Unit 3: Inorganic 2 & Physical 2 Exam Practice

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Section B: Structured Written Questions

Answer all questions in the spaces provided or write answers on paper, then check against the official mark scheme.

The Born-Haber cycle in Table 1 represents the energetic pathways for the formation of solid calcium oxide, CaO(s).

Step	Process	Enthalpy Change / kJ mol^-1
Atomisation of Ca	Ca(s) → Ca(g)	+193
First Ionisation of Ca	Ca(g) → Ca⁺(g) + e^-	+590
Second Ionisation of Ca	Ca⁺(g) → X + e^-	+1150
Atomisation of Oxygen	½ O₂(g) → O(g)	+248
First Electron Affinity of O	O(g) + e^- → O^-(g)	-142
Second Electron Affinity of O	O^-(g) + e^- → O^2-(g)	+844
Lattice Dissociation Enthalpy	CaO(s) → Ca²⁺(g) + O^2-(g)	+3513
Standard Formation	Ca(s) + ½ O₂(g) → CaO(s)	To be calculated

01.1 State the chemical formula, including state symbols, of:
• Species X formed during the second ionisation of calcium.
• The electron affinity product species Y that reacts in the second electron affinity step to form O^2-(g). [2 marks]

01.2 Define the term lattice dissociation enthalpy. [2 marks]

01.3 Calculate the standard enthalpy of formation, Δ_fH°, of calcium oxide, CaO(s), in kJ mol^-1 using the data in Table 1. Show all your working. [3 marks]

01.4 The experimental lattice dissociation enthalpy of silver chloride, AgCl, is +905 kJ mol^-1. The value calculated using a theoretical perfect ionic model is +833 kJ mol^-1.
• Explain why there is a significant difference between these two values.
• Identify the type of bonding character present in silver chloride that is not accounted for in the perfect ionic model. [3 marks]

01.5 Magnesium chloride, MgCl₂, dissolves in water. The process can be investigated experimentally. A student dissolved a sample of 1.57 g of anhydrous magnesium chloride (MgCl₂, M_r = 95.3) in 25.0 g of water. The temperature of the water increased by 14.2 °C.
• Calculate the heat energy released, q, in Joules. Assume the specific heat capacity of water is 4.18 J g^-1 K^-1.
• Calculate the experimental standard enthalpy of solution, Δ_solH°, of magnesium chloride in kJ mol^-1. Show all steps in your calculation. Include the correct mathematical sign. [5 marks]

Show Mark Scheme

01.1

M1: Species X = Ca²⁺(g) [1] (state symbol (g) is required).
M2: Species Y = e^- (or electron) [1].

01.2

M1: Enthalpy change when one mole of solid ionic lattice/compound [1].
M2: Is dissociated completely into gaseous ions [1].

01.3

M1: Working using Hess cycle: Δ_fH°(CaO) = ΔH_at(Ca) + IE1(Ca) + IE2(Ca) + ΔH_at(O) + EA1(O) + EA2(O) - ΔH_LattDiss [1].
M2: Δ_fH° = 193 + 590 + 1150 + 248 + (-142) + 844 - 3513 = 2883 - 3513 [1].
M3: Δ_fH° = -630 kJ mol^-1 [1] (negative sign is essential; allow ECF for minor arithmetic slips; correct answer with no working scores 3 marks).

01.4

M1: The theoretical perfect ionic model assumes ions are perfect, unpolarised spheres with point charges and purely electrostatic attractions [1].
M2: The Ag⁺ cation polarises the large, electron-rich chloride (Cl^-) anion [1].
M3: Introducing covalent character / partial covalent bonding which strengthens the lattice, making the experimental value more endothermic than theoretical [1].

01.5

M1: q = mcΔT = 25.0 × 4.18 × 14.2 = 1483.9 J = 1.484 kJ [1].
M2: Moles of MgCl₂ = mass / M_r = 1.57 / 95.3 = 0.01647 mol [1].
M3: Energy per mole = q / n = 1.4839 / 0.01647 = 90.1 kJ mol^-1 [1].
M4: Exothermic reaction because temperature increased, so sign must be negative: -90.1 kJ mol^-1 (accept -89.9 to -90.2 kJ mol^-1) [1].
M5: Fully correct working with 3 significant figures and units kJ mol^-1 [1].

Examiner tip: In calorimetry calculations for enthalpy of solution, remember that the temperature increase denotes an exothermic dissolution (ΔH < 0). Candidates often lose marks by omitting the negative sign or using the mass of salt instead of mass of water in q = mcΔT.

