IGCSE Chemistry 0620/42 May/June 2025 Past Paper with Answers

Question 1

Using a number only, what is the approximate percentage of nitrogen in clean, dry air?

1. 78
2. 21
3. 0.04
4. 1

Question 2

Using a number only, what is the typical operating temperature, in °C, used in the Contact process for making sulfur trioxide?

1. 450
2. 25
3. 1500
4. 100

Question 3

Period 3 of the Periodic Table runs from sodium to argon. Using a number only, how many of the Period 3 elements are metals?

1. 3
2. 8
3. 5
4. 2

Question 4 · Answer A

Step 1: Going down Group VII the physical state changes from gas to liquid to solid. Step 2: Fluorine and chlorine are gases at r.t.p., bromine is a liquid and iodine is a solid. Step 3: So the number of halogens that are gases at r.t.p. is 2.

Question 5 · Answer A

Step 1: Fermentation uses enzymes in yeast, which work best at warm body-like temperatures. Step 2: A typical fermentation temperature is around 30 °C (the mark scheme accepts any value from 25 to 35 °C). Step 3: At 5 °C the yeast is too slow and at 100 °C the enzymes are denatured, so about 30 °C is correct.

Question 6 · Answer A

Step 1: Decolourising aqueous bromine needs a carbon-carbon double bond, so the isomers must be alkenes. Step 2: The unbranched alkene isomers of $\text{C}_4\text{H}_8$ are but-1-ene and but-2-ene. Step 3: Methylpropene is branched and cyclobutane has no double bond, so the number of unbranched alkene isomers is 2.

Question 7 · Answer A

Step 1: Count each single covalent bond in $\text{CH}_3\text{CH}_2\text{OH}$. Step 2: There are 5 carbon-hydrogen bonds, 1 carbon-carbon bond, 1 carbon-oxygen bond and 1 oxygen-hydrogen bond. Step 3: That gives 5 + 1 + 1 + 1 = 8 covalent bonds.

Question 8 · Answer A

Step 1: An ester name gives the alcohol part first and the acid part second. Step 2: The propyl group comes from propanol and has 3 carbon atoms; the butanoate group comes from butanoic acid and has 4 carbon atoms. Step 3: Adding these gives 3 + 4 = 7 carbon atoms.

Question 9 · Answer A

Step 1: A covalent bond forms between non-metal atoms. Step 2: It is a shared pair of electrons held between the two bonded atoms. Step 3: Transfer of electrons describes ionic bonding and a sea of electrons describes metallic bonding, so a covalent bond is a shared pair of electrons between two atoms.

Question 10 · Answer A

Step 1: A Roman numeral in a name gives the oxidation number of the element it follows. Step 2: Here (I) follows chlorine, so the oxidation number of chlorine is +1. Step 3: It does not give the number of atoms or the oxidation number of oxygen, so (I) means the oxidation number of chlorine is +1.

Question 11 · Answer A

Step 1: $\text{Cl}_2\text{O}$ is covalent, so the atoms share electrons. Oxygen forms one single bond (one shared pair) to each chlorine. Step 2: Oxygen has 6 outer electrons; two are used in the two bonds, leaving 4 non-bonding electrons. Step 3: Each chlorine has 7 outer electrons; one is used in its bond, leaving 6 non-bonding electrons, so the arrangement is one shared pair per bond with 4 and 6 non-bonding electrons respectively.

Question 12 · Answer A

Step 1: When $\text{Cl}_2\text{O}$ boils, only the weak intermolecular attractions between separate molecules are overcome, which needs little energy, so the boiling point is low. Step 2: Thermal decomposition into $\text{Cl}_2$ and $\text{O}_2$ would mean breaking the strong covalent bonds inside the molecules, which needs much more energy. Step 3: Because gentle warming cannot supply enough energy to break those strong covalent bonds, the substance boils at a low temperature without decomposing.

Question 13 · Answer A

Step 1: Electrical conduction needs charged particles that are free to move, either ions or mobile electrons. Step 2: $\text{Cl}_2\text{O}$ is made of neutral molecules, so there are no ions present. Step 3: It also has no mobile (delocalised) electrons, so with neither charge carrier available it cannot conduct, making it a poor conductor.

Question 14 · Answer A

Step 1: In a giant covalent structure of carbon, each atom is bonded to others in a network. Step 2: In graphite each carbon bonds to only three others, leaving one delocalised electron per atom that is free to move and conduct electricity. Step 3: Diamond uses all four electrons in bonds (no free electrons) and silicon(IV) oxide and methane are not forms of carbon, so the conducting form is graphite.

