Chemistry 9701/42 May/June 2020: question paper and mark scheme

Question

EDTA4–, is a polydentate ligand. (a) (i) Explain what is meant by the term polydentate ligand. . …… . …… . …… [2] (ii) When a solution containing EDTA4– is added to a solution containing [Cd(H2O)6]2+ a new complex is formed, [CdEDTA]2–. equilibrium 1 [Cd(H2O)6]2+ + EDTA4– [CdEDTA]2– + 6H2O Circle, on the structure of EDTA4–, the six atoms that form bonds with the metal ion. EDTA4– O O – O CCH2 CH2C O– NCH2CH2N – O CCH2 CH2C O– O O [1] (iii) Write an expression for the stability constant, Kstab1, for equilibrium 1, and state its units. Kstab1 = units = …… [2] (b) Cadmium ions form complexes with methylamine, CH3NH2, and with 1,2‑diaminoethane, H2NCH2CH2NH2, as shown in equilibriums 2 and 3. 1,2‑diaminoethane is shown as en. equilibrium 2 [Cd(H2O)6]2+ + 4CH3NH2 [Cd(CH3NH2)4(H2O)2]2+ + 4H2O Kstab2 = 3.60 × 106 equilibrium 3 [Cd(H2O)6]2+ + 2en [Cd(en)2(H2O)2]2+ + 4H2O Kstab3 = 4.20 × 1010 An equilibrium is set up between these two complexes as shown in equilibrium 4. equilibrium 4 [Cd(CH3NH2)4(H2O)2]2+ + 2en [Cd(en)2(H2O)2]2+ + 4CH3NH2 ∆H o = +0.840 kJ mol–1 ∆S o = +80.9 J K–1 mol–1 (i) Keq4 is the equilibrium constant for equilibrium 4. Write an expression for Keq4 in terms of Kstab2 and Kstab3. Keq4 = [1] (ii) Calculate the value of the standard Gibbs free energy change, ∆G o, for equilibrium 4 at 298 K. ∆G o = …… kJ mol–1 [2] (iii) State how the value of ∆G o changes as the temperature increases. Explain your answer. . …… . …… . …… [1] [Total: 9]

Question

(a) Describe and explain how the solubility of the Group 2 sulfates varies down the group. . …… . …… . …… . …… . …… . …… [4] (b) The trend in the decomposition temperatures of Group 2 peroxides, MO2, is similar to that of Group 2 carbonates. Suggest which of barium peroxide, BaO2, and calcium peroxide, CaO2, will decompose at the lower temperature. Explain your answer. . …… . …… . …… [2] (c) Magnesium iodate(V), Mg(IO3)2, decomposes when heated to form magnesium oxide, oxygen and iodine. Construct an equation for this reaction. . …… [1] (d) Calcium iodate(V), Ca(IO3)2, is sparingly soluble in water. The concentration of its saturated solution is 5.6 × 10–3 mol dm–3 at 298 K. (i) Write an expression for the solubility product, Ksp, of Ca(IO3)2, and state its units. Ksp = units = …… [2] (ii) Calculate the numerical value for Ksp Ca(IO3)2 at 298 K. Ksp = …… [1] (iii) When a few cm3 of concentrated Ca(NO3)2(aq) is added to a saturated solution of Ca(IO3)2 a white precipitate forms. Identify the white precipitate and give an explanation for this observation. . …… . …… . …… [2] (e) Iodised salt is sodium chloride mixed with a small amount of sodium iodate(V), NaIO3. ●● 50.00 g of iodised salt is dissolved in distilled water and the solution made up to 250 cm3 in a volumetric flask with distilled water. ●● 0.0 cm3 of this solution is pipetted into an excess of aqueous acidified potassium iodide. 5 IO3– + 5I– + 6H+ → 3I2 + 3H2O ●● The iodine produced requires 12.40 cm3 of 0.00200 mol dm–3 aqueous sodium thiosulfate solution for complete reaction. I2 + 2S2O32– → 2I– + S4O62– Calculate the mass of sodium iodate(V) present in 50.00 g of iodised salt. mass of NaIO3 = …… g [3] (f) The half-equation for the reduction of iodate(V) ions is shown. IO3– + 6H+ + 5e– → 12 I2 + 3H2O E o = +1.19 V Use data from the Data Booklet to predict whether a reaction is feasible when aqueous solutions of acidified iodate(V) ions and bromide ions are mixed. Explain your answer. . …… . …… [1] (g) Iodate(V) ions react with sulfite ions in acidic solution at pH 5.00 as shown. IO3– + 3SO32– → I– + 3SO42– The initial rate of reaction was found to be first order with respect to IO3–, first order with respect to SO32– and first order with respect to H+. (i) Write the rate equation for this reaction, stating the units of the rate constant, k. rate = …… mol dm–3 s–1 units of k = …… [2] (ii) The rate of reaction depends on the pH of the solution. Assume all other concentrations remain the same. rate at pH 5.00 Use the expression x = to calculate the value of x. rate at pH 4.00 x = …… [1] [Total: 19]

