9701 Chemistry November 2025 Question Paper 42

1 Magnesium nitrate, Mg(NO3)2, and strontium nitrate, Sr(NO3)2, both decompose when heated to form the metal oxide and a mixture of gases.

(a) Write an equation for the thermal decomposition of Mg(NO3)2 [1]

(b) State which of Mg(NO3)2 or Sr(NO3)2 decomposes at a lower temperature.

Explain your answer. compound that decomposes at a lower temperature explanation [2]

(c) Magnesium oxide, MgO, and strontium oxide, SrO, both react with dilute sulfuric acid.

MgO forms a soluble salt, A.

SrO forms an insoluble salt, B.

(i) Identify the products formed when MgO reacts with dilute sulfuric acid [1]

(ii) Explain why A is more soluble than B [3]

[Total: 7] , ,

2 Ethanal, CH3CHO, reacts with nitrogen dioxide, NO2. The products of the first step of this reaction are a CH3C• =O radical and a molecule of nitrous acid, HNO2. CH3CHO + NO2 CH3C• =O + HNO2

(a) (i) Use two words to complete the sentence. This reaction involves of the single covalent bond between a hydrogen atom and a carbon atom in CH3CHO.

[1]

(ii) The hydrogen atom mentioned in (a)(i) forms a covalent bond with one of the oxygen atoms of an NO2 molecule. An NO2 molecule has a single, unpaired electron on the nitrogen atom. All electrons are paired in an HNO2 molecule.

Draw dot-and-cross diagrams of NO2 and HNO2 in the boxes. Show outer shell electrons only. NO2

HNO2

[1]

(iii) Use VSEPR theory to predict the bond angle at the nitrogen atom in an HNO2 molecule.

bond angle = [1]

(b) The rate equation for the reaction between CH3CHO and NO2 is shown. rate = k [CH3CHO][NO2]

Under certain conditions, when the concentrations of both CH3CHO and NO2 are 0.200 mol dm–3, the rate of the reaction is 1.53 × 10–4 mol dm–3 s–1.

Calculate the value of the rate constant, k, under these conditions. Give the units of k.

k = units [2] , ,

(c) The reaction mixture described in (b) is monitored over a period of time.

Predict whether the graph of [NO2] against time shows a constant half-life.

Explain your answer. prediction explanation [1]

(d) NO2 also reacts with ozone, O3.

The rate equation is shown. rate = k1[NO2]

Under certain conditions, the value of k1 is 0.0848 s–1.

The reaction has a constant half-life under these conditions.

Calculate the half-life in seconds. half-life = s [1]

(e) NO2 is present in the exhaust gases of cars. It can react with carbon monoxide, CO, on the surface of a heterogeneous catalyst in the car’s catalytic converter.

Describe the mode of action of this heterogeneous catalyst [2]

[Total: 9] , ,

3 (a) (i) Define conjugate acid–base pair [1]

(ii) Give the formulas of the conjugate acid and the conjugate base of the hydrogen phosphate ion, HPO4 2–.

conjugate acid of HPO4 2– conjugate base of HPO4 2– [1]

(b) The Ka of propanoic acid, CH3CH2COOH, is 1.35 × 10–5 mol dm–3 at 298 K.

Solution C is a solution of CH3CH2COOH with a pH of 3.60 at 298 K.

(i) Calculate the concentration of CH3CH2COOH in solution C.

[CH3CH2COOH] = mol dm–3 [2]

(ii) Calculate the concentration of hydroxide ions in solution C.

[OH–] = mol dm–3 [1]

(iii) Calculate the concentration of a solution of hydrochloric acid with the same pH as solution C.

concentration = mol dm–3 [1] , ,

(iv) Table 3.1 shows three possible values of the Ka of dimethylpropanoic acid, (CH3)3CCOOH.

Place a tick in Table 3.1 to show the correct value. Explain your answer. Table 3.1 value of Ka / mol dm–3 place one tick (✓) in this column 9.33 × 10–6 1.35 × 10–5 3.35 × 10–5 explanation [3]

(c) Solution D is made by mixing 100 cm3 of 0.100 mol dm–3 CH3CH2COOH and 100 cm3 of 0.100 mol dm–3 NaCl.

The pH of solution D is measured as small amounts of H2SO4(aq) are added to it, and when small amounts of NaOH(aq) are added to it.

Solution D only acts as a buffer solution when one of these solutions is added to it.

(i) Complete the sentence and write an equation for the reaction that occurs. Solution D acts as a buffer when is added to it. equation [1]

(ii) Complete the sentence and explain why solution D does not act as a buffer when the other solution is added. Solution D does not act as a buffer when is added to it. explanation [1] , ,

(d) Manganese(II) hydroxide, Mn(OH)2, is only slightly soluble in water.

The solubility of Mn(OH)2 in water is 3.28 × 10–3 g dm–3 at 298 K.

(i) Calculate the concentration of a saturated solution of Mn(OH)2 at 298 K.

[Mn(OH)2] = mol dm–3 [1]

(ii) Write an expression for the Ksp of Mn(OH)2. Give the units of Ksp.

