Mark Scheme

Section A — Module 1

Question 1

(a) (3) — Definition (2): the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions. Both "one mole" and "gaseous" are required; omission of either loses a mark. Equation (1): Mg(g) → Mg⁺(g) + e⁻ State symbols are compulsory; an equation without them scores 0.

(b) (4)

• Identification of the large jump between the 3rd and 4th ionisation energies (2745 → 11 578, a factor of over four) (1).
• This shows three electrons are relatively easily removed from the outer shell, and the fourth must be removed from an inner shell much closer to the nucleus and less shielded (1).
• Therefore X is in Group 13 (Group III) (1).
• X is aluminium (1). Candidates who identify Group 13 but name a different Period 3 element lose only the final mark. Candidates who quote the jump without interpreting it score 1.

(c) (5) — General increase (up to 3):

• Nuclear charge increases across the period as protons are added (1).
• Electrons are added to the same principal quantum shell, so shielding remains approximately constant (1).
• Effective nuclear charge therefore increases and atomic radius decreases, so the outer electron is more strongly attracted and harder to remove (1). Drop Mg → Al (up to 2):
• In magnesium (1s²2s²2p⁶3s²) the electron removed is from the 3s sub-shell; in aluminium (…3s²3p¹) it is from the 3p sub-shell (1).
• The 3p sub-shell is at a higher energy and is partly shielded by the filled 3s sub-shell, so the electron is more easily removed despite the greater nuclear charge (1). Do not credit "aluminium is bigger" — the trend in radius runs the other way.

(d) (3)

• n(CO₂) = 4.4 / 44.0 = 0.10 mol (1)
• Number of molecules = n × L = 0.10 × 6.02 × 10²³ (1)
• = 6.0 × 10²² molecules (1) Award full marks for a correct answer with working. A correct method with an arithmetic slip earns 2. An answer given as "6.0 × 10²² atoms" loses the final mark.

Question 2

(a) (6) — 2 marks per bond type: 1 for the particles, 1 for the nature of the attraction.

• Ionic — between oppositely charged ions formed by electron transfer from metal to non-metal; electrostatic attraction between cations and anions acting in all directions throughout a giant lattice.
• Covalent — between two non-metal atoms; a shared pair of electrons attracted electrostatically by both nuclei, localised between the two atoms.
• Metallic — between metal cations arranged in a lattice and a sea of delocalised outer-shell electrons; electrostatic attraction between the cations and the delocalised electrons. "Sharing" and "transfer" alone score 1; the second mark requires the electrostatic nature to be stated.

(b) (6) — 3 marks each. (i) NH₃ — nitrogen has 3 bonding pairs and 1 lone pair, four electron domains in total (1); lone pair–bond pair repulsion is greater than bond pair–bond pair repulsion, so the bonds are pushed closer together (1); shape is trigonal pyramidal, bond angle 107° (1). (ii) BF₃ — boron has 3 bonding pairs and no lone pairs (1); the three pairs repel equally and arrange themselves as far apart as possible in a plane (1); shape is trigonal planar, bond angle 120° (1). The shape name and angle together carry one mark — a correct angle with the wrong shape name scores 0 for that mark. Candidates who give 107° for NH₃ without mentioning the lone pair cap at 2.

(c) (3)

• Water molecules form hydrogen bonds between the δ⁺ hydrogen of one molecule and a lone pair on the δ⁻ oxygen of another (1).
• Hydrogen bonding occurs because oxygen is highly electronegative and small; sulfur is much less electronegative and larger, so H₂S has only weak permanent dipole–dipole and van der Waals forces (1).
• Hydrogen bonds are considerably stronger, so more energy is required to separate water molecules and its boiling point is much higher (1). Do not credit answers implying the covalent O–H bonds are broken on boiling — this is a common and heavily penalised error; such answers cap at 1.

Section B — Module 2

Question 3

(a) (2) — When a change is imposed on a system at dynamic equilibrium, the position of equilibrium shifts so as to oppose (minimise the effect of) that change. 1 mark for "opposes the change", 1 for reference to a system at equilibrium.

(b) (6) — 2 marks each: 1 for the effect, 1 for the explanation. (i) Increased pressure — yield increases; there are 4 moles of gas on the left and 2 on the right, so the equilibrium shifts to the right, the side with fewer gaseous moles, to reduce the pressure. (ii) Increased temperature — yield decreases; the forward reaction is exothermic (ΔH negative), so the equilibrium shifts in the endothermic (reverse) direction to absorb the added heat. (iii) Catalyst — no change in yield; a catalyst lowers the activation energy of forward and reverse reactions equally, so equilibrium is reached faster but its position is unaffected. Part (iii) is a discriminator: candidates who say a catalyst increases the yield score 0 for it.

