CIE iGCSE Chemistry Paper 4 Prediction - 2025

CIE iGCSE Chemistry Paper 4 Prediction – 2025

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Question 1

This question is about alkanes and alkenes.

(a) Short-chain alkanes and alkenes can be formed from long-chain alkanes in a chemical reaction.

(i) Name the type of chemical reaction which forms short‑chain alkanes and alkenes from long‑chain alkanes.[1]

(ii) Decane has 10 carbon atoms. It forms ethane and ethene as the only products in this type of chemical reaction.

Write the chemical equation for this reaction.[3]

(b) Ethane reacts with chlorine at room temperature to form chloroethane, C₂H₅Cl, and one other product.

(i) Name the other product formed.[1]

(ii) State the condition needed for this reaction to take place.[1]

(c) Ethene reacts with chlorine at room temperature to form dichloroethane, C₂H₄Cl₂.

C₂H₄ + Cl₂ → C₂H₄Cl₂

(i) State why this is an addition reaction.[1]

(ii) The chemical equation for this reaction can be represented as shown.

The energy change for the reaction is –180kJ/mol.

Use the bond energies in the table to calculate the bond energy of a C–Cl bond, in kJ/mol.

Use the following steps.

step 1 Calculate the energy needed to break bonds.

energy needed to break bonds = kJ

step 2 Use your answer in step 1 and the energy change for the reaction to determine the energy released when bonds are formed.

energy released when bonds form = kJ

step 3 Use your answer in step 2 and bond energy values to determine the energy of a C–Cl bond.

bond energy of a C–Cl bond = kJ/mol

[4][Total: 11]

Answer/Explanation

Ans:

5(a)(i) cracking

5(a)(ii) C₁₀H₂₂ → 4C₂H₄ + C₂H₆
C₁₀H₂₂ as only reactant
formulae of ethene and ethane as only products
correct equation

5(b)(i) hydrogen chloride

5(b)(ii) ultraviolet light

5(c)(i) (only) one product is formed

5(c)(ii) M1 Bond energy in breaking bonds
= [(4 × 410) + 610 + 240] = 2490 (kJ / mol)

M2 Use of total E change to find bond energy of C₂H₄Cl₂
= M1 + 180 = 2490 + 180 = 2670 (kJ / mol)

M3 Determination of total C–Cl bond energy
= M2 – [(4 × 410) + 350] = 2670 – 1990 = 680 (kJ / mol)

M4 Determination of each C–Cl bond energy
= M3 / 2 = 680 / 2 = 340 (kJ / mol)

Question 2

(a) Atoms are made of protons, neutrons and electrons. Atoms of the same element are known as isotopes.

(i) Complete the table.[2]

(ii) $_{12}^{24}\textrm{Mg}$ and $_{12}^{25}\textrm{Mg}$ are isotopes of magnesium.

Complete the table to show the numbers of electrons, neutrons and protons in these isotopes of magnesium.[2]

(iii) Explain why magnesium ions have a charge of 2+.[1]

(b) Mg²⁺ ions have the electronic structure 2,8.

Give the formula of the following particles which have the same electronic structure as Mg²⁺ions.

- a cation (positive ion)
- an anion (negative ion)
- an atom

[3] [Total: 8]

Answer/Explanation

Ans:

2(a)(i)

Mark by column

2(a)(ii)

Mark by row

2(a)(iii) (they have) 2 more protons than electrons
OR
(they have) 2 fewer electrons than protons
OR
(they have) 12 protons and 10 electrons

2(b) Na⁺ or Al ³⁺ (1)
F^– or O^2– or N^3– (1)
Ne (1)

Question 3

Potassium reacts with chlorine to form potassium chloride, KCl.
(a) Write a chemical equation for this reaction.
(b) Potassium chloride is an ionic compound.
Complete the diagram to show the electron arrangement in the outer shells of the ions present
in potassium chloride.
Give the charges on both ions.

