S6 Chemistry II 2025 Past Paper: Full Worked Solutions

These are the complete worked solutions for the S6 Chemistry II  NESA Advanced Level Examination, 2024–2025, sat on 10 July 2025. This paper applies to combinations BCG, MCB, PCB, PCM, and ANP.

Every answer below includes a full explanation — not just the letter. In Chemistry exams, understanding why an answer is correct is what prepares you for the next paper.

If any topic here is giving you trouble, book a Chemistry tutor on Mathrone Academy for one-on-one support. You can also revise alongside the S6 Mathematics II 2025 worked solutions and the S6 Physics 2025 worked solutions if you are in PCM or PCB.

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Section A: Attempt ALL Questions (70 marks)

Question 1: Matching Terms (3 marks)

Question 2: Group 15 Elements (3 marks)

i) Lowest electronegativity in Group 15

Answer: D) Bismuth (Bi)

Electronegativity decreases down a group in the periodic table. As atomic number increases, atomic radius increases and the nucleus is further from the bonding electrons. The shielding effect of inner electron shells also increases, reducing the pull on shared electrons. Nitrogen at the top is the most electronegative; Bismuth (Bi, atomic number 83) at the bottom is the least.

ii) Compound NOT typically formed by nitrogen

Answer: B) Dinitrogen heptoxide (N₂O₇)

Nitrogen forms oxides with oxidation states from −3 to +5. The known stable nitrogen oxides include N₂O (+1), NO (+2), N₂O₃ (+3), NO₂ (+4), and N₂O₅ (+5). N₂O₇ would require nitrogen to have an oxidation state of +7, which exceeds nitrogen's maximum valence of 4 bonds  impossible given its electron configuration. It does not exist as a stable compound.

iii) Promoter in the Haber Process

Answer: A) Al₂O₃

In the Haber Process ${\text{N}}_2+3{\text{H}}_2\rightleftharpoons 2{\text{NH}}_3$, iron (Fe) is the catalyst. Al₂O₃ (aluminium oxide) and K₂O act as promoters — they do not catalyse the reaction themselves but increase the activity and longevity of the iron catalyst by preventing sintering (clumping) of iron particles. V₂O₅ is the catalyst in the Contact Process (for SO₃), not the Haber Process. MnO₂ is used in decomposition of hydrogen peroxide.

Question 3: Fertilisers (3 marks)

i) NPK fertiliser labelled 10-20-20

Answer: C) 10% by mass N, 20% by mass P₂O₅ and 20% by mass K₂O

The NPK labelling system is internationally standardised. The three numbers represent: N as elemental nitrogen, P expressed as its oxide equivalent P₂O₅, and K expressed as its oxide equivalent K₂O. The percentages are by mass of the total fertiliser. Pure N, P₂O₅, and K₂O — not the elements P and K on their own, and not ionic compounds like NO₂ or KO₂.

ii) Classification of Diammonium hydrogen phosphate (DAP) — (NH₄)₂HPO₄

Answer: A) NP-type fertilizer

DAP contains two nutrients: nitrogen (from the two ammonium NH₄⁺ groups) and phosphorus (from HPO₄²⁻). It contains no potassium, so it cannot be NK, PK, or NPK. It is classified as an NP-type fertiliser, widely used in agriculture for providing both nitrogen and phosphorus simultaneously.

iii) Consequence of water enriched with nitrates and phosphates

 Answer: D) Eutrophication and harmful algal bloom

When water bodies receive excess nitrates and phosphates — typically from agricultural run-off — algae grow rapidly on the surface (algal bloom). This blocks sunlight from reaching submerged plants, which die and decompose. The decomposition process consumes dissolved oxygen, causing oxygen depletion (hypoxia) that kills aquatic animals. This entire process is called eutrophication. It reduces biodiversity (not increases it), does not affect dissolved nitrogen levels directly, and does not decrease water temperature.

Question 4: True or False: Organic Chemistry (3 marks)

Question 5: Elimination Reaction (3 marks)

The reaction converts compound A (an alcohol) to compound B (an alkene) using concentrated H₂SO₄ at 170°C with loss of water.

i) Type of reaction R

Answer: B) Elimination reaction

An OH group and an H atom from an adjacent carbon are removed together as water (H₂O), forming a C=C double bond. This is dehydration — a type of elimination. No atoms are added (not addition), no atom is swapped (not substitution), and the carbon skeleton is not rearranged.

ii) Rule obeyed by reaction R

Answer: D) Zaitsev's rule

Zaitsev's rule states that in elimination reactions, the major product is the more substituted alkene (the one with more alkyl groups attached to the double bond carbons). Here, the product B is ${\text{CH}}_3\text{C}({\text{CH}}_3)={\text{CHCH}}_2{\text{CH}}_3$ the more substituted alkene, consistent with Zaitsev's rule. Markovnikov's rule applies to addition reactions. Hofmann's rule applies to elimination from ammonium salts. Hückel's rule applies to aromaticity.

iii) Systematic IUPAC name for compound A

Answer: A) 2-methylpentan-2-ol

Compound A is ${\text{CH}}_3\text{C}({\text{CH}}_3)(\text{OH}){\text{CH}}_2{\text{CH}}_2{\text{CH}}_3$. The longest carbon chain containing the OH group has 5 carbons (pentane). The OH is on carbon 2, and there is a methyl group also on carbon 2. Therefore: 2-methylpentan-2-ol.

