 × 24 000 = 30 cm³.
   Method:

   Calculate moles of both reactants, identify the limiting one, then convert moles of CO2 to a gas volume.
   Examiner tips

   • Always check which reactant is limiting before using a mole ratio.

10. Question 10

    1 marksElectroplating with copper
    Step 1: Copper is deposited on the steel cathode, so the steel turns reddish-brown, making statement 1 correct. Step 2: The copper anode dissolves to replace the copper ions removed at the cathode, so the solution stays the same blue colour, making statement 2 wrong. Step 3: Because the copper anode dissolves, it becomes thinner, so statement 3 is correct, giving 1 and 3 only.
    Method:

    Track where copper is deposited and dissolved to judge the colour change and electrode masses.
    Examiner tips

    • In copper plating with a copper anode, the electrolyte concentration and colour stay constant.

11. Question 11

    1 marksAnode half-equation in electrolysis of brine
    Step 1: At the anode negative ions are oxidised, which means they lose electrons. Step 2: With concentrated sodium chloride the chloride ions are discharged, so two chloride ions lose two electrons to form a chlorine molecule. Step 3: This gives $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-$, with electrons on the right because they are lost.
    Method:

    Identify the ion discharged at the anode and write its oxidation half-equation with electrons on the right.
    Examiner tips

    • Anode = oxidation = loss of electrons; balance both atoms and charge.

12. Question 12

    1 marksOverall equation for a fuel cell
    Step 1: A hydrogen-oxygen fuel cell combines hydrogen and oxygen, the reverse of the electrolysis of water. Step 2: The only product is water, and the balanced equation is $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$. Step 3: So the overall reaction makes water from hydrogen and oxygen.
    Method:

    Recall that the fuel cell forms water and choose the correctly balanced equation.
    Examiner tips

    • A fuel cell makes only water: 2H2 + O2 → 2H2O.

13. Question 13

    1 marksExothermic reaction and reaction pathway diagram
    Step 1: The mixture gets hot, so energy is released to the surroundings, which means the reaction is exothermic. Step 2: In an exothermic reaction the products have less stored energy than the reactants, so the reactants are drawn higher than the products. Step 3: So the correct choice is exothermic with the reactants at a higher energy level than the products.
    Method:

    Use the temperature change to decide exothermic or endothermic, then match the energy level diagram.
    Examiner tips

    • Exothermic: heat released, products lower than reactants on the energy level diagram.

14. Question 14

    1 marksRecognising a reversible reaction
    Step 1: The equation uses the $\rightleftharpoons$ symbol, which shows the reaction can go both forwards and backwards. Step 2: A reaction that can proceed in both directions is described as reversible. Step 3: So this equation shows a reversible reaction.
    Method:

    Read the reaction arrow: a double arrow means the reaction is reversible.
    Examiner tips

    • Spot the ⇌ symbol: it always means the reaction is reversible.

15. Question 15

    1 marksEffect of temperature on colliding particles
    Step 1: A lower temperature gives the particles less energy, so their average kinetic energy is lower. Step 2: Slower particles collide less often, so the frequency of collisions is also lower. Step 3: So at 30 °C both the frequency of collisions and the average kinetic energy are lower than at 50 °C.
    Method:

    Use collision theory to see how lowering the temperature affects energy and collision frequency.
    Examiner tips

    • Lower temperature → lower average kinetic energy → fewer collisions per second.

16. Question 16

    1 marksConditions for the Haber process
    Step 1: A lower temperature always slows a reaction, so statement 1 is correct, but because the forward reaction is exothermic a lower temperature actually raises the yield, so statement 2 is wrong. Step 2: Higher pressure favours the side with fewer gas molecules (the ammonia side), so pressures above 200 atm would raise, not lower, the yield, making statement 3 wrong. Step 3: Very high pressures are expensive and difficult to maintain safely, so statement 4 is correct, giving 1 and 4.
    Method:

    Apply rate ideas and equilibrium shifts for an exothermic, fewer-moles-of-gas reaction to each statement.
    Examiner tips

    • For an exothermic equilibrium: lower temperature raises yield; higher pressure raises yield but costs more.

17. Question 17

    1 marksOxidation of ethanol
    Step 1: Acidified potassium manganate(VII) oxidises ethanol to ethanoic acid, so the ethanol undergoes oxidation. Step 2: As it acts as the oxidising agent the manganate(VII) is reduced and its purple colour fades. Step 3: So the reaction is an oxidation and the colour changes from purple to colourless.
    Method:

    Recall that manganate(VII) oxidises ethanol and is decolourised, then match type and colour change.
    Examiner tips

    • Ethanol + acidified manganate(VII): oxidation to ethanoic acid, purple fades to colourless.

