Chemistry 9701/42 February/March 2021: question paper and mark scheme

Question

(a) The most common oxidation states of cobalt are +2 and +3. Complete the electronic configurations of the following free ions. [1] (b) Co2+ and Co3+ both form complexes with edta4–. half-equation Eo/V Co3+ + e– Co2+ +1.82 O2 + 4H+ + 4e– 2H2O +1.23 [Co(edta)]– + e– [Co(edta)]2– +0.38 Co2+ + 2e– Co –0.28 Use the data in the table to predict what happens, if anything, when separate aqueous solutions of Co3+ and [Co(edta)]– are left to stand in the air. aqueous solution of Co3+ . …… . …… aqueous solution of [Co(edta)]– . …… . …… [3] (c) Hydrated cobalt(II) nitrate, Co(NO3)2•6H2O, is a red solid that behaves like hydrated magnesium nitrate, Mg(NO3)2•6H2O, when heated. Describe in detail what you would expect to observe when crystals of Co(NO3)2•6H2O are heated in a boiling tube, gently at first and then more strongly. . …… . …… . …… . …… [2] (d) Explain why the thermal stability of the Group 2 nitrates increases down the group. . …… . …… . …… . …… [2] [Total: 8]

Question

(a) Iron(II) compounds are generally only stable in neutral, non-oxidising conditions. It is difficult to determine the lattice energy of FeO experimentally. (i) Use data from the Data Booklet and this Born–Haber cycle to calculate the lattice energy, ∆H latt, of FeO(s) in kJ mol–1. Fe2+(g) + O2–(g) Fe2+(g) + O(g) + 2e– –141 kJ mol–1 +798 kJ mol–1 Fe2+(g) + O–(g) + e– Fe (g) + O(g) + e + – Fe(g) + O(g) H latt Fe(g) + 12O2(g) Fe(s) + 12O2(g) +416 kJ mol–1 –272 kJ mol–1 FeO(s) ∆H lattFeO(s) = …… kJ mol–1 [2] (ii) Most naturally occurring samples of iron(II) oxide are found as the mineral wüstite. Wüstite has formula Fe20Ox. It contains both Fe2+ and Fe3+ ions. 90% of the iron is present as Fe2+ and 10% is present as Fe3+. Deduce the value of x. x = …… [1] (iii) State and explain how the lattice energy of FeO(s) compares to the lattice energy of CaO(s). . …… . …… . …… . …… . …… [2] (b) Heating of FeO results in the formation of Fe3O4, as shown. reaction 1 4FeO → Fe + Fe3O4 Each formula unit of Fe3O4 contains one Fe2+ and two Fe3+ ions. (i) Show how reaction 1 can be described as a disproportionation reaction. . …… . …… . …… [1] Fe3O4(l) can be electrolysed using inert electrodes to form Fe. (ii) Write the half-equation for the reaction that occurs at the anode during the electrolysis of Fe3O4(l). . …… [1] (iii) Calculate the maximum mass of iron metal formed when Fe3O4(l) is electrolysed for six hours using a current of 50 A. Assume the one Fe2+ and two Fe3+ ions are discharged at the same rate. mass of iron = …… g [3] (c) LiFePO4 can be used in lithium-ion rechargeable batteries. When the cell is charging, lithium reacts with a graphite electrode to form LiC6. When the cell is discharging, the half-equations for the two processes that occur are as follows. anode half-equation LiC6 → 6C + Li+ + e– cathode half-equation Li+ + FePO4 + e– → LiFePO4 (i) State one possible advantage of developing cells such as lithium-ion rechargeable batteries. . …… [1] (ii) Use the cathode half-equation to determine the change, if any, in oxidation states of lithium and iron at the cathode during discharging. change in oxidation state during discharging metal from to lithium iron [1] (iii) Write the equation for the overall reaction that occurs when this cell is discharging. . …… [1] [Total: 13]

