Chemistry 9701/43 May/June 2023: question paper and mark scheme

Question

(a) Group 2 nitrates decompose when heated. Describe how the thermal stability of Group 2 nitrates changes with increasing proton number. Explain your answer. …… [3] (b) Copper(II) nitrate decomposes in a similar manner to Group 2 nitrates. Write an equation for the decomposition of Cu(NO3)2. …… [1] (c) Cu(NO3)2 is added to water to form solution A. Fig. 1.1 shows some reactions of solution A. H 2O Cu(NO3)2(s) solution A solution C excess NH3(aq) NH3(aq) precipitate B excess concentrated HCl solution D Fig. 1.1 Complete Table 1.1 to show the formula and colour of each of the copper‑containing species present in A, B, C and D. Table 1.1 formula of copper‑containing colour of copper‑containing species formed species formed A B C D [4] (d) EDTA4– is a polydentate ligand. (i) Explain what is meant by a polydentate ligand. …… [2] (ii) Group 2 metal ions can form complexes similar to those of transition elements. A solution of EDTA4– is added to water containing [Ca(H2O)6]2+ to form a new complex, [CaEDTA]2–, as shown. equilibrium 1 [Ca(H2O)6]2+ + EDTA4– [CaEDTA]2– + 6H2O Circle on the structure of EDTA4– in Fig. 1.2 the six atoms that form bonds with the metal ion. EDTA4– O O –O CCH2 CH2C O– NCH2CH2N –O CCH2 CH2C O– O O Fig. 1.2 [1] (iii) The calcium ions in [Ca(H2O)6]2+ and [CaEDTA]2– have a coordination number of 6. Explain what is meant by coordination number. …… [1] (iv) The complex [CaEDTA]2– can be used to remove toxic metals from the body. Table 1.2 shows the numerical values for the stability constants, Kstab, for some metal ions with EDTA4–. Table 1.2 complex Kstab [CaEDTA]2– 5.0 × 1010 [CrEDTA] – 2.5 × 1023 [FeEDTA] – 1.3 × 1025 [PbEDTA]2– 1.1 × 1018 An aqueous solution containing [CaEDTA]2– is added to a solution containing equal concentrations of Cr 3+(aq), Fe3+(aq) and Pb2+(aq). The resulting mixture is left to reach a state of equilibrium. State the type of reaction when [CaEDTA]2– reacts with Cr 3+(aq), Fe3+(aq) and Pb2+(aq). …… [1] (v) Deduce the relative concentrations of [CrEDTA]–, [FeEDTA]– and [PbEDTA]2– present in the resulting mixture. Explain your answer. …… > …… > …… highest concentration lowest concentration …… [1] (e) The number of moles of water of crystallisation in a hydrated ionic salt can be determined by titration using aqueous EDTA4– ions with a suitable indicator. • 0.255 g of hydrated chromium(III) sulfate, Cr2(SO4)3•nH2O, is dissolved in water and made up to 100 cm3 in a volumetric flask. • 25.0 cm3 of this solution requires 26.2 cm3 of 0.00800 mol dm–3 aqueous EDTA4– ions to reach the end‑point. The reaction occurs as shown. [Cr(H2O)6]3+ + EDTA4– [CrEDTA] – + 6H2O Use the data to calculate the value of n in the formula of Cr2(SO4)3•nH2O. Show your working. n = …… [3] (f) A solution of Cr 3+(aq) and a solution of Fe3+(aq) have different colours. Explain why the two complexes have different colours. …… [2] [Total: 19]

