0620/12 Oct/Nov 2025 Worked Answers (IGCSE Chemistry 0620 Core)

40 questions · 40 marks · 45 minutes

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1. Question 1

   1 marksCompressibility of states of matter
   Step 1: In a gas the particles are spread far apart with large empty spaces between them, so a gas can be squashed into a much smaller volume. Step 2: In solids and liquids the particles are already touching, so they cannot be compressed. Step 3: Steam is the only gas here, so only steam can easily be compressed.
   Method:

   Pick the state whose particles are far apart, because only that state can be compressed.
   Examiner tips

   • Gases are easily compressed; solids and liquids are not.

2. Question 2

   1 marksParticle arrangement and movement in a solid
   Step 1: At 25 °C, which is below the melting point of 113 °C, sulfur is a solid. Step 2: In a solid the particles are packed in a regular, fixed arrangement. Step 3: The particles cannot move around; they only vibrate about fixed positions.
   Method:

   Decide the state from the temperature, then recall the particle arrangement and movement for that state.
   Examiner tips

   • Solid = regular arrangement + particles vibrating about fixed positions.

3. Question 3

   1 marksClassifying elements, compounds and mixtures
   Step 1: Part of Y dissolves to give a blue solution, so Y contains a soluble compound. Step 2: Part of Y is left behind as a brown solid that conducts electricity, so Y also contains an insoluble metal. Step 3: Because Y contains both a soluble compound and an insoluble metal, it must be a mixture that contains a metal.
   Method:

   Identify each part of Y (a soluble compound and a solid metal) and conclude it is a mixture.
   Examiner tips

   • If a substance separates into different parts, it is a mixture.

4. Question 4

   1 marksMass and charge of subatomic particles
   Step 1: A neutron has a relative mass of 1 and carries no charge. Step 2: A proton has a relative mass of 1 but a positive charge, not a negative one. Step 3: An electron has almost no mass (about 1/2000) and a negative charge, so only the statement about neutrons is correct.
   Method:

   Check the mass and charge in each statement against the standard values for the three particles.
   Examiner tips

   • Proton: mass 1, charge +1. Neutron: mass 1, charge 0. Electron: tiny mass, charge -1.

5. Question 5

   1 marksIdentifying particles from electronic configuration
   Step 1: An argon atom (proton number 18) has 18 electrons arranged 2,8,8 and argon-40 has nucleon number 40, so student 1 is correct. Step 2: A chlorine atom has only 17 electrons (2,8,7) and a nucleon number of about 35, so student 2 is wrong. Step 3: Calcium has 20 electrons, but a Ca²⁺ ion has lost 2 to leave 18 electrons (2,8,8), and calcium-40 has nucleon number 40, so student 3 is correct.
   Method:

   Work out the electron count for each suggested particle and compare it with the 2,8,8 arrangement and nucleon number 40.
   Examiner tips

   • A positive ion has fewer electrons than the neutral atom (Ca has 20, Ca2+ has 18).

6. Question 6

   1 marksDefinition of isotopes
   The term for different atoms of the same element that have the same number of protons but different numbers of neutrons is isotopes.
   Method:

   Match the definition of same protons and different neutrons to the correct term.
   Examiner tips

   • Isotopes: same proton number, different neutron number.

7. Question 7

   1 marksFormula of an ionic compound
   Step 1: Q has 2 electrons in its outer shell, so it loses them to form a Q²⁺ ion. Step 2: R has 6 electrons in its outer shell, so it gains 2 electrons to form an R²⁻ ion. Step 3: The 2+ and 2− charges balance in a 1:1 ratio, so the formula is RQ.
   Method:

   Find the ion charges from the outer electrons, then combine them so the charges cancel.
   Examiner tips

   • Group II atoms form 2+ ions; Group VI atoms form 2- ions; equal charges give a 1:1 formula.

8. Question 8

   1 marksProperties of simple molecular substances
   Step 1: Simple molecular substances have weak forces between their molecules, so they have low melting and boiling points (property 3 is correct). Step 2: They have no free electrons or ions, so they do not conduct electricity. Step 3: Malleability is a property of metals, not simple molecules, so only property 3 is typical.
   Method:

   Check each property against the weak-intermolecular-force model of simple molecules.
   Examiner tips

   • Simple molecular substances: low melting/boiling points, no conductivity.

