0620/41 Oct/Nov 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

39 questions · 80 marks · 75 minutes

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1. Question 1(a)

   2 marksRelative mass and charge of subatomic particles
   Step 1: A neutron has almost the same mass as a proton, so its relative mass is 1, and it carries no charge, so its relative charge is 0. Step 2: An electron is far lighter, with a relative mass of about $\tfrac{1}{1840}$, and it carries one negative charge, so its relative charge is -1. Step 3: Only the first row gives both of these correctly.
   Method:

   Recall the relative masses and charges of the three subatomic particles.
   Examiner tips

   • Protons and neutrons each have relative mass 1; the electron is ~1/1840.

2. Question 1(b)(i)

   2 marksDefinition of isotopes
   Step 1: Isotopes are atoms of the same element, so they all have the same number of protons (the same proton number defines the element). Step 2: They differ only in the number of neutrons in the nucleus, which changes the nucleon number. Step 3: So isotopes have the same number of protons but different numbers of neutrons.
   Method:

   State what is the same (protons) and what differs (neutrons) between isotopes.
   Examiner tips

   • Same proton number, different neutron number.

3. Question 1(b)(ii)

   1 marksWhy isotopes have the same chemical properties
   Step 1: Chemical properties depend on how an atom's outer electrons take part in bonding. Step 2: Isotopes of an element have the same number of protons and therefore the same number of electrons, giving the same electronic configuration. Step 3: With identical outer-electron arrangements they react in the same way, so they have the same chemical properties.
   Method:

   Link chemical properties to the electronic configuration, which is set by electron number.
   Examiner tips

   • Same electrons -> same electronic configuration -> same chemistry.

4. Question 1(c)

   5 marksIdentifying a particle from its subatomic counts
   Step 1: The number of protons is the proton number, and 22 protons means the element is titanium (Ti). Step 2: The nucleon number is protons + neutrons = 22 + 28 = 50. Step 3: There are 2 fewer electrons (20) than protons (22), so the particle has lost 2 electrons and carries a 2+ charge, giving $\text{Ti}^{2+}$.
   Method:

   Use proton number to name the element, then work out charge and nucleon number.
   Examiner tips

   • Element from proton count; charge from (protons - electrons); nucleon number from (protons + neutrons).

5. Question 1(d)

   1 marksTerm for total protons and neutrons
   Nucleon number means the total number of protons and neutrons (nucleons) in the nucleus, so it is another name for mass number.
   Method:

   Recall the alternative term for mass number.
   Examiner tips

   • Nucleon number = mass number = protons + neutrons.

6. Question 1(e)

   2 marksNumber of atoms from a mass
   Step 1: Number of moles of argon = mass / $M_r$ = 2.00 / 40 = 0.0500 mol. Step 2: Number of atoms = moles x Avogadro constant = 0.0500 x $6.02\times10^{23}$. Step 3: This gives $3.01\times10^{22}$ atoms.
   Method:

   Find moles from the mass, then scale by the Avogadro constant.
   Examiner tips

   • Always convert to moles first, then multiply by the Avogadro constant.

7. Question 2(a)

   1 marksGroup II element from number of electron shells
   Step 1: The number of occupied electron shells equals the period (row) number in the Periodic Table. Step 2: Five occupied shells means the element is in Period 5. Step 3: The Group II element in Period 5 is strontium, so strontium is the answer.
   Method:

   Match the number of shells to the period, then read off the Group II element.
   Examiner tips

   • Period number = number of occupied electron shells.

8. Question 2(b)

   4 marksBonding in Group II metals
   Step 1: Group II elements are metals, so they have metallic bonding. Step 2: In a metal the atoms form a regular lattice of positive ions and the outer electrons become delocalised, forming a sea of mobile electrons. Step 3: The bond is the electrostatic attraction between the positive metal ions and the sea of delocalised electrons.
   Method:

   Name the bond as metallic, then describe the ions and delocalised electrons.
   Examiner tips

   • Metallic bond = lattice of cations + sea of delocalised electrons.

