Cambridge A Level Chemistry 9701 — 2024 May/June Paper 4 · Variant 1

This document has 28 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level CHEMISTRY 9701/41 Paper 4 A Level Structured Questions May/June 2024 2 hours You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer all questions. ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● You may use a calculator. ● You should show all your working and use appropriate units. INFORMATION ● The total mark for this paper is 100. ● The number of marks for each question or part question is shown in brackets [ ]. ● The Periodic Table is printed in the question paper. ● Important values, constants and standards are printed in the question paper. DC (LK/SW) 326720/3 © UCLES 2024 * 7 5 3 2 2 8 0 4 9 5 *

2 9701/41/M/J/24 © UCLES 2024 1 (a) (i) Describe the trend in the solubility of the hydroxides of magnesium, calcium and strontium. Explain your answer. … > … > … most soluble least soluble … … … … … … [4] (ii) Suggest the variation in pH of saturated solutions of the hydroxides of magnesium, calcium and strontium. Explain your answer. … … … [1] (b) Barium hydroxide, Ba(OH)2, is a strong base. A 250.0 cm3 solution of Ba(OH)2 with a pH of 12.2 is made by dissolving Ba(OH)2 in distilled water. Calculate the mass of Ba(OH)2 required to make this solution. Show your working. [Mr: Ba(OH)2, 171.3] mass of Ba(OH)2 = … g [4]

3 9701/41/M/J/24 © UCLES 2024 [Turn over (c) The solubility of iron(II) hydroxide, Fe(OH)2, is 5.85 × 10–6 mol dm–3 at 298 K. (i) Write the expression for the solubility product, Ksp, of Fe(OH)2. Ksp = [1] (ii) Calculate the value of Ksp of Fe(OH)2. Include its units. Ksp = … units = … [2] [Total: 12]

4 9701/41/M/J/24 © UCLES 2024 2 (a) (i) Define transition element. … … … [1] (ii) Explain why transition elements can form complex ions. … … [1] (b) The 3d orbitals in an isolated Ag+ ion are degenerate. (i) Define degenerate d orbitals. … … [1] (ii) Sketch the shape of a 3dxy orbital in Fig. 2.1. z y x Fig. 2.1 [1] (c) Tollens’ reagent can be used to distinguish between aldehydes and ketones. Tollens’ reagent contains [Ag(NH3)2]OH, which can be prepared in a two-step process. step 1 Aqueous NaOH is added dropwise to aqueous AgNO3 to form Ag2O as a brown precipitate. step 2 Aqueous NH3 is added dropwise to Ag2O to form a colourless solution containing [Ag(NH3)2]OH. Construct equations for each of the steps in the preparation of [Ag(NH3)2]OH. step 1 … step 2 … [2]

5 9701/41/M/J/24 © UCLES 2024 [Turn over (d) Name the shape of the complex ion [Ag(NH3)2]+. State the bond angle for H-N-Ag and for N-Ag-N. shape … bond angle for H-N-Ag = … ° bond angle for N-Ag-N = … ° [2] (e) An electrochemical cell uses Ag2O as the positive electrode and Zn as the negative electrode immersed in an alkaline electrolyte. The overall cell reaction is shown. Ag2O + Zn + H2O 2Ag + Zn(OH)2 Complete the half-equation for the reaction at each electrode. at the positive electrode Ag2O + … at the negative electrode Zn + … [2]

6 9701/41/M/J/24 © UCLES 2024 (f) Coordination polymers are made when a bidentate ligand acts as a bridge between different metal ions. Under certain conditions Ru3+(aq) and the bidentate ligand dps can form a coordination polymer containing ([Ru(dps)Cl 4]–)n chains. N N S dps Fig. 2.2 The bidentate ligand dps uses each of the nitrogen atoms to bond to a different Ru3+. Complete Fig. 2.3 by drawing the structure for the coordination polymer ([Ru(dps)Cl 4]–)n. Show two repeat units. The dps ligand can be represented using N N . Ru Fig. 2.3 [2] [Total: 12]

