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Chemistry 9701/22 October/November 2021: question paper and mark scheme

Cambridge AS & A Level Chemistry (9701), Paper 2 AS Structured Questions, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9701_w21_qp_229701_w21_ms_229701/22/O/N/21

Paper
Paper 2 AS Structured Questions
Total marks
60
Questions
3
Code on the paper
9701/22/O/N/21
9701_w21_ms_22.pdf

9701/22 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 126 marks

Question

Hydrogen iodide, HI, is a colourless gas at room temperature. (a) (i) Explain why HI has a higher boiling point than HCl and HBr. . …… . …… . …… . …… [2] (ii) The bar chart shows the boiling points of HCl, HBr and HI. The boiling point of HF is not shown. 300 280 260 boiling 240 point / K 220 200 180 160 HF HCl HBr HI Hydrogen bonds form between HF molecules. Draw a bar on the bar chart to predict the boiling point of HF. Explain your answer. . …… . …… [2] (b) The standard enthalpy change of formation, , of HI(g) is +26.5 kJ mol–1. Define the term standard enthalpy change of formation. . …… . …… . …… [2] (c) HI(g) can be formed by reacting H2(g) with I2(g). The reaction is reversible, and an equilibrium forms quickly at high temperatures. H2(g) + I2(g) 2HI(g) (i) Construct an expression for the equilibrium constant, Kp, for the reaction of H2(g) and I2(g) to form HI(g). Kp = [1] (ii) The equilibrium partial pressures of the gases at 200 °C are as follows. pH2(g) = 895 Pa pI2(g) = 895 Pa pHI(g) = 4800 Pa Calculate Kp for this reaction. Kp = …… [1] (iii) State how the value of Kp would change, if at all, if the reaction were carried out at 100 °C rather than 200 °C. Explain your answer. . …… . …… . …… . …… [2] (d) HI reacts with oxygen to form iodine and water. (i) Construct an equation for the reaction of HI with oxygen. . …… [1] (ii) Explain, with reference to oxidation numbers, why this reaction is a redox reaction. . …… . …… . …… . …… [2] (e) HI(g) can also be formed by the reaction of I2(g) with hydrazine, N2H4(g). 2I2(g) + N2H4(g) → 4HI(g) + N2(g) State the change in pressure that would occur when 2 mol I2(g) fully reacts with 1 mol N2H4(g) in a sealed container at constant temperature. Explain your answer. . …… . …… . …… . …… [2] (f) In the laboratory, HI(aq) can be formed in a two-step process. step 1 3I2(s) + 2P(s) → 2PI3(s) step 2 PI3(s) + 3H2O(l) → H3PO3(aq) + 3HI(aq) (i) Draw a ‘dot-and-cross’ diagram of a PI3 molecule. [2] (ii) Name the type of reaction in step 2. . …… [1] (iii) H3PO3(aq) and HI(aq) are both strong Brønsted–Lowry acids. Give the meaning of the term strong Brønsted–Lowry acid. . …… . …… . …… [2] (iv) Give the formula of the conjugate base of H3PO3. . …… [1] (g) HI(g) reacts with propene, CH3CH=CH2(g) to form a mixture of 1-iodopropane and 2-iodopropane. (i) Identify which of 1-iodopropane and 2-iodopropane is the major product of this reaction. Explain your answer. . …… . …… . …… . …… [2] (ii) Complete the diagram to show the mechanism of the reaction between HI and CH3CH=CH2 that forms the major product identified in (g)(i). Include curly arrows, lone pairs of electrons and charges as necessary. H 3C H C C H H H I + – [3] [Total: 26]

