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2025 O-Level Chemistry Paper 2 — Full Worked Solutions

GCE O-Level Chemistry (6092/02) · Paper 2 · October/November 2025 · 80 marks

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Section A Answer all questions · 70 marks

1 Structure & bonding — metals, ionic compounds, graphite 9 marks

The arrangement of particles in a solid metal and a solid ionic compound are shown in Fig. 1.1.

Fig. 1.1: a solid metal (positive ions in a sea of electrons) and a solid ionic compound (alternating positive and negative ions), each with two blank label boxes
a Complete the diagrams of the solid metal and the solid ionic compound by labelling the particles in Fig. 1.1. [2]
Show working
Concept: Giant metallic and giant ionic lattices
Solid metal: a lattice of positive metal ions surrounded by a ‘sea’ of delocalised electrons.
Solid ionic compound: a lattice of alternating positive ions and negative ions.
Fig. 1.1 with the particles labelled: positive metal ions and a sea of delocalised electrons in the solid metal; positive ions and negative ions in the solid ionic compound
Solid metal: positive metal ions and a ‘sea’ of delocalised electrons. Solid ionic compound: positive ions and negative ions.
b Explain why ionic compounds have high melting points. [2]
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Concept: Giant ionic lattice
Ionic compounds have a giant ionic lattice structure.
A large amount of energy is required to overcome the strong electrostatic forces of attraction between the oppositely charged ions.
Giant ionic lattice → strong electrostatic forces of attraction between oppositely charged ions → large amount of energy needed to overcome them.
c Compare the electrical conductivity of a metal and an ionic compound.
Use ideas about structure in your answer. [3]
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Concept: Free-moving charge carriers (ions / electrons)
Conductivity depends on the presence of free-moving ions / electrons to act as charge carriers.
Ionic compound: conducts only in the aqueous or molten state. In the solid state the ions are held in fixed positions by strong electrostatic forces of attraction, so there are no free-moving ions to act as charge carriers.
Metal: conducts in all states, as the ‘sea’ of delocalised electrons in the giant metallic lattice acts as charge carriers.
Ionic compounds conduct only when aqueous or molten (free-moving ions); metals conduct in all states (delocalised electrons).
d(i) Fig. 1.2 shows the structure of a non-metal that conducts electricity. Identify the non-metal. [1]
Fig. 1.2: layers of hexagonally arranged atoms held together by dotted lines between the layers, the structure of graphite
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Concept: Electronic configuration of carbon
Carbon has the electronic configuration 2.4 — four outer electrons, forming layers of hexagonal rings as drawn in Fig. 1.2.
Carbon (graphite)
d(ii) Explain, using Fig. 1.2, how this non-metal conducts electricity. [1]
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Concept: Delocalised electrons in graphite
Each carbon atom is covalently bonded to 3 other carbon atoms, leaving 1 unbonded electron per carbon atom.
This electron is free-moving and acts as a charge carrier.
Each C atom bonds to 3 others, leaving 1 free-moving (delocalised) electron per atom to act as a charge carrier.
2 Ionic & covalent bonding — nitrogen 7 marks

Nitrogen forms both ionic and covalent bonds.
(a) The symbol for the nitride ion is N3−.

a(i) Deduce the formula of calcium nitride. [1]
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Concept: Formula of an ionic compound
Calcium is a metal, so it forms Ca2+; the nitride ion is N3−.
Balance the charges: 3 × (2+) = 6+ and 2 × (3−) = 6−, so 3 Ca2+ : 2 N3−.
Ca3N2
a(ii) State the charge on the nucleus of the N3− ion. [1]
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Concept: Nuclear charge = number of protons
The nucleus contains protons (positive) and neutrons (neutral), so the nuclear charge equals the number of protons.
Nitrogen has 7 protons; gaining electrons to form the ion does not change the nucleus.
+7
a(iii) State the total number of electrons in the N3− ion. [1]
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Concept: Ions: electrons vs protons
The 3− charge means the ion has 3 more electrons than protons.
Electrons = 7 + 3 = 10 (electronic configuration 2.8).
10
a(iv) Give the symbol of one other negative ion with the same electronic configuration as the N3− ion. [1]
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Concept: Isoelectronic ions (2.8)
N3− has the configuration 2.8. Another negative ion with 10 electrons: O has 8 protons, so O2− (8 + 2 = 10 electrons); F has 9 protons, so F− (9 + 1 = 10 electrons).
O2− (or F−)
b(i) (b) Nitrogen forms a covalently bonded oxide, NO. The arrangement of electrons around the nitrogen atom in NO is not usual compared to most covalent compounds. The structure of NO is shown: N=O
Draw a ‘dot-and-cross’ diagram to show the bonding in NO. Show outer electrons only. [2]
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Concept: Dot-and-cross diagram of a double bond
Covalent bond = a shared pair of electrons. A double bond (N=O) shares 2 pairs = 4 electrons.
N has 5 outer electrons: 5 − 2 shared = 3 not shared (one lone pair and one unpaired electron).
O has 6 outer electrons: 6 − 2 shared = 4 not shared (two lone pairs).
Dot-and-cross diagram of NO: two shared pairs in the overlap of two circles, with three non-bonded electrons on nitrogen and four on oxygen