Question 15 · Answer A

Step 1: In graphite each carbon atom bonds to three others, leaving one outer electron that is not held in a bond. Step 2: These delocalised electrons are free to move through the structure. Step 3: Moving electrons carry the current, so the particles responsible are electrons, not protons, ions or neutrons.

Question 16 · Answer A

Step 1: In silicon(IV) oxide each silicon atom is bonded to four oxygen atoms and each oxygen bridges two silicon atoms. Step 2: For this particular 9-atom portion, the correct completion has 5 silicon atoms and 4 oxygen atoms. Step 3: The other combinations break the bonding pattern of the lattice, so the answer is 5 silicon and 4 oxygen atoms.

Question 17 · Answer A

Step 1: Electrolysis breaks down compounds using electricity, and it needs ions that are free to move. Step 2: Only ionic compounds can be electrolysed, and only when they are molten or dissolved in water (aqueous solution) so the ions can move. Step 3: Covalent compounds have no ions and a solid ionic compound has fixed ions, so the definition is ionic compounds, molten or in aqueous solution.

Question 18 · Answer A

Step 1: During electrolysis of copper(II) sulfate, copper ions, $\text{Cu}^{2+}$, are attracted to the negative cathode. Step 2: They gain electrons and are deposited as copper metal on the cathode. Step 3: Adding copper to the cathode increases its mass, so the cathode mass increases.

Question 19 · Answer A

Step 1: The blue colour of the electrolyte comes from copper(II) ions, $\text{Cu}^{2+}$, in solution. Step 2: As electrolysis removes these ions by depositing copper at the cathode, their concentration falls. Step 3: With fewer copper(II) ions the blue colour becomes paler, fading towards colourless.

Question 20 · Answer A

Step 1: With inert electrodes and a dilute aqueous solution, the anode discharges hydroxide ions from the water. Step 2: This produces oxygen gas at the anode. Step 3: The oxygen is seen as bubbles of gas; copper is deposited at the cathode, not the anode, so the observation at the anode is bubbles of gas.

Question 21 · Answer A

Step 1: At the anode hydroxide ions lose electrons (oxidation). Step 2: Four hydroxide ions release four electrons to form two water molecules and one oxygen molecule. Step 3: This balances for atoms and charge as $4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^-$, which is the correct anode half-equation.

Question 22 · Answer A

Step 1: With copper electrodes, copper(II) ions are still attracted to the cathode. Step 2: They gain electrons and are deposited as copper metal on the cathode. Step 3: Copper builds up on the cathode, so its mass increases (while the copper anode dissolves to keep the solution concentration roughly constant).

Question 23 · Answer A

Step 1: At the cathode copper(II) ions are removed as copper is deposited. Step 2: At the copper anode, copper atoms dissolve to form copper(II) ions, replacing those lost. Step 3: Because ions are added at the same rate they are removed, the copper(II) ion concentration stays constant and the blue colour does not change.

Question 24 · Answer A

Step 1: Aluminium is extracted from purified aluminium oxide obtained from its ore. Step 2: The main ore of aluminium is bauxite, which is purified to aluminium oxide. Step 3: Haematite is an iron ore, limestone is calcium carbonate and cryolite is an additive, so the aluminium ore is bauxite.

Question 25 · Answer A

Step 1: Pure aluminium oxide has a very high melting point, which would make electrolysis costly. Step 2: Cryolite is added so the aluminium oxide dissolves in molten cryolite at a much lower temperature. Step 3: This lowers the operating temperature and saves energy, so the substance added is cryolite.

Question 26 · Answer A

Step 1: Oxygen is produced at the positive carbon anodes during the electrolysis. Step 2: At the high operating temperature, the hot carbon reacts with this oxygen to form carbon dioxide gas. Step 3: This reaction gradually burns away the carbon, so the anodes get smaller and must be replaced regularly.

Question 27 · Answer A

Step 1: Count the atoms already balanced: Fe and Cr match on both sides, and there are 4 carbon atoms on the left. Step 2: There are 4 oxygen atoms in $\text{FeCr}_2\text{O}_4$ to account for, and 4 carbon atoms to use. Step 3: Combining 4 carbon with 4 oxygen gives $4\text{CO}$, which balances both carbon and oxygen, so the missing product is $4\text{CO}$.