7 9701/41/M/J/24 © UCLES 2024 [Turn over 3 (a) When a sample of hydrated lithium ethanedioate, Li2C2O4•H2O, is gently heated, two gaseous products are formed and a white solid residue remains. The residue is added to HNO3(aq). A gas is produced that turns limewater milky. Complete the equation for the decomposition of Li2C2O4•H2O. Li2C2O4•H2O … + … + … [1] (b) The trend in the decomposition temperatures of the Group 2 ethanedioates is similar to that of the Group 2 nitrates. Suggest which of CaC2O4 and BaC2O4 will decompose at the lower temperature. Explain your answer. … … … … [2] (c) Potassium iron(III) ethanedioate, K3[Fe(C2O4)3], dissolves in water to form a green solution. Explain why transition elements can form coloured complexes. … … … … … … [3] (d) The anhydrous iron(III) compound K3[Fe(C2O4)3] decomposes on heating to form a mixture of K2[Fe(C2O4)2], K2C2O4 and CO2. Complete the equation for the decomposition of K3[Fe(C2O4)3]. … K3[Fe(C2O4)3] … K2[Fe(C2O4)2] + … K2C2O4 + … CO2 [1]

8 9701/41/M/J/24 © UCLES 2024 (e) The [Fe(C2O4)3]3– complex ion shows stereoisomerism. Complete the three-dimensional diagrams in Fig. 3.1 to show the two stereoisomers of [Fe(C2O4)3]3–. The C2O4 2– ligand can be represented using O O. isomer 1 Fe isomer 2 Fe Fig. 3.1 [2] (f) Buffer solutions are used to regulate pH. Write two equations to describe how a solution containing HC2O4 – ions acts as a buffer solution when small amounts of acid or alkali are added. … … [2] (g) A fuel cell is an electrochemical cell that can be used to generate electrical energy by using oxygen to oxidise a fuel. Ethanedioic acid, (COOH)2, dissolved in an alkaline electrolyte is being investigated as a fuel. The relevant standard electrode potentials, Eo, for the cell are shown. O2(g) + 2H2O(l) + 4e– 4OH–(aq) Eo = +0.40 V 2CO2(g) + 2e– C2O4 2–(aq) Eo = –0.59 V Use these equations to deduce the overall cell reaction. Calculate the value of E o cell. overall cell reaction … E o cell = … V [2] [Total: 13]

9 9701/41/M/J/24 © UCLES 2024 [Turn over 4 (a) Define standard electrode potential, Eo, including a description of standard conditions. … … … [2] (b) (i) An electrochemical cell is set up to measure Eo of the Ag+(aq)/Ag(s) electrode. Draw a labelled diagram of this electrochemical cell. Include all necessary substances. It is not necessary to state conditions used. [3] (ii) A separate electrochemical cell is set up using a lower concentration of Ag+(aq) than that used in (b)(i). Suggest how the electrode potential, E, for the Ag+(aq)/Ag(s) electrode would change from its Eo value. Explain your answer. … … … [1] (c) Define enthalpy change of solution, ΔH o sol . … … [1]

10 9701/41/M/J/24 © UCLES 2024 (d) Some relevant energy changes for AgNO3 are shown in Table 4.1. Table 4.1 energy change value / kJ mol–1 enthalpy change of solution of AgNO3(s) +22.6 enthalpy change of hydration of silver ions –475 enthalpy change of hydration of nitrate ions –314 (i) Complete the energy cycle in Fig. 4.1 to show the relationship between the lattice energy, ΔH o latt , of AgNO3(s) and the energy changes shown in Table 4.1. Include state symbols for all the species. AgNO3(s) Fig. 4.1 [2] (ii) Calculate the lattice energy, ΔH o latt , of AgNO3(s). ΔH o latt = … kJ mol–1 [1]

11 9701/41/M/J/24 © UCLES 2024 [Turn over (e) Suggest the trend in the magnitude of the lattice energies of the metal nitrates, NaNO3(s), Mg(NO3)2(s) and RbNO3(s). Explain your answer. … … … most exothermic least exothermic … … … … … [3] [Total: 13]