(c) (3)

• A lower temperature would give a higher equilibrium yield, since the forward reaction is exothermic (1).
• But at low temperature the rate of reaction is too slow to be economic, as few molecules have energy exceeding the activation energy (1).
• 450 °C is a compromise, giving an acceptable yield at an acceptable rate; unreacted gases are recycled to improve the overall conversion (1). The word "compromise" alone is not sufficient — both the yield and the rate consideration must be stated.

(d) (4)

• Expression: Kc = [NH₃]² / ([N₂][H₂]³) (1)
• Concentrations: since V = 1.00 dm³, [N₂] = 0.20, [H₂] = 0.60, [NH₃] = 0.40 mol dm⁻³ (1)
• Substitution: Kc = (0.40)² / (0.20 × 0.60³) = 0.160 / (0.20 × 0.216) = 0.160 / 0.0432 (1)
• = 3.70 mol⁻² dm⁶ (1) Units are derived from (mol dm⁻³)² / (mol dm⁻³)⁴ = mol⁻² dm⁶. Omission of units loses the final mark. A correct expression with an arithmetic error still earns 3.

Question 4

(a) (4)

• n(HCl) = c × V = 0.100 × (22.5 / 1000) = 2.25 × 10⁻³ mol (1)
• Equation NaOH + HCl → NaCl + H₂O gives a 1 : 1 ratio, so n(NaOH) = 2.25 × 10⁻³ mol (1)
• c(NaOH) = n / V = 2.25 × 10⁻³ / (25.0 / 1000) (1)
• = 0.0900 mol dm⁻³ (1) The commonest error is failure to convert cm³ to dm³; this loses the first mark but subsequent marks are awarded on the candidate's own figures (error carried forward).

(b) (3)

• pH = −log₁₀[H⁺] (1)
• HCl is a strong monobasic acid, fully dissociated, so [H⁺] = 0.0500 mol dm⁻³ (1)
• pH = −log₁₀(0.0500) = 1.30 (1) Accept 1.3. An answer of 1.30 with no working scores 2.

(c) (8) — Definition up to 2:

• A solution that resists a change in pH when small amounts of acid or alkali are added, or when it is diluted (1).
• An acidic buffer consists of a weak acid and its conjugate base — here ethanoic acid CH₃COOH and ethanoate ions CH₃COO⁻ from the sodium ethanoate (1). Equilibrium up to 2:
• CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) (1)
• The weak acid is only slightly dissociated, providing a large reservoir of undissociated CH₃COOH, while the salt is fully dissociated, providing a large reservoir of CH₃COO⁻ (1). On adding acid up to 2:
• Added H⁺ ions are removed by reaction with the large reservoir of ethanoate ions: CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq) (1).
• The equilibrium shifts to the left, so [H⁺] and therefore pH change only slightly (1). On adding alkali up to 2:
• Added OH⁻ ions react with the undissociated acid: CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l) (1). (Accept the two-stage route: OH⁻ removes H⁺ to form water, and the equilibrium shifts right to replace it.)
• The equilibrium shifts to the right, replacing the H⁺ removed, so pH changes only slightly (1). Equations must balance and carry state symbols. Candidates who assert that the buffer "keeps the pH constant" rather than "resists change" lose the definition mark — buffering is a matter of degree.

Section C — Module 3

Question 5

(a) (4)

• Atomic radius increases down the group from Be to Ba (1).
• Each successive element has an additional occupied principal quantum shell (1).
• Shielding of the outer electrons by inner shells therefore increases (1).
• Although nuclear charge also increases, the increased shielding and greater distance outweigh it, so effective nuclear charge on the outer shell falls and the atom is larger (1). A candidate stating only "more shells" scores 2.

(b) (6)

• Magnesium reacts only very slowly with cold water; the reaction is appreciable with steam (1): Mg(s) + 2H₂O(l) → Mg(OH)₂(aq) + H₂(g) or Mg(s) + H₂O(g) → MgO(s) + H₂(g) (1)
• Calcium reacts readily with cold water, effervescing steadily and forming a white suspension (1): Ca(s) + 2H₂O(l) → Ca(OH)₂(aq) + H₂(g) (1)
• Calcium is more reactive because its atomic radius is larger and its outer electrons are better shielded, so its first and second ionisation energies are lower (1).
• The two outer electrons are therefore lost more readily to form the 2+ ion, and reactivity increases down Group 2 (1). Equations without state symbols lose one mark once across the question.