Ksp = units = [2]

(iii) Use your answers to (d)(i) and (d)(ii) to calculate the value of Ksp of Mn(OH)2 at 298 K.

Ksp = [1]

[Total: 15] , ,

4 (a) Define enthalpy change of atomisation, ΔHat [1]

(b) Define first electron affinity, EA [1]

(c) Explain why the first electron affinity of chlorine is more exothermic than the first electron affinity of iodine [2]

(d) The enthalpy change for the reaction Cl 2(g) + 2e– 2Cl –(g) is – 486 kJ mol–1.

The first electron affinity of chlorine is –364 kJ mol–1.

Calculate the enthalpy change of atomisation of chlorine.

ΔHat of chlorine = kJ mol–1 [2]

[Total: 6] , ,

5 Cobalt is a transition element which forms compounds containing Co2+ and Co3+ ions. Cobalt(II) sulfate dissolves in water to form a solution containing the [Co(H2O)6]2+ complex ion.

(a) (i) Complete the electronic configurations of a Co2+ ion and a Co3+ ion.

Co2+ = [Ar] Co3+ = [Ar] [1]

(ii) Explain why transition elements can form complex ions [1]

(iii) An excess of concentrated HCl is added to a solution containing [Co(H2O)6]2+.

Describe the colour change observed and the state of the cobalt-containing product.

The colour changes from to The state of the cobalt-containing product is [2]

(iv) Write an equation for the reaction occurring in (a)(iii) [1]

(v) Name the type of reaction occurring in (a)(iii) [1]

(vi) Write an equation for the reaction that occurs when an excess of NaOH(aq) is added to a solution containing [Co(H2O)6]2+ [1] , ,

(b) Cobalt metal can be oxidised by acidified K2Cr2O7. The relevant half-equations, and their E o– values, are shown.

Co2+ + 2e– Co E o– = –0.28 V

Cr2O7 2– + 14H+ + 6e– 2Cr3+ + 7H2O E o– = +1.33 V

(i) A Co2+/Co electrode is constructed in which [Co2+] is 0.020 mol dm–3 at 298 K.

Use the Nernst equation to show that the E value for this Co2+ / Co electrode is –0.33 V.

[2]

(ii) An electrochemical cell is constructed using the Co2+ / Co electrode described in (b)(i) and a Cr2O7 2– / Cr3+ electrode in which all conditions are standard.

Calculate the value of Ecell.

Ecell = [1]

(iii) A current is drawn from the electrochemical cell described in (b)(ii).

Write an equation for the reaction taking place in the cell [1]

(iv) Complete the sentences to identify the negative electrode and the direction of electron flow when a current is drawn from the cell described in (b)(ii). The electrode is the negative electrode. Electrons flow from the electrode to the electrode.

[1]

(c) A molten Co2+ salt is electrolysed using a current of 0.500 A.

0.547 g of cobalt metal forms at the cathode. Under the conditions used no other reduction reaction occurs at the cathode.

Calculate the time in minutes for which the current flows to produce this mass of cobalt.

Give your answer to three significant figures.

time = min [3]

[Total: 15] , ,

6 (a) Nickel forms complexes.

(i) Give the formula and charge of the tetrahedral complex formed by Ni atoms with carbon monoxide molecules. Carbon monoxide is a monodentate ligand. This is complex E. E = [1]

(ii) Give the formula and charge of the octahedral complex formed by Ni2+ ions with ethanedioate ions. This is complex F. F = [2]

(iii) Identify which complex, E or F, exists as a mixture of two stereoisomers and the type of stereoisomerism involved. The complex which exists as a mixture of two stereoisomers is The type of stereoisomerism involved is [1]

(b) Cadmium forms complexes with methylamine, CH3NH2, and 1,2-diaminoethane, en. The values of the stability constants, Kstab, of these complex ions are given in Table 6.1. Table 6.1 complex Kstab [Cd(CH3NH2)4]2+

3.5 × 106 [Cd(en)2]2+

4.0 × 1010

(i) Explain, by reference to its structure, why CH3NH2 acts as a monodentate ligand [1]

(ii) Some Cd2+(aq) is added to a solution containing equal concentrations of CH3NH2 and en.

Predict which of the two complexes in Table 6.1 forms at the higher concentration.

Explain your answer. complex that forms at the higher concentration explanation [1]

(iii) Complete the expression for the Kstab of [Cd(CH3NH2)4]2+.

Kstab =

[1] [Total: 7] , ,

7 P, Q, R, S, T, U, and V are the seven structural isomers with molecular formula C5H10O that have a carbonyl group.

P CH3(CH2)3CHO Q CH3CH2CH(CH3)CHO R (CH3)2CHCH2CHO

S (CH3)3CCHO T CH3CH2CH2COCH3 U CH3CH2COCH2CH3

V (CH3)2CHCOCH3

(a) Only one of these seven compounds has stereoisomers.

Draw three-dimensional diagrams of the two stereoisomers of this compound.

[2]

(b) P, Q, R, S, T, U, and V are treated separately with alkaline I2(aq) and the product mixture is acidified.

(i) Identify the two compounds that give a positive result with alkaline I2(aq) and [1]

(ii) Describe the observations when one of the compounds you have identified in (b)(i) is treated with alkaline I2(aq) and give the structural formulae of the two carbon-containing products of this reaction. observations two carbon-containing products and [2] , ,

P CH3(CH2)3CHO Q CH3CH2CH(CH3)CHO R (CH3)2CHCH2CHO

S (CH3)3CCHO T CH3CH2CH2COCH3 U CH3CH2COCH2CH3

V (CH3)2CHCOCH3

(c) The proton (1H) NMR spectra of P, Q, R, S, T, U, and V are compared.

(i) Identify the only compound that gives a spectrum with two singlets and no other peaks [1]

Fig. 7.1 shows the spectrum obtained from one of the compounds. 11 10 9 8 7 6 5 4 3 2 1 0 δ / ppm Fig. 7.1

(ii) Identify the compound that gives this spectrum [1]

(iii) Name the splitting pattern of the peak at δ = 1.1 in Fig. 7.1.

Give the reason for this splitting. name reason [1]

(iv) Identify the substance that gives the small peak at δ = 0 in Fig. 7.1 [1] , ,

(d) The carbon-13 NMR spectra of R, S, T and U are compared.

Complete Table 7.1 to state the number of peaks in the spectrum of each compound. Table 7.1 compound number of peaks R (CH3)2CHCH2CHO S (CH3)3CCHO T CH3CH2CH2COCH3 U CH3CH2COCH2CH3

[2]

[Total: 11] , ,

8 Asparagine and aspartic acid are two naturally occurring amino acids. Their structures and isoelectric points are shown in Table 8.1. Table 8.1 amino acid structure isoelectric point asparagine HOOCCH(NH2)CH2CONH2 5.41 aspartic acid HOOCCH(NH2)CH2COOH 2.77

(a) Define isoelectric point [1]

(b) Draw the structures of asparagine and aspartic acid at pH 2. asparagine at pH 2

aspartic acid at pH 2

[2]

(c) Asparagine and aspartic acid are treated separately with an excess of LiAl H4.

Draw the structures of the organic products of these reactions. product of asparagine treated with an excess of LiAl H4

product of aspartic acid treated with an excess of LiAl H4 [2] , ,

(d) Propanedioic acid, HOOCCH2COOH, is treated with an excess of thionyl chloride, SOCl 2. Propanedioyl chloride, Cl OCCH2COCl , is formed.

(i) Write an equation for this reaction [1]

(ii) Propanedioyl chloride reacts with an excess of asparagine to form compound G with molecular formula C11H16N4O8.

Each molecule of compound G has four amide groups.

Draw the structure of compound G. Compound G, C11H16N4O8

[2]

(e) Asparagine is hydrolysed with an excess of hot NaOH(aq).

Draw the structure of the organic product of this reaction.

[2]

(f) A polymer can form from asparagine, HOOCCH(NH2)CH2CONH2, as the only monomer.

Draw a length of the polymer chain containing three monomer residues.

Clearly label the repeat unit of the polymer on your diagram.

[3] , ,

(g) Aspartic acid exists in two optically active forms.

(i) Plane polarised light is passed through pure samples of these two optically active forms in solutions of the same concentration.

Describe two similarities and one difference in their effect on the plane polarised light. similarities difference [2]

(ii) Give the term used to describe a mixture of equal amounts of the two optically active forms [1]

[Total: 16] , , Question 9 starts on page 20. , ,

9 Compound X is made from benzene by the route shown in Fig. 9.1. step 1 CH3 step 2 CH3 NO2 NO2 W NH2 X step 3 COOH step 4 COOH benzene Fig. 9.1

(a) Describe the bonding in benzene, C6H6.

Your answer should include: • the hybridisation of the six carbon atoms • the types of bond between the carbon atoms • the orbitals that overlap to produce the bonds between the carbon atoms • the type of bond between the carbon atoms and the hydrogen atoms • the orbitals that overlap to produce the bonds between the carbon atoms and the hydrogen atoms [3]

(b) Describe the reagents and conditions required for step 1 in Fig. 9.1 [1] , ,

(c) In step 1 of Fig. 9.1 benzene reacts with +CH3.

Complete Fig. 9.2 to show the mechanism for this reaction, including: • the movement of electron pairs using curly arrows • the structure of the intermediate involved. +CH3 CH3 intermediate + Fig. 9.2

[3]

(d) Describe the reagents and conditions required for step 2 of Fig. 9.1 [2]

(e) Identify the reagents required for step 3 of Fig. 9.1. Compound W is the product of this step [1]

(f) Name compound W [1]

(g) The reagents commonly used for step 4 will not reduce the –COOH group.

Identify the reagents and conditions required for step 4 of Fig. 9.1 [1] , ,

(h) Benzene can also be used as a starting material to make compound Y. COOH compound Y NH2

Describe how the route described in Fig. 9.1 (repeated below) can be changed to give compound Y instead of compound X.

Explain your answer [2] step 1 CH3 step 2 CH3 NO2 NO2 W NH2 X step 3 COOH step 4 COOH benzene Fig. 9.1

[Total: 14] , , 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 – , ,