IGCSE Chemistry 0620 Paper 4 (Extended) practice paper — 10 questions, 100 marks, 60 minutes

2026-10-31

IGCSE Chemistry 0620 Paper 4

IGCSE Chemistry 0620 Paper 4 (Extended) practice paper: 10 questions, 100 marks, 60 minutes, on atomic structure, bonding, moles, electrolysis, rates, redox, organic chemistry and ion tests, with worked answers and marking notes.

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Paper details　Level: Grade 9–10 (IGCSE) | Subject: Chemistry (0620) | Time: 60 minutes | Marks: 100 | Questions: 10

Question 1 — Atomic structure and isotopes (10 marks)

1. Answer the following.

#	Question	Answer and marking notes
1	Copper has two natural isotopes: ⁶³Cu with an abundance of 69.2% and ⁶⁵Cu with 30.8%. Calculate the relative atomic mass of copper, giving your answer to 3 significant figures.	63 × 0.692 + 65 × 0.308 = 43.596 + 20.020 = 63.616, so 63.6. Marking: 2 marks for the two products, 1 mark for the sum given to 3 s.f.
2	State the number of protons, neutrons and electrons in a ⁶³Cu atom.	Protons 29; neutrons 63 − 29 = 34; electrons 29 (equal to protons in a neutral atom). Marking: 1 mark each.
3	The electron arrangement of copper is 2, 8, 18, 1. Explain why copper is in Period 4 and why it is a transition element.	Four occupied shells place it in Period 4; the 3d subshell is partly used in bonding (d-block), so it is a transition element. Marking: 2 marks for the period (must mention shells), 2 marks for transition element.

Question 2 — Bonding and structure (10 marks)

2. Compare structures and properties.

#	Question	Answer and marking notes
4	Sodium chloride melts at 801 °C while iodine melts at 114 °C. Explain the large difference.	NaCl is a giant ionic lattice held by strong electrostatic attraction, so melting requires breaking many ionic bonds; iodine is a simple molecular substance with only weak van der Waals forces between molecules, so little energy is needed. Marking: 2 marks NaCl, 2 marks iodine.
5	Potassium reacts with oxygen to form a compound. Describe the electron transfer and give the formula.	Potassium (2, 8, 8, 1) loses one electron to form K⁺; oxygen (2, 6) gains two electrons to form O²⁻; two K⁺ balance one O²⁻, so the formula is K₂O. Marking: 1 mark ions, 2 marks formula.
6	Carbon dioxide sublimes at −78 °C, but silicon dioxide melts above 1,700 °C. Explain why.	CO₂ is a simple molecular substance held by weak van der Waals forces which are easily overcome; SiO₂ is a giant covalent structure in which each silicon atom is bonded to four oxygen atoms, so melting requires breaking many strong covalent bonds. Marking: 1 mark CO₂, 2 marks SiO₂.

Question 3 — Moles and gas volume (12 marks)

3. Relative atomic masses: C 12, O 16, Ca 40; molar gas volume at r.t.p. 24 dm³ mol⁻¹.

#	Question	Answer and marking notes
7	Calculate the number of moles in 25.0 g of calcium carbonate (CaCO₃).	Mr(CaCO₃) = 40 + 12 + 3 × 16 = 100; n = 25.0 ÷ 100 = 0.250 mol. Marking: 1 mark Mr, 1 mark division, 1 mark unit.
8	Calcium carbonate decomposes on heating into calcium oxide and carbon dioxide. Write the balanced equation and state the moles of carbon dioxide formed from 0.250 mol of calcium carbonate.	CaCO₃ → CaO + CO₂; the ratio is 1 : 1, so 0.250 mol of CO₂. Marking: 2 marks equation, 1 mark ratio, 1 mark answer.
9	What volume (dm³) does this carbon dioxide occupy at room temperature and pressure?	V = n × 24 = 0.250 × 24 = 6.00 dm³. Marking: 2 marks formula, 2 marks value, 1 mark unit.

Question 4 — Electrolysis calculations (10 marks)

4. Copper(II) sulphate solution is electrolysed at 2.0 A for 30 minutes (F = 96500 C mol⁻¹, Cu = 63.5).

#	Question	Answer and marking notes
10	Calculate the charge passed, in coulombs.	t = 30 × 60 = 1800 s; Q = It = 2.0 × 1800 = 3600 C. Marking: 2 marks time conversion, 1 mark Q = It.
11	Calculate the number of moles of electrons passed.	n(e⁻) = Q ÷ F = 3600 ÷ 96500 = 0.0373 mol (0.037 mol). Marking: 2 marks formula, 2 marks answer.
12	The cathode reaction is Cu²⁺ + 2e⁻ → Cu. Calculate the mass of copper deposited at the cathode.	n(Cu) = 0.0373 ÷ 2 = 0.0187 mol; m = 0.0187 × 63.5 = 1.18 g. Marking: 1 mark 2 : 1 electron ratio, 2 marks mass.

Question 5 — Rates of reaction (10 marks)

5. 2.0 g of marble chips is added to 50 cm³ of excess dilute hydrochloric acid and the gas is collected.

