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Chemistry 9701/22 October/November 2024: question paper and mark scheme

Cambridge AS & A Level Chemistry (9701), Paper 2 AS Structured Questions, October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9701_w24_qp_229701_w24_ms_229701/22/O/N/24

Paper
Paper 2 AS Structured Questions
Total marks
60
Time
1 hour 15 minutes
Questions
4
Code on the paper
9701/22/O/N/24
9701_w24_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 19 marks

Question

Vanadium, niobium and tantalum are metals in the same group of the Periodic Table. (a) The shorthand electronic configuration of vanadium in the ground state is [Ar]3d34s2. (i) State what is meant by the term ground state. …… [1] (ii) Show the electronic configuration of vanadium using electrons in boxes notation. [Ar] [1] (iii) Deduce the total number of electrons in the p sub-shells of a vanadium atom. …… [1] (b) Pelopium was the suggested name for a new element discovered in a mineral. Pelopium was later found to be a mixture of niobium, Nb, and tantalum, Ta. Only one naturally occurring isotope exists for each of Nb and Ta. (i) Complete Table 1.1. Table 1.1 relative isotopic number of number of isotope mass protons neutrons 93 92.91 41Nb 181 180.95 73Ta [2] (ii) Define relative isotopic mass. …… [2] (iii) A sample of pelopium contains 90.9% by mass 93 181 41Nb and 9.1% by mass 73Ta. Calculate the theoretical relative atomic mass of pelopium based on these data and Table 1.1. Give your answer to two decimal places. Show your working. theoretical relative atomic mass of pelopium = …… [2] [Total: 9]

Mark scheme

1(a)(i) lowest energy state owtte 1 examples • (with) lowest energy • (with) no external energy • (with) no added energy • (where) electrons are not promoted to a higher energy level • most energetically stable 1(a)(ii) ↑ ↑ ↑ ↑↓ 1 [Ar] The 3 × d electrons may go in any of the d orbitals 1(a)(iii) 12 1 1(b)(i) isotope RIM # protons # neutrons 2 93 41Nb 92.91 ● 41 ● 52 181 73Ta 180.95 ● 73 ● 108 1(b)(ii) M1 mass of an (atom of an) isotope 1 M2 relative/compared to (the mass of) the unified atomic mass unit 1 OR on a scale in which a carbon-12 atom / isotope has a mass of exactly 12 units OR 1 divided by / compared to of the mass of a carbon-12 atom / isotope 12 Alternative route using mass of 1 mol throughout M1 mass of one mol of an (atom of an) isotope 1 M2 relative/compared to of the mass of 1 mol of C-12 atom / isotope 12 OR when the mass of one mol C-12 atom / isotope is exactly 12(.000) g 1(b)(iii) 92.91× 0.909 + 180.95 × 0.091 1 = 100.92 1

Question 213 marks

Question

Oxygen is a Group 16 element. (a) (i) Write equations for the following reactions. • sodium and oxygen • sulfur and oxygen [2] (ii) Draw a dot-and-cross diagram to show the species present in Al2O3. Draw outer electrons only. [1] (iii) The maximum oxidation state of the Period 3 elements in their oxides varies across the period. State and explain the variation. (b) H2O reacts with both inorganic and organic compounds. (i) Complete Table 2.1 to give details of the reactions of some Period 3 oxides with H2O. Table 2.1 product of reaction Period 3 oxide pH of solution formed with H2O Mg(OH)2 P4O10 [2] (ii) Write an equation for the reaction of CH3CN with H2O in acidic conditions. CH3CN + ……H2O + ……H+ …… [1] (iii) Draw the structures of the two alcohols formed in the reaction shown in equation 1. H3PO4 equation 1 C3H6(g) + H2O(g) C3H8O(g) [2] (iv) Explain why alcohols are less acidic than water. …… [2] (c) Fig. 2.1 shows the boiling points of H2O and other Group 16 hydrides. 390 340 290 boiling point /K 240 190 140 H 2O H 2S H2Se H2Te Fig. 2.1 (i) Explain the trend in the boiling points of the Group 16 hydrides H2S to H2Te. …… [2] (ii) Explain why the boiling point of H2O is much higher than that of H2S. …… [1]