Question

Iodates are compounds that contain the IO3– anion. (a) The IO3– anion is shown. O O– I O Explain, with reference to the qualitative model of electron-pair repulsion, why the IO3– anion has a pyramidal shape. . …… . …… . …… [1] (b) The reaction of iodine and hot aqueous sodium hydroxide is similar to that of chlorine and hot aqueous sodium hydroxide. Sodium iodate, NaIO3, is formed as one of the products. Suggest an equation for the reaction of iodine and hot aqueous sodium hydroxide. . …… [1] (c) The decomposition of hydrogen peroxide, H2O2, is catalysed by acidified IO3–. H2O2 reduces acidified IO3– as shown. 5H2O2 + 2H+ + 2IO3– → I2 + 5O2 + 6H2O This reaction is followed by the oxidation of I2 by H2O2. half-equation Eo/V H2O2 + 2H+ + 2e– 2H2O +1.77 1 IO3– + 6H+ + 5e– 2 I2 + 3H2O +1.19 O2 + 2H+ + 2e– H2O2 +0.68 (i) Use the data to show that the separate reactions of H2O2 with IO3– and with I2 are both feasible under standard conditions. In your answer, give the equation for the reaction of H2O2 with I2. . …… . …… . …… . …… . …… . …… [3] (ii) Write the overall equation for the decomposition of H2O2 catalysed by acidified IO3–. . …… [1] (d) A student collects some data for the reaction of H2O2 with acidified IO3–, as shown in the table. [H2O2] [IO3–] [H+] initial rate of reaction experiment / mol dm–3 / mol dm–3 / mol dm–3 / mol dm–3 s–1 1 0.0500 0.0700 0.025 1.47 × 10–5 2 0.100 0.0700 0.050 2.94 × 10–5 3 0.100 0.140 0.025 5.88 × 10–5 4 0.150 0.140 0.025 8.82 × 10–5 (i) Use the data to determine the order of reaction with respect to [H2O2], [IO3–] and [H+]. Show your reasoning. order with respect to [H2O2] = …… . …… . …… . …… order with respect to [IO3–] = …… . …… . …… . …… order with respect to [H+] = …… . …… . …… . …… [3] (ii) Use your answer to (d)(i) to write the rate equation for this reaction. (iii) Calculate the value of the rate constant, k, using data from experiment 4 and your answer to (d)(ii). Give the units of k. k = …… units = …… [2] (e) Pb(IO3)2 is only sparingly soluble in water at 25 °C. The solubility product, Ksp, of Pb(IO3)2 is 3.69 × 10–13 mol3 dm–9 at 25 °C. (i) Write an expression for the solubility product of Pb(IO3)2. Ksp = [1] (ii) Calculate the solubility, in mol dm–3, of Pb(IO3)2 at 25 °C. solubility = …… mol dm–3 [2] (f) NH4IO3 is an unstable compound that readily decomposes when warmed. The decomposition reaction is shown. NH4IO3(s) → 12 N2(g) + 12 O2(g) + 12 I2(g) + 2H2O(l) ∆H = –154.6 kJ mol–1 (i) Use the data in the table to calculate the entropy change of reaction, ∆S, of the decomposition of NH4IO3(s). compound S / J K–1 mol–1 NH4IO3(s) 42 N2(g) 192 O2(g) 205 I2(g) 261 H2O(l) 70 ∆S = …… J K–1 mol–1 [2] (ii) This reaction is feasible at all temperatures. Explain why, using the data in (f) and your answer to (f)(i). . …… . …… . …… [1] [Total: 18]

Question

The transition elements are able to form stable complexes with a wide range of molecules and ions. (a) State the meaning of transition element. . …… . …… . …… [1] (b) The d orbitals in an isolated transition metal ion are degenerate. In complexes, the d orbitals occupy two energy levels. (i) Complete the diagram to show the arrangement of d orbital energy levels in octahedral and in tetrahedral complexes. energy degenerate d orbitals octahedral isolated transition tetrahedral complex metal ion complex [1] (ii) Sketch the shape of two d orbitals: ● one d orbital from the lower energy level in an octahedral complex ● one d orbital from the higher energy level in an octahedral complex. Use the axes below. lower energy level higher energy level z z x x y y [2] (c) Edds4– and edta4– are polydentate ligands that form octahedral complexes with Fe3+(aq). edds4– edta4– CO2– CO2– CO2– H – O2C N N CO2– N CO2 – N H CO2– CO2– The formulae of the complexes are [Fe(edds)]– and [Fe(edta)]– respectively. (i) On the diagram of edds4–, circle each atom that forms a bond to the Fe3+ ion in [Fe(edds)]–. [1] (ii) [Fe(edds)]– is red and [Fe(edta)]– is yellow. Explain why the two complexes have different colours. . …… . …… . …… . …… . …… [2] (iii) When edds4–(aq) is added to Fe3+(aq), the following reaction occurs. [Fe(H2O)6]3+(aq) + edds4–(aq) [Fe(edds)]–(aq) + 6H2O(l) State the type of reaction that occurs. . …… [1] (iv) Write an expression for the stability constant, Kstab, of [Fe(edds)]–(aq). Kstab = [1] (v) The table shows the values for the stability constants, Kstab, of both complexes. complex Kstab / mol–1 dm3 [Fe(edds)]– 3.98 × 1020 [Fe(edta)]– 1.26 × 1025 Predict which of the [Fe(edds)]– and [Fe(edta)]– complexes is more stable. Explain your answer with reference to the Kstab value for each complex. . …… . …… . …… [1] (vi) When an excess of edta4–(aq) is added to [Fe(edds)]–(aq), the following equilibrium is established. [Fe(edds)]–(aq) + edta4–(aq) [Fe(edta)]–(aq) + edds4–(aq) Calculate the equilibrium constant, Kc, for this equilibrium, using the Kstab values given in the table in (c)(v). Kc = …… [1] [Total: 11]