12 9701/41/M/J/24 © UCLES 2024 5 (a) In aqueous solution, persulfate ions, S2O8 2–, react with iodide ions, as shown in reaction 1. reaction 1 S2O8 2– + 2I– 2SO4 2– + I2 The rate of reaction 1 is investigated. A sample of S2O8 2– is mixed with a large excess of iodide ions of known concentration. The graph in Fig. 5.1 shows the results obtained. 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 0 50 100 150 200 25 75 125 175 time / min [S2O8 2–] / mol dm–3 Fig. 5.1 (i) Use Fig. 5.1 to determine the initial rate of reaction 1. Show your working. rate = … mol dm–3 min–1 [1] (ii) The rate equation for reaction 1 is rate = k [S2O8 2–] [I–]. Suggest why a large excess of iodide ions allows the rate constant to be determined from the half-life in this investigation. … … … [1]

13 9701/41/M/J/24 © UCLES 2024 [Turn over (b) The reaction of persulfate ions, S2O8 2–, with iodide ions is catalysed by Fe2+ ions. Write two equations to show how Fe2+ catalyses reaction 1. equation 1 … equation 2 … [2] (c) Describe the effect of an increase in temperature on the rate constant and the rate of reaction 1. … … [1] (d) In aqueous solution, thiosulfate ions, S2O3 2–, react with hydrogen ions, as shown in reaction 2. reaction 2 S2O3 2– + 2H+ SO2 + S + H2O The rate of reaction is first order with respect to [S2O3 2–] and zero order with respect to [H+] under certain conditions. The rate constant, k, for this reaction is 1.58 × 10–2 s–1. Calculate the half-life, t 1 2, for reaction 2. t 1 2 = … s [1] (e) The compound nitrosyl bromide, NOBr, can be formed as shown in reaction 3. reaction 3 2NO(g) + Br2(g) 2NOBr(g) The rate is first order with respect to [NO] and first order with respect to [Br2]. The reaction mechanism has two steps. Suggest equations for the two steps of this mechanism. State which is the rate-determining step. step 1 … step 2 … rate-determining step = … [2] [Total: 8]

14 9701/41/M/J/24 © UCLES 2024 6 (a) (i) State what is meant by partition coefficient, Kpc. … … … [1] (ii) The partition coefficient, Kpc, for a compound, X, between carbon disulfide, CS2, and water is 10.5. 1.85 g of X is dissolved in water and made up to 100.0 cm3 in a volumetric flask. 40.0 cm3 of this aqueous solution is shaken with 25.0 cm3 of CS2. The mixture is left to reach equilibrium. Calculate the mass of X, in g, extracted into the CS2 layer. mass of X = … g [2] (b) The compound C6H6 has many structural isomers. Four suggested structures of C6H6 are shown in Fig. 6.1. delocalised benzene Kekulé benzene Dewar benzene Ladenburg benzene Fig. 6.1 Using Fig. 6.1, complete Table 6.1 to predict the number of carbon atoms that have sp, sp2 and sp3 hybridisation in Kekulé benzene, Dewar benzene and Ladenburg benzene. Table 6.1 C6H6 structure sp hybridised sp2 hybridised sp3 hybridised Kekulé benzene Dewar benzene Ladenburg benzene [2]

15 9701/41/M/J/24 © UCLES 2024 [Turn over (c) Describe the shape of delocalised benzene. Include the geometry of each carbon, the C-C-H bond angle and the type of bond(s) between the carbon atoms and between the carbon and hydrogen atoms. … … … … … [2] (d) Suggest why Dewar benzene and Ladenburg benzene are unstable isomers of C6H6. … … [1] (e) Complete Table 6.2 to predict the number of peaks in the proton (1H) NMR spectrum for Dewar benzene, Ladenburg benzene and delocalised benzene. Table 6.2 number of peaks Dewar benzene Ladenburg benzene delocalised benzene [1]

16 9701/41/M/J/24 © UCLES 2024 (f) The reaction of phenylethanone with 1,4-dibromobutane, BrCH2CH2CH2CH2Br, in the presence of FeBr3 is shown in Fig. 6.2. O O BrCH2CH2CH2CH2Br phenylethanone FeBr3 + Br + HBr Fig. 6.2 The mechanism of this reaction is similar to that of the alkylation of benzene. (i) Construct an equation for the formation of the electrophile, BrCH2CH2CH2CH2 +. … [1] (ii) Complete the mechanism in Fig. 6.3 for the reaction of phenylethanone with BrCH2CH2CH2CH2 + ions. Include all relevant curly arrows and charges. Draw the structure of the organic intermediate. O O BrCH2CH2CH2CH2 organic intermediate + Br + … Fig. 6.3 [3]