Question 28 · Answer A

Step 1: The equation shows that both iron and chromium are formed from the chromite. Step 2: This means the chromium is not pure; it is mixed with the iron. Step 3: Extra work is then needed to separate the chromium from the iron, which is the main disadvantage of this method.

Question 29 · Answer A

Step 1: A mixture of a metal with one or more other elements is given a special name. Step 2: This mixture is called an alloy, for example stainless steel. Step 3: A compound has its elements chemically combined in fixed ratios, an isotope is a form of an atom and an electrolyte conducts during electrolysis, so the correct term is alloy.

Question 30 · Answer A

Step 1: The sum of the oxidation numbers of all atoms in an ion equals the overall charge on the ion. Step 2: The ion $\text{Cr}_2\text{O}_7^{2-}$ carries a charge of $2-$. Step 3: Therefore the sum of the oxidation numbers is $-2$.

Question 31 · Answer A

Step 1: The seven oxygen atoms contribute $7 \times (-2) = -14$ to the total. Step 2: The overall charge is $-2$, so the two chromium atoms together must total $-2 - (-14) = +12$. Step 3: Sharing $+12$ between the two chromium atoms gives $+12 \div 2 = +6$ for each chromium.

Question 32 · Answer A

Step 1: Moles of $(\text{NH}_4)_2\text{Cr}_2\text{O}_7 = 1.26 \div 252 = 0.00500\ \text{mol}$. Step 2: One mole of compound gives one mole of nitrogen, so moles of $\text{N}_2 = 0.00500\ \text{mol}$. Step 3: Volume $= 0.00500 \times 24\,000 = 120\ \text{cm}^3$ at r.t.p.

Question 33 · Answer A

Step 1: The alkali metals are the very reactive metals on the far left of the Periodic Table. Step 2: They each have one electron in their outer shell, which places them in Group I. Step 3: Group II is the alkaline earth metals, Group VII the halogens and Group 0 the noble gases, so the alkali metals are in Group I.

Question 34 · Answer A

Step 1: In Group I the melting point decreases as you go down the group. Step 2: Lithium is at the top of the group, so it has the highest melting point. Step 3: Sodium, potassium and caesium are lower down and have progressively lower melting points, so lithium has the highest.

Question 35 · Answer A

Step 1: Density of the Group I metals generally increases going down the group. Step 2: Francium is the lowest member of Group I, so it has the highest density. Step 3: Lithium, sodium and potassium are higher up the group and have lower densities, so francium has the highest density.

Question 36 · Answer A

Step 1: In Group I reactivity increases going down the group, because the outer electron is lost more easily. Step 2: Lithium is at the top of the group, so it loses its outer electron least easily and is the least reactive. Step 3: Sodium, potassium and caesium are more reactive, so lithium has the lowest reactivity.

Question 37 · Answer A

Step 1: Different metal ions give characteristic flame colours. Step 2: Potassium gives a lilac (light purple) flame. Step 3: Sodium gives a yellow flame and lithium gives a red flame, so the metal with a lilac flame is potassium.

Question 38 · Answer A

Step 1: The main nutrients in NPK fertilisers are nitrogen, phosphorus and potassium. Step 2: Of the alkali metals, potassium is the one plants need and that fertilisers supply. Step 3: Lithium, caesium and rubidium are not plant nutrients, so the alkali metal in fertilisers is potassium.

Question 39 · Answer A

Step 1: A sodium atom (2,8,1) loses its single outer electron to form $\text{Na}^+$ with configuration 2,8 and a 1+ charge. Step 2: A sulfur atom (2,8,6) gains two electrons to form $\text{S}^{2-}$ with configuration 2,8,8 and a 2- charge. Step 3: Two sodium ions balance one sulfide ion in $\text{Na}_2\text{S}$, so the ions are $\text{Na}^+$ (2,8) and $\text{S}^{2-}$ (2,8,8).

Question 40 · Answer A

Step 1: $^{85}\text{Rb}$ and $^{87}\text{Rb}$ are atoms of the same element (same protons) with different numbers of neutrons. Step 2: Atoms of the same element with different numbers of neutrons are called isotopes. Step 3: Ions are charged atoms, allotropes are forms of an element and compounds contain different elements, so the correct term is isotopes.

Question 41 · Answer A

Step 1: Protons equal the proton number 37 for both species. Step 2: Neutrons = mass number - protons, so $^{85}\text{Rb}$ has $85-37=48$ neutrons and $^{87}\text{Rb}$ has $87-37=50$ neutrons. Step 3: A neutral atom has 37 electrons; the $1+$ ion has lost one electron, leaving 36, so $^{87}\text{Rb}^+$ has 36 electrons.

