9701 Chemistry November 2025 Question Paper 44

1 (a) Define a transition element [1]

(b) The 3d orbitals in an isolated gaseous Cu2+ ion are degenerate.

(i) Define the term degenerate [1]

(ii) Complete the electronic configuration of Cu2+. 1s2 [1]

(c) (i) State the colours of the aqueous solutions for the two copper(II) complex ions shown. • [Cu(NH3)4(H2O)2]2+(aq) • [CuCl 4]2–(aq) [1]

(ii) Explain why aqueous complex ions of transition elements are usually coloured [3]

(d) (i) When an excess of NH3(aq) is added to a solution of [CuCl 4]2–(aq), [Cu(NH3)4(H2O)2]2+(aq) is formed.

State the type of reaction.

Complete the equation for this reaction. State symbols are not required. type of reaction equation [CuCl 4]2– + [Cu(NH3)4(H2O)2]2+ + [2] , ,

(ii) The [Cu(NH3)4(H2O)2]2+ complex ion shows stereoisomerism.

Complete the three‑dimensional diagrams in Fig. 1.1 to show the two different stereoisomers of [Cu(NH3)4(H2O)2]2+. Cu Cu isomer 1 isomer 2

Fig. 1.1

[2]

(iii) Deduce which stereoisomer in (d)(ii) is polar.

Explain your answer. polar isomer explanation [1]

(e) The dianion P can act as a tridentate ligand. P N O O O O H – –

(i) Suggest how P can form three dative covalent bonds [1]

(ii) 2 moles of dianion P, C4H5NO4 2–, react with 1 mole of aqueous cobalt(III) ions, [Co(H2O)6]3+ to form 1 mole of complex ion Q.

Deduce the formula and charge of Q [1] , ,

(f) Table 1.1 shows values for the stability constants, Kstab, of some silver(I) complexes. Table 1.1 complex value of Kstab [Ag(CN)2]–(aq) 1.1 × 1018 [Ag(NH3)2]+(aq) 1.2 × 107 [Ag(S2O3)2]3–(aq) 2.9 × 1013

(i) Define the stability constant of a complex [1]

(ii) Use the information in Table 1.1 to identify the most stable silver(I) complex.

Explain your answer. most stable explanation [1] , ,

(g) Sodium sulfite, Na2SO3, is used as a food preservative.

A 3.75 g sample of impure Na2SO3 is dissolved in distilled water and made up to 250 cm3 in a volumetric flask.

10.0 cm3 of this solution requires 18.70 cm3 of acidified 0.0150 mol dm–3 MnO4 –(aq) to reach the end‑point.

The equation for the reaction is shown. 2MnO4 – + 5SO3 2– + 6H+ 2Mn2+ + 5SO4 2– + 3H2O

Calculate the percentage by mass of Na2SO3 in the sample.

percentage by mass of Na2SO3 = [3] [Total: 19] , ,

2 (a) The Group 2 sulfates and the Group 2 chromates show similar trends in solubility.

Suggest the trend in the solubility of the Group 2 chromates down the group.

Explain your answer [4]

(b) Silver(I) chromate, Ag2CrO4, is sparingly soluble in water.

(i) Write an ionic equation to show the equilibrium between solid Ag2CrO4 and its aqueous solution.

Include state symbols [1]

(ii) The value of the solubility product, Ksp, of Ag2CrO4 is 1.12 × 10–12 at 298 K.

Calculate the equilibrium concentration of Ag+, in mol dm–3, in a saturated solution of Ag2CrO4 at 298 K.

equilibrium concentration of Ag+ = mol dm–3 [3] , ,

(c) The hydrogenchromate ion, HCrO4 –, is a weak acid. The pKa of HCrO4 – is 6.49.

(i) Calculate the pH of a 0.0250 mol dm–3 HCrO4 – solution.

pH = [2]

(ii) HCrO4 – can show amphoteric behaviour.

State the formula of: • the conjugate acid of HCrO4 – • the conjugate base of HCrO4 – [1] , ,

(d) Table 2.1 shows some energy changes. Table 2.1 energy change value / kJ mol–1 first ionisation energy of silver +731 second ionisation energy of silver +2074 first ionisation energy of sulfur +1000 second ionisation energy of sulfur +2251 first electron affinity of sulfur –200 second electron affinity of sulfur +532 enthalpy change of atomisation of sulfur +279 enthalpy change of formation of silver(I) sulfide, Ag2S(s) –33 lattice energy of silver(I) sulfide, Ag2S(s) –2677

(i) Define the term first electron affinity [1]

(ii) Explain why the value for the second electron affinity of sulfur is positive [1]

(iii) Construct an equation for the lattice energy of Ag2S.