17 9701/41/M/J/24 © UCLES 2024 [Turn over (iii) The reaction shown in Fig. 6.2 forms small amounts of two by-products, Y (C20H22O2) and Z (C12H14O). Suggest structures for Y and Z in the boxes in Fig. 6.4. Fig. 6.4 [2] [Total: 15] Y (C20H22O2) Z (C12H14O)

18 9701/41/M/J/24 © UCLES 2024 7 Four esters, A, B, C and D, with the molecular formula C6H12O2 are shown in Fig. 7.1. O O O O O O O O A B C D Fig. 7.1 (a) Give the systematic name of ester A. … [1] (b) A mixture of these esters, A, B, C and D, is analysed by gas–liquid chromatography. The chromatogram produced is shown in Fig. 7.2. The number above each peak represents the area under the peak. The area under each peak is proportional to the mass of the respective ester in the mixture. detector response ester C ester B ester A ester D 38 52 25 14 time / min Fig. 7.2 (i) State what is meant by retention time. … … [1] (ii) Calculate the percentage by mass of ester D in the original mixture. percentage by mass of ester D = … % [1]

19 9701/41/M/J/24 © UCLES 2024 [Turn over (c) Separate samples of the esters, A, B, C and D, are analysed using proton (1H) NMR and carbon-13 NMR spectroscopy. (i) Complete Table 7.1 to show the number of peaks in each NMR spectrum for esters B and C. Table 7.1 ester number of peaks in proton (1H) NMR spectrum number of peaks in carbon-13 NMR spectrum B C [2] (ii) Identify all of the esters from A, B, C and D that have at least one triplet peak in their proton (1H) NMR spectrum. … [1] Question 7 continues on page 20.

20 9701/41/M/J/24 © UCLES 2024 (d) Compound F, C6H8O3, shows stereoisomerism and effervesces with Na2CO3(aq). Compound F reacts with alkaline I2(aq) to form yellow precipitate G and compound H. Compound F reacts with LiAl H4 to form compound J, C6H12O2. Compound F reacts with SOCl 2 to form compound K, C6H7O2Cl . Compound K reacts with propan-2-ol to form compound L. Draw the structures of compounds F, G, H, J, K and L in the boxes in Fig. 7.3. F, C6H8O3 alkaline I2(aq) + J, C6H12O2 LiAl H4 SOCl 2 K, C6H7O2Cl L propan-2-ol G H Fig. 7.3 [6] [Total: 12]

21 9701/41/M/J/24 © UCLES 2024 [Turn over 8 Neotame is an artificial sweetener added to some foods. N N H H O neotame O O O OH Fig. 8.1 (a) (i) State the number of chiral carbon atoms in a molecule of neotame. … [1] (ii) Neotame contains the arene functional group. Identify all the other functional groups present in neotame. … … [2] (b) Neotame reacts with an excess of hot NaOH(aq) to form three organic products. (i) State the two types of reaction that occur when neotame reacts with hot NaOH(aq). 1 … 2 … [2]

22 9701/41/M/J/24 © UCLES 2024 N N H H O neotame O O O OH Fig. 8.1 (ii) Draw the structures of the three organic products formed from the reaction of neotame with an excess of hot NaOH(aq). [3] [Total: 8]

23 9701/41/M/J/24 © UCLES 2024 [Turn over 9 (a) Samples of phenol, C6H5OH, are reacted separately with sodium and with dilute nitric acid. phenol OH Fig. 9.1 (i) Write the equation for the reaction of C6H5OH with Na. … [1] (ii) Draw the structures of the two major isomeric organic products formed in the reaction of phenol with dilute HNO3. [1]

