Cambridge International AS & A Level

CHEMISTRY 9701/42
Paper 4 A Level Structured Questions May/June 2024
MARK SCHEME
Maximum Mark: 100

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the
details of the discussions that took place at an Examiners’ meeting before marking began, which would have
considered the acceptability of alternative answers.

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.

Cambridge International will not enter into discussions about these mark schemes.

Cambridge International is publishing the mark schemes for the May/June 2024 series for most
Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some
Cambridge O Level components.

This document consists of 15 printed pages.

© Cambridge University Press & Assessment 2024 [Turn over

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
Generic Marking Principles

These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the
specific content of the mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these
marking principles.

GENERIC MARKING PRINCIPLE 1:

Marks must be awarded in line with:

 the specific content of the mark scheme or the generic level descriptors for the question
 the specific skills defined in the mark scheme or in the generic level descriptors for the question
 the standard of response required by a candidate as exemplified by the standardisation scripts.

GENERIC MARKING PRINCIPLE 2:

Marks awarded are always whole marks (not half marks, or other fractions).

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:

 marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond
the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate
 marks are awarded when candidates clearly demonstrate what they know and can do
 marks are not deducted for errors
 marks are not deducted for omissions
 answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the
question as indicated by the mark scheme. The meaning, however, should be unambiguous.

GENERIC MARKING PRINCIPLE 4:

Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level
descriptors.

© Cambridge University Press & Assessment 2024 Page 2 of 15

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
GENERIC MARKING PRINCIPLE 5:

Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may
be limited according to the quality of the candidate responses seen).

GENERIC MARKING PRINCIPLE 6:

Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or
grade descriptors in mind.

Science-Specific Marking Principles

1 Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be present, marks
should not be awarded if the keywords are used incorrectly.

2 The examiner should not choose between contradictory statements given in the same question part, and credit should not be awarded for
any correct statement that is contradicted within the same question part. Wrong science that is irrelevant to the question should be ignored.

3 Although spellings do not have to be correct, spellings of syllabus terms must allow for clear and unambiguous separation from other
syllabus terms with which they may be confused (e.g. ethane / ethene, glucagon / glycogen, refraction / reflection).

4 The error carried forward (ecf) principle should be applied, where appropriate. If an incorrect answer is subsequently used in a scientifically
correct way, the candidate should be awarded these subsequent marking points. Further guidance will be included in the mark scheme
where necessary and any exceptions to this general principle will be noted.

5 ‘List rule’ guidance

For questions that require n responses (e.g. State two reasons …):

 The response should be read as continuous prose, even when numbered answer spaces are provided.
 Any response marked ignore in the mark scheme should not count towards n.
 Incorrect responses should not be awarded credit but will still count towards n.
 Read the entire response to check for any responses that contradict those that would otherwise be credited. Credit should not be
awarded for any responses that are contradicted within the rest of the response. Where two responses contradict one another, this
should be treated as a single incorrect response.
 Non-contradictory responses after the first n responses may be ignored even if they include incorrect science.

© Cambridge University Press & Assessment 2024 Page 3 of 15

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
6 Calculation specific guidance

Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show
your working’.

For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded
by the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values.

For answers given in standard form (e.g. a  10n) in which the convention of restricting the value of the coefficient (a) to a value between 1
and 10 is not followed, credit may still be awarded if the answer can be converted to the answer given in the mark scheme.

Unless a separate mark is given for a unit, a missing or incorrect unit will normally mean that the final calculation mark is not awarded.
Exceptions to this general principle will be noted in the mark scheme.

7 Guidance for chemical equations

Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme.

State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

© Cambridge University Press & Assessment 2024 Page 4 of 15

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks

1(a) M1 magnesium > calcium > strontium 4

M2 Hlatt and Hhyd both become less exothermic / less negative
M3 Hlatt changes less OR Hhyd is dominant factor
M4 Hsol becomes less exothermic / less negative / more positive / more endothermic

1(b) M1H / energy change when 1 mole of an ionic solid / compound is formed 2
M2 from gaseous ions (under standard conditions)

1(c) M1 as ionic radii increases AND Hlatt less exothermic 2

M2 as ionic charge increases AND Hlatt increases/more exothermic

1(d)(i) 3

any two [1] any three [2] all four [3]

© Cambridge University Press & Assessment 2024 Page 5 of 15

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks

1(d)(ii) M1 selection of ONLY six correct values (–381, 89, 419, 279, –200, 640) AND use of  2 as only multiplier with K 2
M2 correct evaluation of data used ecf

–381 = (89  2) + (419  2) + 279 + (–200) + 640 + Holatt

Holatt = –2116 (kJ mol–1)

Question Answer Marks

2(a)(i) 2LiNO3 → Li2O + 2NO2 + ½O2 1

2(a)(ii) M1 radius / size of cation / M+ increases OR charge density of ion decreases 2

M2 less polarisation / distortion of anion / nitrate ion / NO3– / less weakening of NO bond

