Question 1(a)

Step 1: A basic oxide is the oxide of a metal, which reacts with acids to form a salt and water. Step 2: Calcium oxide is a metal oxide, so it is a basic oxide. Step 3: Sulfur dioxide is a non-metal (acidic) oxide, bauxite is an ore and graphite is an element, so the basic oxide is calcium oxide.

Method:

Pick the metal oxide from the list, since metal oxides are basic.

Examiner tips

$Metal oxide = basic; non-metal oxide = acidic.$

Question 1(b)

1 marksSubstance containing carbon atoms only

Step 1: A substance that contains carbon atoms only must be a form of the element carbon. Step 2: Graphite is a giant covalent structure made of carbon atoms only. Step 3: Methane and propene contain hydrogen as well, and ethanol contains hydrogen and oxygen, so the answer is graphite.

Method:

Find the substance that is an element made of carbon only.

Examiner tips

$Graphite and diamond are forms of the element carbon (carbon only).$

Question 1(c)

1 marksSubstance manufactured by fermentation

Step 1: Fermentation uses yeast to convert sugars (glucose) into ethanol and carbon dioxide. Step 2: So ethanol is the substance manufactured by fermentation. Step 3: Methane, sulfur dioxide and propene are made by other processes, so the answer is ethanol.

Method:

Recall which substance is produced when yeast ferments sugars.

Examiner tips

$Fermentation: glucose -> ethanol + carbon dioxide, using yeast.$

Question 1(d)

1 marksGas produced by decomposition of vegetation

Step 1: When vegetation (plant matter) decomposes without much air, methane gas is produced. Step 2: So the gas produced by the decomposition of vegetation is methane. Step 3: Ethanol comes from fermentation, propane from petroleum and nitrogen from the air, so the answer is methane.

Method:

Recall which gas is given off when plant matter decays.

Examiner tips

$Methane (biogas) forms when vegetation decomposes.$

Question 1(e)

1 marksOre containing aluminium oxide

Step 1: Aluminium is extracted from its ore, which contains aluminium oxide. Step 2: Bauxite is the ore of aluminium and contains aluminium oxide. Step 3: Calcium oxide, graphite and sulfur dioxide do not contain aluminium oxide, so the answer is bauxite.

Method:

Identify the aluminium ore, since it contains aluminium oxide.

Examiner tips

$Bauxite -> aluminium oxide -> aluminium (by electrolysis).$

Question 1(f)

1 marksGas that causes acid rain

Step 1: Acid rain is caused by acidic gases that dissolve in rain water. Step 2: Sulfur dioxide dissolves in rain to form an acidic solution, so it causes acid rain. Step 3: Nitrogen and oxygen do not cause acid rain, and methane is a greenhouse gas, so the answer is sulfur dioxide.

Method:

Pick the acidic gas that dissolves in rain water.

Examiner tips

$Sulfur dioxide (and oxides of nitrogen) cause acid rain.$

Question 1(g)

1 marksSimple molecule with 11 atoms

\text{C}_3\text{H}_8

Method:

Work out the formula of each molecule and add up the atoms.

Examiner tips

$Total atoms = sum of all subscript numbers in the formula.$

Question 1(h)

1 marksProduct of limestone decomposition in the blast furnace

Step 1: In the blast furnace, limestone (calcium carbonate) is thermally decomposed. Step 2: This produces calcium oxide and carbon dioxide; the calcium oxide then removes acidic impurities as slag. Step 3: Bauxite, graphite and sulfur dioxide are not made by this decomposition, so the answer is calcium oxide.

Method:

Recall the thermal decomposition of limestone in the blast furnace.

Examiner tips

$CaCO3 -> CaO + CO2 in the blast furnace.$

Question 1(i)

1 marksGas that is about 21% of clean dry air

Step 1: Clean dry air is about 78% nitrogen and 21% oxygen, with small amounts of other gases. Step 2: The gas making up roughly 21% is oxygen. Step 3: Nitrogen is about 78%, while sulfur dioxide and methane are only present in small or trace amounts, so the answer is oxygen.

