Chemistry 9701/42 February/March 2024: question paper and mark scheme

Question

Potassium iodide, KI, is used as a reagent in both inorganic and organic chemistry. (a) KI forms an ionic lattice that is soluble in water. (i) Define enthalpy change of solution, ΔHsol. …… [1] (ii) KI(s) has a high solubility in water although its enthalpy change of solution is endothermic. Explain how this high solubility is possible. …… [2] (b) Table 1.1 gives some data about the halide ions, Cl –, Br – and I–, and their potassium salts. Table 1.1 enthalpy change of hydration, lattice energy of potassium halide, halide ion ΔHhyd / kJ mol–1 ΔHlatt / kJ mol–1 Cl – –364 –701 Br – –335 –670 I– –293 –629 (i) Explain the trend in the enthalpy change of hydration of the halide ions. …… [2] (ii) The ΔHsol values of these potassium halides are almost constant. Use the ΔHhyd and ΔHlatt data in Table 1.1 to suggest why. …… [1] (iii) The enthalpy change of solution of KI(s) is +21.0 kJ mol–1. Use this information and the data in Table 1.1 to calculate the enthalpy change of hydration of the potassium ion, K+(g). ΔHhyd of K+(g) = …… kJ mol–1 [1] (iv) Solid PbI2 forms when KI(aq) is mixed with Pb2+(aq) ions. The solubility product, Ksp, of PbI2 is 7.1 × 10–9 mol3 dm–9 at 25 °C. Calculate the solubility, in mol dm–3, of PbI2(s). solubility of PbI2(s) = ……mol dm–3 [2] (v) The ionic radius of Pb2+ is 0.120 nm compared to 0.133 nm for K+. o o Suggest how the ΔH latt of PbI2(s) differs from ΔH latt of KI(s). Explain your answer. …… [2] (c) KI slowly oxidises in air, forming I2. reaction 1 4KI(s) + 2CO2(g) + O2(g) 2K2CO3(s) + 2I2(s) ΔH o = –203.4 kJ mol–1 Table 1.2 shows some data relevant to this question. Table 1.2 standard entropy, substance S o / J K–1 mol–1 CO2(g) 213.6 I2(s) 116.1 K2CO3(s) 155.5 KI(s) 106.3 O2(g) 205.2 (i) Calculate the standard entropy change, ΔS o , of reaction 1. ΔS o = …… J K–1 mol–1 [2] (ii) Use your answer to (c)(i) to show that reaction 1 is spontaneous at 298 K. [2] (iii) The Group 1 carbonates are much more thermally stable than the Group 2 carbonates. State and explain the trend in the thermal stability of the Group 2 carbonates. …… [2] (d) A student electrolyses a solution of KI(aq) for 8 minutes using a direct current. The half-equation for the reaction that occurs at the anode is given. 2I–(aq) I2(aq) + 2e– (i) Write a half-equation for the reaction that occurs at the cathode. Include state symbols. …… [1] (ii) After the electrolysis, the I2(aq) produced requires 21.35 cm3 of 0.100 mol dm–3 Na2S2O3(aq) to react completely. I2(aq) + 2Na2S2O3(aq) 2NaI(aq) + Na2S4O6(aq) Calculate the average current used in 8 minutes during the electrolysis. current = ……A [3] (e) KI is used as a source of I– ions in organic synthesis. One example of this is shown in the synthetic route in Fig. 1.1. A B C step 1 step 2 NO2 NH2 NaNO2 and HCl step 3 D N2 step 4 I E + N2 I Fig. 1.1 (i) Identify the reagents required for steps 1 and 2. step 1 …… step 2 …… [2] (ii) Step 3 occurs in two stages. stage I NaNO2 and HCl undergo an acid–base reaction to produce HNO2. stage II HNO2 reacts with C, C6H5NH2, to produce D, C6H5N2+. Complete the equations for stage I and for stage II. stage I NaNO2 + HCl …… stage II …… [2] (iii) The I– from KI reacts with D in step 4. The mechanism is shown in Fig. 1.1. Suggest the name for this mechanism. …… [1] [Total: 26]

