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2026 O-Level Chemistry Paper 2 — Blank PDF Download & Full Worked Solutions

GCE O-Level Chemistry (6092/02) · Paper 2 · October/November 2026 · 80 marks

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Section A Answer all questions · 70 marks

1 Air pollution & the carbon cycle 8 marks
a(i) State the names of two pollutant gases that cause acid rain. [2]
Show working
Concept: Acid rain formation
Acid rain is caused by pollutant gases dissolving in atmospheric moisture to form acidic solutions.
Sulfur dioxide and nitrogen dioxide
a(ii) Describe one harmful effect of acid rain. [1]
Show working
Concept: Environmental effects of acid rain
Acid rain falls into rivers and lakes, acidifying water bodies and killing aquatic life.
Alternatively: it damages or corrodes buildings and statues made of limestone or metal.
Acidifies rivers/lakes, killing aquatic life (or corrodes limestone/metal buildings and statues)
b(i) The carbon cycle regulates the amount of carbon dioxide in the atmosphere. State two processes that add carbon dioxide to the atmosphere. [2]
Show working
Concept: The carbon cycle
Processes that release CO2 into the atmosphere include respiration and the combustion of fossil fuels.
Respiration and combustion of fossil fuels
b(ii) State one process that removes carbon dioxide from the atmosphere. [1]
Show working
Concept: The carbon cycle
Green plants remove CO2 from the atmosphere during photosynthesis.
Photosynthesis
c Increased levels of carbon dioxide in the atmosphere may contribute to global warming. Describe two potential effects of global warming. [2]
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Concept: Effects of global warming
Global warming raises the average temperature of the Earth, which can melt polar ice caps, causing rising sea levels, flooding of low-lying coastal areas and loss of habitat for polar bears.
It can also encourage the breeding of mosquitoes, leading to the spread of mosquito-borne diseases such as Zika and dengue.
Any two of: melting polar ice caps & rising sea levels / flooding of low-lying coastal areas / loss of habitat (e.g. polar bears) / increased breeding of mosquitoes & spread of mosquito-borne diseases
2 Rates of reaction — enzymes as catalysts 7 marks

Vanillin is a natural compound found in vanilla food flavouring. Vanillin can be made synthetically from aqueous glucose at 20°C. The reaction uses an enzyme.

a Describe the role of the enzyme in this chemical reaction. [2]
Show working
Concept: Enzymes as biological catalysts
An enzyme is a biological catalyst, so it increases the rate of the reaction.
It does this by providing an alternative reaction pathway with a lower activation energy, and remains chemically unchanged at the end of the reaction.
The enzyme increases the rate of production of vanillin from aqueous glucose by providing an alternative reaction pathway with a lower activation energy, and remains chemically unchanged at the end of the reaction.
b A student states that vanillin will be made more quickly if a higher temperature is used. Explain why the student is only partly correct. [2]
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Concept: Enzymes and optimum temperature
The statement is correct only if the temperature is increased towards the optimum temperature, where the rate of reaction is highest.
If the temperature exceeds the optimum, the enzyme would denature and can no longer catalyse/speed up the reaction, so the statement would not hold true.
Only correct up to the optimum temperature; beyond that the enzyme denatures and can no longer catalyse the reaction.
c Explain the effect of decreasing the concentration of glucose on the rate of the reaction to make vanillin.
Use ideas about particles in your answer. [3]
Show working
Concept: Collision theory — concentration
Decreasing the concentration of glucose decreases the number of glucose molecules per unit volume.
There will be a lower frequency of collisions between the glucose molecules and the enzymes.
This results in a lower frequency of effective collisions, causing a lower rate of reaction to make vanillin.
Lower glucose concentration → fewer glucose molecules per unit volume → lower collision frequency with enzymes → lower frequency of effective collisions → lower rate of reaction.
3 Atomic structure & isotopes 6 marks

A sample of bromine contains two isotopes, 79Br and 81Br.

