A Level Chemistry 9701 Paper 42, Oct/Nov 2025

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9701/42 · Oct/Nov 2025

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1

Magnesium nitrate, Mg(NO3)2, and strontium nitrate, Sr(NO3)2, both decompose when heated to form the metal oxide and a mixture of gases.

a

Write an equation for the thermal decomposition of Mg(NO3)2.

[1]
b

State which of Mg(NO3)2 or Sr(NO3)2 decomposes at a lower temperature. Explain your answer. compound that decomposes at a lower temperature ............................................................... explanation ...............................................................................................................................

[2]
c(i)

Magnesium oxide, MgO, and strontium oxide, SrO, both react with dilute sulfuric acid. MgO forms a soluble salt, A. SrO forms an insoluble salt, B. Identify the products formed when MgO reacts with dilute sulfuric acid.

[1]
c(ii)

Explain why A is more soluble than B.

[3]
2

Ethanal, CH3CHO, reacts with nitrogen dioxide, NO2. The products of the first step of this reaction are a CH3C˙=O radical and a molecule of nitrous acid, HNO2. CH3CHO+NO2→CH3C˙=O+HNO2

a(i)

Use two words to complete the sentence. This reaction involves ........................................ ........................................ of the single covalent bond between a hydrogen atom and a carbon atom in CH3CHO.

[1]
a(ii)

The hydrogen atom mentioned in (a)(i) forms a covalent bond with one of the oxygen atoms of an NO2 molecule. An NO2 molecule has a single, unpaired electron on the nitrogen atom. All electrons are paired in an HNO2 molecule. Draw dot-and-cross diagrams of NO2 and HNO2 in the boxes. Show outer shell electrons only.

[1]
a(iii)

Use VSEPR theory to predict the bond angle at the nitrogen atom in an HNO2 molecule. bond angle = ..............................

[1]
b

The rate equation for the reaction between CH3CHO and NO2 is shown. rate=k[CH3CHO][NO2] Under certain conditions, when the concentrations of both CH3CHO and NO2 are 0.200 mol dm−3, the rate of the reaction is 1.53×10−4 mol dm−3 s−1. Calculate the value of the rate constant, k, under these conditions. Give the units of k. k= .............................. units ..............................

[2]
c

The reaction mixture described in (b) is monitored over a period of time. Predict whether the graph of [NO2] against time shows a constant half-life. Explain your answer. prediction .................................................................................................................. explanation ...............................................................................................................

[1]
d

NO2 also reacts with ozone, O3. The rate equation is shown. rate=k1[NO2] Under certain conditions, the value of k1 is 0.0848 s−1. The reaction has a constant half-life under these conditions. Calculate the half-life in seconds. half-life = .............................. s

[1]
e

NO2 is present in the exhaust gases of cars. It can react with carbon monoxide, CO, on the surface of a heterogeneous catalyst in the car’s catalytic converter. Describe the mode of action of this heterogeneous catalyst.

[2]
3
a(i)

Define conjugate acid–base pair.

[1]
a(ii)

Give the formulas of the conjugate acid and the conjugate base of the hydrogen phosphate ion, HPO42−. conjugate acid of HPO42− .............................. conjugate base of HPO42− ..............................

[1]
b(i)

The Ka of propanoic acid, CH3CH2COOH, is 1.35×10−5 mol dm−3 at 298 K. Solution C is a solution of CH3CH2COOH with a pH of 3.60 at 298 K. Calculate the concentration of CH3CH2COOH in solution C. [CH3CH2COOH]= .............................. mol dm−3

[2]
b(ii)

Calculate the concentration of hydroxide ions in solution C. [OH−]= .............................. mol dm−3

[1]
b(iii)

Calculate the concentration of a solution of hydrochloric acid with the same pH as solution C. concentration = .............................. mol dm−3

[1]
b(iv)

Table 3.1 shows three possible values of the Ka of dimethylpropanoic acid, (CH3)3CCOOH. Place a tick in Table 3.1 to show the correct value. Explain your answer. Table 3.1 value of Ka/mol dm−3 | place one tick (✓) in this column 9.33 ×10−6 1.35 ×10−5 3.35 ×10−5 explanation .......................................................................................................................