Question 42 · Answer A

Step 1: Let the abundance of $^{85}\text{Rb}$ be $x\%$, so $^{87}\text{Rb}$ is $(100-x)\%$. Step 2: The relative atomic mass is $\frac{85x + 87(100-x)}{100} = 85.5$, which gives $85x + 8700 - 87x = 8550$. Step 3: Solving $-2x = -150$ gives $x = 75$, so $^{85}\text{Rb}$ is 75% abundant.

Question 43 · Answer A

Step 1: Organic compounds with the same functional group and general formula form a family. Step 2: This family is called a homologous series, for example the alkenes. Step 3: Allotropes are forms of an element, isotopes are atoms of one element and a mixture is not a chemical family, so the term is homologous series.

Question 44 · Answer A

Step 1: Consecutive members of a homologous series differ by one $\text{CH}_2$ group. Step 2: The relative molecular mass of $\text{CH}_2$ is $12 + (2 \times 1) = 14$. Step 3: So the next alkene ($\text{C}_3\text{H}_6$) has a relative molecular mass 14 greater than $\text{C}_2\text{H}_4$, giving a difference of 14.

Question 45 · Answer A

Step 1: Complete combustion of a hydrocarbon gives carbon dioxide and water only. Step 2: Ethene has 2 carbon and 4 hydrogen atoms, giving $2\text{CO}_2$ and $2\text{H}_2\text{O}$. Step 3: Balancing oxygen: products need $4 + 2 = 6$ oxygen atoms, so $3\text{O}_2$ is required, giving $\text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}$.

Question 46 · Answer A

Step 1: Lowering temperature favours the exothermic forward reaction, using up $\text{C}_2\text{H}_4$, so its concentration decreases. Step 2: Removing the product $\text{C}_2\text{H}_5\text{OH}$ shifts the position forward, so $\text{C}_2\text{H}_4$ decreases; increasing pressure favours the side with fewer gas moles (the product side), so $\text{C}_2\text{H}_4$ decreases. Step 3: A catalyst speeds both directions equally and does not change the equilibrium position, so the concentration shows no change.

Question 47 · Answer A

Step 1: Raising the temperature gives the particles more kinetic energy, so they move faster. Step 2: Faster particles collide more frequently. Step 3: A greater proportion of these collisions now have energy equal to or above the activation energy, so more collisions are successful and the rate increases.

Question 48 · Answer A

Step 1: Count each type of atom in compound B: there are 5 carbon atoms, 6 hydrogen atoms and 4 oxygen atoms. Step 2: Writing these in a molecular formula gives $\text{C}_5\text{H}_6\text{O}_4$. Step 3: The other options miscount the hydrogen, oxygen or carbon atoms, so the molecular formula is $\text{C}_5\text{H}_6\text{O}_4$.

Question 49 · Answer A

Step 1: An unsaturated organic compound contains at least one carbon-carbon double (or triple) bond. Step 2: Compound B has a carbon-carbon bond that is not a single bond, that is a C=C double bond. Step 3: The O-H bond, single bonds and atom ratios do not define unsaturation, so it is unsaturated because of its C=C double bond.

Question 50 · Answer A

Step 1: In addition polymerisation the carbon-carbon double bond opens up. Step 2: This forms a single carbon-carbon bond, and the repeat unit is drawn with a continuation bond extending from each end carbon to show the chain continues. Step 3: No small molecule is lost in addition polymerisation, so the repeat unit keeps all of compound B and shows the opened single bond with continuation bonds.

Question 51 · Answer A

Step 1: Sodium hydroxide neutralises carboxylic acid groups in a 1:1 ratio (one -COOH per NaOH). Step 2: If one mole of compound B reacts with two moles of NaOH, it must contain two acidic groups. Step 3: So compound B has two carboxylic acid (-COOH) groups, each reacting with one NaOH.

Question 52 · Answer A

Step 1: Compound B reacts with NaOH in a 1:2 ratio, so 0.100 mol of B needs $0.100 \times 2 = 0.200\ \text{mol}$ of NaOH. Step 2: Volume in dm³ = moles ÷ concentration = $0.200 \div 0.250 = 0.800\ \text{dm}^3$. Step 3: Converting to cm³: $0.800 \times 1000 = 800\ \text{cm}^3$.