a State how the structure of an atom of 79Br differs from an atom of 81Br. [2]
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Concept: Isotopes
Both isotopes have 35 protons (bromine's proton number), but differ in neutron number.
Mass number − proton number = neutron number: 79 − 35 = 44; 81 − 35 = 46.
79Br has 44 neutrons while 81Br has 46 neutrons.
b Explain why atoms of 79Br have identical chemical properties to atoms of 81Br. [1]
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Concept: Chemical properties depend on electron arrangement
Both atoms of 79Br and 81Br have 35 electrons, since isotopes have the same proton (and electron) number.
Hence they would need to gain or share the same number of electrons (1 electron) and react identically.
Both isotopes have 35 electrons, so they need to gain/share 1 electron and react identically.
c State the charge on the nucleus of an atom of 79Br. [1]
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Concept: Nuclear charge = number of protons
The nuclear charge equals the number of protons, which is based on bromine's group/proton number (bromine has 35 protons) and cannot be read from an electronic configuration such as 2.8.
35+
d Some isotopes are unstable. An isotope of bromine can break down to form isotopes of krypton or selenium. The nuclide symbol for krypton is 8036Kr. Selenium is in Group 16 of the Periodic Table. The isotope of selenium that is formed contains 41 neutrons. Write the nuclide symbol for this isotope of selenium. [2]
Show working
Concept: Nuclide symbols from the Periodic Table
From the Periodic Table, selenium (Group 16) has a proton number of 34.
Mass number = protons + neutrons = 34 + 41 = 75.
7534Se
4 Ionic equations & percentage composition 5 marks

Lead(II) chromate, PbCrO4, is a yellow compound used in some paints. Lead(II) chromate is made by a precipitation reaction between aqueous lead(II) nitrate and aqueous potassium chromate.

a Complete the equation for this precipitation reaction. Include state symbols.
Pb(NO3)2(....) + K2CrO4(aq) → PbCrO4(....) + 2....(aq) [2]
Show working
Concept: Precipitation & spectator ions
Pb2+ and CrO42− combine to form the insoluble solid PbCrO4, while K+ and NO3− remain as spectator ions in solution.
Pb(NO3)2(aq) + K2CrO4(aq) → PbCrO4(s) + 2KNO3(aq)
b Calculate the total percentage by mass of metallic elements in lead(II) chromate. [3]
Show working
Concept: Percentage composition by mass
Metallic elements in PbCrO4 are Pb and Cr (both metals).
Total % by mass of Pb & Cr = (207 + 52) / (207 + 52 + 4(16)) × 100%
= 259 / 323 × 100% = 80.185%
80.2% (3 s.f.)
5 Hydrocarbons — empirical formula, alkanes & bromine water 6 marks
a Explain what is meant by the term hydrocarbon. [1]
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Concept: Hydrocarbons
A hydrocarbon is covalent and has a simple molecular structure.
It is a molecule/compound that contains carbon and hydrogen atoms only.
b A hydrocarbon contains 81.8% by mass of carbon. Determine the empirical formula of the hydrocarbon. Show your working. [3]
Show working
Concept: Empirical formula from percentage composition
Taking a 100 g sample: mass of C = 81.8 g, mass of H = 100 − 81.8 = 18.2 g.
Moles of C = 81.8 / 12 = 6.81667 mol; moles of H = 18.2 / 1 = 18.2 mol.
Mole ratio C : H = 1 : 2.669 ≈ 3 : 8 (simplest ratio).
C3H8 (propane)
c Another hydrocarbon has an empirical formula of C3H7. Predict what is observed when aqueous bromine is added to this hydrocarbon. Explain your reasoning. [2]
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Concept: Bromine water test for saturation
The molecular formula must be a whole-number multiple of C3H7; doubling gives C6H14, which fits the alkane general formula CnH2n+2 (saturated, lacking a C=C bond).
Since the hydrocarbon is saturated, it will not react with aqueous bromine.
observation: red-brown aqueous bromine remains red-brown / does not decolourise.
explanation: the hydrocarbon is saturated (C6H14 fits CnH2n+2, lacking a C=C bond) and will not react with aqueous Br2.
6 Alkenes, addition reactions & energy profiles 8 marks