Method:

Recall the composition of clean dry air and pick the gas at about 21%.

Examiner tips

$Remember the 78% nitrogen / 21% oxygen split of clean dry air.$

Question 1(j)

1 marksMonomer in addition polymerisation

Step 1: Addition polymerisation needs monomers that contain a carbon-carbon double bond (C=C). Step 2: Propene is an alkene with a C=C double bond, so it can act as a monomer. Step 3: Propane and methane are saturated (no C=C) and ethanol has no C=C, so the monomer is propene.

Method:

Pick the unsaturated molecule (one with a C=C double bond).

Examiner tips

$Only molecules with a C=C double bond undergo addition polymerisation.$

Question 2(a)

2 marksRelative charge and mass of subatomic particles

Step 1: An electron carries a relative charge of -1 and has a very small relative mass (1/1840). Step 2: A proton carries a relative charge of +1 and has a relative mass of 1. Step 3: So the electron is -1, and the proton is +1 with a relative mass of 1.

Method:

Recall the relative charge and mass of each subatomic particle.

Examiner tips

$Proton +1 / mass 1; neutron 0 / mass 1; electron -1 / mass 1/1840.$

Question 2(b)(i)

3 marksParticles in isotopes and ions of potassium

^{39}\text{K}

Method:

Use the proton and nucleon numbers, then adjust electrons for the charge.

Examiner tips

$Forming a positive ion removes electrons but never changes protons or neutrons.$

Question 2(b)(ii)

2 marksRelative atomic mass from isotopic abundance

Step 1: Relative atomic mass = [(39 \times 90) + (41 \times 10)] \div 100. Step 2: This is (3510 + 410) \div 100 = 3920 \div 100. Step 3: So the relative atomic mass is 39.2.

Method:

Compute the weighted average of the isotope masses using the abundances.

Examiner tips

$Weighted mean: sum of (mass \times abundance) \div 100.$

Question 2(b)(iii)

1 marksConclusion from matching nucleon number and relative atomic mass

Step 1: The relative atomic mass is the weighted average of all the isotopes of an element. Step 2: If the relative atomic mass (27) equals the nucleon number of the only quoted isotope (27), there can be no other isotope pulling the average away from 27. Step 3: So aluminium must have only one isotope, of nucleon number 27.

Method:

Reason that a single average equal to one mass means one isotope only.

Examiner tips

$RAM = single isotope mass implies only one isotope exists.$

Question 2(c)

2 marksSpecies with the same electronic configuration as a calcium ion

\text{Ca}^{2+}

Method:

Count electrons (18) and pick an atom and a negative ion that both have 18.

Examiner tips

$Isoelectronic with Ca2+ (18 electrons): Ar, K+, Cl-, S2-, P3-.$

Question 3(a)(i)

3 marksBonding in a metallic element

Step 1: In a metal, the atoms lose their outer electrons to form positive metal ions (cations). Step 2: These released electrons become delocalised, forming a sea of mobile electrons around the positive ions. Step 3: There is strong electrostatic attraction between the positive ions and the delocalised electrons, which holds the metal together.

Method:

State the positive ions, the delocalised electron sea, and the attraction between them.

Examiner tips

$Metallic bonding = positive ions + delocalised electron sea + electrostatic attraction.$

Question 3(a)(ii)

1 marksWhy solid copper conducts electricity

Step 1: Solid copper contains a sea of delocalised electrons that are free to move. Step 2: When a voltage is applied, these mobile electrons drift through the metal, carrying the electric charge. Step 3: So copper conducts electricity because of its free-moving (delocalised) electrons.

Method:

Link conduction to the free movement of delocalised electrons.

