9701 Chemistry November 2025 Question Paper 43

1 (a) Solutions of Group 2 hydrogencarbonates, M(HCO3)2, decompose on heating to give the corresponding metal carbonate, carbon dioxide and water.

(i) Write an equation for the decomposition of strontium hydrogencarbonate, Sr(HCO3)2 [1]

(ii) The thermal stability of Group 2 carbonates increases down the group.

Explain this trend [2]

(b) The hydroxides and fluorides of Group 2 elements show similar trends in solubility.

Describe the trend in the solubility of the fluorides of calcium, strontium and barium.

Explain your answer least soluble

most soluble explanation [4] , ,

(c) (i) Define enthalpy change of hydration, ΔHhyd [1]

(ii) State the main factors that affect the magnitude of enthalpy change of hydration.

Explain your answer [2]

(d) Table 1.1 shows various energy changes. Table 1.1 energy change value / kJ mol–1 lattice energy of MgF2 –2957 enthalpy change of hydration, ΔHhyd, of Mg2+ –1926 enthalpy change of hydration, ΔHhyd, of F– –505

Use data from Table 1.1 to calculate the enthalpy change of solution, ΔHsol, for MgF2(s).

It may be helpful to draw a labelled energy cycle. Show your working.

ΔHsol of MgF2(s) = kJ mol–1 [2] , ,

(e) Mercury(I) fluoride, Hg2F2, is sparingly soluble in water.

The cation in Hg2F2 exists as the diatomic ion Hg2 2+ with a covalent Hg–Hg bond.

(i) Write the expression for the solubility product, Ksp, of Hg2F2. Include the units.

Ksp =

units [2]

(ii) The solubility of Hg2F2 is 9.20 × 10–3 mol dm–3 at 298 K.

Calculate the value of Ksp of Hg2F2 at 298 K.

Ksp = [1]

[Total: 15] , ,

2 (a) Iron can form stable ions in the +2 and +3 oxidation states.

Explain why transition elements have variable oxidation states [1]

(b) Aqueous solutions of iron(II) salts contain the complex ion [Fe(H2O)6]2+.

Define complex ion [1]

(c) [Fe(H2O)6]2+ can be converted into [Fe(H2O)4(OH)2].

(i) Suggest a suitable reagent for this conversion. State the type of reaction. reagent type of reaction [1]

(ii) [Fe(H2O)4(OH)2] is a green precipitate that turns brown on standing in air.

Table 2.1 shows electrode potentials for some electrode reactions. Table 2.1 electrode reaction E o / V Fe(H2O)3(OH)3 + H2O + e– Fe(H2O)4(OH)2 + OH– –0.56 O2 + 2H2O + 4e– 4OH– +0.40

Use the information in Table 2.1 to explain why [Fe(H2O)4(OH)2] turns brown on standing in air.

Include an equation for this reaction [3] , ,

(d) The complex [Co(NH3)6]2+ reacts with hydrogen peroxide as shown.

reaction 1 2[Co(NH3)6]2+ + H2O2 2[Co(NH3)6]3+ + 2OH– E cell o = +1.67 V

Calculate ΔG o , in kJ mol–1, for reaction 1.

ΔG o = kJ mol–1 [2]

[Total: 8] , ,

3 (a) Solid manganese(IV) oxide, MnO2, catalyses the decomposition of hydrogen peroxide. 2H2O2(aq) 2H2O(l) + O2(g)

State the type of catalysis for this reaction. Explain your answer [1]

(b) Hydrogen peroxide reacts with iodide ions in acidic conditions as shown. H2O2 + 2I– + 2H+ 2H2O + I2

The initial rate of this reaction is investigated with different concentrations of H2O2, I– and H+. The results obtained are shown in Table 3.1. Table 3.1 experiment [H2O2] / mol dm–3 [I–] / mol dm–3 [H+] / mol dm–3 initial rate / mol dm–3 s–1 1 0.0450 0.0300 0.0125 2.42 × 10–3 2 0.0225 0.0600 0.0125 2.42 × 10–3 3 0.0225 0.120 0.0125 4.84 × 10–3 4 0.0450 0.120 0.0500 9.68 × 10–3

(i) Use the information in Table 3.1 to deduce the rate equation for this reaction.

Explain your reasoning [4]

(ii) Use your rate equation from (b)(i) and the data from Experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k.

k = units [2] , ,

(c) The rate of the thermal decomposition of azomethane, CH3N=NCH3, is investigated. CH3N=NCH3 N2 + C2H6

Fig. 3.1 shows the results obtained. The reaction is first order with respect to CH3N=NCH3. 0.06 0.05 0.04 0.03 0.02 0.01 0 0 50 100 150 200 250 300 350 400 [CH3N=NCH3] / mol dm–3 time / s Fig. 3.1

(i) Use Fig. 3.1 to calculate two half-lives, t 2 1, to show that the reaction is first order [2]

(ii) Use your answer to (c)(i) to calculate the rate constant, k, for the decomposition of azomethane.

k = s–1 [1]

(d) Describe the effect of increasing temperature on the rate constant and on the rate of a reaction [1]

[Total: 11] , ,

4 (a) Define standard cell potential, E cell o . Include a description of standard conditions [2]

(b) The Daniell cell is an electrochemical cell consisting of a Cu2+(aq) / Cu(s) electrode and a Zn2+(aq) / Zn(s) electrode.