2(b) M1 M2 any two for one mark or all four for two marks: 3
 mol total MnO4– = 0.125  0.0500 = 6.25  10–3
 mol Fe2+ = 0.0400  0.0225 = 9.00  10–4
 mol unreacted MnO4– = 9.00  10–4 ÷ 5 = 1.80  10–4 ecf
 mol reacted MnO4– = 6.25  10–3 - 1.80  10–4 = 6.07  10–3 ecf

M3
mol NO2– = 2.5  6.07  10–3 = 1.5175  10–2
conc NaNO2 = 4  1.5175  10–2 = 6.07-6.08  10–2 mol dm–3 ecf min 2sf

2(c)(i) 3MnO42– + 4H+ → 2MnO4– + MnO2 + 2H2O 2

M1 MnO42– as a reactant and MnO4– + MnO2 products identified

M2 correct equation

2(c)(ii) (Ecell) decreases / becomes less positive AND as [H+] decreases AND equilibrium shifts to the left 1
OR in alkali the Ecell = 0.60 – 0.56 = 0.04 V (working required)

© Cambridge University Press & Assessment 2024 Page 6 of 15

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks

3(a) M1 So = (213.8 + 2  248.2) – (237.8 + 3  205.2) 3

So = –143.2 (J K–1 mol–1)

M2 Ho = (–393.5 + 2  –296.8) – (116.7)

Ho = –1103.8 (kJ mol–1)

M3 Go = Ho – TSo

Go = –1103.8 – (298  –0.1432) = –1061.1 to –1061.4 (kJ mol–1) ecf min 3sf

3(b) M1 Go = Ho – TSo AND Go = 0 2

OR T = Ho / So [1]

M2 T = 261.6 ÷ 0.3655 = 715.7 / 716 / 715 K min 3sf

Question Answer Marks

4(a)(i) the (3)d and (4)s (sub-shells/orbitals) are close/similar in energy 1

4(a)(ii) 1

© Cambridge University Press & Assessment 2024 Page 7 of 15

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks

4(b)  precipitation solution A: e.g. NaOH / OH– 3

 observations: (pale) blue ppt. / solid
 product: Cu(OH)2 OR [Cu(OH)2(H2O)4] ecf from A

 ligand substitution solution B: e.g. HCl / Cl –, NH3

 observations: dark/deep blue solution (with NH3) OR yellow solution (with Cl –)
 product: [Cu(NH3)4(H2O)2]2+ OR [CuCl4]2–

any two [1] any four [2] all six [3]

4(c) M1 (Ag+) d-subshell is full / complete OR d10 OR d-orbitals are full 2

M2 no d-d(*) transition OR no d electrons promoted/excited

4(d) M1 species with two lone pairs (of electrons) 2

M2 that form dative (covalent) / co-ordinate bond(s) to a (central) metal atom / ion

4(e)(i) 3

4(e)(ii) optical AND cis-trans / geometical 1

4(e)(iii) trans isomer / trans isomer correctly identified AND dipoles cancel / partial charges cancel / ion dipoles cancel 1

© Cambridge University Press & Assessment 2024 Page 8 of 15

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks

5(a)(i) M1 (using exp 1 and 2) [NO]  3, rate  9 so 2nd order to NO 3

M2 (using exp 1 and 3) [NO]  2, [Cl 2]  4, rate  16 so 1st order to Cl 2

OR (using exp 2 and 3) [NO]  2 / 3, [Cl 2]  4, rate  1.8 so 1st order to Cl 2

M3 (rate =) k [NO]2 [Cl 2] ecf

5(a)(ii) M1 rate = k [NO]2[Cl 2] 2

k = (3.68  10–2) ÷ (0.0252  0.015) = 3925.33 ecf (a)(i) min 2sf

M2 dm6 mol–2 min–1 ecf (a)(i)

5(b) 2

Any two [1] all three [2]

5(c) M1 O3 + NO2 → NO3 + O2 ALLOW O3 + NO2 → NO5 2

M2 NO3 + NO2 → N2O5 ALLOW NO5 + NO2 → N2O5 + O2

Question Answer Marks

6(a)(i) species / molecules / pair that differ (by the presence or absence) of a H+ ion / proton 1

6(a)(ii) HCOOH + H2O ⇌ H3O+ + HCOO– 1

6(b)(i) [H+ ] [CH3CH2CH2COO ] 1

Ka 
[CH3CH2CH2COOH]

© Cambridge University Press & Assessment 2024 Page 9 of 15

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks

6(b)(ii) M1 Ka = 10–4.82 = 1.51  10–5 OR [H+] = 10–3.25 = 5.62  10–4 2

M2 [HA] = (10–3.25)2 ÷ 1.51  10–5 = 0.021 / 0.0209 / 0.02089 (mol dm–3) ecf min 2sf