[3]
c(i)

Solution D is made by mixing 100 cm3 of 0.100 mol dm−3 CH3CH2COOH and 100 cm3 of 0.100 mol dm−3 NaCl. The pH of solution D is measured as small amounts of H2SO4(aq) are added to it, and when small amounts of NaOH(aq) are added to it. Solution D only acts as a buffer solution when one of these solutions is added to it. Complete the sentence and write an equation for the reaction that occurs. Solution D acts as a buffer when ................................................................ is added to it. equation ............................................................................................................................

[1]
c(ii)

Complete the sentence and explain why solution D does not act as a buffer when the other solution is added. Solution D does not act as a buffer when .................................................. is added to it. explanation .......................................................................................................................

[1]
d(i)

Manganese(II) hydroxide, Mn(OH)2, is only slightly soluble in water. The solubility of Mn(OH)2 in water is 3.28×10−3 g dm−3 at 298 K. Calculate the concentration of a saturated solution of Mn(OH)2 at 298 K. [Mn(OH)2]= .............................. mol dm−3

[1]
d(ii)

Write an expression for the Ksp of Mn(OH)2. Give the units of Ksp. Ksp= units = ..............................

[2]
d(iii)

Use your answers to (d)(i) and (d)(ii) to calculate the value of Ksp of Mn(OH)2 at 298 K. Ksp= ..............................

[1]
4
a

Define enthalpy change of atomisation, ΔHat.

[1]
b

Define first electron affinity, EA.

[1]
c

Explain why the first electron affinity of chlorine is more exothermic than the first electron affinity of iodine.

[2]
d

The enthalpy change for the reaction Cl2(g)+2e−→2Cl−(g) is −486 kJ mol−1. The first electron affinity of chlorine is −364 kJ mol−1. Calculate the enthalpy change of atomisation of chlorine. ΔHat of chlorine = .............................. kJ mol−1

[2]
5

Cobalt is a transition element which forms compounds containing Co2+ and Co3+ ions. Cobalt(II) sulfate dissolves in water to form a solution containing the [Co(H2O)6]2+ complex ion.

a(i)

Complete the electronic configurations of a Co2+ ion and a Co3+ ion. Co2+=[Ar] .............................. Co3+=[Ar] ..............................

[1]
a(ii)

Explain why transition elements can form complex ions.

[1]
a(iii)

An excess of concentrated HCl is added to a solution containing [Co(H2O)6]2+. Describe the colour change observed and the state of the cobalt-containing product. The colour changes from .............................. to .............................. . The state of the cobalt-containing product is .............................. .

[2]
a(iv)

Write an equation for the reaction occurring in (a)(iii).

[1]
a(v)

Name the type of reaction occurring in (a)(iii).

[1]
a(vi)

Write an equation for the reaction that occurs when an excess of NaOH(aq) is added to a solution containing [Co(H2O)6]2+.

[1]
b(i)

Cobalt metal can be oxidised by acidified K2Cr2O7. The relevant half-equations, and their E⊖ values, are shown. Co2++2e−⇌CoE⊖=−0.28 V Cr2O72−+14H++6e−⇌2Cr3++7H2OE⊖=+1.33 V A Co2+/Co electrode is constructed in which [Co2+] is 0.020 mol dm−3 at 298 K. Use the Nernst equation to show that the E value for this Co2+/Co electrode is −0.33 V.

[2]
b(ii)

An electrochemical cell is constructed using the Co2+/Co electrode described in (b)(i) and a Cr2O72−/Cr3+ electrode in which all conditions are standard. Calculate the value of Ecell. Ecell= ..............................

[1]
b(iii)

A current is drawn from the electrochemical cell described in (b)(ii). Write an equation for the reaction taking place in the cell.

[1]
b(iv)

Complete the sentences to identify the negative electrode and the direction of electron flow when a current is drawn from the cell described in (b)(ii). The .............................. electrode is the negative electrode. Electrons flow from the .............................. electrode to the .............................. electrode.