9. Question 9

   1 marksStructure and properties of diamond
   Step 1: In diamond each carbon atom forms four strong covalent bonds, giving a rigid giant structure that is very hard. Step 2: All four outer electrons of every carbon are used in bonding, so there are no delocalised (free) electrons. Step 3: With no free electrons to carry charge, diamond does not conduct electricity.
   Method:

   Link hardness to a rigid network of strong bonds and lack of conductivity to no free electrons.
   Examiner tips

   • Diamond: four covalent bonds per carbon, no free electrons.

10. Question 10

    1 marksBalancing combustion equations
    Step 1: Complete combustion of a hydrocarbon produces only carbon dioxide and water. Step 2: Balancing $2\text{C}_4\text{H}_{10}$ gives 8 carbon dioxide and 10 water molecules. Step 3: This needs 13 oxygen molecules, giving $2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}$.
    Method:

    Recall the products of complete combustion, then choose the correctly balanced equation.
    Examiner tips

    • Complete combustion of a hydrocarbon: products are CO2 and H2O only.

11. Question 11

    1 marksRelative formula mass
    Step 1: Add the relative atomic mass of calcium (40) and of sulfur (32) to give 72. Step 2: There are four oxygen atoms, so add $4 \times 16 = 64$. Step 3: The total is $40 + 32 + 64 = 136$.
    Method:

    Count each type of atom, multiply by its relative atomic mass, then add the totals.
    Examiner tips

    • Multiply each relative atomic mass by the number of that atom, then add them up.

12. Question 12

    1 marksElectrolysis of molten compounds
    Step 1: When a molten binary salt is electrolysed, the metal is produced at the cathode, so silver forms at the cathode. Step 2: The non-metal is produced at the anode, so iodine forms at the anode. Step 3: There is no water present, so no hydrogen or oxygen is formed.
    Method:

    Apply the rule that in a molten salt the metal is discharged at the cathode and the non-metal at the anode.
    Examiner tips

    • Molten binary salt: metal at cathode, non-metal at anode; no hydrogen or oxygen.

13. Question 13

    1 marksElectrolysis of dilute sulfuric acid
    Step 1: Electrolysing dilute sulfuric acid produces oxygen at the anode and hydrogen at the cathode; both gases are colourless. Step 2: Twice as much hydrogen is made as oxygen, so the volume of gas at the cathode is double that at the anode. Step 3: This gives 10 cm³ of colourless gas at the anode and 20 cm³ of colourless gas at the cathode.
    Method:

    Recall the gases and their 1:2 volume ratio, then match the colourless gases to the correct electrodes.
    Examiner tips

    • Dilute sulfuric acid: O2 at anode, H2 at cathode, ratio 1:2, both colourless.

14. Question 14

    1 marksIdentifying the most exothermic reaction
    Step 1: An exothermic reaction releases heat, so the temperature rises; the most exothermic gives the biggest rise. Step 2: Experiments A and C have temperatures that fall, so they are endothermic and can be ignored. Step 3: Experiment B rises from 20 °C to 32 °C (a rise of 12 °C) while experiment D rises by only 9 °C, so B has the largest rise and is the most exothermic.
    Method:

    Find the temperature change for each, keep the rises, and choose the largest rise.
    Examiner tips

    • Most exothermic = largest temperature rise (final minus initial).

15. Question 15

    1 marksDistinguishing physical and chemical change
    Step 1: Dissolving sodium chloride in water makes no new substance and the salt can be recovered by evaporation, so it is a physical change. Step 2: An iron nail in water rusts, forming a new substance, which is a chemical change. Step 3: Zinc with acid and calcium carbonate with acid both produce gases (new substances), so those are chemical changes; only test-tube A is physical.
    Method:

    Check each test-tube for a new substance; the one with none is the physical change.
    Examiner tips

    • Dissolving is physical; reactions that make a gas or new solid are chemical.

16. Question 16

    1 marksApparatus for measuring reaction rate
    Step 1: Measuring a rate always needs timing, so a stop-watch is essential. Step 2: To study the effect of temperature you must measure the temperature, so a thermometer is essential. Step 3: A balance or gas syringe is only useful when the reaction changes mass or makes a gas, so they are not needed for every reaction.
    Method:

    Pick the apparatus that any reaction needs (timing and temperature), not the reaction-specific ones.
    Examiner tips

    • Every rate-vs-temperature experiment needs a clock and a thermometer.