The full question

Full question Question 2(b): dot-and-cross diagram of NO with Teacher Dalson's notes on shared and non-shared electrons

Teacher Dalson’s annotated answer

Annotated Question 2(b): dot-and-cross diagram of NO with Teacher Dalson's notes on shared and non-shared electrons

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Two shared pairs between N and O; 3 non-bonded electrons on N; 4 non-bonded electrons (2 lone pairs) on O.
b(ii) Explain how the arrangement of electrons around the nitrogen atom in NO is not usual compared to most covalent compounds. [1]
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Concept: Octet / noble gas configuration
In most covalent compounds every atom achieves a full outer shell by sharing electrons.
The nitrogen atom does not have a fully filled outer electron shell (it has only 7 valence electrons).
3 Condensation polymers & recycling 9 marks

Some types of plastics contain condensation polymers.
(a) The partial structure of a condensation polymer can be represented as shown in Fig. 3.1.

Fig. 3.1: a chain with a black box and a white box alternating, joined by C=O and N-H amide links
a(i) Name a condensation polymer with this structure. [1]
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Concept: Amide linkage → polyamide
The repeating link in Fig. 3.1 is the amide linkage (−CONH−), formed when a diamine reacts with a dicarboxylic acid.
Nylon (a polyamide)
a(ii) Draw the structures of the two monomers used to make the polymer in Fig. 3.1. [2]
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Concept: Monomers of a polyamide
Break the polymer at each amide linkage: the black box sits between two C=O groups (the dicarboxylic acid) and the white box sits between two N−H groups (the diamine).
Add −OH to each C=O end and −H to each N end to restore the monomers.
The two monomers drawn out: a dicarboxylic acid with the black box between two COOH groups, and a diamine with the white box between two NH2 groups
A dicarboxylic acid, HO−C(=O)−[ black box ]−C(=O)−OH, and a diamine, H−N(H)−[ white box ]−N(H)−H.
b(i) (b) The partial structure of another condensation polymer is shown in Fig. 3.2. The key shows the parts represented by boxes. The polymer can be depolymerised into two different monomers: one is a dicarboxylic acid, the other a diol (a molecule containing two alcohol groups).
Complete Table 3.1 to show the molecular and empirical formulae of the monomers that form during depolymerisation. [4]
Fig. 3.2: a polymer chain with C4H8 and C3H6 segments joined by ester links, with a key showing the two shaded boxes
Table 3.1
molecular formulaempirical formula
monomer 1
monomer 2
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Concept: Ester linkage; molecular vs empirical formula
The −COO− link in Fig. 3.2 is an ester linkage (alcohol + carboxylic acid).
Monomer 1 is the dicarboxylic acid HOOC−C4H8−COOH; monomer 2 is the diol HO−C3H6−OH.
Molecular formula = actual number of atoms of each element. Empirical formula = simplest whole-number ratio of atoms.
molecular formulaempirical formula
monomer 1C6H10O4C3H5O2
monomer 2C3H8O2C3H8O2
b(ii) Depolymerisation of the polymers in plastics is a method used to recycle plastics. Describe two benefits of recycling plastics. [2]
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Concept: Environmental impact of plastics
1. Reduces land pollution, as most plastics are non-biodegradable.
2. Reduces air pollution, as plastics release poisonous gases during incineration.
Less land pollution (plastics are non-biodegradable) and less air pollution (no poisonous gases from incineration).
4 Cracking, alcohols, esters & isomers 11 marks

Hydrocarbon molecules from crude oil are used in a series of reactions, as shown in Fig. 4.1.