Examiner tips

$Metals conduct because delocalised electrons are free to move.$

Question 3(b)(i)

1 marksWhy alloys are more useful than pure metals

Step 1: An alloy contains atoms of different sizes, which disrupt the regular layers of metal atoms. Step 2: This makes it harder for the layers to slide over each other, so the alloy is harder and stronger than the pure metal. Step 3: So a key reason alloys are more useful is that they are harder and stronger.

Method:

Recall the main advantage of alloys: greater hardness/strength.

Examiner tips

$Alloys are harder/stronger because different-sized atoms block sliding layers.$

Question 3(b)(ii)

1 marksMetal in brass other than copper

Step 1: Brass is an alloy made from two metals. Step 2: Besides copper, brass contains zinc. Step 3: Tin, lead and iron are not the metal in brass, so the answer is zinc.

Method:

Recall the composition of brass.

Examiner tips

$Brass = copper + zinc; bronze = copper + tin.$

Question 3(c)(i)

2 marksObservations when a carbonate reacts with acid

Step 1: The acid reacts with the copper(II) carbonate, so the solid is used up and dissolves (disappears). Step 2: The carbonate reacts with the acid to release carbon dioxide gas, seen as fizzing/bubbling (effervescence). Step 3: So the two observations are the solid dissolving and effervescence.

Method:

Recall the typical observations for an acid reacting with a carbonate.

Examiner tips

$Carbonate + acid -> salt + water + carbon dioxide (effervescence).$

Question 3(c)(ii)

1 marksWhy the reaction stops when excess carbonate is added

Step 1: Copper(II) carbonate is added until no more reacts, so it is added in excess. Step 2: The reaction stops because all of the acid has been used up (there is no more acid to react with the carbonate). Step 3: So the reaction stops because the sulfuric acid has all reacted.

Method:

Identify which reagent is in excess and which runs out.

Examiner tips

$Adding excess solid until reaction stops means the acid is the limiting reagent.$

Question 3(c)(iii)

1 marksResidue collected by filtration

Step 1: Excess copper(II) carbonate was added, so some did not react. Step 2: Filtration removes this unreacted solid as the residue, while the aqueous copper(II) sulfate passes through as the filtrate. Step 3: So the residue is the unreacted copper(II) carbonate.

Method:

Recall which substance was added in excess and remains undissolved.

Examiner tips

$Residue = insoluble unreacted solid left on the filter paper.$

Question 3(c)(iv)

1 marksAnother substance to make copper(II) sulfate from sulfuric acid

Step 1: Copper(II) sulfate forms when a copper base or carbonate reacts with sulfuric acid. Step 2: Copper(II) oxide is a base that reacts with sulfuric acid to give copper(II) sulfate and water. Step 3: Copper metal is too unreactive to react with dilute sulfuric acid, so the best answer is copper(II) oxide (copper(II) hydroxide would also work).

Method:

Pick a copper base that reacts with the acid to form the salt.

Examiner tips

$Acid + metal oxide/hydroxide/carbonate -> salt (+ water, + CO2 for carbonates).$

Question 3(c)(v)

1 marksTerm for a solution holding the maximum dissolved solute

Step 1: A solution that has dissolved the maximum possible solute at a given temperature cannot dissolve any more. Step 2: The term for this is a saturated solution. Step 3: Dilute and concentrated describe relative amounts but can still dissolve more, so the correct term is saturated.

Method:

Recall the term for a solution that cannot dissolve any more solute.

Examiner tips

$Saturated = no more solute can dissolve at that temperature.$

Question 3(c)(vi)

2 marksEffect of concentration on rate in terms of particles

Step 1: Increasing the concentration packs more acid particles into the same volume. Step 2: With more particles per unit volume, the particles collide more often, so there are more collisions per unit time. Step 3: More frequent collisions mean a faster rate of reaction.

Method:

Link the higher concentration to more particles per volume and more frequent collisions.