17. Question 17

    1 marksReversible reactions and hydration
    Step 1: Heating turns pink into blue, and adding water turns blue back into pink, so the change goes both ways and is reversible. Step 2: The pink form contains water (it gives off water when heated), so it is hydrated. Step 3: Therefore the reaction is reversible and the pink cobalt(II) chloride is hydrated.
    Method:

    See that the colour change reverses, and note the pink form loses water on heating.
    Examiner tips

    • Hydrated = contains water; a change that goes both ways is reversible.

18. Question 18

    1 marksOxidation number
    Step 1: Each oxygen has an oxidation number of −2, and there are three oxygens, giving a total of −6. Step 2: The compound is neutral, so the two iron atoms must total +6. Step 3: That means each iron atom has an oxidation number of +6 ÷ 2 = +3.
    Method:

    Set the oxygen total, balance it against the iron atoms so the compound is neutral.
    Examiner tips

    • The name iron(III) tells you the oxidation number is +3.

19. Question 19

    1 marksReactions of acids
    Step 1: Acids react with metal oxides, carbonates and hydroxides to form a salt, so copper(II) oxide, copper(II) carbonate and copper(II) hydroxide all give copper(II) chloride. Step 2: Copper metal is below hydrogen in the reactivity series, so it does not react with dilute hydrochloric acid. Step 3: Therefore only substances 2, 3 and 4 react.
    Method:

    Apply the reactions of acids and exclude the unreactive copper metal.
    Examiner tips

    • Acid + metal oxide / carbonate / hydroxide gives a salt; unreactive metals like copper do not react.

20. Question 20

    1 marksUsing indicators to identify an alkali
    Step 1: With X alone, methyl orange is red and thymolphthalein is colourless, which shows X is acidic. Step 2: After adding excess Y, thymolphthalein turns blue and methyl orange turns yellow, which are the colours in an alkaline solution. Step 3: Since Y changes the solution from acidic to alkaline, Y must be an alkali, sodium hydroxide.
    Method:

    Read the indicator colours to find the final solution is alkaline, so Y is the alkali.
    Examiner tips

    • Thymolphthalein is blue and methyl orange is yellow in alkaline solutions.

21. Question 21

    1 marksSeparating a soluble from an insoluble solid
    Step 1: Filtration separates a soluble solid (which dissolves and passes through) from an insoluble solid (which stays on the filter). Step 2: Ammonium chloride is soluble and silver chloride is insoluble, so they can be separated this way. Step 3: The other pairs are either both insoluble (sulfates, hydroxides) or both soluble (nitrates), so filtration cannot separate them.
    Method:

    Find the pair where exactly one solid is soluble and the other insoluble.
    Examiner tips

    • Filtration only separates a soluble solid from an insoluble one.

22. Question 22

    1 marksCharacter and ion charge across the Periodic Table
    Step 1: Metals are found on the left of the Periodic Table; they lose electrons to form positive ions. Step 2: Non-metals are found on the right; they gain electrons to form negative ions. Step 3: So the left is metallic forming positive ions and the right is non-metallic forming negative ions.
    Method:

    Recall where metals and non-metals sit and which ion charge each forms.
    Examiner tips

    • Left = metals, positive ions. Right = non-metals, negative ions.

23. Question 23

    1 marksTrends down Group I
    Step 1: Going down Group I, the density of the metals generally increases. Step 2: The melting point decreases down the group, not increases. Step 3: The reactivity increases down the group and each atom has more electrons, so only the density increasing is correct.
    Method:

    Recall the Group I trends and pick the one that is correct.
    Examiner tips

    • Down Group I the metals get denser and more reactive, but melt at lower temperatures.

24. Question 24

    1 marksAppearance of the halogens
    Step 1: At room temperature and pressure iodine is a grey-black solid. Step 2: Chlorine is a yellow-green gas, not colourless and not a liquid. Step 3: Iodine is a solid at room temperature (it only gives a purple vapour when warmed), so the grey-black solid description is correct.
    Method:

    Match each halogen to its correct state and colour at room temperature.
    Examiner tips

    • Halogen states at r.t.p.: chlorine gas, bromine liquid, iodine solid.

