An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V1) Cambridge IGCSE Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.
Section Structured Extended Theory
Answer all questions. Show your working where appropriate. Use of a calculator and the printed Periodic Table is permitted.
15 Question · 80 marks
Question 1 · Short Answer
1 marks
State the number of electrons in one sulfide ion, \(\text{S}^{2-}\).
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Worked solution
The atomic number of sulfur is 16, which means a neutral sulfur atom has 16 protons and 16 electrons. A sulfide ion has a \(2-\)\ charge, indicating it has gained 2 electrons. Therefore, the number of electrons is \(16 + 2 = 18\).
Marking scheme
18 [1]
Question 2 · Concept Identification
1 marks
State the physical state of astatine at room temperature and pressure.
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Worked solution
Down Group VII, the physical state of the halogens changes from gaseous (fluorine and chlorine) to liquid (bromine) and then to solid (iodine). Since astatine lies below iodine in the group, it exists as a solid at room temperature and pressure.
Marking scheme
solid [1] (Reject: any other state)
Question 3 · Short Answer
1 marks
Identify the mobile particles that allow copper to conduct electricity when solid.
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Worked solution
In a metallic lattice, the atoms are arranged in a giant structure with positive metal ions surrounded by a sea of delocalised electrons. These delocalised electrons are free to move and carry electric charge through the metal.
Marking scheme
delocalised electrons / mobile electrons [1]
Question 4 · Concept Identification
1 marks
State the term used to describe the minimum energy that colliding particles must have to react.
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Worked solution
Activation energy is defined as the minimum energy that colliding reactant particles must possess in order for a chemical reaction to occur.
Marking scheme
activation energy [1]
Question 5 · Short Answer
1 marks
A reversible reaction is endothermic in the forward direction. State the effect of increasing temperature on the yield of the forward product.
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Worked solution
According to Le Chatelier's principle, increasing the temperature shifts the equilibrium position in the direction that absorbs heat (the endothermic direction). Since the forward reaction is endothermic, the equilibrium shifts to the right, increasing the yield of the forward product.
Marking scheme
increases / yield increases [1]
Question 6 · Concept Identification
1 marks
State the term used to describe compounds with the same molecular formula but different structural formulae.
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Worked solution
Structural isomers are molecules that have the identical molecular formula (the same number of each type of atom) but different structural arrangements of those atoms.
Marking scheme
structural isomers / isomers [1]
Question 7 · Short Answer
1 marks
Identify the separation technique used to collect a precipitate of silver chloride from a mixture of two aqueous solutions.
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Worked solution
Silver chloride is an insoluble salt formed as a precipitate. To separate an insoluble solid from a liquid or solution, the mixture is filtered, leaving the precipitate behind on the filter paper.
Marking scheme
filtration [1]
Question 8 · Short Answer
1 marks
State the name of a greenhouse gas, other than carbon dioxide, that contributes to global warming.
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Worked solution
Methane (and water vapour) is another major greenhouse gas released into the atmosphere that absorbs thermal radiation emitted from the Earth, trapping heat and contributing to the greenhouse effect.
Marking scheme
methane / water vapour [1] (Accept: chemical formula \(\text{CH}_4\) / \(\text{H}_2\text{O}\))
Question 9 · Short Answer
1 marks
State one chemical property of transition elements that is not typically shown by Group II metals.
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Worked solution
Transition elements exhibit several characteristic properties due to their electronic configuration. One key chemical property is that they have variable oxidation states (meaning they can form ions with different charges), whereas Group II metals only form ions with a +2 oxidation state. Other acceptable answers include their ability to act as catalysts or to form coloured compounds.
Marking scheme
Award 1 mark for any one of: - (have) variable oxidation states / variable oxidation numbers - form coloured compounds / form coloured ions - act as catalysts / have catalytic properties
Ignore: physical properties (e.g., high density, high melting point) Reject: reference to reactivity series position alone
Question 10 · Short Answer
1 marks
Explain, in terms of electronic configuration, why neon is chemically unreactive.
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Worked solution
Neon is a noble gas in Group VIII (0). Elements in this group are extremely stable and unreactive because they have a complete outer shell of electrons (specifically, a stable octet of 8 outer electrons), meaning they do not need to gain, lose, or share electrons to achieve a stable configuration.