Revise this topic: Study the complete topic notes on Thermodynamics & Born-Haber Cycles.

Propanoic acid, CH₃CH₂COOH, is a weak Brønsted-Lowry acid.

02.1 Define the term weak Brønsted-Lowry acid. [1 mark]

02.2 Write an expression for the acid dissociation constant, K_a, of propanoic acid. [1 mark]

02.3 Calculate the pH of a 0.150 mol dm^-3 aqueous solution of propanoic acid at 298 K. At this temperature, K_a = 1.35 × 10^-5 mol dm^-3. Give your answer to two decimal places. [3 marks]

02.4 A buffer solution is prepared by mixing 50.0 cm³ of 0.150 mol dm^-3 propanoic acid with 50.0 cm³ of 0.080 mol dm^-3 sodium propanoate solution. Calculate the pH of this buffer solution at 298 K. Show your working. [4 marks]

02.5 Explain how the buffer solution prepared in 02.4 maintains a nearly constant pH when a small volume of dilute sodium hydroxide is added. Include an ionic equation to support your answer. [3 marks]

02.6 A titration curve is plotted when 0.100 mol dm^-3 aqueous sodium hydroxide is added to 25.0 cm³ of 0.150 mol dm^-3 propanoic acid. State the pH range of a suitable indicator for this titration. Explain your choice. [2 marks]

Show Mark Scheme

02.1

M1: A proton / H⁺ donor that is only partially dissociated / ionised in aqueous solution [1] (reject "weakly dissociated" without specifying "partially").

02.2

M1: K_a = [CH₃CH₂COO^-][H⁺] / [CH₃CH₂COOH] [1] (allow [H₃O⁺] in place of [H⁺]; square brackets required).

02.3

M1: [H⁺] = √(K_a × [HA]) = √(1.35 × 10^-5 × 0.150) [1].
M2: [H⁺] = 1.423 × 10^-3 mol dm^-3 [1].
M3: pH = -log₁₀(1.423 × 10^-3) = 2.85 [1] (mark awarded strictly for 2 decimal places; allow ECF).

02.4

M1: n(acid) = (50.0 / 1000) × 0.150 = 7.50 × 10^-3 mol [1].
M2: n(salt) = (50.0 / 1000) × 0.080 = 4.00 × 10^-3 mol [1].
M3: [H⁺] = K_a × [acid] / [salt] = 1.35 × 10^-5 × (7.50 × 10^-3 / 4.00 × 10^-3) = 2.531 × 10^-5 mol dm^-3 [1].
M4: pH = -log₁₀(2.531 × 10^-5) = 4.60 [1] (2 decimal places required).

02.5

M1: Added OH^- ions react with the weak acid molecules to form water and propanoate ions [1].
M2: Ionic equation: CH₃CH₂COOH + OH^- → CH₃CH₂COO^- + H₂O [1].
M3: The large reservoirs of HA and A^- mean the ratio [HA] / [A^-] changes minimally, so [H⁺] and pH remain almost constant [1].

02.6

M1: Suitable indicator range: 8.0 - 10.0 (allow within 7.5 - 10.5, e.g. phenolphthalein) [1].
M2: Weak acid - strong base titration produces a basic salt at equivalence; the steep vertical pH inflection occurs completely within the alkaline region (pH 7 to 10), matching the indicator colour transition range [1].

Examiner tip: Always quote pH values to exactly 2 decimal places in OxfordAQA assessments. In buffer calculations, notice that since both acid and conjugate base are in the same vessel, volumes cancel: [H⁺] = K_a × (moles acid / moles salt).

Revise this topic: Study the complete topic notes on Acids, Bases, and pH Buffers.

The standard hydrogen electrode is used as the universal reference standard in electrochemistry.

03.1 Draw a fully labelled diagram of the standard hydrogen electrode (SHE). Label the electrode material, the chemical substances present, and state the exact standard conditions required. [4 marks]

03.2 A standard electrochemical cell is represented by the following cell notation:
Pt(s) | H₂(g) | H⁺(aq) || Cr₂O₇^2-(aq), H⁺(aq), Cr³⁺(aq) | Pt(s)
(a) Identify the component represented by the double vertical lines (||) and state its function in the cell. [2 marks]
(b) The standard electrode potential for the chromium/dichromate half-cell is E° = +1.33 V. State the overall cell EMF under standard conditions. [1 mark]
(c) Write a balanced overall ionic equation for the cell reaction. [2 marks]