[1]
c

A molten Co2+ salt is electrolysed using a current of 0.500 A. 0.547 g of cobalt metal forms at the cathode. Under the conditions used no other reduction reaction occurs at the cathode. Calculate the time in minutes for which the current flows to produce this mass of cobalt. Give your answer to three significant figures. time = .............................. min

[3]
6
a(i)

Nickel forms complexes. Give the formula and charge of the tetrahedral complex formed by Ni atoms with carbon monoxide molecules. Carbon monoxide is a monodentate ligand. This is complex E. E = ...............................................................................................................................

[1]
a(ii)

Give the formula and charge of the octahedral complex formed by Ni2+ ions with ethanedioate ions. This is complex F. F = ...............................................................................................................................

[2]
a(iii)

Identify which complex, E or F, exists as a mixture of two stereoisomers and the type of stereoisomerism involved. The complex which exists as a mixture of two stereoisomers is .............................. . The type of stereoisomerism involved is .............................. .

[1]
b(i)

Cadmium forms complexes with methylamine, CH3NH2, and 1,2-diaminoethane, en. The values of the stability constants, Kstab, of these complex ions are given in Table 6.1. Table 6.1 complex | Kstab [Cd(CH3NH2)4]2+ | 3.5×106 [Cd(en)2]2+ | 4.0×1010 Explain, by reference to its structure, why CH3NH2 acts as a monodentate ligand.

[1]
b(ii)

Some Cd2+(aq) is added to a solution containing equal concentrations of CH3NH2 and en. Predict which of the two complexes in Table 6.1 forms at the higher concentration. Explain your answer. complex that forms at the higher concentration ................................................................ explanation .......................................................................................................................

[1]
b(iii)

Complete the expression for the Kstab of [Cd(CH3NH2)4]2+. Kstab=

[1]
7

P, Q, R, S, T, U, and V are the seven structural isomers with molecular formula C5H10O that have a carbonyl group. P CH3(CH2)3CHO Q CH3CH2CH(CH3)CHO R (CH3)2CHCH2CHO S (CH3)3CCHO T CH3CH2CH2COCH3 U CH3CH2COCH2CH3 V (CH3)2CHCOCH3

Figure 1 of 1 for question 7
a

Only one of these seven compounds has stereoisomers. Draw three-dimensional diagrams of the two stereoisomers of this compound.

[2]
b(i)

P, Q, R, S, T, U, and V are treated separately with alkaline I2(aq) and the product mixture is acidified. Identify the two compounds that give a positive result with alkaline I2(aq). ............................................................... and ...............................................................

[1]
b(ii)

Describe the observations when one of the compounds you have identified in (b)(i) is treated with alkaline I2(aq) and give the structural formulae of the two carbon-containing products of this reaction. observations ...................................................................................................................... two carbon-containing products ........................................................................................ and ........................................................................................

[2]
c(i)

The proton (1H) NMR spectra of P, Q, R, S, T, U, and V are compared. Identify the only compound that gives a spectrum with two singlets and no other peaks.

[1]
c(ii)

Fig. 7.1 shows the spectrum obtained from one of the compounds. Identify the compound that gives this spectrum.

[1]
c(iii)

Name the splitting pattern of the peak at δ=1.1 in Fig. 7.1. Give the reason for this splitting. name .............................. reason ...............................................................................................................................

[1]
c(iv)

Identify the substance that gives the small peak at δ=0 in Fig. 7.1.

[1]
d

The carbon-13 NMR spectra of R, S, T and U are compared. Complete Table 7.1 to state the number of peaks in the spectrum of each compound. Table 7.1 compound | number of peaks R (CH3)2CHCH2CHO S (CH3)3CCHO T CH3CH2CH2COCH3 U CH3CH2COCH2CH3

[2]
8

Asparagine and aspartic acid are two naturally occurring amino acids. Their structures and isoelectric points are shown in Table 8.1. Table 8.1 amino acid | structure | isoelectric point asparagine | HOOCCH(NH2)CH2CONH2 | 5.41 aspartic acid | HOOCCH(NH2)CH2COOH | 2.77

a

Define isoelectric point.

[1]
b

Draw the structures of asparagine and aspartic acid at pH 2.

[2]
c

Asparagine and aspartic acid are treated separately with an excess of LiAlH4. Draw the structures of the organic products of these reactions.