Ethene reacts with steam in an addition reaction to form ethanol.
CH2=CH2 + H2O → C2H5OH

a Explain what is meant by an addition reaction. [1]
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Concept: Addition reactions
It is where two (or more, in the case of polymerisation) molecules/reactants join/react together to form a single product.
b Complete the tables to show the type and number of each bond in the reactants and in the product. [3]
Blank tables for the type and number of each bond in the reactants (C=C, C-H, O-H) and the product
Show working
Concept: Counting bonds from structural formulae
Reactants (CH2=CH2 + H2O): 1 C=C, 4 C−H, 2 O−H.
Product (C2H5OH, drawn as H−C(H)(H)−C(H)(H)−O−H): 1 C−C, 5 C−H, 1 C−O, 1 O−H.
Completed bond tables: reactants 1 C=C, 4 C-H, 2 O-H; product 1 C-C, 5 C-H, 1 C-O, 1 O-H
Reactants: C=C ×1, C−H ×4, O−H ×2. Product: C−C ×1, C−H ×5, C−O ×1, O−H ×1.
c(i) The reaction of ethene with steam is exothermic. Complete the energy profile diagram for the reaction. Include labels for the activation energy, Ea, and reaction enthalpy change, ΔH. [3]
Blank energy profile diagram with axes for energy against progress of reaction, with the reactants level marked
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Concept: Energy profile diagrams — exothermic reactions
Since the reaction is exothermic, the product energy level is drawn below the reactants energy level.
Ea is measured from the reactants level up to the highest point (the peak); ΔH is measured from the reactants level down to the products level.
Completed energy profile diagram: a hump from reactants (C2H4 + H2O) up to a peak labelled Ea, then down to a lower products (C2H5OH) level labelled delta H
A curve rising from the reactants (C2H4 + H2O) level to a peak (Ea labelled from the reactants level to the peak), then falling to a lower products (C2H5OH) level (ΔH labelled from reactants level down to products level).
c(ii) The conditions used for the reaction include using a temperature of 300°C. Explain why the reaction does not take place at room temperature. [1]
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Concept: Activation energy & particle energy distribution
At a low (room) temperature, particles have a low average kinetic energy.
The reaction has a high activation energy, hence at room temperature most reactant particles do not have energy equal to or greater than the activation energy to result in effective collisions for the reaction to occur.
7 Electrolysis of aqueous copper(II) sulfate 10 marks

50.0 cm³ of dilute aqueous copper(II) sulfate is electrolysed using inert electrodes, as shown in Fig. 7.1.

Fig. 7.1: a cell with two inert electrodes dipped in aqueous copper(II) sulfate, connected to a battery

The total mass of copper formed at the negative electrode is measured every minute. Fig. 7.2 shows a graph of the results.

Fig. 7.2: graph of mass of copper (g) against time (minutes), rising linearly from 0 to 0.32 g at 6 minutes then levelling off
a At the start of the electrolysis, the aqueous solution is blue. Describe the appearance of the aqueous solution and predict its pH after 6 minutes. Explain your answer. [4]
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Concept: Selective discharge of ions
Cu2+ and OH− ions are selectively discharged over H+ and SO42− ions respectively.
From the graph, the mass of copper formed remained constant at 0.32 g after 6 minutes, indicating that all Cu2+ ions were completely reduced — so the concentration of Cu2+ is zero, giving a colourless solution.
Since OH− ions are discharged preferentially, their concentration becomes lower than that of H+ ions, making the electrolyte more acidic, so pH decreases from 7 (CuSO4 is neutral) to less than 7.
appearance: colourless. pH: less than 7. Cu2+ and OH− ions selectively discharged; after 6 minutes all Cu2+ ions are completely reduced (colourless), while [OH−] < [H+] makes the solution acidic.
b At the start of the electrolysis, 50.0 cm³ of aqueous copper(II) sulfate is used. The solution contains 0.32 g of aqueous Cu2+ ions. Calculate the concentration of Cu2+ ions in the aqueous solution at 2.5 minutes. [3]
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Concept: Reading a graph & concentration calculations
At 2.5 minutes, from the graph, 0.135 g of Cu has formed.
Mass of Cu2+ ions remaining = 0.32 − 0.135 = 0.185 g.
Moles of Cu2+ remaining = 0.185 / 64 = 0.002890625 mol.
Concentration = 0.002890625 / (50/1000) dm³
0.0578 mol/dm³ (3 s.f.)
c(i) The experiment is repeated using the same apparatus. The volume and concentration of aqueous copper(II) sulfate also stay the same. The positive electrode is replaced with a piece of copper with a mass of 5.0 g. Predict the mass of the positive electrode after 6 minutes. [1]
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Concept: Reactive electrodes in electrolysis
5.0 − 0.32 = 4.68 g
4.68 g
c(ii) Explain your answer. Include ionic equations. [2]
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Concept: Reactive copper anode
At the positive (reactive, copper) anode: Cu(s) → Cu2+(aq) + 2e−
At the negative cathode: Cu2+(aq) + 2e− → Cu(s)
With the same current/rate of electron transfer as the previous experiment, 0.32 g of copper still forms at the cathode. From the ionic equations, 0.32 g of copper will be lost due to the oxidation of the copper anode.
Cu(s) → Cu2+(aq) + 2e− at the anode (oxidation) and Cu2+(aq) + 2e− → Cu(s) at the cathode (reduction); the reactive anode loses the same mass that the cathode gains.
8 Redox & lithium-ion batteries 12 marks