03.3 The concentration of dichromate ions, Cr₂O₇^2-(aq), in the right-hand half-cell is increased. Explain, in terms of equilibrium shifts, the effect this change has on the cell EMF. [2 marks]

03.4 State two conditions that must be maintained in the right-hand half-cell for the electrode potential to remain at the standard value of +1.33 V. [2 marks]

Show Mark Scheme

03.1

M1: Platinum foil electrode coated with platinum black [1].
M2: Hydrogen gas entering at a pressure of 100 kPa (do not accept 1 atm) [1].
M3: Aqueous solution containing H⁺(aq) at concentration 1.00 mol dm^-3 (e.g. 1.00 mol dm^-3 HCl or 0.50 mol dm^-3 H₂SO₄) [1].
M4: Temperature maintained at 298 K (25 °C) [1].

03.2

(a) M1: Salt bridge [1].
(a) M2: Completes the electrical circuit / permits migration of ions to preserve electrical neutrality [1] (reject "allows electrons to move").
(b) M1: +1.33 V [1] (E°_cell = E°_red - E°_ox = 1.33 - 0.00 = +1.33 V).
(c) M1: Reduction half-equation: Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O; Oxidation: 3H₂ → 6H⁺ + 6e^- [1].
(c) M2: Overall: Cr₂O₇^2- + 8H⁺ + 3H₂ → 2Cr³⁺ + 7H₂O [1] (fully balanced; state symbols not required).

03.3

M1: Increasing [Cr₂O₇^2-] shifts the reduction equilibrium to the right to oppose the change [1].
M2: This increases the electrode potential of the half-cell, causing the overall cell EMF to increase [1].

03.4

M1: Temperature kept at exactly 298 K (25 °C) [1].
M2: Concentration of each ionic species (Cr₂O₇^2-, H⁺, and Cr³⁺) maintained at 1.00 mol dm^-3 [1].

Examiner tip: The conventional cell notation places the negative electrode (oxidation) on the left and the positive electrode (reduction) on the right: E°_cell = E°_RHS - E°_LHS. In cell equations involving dichromate, remember to cancel the 6H⁺ on both sides to leave 8H⁺ in the overall balanced ionic equation.

Revise this topic: Study the complete topic notes on Electrochemical Cells & Electrode Potentials.

The oxides of Period 3 elements exhibit systematic variations in structure, bonding, and acid-base character across the period.

04.1 Compare the structures and bonding types of sodium oxide, Na₂O, and phosphorus(V) oxide, P₄O₁₀. [2 marks]

04.2 Write a balanced chemical equation for the reaction of sodium oxide, Na₂O, with water. State the approximate pH of the resulting solution. [2 marks]

04.3 Write a balanced chemical equation for the reaction of phosphorus(V) oxide, P₄O₁₀, with water. State the approximate pH of the resulting solution. [2 marks]

04.4
(a) State the term used to describe an oxide that reacts with both acids and bases. [1 mark]
(b) Write a balanced chemical equation for the reaction of aluminium oxide with hydrochloric acid. [1 mark]
(c) Write a balanced chemical equation for the reaction of aluminium oxide with hot, concentrated aqueous sodium hydroxide. [1 mark]

04.5 Silicon dioxide, SiO₂, is insoluble in water. Explain this insolubility in terms of structure and bonding, and state whether it reacts with concentrated sodium hydroxide solution. [2 marks]

Show Mark Scheme

04.1

M1: Na₂O: Giant ionic lattice with ionic bonding between Na⁺ and O^2- ions [1].
M2: P₄O₁₀: Simple molecular structure with covalent bonding between P and O atoms and weak intermolecular forces between molecules [1].

04.2

M1: Na₂O + H₂O → 2NaOH (or Na₂O + H₂O → 2Na⁺ + 2OH^-) [1].
M2: Approximate pH: 13 - 14 (allow pH 12 - 14) [1].

04.3

M1: P₄O₁₀ + 6H₂O → 4H₃PO₄ [1].
M2: Approximate pH: 0 - 2 (allow pH 1 - 2) [1].

04.4

(a) M1: Amphoteric oxide [1].
(b) M1: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O (or Al₂O₃ + 6H⁺ → 2Al³⁺ + 3H₂O) [1].
(c) M1: Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄] (or Al₂O₃ + 2OH^- + 3H₂O → 2[Al(OH)₄]^-; accept 2NaAlO₂ + H₂O) [1].

04.5

M1: SiO₂ has a giant covalent macromolecular structure; breaking the many strong covalent bonds between silicon and oxygen requires too much energy to be compensated by weak hydration interactions [1].
M2: It does react with hot concentrated sodium hydroxide solution as an acidic oxide (forming sodium silicate, Na₂SiO₃) [1].