(c) (5)

• Solubility decreases down the group, by roughly five orders of magnitude from MgSO₄ to BaSO₄ (1 — the mark requires reference to the data, not merely "it decreases").
• Both lattice enthalpy and hydration enthalpy become less exothermic down the group as the cation radius increases (1).
• Because the sulfate anion is large, the inter-ionic distance is dominated by the anion and lattice enthalpy changes relatively little down the group, whereas hydration enthalpy of the cation falls more sharply (1).
• The enthalpy of solution therefore becomes less exothermic (more endothermic) down the group and solubility falls (1).
• Analytical use: the insolubility of BaSO₄ is used in the test for sulfate ions — addition of acidified barium chloride solution to a solution containing sulfate gives a white precipitate; also used as a gravimetric method and as a radiocontrast "barium meal" (1). Accept the converse argument applied to Group 2 hydroxides (solubility increasing) only if the candidate makes clear it is a contrast.

Question 6

(a) (4)

• A transition element is a d-block element that forms at least one stable ion with a partially filled d sub-shell (2 — "partially filled d sub-shell" is essential; "an element in the d block" alone scores 1).
• Zinc has the configuration [Ar]3d¹⁰4s² and forms only the Zn²⁺ ion (1).
• Zn²⁺ is [Ar]3d¹⁰ — a completely filled d sub-shell — so zinc is not a transition element (1). Accept the parallel argument for scandium (Sc³⁺ is 3d⁰) if offered as supporting comparison.

(b) (5) — 1 mark each for any three properties (max 3): variable oxidation states; formation of coloured ions/compounds; catalytic activity; formation of complex ions with ligands; paramagnetism. Up to 2 for explaining one:

• Variable oxidation states — the 4s and 3d sub-shells are very close in energy, so differing numbers of electrons can be lost with similar energy input.
• Coloured ions — ligands split the degenerate d orbitals into two energy levels; an electron absorbs a photon of visible light to be promoted from the lower to the higher set (d–d transition), and the colour observed is the complement of the wavelength absorbed. This requires a partially filled d sub-shell, which is why Zn²⁺ compounds are white.
• Catalytic activity — variable oxidation states allow the catalyst to accept and donate electrons in successive steps, or ligand vacancies allow reactants to adsorb onto the surface. Explanations of colour that stop at "the d orbitals split" without mentioning electron promotion or absorption of visible light score 1.

(c) (6) — 1 mark per correct observation, 1 per correct additional detail, 2 for the equation.

• Fe²⁺ — a green precipitate of Fe(OH)₂, insoluble in excess NaOH; it darkens to brown at the surface on standing, as it is oxidised by air to Fe(OH)₃ (2).
• Fe³⁺ — a red-brown precipitate of Fe(OH)₃, insoluble in excess NaOH (1).
• Cu²⁺ — a pale blue precipitate of Cu(OH)₂, insoluble in excess NaOH; dissolves in excess aqueous ammonia to give a deep blue solution, which distinguishes it further (2).
• Ionic equation, any one, with state symbols (1): Fe²⁺(aq) + 2OH⁻(aq) → Fe(OH)₂(s) Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s) Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s) Colours must be precise: "blue" for Fe(OH)₂ or "brown" for Cu(OH)₂ scores 0 for that observation. Candidates who state that any of these precipitates dissolves in excess sodium hydroxide lose that mark — none is amphoteric.

Level descriptors (extended-response parts)

• Level 4 (top band): Accurate chemical knowledge with correct terminology; equations balanced and carrying state symbols; calculations laid out step by step with units and appropriate significant figures; explanations proceed through a complete causal chain (e.g. nuclear charge → shielding → effective nuclear charge → radius → ionisation energy) rather than asserting the conclusion.
• Level 3: Sound knowledge; equations mostly correct but state symbols or balancing occasionally omitted; calculations correct in method with slips; explanations partially developed.
• Level 2: Recall of facts without explanation; frequent equation errors; calculations attempted without clear method.
• Level 1: Isolated correct statements only.

Sample Answers with Examiner Commentary

Question 1(c) — Sample Answers

Grade I (Distinction) answer (extract)

"Across Period 3 the number of protons in the nucleus rises from 11 in sodium to 18 in argon, so nuclear charge increases steadily. The electrons being added, however, all enter the same third principal quantum shell, and electrons in the same shell shield one another very poorly. Shielding is therefore essentially constant across the period while nuclear charge rises, so the effective nuclear charge experienced by the outer electrons increases. The outer electrons are drawn closer, atomic radius falls, and more energy is required to remove an electron — hence the general rise in first ionisation energy.

The drop from magnesium to aluminium is a departure from this trend and requires a sub-shell explanation. Magnesium is 1s²2s²2p⁶3s², so the electron removed comes from the 3s sub-shell. Aluminium is 1s²2s²2p⁶3s²3p¹, and the electron removed is the single 3p electron. The 3p sub-shell lies at a higher energy than 3s and is somewhat shielded from the nucleus by the filled 3s sub-shell beneath it. This raises the energy of the outermost electron and reduces the attraction it experiences, and the effect is large enough to outweigh the increase of one unit in nuclear charge. Aluminium's first ionisation energy is therefore lower than magnesium's, despite the general trend."