Question

(a) Complete the electronic configuration of an isolated gaseous nickel(II) ion, Ni2+. 3d 4s [Ar] [1] (b) Explain the origin of colour in transition element complexes. . …… . …… . …… . …… . …… . …… [4] (c) Hexaaquanickel(II) ions are green. They form a green precipitate with hydroxide ions, OH–, in equilibrium 1 and a blue complex with ammonia, NH3, in equilibrium 2. equilibrium 1 [Ni(H2O)6]2+ + 2OH– Ni(OH)2 + 6H2O green ppt. equilibrium 2 [Ni(H2O)6]2+ + 6NH3 [Ni(NH3)6]2+ + 6H2O blue Use Le Chatelier’s principle to suggest explanations for the following observations. (i) Explain why when aqueous NH3 is added dropwise to [Ni(H2O)6]2+ a green precipitate is formed. . …… . …… . …… [1] (ii) Explain why when a large excess of aqueous NH3 is added to [Ni(H2O)6]2+, the green precipitate dissolves and a blue solution is formed. . …… . …… . …… [1] (d) The complex ion [NiBr2(CN)2]2– shows stereoisomerism. Draw diagrams to show the two isomers of [NiBr2(CN)2]2–. Name the type of stereoisomerism. isomer 1 isomer 2 type of stereoisomerism …… [2] [Total: 9]

Question

(a) (i) When benzene undergoes nitration a nitro group substitutes at a carbon atom. State the shape (geometry) around the substituted carbon atom ●● in the benzene molecule, . …… ●● in the intermediate complex, . …… ●● in the nitrobenzene product. …… [2] (ii) Naphthalene, C10H8, is an arene hydrocarbon. naphthalene When naphthalene undergoes nitration, a mixture of two organic compounds is formed. Each compound contains one nitro group. Suggest the structures of these compounds. structure 1 structure 2 [1] (b) Naphthalene can be oxidised under certain conditions to phthalic anhydride, C8H4O3, carbon dioxide and water. Construct an equation for this reaction. Use [O] to represent an atom of oxygen from the oxidising agent. . …… [1] (c) The indicator, phenolphthalein, can be synthesised from phthalic anhydride and phenol under certain conditions. phenolphthalein O phenol phthalic anhydride O O 2 + O + H2 O OH O OH HO Deduce the type of reaction shown by this equation. . …… [1] (d) (i) Name the functional groups, in addition to the benzene ring, present in a phenolphthalein molecule. . …… [1] (ii) Phenolphthalein reacts separately with the two reagents shown in the table. Complete the table by: ●● drawing the structures of the organic products formed (part of the structure has been given for you) ●● stating the types of reaction. reagent organic product structure type of reaction an excess of hot NaOH(aq) an excess of Br2(aq) [4] (e) Phenolphthalein is an indicator and is represented by the formula HIn. Phenolphthalein, HIn, is a weak acid. HIn H+ + In– The Ka value for phenolphthalein is 5.0 × 10–10 mol dm–3 at 298 K. This indicator changes colour at a pH of approximately 8.8. [In–] Calculate the ratio at pH 8.8. [HIn] [In–] ratio = …… [2] [HIn] (f) Methyl orange is another acid-base indicator. Its structure in aqueous solution at pH 4.4 is shown. methyl orange H 3C N N N SO3–Na+ H 3C (i) On the structure of methyl orange, circle the bond or bonds that make this compound a dye.[1] The colour of this indicator changes between pH 3.2 and pH 4.4. (ii) Suggest the structure of methyl orange at pH 3.0. Assume the –SO3–Na+ group is unreactive. [1] (g) Methyl orange can be synthesised as shown. O2N SO3–Na+ step 1 R step 2 S H3C N H3C methyl orange H3C N N N SO3–Na+ H3C (i) Deduce the identities of compounds R and S and draw their structures in the boxes. [2] (ii) Suggest reagents and conditions for step 1 and step 2. step 1 . …… step 2 . …… [3] [Total: 19]