Examiner tip: Ensure you know the difference between the reaction of basic metal oxides (which produce OH^- with water), acidic molecular oxides (which produce H⁺), and amphoteric aluminium oxide, which does not dissolve in water due to high lattice energy but dissolves in both acids and bases.

Revise this topic: Study the complete topic notes on Period 3 Elements & Oxides.

Transition metals exhibit characteristic properties including variable oxidation states, catalytic behavior, complex ion formation, and colored solutions resulting from d-orbital splitting.

05.1 Define the term transition metal in terms of electron configuration. [1 mark]

05.2 Cobalt forms a complex ion with three ethanedioate ligands, [Co(C₂O₄)₃]^3-. Draw 3D displayed formulas showing the structures of the two optical isomers of this complex ion. Show the overall charges clearly. [3 marks]

05.3 Define the term bidentate ligand. State one example of a neutral bidentate ligand. [2 marks]

05.4 Aqueous copper(II) sulfate is a pale blue solution. Explain why aqueous copper(II) ions absorb light in the visible region of the spectrum and why the solution appears blue. [2 marks]

05.5 The hexaaquatitanium(III) ion, [Ti(H₂O)₆]³⁺, has a single d-orbital electron. The d-orbital splitting energy needed to excite this electron is ΔE = 3.79 × 10^-19 J. Calculate the wavelength, in nm, of light that is absorbed by this titanium complex.
• Planck constant, h = 6.63 × 10^-34 J s
• Speed of light, c = 3.00 × 10⁸ m s^-1 [3 marks]

05.6 State the difference in phases between reactants and catalysts in homogeneous and heterogeneous catalytic pathways. [1 mark]

05.7 Concentrated hydrochloric acid is added to a solution of cobalt(II) ions. Write a balanced chemical equation for the ligand substitution reaction. State the coordinate geometry and color of the product complex. [2 marks]

Show Mark Scheme

05.1

M1: An element that forms at least one stable ion with a partially filled / incomplete d-subshell / d-orbital [1] (reject reference to atom instead of ion).

05.2

M1: Octahedral geometry around central Co³⁺ with three bidentate ligands [1].
M2: Non-superimposable mirror images showing correct 3D orientation (wedges and dashed bonds) [1].
M3: Charge 3- shown on both enantiomers [1].

05.3

M1: A species that donates two lone pairs of electrons to a central metal ion to form two coordinate (dative covalent) bonds [1].
M2: Neutral example: 1,2-diaminoethane (ethane-1,2-diamine / NH₂CH₂CH₂NH₂ / en) [1].

05.4

M1: Coordination of water ligands causes 3d orbitals to split into two energy levels; d-electrons absorb specific frequencies of visible light (red light) to be promoted from lower to higher d-orbitals [1].
M2: The unabsorbed wavelengths (blue light) are transmitted / reflected, causing the solution to appear blue [1] (do not allow "emits blue light").

05.5

M1: Formula: ΔE = hc / λ ⇒ λ = hc / ΔE [1].
M2: λ = (6.63 × 10^-34 × 3.00 × 10⁸) / (3.79 × 10^-19) = 5.248 × 10^-7 m [1].
M3: λ = 5.248 × 10^-7 × 10⁹ = 525 nm (accept 524.8 nm) [1].

05.6

M1: In homogeneous catalysis, the catalyst is in the same physical state / phase as the reactants; in heterogeneous catalysis, the catalyst is in a different physical state / phase from the reactants [1].

05.7

M1: [Co(H₂O)₆]²⁺ + 4Cl^- ⇔ [CoCl₄]^2- + 6H₂O [1].
M2: Geometry: Tetrahedral; Color: Blue [1] (both required for mark).

Examiner tip: Ligand substitution of 6 water molecules by 4 chloride ligands is accompanied by a change in coordination number from 6 to 4 and geometry from octahedral to tetrahedral, because chloride ions are significantly larger than water molecules and repel each other more strongly.

Revise this topic: Study the complete topic notes on Transition Metals & Complex Ions.

The chemical reactions of metal-aqua ions in aqueous solution illustrate acid-base hydrolysis, precipitation, and amphoteric behavior.