Fig. 4.1: flow diagram showing hydrocarbon molecules from crude oil, cracking to ethene, reaction A to ethanol, reaction B to compound X
a(i) Cracking forms smaller molecules from larger hydrocarbon molecules. Explain why this is essential. [1]
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Concept: Cracking
Large hydrocarbons are less useful and in lower demand than short-chain ones.
Cracking breaks down large hydrocarbons into shorter hydrocarbons, which are more useful and in higher demand.
a(ii) Identify a substance, other than alkenes, that is also made during cracking. [1]
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Concept: Products of cracking
Cracking produces alkanes/alkenes and hydrogen gas.
Hydrogen
b State the type of reaction that occurs in reaction A and state the conditions required. [2]
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Concept: Hydration of ethene
Reaction A: ethene (an alkene) → ethanol (an alcohol), by reacting ethene with steam.
Type of reaction: addition (of steam). Conditions: 300 °C, 60 atm and a H3PO4 catalyst.
c(i) (c) In reaction B, ethanol reacts fully with acidified potassium manganate(VII) to form compound X.
Draw the full structural (displayed) formula of compound X. [1]
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Concept: Oxidation of alcohols
Acidified KMnO4 oxidises an alcohol to a carboxylic acid: ethanol → ethanoic acid, CH3COOH.
Displayed formula of ethanoic acid: CH3 group joined to a C that has a double bond to O and a single bond to O-H
Ethanoic acid, CH3COOH (structure below).
c(ii) State the type of reaction that occurs when ethanol forms compound X. [1]
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Concept: Oxidation of alcohols
Ethanol is converted to ethanoic acid by an oxidising agent (acidified KMnO4).
Oxidation
d Ethanol also reacts with methanoic acid to form an organic product and water. Draw the full structural (displayed) formula of the organic product and state its name. [3]
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Concept: Esterification (condensation reaction)
Alcohol + carboxylic acid → ester + water. The ester linkage is −COO−.
Ethanol + methanoic acid → ethyl methanoate. The name is built from the alcohol (ethyl) first, then the acid (methanoate).
Displayed formula of ethyl methanoate: H-C(=O)-O-CH2-CH3 with every bond drawn out

The full question

Full question Question 4(d) and (e): ester structure and isomer answer with Teacher Dalson's notes (page 1 of 2)
Full question Question 4(d) and (e): ester structure and isomer answer with Teacher Dalson's notes (page 2 of 2)

Teacher Dalson’s annotated answer

Annotated Question 4(d) and (e): ester structure and isomer answer with Teacher Dalson's notes (page 1 of 2)
Annotated Question 4(d) and (e): ester structure and isomer answer with Teacher Dalson's notes (page 2 of 2)

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Ethyl methanoate (structure below).
e Another organic compound has the formula shown in Fig. 4.2. Suggest the relationship between the molecule shown in Fig. 4.2 and ethanol. Explain your answer. [2]
Fig. 4.2: displayed formula of an ether, two CH3 groups joined through an oxygen atom
Show working
Concept: Isomers
Fig. 4.2 is CH3−O−CH3, with molecular formula C2H6O.
Ethanol (C2H5OH) also has molecular formula C2H6O, but the atoms are arranged differently.

The full question

Full question Question 4(d) and (e): ester structure and isomer answer with Teacher Dalson's notes (page 1 of 2)
Full question Question 4(d) and (e): ester structure and isomer answer with Teacher Dalson's notes (page 2 of 2)

Teacher Dalson’s annotated answer

Annotated Question 4(d) and (e): ester structure and isomer answer with Teacher Dalson's notes (page 1 of 2)
Annotated Question 4(d) and (e): ester structure and isomer answer with Teacher Dalson's notes (page 2 of 2)

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They are isomers: same molecular formula but different structural formula.
5 Metal reactivity, displacement & electrolysis 8 marks

An excess of a metal is added to a solution containing an aqueous cation. The maximum temperature change is measured, as shown in Fig. 5.1. The experiment is repeated using different metals and different aqueous cations. All other conditions are kept constant. Table 5.1 shows the results.