18. Question 18

    1 marksStrong and weak acids
    Step 1: A strong acid is fully dissociated, while a weak acid is only partly dissociated. Step 2: Acid R dissociates only 0.0048 of 0.10 mol and acid S only 0.0179 of 0.10 mol, so both are only partly dissociated. Step 3: Because neither acid is fully dissociated, both R and S are weak acids.
    Method:

    Compare the dissociated amount with the dissolved amount for each acid to classify it as strong or weak.
    Examiner tips

    • Compare moles dissociated with moles dissolved; well below full dissociation means a weak acid.

19. Question 19

    1 marksPreparing an insoluble salt
    Step 1: An insoluble salt is made by precipitation, then separated by filtration, washed, and dried. Step 2: Crystallisation is used to obtain a soluble salt from its solution by evaporation, so it is not needed for an insoluble salt. Step 3: So the process not used is crystallisation.
    Method:

    Recall the preparation route for an insoluble salt and find the step that does not belong.
    Examiner tips

    • Insoluble salt route: precipitate, filter, wash, dry — no crystallisation step.

20. Question 20

    1 marksWater of crystallisation in hydrated salts
    Step 1: Hydrated copper(II) sulfate is the familiar blue crystal with five molecules of water of crystallisation, so $x = 5$. Step 2: Hydrated cobalt(II) chloride is the pink solid with six molecules of water of crystallisation, so $y = 6$. Step 3: So $x = 5$ and $y = 6$.
    Method:

    Recall the standard water of crystallisation in each named hydrated salt.
    Examiner tips

    • Remember CuSO4·5H2O (blue) and CoCl2·6H2O (pink) as standard hydrated salts.

21. Question 21

    1 marksTrends across a period
    Step 1: Metals are on the left of the Periodic Table and non-metals on the right. Step 2: So across Period 3 the elements change from metals (such as sodium) to non-metals (such as chlorine). Step 3: The number of shells stays the same across a period and conductivity falls rather than rises, so only the metal-to-non-metal change is correct.
    Method:

    Use the position of metals and non-metals in the table to judge the trend across a period.
    Examiner tips

    • Left to right across a period: metal to non-metal; same number of shells throughout.

22. Question 22

    1 marksTrends down Group VII
    Step 1: Going down Group VII the halogens become less reactive, so reactivity decreases. Step 2: At the same time the atomic number, density and melting point all increase down the group. Step 3: So the property that decreases is reactivity.
    Method:

    Recall the Group VII trends and identify the one quantity that falls down the group.
    Examiner tips

    • Down Group VII: reactivity decreases, but atomic number, density and melting point increase.

23. Question 23

    1 marksProperties shared by transition and Group I metals
    Step 1: Both iron and lithium are metals, so both conduct electricity well and are malleable, making statements 1 and 3 correct for both. Step 2: Lithium is a Group I metal with a low melting point, so statement 2 is not true for both, and only transition elements like iron form ions with variable oxidation numbers, so statement 4 is not true for both. Step 3: So the properties shared by both metals are 1 and 3.
    Method:

    Separate general metallic properties from transition-only properties and keep those shared by both.
    Examiner tips

    • Conductivity and malleability are general metal properties; high mp and variable oxidation are transition-only.

24. Question 24

    1 marksProperties of the noble gases
    Step 1: Noble gases are very unreactive, so describing them as reactive makes statement 1 wrong, although they are indeed monatomic. Step 2: Noble gases do have a full outer shell of electrons, so statement 2 is correct. Step 3: So statement 2 is correct but statement 1 is not.
    Method:

    Check each statement against the known unreactive, full-shell nature of the noble gases.
    Examiner tips

    • Noble gases: monatomic and unreactive, with full outer electron shells.

25. Question 25

    1 marksProperties and uses of copper
    Step 1: Copper is a good electrical conductor and is ductile, so it can be drawn into wires for electrical wiring, making this statement correct. Step 2: Brass is copper mixed with zinc (copper with tin is bronze), copper has a high density so it is not used for overhead cables, and takeaway containers use aluminium. Step 3: So the correct statement is that copper is used in electrical wiring and is ductile.
    Method:

    Match copper to its true properties and uses, and rule out the wrong alloy and density claims.
    Examiner tips

    • Copper: ductile electrical conductor for wiring; brass = copper + zinc.