Include state symbols [1] , ,

(iv) Calculate the enthalpy change of atomisation, ΔHat, in kJ mol–1, of silver using relevant data from Table 2.1.

It may be helpful to draw a labelled Born–Haber cycle.

Show your working.

ΔHat of silver = kJ mol–1 [3]

(e) Suggest how the magnitude for the lattice energy of Ag2S(s) differs from the lattice energy of Cu2S(s).

Explain your answer [1] [Total: 18] , ,

3 (a) Define the term entropy [1]

(b) (i) Place one tick (3) in each row of Table 3.1 to show the sign of the entropy change, ΔS, for each process. Table 3.1 process ΔS is negative ΔS is positive steam condensing into water solid KCl dissolving in water

[1]

(ii) Chlorine trifluoride, Cl F3, decomposes on heating into its elements, as shown.

reaction 1 2Cl F3(g) Cl 2(g) + 3F2(g)

Standard entropies are shown in Table 3.2. Table 3.2 substance Cl F3(g) Cl 2(g) F2(g) S ⦵ / J K–1 mol–1 +281.6 +223.1 +203.0

Calculate the standard entropy change, ΔS ⦵, in J K–1 mol–1, for reaction 1.

ΔS ⦵ for reaction 1 = J K–1 mol–1 [2]

(c) Group 2 carbonates decompose on heating. The decomposition for one of the Group 2 carbonates, MCO3, is shown in reaction 2.

reaction 2 MCO3(s) MO(s) + CO2(g)

(i) Predict the sign of the entropy change, ΔS, for reaction 2.

Explain your answer [1] , ,

(ii) The Gibbs equation is shown. ΔG ⦵ = ΔH  ⦵ – TΔS ⦵

Fig. 3.1 shows values of the Gibbs free energy change, ΔG ⦵, in kJ mol–1, at different temperatures, T, in K, for reaction 2.

Assume ΔH  ⦵ and ΔS ⦵ values for this reaction remain constant over this temperature range. –100 –50 0 50 100 150 200 0 200 400 600 800 1000 1200 1400 1600 G ⦵ J mol–1 T / K Fig. 3.1

Use the gradient and intercept on the y‑axis in Fig. 3.1 and the Gibbs equation to determine: • ΔS ⦵, in J K–1 mol–1, for reaction 2 • the minimum temperature, T, in K, at which the reaction is feasible • ΔH  ⦵, in kJ mol–1, for reaction 2.

ΔS ⦵ for reaction 2 = J K–1 mol–1

minimum temperature, T = K

ΔH  ⦵ for reaction 2 = kJ mol–1

[4] [Total: 9] , ,

4 (a) Nitrogen monoxide, NO, reacts with hydrogen, as shown in reaction 3.

reaction 3 2NO + 2H2 N2 + 2H2O

(i) The rate equation for reaction 3 is shown. rate = k[H2][NO]2

Complete Table 4.1. Table 4.1 the order of reaction with respect to [H2] the order of reaction with respect to [NO] the overall order of the reaction

[1]

(ii) Predict how the initial rate for reaction 3 changes when the concentration of NO is halved [1]

(iii) Predict how the initial rate for reaction 3 changes when the concentrations of NO and H2 are both increased three times [1]

(iv) Suggest why reaction 3 is unlikely to proceed by a mechanism involving only a single step [1]

(v) Suggest equations for the three steps of the reaction mechanism for reaction 3.

Each step involves a reaction between two molecules.

step 1 step 2

• N2O + step 3 N2O + + [2]

(vi) Suggest the role of N2O in this mechanism.

Explain your reasoning [1] , ,

(b) Iodine, I2, reacts with thiosulfate ions, S2O3 2–, as shown in reaction 4.

reaction 4 2S2O3 2– + I2 S4O6 2– + 2I–

Reaction 4 is carried out in the presence of a large excess of I2. Under these conditions, the reaction is first order with respect to [S2O3 2–] and zero order with respect to [I2].

The half‑life, t1 2 , for reaction 4 is 720 s under certain conditions.

Calculate the value of the rate constant, k, for reaction 4. Include the units of k.

k = units [1]

(c) The reaction between iodide ions, I–(aq), and peroxydisulfate ions, S2O8 2–(aq), is catalysed by Co3+(aq). The mechanism is similar to the mechanism of this reaction when Fe3+(aq) is used as the catalyst.

(i) State the type of catalysis that occurs in this reaction.

Explain your reasoning [1]

(ii) Write two equations to show how Co3+(aq) catalyses this reaction. equation 1 equation 2 [2]

(iii) Suggest why this reaction is slow in the absence of Co3+(aq) [1] [Total: 12] , ,

5 (a) Describe and explain the shape of benzene.