25. Question 25

    1 marksProperties of transition elements
    Step 1: Transition elements have high melting points and high densities. Step 2: They typically form coloured compounds. Step 3: So osmium should have a high melting point, a high density and form coloured compounds.
    Method:

    Apply the standard transition-element properties to osmium.
    Examiner tips

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

26. Question 26

    1 marksProperties of the noble gases
    Step 1: Neon is a noble gas in Group VIII (Group 0), so its atoms have a full outer shell of electrons. Step 2: A full outer shell makes neon very unreactive, so it does not react with oxygen. Step 3: Because it is so stable, neon exists as single atoms, not diatomic molecules.
    Method:

    Use the noble-gas full-outer-shell idea to test each statement.
    Examiner tips

    • Noble gases have full outer shells, so they are unreactive and exist as single atoms.

27. Question 27

    1 marksReactions of a reactive metal
    Step 1: Producing hydrogen with dilute acid and forming basic oxides are reactions of a reactive metal. Step 2: Magnesium is a reactive metal, so it reacts with acid, steam and oxygen in these ways. Step 3: Phosphorus and silicon are non-metals (they form acidic oxides) and silver is too unreactive to react with acid or steam, so G is magnesium.
    Method:

    Use the reactions to show G is a reactive metal, then pick magnesium.
    Examiner tips

    • Reactive metals give hydrogen with acids and form basic oxides; non-metals form acidic oxides.

28. Question 28

    1 marksProperties and uses of metals
    Step 1: Aluminium has a low density and conducts electricity well. Step 2: These two properties make it ideal for overhead power cables, which must be light and conduct. Step 3: The other rows give a wrong property, such as a high density for aluminium or low ductility for copper, so they do not match.
    Method:

    Check each row has both correct properties and a use that fits those properties.
    Examiner tips

    • Aluminium is light and conducts well, which suits overhead power cables.

29. Question 29

    1 marksBarrier methods of rust prevention
    Step 1: A barrier method covers the iron to keep out water and oxygen. Step 2: Coating with plastic, greasing and painting all form a physical layer over the surface, so they are barrier methods. Step 3: Alloying changes the metal itself (for example making stainless steel) rather than forming a surface barrier, so it is not a barrier method.
    Method:

    Pick the methods that physically cover the iron, and exclude alloying.
    Examiner tips

    • Barrier methods (paint, plastic, grease) cover the surface; alloying changes the metal.

30. Question 30

    1 marksThermal decomposition in the blast furnace
    Step 1: Thermal decomposition is when a single compound breaks down into simpler substances on heating. Step 2: Calcium carbonate breaks down on heating into calcium oxide and carbon dioxide, which fits this definition. Step 3: The other reactions involve two reactants combining or a reduction (redox), so they are not thermal decompositions.
    Method:

    Find the reaction with a single reactant breaking apart when heated.
    Examiner tips

    • Thermal decomposition: one compound splits into simpler products on heating.

31. Question 31

    1 marksComposition of clean dry air
    Step 1: Clean, dry air is about 78% nitrogen and 21% oxygen. Step 2: It also contains a small amount of carbon dioxide (about 0.04%). Step 3: Carbon monoxide is a pollutant and is not a normal part of clean air, so the correct row is nitrogen 78%, oxygen 21% and a small amount of carbon dioxide.
    Method:

    Recall the proportions of nitrogen and oxygen and that the trace gas is carbon dioxide.
    Examiner tips

    • Clean dry air: about 78% nitrogen, 21% oxygen, ~0.04% carbon dioxide.

32. Question 32

    1 marksHarmful substances in natural water
    Step 1: Microbes are tiny living organisms such as bacteria. Step 2: Sewage contains harmful microbes that can cause disease. Step 3: Metal compounds, nitrates and phosphates are chemical pollutants, not living microbes, so the answer is sewage.
    Method:

    Separate the living microbes (sewage) from the chemical pollutants.
    Examiner tips

    • Sewage carries disease-causing microbes; the others are chemical pollutants.

33. Question 33

    1 marksCombustion products and pollutant gases
    Step 1: Digestion in animals produces methane, so gas X is methane. Step 2: Burning methane with too little oxygen (incomplete combustion) makes carbon monoxide, so gas Y is carbon monoxide. Step 3: Burning methane with plenty of oxygen (complete combustion) makes carbon dioxide, so gas Z is carbon dioxide.
    Method:

    Identify the digestion gas as methane, then apply incomplete and complete combustion.
    Examiner tips

    • Incomplete combustion gives carbon monoxide; complete combustion gives carbon dioxide.