6(c)(i) M1 a solution that resists / opposes / minimises changes in pH 2

M2 when small amounts of acid / H+ and base / alkali / OH– are added to it

6(c)(ii) Method 1 3
M1 [H+] = 10–5.70 OR 2.00  10–6 OR 1.99526  10–6

M2 moles CH3COONa = 0.400  0.125 = 5.00  10–2

moles CH3COOHeqm = (5.00  10–2  2.00  10–6) ÷ 1.78  10–5 OR 5.60  10–3

M3 [CH3COOH]initial = 5.60  10–3  1000 / 600

[CH3COOH]initial = 0.00933–0.00936 (mol dm–3) ecf min 2sf

Method 2
M1 pKa = –log(1.78  10–5) OR 4.75

M2 pH = pKa + log[B–] / [HA]

5.7 = 4.75 + log (0.05 / (0.6x)
0.95 = log (0.05 / (0.6x)

M3 100.95 = (0.05 / (0.6x)

5.348x = 0.05
x = 0.00935-0.00937 (mol dm–3) ecf min 2sf

6(d)(i) HCOOH → CO2 + 2H+ + 2e– 1

6(d)(ii) M1 Q = 3.75  40  60 OR 9000 C AND use of 96500 OR 1.6  10–19  6.02  1023 2
M2 moles of oxygen = 9000 ÷ 386000 = 0.0233
volume of O2 = 0.0233  24000 = 559.6 / 559.8 / 560 cm3 ecf min 2sf
ALLOW [use of Q ÷ (1.6  10–19  6.02  1023)] = 560.6 cm3

© Cambridge University Press & Assessment 2024 Page 10 of 15

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks

7(a)(i) 2-nitrobenzoic acid OR 2-nitrobenzenecarboxylic acid 1

7(a)(ii) hot / reflux / heat AND (alkaline / acidified / neutral) MnO4– / KMnO4 1

7(a)(iii) COOH / carboxyl group is electron-withdrawing / electronegative AND 3- and 5- / meta- directing 1

7(b)(i) 3

7(b)(ii) M1 step 1: Fe / Sn, conc. HCl 3

M2 step 2: HNO2 OR NaNO2 AND HCl
M3 step 1: heat / reflux / hot AND step 2: ⩽10 °C

Question Answer Marks

8(a) M1 benzylamine > ammonia > phenylamine 3

M1 M2 Any two for one mark, all three for two marks:
 (basicity linked to) p orbital of N / lone pair on N AND being able to bond / accept / donate to / coordinate to a proton / H+
 (benzylamine) R / alkyl / CH2 group AND is electron donating / positive inductive effect / has + I effect
 (phenylamine) p orbital of N / lone pair on N is overlaps / incorporated / delocalised in the ring / -bond system

© Cambridge University Press & Assessment 2024 Page 11 of 15

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks

8(b)(i) M1 white ppt. 2

M2 structure

8(b)(ii) M1 lone pair / p-orbital / electrons on the nitrogen / NH2 AND overlap / delocalised / incorporated AND with the (–electrons) 2
ring /  system
M2 increasing its electron density (of the ring) OR it can polarise the electrophile / Br2 better

8(c) lone pair / p-orbital on N is delocalised AND into C=O group / across the two electronegative O and N 1

© Cambridge University Press & Assessment 2024 Page 12 of 15

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks

8(d) 4

M1 / M2 any two for one mark, all four for two marks:
 lone pair on N
 correct arrow from lone pair N to C (of C=O)
 correct dipole on C=O
 correct arrow from the C=O bond to O atom

M3 correct intermediate

M4 arrow from O(–) to C–O bond AND arrow from C–Cl bond to Cl
ALLOW arrow from anywhere on O– to C–O bond

8(e)(i) pH at which a molecule has no overall charge / is neutral / exist as a zwitterion 1

8(e)(ii) C6H5CH2CH(NH2)COO– 1

8(f) 2

OR

M1 correct peptide bond displayed (there must be a saturated carbon attached to either side of the peptide group)
M2 rest of the dipeptide correct

© Cambridge University Press & Assessment 2024 Page 13 of 15

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks

9(a) due to the electron-withdrawing effect/electronegative / –I effect of chlorine 1

AND stabilising the anion / carboxylate ion OR weakening the O-H bond

9(b)(i) (COOH)2 + 2SOCl2 → (COCl)2 + 2HCl + 2SO2 1

9(b)(ii) 2

9(b)(iii) 2

M1 one correct repeat unit (within their structure)

M2 the rest of the structure correct
ecf on one incorrect monomer used
[If structure partly or fully skeletal: continuation bonds must be dashed / wavy / different or have through brackets]

9(c)(i) CHI3 / triiodomethane 1

© Cambridge University Press & Assessment 2024 Page 14 of 15

9701/42 Cambridge International AS & A Level – Mark Scheme May/June 2024
PUBLISHED
Question Answer Marks

9(c)(ii) 4
chemical shift splitting number of 1H atoms number of protons on
() pattern responsible for the peak adjacent carbon atoms

1.15 triplet 3 2

2.25 singlet 3 0

3.60 singlet 2 0

3.95 quartet / quad 2 3

ruplet

mark as any three [1] any six [2] any nine [3] all twelve [4]

9(c)(iii) 1

OR

© Cambridge University Press & Assessment 2024 Page 15 of 15