Marking scheme
Award 1 mark for: - (it has a) full outer shell of electrons / complete outer shell / 8 electrons in its outer shell
Reject: 'it is stable' / 'it is in Group VIII' on its own (must refer to electronic configuration/outer shell)
Question 11 · theory
14 marks
Sulfur dioxide reacts with oxygen to form sulfur trioxide in a reversible reaction. The equation for the reaction is shown: \[2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)} \quad \Delta H = -197\text{ kJ/mol}\]
(a) State and explain the effect of increasing the pressure on: (i) the rate of the forward reaction [2] (ii) the position of equilibrium [2]
(b) State and explain the effect of increasing the temperature on: (i) the rate of the reverse reaction [2] (ii) the yield of sulfur trioxide, \text{SO}_3 [2]
(c) A catalyst of vanadium(V) oxide is used in this reaction. (i) State the effect of vanadium(V) oxide on the position of equilibrium. Explain your answer. [2] (ii) Define the term activation energy. [1] (iii) Explain, in terms of collision theory, how the vanadium(V) oxide catalyst increases the rate of reaction. [3]
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Worked solution
(a) (i) The rate of the forward reaction increases because particles are closer together / more concentrated, resulting in more frequent successful collisions. (ii) The position of equilibrium shifts to the right (the product side) because there are fewer moles of gas on the right-hand side (2 moles) than on the left-hand side (3 moles).
(b) (i) The rate of the reverse reaction increases because particles have more kinetic energy, so they move faster and collide more frequently with energy greater than or equal to the activation energy. (ii) The yield of sulfur trioxide decreases because the forward reaction is exothermic, so increasing the temperature shifts the equilibrium in the endothermic direction (to the left).
(c) (i) No effect on the position of equilibrium. A catalyst increases the rate of both the forward and reverse reactions by the same factor. (ii) Activation energy is the minimum energy colliding particles must have to react. (iii) The catalyst provides an alternative reaction pathway with a lower activation energy, meaning a greater proportion of collisions have energy greater than the activation energy, increasing the rate of successful collisions.
Marking scheme
(a) (i) Rate increases [1]; particles are closer together / more frequent collisions [1]. (ii) Shifts to the right [1]; to the side with fewer gas moles [1]. (a) (b) (i) Rate increases [1]; particles have more kinetic energy / more frequent collisions [1]. (ii) Yield decreases [1]; equilibrium shifts in endothermic direction [1]. (c) (i) No effect [1]; increases forward and reverse rates equally [1]. (ii) Minimum energy required to react [1]. (iii) Alternative pathway with lower activation energy [1]; more particles have energy greater than activation energy [1]; higher frequency of successful collisions [1].
Question 12 · theory
14 marks
Lead(II) iodide, \text{PbI}_2, is an insoluble salt prepared by a precipitation reaction.
(a) (i) Suggest the names of two soluble salts that can be reacted together to prepare a pure, dry sample of lead(II) iodide. [2] (ii) Write a balanced chemical equation, including state symbols, for the reaction between the two soluble salts you named in (a)(i). [3] (iii) Write the ionic equation, including state symbols, for this precipitation reaction. [2]
(b) Describe a step-by-step method to prepare a pure, dry sample of lead(II) iodide from the mixture of the two solutions named in (a)(i). [5]
(c) State the observations when aqueous sodium hydroxide is added dropwise and then in excess to an aqueous solution containing lead(II) ions. [2]
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(b) Step 1: Mix the two solutions together to form the precipitate. Step 2: Filter the mixture to separate the precipitate (residue) of lead(II) iodide from the solution. Step 3: Wash the residue with distilled water to remove any soluble impurities. Step 4: Dry the precipitate using filter paper or in a warm oven.
(c) Dropwise: A white precipitate forms. In excess: The white precipitate dissolves to form a colourless solution.
Marking scheme
(a) (i) Lead(II) nitrate [1] and potassium iodide / sodium iodide [1] (Accept any other soluble iodide). (ii) Correct formulas [1], balanced [1], state symbols correct [1]. (iii) Correct reactant and product ions [1], state symbols correct [1]. (b) Mix solutions [1]; Filter [1]; Wash residue with distilled water [1]; Dry with filter paper / in warm oven [1]; Quality of description / logical sequence [1]. (c) Dropwise: white precipitate [1]; Excess: precipitate dissolves / soluble in excess to form colourless solution [1].
Question 13 · theory
14 marks
This question is about the Group VII elements (halogens) and their displacement reactions.
