0620/22 Oct/Nov 2025 Worked Answers (IGCSE Chemistry 0620 Extended)
40 questions · 40 marks · 45 minutes
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Worked answers for 40 questions
- Step 1: Heating a gas gives its molecules more kinetic energy so they move faster and spread further apart, increasing the volume. Step 2: Reducing the pressure on a gas lets the molecules push outwards, so the gas expands and its volume increases. Step 3: So both increasing the temperature and decreasing the pressure increase the volume of the gas.Method:Treat the two changes separately and recall that both heating and reducing pressure expand a gas.Examiner tips
- Higher temperature or lower pressure both make a gas occupy a larger volume.
- Step 1: Particles are held in fixed positions in a solid and become mobile when the solid melts into a liquid. Step 2: Melting happens at a constant temperature, which is the first flat section of the heating curve, between Q and R. Step 3: The second flat section (S to T) is boiling (liquid to gas), and the sloping sections are just warming, so the change from fixed to mobile is between Q and R.Method:Identify melting as the first flat section of the heating curve and read off the two points.Examiner tips
- First plateau = melting (solid to liquid); second plateau = boiling (liquid to gas).
- Step 1: Lighter gas molecules diffuse faster than heavier ones, so a heavier gas takes longer to travel the same distance. Step 2: Hydrogen chloride (relative molecular mass 36.5) is much heavier than ammonia (17), so it diffuses more slowly and must take longer than the 30 seconds ammonia took. Step 3: Of the options, only 64 seconds is longer than 30 seconds, so the hydrogen chloride takes about 64 seconds.Method:Compare the relative molecular masses, decide which gas is slower, then pick the longer time.Examiner tips
- Heavier molecules diffuse more slowly, so they take more time over the same distance.
- Step 1: An element is made of only one type of atom, so all its atoms are the same. Step 2: A compound has two or more different atoms chemically joined (bonded) together, and can only be split up by chemical processes. Step 3: A mixture is not chemically joined, so it can be separated by physical processes such as filtering or distillation, which matches the first row.Method:Check each of the three definitions against the standard meanings of element, compound and mixture.Examiner tips
- Compounds need chemical processes to separate; mixtures need only physical processes.
- Step 1: The first electron shell can hold up to 2 electrons; hydrogen has just 1 electron, so its first shell is incomplete, while every other atom either fills it or has further shells. Step 2: The mass number is the number of nucleons (protons + neutrons), not nucleons plus protons, and an Al³⁺ ion has lost 3 electrons so its configuration is 2,8, not 2,8,3. Step 3: An atom is mostly empty space with a tiny dense nucleus, so the correct statement is that hydrogen is the only atom with an incomplete first shell.Method:Test each statement against the rules for mass number, ion configuration and atomic structure.Examiner tips
- Hydrogen has one electron in a first shell that can hold two, so the shell is incomplete.
- Step 1: Relative atomic mass is the average mass of the atoms, weighted by how common each isotope is. Step 2: To work it out you multiply the relative mass of each isotope by its relative abundance, add these up and divide by the total abundance. Step 3: So you need the relative mass of each isotope and the relative abundance of each isotope, not bare totals of atoms.Method:Recall the weighted-average calculation and identify the two quantities it requires.Examiner tips
- Relative atomic mass = weighted average of isotope masses using their abundances.
- Step 1: Sodium chloride is an ionic compound made of Na⁺ and Cl⁻ ions. Step 2: These oppositely charged ions are arranged in a regular giant lattice with positive and negative ions alternating. Step 3: It is not metallic (no sea of electrons) and contains ions, not neutral atoms, so the correct description is a lattice of alternating positive and negative ions.Method:Recognise sodium chloride as ionic and pick the alternating-ion lattice description.Examiner tips
- Ionic solids are giant lattices of alternating + and − ions.
- Step 1: In graphite each carbon atom is covalently bonded to 3 others, forming flat layers of hexagons with a spare delocalised electron. Step 2: In diamond each carbon atom is covalently bonded to 4 others in a rigid three-dimensional structure. Step 3: So graphite is 3 and diamond is 4.Method:Recall the layered structure of graphite and the tetrahedral network of diamond.Examiner tips
- Graphite = 3 bonds per carbon (layers); diamond = 4 bonds per carbon (3D network).