Many devices, such as mobile phones, use lithium-ion batteries. A new battery for a mobile phone has a mass of 30.0 g. Lithium-ion batteries contain carbon, lithium and cobalt.

Table 8.1 — percentage composition by mass
percentage composition by mass
cobalt15.0
lithium9.0
a(i) Cobalt is present in the battery as an oxide of cobalt, CoO2. Calculate the mass of CoO2 in a new battery. [3]
Show working
Concept: Mass → % composition → compound mass
Mass of Co in a new battery = 15/100 × 30.0 g = 4.5 g.
% by mass of Co in CoO2 = 59 / (59 + 2(16)) × 100% = 64.835%.
Mass of CoO2 = 4.5 / 64.835 × 100 = 6.94 g
6.94 g (3 s.f.)
a(ii) Explain why the mass ratio of cobalt to lithium in a battery is a much higher value than the mole ratio of cobalt to lithium. [1]
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Concept: Converting mass ratio to mole ratio
Mole ratio = mass ÷ molar mass for each element. Mass ratio Co:Li = 15:9; mole ratio = (15/59):(9/7) ≈ 1.667:1 vs 0.198:1.
Co has a much higher molar mass (59 g/mol) compared to Li's molar mass (7 g/mol).
The mole ratio is obtained by dividing each element's mass by its molar mass; since Co's molar mass (59) is much greater than Li's (7), dividing reduces Co's share far more, making the mole ratio much lower than the mass ratio.

How a lithium-ion battery works. Lithium-ion batteries have two electrodes, each in contact with an electrolyte. A semi-permeable barrier separates the electrodes. When discharging, electrons flow through the device from one electrode to another; when charging, the flow of electrons is reversed.

Fig. 8.1: electrode 1 (graphite containing lithium metal, releases electrons) and electrode 2 (cobalt oxide CoO2, uses electrons), separated by a semi-permeable barrier and electrolyte, with Li+ ions flowing across

Reactions at the electrodes when discharging: electrode 1: Li → Li+ + e−. The lithium ions travel from electrode 1 through the electrolyte and the semi-permeable barrier (which does not allow any other substances through) to electrode 2: electrode 2: CoO2 + Li+ + e− → LiCoO2.