Examiner tips

$Concentration ↑ -> particles per volume ↑ -> collision frequency ↑ -> rate ↑.$

Question 3(c)(vii)

1 marksTerm for a substance chemically combined with water

\text{CuSO}_4{\cdot}5\text{H}_2\text{O}

Method:

Recall the term for a solid containing water of crystallisation.

Examiner tips

$Hydrated = contains water of crystallisation; anhydrous = none.$

Question 4(a)

2 marksDot-and-cross diagram of a nitrogen molecule

Step 1: Each nitrogen atom has 5 outer electrons and needs 3 more to reach a full octet. Step 2: The two atoms share three pairs of electrons (a triple bond), giving each atom a share of 8 outer electrons. Step 3: After the three shared pairs, each nitrogen still has one non-bonding (lone) pair, so the diagram shows a triple bond plus one lone pair on each atom.

Method:

Form enough shared pairs to complete each octet, then place the remaining electrons as lone pairs.

Examiner tips

$N2 has a triple bond (3 shared pairs) and one lone pair on each atom.$

Question 4(b)(i)

3 marksConditions for the Haber process

Step 1: The Haber process uses a moderately high temperature of about 450 °C to give a reasonable rate and yield. Step 2: It uses a high pressure of about 200 atmospheres to favour the side with fewer gas molecules (ammonia). Step 3: An iron catalyst speeds up the reaction, so the conditions are about 450 °C, about 200 atmospheres and an iron catalyst.

Method:

Recall the standard industrial conditions for the Haber process.

Examiner tips

$Memorise the Haber conditions: 450 °C, 200 atm, iron.$

Question 4(b)(ii)

1 marksSymbol equation for the Haber process

\text{NH}_3

Method:

Write the reactants and product, then balance N and H.

Examiner tips

$N2 + 3H2 -> 2NH3 balances both nitrogen and hydrogen.$

Question 4(c)(i)

1 marksBalancing the oxidation of ammonia

\text{NH}_3

Method:

Balance N and H from the 4 NH3, then balance O to find the O2 coefficient.

Examiner tips

$Balance metals/non-metals, then hydrogen, then oxygen last.$

Question 4(c)(ii)

2 marksOxidation number of nitrogen in two compounds

\text{NH}_3

Method:

Apply H = +1 and O = -2, then solve for nitrogen so the molecule is neutral.

Examiner tips

$Set the sum of oxidation numbers in a neutral molecule to zero.$

Question 4(c)(iii)

1 marksDefinition of oxidation in terms of oxidation number

Step 1: Oxidation number tracks how oxidised or reduced an element is. Step 2: In terms of oxidation number, oxidation is defined as an increase in oxidation number. Step 3: A decrease in oxidation number is reduction, so oxidation is an increase.

Method:

Recall the oxidation-number definition of oxidation.

Examiner tips

$Oxidation = increase in oxidation number; reduction = decrease.$

Question 4(c)(iv)

2 marksEquation for the formation of nitric acid

\text{HNO}_3

Method:

Balance nitrogen, then hydrogen via water, then oxygen via O2.

Examiner tips

$When water is a reactant, balance H via water and O last.$

Question 4(d)

3 marksMass of product from a gas volume

\text{NH}_3

Method:

Find moles of gas, apply the equation ratio, then convert moles to mass.

Examiner tips

$Three steps: gas moles, mole ratio, then mass = moles \times Mr.$

Question 5(a)

2 marksConcentration of a product at equilibrium

\text{CH}_3\text{OH}

Method:

Explain constant concentration using equal forward and reverse rates.

Examiner tips

$Dynamic equilibrium: constant concentrations, equal opposing rates, reactions still going.$

Question 5(b)(i)

2 marksDisadvantages of using a lower pressure

Step 1: Lowering the pressure means the particles are further apart, so they collide less often and the rate of reaction decreases. Step 2: The forward reaction makes fewer gas molecules, so a lower pressure shifts the equilibrium towards the reactants, decreasing the yield of methanol. Step 3: So both the rate and the yield of methanol decrease at lower pressure.