Question

(a) Carboplatin and satraplatin are used as anticancer drugs instead of cisplatin. carboplatin satraplatin O O O NH3 O Cl NH3 Pt Pt Cl N O NH3 O H2 O O (i) Describe the action of cisplatin as an anticancer drug. . …… . …… . …… [2] (ii) Suggest the geometry of the platinum centre in the carboplatin complex. . …… [1] (iii) Suggest why carboplatin does not show cis-trans isomerism. . …… . …… . …… [1] (iv) Satraplatin is a neutral complex, containing the ligands CH3CO2–, C6H11NH2, Cl – and NH3. Deduce the oxidation state of platinum in satraplatin. . …… [1] (b) Compound M is made from 1,3-dimethylbenzene in a two-step synthesis. 1,3-dimethylbenzene L M O O step 1 step 2 C8H6O4 Cl Cl (i) Draw the structure of L. [1] (ii) Suggest reactants and conditions for each step of this synthesis. step 1 . …… step 2 . …… [2] (iii) Write an equation for step 2. . …… [1] (iv) A student investigates a possible synthesis of M directly from benzene using COCl 2 in the presence of an Al Cl 3 catalyst. Benzene initially reacts with COCl 2 as shown. + reaction 1 COCl 2 + Al Cl 3 → Al Cl 4– + Cl C O + reaction 2 C6H6 + Cl C O → C6H5COCl + H+ + Reaction 2 is the electrophilic substitution of Cl C O for H+ in benzene. Suggest a mechanism for reaction 2. [3] [Total: 12]

Question

Fumaric acid is a naturally occurring dicarboxylic acid. fumaric acid CO2H HO2C (a) Identify the products of the reaction between fumaric acid and an excess of hot, concentrated, acidified manganate(VII). . …… [1] (b) Fumaric acid can form addition and condensation polymers. (i) Draw the repeat unit of the addition polymer poly(fumaric acid). [1] (ii) Draw the repeat unit of the polyester formed when fumaric acid reacts with ethane‑1,2‑diol, (CH2OH)2. The ester bond should be shown fully displayed. [2] (iii) Explain why polyesters normally biodegrade more readily than polyalkenes. . …… . …… . …… [1] (c) Fumaric acid reacts with cold, dilute, acidified manganate(VII) to form compound P. P OH CO2H HO2C OH Only three stereoisomers of P exist. One of the stereoisomers is shown. OH H CO2H HO2C H OH Complete the three-dimensional diagrams in the boxes to show the other two stereoisomers of P. HO2C HO2C [2] (d) The enzyme fumarase catalyses the reaction of fumarate ions, C4H2O42–, with water to form malate ions, C4H4O52–. C4H2O42– + H2O C4H4O52– Describe, with the aid of a suitably labelled diagram, how an enzyme such as fumarase can catalyse a reaction. . …… . …… . …… . …… [3] [Total: 10]