- Step 1: Cobalt is a metallic element, so it is held together by metallic bonding (a lattice of positive ions in a sea of delocalised electrons). Step 2: Metals form giant structures, not small simple molecules. Step 3: So cobalt has a giant structure with metallic bonding.Method:Identify cobalt as a metal and recall that metals are giant metallic structures.Examiner tips
- Metals: giant structure + metallic bonding.
- Step 1: 184 g of ethanol is 184 ÷ 46 = 4 mol. Step 2: The equation makes 3 mol of water per mole of ethanol, so 4 mol of ethanol makes 12 mol of water, which is 12 × 18 = 216 g. Step 3: So 184 g of ethanol produces 216 g of water, while the other options give masses that do not match the mole ratios.Method:Convert each mass to moles and check it against the mole ratios in the equation.Examiner tips
- Always convert masses to moles first, then apply the balancing ratios.
- Step 1: Divide each percentage by the relative atomic mass: C 40.0 ÷ 12 = 3.33, H 6.7 ÷ 1 = 6.7, O 53.3 ÷ 16 = 3.33. Step 2: Dividing by the smallest (3.33) gives a ratio of 1 : 2 : 1, so the empirical formula is CH₂O (empirical mass 30). Step 3: The relative molecular mass is 60, which is 2 × 30, so the molecular formula is twice the empirical formula, C₂H₄O₂.Method:Find the simplest mole ratio for the empirical formula, then scale it to the given Mr.Examiner tips
- Empirical formula mass of CH₂O is 30; Mr 60 means multiply by 2.
- Step 1: In a fuel cell hydrogen and oxygen react together to make water and release electrical energy, so the only chemical product is water (statement 3 correct). Step 2: This is the opposite of electrolysis, which uses electricity to split water, so statement 1 is wrong and statement 2 (water being broken down) describes electrolysis, not a fuel cell. Step 3: So only statement 3 is correct.Method:Compare each statement with how a fuel cell works versus electrolysis.Examiner tips
- Fuel cell: H₂ + O₂ → H₂O + electricity; only product is water.
- Step 1: With copper electrodes the copper anode dissolves: copper atoms lose electrons to form Cu²⁺ ions, so statement 2 is correct and no oxygen forms, making statement 1 wrong. Step 2: At the cathode (with either electrode) Cu²⁺ ions gain electrons and copper metal is deposited, so statements 3 and 4 are correct. Step 3: So statements 2, 3 and 4 are correct.Method:Decide the anode behaviour for each electrode type, then the common cathode reaction.Examiner tips
- Reactive (copper) anode dissolves; inert (carbon) anode gives oxygen; cathode always deposits copper here.
- Step 1: Bonds broken (reactants): 4 N–H (4 × 391 = 1564) + 1 N–N (409) + 1 O=O (496) = 2469 kJ. Step 2: Bonds made (products): 1 N≡N (944) + 4 O–H (4 × 463 = 1852) = 2796 kJ. Step 3: Enthalpy change = bonds broken − bonds made = 2469 − 2796 = −327 kJ/mol, which is exothermic.Method:Add up all the bonds broken, all the bonds made, then subtract to get ΔH.Examiner tips
- Count every bond in the molecules carefully; hydrazine has 4 N–H and 1 N–N bond.
- Step 1: In an endothermic change energy is taken in (absorbed) from the surroundings. Step 2: Because the mixture loses energy to the dissolving process, its temperature falls. Step 3: So energy is absorbed and the temperature decreases.Method:Link "endothermic" to energy absorbed and a temperature decrease.Examiner tips
- Endothermic: energy absorbed, temperature of the surroundings drops.
- Step 1: Heating gives the particles more kinetic energy, so they move faster and collide more frequently (statement 1 correct). Step 2: The faster particles also collide with more energy, so more collisions are successful (statement 2 correct). Step 3: Heating does not change the activation energy (only a catalyst lowers it), so statement 3 is wrong, leaving 1 and 2.Method:Recall the collision theory effects of temperature and reject the activation-energy statement.Examiner tips
- Higher temperature: more frequent AND more energetic collisions; activation energy is unchanged.