(i) Draw a labelled diagram of this electrochemical cell.

Include all necessary substances and relevant pieces of apparatus needed to measure the E cell o .

It is not necessary to state the conditions used.

[3]

(ii) State the charge carriers that transfer current through the solutions and through the wire.

the solutions the wire [1]

(iii) The standard electrode potential, E o , for the Zn2+(aq) / Zn(s) electrode is –0.76 V.

Water is added to a standard Zn2+(aq) / Zn(s) electrode.

The new concentration of Zn2+(aq) is 0.25 mol dm–3.

Use the Nernst equation to calculate the electrode potential, E, for this new Zn2+(aq) / Zn(s) electrode.

E (Zn2+(aq) / Zn(s)) = V [2]

(c) An electrochemical cell consists of a ZnO / Zn electrode and a MnO2 / Mn2O3 electrode in an alkaline electrolyte.

The standard cell potential, E cell o , for this cell is +1.47 V.

The half-equation at each electrode when this cell is discharging is shown. Zn + 2OH– ZnO + H2O + 2e– 2MnO2 + H2O + 2e– Mn2O3 + 2OH–

(i) Use this information to determine the change in oxidation state of manganese when this cell is discharging.

from to [1]

(ii) Write the equation for the overall reaction that occurs when this cell is discharging [1]

(iii) The E o for the ZnO / Zn electrode is –1.28 V.

Calculate the standard electrode potential, E o, for the MnO2 / Mn2O3 electrode.

E o (MnO2 / Mn2O3) = V [1]

[Total: 11] , ,

5 (a) Copper shows typical properties of transition elements, including its behaviour as a catalyst.

Complete Table 5.1 to show the total number of unpaired electrons in the 3d and 4s orbitals of an isolated gaseous Cu atom and a Cu2+ ion. Table 5.1 species number of unpaired electrons 3d 4s Cu Cu2+

[1]

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

Complete the diagram to show the relative energies of the 3d orbitals in an isolated Cu2+ ion and in Cu2+ in a tetrahedral complex. energy isolated Cu2+ ion Cu2+ in a tetrahedral complex

[2]

(c) Explain why transition elements behave as catalysts [2] , ,

(d) CN– is a monodentate ligand.

Table 5.2 shows information about two complex ions that contain only CN– ions as ligands.

Complete Table 5.2. Table 5.2 metal ion coordination number formula of complex ion charge of complex ion Ag+ 2 Fe2+ 4–

[2]

(e) The complex ion [Au(CN)2Br2]– displays geometrical (cis / trans) isomerism.

Draw the structure of trans-[Au(CN)2Br2]–. State its shape and the Br-Au-Br bond angle.

shape Br-Au-Br bond angle = [2] , ,

(f) An impure sample of a vanadium(V) compound of mass 0.250 g is dissolved in aqueous acid. This solution contains VO3 – ions.

An excess of zinc is added to this solution. All the VO3 – ions are reduced to V2+ ions and Zn atoms are oxidised to Zn2+ ions.

The unreacted zinc is removed and the resulting solution is titrated with acidified MnO4 –.

The end-point is reached when 22.5 cm3 of 0.0750 mol dm–3 MnO4 – is added.

A redox reaction takes place and all the V2+ reacts forming VO3 –. 3MnO4 – + 5V2+ + 3H2O 3Mn2+ + 5VO3 – + 6H+

(i) Calculate the percentage by mass of vanadium in the 0.250 g of impure sample.

Assume the impurities do not contain any vanadium ions.

Show your working.

percentage of vanadium = [3]

(ii) Complete the equation for the reaction between acidified VO3 – ions and Zn metal VO3 – + Zn + V2+ + Zn2+ + [2]

[Total: 14] , ,

6 (a) Thin-layer and gas / liquid chromatography can be used to separate mixtures into their individual components.

(i) Define the following terms used in chromatography. Rf value retention time [2]

(ii) Each type of chromatography makes use of a stationary phase and a mobile phase.

Complete Table 6.1 with a description of each of these. Table 6.1 stationary phase mobile phase thin-layer chromatography gas / liquid chromatography

[1]

(b) A mixture of two substances A and B is analysed by thin-layer chromatography.

The Rf value of substance A is larger than that of substance B.