Fig. 5.1: a beaker containing a solution with an aqueous cation, a piece of metal at the bottom and a thermometer
a Explain why two of the experiments result in no change in temperature. [2]
Table 5.1
experimentsolid metalaqueous cationmaximum temperature change / °C
1ZnPb2+6.0
2ZnH+8.5
3CuPb2+0.0
4CuAg+4.0
5PbH+2.5
6AgH+0.0
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Concept: Reactivity series & metal displacement
Reactivity series (mnemonic): P S C M A C Z I L H C S G — potassium, sodium, calcium, magnesium, aluminium, carbon, zinc, iron, lead, hydrogen, copper, silver, gold.
Experiment 3: Cu is less reactive than Pb, so Cu cannot displace Pb from its salt solution.
Experiment 6: Ag is below H in the reactivity series, so Ag cannot react with the acid — it is too unreactive.
Expt 3: Cu is less reactive than Pb, so no displacement. Expt 6: Ag is below H in the reactivity series, so it does not react with acid.
b Metal nitrates are used to provide the aqueous metal ions in this experiment. Explain why metal nitrates, rather than other salts, are most suitable to provide aqueous metal ions. [1]
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Concept: Solubility rules
The ions must be in solution, so the salt needs to be soluble.
All nitrates are soluble in water.
c Write an ionic equation with state symbols for the reaction in experiment 4. [2]
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Concept: Ionic equations for displacement
Experiment 4: Cu + Ag+. Cu is more reactive than Ag, so Cu displaces Ag.
Cu loses 2 electrons; each Ag+ gains 1, so 2 Ag+ are needed to balance the charge.
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
d The metals in Table 5.1 can be used as electrodes in contact with an electrolyte to generate electrical energy. Deduce which two metals in Table 5.1 generate the largest amount of electrical energy when they are used in this way. [1]
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Concept: Simple cell
In a simple cell, the larger the difference in reactivity between the two metals, the larger the voltage (electrical energy).
Zn is the most reactive metal in the table and Ag is the least reactive.
Zinc and silver
e Separate dilute aqueous solutions containing zinc, copper, lead and silver ions are electrolysed using inert electrodes. Which two metals do not form at the negative electrode during electrolysis of their aqueous salts? Explain your answer. [2]
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Concept: Electrolysis of aqueous solutions (cathode)
Every aqueous solution also contains H+ (and OH−) ions from water.
At the negative electrode (cathode), the less reactive ion is discharged first. Ag+ and Cu2+ are below hydrogen, so Ag(s) and Cu(s) form. Pb2+ and Zn2+ are above hydrogen, so H+ is discharged instead.

The full question

Full question Question 5(e): electrolysis answer with Teacher Dalson's reactivity-series mnemonic and cathode sketch (page 1 of 2)
Full question Question 5(e): electrolysis answer with Teacher Dalson's reactivity-series mnemonic and cathode sketch (page 2 of 2)

Teacher Dalson’s annotated answer

Annotated Question 5(e): electrolysis answer with Teacher Dalson's reactivity-series mnemonic and cathode sketch (page 1 of 2)
Annotated Question 5(e): electrolysis answer with Teacher Dalson's reactivity-series mnemonic and cathode sketch (page 2 of 2)

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Zinc and lead. Both are above hydrogen in the electrochemical series; since all aqueous solutions contain H+ ions, H+ is selectively discharged to form H2(g) instead of Zn2+ and Pb2+.
6 Atomic structure, isotopes & bonding 6 marks

The relative atomic mass of chlorine is 35.5.

a State the meaning of relative atomic mass. [1]
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Concept: Relative atomic mass (Ar)
The word ‘average’ matters: chlorine has two isotopes, which is why its Ar is 35.5.
The average mass of an atom of an element relative to 1/12 the mass of an atom of carbon-12 (12C).
b(i) (b) The symbols for two isotopes of chlorine are 3517Cl and 3717Cl.
Explain, in terms of atomic structure, why these two atoms of chlorine are isotopes. [2]
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Concept: Protons, neutrons, electrons
Compare protons, neutrons and electrons for the two atoms.
Neutrons = nucleon number − proton number: 35 − 17 = 18 and 37 − 17 = 20.
Both atoms have the same number of protons (17) but different mass numbers (35 and 37), hence a different number of neutrons (18 and 20 respectively).
b(ii) Explain why the two isotopes of chlorine have identical chemical properties. [1]
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Concept: Chemical properties depend on valence electrons
Chemical properties are determined by the number of valence (outer-shell) electrons.
Both atoms have the same number of valence electrons (7) / the same electronic configuration.
c Chlorine forms a bond with sodium and a different type of bond with hydrogen. Explain, in terms of electrons, how each type of bond is formed. [2]
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Concept: Ionic vs covalent bond formation
Ionic (Na and Cl): a sodium atom loses an electron to form Na+, while a chlorine atom gains an electron to form Cl−. The oppositely charged Na+ and Cl− ions are attracted by strong electrostatic forces of attraction, forming an ionic bond.
Covalent (H and Cl): a Cl atom and an H atom each share an electron to achieve a noble gas electronic configuration, forming a covalent bond.
Ionic: transfer of an electron from Na to Cl. Covalent: sharing of an electron pair between H and Cl.
7 Energetics, mole calculations & states of matter — carbon capture 12 marks

Read the passage below, then answer the questions that follow.