34. Question 34

    1 marksPhotosynthesis
    Step 1: Photosynthesis uses carbon dioxide and water, with light energy, to make glucose and oxygen. Step 2: It releases oxygen, so it increases (not decreases) the oxygen in the atmosphere. Step 3: It removes carbon dioxide (a greenhouse gas) and does not need such high temperatures, so the correct statement is that it is a reaction between carbon dioxide and water.
    Method:

    Recall the photosynthesis reactants and products to test each statement.
    Examiner tips

    • Photosynthesis: carbon dioxide + water gives glucose + oxygen, using light.

35. Question 35

    1 marksGeneral formulae of homologous series
    Step 1: Decolourising aqueous bromine shows series L is unsaturated, so L is the alkenes with general formula $\text{C}_n\text{H}_{2n}$. Step 2: A pH below 7 shows series M is acidic, so M is the carboxylic acids with general formula $\text{C}_n\text{H}_{2n+1}\text{COOH}$. Step 3: Alcohols ($\text{C}_n\text{H}_{2n+1}\text{OH}$) are neutral and alkanes ($\text{C}_n\text{H}_{2n+2}$) do not decolourise bromine, so the correct pair is $\text{C}_n\text{H}_{2n}$ and $\text{C}_n\text{H}_{2n+1}\text{COOH}$.
    Method:

    Use the bromine test and the pH to identify each series, then give its general formula.
    Examiner tips

    • Alkenes: CnH2n (decolourise bromine). Carboxylic acids: CnH2n+1COOH (acidic).

36. Question 36

    1 marksIdentifying an unsaturated hydrocarbon
    Step 1: A hydrocarbon contains only carbon and hydrogen, which rules out structure B because it contains oxygen. Step 2: Unsaturated means there is a carbon to carbon double bond. Step 3: Only structure D has a carbon to carbon double bond (it is ethene), so structure D is the unsaturated hydrocarbon; A and C are saturated.
    Method:

    Remove any molecule with oxygen, then find the one with a carbon to carbon double bond.
    Examiner tips

    • Look for a C=C double bond in a molecule made only of carbon and hydrogen.

37. Question 37

    1 marksComposition of petroleum
    Step 1: Petroleum is a mixture of hydrocarbons. Step 2: Hydrocarbons are compounds that contain only carbon and hydrogen. Step 3: Oxygen is not present, so the elements in all of these compounds are carbon and hydrogen only.
    Method:

    Recall that petroleum is hydrocarbons, which contain just carbon and hydrogen.
    Examiner tips

    • Hydrocarbons contain only carbon and hydrogen.

38. Question 38

    1 marksEthanol as a fuel
    Step 1: A fuel is a substance that is burned to release useful energy. Step 2: Ethanol releases energy when it is burned, which is exactly why it is chosen as a fuel. Step 3: Being colourless, having a low density, or forming carbon dioxide do not explain why it is useful as a fuel.
    Method:

    Choose the property that describes the energy released when ethanol burns.
    Examiner tips

    • Fuels are used because burning them releases energy.

39. Question 39

    1 marksTests for ions
    Step 1: A white precipitate with acidified silver nitrate shows the chloride ion is present (bromide would give a cream precipitate). Step 2: A green precipitate with sodium hydroxide that dissolves in excess shows chromium(III) ions (iron(II) gives a green precipitate that does not dissolve in excess). Step 3: Putting the two results together, the solid is chromium(III) chloride.
    Method:

    Use the silver nitrate test for the anion and the sodium hydroxide test for the cation.
    Examiner tips

    • Chloride: white precipitate with acidified silver nitrate. Chromium(III): green precipitate, dissolves in excess NaOH.

40. Question 40

    1 marksInterpreting a chromatogram
    Step 1: In chromatography, a dye in the mixture gives a spot at the same height as its pure reference sample. Step 2: W has two spots, one matching P and one matching S, so W contains P and S. Step 3: The spots of Q and R do not line up with any of W's spots, so they are not present; W contains P and S only.
    Method:

    Compare the heights of W's spots with each pure dye and keep only those that match.
    Examiner tips

    • Matching spot heights between the mixture and a pure sample means that dye is present.