Mark scheme

1(a)(i) M1 (HI / I / iodine / hydrogen iodide has a) greater number of electrons 2 M2 greater induced dipoles (between molecules) 1(a)(ii) M1 bar at HF shows any boiling point above HI on graph 2 M2 explanation of difference in boiling point of a sample of HF in terms of strength (sum of) hydrogen bonds (and induced dipoles in HF) are stronger than (sum of) induced dipoles (and permanent dipoles in HCl / HBr / HI) 1(b) (enthalpy / energy change) when one mole of a compound is formed 2 from its elements in their standard states 1(c)(i) 𝑝HI 1 𝐾 = 𝑝H 𝑝 I 1(c)(ii) 28.76 OR 28.8 OR 29 1 1(c)(iii) EITHER 2 option 1 which assumes ∆Hreaction is (still) endothermic (using the value shown in (b)). M1 (Kp) decreases AND endothermic / ∆H(f) = + / positive M2 reaction favours formation of reactants / hydrogen and iodine OR (product) yield / partial pressure of HI decreases / equilibrium shifts to the left OR option 2 which realises that ∆Hreaction is in fact exothermic (using bond energy data in Data Booklet) M1 (Kp) increases AND exothermic / ∆H(f) = + / negative M2 reaction favours formation of products / hydrogen iodide yield increases / partial pressure of HI increases / equilibrium shifts to the right 1(d)(i) 4HI + O2 → 2I2 + 2H2O 1 1(d)(ii) M1 I / iodine (increases) oxidation number –1 → 0 so oxidation / is oxidised 2 OR HI / is oxidised as I (increases) oxidation number –1 → 0 M2 O (decreases) oxidation number 0 → –2 so reduction / is reduced 1(e) M1 pressure increases 2 M2 (pressure goes up as) number of moles/molecules increases in ratio 3 (gas) reactants to 5 (gas) products OR pressure is (directly) proportional to number of moles/molecules 1(f)(i) 2 M1 correct bonding pairs M2 correct number of remaining outer electrons on each atom 1(f)(ii) hydrolysis 1 1(f)(iii) proton donor / H+ donor 2 fully dissociates / fully ionises 1(f)(iv) H2PO3– 1 1(g)(i) M1 2-iodopropane – formed from a (more) stable (secondary) (carbo)cation/intermediate 2 M2 (because of) greater (positive) inductive effect / (+)I of two alkyl groups OR (because of positive) inductive effect / (+)I of more R / more methyl / more alkyl groups 1(g)(ii) 3 M1 curly arrow from = of C=C to H AND curly arrow from bond of H—I to I M2 curly arrow from lone pair of I– to C(+) of their intermediate M3 correct carbocation AND product for 2-iodopropane

Question 217 marks

Question

(a) Table 1 gives physical data for some of the Period 3 elements. Table 1 atomic number, Z 11 12 13 14 15 16 17 bonding present in element M C first ionisation energy / kJ mol–1 494 736 577 786 1060 1000 1260 maximum oxidation number +7 anionic radius / nm – – – 0.271 0.212 0.184 0.181 (i) Complete the row in the table labelled ‘bonding present in element’. Use C = covalent, I = ionic, M = metallic, as appropriate. [1] (ii) Explain the difference between the first ionisation energies of the elements with atomic numbers 11 and 17. . …… . …… . …… . …… . …… [2] (iii) Explain the difference between the first ionisation energies of the elements with atomic numbers 15 and 16. . …… . …… . …… . …… . …… [2] (iv) Complete the row in the table labelled ‘maximum oxidation number’. [1] (v) Explain the variation in anionic radius for the elements with atomic numbers 14 to 17. . …… . …… . …… . …… [2] (b) Use the axes to sketch a graph that shows the trend in melting points of the elements with atomic numbers 11 to 17. melting point of element 11 12 13 14 15 16 17 atomic number, Z [2] (c) Dmitri Mendeleev published the first Periodic Table in 1869. Mendeleev used his knowledge of chemical periodicity to propose the properties of gallium, 31Ga, a Group 13 element. Table 2 gives some chemical and physical data of elements in Group 13. Table 2 density boiling point cationic radius element / g cm–3 /K / nm 5 B 2.34 3930 0.020 13 Al 2470 0.050 31 Ga 5.91 2400 49 In 7.30 0.081 81 Tl 11.8 1460 0.095 Complete the table by predicting values for the missing data. [3] (d) Indium and aluminium are elements in Group 13 of the Periodic Table. Indium has very similar chemical properties to aluminium. ● Indium reacts vigorously with hydrochloric acid to form a colourless gas and a salt in solution. ● I ndium oxide, In2O3, is amphoteric. ● Gaseous indium bromide has the formula In2Br6. This molecule contains coordinate bonds. (i) Identify the formula of the salt formed when indium reacts with hydrochloric acid. . …… [1] (ii) Construct an equation for the reaction of In2O3 with excess aqueous NaOH. . …… [1] (iii) Draw a diagram that clearly shows the types of bond present in In2Br6(g). [2] [Total: 17]