b Redox reactions occur at the electrodes when a battery is discharging. For each reaction, identify the element that is oxidised or reduced and state the changes in oxidation state. [3]
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Concept: Oxidation states in redox reactions
Electrode 1: Li → Li+ + e− — lithium loses an electron (oxidation), oxidation state 0 → +1.
Electrode 2: CoO2 + Li+ + e− → LiCoO2 — cobalt gains an electron (reduction), oxidation state +4 → +3.
Element oxidised: lithium (0 → +1). Element reduced: cobalt (+4 → +3).
c Write an overall equation for the reaction that happens in a battery when it is charging. [2]
Show working
Concept: Reversing the discharge reactions
When charging, both electrode reactions reverse: LiCoO2 → CoO2 + Li+ + e− and Li+ + e− → Li.
LiCoO2 → CoO2 + Li
d Electrode 1 contains carbon in the form of graphite. Lithium ions fit inside spaces between the layers in graphite. Explain why the structure of graphite makes it suitable for use as an electrode. [1]
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Concept: Structure and bonding of graphite
Each carbon atom is only covalently bonded to three other carbon atoms, so there is one electron per carbon atom that can be delocalised.
Each carbon atom is only covalently bonded to three other carbon atoms, where there is one electron per carbon atom that can be delocalised and act as a mobile charge carrier to conduct electricity.
e Explain why lithium ions are smaller than most other metal ions. [1]
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Concept: Ionic radius & electron shells
Fewer electron shells means a smaller distance between the nucleus and the valence shell.
Lithium ions have one electron shell, resulting in a smaller ionic radius compared to most other metal ions with two or more electron shells.
f The lithium metal in electrode 1 is in contact with the electrolyte, which contains an organic solvent. Explain why water is not used for the solvent in the electrolyte. [1]
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Concept: Reactivity of lithium with water
Lithium metal reacts quickly with water to produce lithium hydroxide and hydrogen — this would prevent oxidation of Li atoms at electrode 1 when the battery is discharging.
Alternatively: H+ ions would be selectively discharged over Li+ ions instead, preventing reduction of Li+ ions at electrode 1 when the battery is charging.
Lithium metal reacts quickly with water to form lithium hydroxide and hydrogen, which would prevent the oxidation of Li atoms at electrode 1 during discharging (or: H+ ions would be selectively discharged over Li+ ions, preventing reduction of Li+ during charging).
9 Addition polymers — PAA, disposal & weak acids 8 marks

PAA is an addition polymer used in diapers. The partial structure of PAA is shown in Fig. 9.1.

Fig. 9.1: partial structure of PAA, a repeat unit -C(COOH)(H)-C(H)(H)- with a carboxylic acid group
a(i) PAA is non-biodegradable. One method of disposal of PAA is incineration. The PAA completely combusts. Name two products of the complete combustion of PAA. [1]
Show working
Concept: Complete combustion
Carbon dioxide and water
a(ii) Describe the pollution problems caused by other methods of disposal of non-biodegradable plastics. [2]
Show working
Concept: Pollution from non-biodegradable plastics
Dumping at a landfill site causes land pollution; it takes up valuable land space as the plastics cannot be broken down by micro-organisms/bacteria.
Littering/dumping in the sea harms or kills aquatic wildlife, e.g. through ingestion (mistaken for food) or entanglement in the plastics.
Landfill: land pollution, takes up space, cannot be broken down by microorganisms. Sea littering: harms/kills aquatic wildlife via ingestion or entanglement.
b Waste PAA can be depolymerised into its monomers. The monomer of PAA has the formula C2H3COOH (acrylic acid). Complete the ‘dot-and-cross’ diagram of this monomer. Add the hydrogen atoms and show the arrangement of electrons. Show outer electrons only. [3]
Blank dot-and-cross skeleton for the monomer of PAA: three linked carbon circles and two oxygen circles, hydrogen atoms not yet added
Show working
Concept: Dot-and-cross diagrams for covalent molecules
The monomer is CH2=CH−COOH; two fewer hydrogens than the saturated carboxylic acid (general formula CnH2n+1COOH, so propanoic acid is C2H5COOH) confirms the presence of a C=C bond.
Add 4 hydrogen atoms (one per terminal CH2 position and one on the O−H of the −COOH group), each sharing a single pair of electrons with carbon or oxygen; the C=C and C=O bonds each share two pairs.
Completed dot-and-cross diagram of the PAA monomer with all outer electrons and hydrogen atoms shown
H−C(H)=C(H)−C(=O)−O−H, with shared electron pairs (dots and crosses) shown at every bond, outer electrons only.
c(i) The monomer of PAA, C2H3COOH, acts as a weak acid. Explain what is meant by weak acid. [1]
Show working
Concept: Weak acids
It is a substance that dissolves and ionises partially in water to produce a low concentration of H+ ions (and anions).
c(ii) Deduce the formula of the anion formed when the monomer of PAA acts as an acid. [1]
Show working
Concept: Ionisation of carboxylic acids
A carboxylic acid ionises by losing the acidic H from −COOH as H+, leaving the carboxylate anion.
C2H3COO−

Section B Answer one question from this section · 10 marks

10 Bonding & structure — simple molecular vs giant covalent vs ionic 10 marks

Table 10.1 shows information about element X and the compounds silicon dioxide and sodium chloride.