- Step 1: Both processes run at a moderately high temperature of about 450 °C, so their temperatures are similar (correct). Step 2: Their pressures differ greatly: the Haber process uses about 200 atmospheres while the Contact process uses only a low pressure, so the pressures are not similar. Step 3: The catalysts are the other way round (Haber uses iron, Contact uses vanadium(V) oxide), and the Haber process makes ammonia (an alkaline gas), so the correct statement is that the temperatures are similar.Method:Compare temperature, pressure, catalyst and products of the two processes one at a time.Examiner tips
- Haber and Contact share a similar temperature (~450 °C) but differ in pressure and catalyst.
- Step 1: Increasing the pressure shifts the equilibrium towards the side with fewer gas molecules, which is N₂O₄ (1 molecule) rather than NO₂ (2 molecules), so the mixture becomes paler, not darker; this statement is NOT correct. Step 2: At equilibrium the concentrations are constant and the forward and reverse rates are equal, so those statements are correct. Step 3: Raising the temperature favours the endothermic forward reaction, making more brown NO₂, so that statement is correct, leaving the pressure statement as the false one.Method:Apply the equilibrium rules for pressure and temperature, then find the statement that breaks them.Examiner tips
- Increasing pressure shifts to fewer gas molecules; here that is the colourless N₂O₄ side.
- Step 1: The seven oxygen atoms contribute 7 × (−2) = −14 to the charge. Step 2: The whole ion has a charge of −2, so the two chromium atoms together must contribute +12 (because +12 − 14 = −2). Step 3: Sharing +12 between two chromium atoms gives +6 on each chromium.Method:Balance the oxidation numbers to the ion charge, then divide the chromium total by 2.Examiner tips
- Sum of oxidation numbers = overall ion charge; remember there are TWO chromium atoms.
- Step 1: Lead(II) chloride is insoluble, so it is made by precipitation: mixing two soluble solutions that supply Pb²⁺ ions and Cl⁻ ions, here lead(II) nitrate and hydrochloric acid. Step 2: The insoluble lead(II) chloride forms as a solid (the residue), so it is collected by filtering, then washed and dried. Step 3: The salt is the residue, not the filtrate, so the correct method filters, then washes and dries the residue.Method:Recognise the insoluble salt, choose two soluble reactants, then collect and dry the residue.Examiner tips
- Insoluble salt: mix two soluble solutions, filter, wash and dry the residue.
- Step 1: Phosphorus is a non-metal, so its oxide is a non-metal oxide. Step 2: The oxide dissolves in water to give a pH of 1, which is strongly acidic, so it is an acidic oxide. Step 3: So the oxide is a non-metal oxide and an acidic oxide, not a metal or basic oxide.Method:Use the element type for metal/non-metal and the pH for acidic/basic.Examiner tips
- Non-metal oxides are usually acidic; a low pH confirms an acidic oxide.
- Step 1: Hydrogen bromide gives away an H⁺ ion (a proton), which the ammonia takes to become NH₄⁺. Step 2: A species that gives away a proton is an acid acting as a proton donor. Step 3: So aqueous hydrogen bromide is acting as a proton donor.Method:Track the H⁺ ion: the species that gives it away is the proton donor.Examiner tips
- Acid = proton (H⁺) donor; base = proton acceptor.
- Step 1: Group I metals react with water to form a metal hydroxide and hydrogen gas, for example 2Na + 2H₂O → 2NaOH + H₂. Step 2: Their ions have a 1+ charge (not 1−), and reactivity increases down the group, so potassium is more reactive than sodium. Step 3: Sodium is a metal and conducts electricity well, so the only correct statement is that hydrogen is produced when they react with water.Method:Check each statement against the standard properties of the alkali metals.Examiner tips
- Group I: 1+ ions, reactivity increases down the group, react with water to give hydrogen.
- Step 1: Iodine is a Group VII element that is a solid at room temperature and pressure. Step 2: Solid iodine is a shiny grey-black colour (its vapour is purple, but the solid is grey-black). Step 3: So at room temperature and pressure iodine is a grey-black solid.Method:Recall the trend in Group VII states and colours and place iodine.Examiner tips
- Down Group VII: chlorine (green gas), bromine (red-brown liquid), iodine (grey-black solid).