Carbon capture storage (CCS)

Some power stations that generate electricity use methane as a fuel. Carbon capture storage (CCS) reduces the amount of carbon dioxide entering the atmosphere. This is done by capturing carbon from methane in the form of carbon dioxide. Carbon from methane can be captured either before the methane burns (pre-combustion) or after combustion (post-combustion) as carbon dioxide.

Carbon dioxide can be stored deep underground in the space left when crude oil is extracted.

Pre-combustion capture

Steam methane reforming (SMR) and water-gas shift (WGS) are two reactions involved in pre-combustion capture of carbon from methane.

In SMR, steam reacts with methane at a pressure of 3–25 atmospheres and at a temperature of 700–1000 °C.

SMR reaction equation: CH4 + H2O ⇌ CO + 3H2    ΔH = +206 kJ/mol

The SMR reaction is reversible and therefore the products contain small amounts of unreacted methane and water. The reaction has a percentage yield of 74%.

This is followed by the WGS reaction at 200–400 °C using a catalyst.

WGS reaction equation: CO + H2O ⇌ CO2 + H2    ΔH = −41 kJ/mol

The carbon dioxide produced in the WGS reaction is separated from hydrogen and goes to be stored. The hydrogen is burned as a fuel to generate electricity. It does not cause environmental harm when it is combusted.

Post-combustion capture

Some power stations capture the carbon dioxide post-combustion. In order to easily separate the carbon dioxide after the reaction, the methane is burned in pure oxygen rather than air.

CH4 + 2O2 → CO2 + 2H2O

The water vapour that is produced is separated from the mixture of products. The carbon dioxide then goes to be stored.

Storing carbon dioxide

The carbon dioxide collected from the power station is pressurised. This produces liquid carbon dioxide which can then be pumped directly to be stored deep underground.

The fuel industry thinks that the space made available from the extraction of crude oil will be enough to store all the carbon dioxide produced in the world from fossil fuel combustion for many years.

a Draw energy profile diagrams to represent the steam methane reforming (SMR) reaction and the water-gas shift (WGS) reaction. For each diagram, label the activation energy and the enthalpy change of reaction. [3]
Two blank sets of axes, labelled SMR reaction and WGS reaction, for drawing energy profile diagrams
Show working
Concept: Energy profile diagrams
SMR is endothermic (ΔH = +206 kJ/mol): the products (CO + 3H2) sit higher than the reactants (CH4 + H2O).
WGS is exothermic (ΔH = −41 kJ/mol): the products (CO2 + H2) sit lower than the reactants (CO + H2O).
In both: the activation energy is the arrow from the reactant level up to the peak; ΔH is the arrow between the reactant and product levels.
Completed energy profile diagrams: SMR with products higher than reactants (+206 kJ/mol) and WGS with products lower than reactants (-41 kJ/mol), each with the activation energy labelled
SMR: products higher than reactants, ΔH = +206 kJ/mol. WGS: products lower than reactants, ΔH = −41 kJ/mol. Label Ea on each.
b(i) (b) The SMR and WGS reactions produce hydrogen fuel that does not cause environmental harm.
Explain why both the SMR and WGS reactions are required to produce the hydrogen fuel without causing environmental harm. [1]
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Concept: Linking the two reactions
SMR converts CH4 and steam into H2, but also produces CO.
The WGS reaction then removes that CO.
SMR removes harmful CH4 while producing harmful CO, which is then removed by the WGS reaction.
b(ii) Explain why the hydrogen fuel produced does not cause environmental harm when combusted. [1]
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Concept: Combustion of hydrogen
2H2 + O2 → 2H2O
It produces water as the only by-product, which does not cause any environmental harm.
c Calculate the mass of hydrogen that is made in the SMR reaction from 1 tonne of methane. (1 tonne = 106 g) [3]
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Concept: Mole calculation with percentage yield
CH4 + H2O ⇌ CO + 3H2
Moles of CH4 = 106 ÷ (12 + 4) = 62 500 mol
Assuming 100% yield: moles of H2 = 3 × 62 500 = 187 500 mol
The percentage yield is 74% (from the passage): actual moles of H2 = 0.74 × 187 500 = 138 750 mol
Mass of H2 = 138 750 × 2 = 277 500 g
277 500 g (2.775 × 105 g)
d(i) (d) Suggest how carbon dioxide is separated from the mixture of combustion products in the post-combustion capture process. [1]
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Concept: Separating mixtures
Carbon dioxide and water vapour have different boiling points.
Carbon dioxide and water vapour have different boiling points, so they can be separated using fractional distillation.
d(ii) Explain why it is more difficult to separate carbon dioxide from the mixture of combustion products if air is used for combustion of methane rather than pure oxygen. [1]
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Concept: Composition of air
Air is a mixture of gases (N2, O2, Ar, CO2).
Air is a mixture of oxygen and other gases, so the products will also contain other gases such as argon and nitrogen.
e Describe how the arrangement and movement of carbon dioxide molecules change when it is pressurised for storage. [2]
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Concept: Gas → liquid: particle arrangement and movement
Arrangement: from being far apart and in a random arrangement, to more closely packed in a disordered manner.
Movement: from moving in all directions at high speed, to sliding past each other across the liquid.
Gas → liquid: particles go from far apart/random to closely packed/disordered, and from fast random motion to sliding past one another.
8 Activation energy, oxidation states & covalent bonding 8 marks