Question

Water is an amphoteric compound that also acts as a good solvent of polar and ionic compounds. (a) Equation 1 shows water acting as a Brønsted–Lowry acid. equation 1 H2O + NO2– HNO2 + OH– (i) Identify the two conjugate acid–base pairs in equation 1. acid I H2O conjugate base of acid I acid II conjugate base of acid II [1] (ii) Water also behaves as a Brønsted–Lowry acid when it dissolves CH3NH2. Explain the ability of CH3NH2 to act as a base. …… [1] (iii) Write an equation to show water acting as a base with CH3COOH. …… [1] (b) The ionic product of water, Kw, measures the extent to which water dissociates. H2O(l) H+(aq) + OH–(aq) Fig. 2.1 shows how Kw varies with temperature. 6.00 5.00 4.00 Kw 3.00 / 10–14 mol2 dm–6 2.00 1.00 0.00 0 10 20 30 40 50 temperature / °C Fig. 2.1 (i) Write an expression for Kw. …… [1] (ii) Use information from Fig. 2.1 to deduce whether the dissociation of water is an exothermic or an endothermic process. Explain your answer. …… [1] (iii) An aqueous solution has pH = 7.00 at 30 °C. Use information from Fig. 2.1 to explain why this solution can be considered to be alkaline at 30 °C. …… [2] (c) The three physical states of H2O have different standard entropies, S o , associated with them. Table 2.1 shows these S o values. Table 2.1 standard entropy, state of H2O S o / J K–1 mol–1 solid +48.0 liquid +70.1 gas +188.7 (i) Explain the difference in the S o values of H2O(s) and H2O(l). …… [1] (ii) Explain why the increase in S o is much greater when H2O boils than when it melts. …… [1] (iii) The energy changes for H2O(s) → H2O(l) are shown. ΔG = 0.00 kJ mol–1 ΔH = +6.03 kJ mol–1 Use these data to show that the melting point of H2O(s) is 0 °C. [1] (d) Metal–air batteries are electrochemical cells that generate electrical energy from the reaction of metal anodes with air. The standard electrode potentials for the zinc–air battery are shown. [Zn(OH)4]2– + 2e– Zn + 4OH– E o = –1.22 V 1 2 O2 + H2O + 2e – 2OH– E o = +0.40 V o (i) Calculate the standard cell potential, E cell , of the zinc–air battery. o E cell = ……V [1] (ii) The zinc–air battery usually operates at pH 11 and 298 K. The overall cell potential is dependent on [OH–]. The Nernst equation shows how the electrode potential at the cathode changes with [OH–]. E = 0.40 – (0.059 z ) log([OH ] ) –2 Calculate the electrode potential, E, at pH 11. E = ……V [2] [Total: 13]

Question

Iron is a transition metal in Group 8 of the Periodic Table. (a) (i) Explain why iron has variable oxidation states. …… [1] (ii) Complete the shorthand electronic configurations of Fe and Fe3+. Fe [Ar] …… Fe3+ [Ar] …… [1] (b) An aqueous solution of Fe(NO3)3 contains the complex [Fe(H2O)6]3+. When solutions of KSCN(aq) and [Fe(H2O)6]3+(aq) are mixed, a colour change is observed. The red complex [Fe(H2O)5SCN]2+ forms. (i) Define complex. …… [1] (ii) State the coordination number of Fe in [Fe(H2O)6]3+. …… [1] (iii) The H—O—H bond angle in water is 104.5°. Suggest the H—O—H bond angle in [Fe(H2O)6]3+. Explain your answer. …… [1] (iv) Explain why iron complexes are coloured. …… [3] (v) Aqueous solutions of complexes [Fe(H2O)6]3+ and [Fe(H2O)5SCN]2+ are different colours. Explain why these complexes are different colours. …… [2] (c) Table 3.1 gives values for the stability constants, Kstab, of different complexes of iron. Table 3.1 complex stability constant, Kstab [Fe(H2O)5(H2PO4)]2+ 5.90 × 101 [Fe(H2O)5SCN]2+ 1.30 × 102 (i) [Fe(H2O)5(H2PO4)]2+ can form when H3PO4 reacts with [Fe(H2O)6]3+. Write an equation for this reaction. …… [1] (ii) Write an expression for Kstab of [Fe(H2O)5SCN]2+ and give its units. Kstab = units = …… [2] (iii) Use the stability constant data in Table 3.1 to calculate the value of the equilibrium constant, Kc, for the following equilibrium. [Fe(H2O)5(H2PO4)]2+ + SCN– [Fe(H2O)5SCN]2+ + H2PO4– value of Kc = …… [1] [Total: 14]