Question

(a) Some transition element complexes can show stereoisomerism. State two types of stereoisomerism shown by transition element complexes. 1 …… 2 …… [1] (b) The complexes [Pt(NH3)2Cl2] and [Pt(en)2]2+ have the same geometry (shape) around the metal ion. [Pt(NH3)2Cl2] exists as two stereoisomers whereas [Pt(en)2]2+ only has one possible structure. State the geometry around the metal ion. …… [1] (c) The complex [Cr(en)3 ]2+ exists as two stereoisomers whereas the complex [Cr(OCH2CH2NH2)3 ]− exists as four stereoisomers. Complete the three‑dimensional diagrams in Fig. 2.1 to show the four stereoisomers of [Cr(OCH2CH2NH2)3 ]−. Represent the ligand –OCH2CH2NH2 by using . O N Cr Cr Cr Cr Fig. 2.1 [3] (d) The complex [Cr(OCH2CH2NH2)3]− is formed by reacting Cr 2+(aq) with the conjugate base of 2‑aminoethanol. A synthesis of 2‑aminoethanol is shown in Fig. 2.2. oxirane 2-aminoethanol O OH H C C H + NH3 H2 N H H Fig. 2.2 (i) Suggest the mechanism for step 1 of the reaction of oxirane with ammonia in Fig. 2.3. Include all relevant curly arrows, lone pairs of electrons, charges and partial charges. Draw the structure of the organic intermediate. oxirane intermediate 2-aminoethanol O step 1 step 2 OH H C C H H 2N H H H3N Fig. 2.3 [3] (ii) A small amount of by‑product E, shown in Fig. 2.4, is produced during the reaction shown in Fig. 2.2. E HO OH N H Fig. 2.4 Suggest how the formation of by‑product E can be minimised. …… [1] (iii) Compound F, C4H9NO, can be formed from the reaction of by‑product E, C4H11NO2, with concentrated H2SO4. Compound F is a saturated and basic organic compound. Suggest a structure for compound F. State the type of reaction undergone by E to form F. F type of reaction …… [2] [Total: 11]

Question

(a) Aqueous acidified iodate(V) ions, IO3–, react with iodide ions, as shown. IO3– + 6H+ + 5I– 3I2 + 3H2O The initial rate of this reaction is investigated. Table 3.1 shows the results obtained. Table 3.1 experiment [IO3–] / mol dm–3 [H+] / mol dm–3 [I–] / mol dm–3 initial rate / mol dm–3 min–1 1 0.0400 0.0150 0.0250 4.20 × 10–2 2 0.120 to be calculated 0.0125 7.09 × 10–2 The rate equation for this reaction is rate = k [IO3–][H+]2[I–]2. (i) Explain what is meant by order of reaction. …… [1] (ii) Complete Table 3.2. Table 3.2 the order of reaction with respect to [IO3–] the order of reaction with respect to [H+] the order of reaction with respect to [I–] the overall order of reaction [1] (iii) Use your answer to (a)(ii) to sketch lines in Fig. 3.1 to show the relationship between the initial rates and the concentrations of [IO3–] and [I–]. initial rate initial rate 0 0 0 [IO3–] 0 [I–] Fig. 3.1 [1] (iv) Use data from Table 3.1 to calculate the rate constant, k, for this reaction. Include the units of k. k = …… units …… [2] (v) Use data from Table 3.1 to calculate the concentration of hydrogen ions, [H+], in experiment 2. [H+] = …… mol dm–3 [1] (vi) This reaction is repeated in two separate experiments. The experiments are carried out at the same temperature and with the same concentrations of I– and IO3–. One experiment takes place at pH 1.0 and the other experiment takes place at pH 2.0. rate at pH 1.0 Calculate the value of . rate at pH 2.0 rate at pH 1.0 value of = …… [1] rate at pH 2.0 (b) In aqueous solution, iron(III) ions react with iodide ions, as shown. 2Fe3+ + 2I– 2Fe2+ + I2 The initial rate of reaction is first order with respect to Fe3+ and second order with respect to I–. The mechanism for this reaction has three steps. Each step involves only two ions reacting together. Suggest equations for the three steps of this mechanism. Identify the rate‑determining step. step 1 …… step 2 …… step 3 …… rate‑determining step = …… [3] [Total: 10]

Question

(a) State the hybridisation of the carbon atoms and the C–C–H bond angle in benzene, C6H6. Explain how orbital overlap leads to the formation of σ and π bonds in benzene. …… [3] (b) Compound Z can be synthesised from benzene in three steps by the route shown in Fig. 4.1. benzene Z NH2 step 1 step 3 X Y step 2 Fig. 4.1 (i) Draw structures for X and Y in Fig. 4.1. [2] (ii) Give the reagents and conditions for steps 1, 2 and 3. step 1 …… step 2 …… step 3 …… [3] (c) Compound W is an isomer of Z. W NH2 Fig. 4.2 Give the systematic name of W. …… [1] (d) Complete Table 4.1 to show the number of peaks observed in the carbon‑13 NMR spectrum for W and Z. Table 4.1 compound number of peaks observed W Z [1] [Total: 10]