(a) Describe the trend in the colour and physical state of the Group VII halogens down the group from fluorine to iodine at room temperature and pressure. [3]
(b) Chlorine gas is bubbled into an aqueous solution of potassium bromide. (i) State the observation for this reaction. [1] (ii) Write a balanced chemical equation for this reaction. [2] (iii) Identify the species that is oxidised and explain your answer in terms of electron transfer. [3] (iv) Explain why no reaction occurs when aqueous iodine is added to an aqueous solution of sodium chloride. [2]
(c) Describe the test used to identify halide ions in separate aqueous solutions of chloride, bromide, and iodide. State the expected observation for each ion. [3]
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Worked solution
(a) Down the group, the halogens become darker in colour (from pale yellow gas fluorine to yellow-green gas chlorine, red-brown liquid bromine, and grey-black solid iodine) and the physical state changes from gas to liquid to solid.
(b) (i) The solution turns orange-brown. (ii) \(\text{Cl}_2\text{(g)} + 2\text{KBr(aq)} \rightarrow 2\text{KCl(aq)} + \text{Br}_2\text{(aq)}\) (iii) The bromide ion (\(\text{Br}^-\)) is oxidised because it loses electrons to form bromine molecules (\(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\)). (iv) Iodine is less reactive than chlorine, so it cannot displace chloride ions from sodium chloride.
(c) Add dilute nitric acid followed by aqueous silver nitrate. Chloride ions give a white precipitate; bromide ions give a cream precipitate; iodide ions give a yellow precipitate.
Marking scheme
(a) State changes from gas to liquid to solid [1]; Colour gets darker down the group [1]; Mention of specific colours/states (e.g. chlorine gas, bromine liquid, iodine solid) [1]. (b) (i) Turn orange / brown / orange-brown [1]. (ii) Correct formulas [1]; balanced [1]. (iii) Bromide / \(\text{Br}^-\)[1]; oxidation is loss of electrons [1]; bromide loses electrons to form bromine [1]. (iv) Iodine is less reactive than chlorine [1]; cannot displace it [1]. (c) Add dilute nitric acid and silver nitrate [1]; White ppt with chloride, cream ppt with bromide [1]; Yellow ppt with iodide [1].
Question 14 · theory
14 marks
This question is about chemical formulae, calculations, and gas stoichiometry.
(a) Hydrated iron(II) sulfate crystals have the formula \(\text{FeSO}_4 \cdot x\text{H}_2\text{O}\). A student heated 5.56 g of these crystals to constant mass. 3.04 g of anhydrous iron(II) sulfate, \(\text{FeSO}_4\), remained. (i) Calculate the mass of water lost from the crystals. [1] (ii) Calculate the number of moles of anhydrous \(\text{FeSO}_4\) formed. (\(M_r\): \(\text{FeSO}_4 = 152\)) [1] (iii) Calculate the number of moles of water lost. (\(M_r\): \(\text{H}_2\text{O} = 18\)) [1] (iv) Determine the value of \(x\) in \(\text{FeSO}_4 \cdot x\text{H}_2\text{O}\). [1]
(b) Hydrocarbon Y contains 85.7% carbon and 14.3% hydrogen by mass. (i) Calculate the empirical formula of hydrocarbon Y. [3] (ii) The relative molecular mass, \(M_r\), of Y is 56. Deduce the molecular formula of Y. [2]
(c) Nitrogen monoxide reacts with oxygen as shown: \(2\text{NO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{NO}_2\text{(g)}\). (i) Calculate the volume of oxygen gas, in \(\text{dm}^3\) (at r.t.p.), required to react completely with 12.0 \(\text{dm}^3\) of nitrogen monoxide. [2] (ii) Calculate the mass of nitrogen dioxide formed when 12.0 \(\text{dm}^3\) of nitrogen monoxide reacts completely. (Assume 1 mole of gas occupies 24 \(\text{dm}^3\) at r.t.p.; \(M_r\): \(\text{NO}_2 = 46\)) [3]
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Worked solution
(a) (i) Mass of water lost = \(5.56\text{ g} - 3.04\text{ g} = 2.52\text{ g}\). (ii) Moles of \(\text{FeSO}_4 = 3.04 / 152 = 0.02\text{ mol}\). (iii) Moles of \(\text{H}_2\text{O} = 2.52 / 18 = 0.14\text{ mol}\). (iv) Ratio \(x = 0.14 / 0.02 = 7\). Thus, \(x = 7\).