[2]
d(i)

Propanedioic acid, HOOCCH2COOH, is treated with an excess of thionyl chloride, SOCl2. Propanedioyl chloride, ClOCCH2COCl, is formed. Write an equation for this reaction.

[1]
d(ii)

Propanedioyl chloride reacts with an excess of asparagine to form compound G with molecular formula C11H16N4O8. Each molecule of compound G has four amide groups. Draw the structure of compound G.

[2]
e

Asparagine is hydrolysed with an excess of hot NaOH(aq). Draw the structure of the organic product of this reaction.

[2]
f

A polymer can form from asparagine, HOOCCH(NH2)CH2CONH2, as the only monomer. Draw a length of the polymer chain containing three monomer residues. Clearly label the repeat unit of the polymer on your diagram.

[3]
g(i)

Aspartic acid exists in two optically active forms. Plane polarised light is passed through pure samples of these two optically active forms in solutions of the same concentration. Describe two similarities and one difference in their effect on the plane polarised light. similarities ......................................................................................................................... ........................................................................................................................................... difference ........................................................................................................................... ...........................................................................................................................

[2]
g(ii)

Give the term used to describe a mixture of equal amounts of the two optically active forms.

[1]
9

Compound X is made from benzene by the route shown in Fig. 9.1.

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a

Describe the bonding in benzene, C6H6. Your answer should include: • the hybridisation of the six carbon atoms • the types of bond between the carbon atoms • the orbitals that overlap to produce the bonds between the carbon atoms • the type of bond between the carbon atoms and the hydrogen atoms • the orbitals that overlap to produce the bonds between the carbon atoms and the hydrogen atoms.

[3]
b

Describe the reagents and conditions required for step 1 in Fig. 9.1.

[1]
c

In step 1 of Fig. 9.1 benzene reacts with +CH3. Complete Fig. 9.2 to show the mechanism for this reaction, including: • the movement of electron pairs using curly arrows • the structure of the intermediate involved.

[3]
d

Describe the reagents and conditions required for step 2 of Fig. 9.1.

[2]
e

Identify the reagents required for step 3 of Fig. 9.1. Compound W is the product of this step.

[1]
f

Name compound W.

[1]
g

The reagents commonly used for step 4 will not reduce the −COOH group. Identify the reagents and conditions required for step 4 of Fig. 9.1.

[1]
h

Benzene can also be used as a starting material to make compound Y. Describe how the route described in Fig. 9.1 (repeated below) can be changed to give compound Y instead of compound X. Explain your answer.

[2]

Mark scheme

9701/42 · Oct/Nov 2025

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1a · 1 mark

  • 1 mark

    Mg(NO3)2 → MgO + 2NO2 + ½O2 OR 2Mg(NO3)2 → 2MgO + 4NO2 + O2

1b · 2 marks

  • M1

    Mg(NO3)2 / magnesium nitrate AND magnesium ion / Mg2+ AND is smaller / has higher charge density