Question

(a) Define the term partition coefficient, Kpc. . …… . …… . …… [2] (b) Kpc of benzoic acid between octan‑1‑ol and water is 79.4. (i) A solution of 0.400 g of benzoic acid in 25.0 cm3 octan‑1‑ol is shaken with 125 cm3 of water. Calculate the mass of benzoic acid extracted into the water layer. mass of benzoic acid extracted = …… g [2] (ii) Kpc of benzophenone, C6H5COC6H5, between octan‑1‑ol and water is different from the value of Kpc of benzoic acid given in (b)(i). Explain why. . …… . …… . …… [1] (c) Benzophenone can be synthesised from benzoic acid in two steps as shown. In step 1 compound J, a reactive reaction intermediate, is formed. Compound J then reacts with an organic compound, K, to form benzophenone. benzoic acid CO2H J step 1 benzophenone O Al Cl 3 K (i) Deduce the identities of organic compounds J and K and draw their structures in the boxes.[2] (ii) Suggest reagents and conditions for step 1. . …… [1] (d) Benzophenone can also be synthesised in two steps from bromomethylbenzene. bromomethylbenzene L, C13H12 Br step 3 step 4 KMnO4, heat benzophenone O (i) Deduce the identity of compound L and draw its structure in the box. [1] (ii) Name the mechanism of step 3 and suggest reagents and conditions for step 3. mechanism of step 3 …… reagents and conditions …… [2] (iii) Deduce the type of reaction in step 4. . …… [1] (e) (i) Deduce the number of peaks that would be present in the carbon‑13 NMR spectrum of benzophenone. number of peaks . …… [1] (ii) Identify two different environments of carbon atom that would result in different chemical shift ranges in this carbon‑13 NMR spectrum of benzophenone. environment of carbon atom chemical shift range (δ) [2] [Total: 15]

Question

The class of polymers called polycarbonates are made by the reaction of carbonyl dichloride, COCl 2, with a diol. a polycarbonate O a diol n HOROH + n COCl 2 O R O C + 2nHCl n (a) (i) Deduce the type of polymerisation shown here. . …… [1] Nalgene® is a polycarbonate formed from the diol P and COCl 2. P H3C CH3 C HO OH (ii) Draw one repeat unit of Nalgene®. [1] (iii) Nalgene® is a strong and tough polymer. Identify two types of intermolecular force that are responsible for these properties of Nalgene®. 1 . …… 2 . …… [1] (b) Proteins are polymers of amino acids. Complete the table to show how the secondary and tertiary structures of proteins are stabilised. one intermolecular groups involved force responsible secondary structure tertiary structure [2] (c) Explain the significance of hydrogen bonding in DNA in relation to the accurate replication of genetic information. . …… . …… . …… [2] (d) Many polymers are degradable. State two different processes by which some polymers can be degraded. . …… . …… [1] (e) The cyclic peptide B is shown. B O O N H HN NH O O HN NH HN H O O N HO O Cyclic peptide B is broken into its monomers by heating under reflux with dilute hydrochloric acid. The amino acid threonine, Thr, and two other organic products are formed. threonine OH H2N CO2H (i) Draw the structures of the two other organic products formed. [2] (ii) Using the 3-letter abbreviations for the amino acids as given in the Data Booklet, complete the sequence for the cyclic peptide, B. Thr …… [1] (iii) Name two analytical techniques that could be used to separate these amino acids. …… and . …… [1] [Total: 12]

Question

(a) (i) Define the term electron affinity. . …… . …… . …… [2] (ii) Define the term lattice energy. . …… . …… . …… [2] (b) Use the following data and relevant data from the Data Booklet to calculate a value for the enthalpy change of formation of zinc bromide, ZnBr2(s). You might find it helpful to construct an energy cycle. electron affinity of Br(g) = –325 kJ mol–1 enthalpy change of atomisation of Zn(s) = +131 kJ mol–1 enthalpy change of vaporisation of Br2(l) = +31 kJ mol–1 lattice energy of ZnBr2(s) = –2678 kJ mol–1 enthalpy change of formation of ZnBr2(s) = …… kJ mol–1 [4] (c) The lattice energies of ZnBr2, ZnCl 2 and ZnO are shown. compound lattice energy / kJ mol–1 ZnBr2 –2678 ZnCl 2 –2734 ZnO –3971 (i) Explain why there is a difference between the lattice energies of ZnBr2 and ZnCl 2. . …… . …… [1] (ii) Explain why there is a difference between the lattice energies of ZnCl 2 and ZnO. . …… . …… [1] [Total: 10]

Question

(a) (i) Define the term standard cell potential. . …… . …… . …… [2] An electrochemical cell is set up to measure the standard electrode potential of a cell, , made of a Co3+ / Co2+ half-cell and a Cl 2 / Cl – half-cell. (ii) Complete the table with the substance used to make the electrode in each of these half‑cells. half-cell electrode Co3+ / Co2+ Cl 2 / Cl – [1] (iii) Use data from the Data Booklet to calculate the . = …… V [1] (iv) Write the equation for the overall cell reaction. . …… [1] (b) A fuel cell is an electrochemical cell that can be used to generate electrical energy. A methanol-oxygen fuel cell can be used as an alternative to a hydrogen-oxygen fuel cell. When the cell operates, the carbon atoms in the methanol molecules are converted into carbon dioxide. CH3OH + H2O → CO2 + 6H+ + 6e– Calculate the volume of CO2, in cm3, formed when a current of 2.5 A is delivered by the cell for 30 minutes. Assume the cell is operated at room conditions. volume of CO2 = …… cm3 [2] [Total: 7]