9. Question 2(c)(i)

   3 marksObservations when calcium reacts with cold water
   Step 1: Calcium reacts with water to release hydrogen gas, so bubbles (effervescence) are seen. Step 2: The calcium is used up in the reaction, so the solid dissolves and gets smaller. Step 3: The other product is calcium hydroxide, an alkali, so the universal indicator turns blue/purple.
   Method:

   Recall the products (hydrogen + alkali) and describe what each would look like.
   Examiner tips

   • Look for gas bubbles, a shrinking solid, and an alkaline colour change.

10. Question 2(c)(ii)

    2 marksProducts of calcium reacting with water
    Step 1: A reactive metal reacting with water gives a metal hydroxide and hydrogen gas. Step 2: For calcium this gives calcium hydroxide and hydrogen. Step 3: So the two products are calcium hydroxide and hydrogen.
    Method:

    Apply the general metal + water reaction to calcium.
    Examiner tips

    • Reactive metal + water always gives the hydroxide plus hydrogen.

11. Question 2(d)(i)

    1 marksFlame colour when calcium burns
    Step 1: Burning calcium produces a characteristic brick-red, orange-red flame. Step 2: Lilac, green and yellow flames belong to potassium, copper and sodium respectively. Step 3: So calcium burns with an orange-red flame.
    Method:

    Recall the flame colour associated with burning calcium.
    Examiner tips

    • Calcium = orange-red / brick-red flame.

12. Question 2(d)(ii)

    2 marksSymbol equation for calcium burning in oxygen
    Step 1: Calcium forms $\text{Ca}^{2+}$ and oxygen forms $\text{O}^{2-}$, so calcium oxide is CaO. Step 2: Oxygen gas is $\text{O}_2$, so two calcium atoms are needed to use both oxygen atoms. Step 3: Balancing gives $2\text{Ca} + \text{O}_2 \rightarrow 2\text{CaO}$.
    Method:

    Deduce the formula of the oxide from the ion charges, then balance.
    Examiner tips

    • Write CaO first, then balance using O2.

13. Question 2(e)(i)

    1 marksWater in hydrated crystals
    Water held within a crystal structure in a fixed ratio is called water of crystallisation.
    Method:

    Recall the term for water built into a crystal structure.
    Examiner tips

    • Hydrated salt crystals hold water of crystallisation.

14. Question 2(e)(ii)

    3 marksFinding x in a hydrated salt formula
    Step 1: Moles of $\text{Ca(NO}_3)_2$ = 3.28 / 164 = 0.0200 mol. Step 2: Moles of water = 1.44 / 18 = 0.0800 mol. Step 3: x = moles of water / moles of salt = 0.0800 / 0.0200 = 4.
    Method:

    Convert each mass to moles, then divide water moles by salt moles.
    Examiner tips

    • Find moles of each substance, then take the ratio water:salt.

15. Question 3(a)

    2 marksCompleting a neutralisation equation with state symbols
    Step 1: Acid + alkali gives a salt plus water, so the products are sodium sulfate and water. Step 2: The sodium sulfate stays dissolved, so it is (aq), and the water is a liquid, so it is (l). Step 3: Two water molecules are needed to balance the hydrogen and oxygen, giving $\text{Na}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})$.
    Method:

    Identify salt + water as products, balance, then add the correct state symbols.
    Examiner tips

    • Neutralisation: salt (aq) + water (l).

16. Question 3(b)

    1 marksType of exothermic reaction between acid and alkali
    A reaction between an acid and an alkali to form a salt and water is called neutralisation.
    Method:

    Name the reaction type for an acid reacting with an alkali.
    Examiner tips

    • Acid neutralises alkali -> neutralisation.

