Chemistry 2020 May/June | Past Paper Questions & Answers

Q1a(iii).

Short Answer 2 marks

Write an expression for the stability constant, Kstab1, for equilibrium 1, and state its units.

Model Answer:

Kstab1 = [[Cd(CH3NH2)4(H2O)2]2+] / ([[Cd(H2O)6]2+][CH3NH2]^4)

Explanation: 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.

Model Answer:

K_eq4 = K_stab3 / K_stab2

Explanation: 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.

Model Answer:

∆G° = -4.17 kJ mol^-1

Explanation: 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.

Model Answer:

∆G° becomes more negative as temperature increases because ∆S° is positive, so the -T∆S° term becomes more negative.

Explanation: 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.

Model Answer:

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.

Explanation: 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.

Model Answer:

Calcium peroxide, CaO2, decomposes at the lower temperature.

Explanation: 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.

Model Answer:

2Mg(IO3)2 -> 2MgO + 3O2 + 2I2

Explanation: 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.

Model Answer:

Ksp = [Ca2+][IO3-]^2; units = mol^3 dm^-9

Explanation: 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.

Model Answer:

Ksp = 1.76 × 10^-7

Explanation: 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.

Model Answer:

The precipitate is Ca(IO3)2. Adding Ca2+ increases the common ion concentration, shifting the dissolution equilibrium to the left and causing precipitation.

Explanation: 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.

Model Answer:

mass of NaIO3 = 0.0410 g

Explanation: 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.

Model Answer:

Yes, the reaction is feasible because iodate(V) is a stronger oxidising agent than bromine/bromide couple; the overall E°cell is positive.

Explanation: 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.

Model Answer:

rate = k[IO3-][SO3^2-][H+] ; units of k = dm^6 mol^-2 s^-1

Explanation: 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.

Model Answer:

x = 0.10

Explanation: 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+.

Model Answer:

[Ar] 3d8

Explanation: 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.

Model Answer:

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.

Explanation: 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.

Model Answer:

NH3 acts as a base and removes H+ from coordinated water, producing OH-; Ni(OH)2 precipitates because the equilibrium shifts to the right.

Explanation: 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.

Model Answer:

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.

Explanation: 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.

Model Answer:

The two isomers are cis-[NiBr2(CN)2]2- and trans-[NiBr2(CN)2]2-; type of stereoisomerism: geometrical isomerism.

Explanation: 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.

Model Answer:

Benzene molecule: trigonal planar; intermediate complex: tetrahedral; nitrobenzene product: trigonal planar.

Explanation: 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.

Model Answer:

1-nitronaphthalene and 2-nitronaphthalene

Explanation: 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.

Model Answer:

C10H8 + 9[O] -> C8H4O3 + 2CO2 + 2H2O

Explanation: 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.

Model Answer:

Condensation reaction

Explanation: 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.

Model Answer:

Ketone and alcohol (phenol) groups

Explanation: 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.

Model Answer:

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.

Explanation: 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.

Model Answer:

[In-]/[HIn] = 1.6 × 10^4

Explanation: 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.

Model Answer:

The azo bond, N=N

Explanation: 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.

Model Answer:

The dimethylamino group is protonated to give the cationic form, with the azo structure retained and the sulfonate group unchanged.

Explanation: 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.

Model Answer:

R = benzenediazonium chloride; S = N,N-dimethylaniline.

Explanation: 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.

Model Answer:

Step 1: NaNO2 and HCl at 0-5 °C. Step 2: alkaline conditions for azo coupling, typically cold aqueous solution.

Explanation: Diazotisation requires nitrous acid generated in situ under cold conditions; coupling is carried out in alkaline solution.

CAIE Chemistry (9701)

Questions & Answers

CAIE Chemistry `(9701)`