Free until 14 Nov · 80% off for the first 100 · Pricing

Chemistry 0620/42 October/November 2023: question paper and mark scheme

Cambridge IGCSE Chemistry (0620), Paper 4 Theory (Extended), October/November 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

0620_w23_qp_420620_w23_ms_420620/42/O/N/23

Paper
Paper 4 Theory (Extended)
Total marks
80
Questions
6
Code on the paper
0620/42/O/N/23
0620_w23_ms_42.pdf

0620/42 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 18 marks

Question

Table 1.1 gives the electronic configurations of some atoms and ions, A to G. Table 1.1 electronic configuration A 2,5 B 2,8 C 2,8,2 D 2,8,4 E 2,8,5 F 2,8,6 G 2,8,18,7 Answer the following questions about A to G. Each letter may be used once, more than once or not at all. State which of the atoms or ions, A to G, could be: (a) a noble gas atom . …… [1] (b) an atom of an element in Group VI . …… [1] (c) an atom with an atomic number of 14 . …… [1] (d) atoms from the same group …… and . …… [1] (e) a halogen atom . …… [1] (f) an atom of an element which is a good conductor of electricity . …… [1] (g) a stable ion of a Group V element . …… [1] (h) an atom that forms an ion with a 2– charge. . …… [1] [Total: 8]

Mark scheme

1(a) B 1 1(b) F 1 1(c) D 1 1(d) A AND E 1 1(e) G 1 1(f) C 1 1(g) B 1 1(h) F 1

Question 214 marks

Question

Cobalt and copper are transition elements. (a) Copper has two naturally occurring isotopes, 63Cu and 65Cu. Cobalt has only one naturally occurring isotope, 59Co. (i) Complete Table 2.1 to show the number of protons, neutrons and electrons in the 59Co atom and the 65Cu2+ ion. Table 2.1 59 Co 65 Cu2+ protons neutrons electrons [3] (ii) Table 2.2 shows the relative abundance of the two naturally occurring isotopes of copper. Table 2.2 isotope 63 Cu 65 Cu relative abundance 70% 30% Calculate the relative atomic mass of copper to one decimal place. relative atomic mass = …… [2] (b) One physical property of transition elements such as copper and cobalt is that they are hard. Other metals such as lithium are softer. State two other physical properties of copper and cobalt which are significantly different from lithium. 1 . …… 2 . …… [2] (c) Both copper and cobalt can form coloured compounds. Some of these compounds contain water of crystallisation. (i) Define the term water of crystallisation. . …… . …… [2] (ii) State the colour and formula of hydrated cobalt(II) chloride crystals. colour . …… formula . …… [2] (iii) State the colour change seen when a few drops of water are added to anhydrous copper(II) sulfate. (iv) State how this colour change can be reversed. . …… [1] [Total: 14]

Mark scheme

2(a)(i) 59 65 3 Co Cu2+ protons 27 29 neutrons 32 36 electrons 27 27 one mark for each correct row 2(a)(ii) M1 (63 × 70) + (65 × 30) = 6360 (1) 2 M2 6360 / 100 = 63.6 (1) 2(b) M1 (high) density (1) 2 M2 (high) melting point (1) 2(c)(i) M1 water molecules (1) 2 M2 (the water present) in hydrated crystals (1) 2(c)(ii) M1 pink (1) 2 M2 CoCl2∙6H2O (1) 2(c)(iii) M1 white (1) 2 M2 blue (1) 2(c)(iv) heating (the hydrated copper(II) sulfate) 1

Question 317 marks

Question

Iron is manufactured in a blast furnace. (a) Three of the starting materials added to the blast furnace are coke, iron ore and limestone. Name the other starting material added to the blast furnace. . …… [1] (b) The source of iron in the blast furnace is Fe2O3. Fe2O3 is found in iron ore. (i) Name the main ore of iron which contains Fe2O3. . …… [1] (ii) The iron in Fe2O3 is reduced by reaction with carbon monoxide. The unbalanced symbol equation is shown. Complete the equation. Fe2O3 + ……CO → ……CO2 + ……Fe [1] (iii) State the change in oxidation number of iron in the reaction in (ii). (iv) Explain how the change of oxidation number shows that iron has been reduced. . …… [1] (c) The major impurity in iron ore is silicon(IV) oxide. Limestone is added to the blast furnace to remove this impurity. Write two symbol equations to show how silicon(IV) oxide is removed. For each equation, state the type of chemical reaction that takes place. equation 1 . …… type of chemical reaction . …… equation 2 . …… type of chemical reaction . …… [4] (d) Iron is converted to steel by mixing it with carbon and other elements. (i) State the term given to a substance which is a mixture of a metal and other elements. . …… [1] (ii) Name one element, other than carbon, mixed with iron in the making of stainless steel. . …… [1] (e) Preventing the rusting of steel is important. State the chemical name of rust. . …… [1] (f) Steel can be coated with zinc to prevent rusting. This provides both a barrier method and sacrificial protection. (i) State the term used for coating steel with zinc. . …… [1] (ii) Describe another barrier method for preventing rusting. . …… [1] (iii) Explain how zinc provides sacrificial protection. . …… . …… [2] [Total: 17]