Question

(a) The exhaust systems of most modern gasoline‑fuelled cars contain a catalytic converter with three metal catalysts. These metals act as heterogeneous catalysts. (i) Name three metal catalysts used in catalytic converters. 1 …… 2 …… 3 …… [1] (ii) Explain what is meant by a heterogeneous catalyst. …… [1] (b) The exhaust systems of many diesel‑fuelled cars contain an additional system to reduce vehicle emissions. This uses a liquid that is added to the exhaust system. This liquid contains urea, (NH2)2CO, which decomposes on heating to isocyanic acid, HNCO, and ammonia. reaction 1 (NH2)2CO HNCO + NH3 Isocyanic acid reacts with water vapour to form ammonia and carbon dioxide. reaction 2 HNCO(g) + H2O(g) NH3(g) + CO2(g) Some values for standard enthalpy changes of formation, ΔHf o , and standard entropies, S o , are given in Table 5.1. Table 5.1 compound ΔHf o / kJ mol–1 S o / J K–1 mol–1 HNCO(g) –101.7 +238.2 H2O(g) –241.8 +188.8 NH3(g) –45.9 +192.8 CO2(g) –393.5 +213.8 (i) Explain what is meant by the term entropy of a system. …… [1] (ii) Use the data in Table 5.1 to calculate ΔG o for reaction 2 at 25 °C. Show your working. ΔG o = …… kJ mol–1 [4] (c) The ammonia formed in reactions 1 and 2 can be used to remove nitrogen dioxide from exhaust emissions, as shown. reaction 3 8NH3 + 6NO2 7N2 + 12H2O Use the equations for reactions 1, 2 and 3 to construct an overall equation for the reduction of NO2 by (NH2)2CO. …… [1] (d) Isocyanic acid, HNCO, can form cyanuric acid, C3H3N3O3, under certain conditions. C3H3N3O3 has a cyclic structure containing alternating carbon and nitrogen atoms in the ring system. Suggest a structure for cyanuric acid. [1] (e) Isocyanic acid, HNCO, is a weak acid. HNCO + H2O H3O+ + NCO– pKa = 3.70 at 25 °C (i) Write the mathematical expressions for pKa and pH. pKa = …… pH = …… [1] (ii) Calculate the pH of 0.120 mol dm–3 HNCO(aq). Give your answer to three significant figures. pH = …… [2] (iii) Calculate the percentage of HNCO molecules that are ionised in 0.120 mol dm–3 HNCO. percentage ionisation of HNCO = …… [1] [Total: 13]

Question

(a) Compound H has the structural formula CH2=CHCH(NH2)COOH. (i) Name all the functional groups in H. …… [2] (ii) Compound H exhibits stereoisomerism. Draw three‑dimensional structures for the two stereoisomers of H. Name this type of stereoisomerism. type of stereoisomerism …… [2] (b) Compound H can be prepared from the reaction of J with an excess of hot aqueous acid. J O O N H O Fig. 6.1 (i) Complete Fig. 6.2 to show the equation for this reaction. O O + CH2=CHCH(NH2)COOH + N H O Fig. 6.2 [1] (ii) Name the type of reaction in (b)(i). …… [1] (c) Polymers consist of monomers joined together by undergoing either addition or condensation polymerisation. Compound H can react to form an addition polymer, K, or a condensation polymer, L, depending on the conditions. (i) Draw one repeat unit of addition polymer K. [1] (ii) Draw two repeat units of condensation polymer L. The new functional group formed should be displayed. [2] (iii) Explain why condensation polymers can normally biodegrade more readily than addition polymers. …… [1] [Total: 10]