(c) Molten potassium chloride undergoes electrolysis.
(i) State what is meant by the term electrolysis.
(ii) Name the products formed at the positive electrode (anode) and negative electrode
(cathode) when molten potassium chloride undergoes electrolysis.
anode ………………………………………………………………………………………………………………….
cathode ……………………………………………………………………………………………………………….
(d) Concentrated aqueous potassium chloride undergoes electrolysis.
(i) Write an ionic half-equation for the reaction at the negative electrode (cathode).
(ii) Name the product formed at the positive electrode (anode).
(iii) Name the potassium compound that remains in the solution after electrolysis.
(e) Complete the dot-and-cross diagram to show the electron arrangement in a molecule of
chlorine, $Cl_2$.
Show the outer electrons only.

(f) The melting points and boiling points of chlorine and potassium chloride are shown.

(i) Deduce the physical state of chlorine at –75°C. Use the data in the table to explain your
answer.
physical state ……………………………………………………………………………………………………….
explanation ………………………………………………………………………………………………………….
(ii) Explain, in terms of structure and bonding, why potassium chloride has a much higher melting point than chlorine.
Your answer should refer to the:
● types of particle held together by the forces of attraction
● types of forces of attraction between particles
● relative strength of the forces of attraction.

Answer/Explanation

Answer:

(a) $2K + Cl_2 → 2KCl$
$Cl_2$ on left hand side (1)
equation fully correct (1)

(b) K outer shell with 8 crosses (1)
Cl outer shell with 7 dots and 1 cross (1)
$^+$ and – (1)

(c) (i) breakdown by (the passage of) electricity (1)
of an ionic compound in molten or aqueous (state) (1)
(ii) (anode) chlorine
(cathode)potassium

(d) (i) $2H^+ + 2e(^–) → H_2$
$H^+$ and $e(^–)$ on left hand side (1)
equation fully correct (1)
(ii) chlorine
(iii) potassium hydroxide (1)

(e) one shared pair of electrons and 6 non-bonding electrons on each chlorine atom

(f) (i) liquid (1)
BOTH melting point is below –75 oC AND boiling point is above –75 oC
OR
BOTH –75 oC is higher than –101 oC / melting point AND lower than –35 oC / boiling point
OR
–75 oC is between melting point or –101 oC and boiling point or –35 oC
(ii) ionic bonds in KCl (1)
attraction between molecules in $Cl_2$ (1)
weaker attraction (between particles) in $Cl_2$ ORA (1)

Question 4

Ethanoic acid is manufactured by the reaction of methanol with carbon monoxide.

An equilibrium mixture is produced.

$CH_{3}OH\left ( g \right )+CO\left ( g \right )\rightleftharpoons CH_{3}COOH\left ( g \right )$

(a) State two characteristics of an equilibrium.[2]

(b) The purpose of the industrial process is to produce a high yield of ethanoic acid at a high rate of reaction.

The manufacture is carried out at a temperature of 300°C.

The forward reaction is exothermic.

Use this information to state why the manufacture is not carried out at temperatures:[2]

- below 300°C
- above 300°C.

(c) Complete the table using only the words increases, decreases or no change.[3]

(d) Suggest which of the following metals is a suitable catalyst for the reaction. Give a reason for your answer.[2]

aluminium calcium cobalt magnesium potassium

suitable catalyst

reason

(e) Ethanoic acid is a member of the homologous series of carboxylic acids.

State the general formula of this homologous series.[1]

(f) Draw the structure of the carboxylic acid containing three carbon atoms. Show all of the atoms and all of the bonds. [2]

(g) When carboxylic acids react with alcohols, esters are produced.

The formula of ester X is CH₃CH₂CH₂COOCH₃.

(i) Name ester X. [1]

(ii) Give the name of the carboxylic acid and the alcohol that react together to produce ester X.[2]

carboxylic acid

alcohol

(h) Ester Y has the following composition by mass:

C, 48.65%; H, 8.11%; O, 43.24%.

Calculate the empirical formula of ester Y.

empirical formula = [3]

(i) Ester Z has the empirical formula C₂H₄O and a relative molecular mass of 88.