24 9701/41/M/J/24 © UCLES 2024 (b) Salicylic acid can be synthesised from phenol. salicylic acid OH COOH Fig. 9.2 One of the steps in this synthesis is the electrophilic substitution reaction of carbon dioxide with the phenoxide ion, C6H5O–. Complete the mechanism in Fig. 9.3 for the reaction of C6H5O– with CO2. Include all relevant curly arrows, dipoles and charges. Draw the structure of the organic intermediate. O– O– + … C O O COO– organic intermediate Fig. 9.3 [3]

25 9701/41/M/J/24 © UCLES 2024 (c) Some syntheses use Diels–Alder reactions, which normally involve a diene and an alkene reacting together to form a cyclohexene. (i) Draw three curly arrows in Fig. 9.4 to complete the mechanism for the Diels–Alder reaction between buta-1,3-diene and ethene. buta-1,3-diene cyclohexene ethene + Fig. 9.4 [1] (ii) Another Diels–Alder reaction of buta-1,3-diene is shown in Fig. 9.5. Predict the product formed in this reaction. + Fig. 9.5 [1] [Total: 7]

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27 9701/41/M/J/24 © UCLES 2024 Important values, constants and standards molar gas constant R = 8.31 J K–1 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Avogadro constant L = 6.022 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K) Vm = 24.0 dm3 mol–1 at room conditions ionic product of water K w = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1)

28 9701/41/M/J/24 © UCLES 2024 Group The Periodic Table of Elements 1 H hydrogen 1.0 2 He helium 4.0 1 2 13 14 15 16 17 18 3 4 5 6 7 8 9 10 11 12 3 Li lithium 6.9 4 Be beryllium 9.0 atomic number atomic symbol Key name relative atomic mass 11 Na sodium 23.0 12 Mg magnesium 24.3 19 K potassium 39.1 20 Ca calcium 40.1 37 Rb rubidium 85.5 38 Sr strontium 87.6 55 Cs caesium 132.9 56 Ba barium 137.3 87 Fr francium – 88 Ra radium – 5 B boron 10.8 13 Al aluminium 27.0 31 Ga gallium 69.7 49 In indium 114.8 81 Tl thallium 204.4 6 C carbon 12.0 14 Si silicon 28.1 32 Ge germanium 72.6 50 Sn tin 118.7 82 Pb lead 207.2 22 Ti titanium 47.9 40 Zr zirconium 91.2 72 Hf hafnium 178.5 104 Rf rutherfordium – 23 V vanadium 50.9 41 Nb niobium 92.9 73 Ta tantalum 180.9 105 Db dubnium – 24 Cr chromium 52.0 42 Mo molybdenum 95.9 74 W tungsten 183.8 106 Sg seaborgium – 25 Mn manganese 54.9 43 Tc technetium – 75 Re rhenium 186.2 107 Bh bohrium – 26 Fe iron 55.8 44 Ru ruthenium 101.1 76 Os osmium 190.2 108 Hs hassium – 27 Co cobalt 58.9 45 Rh rhodium 102.9 77 Ir iridium 192.2 109 Mt meitnerium – 28 Ni nickel 58.7 46 Pd palladium 106.4 78 Pt platinum 195.1 110 Ds darmstadtium – 29 Cu copper 63.5 47 Ag silver 107.9 79 Au gold 197.0 111 Rg roentgenium – 30 Zn zinc 65.4 48 Cd cadmium 112.4 80 Hg mercury 200.6 112 Cn copernicium – 114 Fl flerovium – 116 Lv livermorium – 7 N nitrogen 14.0 15 P phosphorus 31.0 33 As arsenic 74.9 51 Sb antimony 121.8 83 Bi bismuth 209.0 8 O oxygen 16.0 16 S sulfur 32.1 34 Se selenium 79.0 52 Te tellurium 127.6 84 Po polonium – 9 F fluorine 19.0 17 Cl chlorine 35.5 35 Br bromine 79.9 53 I iodine 126.9 85 At astatine – 10 Ne neon 20.2 18 Ar argon 39.9 36 Kr krypton 83.8 54 Xe xenon 131.3 86 Rn radon – 113 Nh nihonium – 115 Mc moscovium – 117 Ts tennessine – 118 Og oganesson – 21 Sc scandium 45.0 39 Y yttrium 88.9 57–71 lanthanoids 89–103 actinoids 57 La lanthanum 138.9 89 Ac lanthanoids actinoids actinium – 58 Ce cerium 140.1 90 Th thorium 232.0 59 Pr praseodymium 140.9 91 Pa protactinium 231.0 60 Nd neodymium 144.2 92 U uranium 238.0 61 Pm promethium – 93 Np neptunium – 62 Sm samarium 150.4 94 Pu plutonium – 63 Eu europium 152.0 95 Am americium – 64 Gd gadolinium 157.3 96 Cm curium – 65 Tb terbium 158.9 97 Bk berkelium – 66 Dy dysprosium 162.5 98 Cf californium – 67 Ho holmium 164.9 99 Es einsteinium – 68 Er erbium 167.3 100 Fm fermium – 69 Tm thulium 168.9 101 Md mendelevium – 70 Yb ytterbium 173.1 102 No nobelium – 71 Lu lutetium 175.0 103 Lr lawrencium – To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series.