Mark scheme

2(a)(i) 1 M M M C C C C 2(a)(ii) All 3 points correct scores two marks 2 Any 2 points scores one mark • nuclear charge increases OR increasing proton number e.g. 17 / Cl has a greater nuclear charge • describe the similarity in shielding between the two elements e.g. they have almost the same shielding • describe the overall effect in terms of greater nuclear attraction for (outer) electrons e.g. (outer) electrons are attracted more (strongly) to the nucleus 2(a)(iii) M1 describes the difference between 1st IE of elements 15 and 16 (P and S) in terms of either: 2 spin-pair repulsion (in element 16 / S) OR electron pair repulsion (in element 16 / S) M2 describes the location of the electron pair in the (3)p orbital which repel each other 2(a)(iv) 1 (+)1 (+)2 (+)3 (+)4 (+)5 (+)6 2(a)(v) M1 (anions have) same number of electrons (but increasing proton number) 2 M2 increasing proton number / nuclear charge AND increasing attraction of nucleus for (outer) electrons OR (outer) electrons attracted more (strongly) to the nucleus AND because of increasing proton number / nuclear charge 2(b) 2 2(c) M1 density of 13Al: value within range 2.5–5.0 (g cm–3) 3 M2 cationic radius of 31Ga: value within range 0.055–0.075 (nm) M3 boiling point of 49In: value within range 1500–2300 (K) 2(d)(i) InCl3 1 2(d)(ii) In2O3 + 2NaOH + 3H2O → 2NaIn(OH)4 1 2(d)(iii) 2 M1 correct connectivity of In2Br6 M2 showing the two correct dative covalent bonds

Question 317 marks

Question

Compound T is an isomer of C6H12. T (a) Name T. . …… [1] (b) Draw the skeletal formula of a structural isomer of T that shows cis-trans (geometrical) isomerism. [1] (c) Each carbon atom in T forms a sigma (σ) bond to at least one other carbon atom, as shown. H 3C CH3 C C H3C CH3 (i) On the diagram, draw the orbitals that represent the pi (π) bond that is also present in T. [1] (ii) State the hybridisation of the two carbon atoms between which the pi (π) bond forms. . …… [1] (d) A reaction scheme starting with T is shown. Reaction 2 occurs in the presence of a catalyst; knowledge of the mechanism for this reaction is not required. T U V reaction 1 HO OH reaction 2 O catalyst (i) Give the reagent(s) and conditions for reaction 1. . …… [1] (ii) State and explain how 2,4-dinitrophenylhydrazine (2,4-DNPH) can be used to detect the presence of V as a product of reaction 2. . …… . …… . …… [2] (iii) The progress of reaction 2 can be monitored by infrared spectroscopy. The absorption caused by O–H bonds is always present because water is used as a solvent. Identify two absorptions, and the bonds responsible for these absorptions, whose appearance will change significantly during the reaction. 1 . …… . …… 2 . …… . …… [2] (e) V is used in a wide range of organic reactions. Some reactions of V are shown. V W reaction 3 O O alkaline aqueous I2 O– reaction 4 NaBH4 X Y reaction 5 reaction 6 OH Z dehydration addition polymerisation (i) V and W are colourless and soluble in water. State what you would observe in reaction 3. . …… [1] (ii) Reaction 3 is a redox reaction. Identify which of the reactants is reduced in this reaction. . …… [1] (iii) Construct an equation for reaction 4. Use [H] in the equation to represent an atom of hydrogen from NaBH4. (iv) X is a mixture of two optical isomers. Draw the two optical isomers in the boxes provided. [2] (v) Both optical isomers of X can be dehydrated to form a single product, Y. Give the reagent(s) and conditions required for reaction 5. . …… [1] (vi) Y can form an addition polymer Z. Draw one repeat unit of Z. [1] (vii) Reaction 6 does not proceed quickly at room temperature. Suggest why this is the case. . …… . …… [1] [Total: 17]

Mark scheme

3(a) (2,3-)dimethylbut-2-ene 1 3(b) 1 OR OR OR 3(c)(i) 1 3(c)(ii) sp2 1 3(d)(i) cold dilute acidified potassium manganate(VII) 1 3(d)(ii) M1 (2,4-DNPH will produce a) red / orange / yellow precipitate 2 M2 V has a carbonyl group 3(d)(iii) M1 C—O in range 1040–1300 (cm–1) 2 M2 C=O in range 1670–1740 (cm–1) 3(e)(i) yellow precipitate 1 3(e)(ii) Iodine / I2 1 3(e)(iii) C6H12O + 2[H] → C6H14O 1 3(e)(iv) Draw two optical isomers of X. 1 eg 1 OR M1 One correct 3-dimensional representation of X M2 3-d structure which represents the enantiomer of their X 3(e)(v) heat AND concentrated H2SO4 / concentrated H3PO4 / concentrated sulfuric [(VI)] acid / concentrated phosphoric[(V)] acid 1 3(e)(vi) 1 OR 3(e)(vii) high activation energy 1

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Where can I find the 9701/22 October/November 2021 mark scheme?

On this page, free: the mark scheme for AS & A Level Chemistry 9701/22, October/November 2021 (9701_w21_ms_22), and the question paper (9701_w21_qp_22), to read or download, with each question's marking points written out underneath.

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Chemistry 9701/22 for October/November 2021 is out of 60 marks, across 3 questions.

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