Mark scheme

2(a)(i) M1 4Na + O2 → 2Na2O 1 M2 S + O2 → SO2 1 2(a)(ii) 1 2(a)(iii) it increases 1 as number of valence electrons (which can be used in bonding / lost / shared / donated) increases 1 2(b)(i) 2 Period 3 oxide product of rxn with water pH of solution formed ● MgO Mg(OH)2 ● 8 ⩽ pH ⩽ 12 P4O10 ● H3PO4 ● 1 ⩽ pH ⩽ 4 ●✓●✓ 2(b)(ii) CH3CN + 2H2O + (1)H+ → CH3COOH + NH4+ 1 2(b)(iii) 2 2(b)(iv) alkyl groups are electron donating / have positive inductive effect. 1 strengthens the O—H bond / makes H+ less likely to be donated 1 2(c)(i) (molecules) H2S to H2Te / they have greater number of electrons 1 stronger instantaneous dipole–induced dipole / London dispersion forces 1 OR more energy required to overcome the instantaneous dipole–induced dipole / London dispersion forces 2(c)(ii) (only) H2O has hydrogen bonding AND hydrogen bonding (much) stronger than the other intermolecular forces 1 OR H2O has hydrogen bonding AND because of higher electronegativity of O compared to S OR stronger van der waals’ forces because (only) H2O has hydrogen bonding OR hydrogen bond AND increased strength of permanent dipoles in H2O outweighs the increase in strength of id-id in the others

Question 319 marks

Question

Nitrogen and phosphorus are elements in Group 15 of the Periodic Table. (a) Nitrogen is found in inorganic compounds such as nitrogen oxides (NOx), nitrates and nitric acid. (i) Identify one natural and one man-made occurrence of nitrogen oxides in the atmosphere. natural …… man-made …… [2] (ii) Write an equation to describe the role of NO2 in the direct formation of acid rain. …… [1] (iii) Peroxyacetyl nitrate, PAN, is a component of photochemical smog. Describe how PAN forms from NO2. …… [1] (iv) Nitric acid reacts with basic oxides to form nitrates. Write an equation for the reaction of nitric acid with calcium oxide. …… [1] (v) Describe what is seen when solid calcium nitrate is heated strongly. …… [1] (b) A common test for nitrates is the reaction with NaOH and Al. Equation 1 shows the reaction. equation 1 3NO3– + 8Al + 5OH– + 18H2O 3NH3 + 8[Al(OH)4]– (i) Deduce the oxidation state of nitrogen in NO3–. …… [1] (ii) Identify the species that is oxidised in equation 1. …… [1] (iii) NH3 is a basic gas. Describe how NH3 is able to act as a base. …… [1] (iv) Suggest the shape of the [Al(OH)4]– ion. …… [1] (c) Fig. 3.1 shows a sketch of some of the ionisation energies of phosphorus, P. ionisation energy 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th ionisation Fig. 3.1 (i) Construct an equation to represent the third ionisation energy of P. …… [1] (ii) Complete the graph in Fig. 3.1 to show the third to sixth ionisation energies of P. [2] (d) Complete Table 3.1 to show the properties of nitrogen and phosphorus in their standard states. Table 3.1 nitrogen phosphorus state and appearance colourless gas white solid of standard state electrical conductivity poor type of bonding type of structure simple [2] (e) A form of solid nitrogen has a lattice structure similar to solid iodine. Identify the type of lattice structure of solid nitrogen. …… [1] (f) At very high temperatures, phosphorus can form P2 molecules. P2 contains a triple bond, P P. (i) Describe the formation of the P P bond in terms of orbital overlap. …… [2] (ii) The bond energy of P P is 485 kJ mol–1. The bond energy of N N is 944 kJ mol–1. Compare the reactivity of P2 and N2. Explain your answer. …… [1] [Total: 19]

Mark scheme

3(a)(i) (natural = )lightning 1 (man-made = )internal combustion engines 1 3(a)(ii) 2NO2 + H2O → HNO2 + HNO3 OR 4NO2 + O2 + 2H2O → 4HNO3 1 3(a)(iii) It / NO2 reacts with (unburned) hydrocarbons / VOCs 1 ALLOW reaction of unburned hydrocarbons / VOCs in presence of NO2 3(a)(iv) 2HNO3 + CaO → Ca(NO3)2 + H2O 1 3(a)(v) brown fumes given off 1 3(b)(i) (+)5 / V 1 3(b)(ii) aluminium / Al 1 3(b)(iii) (NH3 is an) H+ acceptor 1 3(b)(iv) tetrahedral 1 3(c)(i) P2+(g) → P3+(g) + e(–) 1 3(c)(ii) general increase 1 increase in IE between IE5 and IE6 is noticeably the largest increase between IEs shown 1 3(d) 2 element nitrogen phosphorus state and appearance colourless gas white solid electrical conductivity ● poor poor type of bonding ● covalent ● covalent type of structure simple ● simple 3(e) simple molecular (lattice structure) 1 3(f)(i) M1 one sigma / σ bond AND head-on (overlap of) p / sp (orbitals) 1 M2 two pi / π bond(s) AND side-on (overlap of / involving) p (orbitals) 1 3(f)(ii) P≡P is much weaker so P2 is more reactive (than N2) 1