17. Question 3(c)

    1 marksConcentration from moles and volume
    Step 1: Concentration = moles / volume in $\text{dm}^3$, and 40.0 $\text{cm}^3$ = 0.0400 $\text{dm}^3$. Step 2: Concentration = 0.0100 / 0.0400. Step 3: This gives 0.25 $\text{mol/dm}^3$.
    Method:

    Convert the volume to dm^3 and divide the moles by it.
    Examiner tips

    • Concentration (mol/dm^3) = moles / (volume in cm^3 / 1000).

18. Question 3(d)

    1 marksApparatus for adding acid in a titration
    A burette is used to add an accurately measured, variable volume of acid during a titration.
    Method:

    Recall the standard titration apparatus for delivering acid.
    Examiner tips

    • Acid is delivered from a burette in a titration.

19. Question 3(e)

    2 marksVolume of acid from a titration calculation
    Step 1: From the equation, 2 mol NaOH react with 1 mol $\text{H}_2\text{SO}_4$, so moles of acid = 0.0100 / 2 = 0.00500 mol. Step 2: Volume in $\text{dm}^3$ = moles / concentration = 0.00500 / 0.200 = 0.0250 $\text{dm}^3$. Step 3: Converting to $\text{cm}^3$ gives 25 $\text{cm}^3$.
    Method:

    Apply the mole ratio, find acid moles, then convert moles and concentration to a volume.
    Examiner tips

    • Account for the 2:1 mole ratio before dividing by concentration.

20. Question 3(f)

    2 marksColour change of methyl orange at the end-point
    Step 1: Methyl orange is yellow in an alkali, so it starts yellow in the sodium hydroxide. Step 2: At the end-point the solution becomes neutral/just acidic and methyl orange turns orange. Step 3: So the colour change is from yellow to orange.
    Method:

    Recall the methyl orange colours in alkali and acid, then state the change.
    Examiner tips

    • Methyl orange is yellow (alkali) -> orange/red (acid).

21. Question 3(g)

    1 marksObtaining white salt crystals after a titration
    Step 1: The crystals are coloured because the indicator stays in the solution and is left behind on evaporation. Step 2: Once the correct acid volume is known from the titration, the reaction can be repeated with those exact volumes but no indicator. Step 3: Evaporating this indicator-free solution gives white crystals.
    Method:

    Explain that the indicator causes the colour, so repeat without it using the measured volumes.
    Examiner tips

    • Find the titre first, then repeat with the same volumes and no indicator.

22. Question 3(h)(i)

    2 marksMeaning of a saturated solution
    Step 1: A saturated solution contains as much dissolved solute as it can hold at a particular temperature. Step 2: Because it already holds the maximum amount, no further solute can dissolve at that temperature. Step 3: So a saturated solution is one with the maximum concentration of solute dissolved at that temperature.
    Method:

    Define saturation in terms of maximum dissolved solute at a fixed temperature.
    Examiner tips

    • Saturated means no more solute will dissolve at that temperature.

23. Question 3(h)(ii)

    1 marksWhy crystals form as a hot solution cools
    Step 1: The amount of solute that can dissolve (its solubility) decreases as the temperature falls. Step 2: The cooling solution can no longer hold all the dissolved solute, so the excess separates out. Step 3: This excess solute appears as crystals.
    Method:

    Link falling temperature to falling solubility and excess solute crystallising.
    Examiner tips

    • Solubility falls with temperature, so cooling produces crystals.

24. Question 3(h)(iii)

    1 marksEffect of drying without gentle heating on crystal mass
    Step 1: The mass of salt dissolved in the solution is fixed, whatever method is used to remove the water. Step 2: Whether the water is removed by gentle heating or by leaving it to dry, the same dissolved salt is left behind. Step 3: So drying without gentle heating has no effect on the final mass of crystals collected.
    Method:

    Reason that the salt mass is conserved regardless of how the water is removed.
    Examiner tips

    • Removing water more slowly does not change how much solute is left.