Question 6: Ester Structure (3 marks)

Compound C is: H₃C–C(=O)–O–CH₂CH₂CH₃ (methyl group on one side, propyl on the other via the ester linkage).

Looking at the structure carefully: the acyl side is CH₃CO– (from ethanoic acid) and the alkoxy side is –CH₂CH₂CH₃ (propyl group from propanol).

i) General molecular formula for esters

Answer: A) CₙH₂ₙO₂

Esters have the functional group –COO–. The general formula for saturated esters with no rings or other functional groups is $C_n{H}_{2n}{O}_2$. For example, methyl ethanoate (CH₃COOCH₃) is C₃H₆O₂, which fits $C_n{H}_{2n}{O}_2$ with n=3.

ii) Systematic IUPAC name for compound C

Answer: A) Propyl ethanoate

IUPAC naming of esters: the alkyl group from the alcohol is named first, then the acid part with the suffix "-anoate". The acid portion is ethanoic acid (CH₃COOH → ethanoate), and the alcohol portion is propanol (–CH₂CH₂CH₃ → propyl). Therefore: propyl ethanoate.

iii) Reactant molecules that formed compound C

Answer: B) Propanol and ethanoic acid

Esters are formed by condensation (esterification) between a carboxylic acid and an alcohol with loss of water: $${\text{CH}}_3\text{COOH}+{\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{OH}\stackrel{{\text{conc. H}}_2{\text{SO}}_4,\mathrm{\Delta }}{\to }{\text{CH}}_3{\text{COOCH}}_2{\text{CH}}_2{\text{CH}}_3+{\text{H}}_2\text{O}$$ Ethanoic acid + propanol → propyl ethanoate + water.

Question 7:Colligative Properties: True or False (4 marks)

Question 8: Group 17 Hydride Boiling Points (4 marks)

Atomic masses given: H=1, F=19, Cl=35.5, Br=80, I=127. Molecular masses: HF=20, HCl=36.5, HBr=81, HI=128.

Generally, boiling point increases with molecular mass (stronger London dispersion forces). However, HF is anomalous,  it has an unusually HIGH boiling point due to strong hydrogen bonding between F and H atoms.

⚠️ Note on HI and HBr ordering: The actual boiling points are HF (+19.5°C) > HI (−35.4°C) > HBr (−66.8°C) > HCl (−85°C). Matching to the options given: HBr → D (−35.4°C), HI → C (−66.8°C). So the correct full matching is: i)HBr = D, ii)HF = B, iii)HCl = A, iv)HI = C.

Question 9: Organic Reaction Flow Chart (4 marks)

Starting material: ${\text{CH}}_3\text{CH}({\text{CH}}_3)\text{CBr}({\text{CH}}_3{)}_2$ — a tertiary alkyl bromide.

i) Type of reaction X (addition of HBr to an alkene)

Answer: B) Addition reaction

Reaction X converts the alkene ${\text{CH}}_3\text{CH}=\text{C}({\text{CH}}_3{)}_2$ to the tertiary bromide by adding HBr across the double bond. Two reactant molecules combine to give one product — this is electrophilic addition.

ii) IUPAC name of product D

Answer: B) 2-Methylbutan-2-ol

Product D is formed from the tertiary bromide reacting with dilute KOH/heat (Reaction Y nucleophilic substitution, OH⁻ replaces Br⁻). The product is ${\text{CH}}_3\text{C}(\text{OH})({\text{CH}}_3){\text{CH}}_2{\text{CH}}_3$. The longest chain through the OH carbon is 4 carbons (butane), with a methyl substituent on C2 and the OH on C2: 2-methylbutan-2-ol.

iii) Reaction mechanism for reaction Y (dilute KOH/heat on a tertiary halide)

Answer: C) Unimolecular Nucleophilic Substitution (SN1)

The starting alkyl bromide is tertiary (three alkyl groups on the carbon bearing Br). Tertiary halides undergo SN1 — a two-step mechanism where the C–Br bond breaks first to form a stable tertiary carbocation, and then the nucleophile (OH⁻) attacks. SN2 is favoured by primary halides, not tertiary ones, due to steric hindrance.

iv) Suitable reagent for reaction Z (converting alcohol to alkene)

Answer: A) H₂O/conc. H₂SO₄, heat

Reaction Z converts 2-methylbutan-2-ol back to an alkene via acid-catalysed dehydration. Concentrated H₂SO₄ protonates the OH group, converting it to a good leaving group (H₂O). Heat drives the elimination. This regenerates the alkene, consistent with Zaitsev's rule.