The first self-lighting matches were small wooden sticks with a tip that was coated with compounds called an ignition mixture. The matches were safe to carry because the compounds in the ignition mixture only start to react when rubbed against a rough surface. The rubbing generated heat to start the chemical reaction.

a Suggest why the compounds only react when heated. Use ideas about bonds in your answer. [1]
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Concept: Activation energy; bond breaking and forming
Heat (thermal energy) provides the activation energy needed to break existing bonds in the reactants.
Once bonds are broken, the atoms rearrange and new, stable bonds can form to make the products.
Heat provides the activation energy required to break the existing bonds in the reactants, so new bonds can form in the products.
b(i) (b) The ignition mixture contained a compound of antimony, Sb. The equation for a reaction in the ignition mixture is:
Sb2S3(s) + 3KClO3(s) → 3KCl(s) + 3SO2(g) + solid product X
State the name of solid product X. [1]
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Concept: Naming compounds with oxidation states
In Sb2S3, sulfur is S2−, so antimony is Sb3+ (2 × 3+ = 6+ balances 3 × 2−).
Antimony ends up combined with oxygen (check: 9 O on the left = 6 O in 3SO2 + 3 O in X), so X is Sb2O3. The oxidation state (III) goes in the name.

The full question

Full question Question 8(b): oxidation state workings for Sb2S3, KClO3 and SO2 with Teacher Dalson's notes (page 1 of 2)
Full question Question 8(b): oxidation state workings for Sb2S3, KClO3 and SO2 with Teacher Dalson's notes (page 2 of 2)

Teacher Dalson’s annotated answer

Annotated Question 8(b): oxidation state workings for Sb2S3, KClO3 and SO2 with Teacher Dalson's notes (page 1 of 2)
Annotated Question 8(b): oxidation state workings for Sb2S3, KClO3 and SO2 with Teacher Dalson's notes (page 2 of 2)

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Antimony(III) oxide
b(ii) State the change in oxidation state of sulfur during this reaction. [1]
Show working
Concept: Oxidation states
In Sb2S3: S is −2 (as S2−).
In SO2: S + 2(−2) = 0, so S = +4.
From −2 to +4
b(iii) Identify the other element whose oxidation state changes and give the change in oxidation state that occurs. [1]
Show working
Concept: Oxidation states
In KClO3: K is +1 and O is −2, so +1 + x + 3(−2) = 0, giving Cl = +5.
In KCl: Cl is −1.
Chlorine: from +5 to −1
b(iv) The gaseous product of the reaction is passed through water to form an aqueous solution. Predict the pH of this solution. Explain your reasoning. [2]
Show working
Concept: Acidic oxides
The gaseous product is SO2, an acidic oxide, so the solution is acidic (pH below 7).
SO2 + H2O → H2SO3 (sulfurous acid), which is a weak acid.
pH 3. SO2 dissolves in water to form a weak acid, sulfurous acid.
c(i) (c) The proton (atomic) number of antimony is 51. Antimony hydride is a compound that contains antimony and hydrogen.
Suggest the formula for antimony hydride. [1]
Show working
Concept: Valency and covalent formulae
Antimony is in Group 15 (5 outer electrons), so it needs 3 more electrons to achieve a noble gas configuration: valency 3. Hydrogen has valency 1.
SbH3
c(ii) Explain your reasoning. [1]
Show working
Concept: Valency and covalent formulae
Valency = the number of electrons an atom needs to share to achieve a noble gas configuration.
Sb is in Group 15 and has a valency of 3, while H has a valency of 1. Therefore 1 Sb atom requires 3 H atoms to share electrons to achieve noble gas configuration.