  • M2

    nitrate ion / anion is more distorted / polarised

1c(i) · 1 mark

  • 1 mark

    magnesium sulfate and water / MgSO4 and H2O

1c(ii) · 3 marks

  • M1

    magnesium sulfate / A has a more exothermic Hlatt and Hhyd

  • M2

    difference in Hhyd is greater

  • M3

    magnesium sulfate / A has a more exothermic Hsol

2a(i) · 1 mark

  • 1 mark

    homolytic fission

2a(ii) · 1 mark

  • 1 mark

    •O• :N: •O• H:O• :N: •O•

2a(iii) · 1 mark

  • 1 mark

    115–120°

2b · 2 marks

  • M1

    3.825  10–3 / 0.003825 / 0.00383 / 0.0038 / 3.8  10–3

  • M2

    units = mol–1 dm3 s–1

2c · 1 mark

  • 1 mark

    not constant half-life because overall second order

2d · 1 mark

  • 1 mark

    8.17 / 8.2

2e · 2 marks

  • 1 mark

    reactants are adsorbed onto the catalyst surface

  • 1 mark

    bonds within reactant molecules are weakened

  • 0 marks

    products are desorbed from catalyst surface

3a(i) · 1 mark

  • 1 mark

    two species that differ by one proton / H+

3a(ii) · 1 mark

  • 1 mark

    H2PO4– AND PO43–

3b(i) · 2 marks

  • M1

    [H+] = 2.51  10–4

  • M2

    [CH3CH2COOH] = 4.67  10–3

3b(ii) · 1 mark

  • 1 mark

    3.98  10–11

3b(iii) · 1 mark

  • 1 mark

    [HCl] = 2.51  10–4

3b(iv) · 3 marks

  • M1

    top box ticked AND weaker acid

  • M2

    alkyl group is electron donating

  • M3

    O–H bond strengthened / anion is destabilised

3c(i) · 1 mark

  • 1 mark

    NaOH AND CH3CH2COOH + OH– → CH3CH2COO– + H2O

3c(ii) · 1 mark

  • 1 mark

    H2SO4 AND conjugate base of CH3CH2COOH is not present

3d(i) · 1 mark

  • 1 mark

    3.69  10–5

3d(ii) · 2 marks

  • M1

    Ksp = [Mn2+][OH–]2

  • M2

    units = mol3 dm–9

3d(iii) · 1 mark

  • 1 mark

    2.0  10–13

4a · 1 mark

  • 1 mark

    energy required when one mole of gaseous atoms is formed from the element

4b · 1 mark

  • 1 mark

    energy released when one mole of gaseous atoms gains one mole of electrons and becomes one mole of gaseous 1– ions

4c · 2 marks

  • M1

    iodine atom has greater radius

  • M2

    less attraction between nucleus and incoming electron in iodine

4d · 2 marks

  • M1

    ½ (–486 + (2  364) )

  • M2

    +121

5a(i) · 1 mark

  • 1 mark

    Co2+ = [Ar] 3d7 AND Co3+ = [Ar] 3d6

5a(ii) · 1 mark

  • 1 mark

    has vacant d orbitals that are energetically accessible

5a(iii) · 2 marks

  • 1 mark

    pink

  • 1 mark

    blue

  • 0 marks

    aqueous

5a(iv) · 1 mark

  • 1 mark

    [Co(H2O)6]2+ + 4HCl → [CoCl4]2– + 6H2O + 4H+ OR [Co(H2O)6]2+ + 4Cl– → [CoCl4]2– + 6H2O

5a(v) · 1 mark

  • 1 mark

    ligand exchange

5a(vi) · 1 mark

  • 1 mark

    [Co(H2O)6]2+ + 2OH– → Co(OH)2(H2O)4 + 2H2O OR [Co(H2O)6]2+ + 2OH– → Co(OH)2 + 6H2O

5b(i) · 2 marks

  • M1

    Nernst equation E = E° + (0.059 / z) × log([oxidised]/[reduced])

  • M2

    E = –0.28 + (0.059 / 2) × log(0.02)

5b(ii) · 1 mark

  • 1 mark

    1.66 V

5b(iii) · 1 mark

  • 1 mark

    3Co + Cr2O72– + 14H+ → 3Co2+ + 2Cr3+ + 7H2O

5b(iv) · 1 mark

  • 1 mark

    Co2+ / Co, Co2+/ Co, Cr2O72– / Cr3+

5c · 3 marks

  • M1

    moles of cobalt = 0.547 / 58.9 = 0.00929

  • M2

    coulombs required = 0.00929  96500  2 = 1792

  • M3

    time required = 1792 / 0.50 = 3584.8 s = 59.7 minutes 3sf

6a(i) · 1 mark

  • 1 mark

    Ni(CO)4

6a(ii) · 2 marks

  • 1 mark

    ethanedioate given as C2O4

  • 1 mark

    formula shows one Ni and three ligands

  • 0 marks

    charge = 4–

6a(iii) · 1 mark

  • 1 mark

    F AND optical

6b(i) · 1 mark

  • 1 mark

    N atom can donate one lone pair of electrons

6b(ii) · 1 mark

  • 1 mark

    [Cd(en)2]2+ AND has a larger Kstab / is more stable

6b(iii) · 1 mark

  • 1 mark

    [[Cd(CH3NH2)4]2+] / ([Cd2+][CH3NH2]4)