Table 10.1
formulamelting point/°Cboiling point/°Celectrical conductivity at room temperaturetype of bondingtype of structure
element XX2−218−183…covalent…
silicon dioxideSiO217132950does not conduct…giant
sodium chlorideNaCℓ8011465does not conductionic…
a Complete Table 10.1. [3]
Show working
Concept: Simple molecular vs giant covalent structures
Element X exists as a gas at room temperature (boiling point −183°C) and is diatomic (X2), covalent — so it has a simple molecular structure and does not conduct (no mobile ions/electrons).
Silicon dioxide (SiO2) has a giant structure, so its bonding is covalent (giant covalent/macromolecular).
Sodium chloride is ionic with a 1:1 ratio of ions, so it has a giant (ionic lattice) structure.
Table 10.1 — completed
formulam.p./°Cb.p./°Cconductivitybondingstructure
element XX2−218−183does not conductcovalentsimple
silicon dioxideSiO217132950does not conductcovalentgiant
sodium chlorideNaCℓ8011465does not conductionicgiant
Element X: does not conduct, simple structure. Silicon dioxide: covalent bonding. Sodium chloride: giant structure.
b Silicon dioxide and sodium chloride are each heated from room temperature to 1000°C. Predict the state and electrical conductivity of each substance at 1000°C. Explain your reasoning. [4]
Show working
Concept: Comparing melting points to a given temperature
At 1000°C, silicon dioxide still exists as a solid since its melting point is 1713°C. It cannot conduct electricity as there are no mobile ions/electrons to act as charge carriers.
At 1000°C, sodium chloride exists in the molten (liquid) state, as 1000°C is between its melting (801°C) and boiling (1465°C) points. It is able to conduct electricity due to the presence of free-moving ions that can act as mobile charge carriers.
Silicon dioxide: solid, does not conduct (m.p. 1713°C not reached, no mobile charge carriers). Sodium chloride: molten/liquid, conducts (between its m.p. and b.p., free-moving ions present).
c Explain why element X and sodium chloride have different melting points. In your answer, include ideas about energy, bonds and forces between particles. [3]
Show working
Concept: Simple molecular vs ionic lattice melting
Element X: weak intermolecular forces of attraction between X2 molecules require very little energy to overcome during melting, hence a low melting point.
Sodium chloride: strong electrostatic forces of attraction between Na+ and Cℓ− ions (ionic bonds) require a lot of energy to overcome during melting, hence a high melting point.
Element X has only weak intermolecular forces between X2 molecules (little energy to overcome, low m.p.); NaCℓ has strong electrostatic forces between oppositely charged ions (a lot of energy to overcome, high m.p.).
11 Metallic bonding, alloys & ionic lattices 10 marks

Element Y is a metal. Table 11.1 shows some properties of element Y, lead(II) bromide and element Z.

Table 11.1
bondingstate at r.t.p.melting point/°Celectrical conductivity at r.t.p.other information
element Ymetallicsolid420conductsY is malleable
lead(II) bromideionicsolid373does not conductformula is PbBr2
element Zmetallicsolid1455conductsZ forms coloured compounds

Fig. 11.1 represents the arrangement of particles in a metallic lattice structure (key: ⊕ positively charged ions; ⊖ delocalised electrons).