Section B Answer one question · 10 marks

9 Titration, acid strength & basicity 10 marks

A student has a bottle of aqueous sodium carbonate of unknown concentration. The student titrates 25.0 cm3 of the solution against dilute acid A until the sodium carbonate is completely neutralised. The student repeats the titration with other dilute acids, B, C, D, E and F, and calculates a mean titre for each acid. The acids used and the mean titres are shown in Table 9.1.

a The student says that an error was made in the recording of the titre for acid B. Explain why the student is correct. [1]
Table 9.1
acidformulaconcentration of acid in mol/dm3mean titre / cm3
AHCl(aq)0.1039.50
BHNO3(aq)0.2039.55
CH2SO4(aq)0.1019.75
DHCl(aq)0.2019.80
EHNO3(aq)0.507.90
FH2SO4(aq)0.504.00
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Concept: Strong monobasic acids
Compare B and D: HNO3 and HCl are both strong, monobasic acids, and both are at 0.20 mol/dm3.
Same concentration of the same type of acid should need the same titre — D needs 19.80 cm3, but B is recorded as 39.55 cm3 (the value expected for the 0.10 mol/dm3 acid A).

The full question

Full question Question 9(a) and (b): acid titration table with Teacher Dalson's notes on strong, weak, monobasic and dibasic acids

Teacher Dalson’s annotated answer

Annotated Question 9(a) and (b): acid titration table with Teacher Dalson's notes on strong, weak, monobasic and dibasic acids

Tap either image to enlarge

B and D are both strong monobasic acids at the same concentration, so their titres should be about the same; B’s 39.55 cm3 does not match D’s 19.80 cm3.
b(i) (b) Explain the difference in the titres for acids A and C in Table 9.1. [2]
Show working
Concept: Monobasic vs dibasic acids
HCl → H+ + Cl− (monobasic: 1 H+ per molecule).
H2SO4 → 2H+ + SO42− (dibasic: 2 H+ per molecule).
H2SO4 is dibasic while HCl is monobasic. For the same amount of Na2CO3 and the same acid concentration, H2SO4 produces twice the moles of H+, so it needs half the titre.
b(ii) Describe the trend in the concentration of acid and the mean titre as shown in Table 9.1. [1]
Show working
Concept: Reading trends from data
Example: HCl rises from 0.10 to 0.20 mol/dm3 while the mean titre falls from 39.50 to 19.80 cm3.
As the concentration of the acid increases, the mean titre decreases.
c Propanoic acid, C2H5COOH, is a weak acid. Predict the mean titre when a solution of 0.10 mol/dm3 propanoic acid is used. Explain your answer. [2]
Show working
Concept: Weak acids and basicity
Propanoic acid has only one acidic H, so it is monobasic, just like HCl.
In a titration to neutralisation, a weak acid eventually releases all of its H+, so strength does not change the volume needed.

The full question

Full question Question 9(c) and (d): weak acid reasoning and titration calculation with Teacher Dalson's notes

Teacher Dalson’s annotated answer

Annotated Question 9(c) and (d): weak acid reasoning and titration calculation with Teacher Dalson's notes

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Mean titre = 39.50 cm3. It is monobasic just like HCl, so at the same concentration (0.10 mol/dm3) it provides the same number of moles of H+ as HCl.
d(i) (d) Use the mean titre for acid A to calculate the concentration of the aqueous sodium carbonate. [3]
Show working
Concept: Titration calculation
2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + CO2(g) + H2O(l)
Moles of HCl = 0.10 × 39.50/1000 = 0.00395 mol
Moles of Na2CO3 = ½ × 0.00395 = 0.001975 mol
Concentration = 0.001975 ÷ (25.0/1000) = 0.0790 mol/dm3
0.0790 mol/dm3
d(ii) In each reaction, hydrogen ions react with carbonate ions. Construct an ionic equation for the reaction of hydrogen ions with carbonate ions. [1]
Show working
Concept: Ionic equations
Acid + carbonate → water + carbon dioxide.
2H+(aq) + CO32−(aq) → H2O(l) + CO2(g)
10 Limiting reactant, percentage yield & salt preparation 10 marks