Method:

Use collision theory for rate and Le Chatelier (gas-molecule count) for yield.

Examiner tips

$Fewer gas molecules on the product side -> high pressure favours product.$

Question 5(b)(ii)

2 marksEffect of a catalyst on yield and reverse rate

\text{CH}_3\text{OH}

Method:

Recall that a catalyst speeds both directions equally and does not move the equilibrium.

Examiner tips

$Catalyst: rates up (both directions), equilibrium position and yield unchanged.$

Question 5(b)(iii)

1 marksDeducing the energy change from the yield-temperature trend

Step 1: Raising the temperature shifts an equilibrium in the endothermic direction. Step 2: Since a higher temperature decreases the yield of methanol, the reverse reaction must be endothermic, so the forward reaction is exothermic. Step 3: So the forward reaction is exothermic.

Method:

Use the temperature effect on equilibrium to deduce the energy change of the forward reaction.

Examiner tips

$Yield falls as T rises => forward reaction is exothermic.$

Question 5(b)(iv)

2 marksChoosing a suitable catalyst element

Step 1: Transition elements and their compounds are widely used as catalysts. Step 2: From the list, copper is the only transition element, so it is a suitable catalyst. Step 3: Barium, potassium, carbon and sulfur are not transition elements, so the answer is copper because it is a transition element.

Method:

Pick the transition element from the list and justify it as a catalyst.

Examiner tips

$Transition metals (and their compounds) are common catalysts.$

Question 5(c)(i)

2 marksCharacteristics of a homologous series

\text{CH}_2

Method:

Recall the defining features that all members of a homologous series share.

Examiner tips

$Homologous series: same general formula, same functional group, differ by CH2.$

Question 5(c)(ii)

1 marksMolecular formula of a five-carbon alcohol

\text{C}_n\text{H}_{2n+1}\text{OH}

Method:

Apply the alcohol general formula for n = 5 and combine the hydrogen atoms.

Examiner tips

$Alcohols: CnH(2n+1)OH = CnH(2n+2)O as a molecular formula.$

Question 5(d)(i)

2 marksDisplayed formula of an ester with three carbon atoms

Step 1: An ester contains the ester linkage, a carbon double-bonded to one oxygen and single-bonded to a second oxygen (-C(=O)-O-). Step 2: Methyl ethanoate has a CH3 group joined to this ester carbon, whose second oxygen is joined to another CH3 group, giving three carbon atoms in total. Step 3: So the correct displayed formula shows the CH3-C(=O)-O-CH3 arrangement with all bonds drawn.

Method:

Draw the ester linkage with carbon groups on each side, showing all bonds.

Examiner tips

$Ester linkage = -C(=O)-O-; methyl ethanoate is CH3COOCH3.$

Question 5(d)(ii)

2 marksAlcohol and acid that form a named ester

Step 1: In an ester name, the first part comes from the alcohol and the second part from the acid. Step 2: "Butyl" comes from butan-1-ol, and "ethanoate" comes from ethanoic acid. Step 3: So butyl ethanoate is made from butan-1-ol and ethanoic acid.

Method:

Split the ester name: the -yl part names the alcohol, the -oate part names the acid.

Examiner tips

$"Butyl" = from butanol; "ethanoate" = from ethanoic acid.$

Question 5(e)

2 marksEmpirical formula from percentage composition

A_r

Method:

Convert percentages to moles, then scale to the simplest whole-number ratio.

Examiner tips

$Empirical formula: % \div Ar, then \div smallest to get whole-number ratio.$

Question 6(a)

1 marksName of the Group VII elements

Step 1: Group VII is the second-to-last group of the Periodic Table. Step 2: These elements (fluorine, chlorine, bromine, iodine) are called the halogens. Step 3: Noble gases are Group VIII/0, alkali metals are Group I, so Group VII are the halogens.