- Step 1: The inert (noble) gases are in Group VIII (Group 0), the last column on the far right of the Periodic Table. Step 2: Only position C is at the far right-hand edge in the last column. Step 3: So position C is the inert gas.Method:Recall that inert gases are the far-right group and pick the matching position.Examiner tips
- Group VIII (Group 0) on the far right = the noble/inert gases.
- Step 1: Metals whose oxides ARE reduced by carbon (a metal forms) are below carbon and less reactive: Q (no water reaction) and T (a few bubbles) both give a solid metal, so Q is least reactive and T is next. Step 2: Metals whose oxides are NOT reduced by carbon are above carbon and more reactive: S (a few bubbles, oxide unreduced) and R (rapid fizzing). Step 3: R reacts fastest with water so it is the most reactive, giving the order least-first Q, T, S, R.Method:Rank by water-reaction vigour and by whether carbon can reduce each oxide.Examiner tips
- Use both clues: water-reaction vigour and whether carbon reduces the oxide.
- Step 1: The zinc layer coats the iron, keeping out air and water, so it is a barrier method. Step 2: Zinc is more reactive than iron, so even if the layer is scratched the zinc is oxidised in preference to the iron, which is sacrificial protection. Step 3: So it is both a barrier method and sacrificial protection, working because zinc is more reactive than iron.Method:Identify both the physical coating and the sacrificial role, then the reactivity reason.Examiner tips
- Galvanising = barrier + sacrificial protection; zinc is more reactive than iron.
- Step 1: The very high temperature in a blast furnace comes from burning the coke (carbon) in the hot air blast. Step 2: Carbon + oxygen → carbon dioxide is a strongly exothermic combustion reaction that releases a lot of heat. Step 3: The other reactions are either endothermic (decomposing limestone, and carbon dioxide + carbon → carbon monoxide) or slag formation, so the heating reaction is carbon + oxygen → carbon dioxide.Method:Pick the exothermic combustion reaction; the others take in heat or form slag.Examiner tips
- Combustion of coke (C + O₂ → CO₂) is the exothermic reaction that heats the furnace.
- Step 1: Oxidation (loss of electrons) happens at the anode, so the negative O²⁻ ions give up electrons: 2O²⁻ → O₂ + 4e⁻; reduction happens at the cathode, where Al³⁺ + 3e⁻ → Al. Step 2: The oxygen made at the hot carbon anode reacts with it (the anode burns away as carbon dioxide), so the anode wears down. Step 3: So the anode must be replaced regularly because it burns away in the oxygen produced.Method:Assign oxidation to the anode and reduction to the cathode, then explain the anode wearing away.Examiner tips
- Anode = oxidation (O²⁻ → O₂); cathode = reduction (Al³⁺ → Al); carbon anode burns away.
- Step 1: First the water is left to stand so large particles settle out (sedimentation). Step 2: Then it is passed through filters to remove smaller suspended solids (filtration). Step 3: Finally chlorine is added to kill bacteria (chlorination), giving the order sedimentation → filtration → chlorination.Method:Order the steps from removing the largest solids to final disinfection.Examiner tips
- Order: sedimentation → filtration → chlorination.
- Step 1: Oxides of nitrogen form when nitrogen and oxygen from the AIR react together in the heat of the engine, not from incomplete combustion or nitrogen in the fuel. Step 2: In a catalytic converter they react with carbon monoxide, for example 2CO + 2NO → 2CO₂ + N₂, removing both pollutants. Step 3: This produces nitrogen and carbon dioxide (not sulfur dioxide), so the correct statement is that they react with carbon monoxide in catalytic converters.Method:Separate how NOₓ forms from how it is removed, and check the products.Examiner tips
- NOₓ forms from air nitrogen + oxygen; catalytic converters convert it with CO to N₂ + CO₂.
- Step 1: Members of a homologous series share the same functional group; for alkenes this is the carbon-carbon double bond, C=C. Step 2: Because they share the functional group they have similar chemical properties and the same general formula (CₙH₂ₙ for alkenes). Step 3: Their relative molecular masses differ (each member adds CH₂), so the correct statement is that they have the same functional group.Method:Recall the defining features of a homologous series and test each statement.Examiner tips
- Homologous series: same functional group, same general formula, similar chemistry.