Suggest why substance A has a larger Rf value [1] , ,

(c) The two isomeric compounds Y and Z are analysed by proton (1H) NMR spectroscopy. O O O O Y Z

(i) Complete Table 6.2 to predict the number of peaks observed in the proton (1H) NMR spectra for Y and Z. Table 6.2 compound number of peaks observed Y Z

[1]

(ii) Name all the different splitting patterns observed in the proton (1H) NMR spectra for Y and Z. Y Z [2]

[Total: 7] , ,

7 (a) State the relative acidities of bromoethanoic acid, BrCH2COOH, chloroethanoic acid, Cl CH2COOH, ethanoic acid, CH3COOH and ethanol, CH3CH2OH.

Explain your answer most acidic least acidic [4]

(b) Fig. 7.1 shows the reaction of methylbenzene and ethanedioic acid with KMnO4.

Predict the major carbon-containing product for each of these reactions. hot alkaline KMnO4 hot acidified KMnO4 HO O O OH methylbenzene ethanedioic acid Fig. 7.1

[2] , ,

(c) Polyamide X can be synthesised from ethanedioic acid and benzene-1,4-diamine. benzene-1,4-diamine H2N NH2

(i) Draw the repeat unit of polyamide X in the box.

The new functional group formed should be shown displayed. polyamide X

[2]

(ii) Benzene-1,4-diamine can be formed by reduction of 1,4-dinitrobenzene.

Complete the equation for this reduction.

[H] represents one atom of hydrogen from a reducing agent. H2N

• [H] NH2 + [1] , ,

(d) Fig. 7.2 shows the two-step synthesis of the azo compound W. H2N NH2 step 1 step 2 V C6H4N4Cl2 W C18H14N4O2 OH and NaOH(aq) Fig. 7.2

(i) Suggest structures for compounds V and W and draw them in the boxes in Fig. 7.2. [2]

(ii) Give the reagents and conditions for step 1 [1]

[Total: 12] , ,

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

(i) Describe the directing effect of the –NO2 group. Explain your answer [1]

(ii) The nitration of arenes uses a mixture of concentrated HNO3 and concentrated H2SO4 to generate the NO2

• electrophile.

Write an equation for the formation of the NO2

• electrophile [1]

(b) Carbon-carbon bond formation is an important reaction in organic synthesis.

Fig. 8.1 shows the synthesis of compound Q from benzene in two reaction steps. reaction 1 reaction 2 compound P compound Q O Fig. 8.1

(i) Draw the structure of compound P in the box in Fig. 8.1. [1]

(ii) Suggest reagents and conditions for reactions 1 and 2 in Fig. 8.1. reaction 1 reaction 2 [2] , ,

(c) Separate samples of C6H5Br and C6H5CH2Br are added to warm AgNO3(aq).

State the expected observations, if any. Explain your answer.

C6H5Br with AgNO3(aq) C6H5CH2Br with AgNO3(aq) explanation [3]

(d) Acyl bromides, RCOBr, react readily with H2O.

The mechanism of this reaction is similar to that of the reaction of H2O with acyl chlorides, RCOCl.

(i) Name the mechanism of this reaction [1]

(ii) Complete the mechanism in Fig. 8.2 for the reaction of RCOBr with H2O.

Include all relevant lone pairs of electrons, curly arrows, charges and dipoles.

Draw the structure of the intermediate. intermediate products O C R Br O H H Fig. 8.2 [4]

[Total: 13] , ,

9 (a) Explain why amides are much weaker bases than amines [2]

(b) Fig. 9.1 shows the preparation of 2-phenylethylamine, C6H5CH2CH2NH2, by three different routes. H2/Ni M N O NH2 NH2 reaction 1 2-phenylethylamine NH3 Fig. 9.1

(i) Suggest structures for compounds M and N and draw them in the boxes in Fig. 9.1. [2]

(ii) Give the reagents and conditions for reaction 1 [1] , ,

(c) Fig. 9.2 shows compound H which is a useful starting material in organic synthesis. O N H H Fig. 9.2

H contains an alkene and an amine functional group.

Name the other functional group and give the classification of the amine group in H.

other functional group in H classification of amine [1]

(d) Ozonolysis involves the oxidative cleavage of a C=C bond in alkenes using ozone, O3, as shown in Fig. 9.3. R1 R2 R3 O3 R2 R1 R3 R4 R4 O O + Fig. 9.3

Fig. 9.4 shows the first step in this reaction which involves the formation of an ozonide intermediate. R1 R2 R3 R4 R2 R1 R4 R3 O O O O O O +

ozonide Fig. 9.4

(i) On Fig. 9.4, draw three curly arrows to complete the mechanism of this step. [2] , ,

(ii) L is formed from alkene K, C8H14, by a similar reaction to that shown in Fig. 9.3. O O L

Suggest the structure of K. K C8H14

[1]

[Total: 9] , , 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 – , ,