Mark scheme

3(a) (hot) air 1 3(b)(i) hematite 1 3(b)(ii) Fe2O3 + 3CO → 3CO2 + 2Fe 1 3(b)(iii) M1 from +3 (1) 2 M2 to 0 (1) 3(b)(iv) decrease (in oxidation number) 1 3(c) M1 CaCO3 → CaO + CO2 (1) 4 M2 thermal decomposition (1) M3 SiO2 + CaO → CaSiO3 (1) M4 acid-base reaction (1) 3(d)(i) alloy 1 3(d)(ii) nickel / chromium 1 3(e) hydrated iron(III) oxide 1 3(f)(i) galvanising 1 3(f)(ii) painting / greasing / coating with plastic 1 3(f)(iii) M1 zinc more reactive (than iron) (1) 2 M2 zinc corrodes / oxidises / reacts in preference to / instead of iron (1)

Question 414 marks

Question

This question is about lead(II) chloride, PbCl 2. (a) A student prepares a sample of insoluble lead(II) chloride, PbCl 2, by mixing aqueous solutions of two salts in a beaker. (i) Identify two soluble salts suitable for making lead(II) chloride when mixed together. . …… . …… [2] (ii) Write the ionic equation for the formation of lead(II) chloride by mixing aqueous solutions. Include state symbols. . …… [3] (iii) List the steps the student should take in preparing a pure sample of lead(II) chloride from the mixture in the beaker. . …… . …… . …… [3] (b) The student carries out an electrolysis experiment on molten lead(II) chloride using the apparatus shown in Fig. 4.1. Chlorine gas forms at the anode and escapes from the apparatus. power supply molten lead(II) chloride Fig. 4.1 (i) Explain why lead(II) chloride needs to be molten before it will conduct electricity. . …… . …… [1] (ii) Write the ionic half-equation for the reaction occurring at the anode. . …… [2] (iii) State the test for chlorine gas. test . …… observations …… [2] (iv) Describe what is observed at the cathode. . …… [1] [Total: 14]

Mark scheme

4(a)(i) M1 lead nitrate (1) 2 M2 soluble chloride e.g. sodium chloride (1) 4(a)(ii) Pb2+(aq) + 2Cl–(aq) → PbCl2(s) 3 M1 PbCl2 as only product (1) M2 Pb2+ + 2Cl– as only reactants (1) M3 state symbols (1) 4(a)(iii) M1 filter (1) 3 M2 wash (the residue) using water (1) M3 dry the residue between filter papers / in a warm place (1) 4(b)(i) mobile ions 1 4(b)(ii) 2Cl– → Cl2 + 2e– 2 M1 any negative Cl species losing electron(s) (1) M2 correct ionic half equation (1) 4(b)(iii) M1 (damp) litmus (paper) (1) 2 M2 is bleached / goes white (1) 4biv (shiny) grey AND solid 1

Question 516 marks

Question

Chemical reactions can involve transfer of thermal energy. (a) State the term used for the transfer of thermal energy during a reaction. . …… [1] (b) Tetrachloromethane gas, CCl 4(g), reacts with steam as shown. CCl 4(g) + 2H2O(g) CO2(g) + 4HCl (g) The reaction is reversible. The forward reaction is exothermic. (i) State what happens, if anything, to the rate of the forward reaction if the concentration of CCl 4 is increased. Explain your answer in terms of collision theory. . …… . …… . …… . …… [3] (ii) State what happens to the position of equilibrium, if anything, when the pressure is increased. Explain your answer. . …… . …… [2] (iii) Fig. 5.1 shows an incomplete reaction pathway diagram for the forward reaction. CCl 4(g) + 2H2O(g) CO2(g) + 4HCl (g) energy progress of reaction Fig. 5.1 On Fig. 5.1: ● insert the formulae of the reactants and products ● draw an arrow, labelled Ea, to show the activation energy ● draw an arrow, labelled ∆H, to show the transfer of energy in the reaction. [3] (iv) Define the term activation energy. . …… . …… [2] (v) State one way in which the activation energy of a reaction can be changed. . …… [1] (c) The equation for the reaction between tetrachloromethane gas and steam can be represented as shown in Fig. 5.2. Cl H Cl H O H H Cl Cl C Cl + O C O + H = –130 kJ / mol H O H H Cl H Cl Cl Fig. 5.2 Table 5.1 shows some bond energies. Table 5.1 bond C–Cl H–O C=O bond energy in kJ / mol 340 460 805 Use the bond energies in Table 5.1 and the ∆H value for the reaction to calculate the H–Cl bond energy using the following steps. ● Calculate the energy needed to break the bonds in the reactants. …… kJ ● Calculate the energy released when the bonds in carbon dioxide form. …… kJ ● Calculate the H–Cl bond energy. …… kJ / mol [4] [Total: 16]