Question

Ruthenium and osmium are transition metals below iron in Group 8 of the Periodic Table. (a) Two different complex ions, X and Y, can form when anhydrous RuCl 3 reacts with water under certain conditions. X and Y have octahedral geometry. Aqueous samples of X and Y react separately with an excess of AgNO3(aq). Different amounts of AgCl are precipitated: • 1 mole of complex ion X produces 2 moles of AgCl • 1 mole of complex ion Y produces 1 mole of AgCl. (i) Complete Table 4.1 to suggest formulae for X and Y. Table 4.1 X Y formula of complex [2] (ii) Both complexes react with an excess of bipyridine, bipy, to form a mixture of two stereoisomers of [Ru(bipy)3]3+. bipy N N Bipyridine is a bidentate ligand. Draw three-dimensional diagrams of the two stereoisomers of [Ru(bipy)3]3+. Use N N to represent the bipy ligand in your structures. Ru Ru [2] (b) Fig. 4.1 shows another ruthenium complex. 5+ (H3N)5Ru N N Ru(NH3)5 Fig. 4.1 This complex contains the neutral ligand pyrazine. pyrazine N N (i) Suggest how pyrazine is able to bond to two separate ruthenium ions. …… [1] (ii) Pyrazine is an aromatic compound. The bonding and structure of pyrazine is similar to that of benzene. Describe and explain the shape of pyrazine. In your answer, include: • the hybridisation of the nitrogen and carbon atoms • how orbital overlap forms π bonds between the atoms in the ring. …… [2] (iii) Predict the number of peaks seen in the carbon–13 NMR spectrum of pyrazine. Explain your answer. …… [2] (iv) The overall charge of the ruthenium complex in Fig. 4.1 is 5+. Deduce the possible oxidation states of the two ruthenium ions in the complex. …… [1] (c) Osmium tetroxide, OsO4, reacts with alkenes in a similar manner to cold dilute acidified MnO4–. Fig. 4.2 shows a proposed synthesis of a condensation polymer G. step 1 HOOC COOH ClOC COCl G step 3 OsO4 C6H12O2 step 2 Fig. 4.2 (i) Suggest a reagent for step 1. …… [1] (ii) Draw the structure of exactly one repeat unit of the condensation polymer G. The ester linkage should be shown fully displayed. [2] [Total: 13]