Question

(a) State the relative basicities of ethanamide, diethylamine and ethylamine in aqueous solution. Explain your answer. …… > …… > …… most basic least basic [4] (b) The amino acid alanine, H2NCH(CH3)COOH, can act as a buffer. (i) Define a buffer solution. …… [2] (ii) Write two equations to show how an aqueous solution of alanine can act as a buffer solution. …… [2] (c) Glutamic acid is another amino acid that acts as a buffer. glutamic acid CH2CH2COOH H 2N C COOH H Fig. 7.1 (i) Draw the skeletal formula for glutamic acid. [1] (ii) Draw the structure for the dipeptide, ala‑glu, formed from one molecule of alanine and one molecule of glutamic acid. The peptide bond formed should be displayed. [2] (d) The isoelectric point of alanine is 6.0 and of glutamic acid is 3.2. A mixture of the dipeptide, ala‑glu, and its two constituent amino acids, alanine and glutamic acid, is analysed by electrophoresis using a buffer at pH 6.0. + – mixture applied here Fig. 7.2 Draw and label three spots on Fig. 7.2 to indicate the predicted position of each of these three species after electrophoresis. Explain your answer. …… [3] (e) Alanine, H2NCH(CH3)COOH, reacts with methanol to form the ester G under certain conditions. The proton (1H) NMR spectrum of G dissolved in D2O is shown in Fig. 7.3. G O H 2N O 7 6 5 4 3 2 1 0 chemical shift δ / ppm Fig. 7.3 Table 7.1 chemical shift range, environment of proton example δ / ppm alkane –CH3, –CH2–, >CH– 0.9–1.7 alkyl next to C=O CH3–C=O, –CH2–C=O, >CH–C=O 2.2–3.0 alkyl next to aromatic ring CH3–Ar, –CH2–Ar, >CH–Ar 2.3–3.0 alkyl next to electronegative atom CH3–O, –CH2–O, –CH2–Cl 3.2–4.0 attached to alkene =CHR 4.5–6.0 attached to aromatic ring H–Ar 6.0–9.0 aldehyde HCOR 9.3–10.5 alcohol ROH 0.5–6.0 phenol Ar–OH 4.5–7.0 carboxylic acid RCOOH 9.0–13.0 alkyl amine R–NH– 1.0–5.0 aryl amine Ar–NH2 3.0–6.0 amide RCONHR 5.0–12.0 (i) Complete Table 7.2 for the proton (1H) NMR spectrum of G. Table 7.2 chemical number of 1H atoms number of protons on splitting pattern shift (δ) responsible for the peak adjacent carbon atoms 1.4 3.5 4.0 [3] (ii) The proton (1H) NMR spectrum of G dissolved in CDCl3 is obtained. Describe the difference observed between this spectrum and the proton NMR spectrum in D2O shown in Fig 7.3. Explain your answer. …… [1] [Total: 18]

Question

(a) Complete Table 8.1 by placing one tick (✓) in each row to indicate the sign of each type of energy change under standard conditions. Table 8.1 can be either energy change always positive always negative negative or positive lattice energy enthalpy change of hydration enthalpy change of solution [1] (b) Define enthalpy change of hydration. …… [1] (c) Table 8.2 shows various energy changes which can be used in the following questions. Table 8.2 energy change value / kJ mol–1 standard enthalpy change of atomisation of calcium +178.2 first ionisation energy of calcium +590 second ionisation energy of calcium +1145 standard enthalpy change of atomisation of bromine +111.9 Br–Br bond energy +192.9 standard enthalpy change of solution of calcium bromide, CaBr2(s) –103.1 standard enthalpy change of formation of calcium bromide, CaBr2(s) –682.8 standard enthalpy change of hydration of Ca2+ –1579 first electron affinity of bromine –324.6 first ionisation energy of bromine +1140 o (i) Select and use relevant data from Table 8.2 to calculate the lattice energy, ΔH latt , of CaBr2(s). It may be helpful to draw a labelled energy cycle. Show your working. o ΔH latt of CaBr2(s) = …… kJ mol–1 [3] (ii) Select and use relevant data from Table 8.2 and your answer to (c)(i) to calculate the o standard enthalpy change of hydration, ΔH hyd , of Br –. It may be helpful to draw a labelled energy cycle. o If you were not able to answer (c)(i), use –2500 kJ mol–1 as your value for ΔH latt of CaBr2(s). This is not the correct value. Show your working. o ΔH hyd of Br – = …… kJ mol–1 [2] (iii) The enthalpy change of hydration of the Br – ion is more negative than the enthalpy change of hydration of the I– ion. Explain why. …… [2] [Total: 9] 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 138.9 140.1 140.9 144.4 – 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