25. Question 4(a)

    4 marksReaction pathway diagram for an exothermic reaction
    Step 1: Because $\Delta H$ is negative the reaction is exothermic, so the products are drawn at a lower energy level than the reactants. Step 2: The activation energy $E_a$ is the energy rise from the reactant level up to the top of the energy barrier (peak), shown as an upward arrow. Step 3: $\Delta H$ is the overall fall in energy from reactants to products, shown as a downward arrow from the reactant level to the lower product level.
    Method:

    Use the sign of dH to place the product level, then describe Ea and dH as arrows.
    Examiner tips

    • Exothermic diagram: reactants high, peak in the middle, products low.

26. Question 4(b)

    4 marksBond energy calculation
    Step 1: Energy to break the reactant bonds = (4 x 330) + (4 x 460) = 1320 + 1840 = 3160 kJ. Step 2: $\Delta H$ = energy in - energy out, so energy released forming products = 3160 - (-54) = 3214 kJ, and this equals (2 x S=O) + (4 x 570). Step 3: (2 x S=O) = 3214 - 2280 = 934, so the S=O bond energy = 934 / 2 = 467 kJ/mol.
    Method:

    Sum bonds broken, use dH to find bonds formed, subtract the H-F bonds, then halve.
    Examiner tips

    • Total bonds formed = bonds broken - dH; then subtract the H-F bonds and divide by 2.

27. Question 4(c)

    5 marksEffect of changes on an equilibrium position
    Step 1: The forward reaction is exothermic, so raising the temperature favours the endothermic (reverse) direction and the equilibrium moves to the left. Step 2: There are fewer gas moles on the left (3) than the right (5), so increasing the pressure shifts the equilibrium towards the side with fewer moles, the left. Step 3: A catalyst speeds up the forward and reverse reactions equally, so it has no effect on the position of equilibrium.
    Method:

    Treat each change separately using exothermicity, the gas-mole counts, and catalyst behaviour.
    Examiner tips

    • Apply Le Chatelier: temperature, pressure (gas moles), and catalyst each have a defined effect.

28. Question 4(d)

    3 marksCollision theory and lower temperature
    Step 1: Lowering the temperature gives the particles less kinetic energy, so they move more slowly. Step 2: Slower particles collide less frequently, so there are fewer collisions each second. Step 3: Fewer particles also have energy greater than or equal to the activation energy, so a smaller proportion of collisions are successful, and the rate falls.
    Method:

    Use collision theory: link lower temperature to collision frequency and energy.
    Examiner tips

    • Mention collision frequency and the proportion exceeding the activation energy.

29. Question 5(a)(i)

    1 marksGeneral formula of the alcohols
    The general formula of the alcohols is $\text{C}_n\text{H}_{2n+1}\text{OH}$, showing the -OH functional group.
    Method:

    Recall the general formula that includes the alcohol -OH group.
    Examiner tips

    • Alcohols = CnH2n+1OH (the -OH group is the key).

30. Question 5(a)(ii)

    1 marksCharacteristics of a homologous series
    Step 1: A homologous series is a family of compounds with the same general formula and the same functional group. Step 2: Because they share the functional group, members have similar chemical properties and show a steady trend in physical properties. Step 3: So another shared characteristic is having the same functional group.
    Method:

    Recall a defining feature of a homologous series other than the general formula.
    Examiner tips

    • Same functional group -> similar chemical reactions.

31. Question 5(b)(i)

    6 marksConditions for manufacturing ethanol
    Step 1: Fermentation (method 1) uses yeast at a warm temperature of about 25-35 C in the absence of oxygen (anaerobic conditions). Step 2: Catalytic hydration of ethene (method 2) uses a temperature of about 300 C, an acid (phosphoric acid) catalyst, and a pressure of about 60 atm. Step 3: Only the first row gives both sets of conditions correctly.
    Method:

    Recall the standard conditions for fermentation and for hydration of ethene.
    Examiner tips

    • Know both ethanol routes: fermentation vs catalytic hydration of ethene.