- Step 1: Thermal energy leaving the Earth is first absorbed by greenhouse gases in the atmosphere (X = absorption by greenhouse gases). Step 2: The greenhouse gases then re-emit some of this thermal energy back towards the Earth (Y = emission back to the Earth). Step 3: The overall effect is that less thermal energy escapes to space (Z = reduced thermal energy loss to space), which warms the planet.Method:Trace the heat: absorbed by gases, sent back to Earth, so less reaches space.Examiner tips
- Greenhouse effect: absorb outgoing heat → re-emit to Earth → less heat lost to space.
- Step 1: Structural isomers have the SAME molecular formula but a different arrangement of atoms. Step 2: Structures 1 and 3 are both C₄H₁₀ (one unbranched, one branched), so they are structural isomers of each other. Step 3: Structures 2 and 4 have different molecular formulae, so they are not isomers of 1, 3 or each other, leaving 1 and 3 only.Method:Work out each molecular formula and pick the pair that matches with a different arrangement.Examiner tips
- Same molecular formula + different structure = structural isomers.
- Step 1: An ester is named as (alcohol part)yl (acid part)oate. The part attached to the single-bonded oxygen comes from the alcohol: here it is a CH₃ group, so methyl. Step 2: The part containing the C=O carbon comes from the acid: CH₃CH₂COO has three carbons, so it comes from propanoic acid, giving propanoate. Step 3: Combining the two parts gives methyl propanoate.Method:Split the ester at the –COO– link, name the alcohol part then the acid part.Examiner tips
- (alcohol)yl (acid)oate: count carbons on each side of the –COO– link.
- Step 1: Cracking breaks a large alkane into a smaller alkane plus an alkene, so an alkene is always produced. Step 2: Cracking can also produce hydrogen as one of the small molecules. Step 3: No oxygen is involved in cracking, so water is not produced, giving alkene yes, hydrogen yes, water no.Method:Recall what cracking does and check whether each listed product is possible.Examiner tips
- Cracking: large alkane → smaller alkane + alkene (+ sometimes hydrogen); no water.
- Step 1: Fermentation uses glucose (from plants), which is a renewable raw material (advantage), but it is slow and gives only a dilute aqueous solution of ethanol that must be distilled (disadvantage). Step 2: Catalytic addition of steam to ethene is fast and continuous (advantage), but ethene comes from petroleum, a non-renewable raw material (disadvantage). Step 3: So the matching set is fermentation = renewable / dilute product, and catalytic addition = fast / non-renewable.Method:List the pros and cons of each method, then match them to the correct row.Examiner tips
- Fermentation: renewable, slow, dilute, batch. Hydration of ethene: fast, continuous, pure, non-renewable.
- Step 1: In addition polymerisation the C=C double bond of the monomer opens up to form single bonds that link the units together. Step 2: Propenoic acid CH₂=CH–COOH therefore gives a repeat unit with a two-carbon backbone, –CH₂– from one end and –CH(COOH)– from the other, with an open bond at each end. Step 3: So the repeat unit is –CH(COOH)–CH₂–, keeping all the atoms of the monomer but with the double bond now single.Method:Open the double bond of the monomer and draw the two-carbon repeat unit with open ends.Examiner tips
- Addition polymer repeat unit = monomer with the C=C opened to a single bond and open ends.
- Step 1: Measuring 13.7 cm³ accurately needs apparatus with a fine scale that can deliver any chosen volume. Step 2: A burette has a graduated scale to 0.1 cm³ and a tap, so it can deliver an exact variable volume such as 13.7 cm³. Step 3: A balance measures mass, a conical flask is not accurately graduated, and a volumetric pipette only delivers one fixed volume, so the correct choice is a burette.Method:Match the need for a precise, variable volume to the burette.Examiner tips
- Burette = accurate, variable volumes; pipette = one fixed volume only.
- Step 1: Colourless substances leave invisible spots on the chromatogram, so they must first be made visible. Step 2: A locating agent is sprayed on (or the paper is viewed under UV) to react with the spots and make them show up. Step 3: Only after the spots are visible can you measure Rf values, count spots or compare with known substances, so the next step is to use a locating agent.Method:Recognise the spots are invisible and that a locating agent must reveal them next.Examiner tips
- Colourless spots → use a locating agent (or UV) to make them visible first.
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