Determine the molecular formula of ester Z.

molecular formula = [1] [Total: 19]

Answer/Explanation

Ans:

5(a) the rate of forward reaction equals the rate of the reverse reaction (1)
concentrations of reactants and products are constant (1)

5(b) reaction too slow (1)
yield of ethanoic acid too low (1)

5(c)

5(d) cobalt (1)
transition element (1)

5(e) C_nH_2n+1COOH

5(f) COOH fully displayed (1)

whole molecule completely correct (1)

5(g)(i) methyl butanoate

5(g)(ii) butanoic acid (1)
methanol (1)

5(h) C 48.65 / 12
H 8.11 / 1
O 43.24 / 16
OR
4.05:8.11:2.70 (1)

fractions shown dividing all by smallest
OR
1.5:3:1
OR
3:6:2 (1)

C₃H₆O₂ (1)

5(i) C₄H₈O₂

Question 5

This question is about salts.
(a) Salts that are insoluble in water are made by precipitation.
● Lead(II) iodide, $PbI_2$, is insoluble in water.
● All nitrates are soluble in water.
● All sodium salts are soluble in water.
You are provided with solid lead(II) nitrate, $Pb(NO_3)_2$, and solid sodium iodide, NaI.
Describe how you would make a pure sample of lead(II) iodide by precipitation.
Your answer should include:
● practical details
● a chemical equation for the precipitation reaction.
(b) Nitrates decompose when heated.
(i) When hydrated zinc nitrate is heated, oxygen gas is given off.
Describe a test for oxygen.
test ……………………………………………………………………………………………………………………..
observations ………………………………………………………………………………………………………..

(ii) Complete the equation for the decomposition of hydrated zinc nitrate.

$2Zn(NO_3)_2•6H_2O → …..ZnO + …..NO_2 + O_2 + …..H_2O$

$Na_2SO_4•xH_2O(s) → Na_2SO_4(s) + xH_2O(g)$

A student carries out an experiment to determine the value of x in $Na_2SO_4•xH_2O$.
step 1 Hydrated sodium sulfate crystals are weighed.
step 2 The hydrated sodium sulfate crystals are then heated.
step 3 The remaining solid is weighed.
(i) Describe how the student can check that all the water has been given off.

(ii) In an experiment, 1.61g of $Na_2SO_4•xH_2O$ is heated until all the water is given off. The mass of $Na_2SO_4$ remaining is 0.71g.
$[M_r : Na_2SO_4,142; H_2O,18]$
Determine the value of x using the following steps.
● Calculate the number of moles of $Na_2SO_4$ remaining.
………………………… mol

● Calculate the mass of $H_2O$ given off.
………………………… g

● Calculate the number of moles of $H_2O$ given off.
………………………… mol

● Determine the value of x.
x = …………………………

Answer/Explanation

Answer:

(a) (add) water (to both salts) (1)
dissolve both salts / make solutions (1)
filter (lead(II) iodide)(1)
wash (residue of lead(II) iodide) with water AND dry e.g. with filter paper / description of washing and drying (1)
$Pb(NO_3)_2 + 2 NaI → 2NaNO_3 + PbI_2$
OR $Pb^{2+} + 2I^– → PbI_2$ (1)

(b) (i) glowing splint (1)
relights / rekindles (1)
(ii) 2ZnO(s) and $4NO_2$(g) (1)
$12H_2O$(g) (1)

(c) (i) heat again and weigh again / repeat steps 2 and 3 (1)
until mass is constant (1)
(ii) 0.005 (1)
0.9 (1)
(0.9 ÷ 18 =) 0.05 (1)
(0.05 ÷ 0.005 =) 10 (1)

Question 6

Zinc is extracted from an ore containing zinc sulfide.
(a) State the name of this zinc ore…………………………………………………………………………………………………………………………….. [1]