#	Question	Answer and marking notes
13	At room temperature, 30 cm³ of gas is collected in 40 seconds. Calculate the average rate in cm³ s⁻¹.	30 ÷ 40 = 0.75 cm³ s⁻¹. Marking: 2 marks expression, 1 mark answer with unit.
14	The experiment is repeated with the same mass of powdered marble, all else unchanged. How does the rate change? Explain.	The rate increases; powder has a larger surface area, so acid particles collide with the solid more often and the collision frequency rises. Marking: 2 marks direction, 2 marks reason.
15	State two other ways to increase the rate of this reaction (not surface area; a catalyst scores no mark because none applies here).	Raise the temperature; increase the concentration of the acid. Marking: 1.5 marks each, 3 marks total.

Question 6 — Neutralisation calculations (8 marks)

6. Hydrochloric acid is titrated with sodium hydroxide solution: NaOH + HCl → NaCl + H₂O.

#	Question	Answer and marking notes
16	How many moles of HCl are needed to neutralise 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution?	n(NaOH) = 0.100 × 0.0250 = 0.00250 mol; the ratio is 1 : 1, so 0.00250 mol of HCl. Marking: 2 marks moles of NaOH, 2 marks ratio.
17	If the acid used is 0.125 mol dm⁻³, what volume in cm³ is needed?	V = n ÷ c = 0.00250 ÷ 0.125 = 0.0200 dm³ = 20.0 cm³. Marking: 3 marks calculation, 1 mark dm³ to cm³.

Question 7 — Oxidation and reduction (10 marks)

7. In acidic solution, K₂Cr₂O₇ reacts with iron(II) ions.

#	Question	Answer and marking notes
18	Determine the oxidation number of chromium in K₂Cr₂O₇.	Let x be the oxidation number of Cr: 2(+1) + 2x + 7(−2) = 0, so 2x = 12 and x = +6. Marking: 2 marks equation, 2 marks answer.
19	Cr₂O₇²⁻ is reduced to Cr³⁺. State the change in oxidation number of chromium and say whether K₂Cr₂O₇ is the oxidising or the reducing agent.	It falls from +6 to +3 (each Cr gains 3 electrons); K₂Cr₂O₇ is itself reduced, so it is the oxidising agent. Marking: 2 marks change, 1 mark oxidising agent.
20	Write the half-equation for the oxidation of iron(II) to iron(III).	Fe²⁺ → Fe³⁺ + e⁻. Marking: 2 marks species, 1 mark electron.

Question 8 — Organic chemistry (12 marks)

8. Propane (C₃H₈) and propene (C₃H₆) are both three-carbon hydrocarbons.

#	Question	Answer and marking notes
21	Write the balanced equation for the complete combustion of propane.	C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Check: C 3 = 3, H 8 = 8, O 10 = 6 + 4. Marking: 4 marks for all coefficients correct, 1 mark deducted per error.
22	0.100 mol of propane burns completely. What volume of carbon dioxide is formed at r.t.p. (24 dm³ mol⁻¹)?	n(CO₂) = 3 × 0.100 = 0.300 mol; V = 0.300 × 24 = 7.20 dm³. Marking: 2 marks ratio, 2 marks volume with unit.
23	Propene decolourises bromine water. Name the product and the type of reaction.	1,2-dibromopropane is formed; this is an addition reaction (the double bond opens and one bromine atom adds to each carbon). Marking: 2 marks product, 2 marks reaction type.

Question 9 — Polymers (8 marks)

9. Compare addition and condensation polymerisation.

#	Question	Answer and marking notes
24	Ethene polymerises to poly(ethene). Draw the repeat unit and state which type of polymerisation occurs.	The repeat unit is —CH₂—CH₂—; no small molecule is eliminated, so it is addition polymerisation. Marking: 2 marks repeat unit, 2 marks type.
25	Polyesters such as PET are made from a diol and a dicarboxylic acid. Name this type of polymerisation and state the small molecule eliminated.	Condensation polymerisation; one water molecule (H₂O) is eliminated for each ester link formed. Marking: 2 marks name, 2 marks small molecule.

Question 10 — Tests for ions (10 marks)

10. Use flame tests and precipitation reactions to identify ions.

#	Question	Answer and marking notes
26	A potassium salt gives a lilac flame through cobalt glass and a sodium salt gives a yellow flame. Name the metal ions involved.	Lilac → K⁺; yellow → Na⁺. Marking: 2 marks K⁺, 2 marks Na⁺.
27	How would you test a solution for sulphate ions (SO₄²⁻)? Give the reagent and the observation.	Acidify with dilute hydrochloric acid (to remove carbonate interference), then add barium chloride solution; a white precipitate (BaSO₄) forms. Marking: 2 marks reagent, 1 mark observation.
28	How could one reagent distinguish a chloride solution from an iodide solution? Give the reagent and both observations.	Add silver nitrate solution: chloride gives a white precipitate (AgCl) and iodide gives a yellow precipitate (AgI). Marking: 2 marks reagent, 1 mark observations.

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