(b) (i) Carbon: \(85.7 / 12 = 7.14\text{ mol}\). Hydrogen: \(14.3 / 1 = 14.3\text{ mol}\). Ratio: Carbon = \(7.14 / 7.14 = 1\); Hydrogen = \(14.3 / 7.14 = 2\). Empirical formula = \(\text{CH}_2\). (ii) Empirical formula mass of \(\text{CH}_2 = 12 + (2 \times 1) = 14\). Factor = \(56 / 14 = 4\). Molecular formula = \(\text{C}_4\text{H}_8\).
(c) (i) According to the equation, 2 moles of \(\text{NO}\) react with 1 mole of \(\text{O}_2\). Therefore, volume of \(\text{O}_2 = 12.0 / 2 = 6.0\text{ dm}^3\). (ii) Moles of \(\text{NO} = 12.0 / 24 = 0.5\text{ mol}\). Moles of \(\text{NO}_2\) formed = 0.5 mol. Mass of \(\text{NO}_2\) formed = \(0.5 \times 46 = 23.0\text{ g}\).
Marking scheme
(a) (i) 2.52 g [1]. (ii) 0.02 mol [1]. (iii) 0.14 mol [1]. (iv) x = 7 [1]. (b) (i) Dividing percentages by relative atomic masses (7.14 and 14.3) [1]; finding simplest ratio (1:2) [1]; empirical formula \(\text{CH}_2\) [1]. (ii) Empirical formula mass = 14 [1]; molecular formula \(\text{C}_4\text{H}_8\) [1]. (c) (i) Mole ratio from equation is 2:1 [1]; volume = 6.0 \(\text{dm}^3\) [1]. (ii) Moles of \(\text{NO}\) (0.5 mol) [1]; moles of \(\text{NO}_2\) (0.5 mol) [1]; mass of \(\text{NO}_2\) (23.0 g) [1].
Question 15 · theory
14 marks
This question is about alcohols, carboxylic acids, and esters.
(a) Ethanol, \(\text{C}_2\text{H}_5\text{OH}\), can be manufactured industrially by two different methods. (i) State the reagent and catalyst required for the catalytic hydration of ethene to produce ethanol. [2] (ii) Name the alternative process that produces ethanol from glucose using yeast. [1] (iii) State one advantage of the catalytic hydration of ethene compared to fermentation. [1] (iv) State one advantage of fermentation compared to the catalytic hydration of ethene. [1]
(b) Ethanol can be oxidised to ethanoic acid. (i) Name a suitable oxidising agent for this reaction and state the colour change observed. [2] (ii) Write the chemical equation for this oxidation. Use [O] to represent oxygen from the oxidising agent. [2]
(c) Ethanoic acid reacts with propan-1-ol in the presence of an acid catalyst to form an ester. (i) Name the ester formed and write its molecular formula. [2] (ii) Draw the displayed formula of this ester, showing all atoms and all bonds. [3]
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Worked solution
(a) (i) Reagent: steam (water). Catalyst: phosphoric acid (acid catalyst). (ii) Fermentation. (iii) It is a continuous / fast process / produces pure ethanol. (iv) It uses renewable resources (glucose/sugar) / uses lower temperature/energy.
(b) (i) Acidified potassium manganate(VII) / acidified potassium dichromate(VI). For manganate(VII), the colour change is purple to colourless. (For dichromate(VI), orange to green). (ii) \(\text{C}_2\text{H}_5\text{OH} + 2[\text{O}] \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}\)
(c) (i) Propyl ethanoate. Molecular formula: \(\text{C}_5\text{H}_{10}\text{O}_2\). (ii) The displayed formula shows all covalent bonds: \(\text{CH}_3-\text{C}(=\text{O})-\text{O}-\text{CH}_2-\text{CH}_2-\text{CH}_3\). Specifically: H O H H H | || | | | H-C - C - O-C - C - C-H | | | | H H H H
Marking scheme
(a) (i) Steam [1] and phosphoric acid catalyst [1]. (ii) Fermentation [1]. (iii) Fast / continuous / produces pure product [1]. (iv) Renewable resources / low temperature/pressure/energy [1]. (b) (i) Acidified potassium manganate(VII) [1] and purple to colourless [1] (or acidified potassium dichromate(VI) and orange to green). (ii) Correct reactants [1], correct products [1]. (c) (i) Propyl ethanoate [1]; \(\text{C}_5\text{H}_{10}\text{O}_2\) [1]. (ii) Correct ester linkage (\(-\text{C}(=\text{O})-\text{O}-\)) shown [1]; correct propyl group [1]; correct methyl group and all bonds/atoms shown [1].
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