(b) This ore is converted to zinc oxide, ZnO.
Zinc oxide is then reacted with carbon.
(i) Write a chemical equation for the reaction of zinc oxide with carbon………………………………………………………………………………………………………………………. [1]
(ii) State what type of chemical change happens to the zinc in zinc oxide in this reaction.
Explain your answer.
chemical change …………………………………………………………………………………………………..
explanation ……………………………………………………………………………………………………………………………………………………………………………………………………………………………………….[2]
(iii) Explain why aluminium is not extracted from aluminium oxide by heating with carbon………………………………………………………………………………………………………………………. [1]
(iv) Suggest an alternative method for the extraction of zinc from zinc oxide……………………………………………………………………………………………………………………… [1]

(c) Brass is an alloy of zinc.
Explain, in terms of particles, why brass is harder than pure zinc…………………………………………………………………………………………………………………………….. [3]

▶️Answer/Explanation

Ans:

2(a) zinc blende

2(b)(i) $\begin{aligned} & \mathrm{ZnO}+\mathrm{C} \rightarrow \mathrm{Zn}+\mathrm{CO} \\ & \text { or } \\ & 2 \mathrm{ZnO}+\mathrm{C} \rightarrow 2 \mathrm{Zn}+\mathrm{CO}_2\end{aligned}$

2(b)(ii) chemical change: reduction (1)
explanation: oxygen is lost (1)

2(b)(iii) aluminium is more reactive than carbon
2(b)(iv) electrolysis

2(c) exists as layers (1)
(alloy) contains different sized (copper) atoms (1)
makes it more difficult for layers (of atoms) to slide over each slip/shift other (1)

Question 7

A student adds excess large pieces of magnesium carbonate, MgCO₃, to dilute hydrochloric acid, HCl, and measures the volume of carbon dioxide gas, CO₂, given off.

(a) Add the missing state symbols to the chemical equation for the reaction.

MgCO₃ ….. + 2HCl….. → MgCl 2(aq) + H₂O ….. + CO₂ ….. [1]

(b) Complete the dot-and-cross diagram to show the electron arrangement of the ions in magnesium chloride.

The inner shells have been drawn.

Give the charges on the ions.[3]

(c) Complete the dot-and-cross diagram to show the electron arrangement in a molecule of carbon dioxide.

Show outer shell electrons only.[2]

(d) The graph shows how the volume of carbon dioxide gas changes with time.

(i) Describe how the graph shows that the rate of this reaction decreases as time increases.[1]

(ii) Explain, in terms of particles, why the rate of this reaction decreases as time increases.[2]

(iii) The student repeats the experiment using powdered MgCO3 instead of large pieces.

All other conditions stay the same.

On the grid, draw the line expected when powdered MgCO3 is used instead of large pieces. [2]

(e) Determine the volume of CO₂ gas given off when excess MgCO₃ is added to 25.0cm³ of 0.400mol/dm³ HCl at room temperature and pressure.

MgCO₃ + 2HCl → MgCl₂ + H2O + CO₂

Use the following steps.

- Calculate the number of moles of HCl in 25.0cm³ of 0.400mol/dm³ of acid.

mol

- Determine the number of moles of CO₂ gas given off.

mol

- Calculate the volume of CO₂ gas given off in cm³.

cm³[3] [Total: 14]

Answer/Explanation

Ans:

2(a) MgCO₃(s) + 2HCl(aq) → MgCl₂(q) + H₂O(l) + CO₂(g)

2(b) eight crosses in second shell of Mg
7 dots and 1 cross in third shell of both Cl
2+ charge on Mg and – charge on both Cl ions on correct answer line

2(c) C atom double bonded to 2 O atoms
4 non-bonding e^– on each O and no non-bonding e^– on C and both octets complete

2(d)(i) gradient (of line) decreases

2(d)(ii) concentration of particles (of acid) decreases
lower rate of collisions of particles

2(d)(iii) a new line steeper than printed line and starts at origin and levels off earlier than printed line
levels off at the same volume

2(e) M1 mol HCl = 0.400 × 25.0 / 1000 = 0.01(00)
M2 mol CO₂ = M1 / 2 = 0.0100 / 2 = 0.005(00)
M3 volume CO₂ = M2 × 24000 = 120 (cm³)

CIE iGCSE Chemistry Paper 4 Prediction

CIE iGCSE Chemistry Paper 4 Prediction - 2025