Fig. 11.1: a metallic lattice of positive ions surrounded by delocalised electrons
a(i) Explain the properties of element Y given in Table 11.1 in terms of the particles shown in Fig. 11.1. [3]
Show working
Concept: Metallic bonding explains metallic properties
There are strong electrostatic forces of attraction between metal cations and delocalised electrons, which require a large amount of energy to overcome, hence element Y exists as a solid / has a high melting point.
The presence of a sea of delocalised electrons that act as mobile charge carriers allows element Y to conduct electricity.
Regular layers of metal cations can slide over each other easily when a force is applied, hence element Y is malleable.
Strong electrostatic forces between cations and delocalised electrons (solid/high m.p.); delocalised electrons as mobile charge carriers (conducts); regular layers of cations slide over each other (malleable).
a(ii) An alloy of element Y is used in building materials. Describe how the structure of this alloy differs from the structure shown in Fig. 11.1. [2]
Show working
Concept: Alloys
An alloy consists of two (or more) different types of atoms (can be metal or non-metal) of different sizes, which results in irregular layers of atoms.
Instead of having regular layers of atoms (cations) of the same size, the alloy consists of two or more different types of atoms of different sizes, which results in irregular layers of atoms present.
b Describe how the structure of solid lead(II) bromide differs from the metallic lattice structure shown in Fig. 11.1. [3]
Show working
Concept: Metallic lattice vs ionic lattice
Particles present: metallic lattice has metal cations and delocalised electrons (1:1 ratio of cation to electron shown); PbBr2 has Pb2+ cations and Br− anions in a 1:2 ratio.
Movement of particles: in the metallic lattice, cations stay in the lattice while electrons are free-moving; in PbBr2, both cations and anions are held in fixed positions.
Metallic lattice: metal cations + delocalised electrons, 1:1, cations fixed/electrons free-moving. PbBr2 (ionic lattice): Pb2+ cations + Br− anions, 1:2 ratio, both ions held in fixed positions.
c State two reasons how the information in Table 11.1 shows that element Z is not a Group 1 element. [2]
Show working
Concept: Group 1 metals vs transition elements
Z has a very high melting point (1455°C) while Group 1 metals have low melting points.
Z forms coloured compounds while Group 1 metals form white solids/colourless solutions.
1. Z has a very high melting point, while Group 1 metals have low melting points. 2. Z forms coloured compounds, while Group 1 metals form white solids/colourless solutions.

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FAQ

Common questions about this paper

Where can I download the 2026 O-Level Chemistry Paper 2 blank PDF for free?

You can download the blank 2026 O-Level Chemistry (6092/02) question paper, along with the annotated answer key, for free from this page. Click the ‘Download for Free’ button, fill in your details, and the PDF download link will be emailed to you instantly.

What topics are tested in the 2026 O-Level Chemistry Paper 2 (6092/02)?

Paper 2 tests air pollution and the carbon cycle, enzymes as catalysts, isotopes and nuclide symbols, ionic equations and percentage composition, hydrocarbons and the bromine water test for saturation, addition reactions and energy profile diagrams, electrolysis of aqueous copper(II) sulfate with inert and reactive electrodes, redox reactions in lithium-ion batteries, addition polymers and their disposal, and then either a question on simple molecular vs giant covalent vs ionic structures, or a question on metallic bonding, alloys and ionic lattices in Section B.

Why does a reactive copper anode lose a different mass to an inert electrode during electrolysis?

With an inert (e.g. graphite or platinum) anode, only the cathode gains mass as Cu2+ ions are reduced to copper. With a reactive copper anode, the anode itself is oxidised (Cu → Cu2+ + 2e−) at the same rate copper is deposited at the cathode (Cu2+ + 2e− → Cu), since the same quantity of charge passes through the circuit. So the anode loses exactly as much mass as the cathode gains.

Why is water not used as the solvent in a lithium-ion battery's electrolyte?

Lithium metal reacts quickly with water to form lithium hydroxide and hydrogen gas, which would consume the lithium at electrode 1 and prevent the proper discharging reaction. H+ ions from water would also be selectively discharged over Li+ ions, blocking the reduction of Li+ needed when the battery charges. An organic solvent avoids both problems.

Why do ionic compounds and giant covalent structures have very different melting points even though both involve strong bonds?

Both giant ionic lattices and giant covalent (macromolecular) structures need a large amount of energy to melt because of strong bonds throughout the structure. Simple molecular substances, by contrast, have only weak intermolecular forces between molecules, which need very little energy to overcome — so they have much lower melting points than either ionic or giant covalent structures.