26. Question 26

    1 marksIdentifying a metal from its reactions
    Step 1: Reacting with dilute acid to give hydrogen, and only slowly with cold water but quickly with steam, fits a moderately reactive metal. Step 2: Magnesium reacts with acids, reacts very slowly with cold water and vigorously with steam, which matches Z. Step 3: Gold and silver are too unreactive to react with acid, and potassium reacts violently with cold water, so Z is magnesium.
    Method:

    Match the reaction behaviour to a metal of moderate reactivity in the reactivity series.
    Examiner tips

    • Magnesium: reacts with acids, very slow with cold water, vigorous with steam.

27. Question 27

    1 marksReactivity order from displacement results
    Step 1: A metal displaces another from solution when it is more reactive (more able to form positive ions). Step 2: J displaces K, L and M, so J is the most reactive; L displaces K and M; K displaces only M; and M displaces none, giving the order J > L > K > M. Step 3: Since J is above K in this order, J has a greater tendency to form positive ions than K, so that statement is correct.
    Method:

    Use the displacement results to rank the metals, then test each statement against that order.
    Examiner tips

    • Order the metals by what each can displace; the one that displaces most is most reactive.

28. Question 28

    1 marksExtraction of iron from hematite
    Step 1: Iron is less reactive than carbon, so it can be extracted by reducing its oxide with carbon (as carbon monoxide) in the blast furnace. Step 2: Removing oxygen from iron(III) oxide is a reduction. Step 3: So the extraction of iron from hematite uses reduction.
    Method:

    Recall how iron sits below carbon in reactivity and is extracted by reduction.
    Examiner tips

    • Metals below carbon, like iron, are extracted by reduction with carbon.

29. Question 29

    1 marksGalvanising and rust prevention
    Step 1: Galvanising coats iron with zinc, not silver, so statement 1 is wrong. Step 2: The zinc layer acts as a barrier that keeps oxygen and water away from the iron, so statement 2 is correct. Step 3: Zinc is more reactive than iron, so it loses electrons more readily and protects the iron sacrificially, making statement 3 correct, giving 2 and 3 only.
    Method:

    Recall that galvanising uses zinc and protects both by barrier and sacrificial action.
    Examiner tips

    • Galvanising = zinc coating: a barrier plus sacrificial protection.

30. Question 30

    1 marksNPK fertilisers
    Step 1: Ammonium phosphate supplies both nitrogen (in the ammonium ion) and phosphorus (in the phosphate ion). Step 2: Potassium chloride supplies potassium, so together with ammonium phosphate all three elements are present. Step 3: The other mixtures miss either phosphorus or potassium, so the correct mixture is ammonium phosphate with potassium chloride.
    Method:

    Identify which elements each compound provides and find the mixture covering N, P and K.
    Examiner tips

    • Check each mixture for all of N, P and K; ammonium phosphate gives N and P, KCl gives K.

31. Question 31

    1 marksGreenhouse gases and global warming
    Step 1: Greenhouse gases trap thermal energy that the Earth radiates back outwards. Step 2: Statement 1 (carbon dioxide absorbs thermal energy from the Earth) and statement 4 (methane reflects thermal energy from the Earth back to the Earth) both describe trapping of the Earth's heat, so they explain the warming. Step 3: Statements 2 and 3 are about energy from the Sun and do not explain how the gases trap the Earth's heat, so the answer is 1 and 4.
    Method:

    Select the statements that describe trapping the Earth's radiated heat rather than the Sun's energy.
    Examiner tips

    • The greenhouse effect is about trapping outgoing heat from the Earth, not incoming sunlight.

32. Question 32

    1 marksFormation of oxides of nitrogen
    Step 1: Air is drawn into the engine, and at the high temperatures inside the engine the nitrogen and oxygen from the air react together. Step 2: This reaction forms oxides of nitrogen, which then leave in the exhaust. Step 3: So the oxides form from nitrogen and oxygen in the air reacting at the high temperatures in the engine.
    Method:

    Trace where the nitrogen and oxygen come from and where the high temperature occurs.
    Examiner tips

    • Oxides of nitrogen form when air's nitrogen and oxygen react in the high-temperature engine.

33. Question 33

    1 marksStructural isomers
    Step 1: Structural isomers have the same molecular formula but different structures, so they must match $\text{C}_3\text{H}_6\text{O}_2$. Step 2: Compound 1 ($\text{HCOOCH}_2\text{CH}_3$) and compound 2 ($\text{CH}_3\text{CH}_2\text{COOH}$) are both $\text{C}_3\text{H}_6\text{O}_2$, so they are isomers. Step 3: Compound 3 ($\text{HOCH}_2\text{CH}_2\text{CH}_2\text{OH}$) is $\text{C}_3\text{H}_8\text{O}_2$, which has two extra hydrogen atoms, so it is not an isomer, giving 1 and 2 only.
    Method:

    Work out each molecular formula and keep only those that match C3H6O2.
    Examiner tips

    • Structural isomers share the molecular formula but differ in structure; check the atom counts.