Question

Proline (Pro) is a naturally occurring amino acid. proline O OH N H (a) Proline is often found bonded to glycine (Gly) in a protein. (i) Draw the dipeptide Pro-Gly. The peptide bond must be shown fully displayed. [2] (ii) Name the type of reaction that forms a dipeptide from two amino acids. . …… [1] (iii) Proline is able to form a poly(proline) peptide chain. A section of a poly(proline) chain is shown. N N N O O O Suggest why the secondary structure of poly(proline) cannot be stabilised by hydrogen bonding. . …… . …… [1] (b) The reaction scheme shows several reactions of proline. proline O reaction 1 Q OH NaOH(aq) N H reaction 2 reaction 3 CH3COCl prolinol R C7H11NO3 OH N H (i) Write an equation for the reaction of proline with NaOH(aq) in reaction 1. (ii) Proline has a secondary amine functional group. Secondary amines react with acyl chlorides. For example, dimethylamine reacts with RCOCl according to the following equation. dimethylamine H 3C H 3C NH + RCOCl NCOR + HCl H3C H3C Suggest the skeletal structure of R, C7H11NO3, the product of reaction 2. [1] (iii) Suggest the reagent required for reaction 3. . …… [1] (c) Proline was first synthesised in the laboratory using a multi-stage synthetic route. In stage 1, CH2(CO2C2H5)2 and CH2=CHCN react to form a single product U. U CO2C2H5 stage 1 CH2(CO2C2H5)2 + CH2 CHCN H C CO2C2H5 CH2CH2CN (i) Name all the functional groups present in the reactants of stage 1. CH2(CO2C2H5)2 . …… CH2=CHCN . …… [2] (ii) Suggest the type of reaction that occurs in stage 1. . …… [1] In stage 2, U reacts with reagent V to form W. U W CO2C2H5 CO2C2H5 reagent V stage 2 H C CO2C2H5 H C CO2C2H5 CH2CH2CN CH2CH2CH2NH2 (iii) Suggest a suitable reagent V. . …… [1] Stage 3 takes place in the presence of an acid catalyst. X and Y are the only products of the reaction. W X CO2C2H5 O CO2C2H5 stage 3 H C CO2C2H5 HN + Y CH2CH2CH2NH2 (iv) Suggest the type of reaction that occurs in stage 3. . …… [1] (v) Deduce the identity of Y. . …… [1] After several further stages, Z is produced. Z Cl CO2H NH2 In the final stage of the synthesis, Z reacts via a nucleophilic substitution mechanism to form proline. (vi) Complete the diagram to describe the reaction mechanism of the final stage. Draw curly arrows, ions and charges, partial charges and lone pairs of electrons, as appropriate. Draw the structure of any organic intermediate ion. Z proline Cl CO2H CO2H N NH2 H [3] (vii) Identify with an asterisk (*) the chiral centre in proline. CO2H N H [1] (d) Part of the structure of gelatin is shown. O O H O O H H H H H C N C C N C C N C C N O H O CH2 H H 2C H H H H C N C C N C C N O N C C N CH2 H 2C H CH3 H H O CH2 CO2– C NH O + C NH2 NH2 Identify the number of amino acid units in the structure shown. . …… [1] (e) (i) At pH 6.5, proline exists in aqueous solution as a zwitterion. Draw the structure of the zwitterion of proline. Explain how the zwitterion of proline forms. . …… . …… . …… [2] (ii) The isoelectric point of an amino acid is the pH at which it exists as a zwitterion. Three of the amino acids in gelatin are proline, alanine and glutamic acid. Their isoelectric points are shown. proline alanine glutamic acid CO2H CO2H amino acid N H 2N CO2H H H 2N CO2H isoelectric point 6.5 6.0 3.1 A mixture of these amino acids was analysed by electrophoresis using a buffer solution at pH 4.0. Draw and label three spots on the diagram of the electropherogram to indicate the likely position of each of these three species after electrophoresis. Explain your answer. + – mixture applied here . …… . …… . …… . …… . …… . …… . …… . …… [4] (f) The weak acid ACES is a compound that can be used to make a buffer solution for electrophoresis experiments. ACES O H N OH H2 N S O O The anion of the sodium salt of ACES, C4H9N2O4SNa, is a strong base. A buffer solution is prepared by the following steps. ● 3.50 g of C4H9N2O4SNa is dissolved in 100 cm3 of distilled water. ● 50.0 cm3 of 0.200 mol dm–3 dilute hydrochloric acid is added to the solution. ● The resulting mixture is transferred to a 250.0 cm3 volumetric flask, and the solution made up to the mark. C4H9N2O4SNa reacts with HCl with a 1 : 1 stoichiometry. The pKa of ACES is 6.88 at 298 K. Calculate the pH of the buffer solution formed at 298 K. [Mr: C4H9N2O4SNa, 204.1] pH = …… [4] [Total: 28]