This document consists of 18 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/41 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.

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 2 of 18 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.

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 3 of 18 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.

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 4 of 18 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.

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 5 of 18 Question Answer Marks 1(a)(i) M1 strontium > calcium > magnesium M2Hlatt and Hhyd both become less exothermic / less negative M3Hlatt changes more M4Hsol becomes more exothermic / more negative 4 1(a)(ii) pH increases as the concentration of [OH–] ions increases 1 1(b) M1 [H+] = 10–12.2 OR 6.31  10–13 M2 [OH–] = 0.0158 mol dm–3 ecf M3 [Ba(OH)2] = 0.01585 ÷ 2 = 7.92 ecf M4 mol of Ba(OH)2 = 0.00792 ÷ 4 = 1.98  10–3 mass Ba(OH)2 = 0.339 g ecf 4 1(c)(i) Ksp = [Fe2+][OH–]2 1 1(c)(ii) M1 Ksp = 4  (5.85  10–6)3 = 8.01  10–16 M2 mol3 dm–9 2 Question Answer Marks 2(a)(i) element which forms one or more stable ion with incomplete d orbitals 1 2(a)(ii) they have vacant d orbitals that are energetically accessible 1 2(b)(i) orbitals of the same energy 1 2(b)(ii) 1

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 6 of 18 Question Answer Marks 2(c) M1 2AgNO3 + 2NaOH → Ag2O + 2NaNO3 + H2O OR 2Ag+ + 2OH– → Ag2O + H2O M2 Ag2O + 4NH3 + H2O → 2[Ag(NH3)2]OH OR Ag+ + 2NH3 + OH– → [Ag(NH3)2]OH 2 2(d) M1 bond angle H–N–Ag is 109.5° M2 shape is linear AND bond angle for N–Ag–N is 180° 2 2(e) M1 Ag2O + H2O + 2e– → 2Ag + 2OH– M2 Zn + 2OH– → Zn(OH)2 + 2e– 2 2(f) For example: M1 presence of dps ligand bonded to two Ru M2 the rest of the structure correct 2

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 7 of 18 Question Answer Marks 3(a) Li2C2O4H2O → Li2CO3 + H2O + CO 1 3(b) M1 CaC2O4 AND Ca2+ has a smaller ionic radius / Ca2+ has a higher charge density M2 anion / C2O42– becomes more polarised / distorted 2 3(c) M1 d orbital(s) of different energy / d-d splitting occurs M2 electron(s) promoted / excited M3 light is absorbed AND colour seen is complementary 3 3(d) 2K3Fe(C2O4)3 → 2K2Fe(C2O4)2 + K2C2O4 + 2CO2 1 3(e) 2 3(f) M1 HC2O4– + H+ → H2C2O4 M2 HC2O4– + OH– → C2O42– + H2O 2 3(g) M1 2C2O42– + O2(g) + 2H2O(l) → 4CO2(g) + 4OH–(aq) M2 Ecell = 0.40 – (–0.59) = 0.99 V 2

