Chemistry 9701/41 May/June 2024: question paper and mark scheme

Question

(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] (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]

Question

(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. y z 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] (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] (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. dps S N N 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]

Question

(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] (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 C2O42– ligand can be represented using O O. isomer 1 isomer 2 Fe 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, E o , for the cell are shown. O2(g) + 2H2O(l) + 4e– 4OH–(aq) E o = +0.40 V 2CO2(g) + 2e– C2O42–(aq) E o = –0.59 V o Use these equations to deduce the overall cell reaction. Calculate the value of Ecell. overall cell reaction …… o Ecell = …… V [2] [Total: 13]

Question

(a) Define standard electrode potential, E o , including a description of standard conditions. …… [2] (b) (i) An electrochemical cell is set up to measure E o 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 E o value. Explain your answer. …… [1] o (c) Define enthalpy change of solution, ΔH sol . …… [1] (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, o ΔHlatt , 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] o (ii) Calculate the lattice energy, ΔHlatt , of AgNO3(s). o ΔHlatt = …… kJ mol–1 [1] (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]

Question

(a) In aqueous solution, persulfate ions, S2O82–, react with iodide ions, as shown in reaction 1. reaction 1 S2O82– + 2I– 2SO42– + I2 The rate of reaction 1 is investigated. A sample of S2O82– is mixed with a large excess of iodide ions of known concentration. The graph in Fig. 5.1 shows the results obtained. 0.900 0.800 0.700 0.600 0.500 [S2O82–] / mol dm–3 0.400 0.300 0.200 0.100 0.000 0 25 50 75 100 125 150 175 200 time / min 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 [S2O82–] [I–]. Suggest why a large excess of iodide ions allows the rate constant to be determined from the half-life in this investigation. …… [1] (b) The reaction of persulfate ions, S2O82–, 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, S2O32–, react with hydrogen ions, as shown in reaction 2. reaction 2 S2O32– + 2H+ SO2 + S + H2O The rate of reaction is first order with respect to [S2O32–] 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, for reaction 2. 2 t 1 = …… s [1] 2 (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]

Question

(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. Kekulé benzene Dewar benzene Ladenburg benzene delocalised 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] (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] (f) The reaction of phenylethanone with 1,4-dibromobutane, BrCH2CH2CH2CH2Br, in the presence of FeBr3 is shown in Fig. 6.2. phenylethanone O O FeBr3 + BrCH2CH2CH2CH2Br + HBr Br 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. organic intermediate O + BrCH2CH2CH2CH2 O + …… Br Fig. 6.3 [3] (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. Y (C20H22O2) Z (C12H14O) Fig. 6.4 [2] [Total: 15]

Question

Four esters, A, B, C and D, with the molecular formula C6H12O2 are shown in Fig. 7.1. A B O O O O C D O O O O 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. ester B detector response 52 ester A 38 ester D 25 14 ester C 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] (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 number of peaks in number of peaks in ester proton (1H) NMR spectrum 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. (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 G H alkaline I2(aq) + SOCl 2 LiAl H4 K, C6H7O2Cl J, C6H12O2 propan-2-ol L Fig. 7.3 [6] [Total: 12]

Question

Neotame is an artificial sweetener added to some foods. neotame H O N O N O H 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] neotame H O N O N O H 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]

Question

(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] (b) Salicylic acid can be synthesised from phenol. salicylic acid COOH OH 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. organic intermediate O C O O– COO– O– + …… Fig. 9.3 [3] (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 ethene cyclohexene + 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] 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) The Periodic Table of Elements Group 1 2 13 14 15 16 17 18 1 2 H He hydrogen helium Key 1.0 4.0 3 4 atomic number 5 6 7 8 9 10 Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon 6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon 23.0 24.3 3 4 5 6 7 8 9 10 11 12 27.0 28.1 31.0 32.1 35.5 39.9 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 at www.cambridgeinternational.org after the live examination series. K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton 39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 28 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon 85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon 132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – – 87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson – – – – – – – – – – – – – – – – – 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium 138.9 140.1 140.9 144.2 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium – 232.0 231.0 238.0 – – – – – – – – – – – Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge