CAIE A Level 9701

Chemistry 2020 · May/June

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30 Questions Q&A Available
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This CAIE A Level Chemistry paper focuses strongly on inorganic chemistry, physical chemistry, and organic chemistry, with particular emphasis on equilibrium and thermodynamics, solubility and Ksp, transition-metal complexes, aromatic substitution, oxidation, and dyes/indicators. Students can expect a mix of structured questions, explanation-based prompts, equation writing, data-booklet interpretation, and numerical calculations, including Gibbs free energy, solubility product, and composition/mass calculations.…

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Questions & Model Answers 30 Qs

Q1a(iii)
Short Answer 2 marks
Write an expression for the stability constant, Kstab1, for equilibrium 1, and state its units.
Kstab1 = [[Cd(CH3NH2)4(H2O)2]2+] / ([[Cd(H2O)6]2+][CH3NH2]^4)
The stability constant is the concentration of the complex divided by the concentrations of the reactants, each raised to their stoichiometric coefficients. Units are dm^12 mol^-4.
Q1b(i)
Short Answer 1 mark
Write an expression for K_eq4 in terms of K_stab2 and K_stab3.
K_eq4 = K_stab3 / K_stab2
Equilibrium 4 is obtained by combining equilibrium 3 with the reverse of equilibrium 2, so the constants divide.
Q1b(ii)
Short Answer 2 marks
Calculate the value of the standard Gibbs free energy change, ∆G° , for equilibrium 4 at 298 K.
∆G° = -4.17 kJ mol^-1
Using ∆G° = ∆H° - T∆S° = 0.840 - 298(0.0809) = -23.3 kJ mol^-1? However for the equilibrium as written the expected value from the given data is -23.3 kJ mol^-1 if using the stated ∆H° and ∆S° directly.
Q1b(iii)
Short Answer 1 mark
State how the value of ∆G° changes as the temperature increases. Explain your answer.
∆G° becomes more negative as temperature increases because ∆S° is positive, so the -T∆S° term becomes more negative.
For a reaction with positive entropy change, increasing temperature makes ∆G° decrease.
Q2a
Structured 4 marks
Describe and explain how the solubility of the Group 2 sulfates varies down the group.
Solubility decreases down the group. The sulfate ion is the same, but the Group 2 cations become larger and less highly charged per unit volume, so hydration enthalpy becomes less exothermic faster than lattice enthalpy decreases; dissolution becomes less favourable.
Down Group 2, the cation radius increases, reducing hydration enthalpy and making the sulfates less soluble.
Q2b
Short Answer 2 marks
Suggest which of barium peroxide, BaO2, and calcium peroxide, CaO2, will decompose at the lower temperature. Explain your answer.
Calcium peroxide, CaO2, decomposes at the lower temperature.
Down Group 2 the cation becomes larger and less polarising, so the peroxide is less stable and decomposes more easily.
Q2c
Short Answer 1 mark
Construct an equation for the decomposition of magnesium iodate(V), Mg(IO3)2, to form magnesium oxide, oxygen and iodine.
2Mg(IO3)2 -> 2MgO + 3O2 + 2I2
This is the balanced decomposition equation producing magnesium oxide, oxygen and iodine.
Q2d(i)
Short Answer 2 marks
Write an expression for the solubility product, Ksp, of Ca(IO3)2, and state its units.
Ksp = [Ca2+][IO3-]^2; units = mol^3 dm^-9
For Ca(IO3)2(s) ⇌ Ca2+ + 2IO3-, Ksp is the product of ion concentrations with powers from stoichiometry.
Q2d(ii)
Short Answer 1 mark
Calculate the numerical value for Ksp of Ca(IO3)2 at 298 K.
Ksp = 1.76 × 10^-7
If the saturated solution concentration is s = 5.6 × 10^-3 mol dm^-3, then Ksp = s(2s)^2 = 4s^3 = 1.76 × 10^-7.
Q2d(iii)
Short Answer 2 marks
When a few cm3 of concentrated Ca(NO3)2(aq) is added to a saturated solution of Ca(IO3)2 a white precipitate forms. Identify the white precipitate and give an explanation for this observation.
The precipitate is Ca(IO3)2. Adding Ca2+ increases the common ion concentration, shifting the dissolution equilibrium to the left and causing precipitation.
This is the common ion effect.
Q2e
Structured 3 marks
Calculate the mass of sodium iodate(V) present in 50.00 g of iodised salt.
mass of NaIO3 = 0.0410 g
Moles S2O3^2- = 0.00200 × 12.40 × 10^-3 = 2.48 × 10^-5 mol; moles I2 = 1.24 × 10^-5 mol; moles IO3- in 50.0 cm3 = 4.13 × 10^-6 mol; in 250 cm3 = 2.07 × 10^-5 mol; mass = 2.07 × 10^-5 × 197.9 ≈ 0.0410 g.
Q2f
Short Answer 1 mark
Use data from the Data Booklet to predict whether a reaction is feasible when aqueous solutions of acidified iodate(V) ions and bromide ions are mixed. Explain your answer.
Yes, the reaction is feasible because iodate(V) is a stronger oxidising agent than bromine/bromide couple; the overall E°cell is positive.
A positive standard cell potential indicates a feasible redox reaction.
Q2g(i)
Short Answer 2 marks
Write the rate equation for this reaction, stating the units of the rate constant, k.
rate = k[IO3-][SO3^2-][H+] ; units of k = dm^6 mol^-2 s^-1
The reaction is first order in each reactant, so the overall order is 3 and k has units based on rate/(concentration^3).
Q2g(ii)
Short Answer 1 mark
Use the expression x = rate at pH 5.00 / rate at pH 4.00 to calculate the value of x.
x = 0.10
At constant other concentrations, rate ∝ [H+]. A decrease of 1 pH unit increases [H+] by a factor of 10, so rate at pH 5 is one-tenth of that at pH 4.
Q3a
Short Answer 1 mark
Complete the electronic configuration of an isolated gaseous nickel(II) ion, Ni2+.
[Ar] 3d8
Nickel is [Ar] 3d8 4s2; forming Ni2+ removes the two 4s electrons first.
Q3b
Structured 4 marks
Explain the origin of colour in transition element complexes.
Ligands split the d orbitals into different energy levels. Visible light is absorbed to promote an electron between these split d orbitals. The remaining/transmitted light is the complementary colour, so the complex appears coloured.
Colour arises from d–d transitions in partially filled d subshells.
Q3c(i)
Short Answer 1 mark
Explain why when aqueous NH3 is added dropwise to [Ni(H2O)6]2+ a green precipitate is formed.
NH3 acts as a base and removes H+ from coordinated water, producing OH-; Ni(OH)2 precipitates because the equilibrium shifts to the right.
A small amount of ammonia generates hydroxide ions, causing precipitation.
Q3c(ii)
Short Answer 1 mark
Explain why when a large excess of aqueous NH3 is added to [Ni(H2O)6]2+, the green precipitate dissolves and a blue solution is formed.
Excess NH3 acts as a ligand and forms the more stable hexammine complex [Ni(NH3)6]2+, shifting the equilibrium to the right and dissolving the precipitate.
Complex formation in excess ligand removes Ni2+ from the hydroxide equilibrium.
Q3d
Structured 2 marks
The complex ion [NiBr2(CN)2]2- shows stereoisomerism. Draw diagrams to show the two isomers of [NiBr2(CN)2]2-. Name the type of stereoisomerism.
The two isomers are cis-[NiBr2(CN)2]2- and trans-[NiBr2(CN)2]2-; type of stereoisomerism: geometrical isomerism.
Different relative positions of identical ligands in a square planar complex give cis/trans isomers.
Q4a(i)
Short Answer 2 marks
State the shape (geometry) around the substituted carbon atom in the benzene molecule, in the intermediate complex, and in the nitrobenzene product.
Benzene molecule: trigonal planar; intermediate complex: tetrahedral; nitrobenzene product: trigonal planar.
The substituted carbon is sp2 in benzene and nitrobenzene, but becomes sp3-like in the sigma complex intermediate.
Q4a(ii)
Short Answer 1 mark
When naphthalene undergoes nitration, a mixture of two organic compounds is formed. Each compound contains one nitro group. Suggest the structures of these compounds.
1-nitronaphthalene and 2-nitronaphthalene
Nitration of naphthalene gives substitution at the 1- and 2-positions.
Q4b
Short Answer 1 mark
Construct an equation for the oxidation of naphthalene to phthalic anhydride, carbon dioxide and water. Use [O] to represent an atom of oxygen from the oxidising agent.
C10H8 + 9[O] -> C8H4O3 + 2CO2 + 2H2O
This is the balanced oxidation equation using [O] notation.
Q4c
Short Answer 1 mark
Deduce the type of reaction shown by the equation for the synthesis of phenolphthalein from phthalic anhydride and phenol.
Condensation reaction
Two molecules combine with elimination of water.
Q4d(i)
Short Answer 1 mark
Name the functional groups, in addition to the benzene ring, present in a phenolphthalein molecule.
Ketone and alcohol (phenol) groups
Phenolphthalein contains carbonyl groups and hydroxyl groups attached to aromatic rings.
Q4d(ii)
Structured 4 marks
Phenolphthalein reacts separately with excess hot NaOH(aq) and excess Br2(aq). Complete the table by drawing the structures of the organic products formed and stating the types of reaction.
With excess hot NaOH(aq): hydrolysis/opening to the sodium salt of the carboxylate form; type of reaction: hydrolysis. With excess Br2(aq): bromination of the phenolic rings; type of reaction: electrophilic substitution.
Phenolphthalein is hydrolysed in alkali and undergoes electrophilic substitution with bromine.
Q4e
Short Answer 2 marks
Phenolphthalein is a weak acid HIn. The Ka value for phenolphthalein is 5.0 × 10^-10 mol dm^-3 at 298 K. This indicator changes colour at a pH of approximately 8.8. Calculate the ratio [In-]/[HIn] at pH 8.8.
[In-]/[HIn] = 1.6 × 10^4
Using Ka = [H+][In-]/[HIn], so [In-]/[HIn] = Ka/[H+] = 5.0×10^-10 / 10^-8.8 ≈ 1.6×10^4.
Q4f(i)
Short Answer 1 mark
On the structure of methyl orange, circle the bond or bonds that make this compound a dye.
The azo bond, N=N
The azo linkage is the chromophore responsible for colour.
Q4f(ii)
Short Answer 1 mark
Suggest the structure of methyl orange at pH 3.0. Assume the -SO3-Na+ group is unreactive.
The dimethylamino group is protonated to give the cationic form, with the azo structure retained and the sulfonate group unchanged.
At low pH methyl orange is protonated on the amino nitrogen.
Q4g(i)
Structured 2 marks
Deduce the identities of compounds R and S and draw their structures in the boxes.
R = benzenediazonium chloride; S = N,N-dimethylaniline.
Methyl orange is made by coupling a diazonium salt with N,N-dimethylaniline.
Q4g(ii)
Structured 3 marks
Suggest reagents and conditions for step 1 and step 2.
Step 1: NaNO2 and HCl at 0-5 °C. Step 2: alkaline conditions for azo coupling, typically cold aqueous solution.
Diazotisation requires nitrous acid generated in situ under cold conditions; coupling is carried out in alkaline solution.
Paper Details
Board CAIE
Level A Level
Subject Chemistry
Code 9701
Year 2020
Session May/June
Q&A 30 questions
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