In your answer, include: • the shape and bond angle in the ring • the hybridisation of the carbon atoms • how orbital overlap forms σ and π bonds between the carbon atoms in the ring [4]

(b) Fig. 5.1 shows two reactions of benzoic acid. O COOH COOH COOH reaction 5 reaction 6

Fig. 5.1

(i) Suggest reagents and conditions for reaction 5 and for reaction 6 in Fig. 5.1. reaction 5 reaction 6 [2]

(ii) State the type of reaction for reaction 5 in Fig. 5.1 [1] , ,

(c) In the electrophilic substitution of arenes, different substituents can direct to different ring positions.

(i) Describe the directing effect of the –CH2CH3 group.

Explain your answer [1]

(ii) The alkylation of arenes uses a mixture of CH3CH2Br and FeBr3 to generate the CH3CH2

• electrophile.

Write an equation for the formation of the CH3CH2

• electrophile [1]

(iii) Complete the mechanism in Fig. 5.2.

Include all relevant curly arrows and charges. Draw the structure of the organic intermediate. COOH COOH +CH2CH3 + organic intermediate

Fig. 5.2

[3]

(iv) Write an equation to show how FeBr3 is regenerated after the reaction in Fig. 5.2 [1] [Total: 13] , ,

6 (a) Compound Z is used in organic synthesis. Z O NH2 CN H2N

Complete Table 6.1 to show the number of sp, sp2 and sp3 hybridised carbon atoms present in one molecule of Z. Table 6.1 type of hybridisation sp sp2 sp3 number of carbon atoms

[1]

(b) Z can undergo different reactions, as shown in Fig. 6.1. Z O NH2 CN H2N excess HCl (aq) heat CH3COCl excess LiAl H4 reaction 7 reaction 8 reaction 9

Fig. 6.1

(i) Name the two types of reaction occurring in reaction 7 in Fig. 6.1 and [1]

(ii) Draw the structures of the organic products of reactions 7, 8 and 9 in Fig. 6.1. [4] , ,

(c) Compound Z is dissolved in D2O and analysed by carbon‑13 NMR and proton (1H) NMR spectroscopy.

(i) Predict the number of peaks in the carbon‑13 NMR spectrum of Z [1]

(ii) The proton (1H) NMR spectrum of Z in D2O gives three peaks for the proton environments, labelled a, b and c, as shown on Fig. 6.2. Z O NH2 CN H2N C CH CH2 CH2 a b c Fig. 6.2

Complete Table 6.2 for the proton (1H) NMR spectrum of Z in D2O. Table 6.2 proton environment a b c name of splitting pattern chemical shift range, δ / ppm

[2] Table 6.3 environment of proton example chemical shift range, δ / ppm alkane –CH3, –CH2–, >CH– 0.9–1.7 alkyl next to C=O CH3–C=O, –CH2–C=O,

CH–C=O 2.2–3.0 alkyl next to nitrile –CH2–CN 2.0–3.0 alkyl next to electronegative atom CH3–O, –CH2–O, –CH2–N 3.2– 4.0 attached to alkene =CHR 4.5–6.0 alkyl amine R–NH– 1.0–5.0 amide RCONHR 5.0–12.0 [Total: 9] , ,

7 Phenylmethanol and 4‑methylphenol are isomers. CH2OH OH phenylmethanol 4-methylphenol

(a) Complete Table 7.1 to show the relative acidities of benzoic acid (C6H5COOH), phenylmethanol, 4‑methylphenol and water.

Explain your answer. Table 7.1 name of compound most acidic least acidic explanation [4]

(b) 4‑methylphenol reacts readily with sodium.

Complete the equation for this reaction. OH

• [1] , ,

(c) Under certain conditions, ethane‑1,2‑diol, HOCH2CH2OH, reacts with propane‑1,3‑dioic acid, HOOCCH2COOH, to form different organic products, as shown in Fig. 7.1. O O HO HO OH OH + reaction 10 reaction 11 polymer Y X, C5H6O4 Fig. 7.1

(i) X does not react with Na metal.

Draw the structure of the organic product X, C5H6O4, shown in Fig. 7.1.

[1]

(ii) Reactions 10 and 11 in Fig. 7.1 are different types of reaction.

Name the type of reaction for reaction 10 and for reaction 11. reaction 10 reaction 11 [1]

(iii) Draw a section of polymer Y showing only one repeat unit.

The new functional group formed should be displayed.

[2] [Total: 9] , ,

8 (a) Describe the difference in reactivity between ethanoyl chloride and chlorobenzene with water.

Explain your answer [2]

(b) The structure of compound V is shown. V CN O O O N H O

(i) Name all the functional groups in V [2]

(ii) Deduce the number of possible optical isomers for V [1]

(iii) Suggest one reason, other than better biological activity and lower dosage required, why it is beneficial to synthesise a single optical isomer of V for use as a drug [1] , ,

(c) A sample of V is hydrolysed with an excess of hot aqueous alkali.