Mark scheme

5(a) enthalpy change (of reaction) 1 5(b)(i) M1 rate of reaction increases (1) 3 M2 more (CCl4) particles per unit volume (1) M3 frequency of collisions between (CCl4 and H2O) particles increases (1) 5(b)(ii) M1 (position of) equilibrium moves to left hand side (1) 2 M2 more (gaseous) moles / molecules on right hand side (1) 5(b)(iii) M1 Labels mark 3 CCl4(g) + 2H2O(g) on reactant line AND CO2(g) + 4HCl(g) on product line M2 Activation E mark upward arrow labelled Ea from energy level of reactants to top of ‘hump’ M3 Energy change mark one downward arrow labelled ΔH AND energy change starting from E level of reactants and finishing at E level of products 5(b)(iv) M1 minimum energy (1) 2 M2 that colliding particles must have to react (1) 5(b)(v) (decreased by adding) catalyst 1 5(c) energy needed to break bonds in reactants 4 M1 [(4 × 340) + (4 × 460) = 3200 (kJ / mol) (1) energy released when bond in carbon dioxide form M2 2 × 805 = 1610 (kJ / mol) (1) calculate H–Cl bond energy M3 3200 – (1610 + E(4H–Cl)) = – 130 E(4H–Cl) = 3330 – 1610 = 1720 (kJ / mol) (1) M4 1720 / 4 = 430 (kJ / mol) (1)

Question 611 marks

Question

A homologous series is a family of organic compounds whose members have similar chemical properties. (a) Give two characteristics that are the same for all members of a homologous series. 1 . …… 2 . …… [2] (b) In terms of structure, state how one member of a homologous series differs from the next member of that homologous series. . …… [1] (c) A, B and C are organic compounds. A has the molecular formula C12H24. B has the name tetradecane. C has three carbon atoms and is in the homologous series with the general formula CnH2n+1COOH. (i) Name the homologous series each organic compound belongs to. A …… B …… C …… [3] (ii) Name C and draw its displayed formula. name . …… displayed formula [2] (d) Amino acids are a homologous series where each member has the general structure shown in Fig. 6.1. The R side chain contains carbon and hydrogen atoms only. H R O N C C H H O H Fig. 6.1 (i) An amino acid has a relative molecular mass of 103. Deduce the formula of the R side chain in this amino acid. Show your working. . …… [2] (ii) State the name given to the natural polyamides formed from amino acid monomers. . …… [1] [Total: 11] The Periodic Table of Elements Group I II III IV V VI VII VIII 1 2 H He hydrogen helium Key 1 4 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 7 9 relative atomic mass 11 12 14 16 19 20 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 24 27 28 31 32 35.5 40 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 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 40 45 48 51 52 55 56 59 59 64 65 70 73 75 79 80 84 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 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 16 85 88 89 91 93 96 – 101 103 106 108 112 115 119 122 128 127 131 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 133 137 178 181 184 186 190 192 195 197 201 204 207 209 – – – 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 139 140 141 144 – 150 152 157 159 163 165 167 169 173 175 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 – 232 231 238 – – – – – – – – – – – The volume of one mole of any gas is 24 dm3 at room temperature and pressure (r.t.p.).

Mark scheme

6(a) M1 (same) general formula (1) 2 M2 (contain the same) functional group (1) 6(b) -CH2- unit 1 M1 alkenes (1) 3 6(c)(i) M2 alkanes (1) M3 carboxylic acids (1) 6(c)(ii) M1 propanoic acid (1) 2 M2 displayed formula of propanoic acid (1) 6(d)(i) M1 working to give the mass of atoms shown in Fig 6.1 2 103 – 14 – (2 × 12) – (2 × 16) – (4 × 1) (= 29) M2 ‘R’ = C2H5 6(d)(ii) proteins 1

Checked your answers against these? Let AI mark the whole paper, on your own working.

Mark my 0620/42

How this paper is marked

Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.

Check your answers against it

Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.

0620/42 in other series

Find another Chemistry paper

Every IGCSE Chemistry mark scheme

Question papers and mark schemes © Cambridge University Press & Assessment. CIE Buddy is independent and not connected to Cambridge.

0620/42 questions

Where can I find the 0620/42 October/November 2023 mark scheme?

On this page, free: the mark scheme for IGCSE Chemistry 0620/42, October/November 2023 (0620_w23_ms_42), and the question paper (0620_w23_qp_42), to read or download, with each question's marking points written out underneath.

How many marks is 0620/42 out of?

Chemistry 0620/42 for October/November 2023 is out of 80 marks, across 6 questions.

Can I get my 0620/42 answers marked?

Yes. Upload your answers to 0620/42/O/N/23 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.