7a · 2 marks

  • M1

    Q CH3CH2CH(CH3)CHO

  • M2

    3D enantiomers of CH3CH2CH(CH3)CHO drawn correctly

7b(i) · 1 mark

  • 1 mark

    T CH3CH2CH2COCH3 AND V (CH3)2CHCOCH3

7b(ii) · 2 marks

  • 1 mark

    yellow solid

  • 1 mark

    CH3CH2CH2COOH OR (CH3)2CHCOOH

  • 0 marks

    CHI3

7c(i) · 1 mark

  • 1 mark

    S (CH3)3CCHO

7c(ii) · 1 mark

  • 1 mark

    U CH3CH2COCH2CH3

7c(iii) · 1 mark

  • 1 mark

    triplet AND two hydrogen atoms on neighbouring carbon atoms

7c(iv) · 1 mark

  • 1 mark

    TMS / tetramethylsilane / Si(CH3)4

7d · 2 marks

  • 1 mark

    4

  • 1 mark

    3

  • 0 marks

    5

  • 0 marks

    3

8a · 1 mark

  • 1 mark

    the pH at which an amino acid exists as a zwitterion

8b · 2 marks

  • M1

    [HOOCCH(NH3)CH2CONH2]+

  • M2

    [HOOCCH(NH3)CH2COOH]+

8c · 2 marks

  • M1

    HOCH2CH(NH2)CH2CH2NH2

  • M2

    HOCH2CH(NH2)CH2CH2OH

8d(i) · 1 mark

  • 1 mark

    HOOCCH2COOH + 2SOCl2 → ClOCCH2COCl + 2SO2 + 2HCl

8d(ii) · 2 marks

  • M1

    three molecules joined, with four correct amide groups only

  • M2

    whole structure correct as above, displayed formula accepted, skeletal formula as below accepted

8e · 2 marks

  • M1

    hydrolysis of amide to carboxylate ion

  • M2

    hydrolysis of –COOH to carboxylate ion and rest of molecule / –OOCCH(NH2)CH2COO–

8f · 3 marks

  • 1 mark

    three monomer residues

  • 1 mark

    all linkages correct

  • 1 mark

    correct trailing bonds

  • 0 marks

    repeat unit marked contains correct atoms

8g(i) · 2 marks

  • 1 mark

    rotate the plane of the plane polarised light

  • 1 mark

    by the same angle

  • 0 marks

    in opposite directions

8g(ii) · 1 mark

  • 1 mark

    racemic mixture

9a · 3 marks

  • 1 mark

    all C atoms are sp2 hybridised

  • 1 mark

    bonds between C atoms are  and 

  • 1 mark

    C–C  bonds are formed by overlap of sp2 hybrid orbitals

  • 0 marks

    C–C  bonds are formed by overlap of p orbitals

  • 0 marks

    bonds between C and H atoms are 

  • 0 marks

    C–H bonds are formed by overlap of sp2 hybrid orbitals and s orbitals

9b · 1 mark

  • 1 mark

    CH3Cl and AlCl3

9c · 3 marks

  • M1

    curly arrow from within hexagon towards +C

  • M2

    intermediate, usual rules for horseshoe and + charge

  • M3

    curly arrow from C–H bond into ring AND H+ product

9d · 2 marks

  • M1

    HNO3 AND H2SO4

  • M2

    concentrated AND 25 °C ⩽ T ⩽ 60 °C

9e · 1 mark

  • 1 mark

    alkaline KMnO4

9f · 1 mark

  • 1 mark

    4-nitrobenzoic acid

9g · 1 mark

  • 1 mark

    (hot) concentrated HCl and Sn

9h · 2 marks

  • M1

    perform step 3 before step 2 OR perform step 2 before step 1

  • M2

    COOH is 3-directing OR NO2 is 3-directing

Questions

  1. 1
    Question 17 marks
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  2. 2
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  3. 3
    Question 315 marks
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  4. 4
    Question 46 marks
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  5. 5
    Question 515 marks
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  6. 6
    Question 67 marks
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  7. 7
    Question 711 marks
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  8. 8
    Question 816 marks
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  9. 9
    Question 914 marks
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