Question 417 marks

Question

Bromoalkanes are used widely in industry, although there is increasing concern about their environmental impact. Fig. 4.1 shows a reaction scheme involving 1,2-dibromoethane. A Br2 Br NaOH(aq) H2C=CH2 Br reaction 1 reaction 2 NaOH in reaction 3 ethanol B Br polymer C reaction 4 name …… an excess of reaction 5 NaOH in ethanol HBr, H2O and D Fig. 4.1 (a) Complete Fig. 4.2 to show the mechanism for the formation of 1,2-dibromoethane in reaction 1. Include charges, dipoles, lone pairs of electrons and curly arrows as appropriate. H 2C = Br CH2 Br Br Br Fig. 4.2 [3] (b) The enthalpy change of reaction 1, ΔHr = –90.0 kJ mol–1. Br2 H2C=CH2 Br reaction 1 Br The enthalpy change of formation of ethene, ΔHf = +52.2 kJ mol–1. Calculate the enthalpy change of formation of 1,2-dibromoethane. ΔHf of 1,2-dibromoethane = …… kJ mol–1 [1] (c) (i) Complete Fig. 4.1 to: • draw the structure of compound A • name compound B. [2] (ii) Draw the structure of one repeat unit of polymer C in the box. one repeat unit of polymer C [1] (iii) In reaction 5, compound B reacts with an excess of NaOH dissolved in ethanol. The products are HBr, H2O and an unsaturated hydrocarbon D. Suggest the identity of D. …… [1] (d) Compound E is the only isomer of 1,2-dibromoethane. Alkaline hydrolysis of E gives compound F. E F Br O Br H (i) Identify the type of isomerism shown by E and 1,2-dibromoethane. …… [1] (ii) Name the homologous series that F belongs to. …… [1] (iii) Complete Table 4.1 to state what is observed when F reacts with the reagents listed. Table 4.1 reagent observation with F 2,4-dinitrophenylhydrazine (2,4-DNPH reagent) Tollens’ reagent alkaline I2(aq) [3] Question 4 continues on page 14. (e) Compound F reacts with reagent G to form compound H. F O G H H The infrared spectrum of H is shown in Fig. 4.3. 100 transmittance / % 50 0 4000 3000 2000 1500 1000 500 wavenumber / cm–1 Fig. 4.3 Table 4.2 characteristic infrared absorption range bond functional groups containing the bond (in wavenumbers) / cm–1 C–O hydroxy, ester 1040–1300 C=C aromatic compound, alkene 1500–1680 amide 1640–1690 C=O carbonyl, carboxyl 1670–1740 ester 1710–1750 C N nitrile 2200–2250 C–H alkane 2850–2950 N–H amine, amide 3300–3500 carboxyl 2500–3000 O–H hydroxy 3200–3600 H also shows a molecular ion peak at m/e = 60 in its mass spectrum. (i) Use the information in (e), Fig. 4.3 and Table 4.2 to deduce the structure of H. Explain your answer fully. H [3] (ii) Suggest the role of reagent G. [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 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 16 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

Mark scheme

4(a) 3 pt 1 dipole on Br2 pt 2 curly arrow from C=C bond to a Brδ+ pt 3 curly arrow from Br–Br bond to the other Br pt 4 correct intermediate pt 5 curly arrow from lone pair on Br – to C+ 5• = 3 marks. 4• = 2 marks. 3• = 2 marks. 2• = 1 marks. 1• = 0 marks. 4(b) = –37.8 (kJ mol–1) 1 4(c)(i) 1 bromoethene 1 4(c)(ii) 1 OR 4(c)(iii) C2H2 1 4(d)(i) structural / positional 1 4(d)(ii) aldehyde 1 4(d)(iii) 3 reagent observation with F 2,4-DNPH red / orange / yellow ppt Tollens’ reagent silver mirror OR grey ppt OR black ppt alkaline I2(aq) yellow ppt 4(e)(i) M1 identify H = M2 (broad) absorption within the range 3600–2500 cm–1 so O-H (bond) O-H (bond) is equivalent to OH bond M3 pt 1 and pt 2 •✓ OR pt 1 and pt 3 •✓ •pt 1 absorption within the range 1670–1750 cm–1 so C=O (bond) •pt 2 absorption within the range 1040–1300 cm–1 so C-O (bond) •pt 3 (M+ at m / e = 60 so it has) molecular mass / Mr = 60 4(e)(ii) oxidising agent 1

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Where can I find the 9701/22 October/November 2024 mark scheme?

On this page, free: the mark scheme for AS & A Level Chemistry 9701/22, October/November 2024 (9701_w24_ms_22), and the question paper (9701_w24_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 2024 is out of 60 marks, with 1 hour 15 minutes to answer it, across 4 questions.

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