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 8 of 18 Question Answer Marks 4(a) M1 voltage of a half-cell compared to SHE M2 1 mol dm–3 AND 101 kPa AND 298 K 2 4(b)(i) M1 Pt Ag Ag+ M2 gas delivery system H2 H+ M3 voltmeter salt bridge wiring Pt to V to Ag 3 4b(ii) E value would be more negative AND shifts Ag+ (+ e–) ⇋ Ag to the left 1 4(c) enthalpy change when one mole of a solute dissolves in water 1 4(d)(i) M1 two arrows in blue M2 correct species in red with all state symbols 2

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 9 of 18 Question Answer Marks 4(d)(ii) Hlatt = –811.6 (kJ mol–1) 1 4(e) M1 Mg(NO3)2(s), NaNO3(s), RbNO3(s) M2 Mg2+ has a higher charge than Na+ OR Rb+ AND Na+ has a smaller radius than Rb+ M3 correct statement relating magnitude of LE to attraction between ions OR strength of ionic bonds 3 Question Answer Marks 5(a)(i) tangent drawn at t = 0 AND gradient of tangent calculated at t = 0 answer between 0.016 to 0.040 1 5(a)(ii) [I–] stays constant / [I–] does not change OR the overall order is one under these conditions 1 5(b) M1 equation 1: 2Fe2+ + S2O82– → 2Fe3+ + 2SO42– M2 equation 2: 2Fe3+ + 2I– → 2Fe2+ + I2 2 5(c) the rate constant and rate of reaction 1 will both increase 1 5(d) k = 0.693 ÷ t½ t½ = 0.693 ÷ 0.0158 = 43.9 s 1 5(e) NO + Br2 → NOBr2 OR NO + Br2 → NOBr + Br NOBr2 + NO → 2NOBr OR Br + NO → NOBr M1 two equations adding up to the overall equation M2 step 1 has one NO and one Br2 only and is identified as slow step 2

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 10 of 18 Question Answer Marks 6(a)(i) ratio of the concentrations of solute in two solvents 1 6(a)(ii) M1 10.5 = (y ÷ 25) ÷ ((0.74–y) ÷ 40) M2 y = 0.642 g 2 6(b) sp sp2 sp3 Kekulé benzene 0 6 0 Dewar benzene 0 4 2 Ladenburg benzene 0 0 6 M1 row 1 and 2 correct M2 row 3 correct 2 6(c) M1 120° AND hexagonal/trigonal planar M2 C-C has -bonds and -bonds AND C-H have -bonds only 2 6(d) bond strain OR ring strain 1 6(e) peaks Dewar benzene 2 Ladenburg benzene 1 delocalised benzene 1 1 6(f)(i) BrCH2CH2CH2CH2Br + FeBr3 → BrCH2CH2CH2CH2+ + FeBr4– 1

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 11 of 18 Question Answer Marks 6(f)(ii) M1 curly arrow from inside the hexagon towards C+ M2 structure of the intermediate M3 curly arrow from C–H bond into the ring AND H+ 3

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 12 of 18 Question Answer Marks 6(f)(iii) M1 Y M2 Z O OR 2 Question Answer Marks 7(a) methyl pentanoate 1 7(b)(i) time between injection and detection 1 7(b)(ii) 29.5% 1

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 13 of 18 Question Answer Marks 7(c)(i) ester number of peaks seen in proton (1H) NMR number of peaks seen in carbon-13 NMR B 5 6 C 4 5 M1 row 1 correct M2 row 2 correct 2 7(c)(ii) A and B 1

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 14 of 18 Question Answer Marks 7(d) one mark for each structure: F, G, H, J, K and L 6

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 15 of 18 Question Answer Marks 7(d) six alternatives for F

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 16 of 18 Question Answer Marks 8(a)(i) two 1 8(a)(ii) amide amine ester carboxylic acid / carboxyl any two [1] all four [2] 2 8(b)(i) M1 hydrolysis M2 acid–base 2 8(b)(ii) one mark for each structure 3

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 17 of 18 Question Answer Marks 9(a)(i) C6H5OH + Na → C6H5ONa + ½H2 1 9(a)(ii) both structures for one mark 1 9(b) M1 correct dipole AND curly arrow from inside the hexagon to the carbon atom M2 structure of the intermediate M3 curly arrow from C-H bond into the ring AND H+ 3

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 18 of 18 Question Answer Marks 9(c)(i) All three curly arrows for one mark 1 9(c)(ii) 1