A student prepares calcium chloride by reacting 1.0 g of solid calcium carbonate with 25.0 cm3 of 0.5 mol/dm3 hydrochloric acid.

a Deduce the limiting reactant in this reaction. Show your working and include a balanced equation. [3]
Show working
Concept: Limiting reactant
2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l)
Moles of CaCO3 = 1.0 ÷ 100 = 0.01 mol
Moles of HCl = (25.0/1000) × 0.5 = 0.0125 mol
From the equation, 0.01 mol of CaCO3 needs 0.02 mol of HCl. Only 0.0125 mol of HCl is present, so HCl runs out first.
Limiting reactant: hydrochloric acid
b The student separates solid calcium chloride at the end of the reaction. The mass of solid calcium chloride is 0.6 g. Calculate the percentage yield for the reaction. Show your working. [3]
Show working
Concept: Percentage yield
Use the limiting reactant (HCl): theoretical moles of CaCl2 = ½ × 0.0125 = 0.00625 mol
Mr of CaCl2 = 40 + 2(35.5) = 111, so theoretical mass = 0.00625 × 111 = 0.69375 g
% yield = (actual mass ÷ theoretical mass) × 100% = (0.6 ÷ 0.69375) × 100%
86.5%
c(i) (c) The student adds a lump of solid calcium carbonate to dilute sulfuric acid in order to prepare calcium sulfate. The reaction stops after a few seconds and the percentage yield is very low.
Explain why this happens. [2]
Show working
Concept: Insoluble salts coating a reactant
CaCO3 + H2SO4 → CaSO4(s) + H2O + CO2
An insoluble salt, calcium sulfate, forms on the surface of the solid calcium carbonate, acting as a protective layer that prevents the acid from reacting further with it.
c(ii) Suggest a more suitable method and reagents to prepare calcium sulfate. Explain your reasoning. [2]
Show working
Concept: Preparing an insoluble salt: precipitation
Calcium sulfate is insoluble, so it is best made by precipitation: mix a soluble calcium salt with a soluble sulfate.
1. Add aqueous calcium nitrate to dilute sulfuric acid, and stir until a precipitate forms.
2. Filter to obtain the CaSO4(s) precipitate as the residue.
3. Wash it with distilled water and dry it between filter papers.
Method: precipitation (calcium nitrate + dilute sulfuric acid, then filter, wash and dry).

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Teacher Dalson
Teacher Dalson
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FAQ

Common questions about this paper

What topics are tested in the 2025 O-Level Chemistry Paper 2 (6092/02)?

Paper 2 tests ionic, metallic and covalent bonding (including graphite and dot-and-cross diagrams), ions and electronic configuration, condensation polymers and recycling, cracking, alcohols, esters and isomers, metal reactivity and electrolysis, isotopes and relative atomic mass, energetics with energy profile diagrams, oxidation states and acidic oxides, and then either a titration question (acid strength, basicity, concentration) or a limiting reactant, percentage yield and salt preparation question in Section B.

How do you find the limiting reactant and the percentage yield?

Convert each reactant to moles (mass ÷ Mr, or concentration × volume in dm³), then use the mole ratio in the balanced equation to see which reactant runs out first — that is the limiting reactant. The theoretical yield is calculated from the limiting reactant, and percentage yield = (actual mass ÷ theoretical mass) × 100%.

Why do ionic compounds conduct electricity only when molten or dissolved?

In the solid state the ions in a giant ionic lattice are held in fixed positions by strong electrostatic forces of attraction, so there are no free-moving charge carriers. When molten or dissolved in water, the ions are free to move and can carry charge. Metals conduct in every state because of their ‘sea’ of delocalised electrons.

How can you tell whether a polymer is a polyamide or a polyester?

Look at the linkage between the repeat units. An amide linkage (−CONH−) means a polyamide, such as nylon, made from a diamine and a dicarboxylic acid. An ester linkage (−COO−) means a polyester, made from a diol and a dicarboxylic acid. To find the monomers, break the polymer at each linkage and restore the −OH / −H groups.

Why does sulfuric acid need only half the titre of hydrochloric acid at the same concentration?

Sulfuric acid is dibasic: each molecule releases 2 H+ ions, whereas hydrochloric acid is monobasic and releases only 1. At the same concentration, sulfuric acid supplies twice the moles of H+ per cm³, so it neutralises the same amount of sodium carbonate with half the volume.