34. Question 34

    1 marksCounting covalent bonds in propane
    Step 1: Propane has a chain of three carbon atoms, giving 2 carbon-carbon bonds. Step 2: There are 8 hydrogen atoms, each joined to a carbon by 1 bond, giving 8 carbon-hydrogen bonds. Step 3: The total number of covalent bonds is 2 + 8 = 10.
    Method:

    Draw the displayed structure of propane and count every line (bond).
    Examiner tips

    • For an alkane chain, bonds = (C-C bonds) + (number of hydrogen atoms).

35. Question 35

    1 marksManufacture of ethanol
    Step 1: Fermentation relies on enzymes in yeast and is a slow batch process, while the catalytic reaction of ethene with steam is fast and continuous, so that comparison is correct. Step 2: Fermentation works at around 30 to 40 °C (not 100 °C), and the ethene-with-steam route does need a catalyst, so those statements are wrong. Step 3: The ethene route gives a high yield, so the yield comparison is also wrong, leaving the slow-versus-fast statement.
    Method:

    Compare the speed, temperature, catalyst and yield of the two ethanol routes statement by statement.
    Examiner tips

    • Fermentation: slow, ~30 °C, yeast; ethene + steam: fast, catalysed, high yield.

36. Question 36

    1 marksIdentifying a carboxylic acid
    Step 1: A solution with a pH of 3 that reacts with calcium carbonate to give carbon dioxide is acidic. Step 2: Among the organic families, only carboxylic acids are acidic and react with carbonates to release carbon dioxide. Step 3: So J is a carboxylic acid.
    Method:

    Use the acidic behaviour (low pH, reaction with carbonate) to pick the carboxylic acid family.
    Examiner tips

    • Reacting with a carbonate to give CO2 and a low pH are the signs of a carboxylic acid.

37. Question 37

    1 marksRepeat unit of a condensation polymer
    Step 1: A carboxylic acid and an alcohol form an ester link, $\text{C(=O)}{-}\text{O}$, with the loss of water, and this is repeated for a condensation polymer. Step 2: Each -COOH of the diacid joins to an -OH of the diol, losing water and forming an ester linkage, so both carbonyl groups become ester links to the three-carbon diol chain. Step 3: So the correct repeat unit is the polyester linking the diacid and diol through ester groups, which is "a repeat unit joined by simple carbon-carbon bonds with no ester linkages, as if the monomers had added together without losing water".
    Method:

    Join a -COOH to an -OH to make an ester link with loss of water, repeating to build the polyester repeat unit.
    Examiner tips

    • Diacid + diol → polyester: each link is an ester group (C(=O)-O) formed by losing water.

38. Question 38

    1 marksApparatus for measuring an accurate volume
    Step 1: Measuring 25.0 cm³ to one decimal place needs apparatus with a fine, accurate scale. Step 2: A burette has a precise graduated scale and a tap, so it can deliver an accurate volume such as 25.0 cm³. Step 3: A beaker and conical flask are only for holding liquids, and a measuring cylinder is far less precise, so the burette is the correct choice.
    Method:

    Choose the apparatus whose fine graduations allow an accurate volume to be measured.
    Examiner tips

    • For volumes quoted to 0.1 cm³, use a burette (or pipette), not a beaker or measuring cylinder.

39. Question 39

    1 marksObtaining crystals from a solution
    Step 1: The copper(II) sulfate is dissolved in the solution, so it must be brought out of solution to form crystals. Step 2: Evaporating some of the water makes the solution saturated, and on cooling the dissolved salt forms crystals. Step 3: So the crystals are obtained by evaporation (to crystallisation), not by chromatography, condensation or filtration.
    Method:

    Recall that a dissolved salt is recovered as crystals by evaporation and crystallisation.
    Examiner tips

    • Crystals from a solution: evaporate to a saturated solution, then let it crystallise.

40. Question 40

    1 marksInterpreting a chromatogram with overlapping spots
    Step 1: Each amino acid travels a characteristic distance set by its Rf value. Step 2: If two of the amino acids have the same Rf value, their spots travel the same distance and land in the same place, so they overlap into a single spot. Step 3: That makes four amino acids show as only three spots, so the correct reason is that two of them have the same Rf value.
    Method:

    Use the idea of Rf values to explain how two substances can land at the same height as one spot.
    Examiner tips

    • Overlapping spots in chromatography means two substances share the same Rf value.