Question

Compound Q can be synthesised from chlorobenzene in seven steps, using the route shown in Fig. 5.1. chlorobenzene J K Cl Cl Cl AlCl3 and CH3 CHO CH3Cl [O] step 1 step 2 step 3 L Cl CN OH step 4 P N M Cl COOCH3 Cl COOCH3 Cl COOH acidified CH3OH SOCl2 Cl OH step 5 OH step 6 step 7 HN and K2CO3(aq) S Q Cl COOCH3 N S Fig. 5.1 (a) (i) Write an equation for the formation of the electrophile for step 1. …… [1] (ii) Complete the mechanism in Fig. 5.2 for step 1, the alkylation of chlorobenzene. Include all relevant curly arrows and charges. Draw the structure of the intermediate. intermediate Cl Cl CH3 +…… Fig. 5.2 [3] (iii) Step 2 is an oxidation reaction. Construct an equation for the reaction in step 2. Use [O] to represent an atom of oxygen from an oxidising agent. …… [1] (iv) Suggest reagents for the conversion of K to M in steps 3 and 4. step 3 …… step 4 …… [2] (v) Identify the type of reaction that occurs in step 5. …… [1] (vi) Step 7 takes place when P is heated with a weak base such as K2CO3(aq). HN P S Q Cl COOCH3 Cl COOCH3 and K2CO3(aq) Cl N step 7 S Suggest why a strong base such as NaOH(aq) is not used for this reaction. …… [1] (vii) Q is optically active. Explain the meaning of optically active. …… [1] (viii) Give two reasons why it might be desirable to synthesise a single optical isomer of Q for use as a drug. 1 …… 2 …… [2] (b) Q is commonly used in conjunction with aspirin. aspirin COOH O O Aspirin is a weak Brønsted–Lowry acid. (i) The pKa of aspirin is 3.49. 75 mg of aspirin dissolves in water to form 100 cm3 of an aqueous solution. Calculate the pH of this solution. [Mr: aspirin, 180.0] pH = …… [3] (ii) Aspirin undergoes acid hydrolysis in the stomach. Give the structures of the organic products of this acid hydrolysis. [2] [Total: 17]

Question

Amino acids are molecules that contain —NH2 and —COOH functional groups. Glycine, H2NCH2COOH, is the simplest stable amino acid. (a) The isoelectric point of glycine is 6.2. (i) Define isoelectric point. …… [1] (ii) Draw the structure of glycine at pH 4. [1] (b) Fig. 6.1 shows two syntheses starting with glycine. glycine C2H5Br N COOH H 2N COOH reaction 1 reaction 2 hippuric acid O N COOH H an excess of LiAl H4 reaction 3 U Fig. 6.1 (i) State the essential conditions for reaction 1. …… [1] (ii) Identify the reagent used in reaction 2. …… [1] (iii) Draw the structure of the organic product U that forms when hippuric acid reacts with an excess of LiAl H4 in reaction 3. [2] (iv) A molecule of phenylalanine, R, can react with a molecule of glycine to form two dipeptides, S and T. S and T are structural isomers. R glycine H2N COOH S and T + H 2N COOH Draw the structures of these dipeptides. The peptide bond formed should be shown fully displayed. S T [2] (c) A student proposes a synthesis of hippuric acid by the reaction of benzamide, C6H5CONH2, and chloroethanoic acid, Cl CH2COOH. The reaction does not work well because benzamide is a very weak base. (i) Explain why amides are weaker bases than amines. …… [2] (ii) The pKa of chloroethanoic acid is 2.86 whereas the pKa of ethanoic acid is 4.76. Explain the difference between these two pKa values. …… [2] (d) Compound V is another amino acid. The proton (1H) NMR spectrum of V shows hydrogen atoms in five different environments, a, b, c, d and e, as shown in Fig. 6.2. V H H HOOC CH2CH2NH2 H H a b c d e Fig. 6.2 Table 6.1 environment of proton example chemical shift range, d / ppm alkane –CH3, –CH2–, >CH– 0.9–1.7 alkyl next to C=O CH3–C=O, –CH2–C=O, 2.2–3.0 >CH–C=O alkyl next to aromatic ring CH3–Ar, –CH2–Ar, 2.3–3.0 >CH–Ar alkyl next to electronegative atom CH3–O, –CH2–O, 3.2–4.0 –CH2–Cl, –CH2–N 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 6.2 for the proton (1H) NMR spectrum of V taken in CDCl 3. Table 6.1 gives some relevant data. Table 6.2 proton a b c d e chemical shift range, d / ppm name of splitting multiplet pattern [4] (ii) Complete Table 6.3 by placing a tick (✓) to indicate any protons whose peaks are still present in the proton (1H) NMR spectrum of V taken in D2O. Table 6.3 proton a b c d e present in D2O [1] [Total: 17] 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 Kw = 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