32. Question 5(b)(ii)

    2 marksEquation for fermentation of glucose
    Step 1: In fermentation, glucose breaks down to ethanol and carbon dioxide with no other reactant. Step 2: One glucose molecule, $\text{C}_6\text{H}_{12}\text{O}_6$, forms two ethanol and two carbon dioxide molecules. Step 3: The balanced equation is $\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$.
    Method:

    Recall the fermentation products and balance the 1:2:2 ratio.
    Examiner tips

    • Fermentation: C6H12O6 -> 2C2H5OH + 2CO2.

33. Question 5(b)(iii)

    2 marksEquation for hydration of ethene
    Step 1: Ethene, $\text{C}_2\text{H}_4$, reacts with steam (water) across its C=C double bond. Step 2: One molecule of water adds on to give one molecule of ethanol. Step 3: The balanced equation is $\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}$.
    Method:

    Add one water molecule across the ethene double bond and balance.
    Examiner tips

    • Hydration of ethene adds H2O across the double bond.

34. Question 5(c)(i)

    1 marksMolecular formula from a structural formula
    Step 1: Count the carbon atoms: there are four CH$_2$ groups, giving 4 carbons. Step 2: Count the hydrogens: 4 x 2 = 8 from the CH$_2$ groups plus 2 from the two -OH groups = 10 hydrogens. Step 3: There are 2 oxygen atoms (one in each -OH), so the molecular formula is $\text{C}_4\text{H}_{10}\text{O}_2$.
    Method:

    Total each kind of atom in the structural formula.
    Examiner tips

    • Count atoms directly from the drawn structure.

35. Question 5(c)(ii)

    1 marksMoles of acid reacting with a diol
    Step 1: An -OH group reacts with one molecule of a carboxylic acid to make one ester linkage. Step 2: Butane-1,4-diol has two -OH groups, so it can form two ester linkages. Step 3: Therefore one mole of the diol reacts with two moles of ethanoic acid.
    Method:

    Match the number of acid molecules to the number of -OH groups.
    Examiner tips

    • Number of acid moles = number of -OH groups on the alcohol.

36. Question 5(d)(i)

    1 marksEmpirical formula from a structural formula
    Step 1: The molecular formula is $\text{C}_4\text{H}_6\text{O}_4$ (4 C, 6 H, 4 O). Step 2: The empirical formula is the simplest whole-number ratio, so divide each number by the highest common factor, 2. Step 3: This gives $\text{C}_2\text{H}_3\text{O}_2$.
    Method:

    Find the molecular formula, then simplify the ratio to the smallest whole numbers.
    Examiner tips

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

37. Question 5(d)(ii)

    1 marksGas formed when a carboxylic acid reacts with sodium
    Step 1: A carboxylic acid reacting with a reactive metal behaves like an acid + metal reaction. Step 2: Acid + reactive metal produces a salt plus hydrogen gas. Step 3: So the gas formed when butanedioic acid reacts with sodium is hydrogen.
    Method:

    Apply the acid + metal reaction to a carboxylic acid.
    Examiner tips

    • Reactive metal + acid always gives hydrogen gas.

38. Question 5(e)(i)

    3 marksLinkage in a polyester repeat unit
    Step 1: A diol and a dicarboxylic acid react by condensation, where an -OH group and a -COOH group join and a small molecule of water is removed. Step 2: The new linkage formed between the carbon of the acid group and the oxygen of the alcohol group is an ester linkage, -C(=O)-O-. Step 3: Repeating this joins many monomers into a polyester, so the chain is held together by ester linkages.
    Method:

    Identify the functional groups, then name the linkage they form on condensation.
    Examiner tips

    • Alcohol + carboxylic acid -> ester linkage, releasing water.

39. Question 5(e)(ii)

    1 marksType of polymerisation forming a polyester
    Joining monomers with the loss of a small molecule such as water is called condensation polymerisation.
    Method:

    Recall the name for polymerisation that releases a small molecule.
    Examiner tips

    • Water lost on joining = condensation polymerisation.