Mark: 5/5. Examiner commentary: The general trend is developed as a complete causal chain — nuclear charge, then constant shielding, then effective nuclear charge, then radius, then ionisation energy. Many candidates give the first and last links and omit the middle. The anomaly is handled exactly as the mark scheme requires: full electron configurations are quoted, the sub-shell of the removed electron is identified in each case, and the candidate explicitly weighs the sub-shell effect against the increased nuclear charge rather than simply asserting that 3p is higher in energy. Full marks.

Grade III (Pass) answer (extract)

"The ionisation energy goes up across the period because the atoms get smaller and the nuclear charge gets bigger, so the electrons are held on more tightly and need more energy to remove. Magnesium to aluminium goes down because aluminium is in a different group and has a 3p electron which is easier to remove."

Mark: 2/5. Examiner commentary: Two marks are earned — increasing nuclear charge, and the identification of the 3p electron. The response falls short in two ways that are typical of this band. First, "the atoms get smaller" is offered as a cause of the trend when it is a consequence of the same underlying factor; there is no reference to shielding remaining constant or to effective nuclear charge, which is the mechanism the question is testing. Second, the anomaly is asserted rather than explained: the candidate says the 3p electron is easier to remove but never says why — that 3p is higher in energy than 3s and shielded by the filled 3s sub-shell. Neither electron configuration is quoted. Adding the configurations and one sentence of reasoning on each half would take this to full marks.

Question 3(d) — Sample Answers

Grade I (Distinction) answer (extract)

"Kc = [NH₃]² / ([N₂][H₂]³)

Since the vessel has a volume of 1.00 dm³, the equilibrium concentrations are numerically equal to the number of moles: [N₂] = 0.20 mol dm⁻³, [H₂] = 0.60 mol dm⁻³, [NH₃] = 0.40 mol dm⁻³

Kc = (0.40)² / (0.20 × (0.60)³) = 0.160 / (0.20 × 0.216) = 0.160 / 0.0432 = 3.70

Units: (mol dm⁻³)² / [(mol dm⁻³) × (mol dm⁻³)³] = (mol dm⁻³)² / (mol dm⁻³)⁴ = (mol dm⁻³)⁻² = mol⁻² dm⁶

Kc = 3.70 mol⁻² dm⁶"

Mark: 4/4. Examiner commentary: Everything the mark scheme asks for, laid out so that every step is separately creditable. Two details lift this above a merely correct answer: the candidate states explicitly why moles may be used as concentrations here (the volume is 1.00 dm³) rather than silently substituting, and the units are derived from the expression rather than recalled. Deriving the units is the safer method and is what examiners look for, since candidates who memorise units routinely attach mol⁻² dm⁶ to equilibria with different stoichiometry. Full marks.

Grade II (Merit) answer (extract)

"Kc = [NH₃]² / ([N₂][H₂]³)

Kc = (0.40)² / (0.20 × 0.60 × 3) = 0.160 / 0.36 = 0.44"

Mark: 2/4. Examiner commentary: The expression is correct, earning the first mark, and the candidate correctly recognises that the concentrations may be read directly from the moles in a 1.00 dm³ vessel, earning the second. The error is in the substitution: [H₂]³ has been evaluated as 0.60 × 3 rather than 0.60³. This is a mathematical slip rather than a chemical misunderstanding, but it costs both remaining marks, since the final value is wrong and no units are given at all. Note that units alone would have earned a mark here even with the wrong number — candidates should always state units on an equilibrium constant, and should derive them from the expression, which in this case would also have prompted a check of the cubed term.

How Unit 1 is assessed

Paper 02 carries 90 marks and contributes 50% of the Unit 1 external assessment mark, alongside Paper 01 (multiple choice, 30%) and the Internal Assessment (20%). All six questions are compulsory and the three Modules carry equal weight. Grades are reported Grade I – Grade VII, with Grade I the highest.

Three habits separate the top band. First, lay calculations out line by line with units — this paper reserves method marks at every stage, and a candidate who writes only a final answer forfeits them all if that answer is wrong, while a candidate who shows each step keeps most of the marks after a slip. Always derive the units of an equilibrium constant from its expression rather than recalling them. Second, write complete causal chains in explanations: trends in ionisation energy, radius and reactivity are all examined through the sequence nuclear charge → shielding → effective nuclear charge → observed property, and answers that jump from the first term to the last score roughly half. Third, include state symbols in every equation — they are required by the rubric, and marks are lost for their omission across all three Modules on almost every paper.