Method:

Recall the family name of Group VII.

Examiner tips

$Group I = alkali metals; Group VII = halogens; Group VIII/0 = noble gases.$

Question 6(b)

1 marksMost reactive Group VII element

Step 1: In Group VII, reactivity decreases going down the group. Step 2: Fluorine is at the top of the group, so it is the most reactive halogen. Step 3: Iodine and astatine are lower down and less reactive, so the most reactive is fluorine.

Method:

Use the Group VII reactivity trend to pick the topmost element.

Examiner tips

$Halogens: reactivity decreases down the group (opposite to Group I).$

Question 6(c)

2 marksState and colour of iodine at room temperature

Step 1: Down Group VII the elements change from gases to liquid to solid. Step 2: Iodine is near the bottom, so it is a solid at room temperature and pressure. Step 3: Its colour is grey-black, so iodine is a grey-black solid.

Method:

Recall the state and colour of iodine from the Group VII trend.

Examiner tips

$Group VII states: chlorine gas, bromine liquid, iodine solid.$

Question 6(d)(i)

2 marksIonic equation for a halogen displacement

\text{K}^+

Method:

Remove spectator ions, then balance atoms and charges.

Examiner tips

$Ionic equations drop spectator ions (here K+) and must balance.$

Question 6(d)(ii)

3 marksEnthalpy change from bond energies

\Delta H

Method:

Add the bond energies broken, subtract those formed, and keep the sign.

Examiner tips

$deltaH = (sum bonds broken) - (sum bonds formed); negative = exothermic.$

Question 6(e)(i)

2 marksObservations when sodium reacts with water

Step 1: Sodium is less dense than water, so it floats on the surface. Step 2: The reaction releases heat and hydrogen gas, so the sodium melts into a ball, fizzes (effervescence) and moves around the surface. Step 3: So two valid observations are that the sodium floats and moves around on the surface.

Method:

Recall the visible features of a Group I metal reacting with water.

Examiner tips

$Group I + water: float, fizz, melt into a ball, move around the surface.$

Question 6(e)(ii)

1 marksColour of thymolphthalein in the product solution

Step 1: Sodium reacts with water to form sodium hydroxide, which is an alkaline solution. Step 2: Thymolphthalein is blue in alkaline solutions and colourless in neutral or acidic ones. Step 3: Since the solution is alkaline (sodium hydroxide), the thymolphthalein turns blue.

Method:

Identify the solution as alkaline, then recall the indicator colour.

Examiner tips

$Thymolphthalein: colourless in acid/neutral, blue in alkali.$

0620/41 May/June 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 1(f)

Question 1(g)

Question 1(h)

Question 1(i)

Question 1(j)

Question 2(a)

Question 2(b)(i)

Question 2(b)(ii)

Question 2(b)(iii)

Question 2(c)

Question 3(a)(i)

Question 3(a)(ii)

Question 3(b)(i)

Question 3(b)(ii)

Question 3(c)(i)

Question 3(c)(ii)

Question 3(c)(iii)

Question 3(c)(iv)

Question 3(c)(v)

Question 3(c)(vi)

Question 3(c)(vii)

Question 4(a)

Question 4(b)(i)

Question 4(b)(ii)

Question 4(c)(i)

Question 4(c)(ii)

Question 4(c)(iii)

Question 4(c)(iv)

Question 4(d)

Question 5(a)

Question 5(b)(i)

Question 5(b)(ii)

Question 5(b)(iii)

Question 5(b)(iv)

Question 5(c)(i)

Question 5(c)(ii)

Question 5(d)(i)

Question 5(d)(ii)

Question 5(e)

Question 6(a)

Question 6(b)

Question 6(c)

Question 6(d)(i)

Question 6(d)(ii)

Question 6(e)(i)

Question 6(e)(ii)

Sit this paper in the app