06.1 Aqueous ammonia is added dropwise, and then in excess, to an aqueous solution of hexaaquacopper(II) ions, [Cu(H₂O)₆]²⁺. Describe what is observed at each stage and write balanced chemical equations for the reactions that occur. [4 marks]

06.2 Aqueous sodium hydroxide is added dropwise, and then in excess, to an aqueous solution of hexaaquaaluminium(III) ions, [Al(H₂O)₆]³⁺. Describe what is observed at each stage and write balanced chemical equations for the reactions that occur. [4 marks]

06.3 Explain why an aqueous solution of hexaaquaaluminium(III) ions, [Al(H₂O)₆]³⁺, is significantly more acidic than an aqueous solution of hexaaquacopper(II) ions, [Cu(H₂O)₆]²⁺. In your answer, refer to:
• The sizes and charges of the metal ions.
• The strength and polarization of the O-H bonds in the coordinated water molecules. [4 marks]

Show Mark Scheme

06.1

M1: Dropwise observation: Pale blue precipitate forms [1].
M2: Dropwise equation: [Cu(H₂O)₆]²⁺ + 2NH₃ → [Cu(H₂O)₄(OH)₂] + 2NH₄⁺ [1].
M3: Excess observation: Precipitate dissolves to form a deep / dark blue solution [1].
M4: Excess equation: [Cu(H₂O)₄(OH)₂] + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 2OH^- + 2H₂O (or overall [Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O) [1].

06.2

M1: Dropwise observation: White precipitate forms [1].
M2: Dropwise equation: [Al(H₂O)₆]³⁺ + 3OH^- → [Al(H₂O)₃(OH)₃] + 3H₂O [1].
M3: Excess observation: Precipitate dissolves to yield a colourless solution [1].
M4: Excess equation: [Al(H₂O)₃(OH)₃] + OH^- → [Al(OH)₄]^- + 3H₂O (or [Al(OH)₆]^3-) [1].

06.3

M1: The Al³⁺ ion has a smaller ionic radius and higher ionic charge (+3 vs +2) than the Cu²⁺ ion / Al³⁺ has a significantly higher charge density [1].
M2: The Al³⁺ ion is much more polarising than the Cu²⁺ ion [1].
M3: It withdraws electron density more strongly from the oxygen atom of the coordinated water molecules [1].
M4: Weakening the O-H bonds and facilitating the donation/release of a proton (H⁺) to surrounding water molecules: [Al(H₂O)₆]³⁺ + H₂O ⇔ [Al(H₂O)₅(OH)]²⁺ + H₃O⁺ [1].

Examiner tip: When aqueous sodium carbonate (Na₂CO₃) is added to 3+ aqua ions such as Al³⁺ or Fe³⁺, the high acidity of the solution results in carbon dioxide gas effervescence and a hydroxide precipitate. Contrast this with 2+ ions (Cu²⁺, Fe²⁺), which form simple insoluble metal carbonates (CuCO₃, FeCO₃) without effervescence.

Revise this topic: Study the complete topic notes on Reactions of Aqueous Ions.

Data Insert & Formulae Reference

Physical constants and formulas for OxfordAQA International A-Level Chemistry Unit 3 (CH03).

Physical Constants:
• Gas constant, R = 8.31 J K^-1 mol^-1
• Avogadro constant, L = 6.022 × 10²³ mol^-1
• Specific heat capacity of water, c = 4.18 J g^-1 K^-1
• Planck constant, h = 6.63 × 10^-34 J s
• Speed of light, c = 3.00 × 10⁸ m s^-1

Key Equations:
• Ideal gas: PV = nRT
• Heat energy: q = mcΔT
• Photon energy: ΔE = hν = hc / λ
• Gibbs free energy: ΔG = ΔH - TΔS
• Acid dissociation: K_a = [H⁺][A^-] / [HA]
• Buffer equation: [H⁺] = K_a × ([HA] / [A^-])

Relative Atomic Masses (A_r):
H: 1.0 • C: 12.0 • N: 14.0 • O: 16.0
Na: 23.0 • Mg: 24.3 • Al: 27.0 • Si: 28.1
P: 31.0 • S: 32.1 • Cl: 35.5 • Ca: 40.1
Ti: 47.9 • Cr: 52.0 • Mn: 54.9 • Fe: 55.8
Co: 58.9 • Cu: 63.5 • Zn: 65.4 • Ag: 107.9

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Section B: Structured Written Questions

Question 1: Thermodynamics and Born-Haber Cycles

Question 2: Acids, Bases, and pH Buffers

Question 3: Electrochemical Cells and Electrode Potentials

Question 4: Period 3 Elements and Oxides

Question 5: Transition Metals and Complex Ions

Question 6: Reactions of Ions in Aqueous Solution

Data Insert & Formulae Reference