The products are isolated from the reaction mixture at pH 12.

Draw the structures of the two organic products of the complete alkaline hydrolysis of V in Fig. 8.1.

Fig. 8.1

[3] , ,

(d) A polypeptide formed from four amino acids, A, B, C and D, is completely hydrolysed and then analysed by gas–liquid chromatography.

The chromatogram produced is shown in Fig. 8.2. intensity time / minutes A B C D 28 58 13 42 Fig. 8.2

The number above each peak represents the area under the peak.

The area under each peak is proportional to the mass of the respective amino acid in the mixture.

(i) Calculate the percentage by mass of amino acid A in the original mixture.

percentage by mass of amino acid A = [1]

(ii) The retention time for amino acid D is the longest.

Explain why D has a longer retention time than the other amino acids A, B and C [1] [Total: 11] , , 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 = 6.02 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at 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) , , Group The Periodic Table of Elements 1 H hydrogen 1.0 2 He helium 4.0 1 2 13 14 15 16 17 18 3 4 5 6 7 8 9 10 11 12 3 Li lithium 6.9 4 Be beryllium 9.0 atomic number atomic symbol Key name relative atomic mass 11 Na sodium 23.0 12 Mg magnesium 24.3 19 K potassium 39.1 20 Ca calcium 40.1 37 Rb rubidium 85.5 38 Sr strontium 87.6 55 Cs caesium 132.9 56 Ba barium 137.3 87 Fr francium – 88 Ra radium – 5 B boron 10.8 13 Al aluminium 27.0 31 Ga gallium 69.7 49 In indium 114.8 81 Tl thallium 204.4 6 C carbon 12.0 14 Si silicon 28.1 32 Ge germanium 72.6 50 Sn tin 118.7 82 Pb lead 207.2 22 Ti titanium 47.9 40 Zr zirconium 91.2 72 Hf hafnium 178.5 104 Rf rutherfordium – 23 V vanadium 50.9 41 Nb niobium 92.9 73 Ta tantalum 180.9 105 Db dubnium – 24 Cr chromium 52.0 42 Mo molybdenum 95.9 74 W tungsten 183.8 106 Sg seaborgium – 25 Mn manganese 54.9 43 Tc technetium – 75 Re rhenium 186.2 107 Bh bohrium – 26 Fe iron 55.8 44 Ru ruthenium 101.1 76 Os osmium 190.2 108 Hs hassium – 27 Co cobalt 58.9 45 Rh rhodium 102.9 77 Ir iridium 192.2 109 Mt meitnerium – 28 Ni nickel 58.7 46 Pd palladium 106.4 78 Pt platinum 195.1 110 Ds darmstadtium – 29 Cu copper 63.5 47 Ag silver 107.9 79 Au gold 197.0 111 Rg roentgenium – 30 Zn zinc 65.4 48 Cd cadmium 112.4 80 Hg mercury 200.6 112 Cn copernicium – 114 Fl flerovium – 116 Lv livermorium – 7 N nitrogen 14.0 15 P phosphorus 31.0 33 As arsenic 74.9 51 Sb antimony 121.8 83 Bi bismuth 209.0 8 O oxygen 16.0 16 S sulfur 32.1 34 Se selenium 79.0 52 Te tellurium 127.6 84 Po polonium – 9 F fluorine 19.0 17 Cl chlorine 35.5 35 Br bromine 79.9 53 I iodine 126.9 85 At astatine – 10 Ne neon 20.2 18 Ar argon 39.9 36 Kr krypton 83.8 54 Xe xenon 131.3 86 Rn radon – 113 Nh nihonium – 115 Mc moscovium – 117 Ts tennessine – 118 Og oganesson – 21 Sc scandium 45.0 39 Y yttrium 88.9 57–71 lanthanoids 89–103 actinoids 57 La lanthanum 138.9 89 Ac lanthanoids actinoids actinium – 58 Ce cerium 140.1 90 Th thorium 232.0 59 Pr praseodymium 140.9 91 Pa protactinium 231.0 60 Nd neodymium 144.2 92 U uranium 238.0 61 Pm promethium – 93 Np neptunium – 62 Sm samarium 150.4 94 Pu plutonium – 63 Eu europium 152.0 95 Am americium – 64 Gd gadolinium 157.3 96 Cm curium – 65 Tb terbium 158.9 97 Bk berkelium – 66 Dy dysprosium 162.5 98 Cf californium – 67 Ho holmium 164.9 99 Es einsteinium – 68 Er erbium 167.3 100 Fm fermium – 69 Tm thulium 168.9 101 Md mendelevium – 70 Yb ytterbium 173.1 102 No nobelium – 71 Lu lutetium 175.0 103 Lr lawrencium – , ,