Question 10: True or False: Organic Acids and Isomers (4 marks)

Question 11: Base Strengths (4 marks)

A larger Kb means the base ionises more in water  it is a stronger base.

i) Strongest base

Answer: B) Methylamine (CH₃NH₂)

Methylamine has the highest Kb (4.6×10⁻⁴). The electron-donating methyl group increases electron density on the nitrogen, making it more willing to accept a proton.

ii) Weakest base

Answer: D) Phenylamine (C₆H₅NH₂)

Phenylamine has the lowest Kb (4.0×10⁻¹⁰). The lone pair on nitrogen is delocalised into the benzene ring through resonance, making it far less available to accept a proton.

iii) Dissociation of CH₃NH₃⁺(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃NH₂(aq)

a) Species acting as acid: CH₃NH₃⁺
It donates a proton (H⁺) to water, acting as a Brønsted-Lowry acid.

b) Species acting as base: H₂O
Water accepts the proton from CH₃NH₃⁺, forming H₃O⁺. It is the Brønsted-Lowry base in this equilibrium.

Question 12: Electrolysis (4 marks)

i) Substance NOT decomposed by electricity

Answer: A) Molten silver

Electrolysis decomposes ionic compounds (which contain ions that can move and discharge). Molten silver is a pure metal — it consists of silver atoms, not ions in a dissociated compound. It conducts electricity but is not decomposed by it. Dilute H₂SO₄ (aq), aqueous AgNO₃, and NaCl solution all contain free ions and are electrolysed.

ii) Equation at positive electrode when molten Al₂O₃ is electrolysed

Answer: C) 2O²⁻ → O₂ + 4e⁻

The positive electrode is the anode — oxidation occurs here. Oxide ions (O²⁻) lose electrons and are oxidised to oxygen gas. At the cathode (negative electrode), Al³⁺ ions gain electrons and are reduced to aluminium metal: Al³⁺ + 3e⁻ → Al.

iii) Electrolysis of aqueous CuSO₄ using copper electrodes

Answer: D) Anode becomes smaller | Cathode becomes bigger | Blue colour remains

With copper electrodes in CuSO₄ solution: at the anode, copper metal dissolves (Cu → Cu²⁺ + 2e⁻), so the anode gets smaller. At the cathode, Cu²⁺ ions are deposited (Cu²⁺ + 2e⁻ → Cu), so the cathode grows. Because copper dissolves from the anode at the same rate it deposits at the cathode, the Cu²⁺ concentration stays constant — the blue colour of the solution remains unchanged. This is the principle behind copper refining and electroplating.

iv) Silver plating setup

Answer: B) Metal statue = Cathode | Electrolyte = AgNO₃(aq)

In electroplating, the object to be plated is always the cathode (negative electrode) — silver ions (Ag⁺) from the electrolyte are reduced and deposited onto it. AgNO₃ solution provides the Ag⁺ ions needed. AgCl is sparingly soluble and would not provide sufficient free Ag⁺ ions for effective plating.

Question 13: Nuclear Equations (4 marks)

i) Symbol for a positron

 Answer: C) ${}_{+1}^{0}e$

A positron is the antiparticle of an electron. It has the same mass as an electron (mass number 0) but a positive charge (+1). Written as ${}_{+1}^{0}e$. An alpha particle is ${}_2^{4}He$, a beta particle (electron) is ${}_{-1}^{0}e$, and a proton is ${}_1^{1}H$.

ii) Completing nuclear equations

a) ${}_7^{14}N{+}_2^{4}He{\to }_8^{17}O+$ ?

Check mass numbers: 14 + 4 = 18. Product O has mass 17. Missing particle mass = 18 − 17 = 1.
Check atomic numbers: 7 + 2 = 9. O has atomic number 8. Missing particle charge = 9 − 8 = 1.
Missing particle: ${}_1^{1}H$ (a proton)

b) ${}_{95}^{241}Am{+}_2^{4}He{\to }_{97}^{243}Bk+$ ?

Mass numbers: 241 + 4 = 245. Bk has mass 243. Missing mass = 245 − 243 = 2.
Atomic numbers: 95 + 2 = 97. Bk has atomic number 97. Missing charge = 97 − 97 = 0.
Missing particle: ${}_0^{2}n$ — but from the list, two neutrons $2{\times }_0^{1}n$. Answer: $2_0^{1}n$ (2 neutrons)

c) ${}_{92}^{238}U{+}_0^{1}n{\to }_{93}^{239}Np+$ ?

Mass numbers: 238 + 1 = 239. Np has mass 239. Missing mass = 0.
Atomic numbers: 92 + 0 = 92. Np has atomic number 93. Missing charge = 92 − 93 = −1.
Missing particle: ${}_{-1}^{0}e$ (a beta particle / electron)

Question 14: Periodic Trends: True or False (4 marks)

Question 15  Phase Diagram of H₂O (4 marks)

i) Point where liquid and gas phases CANNOT be distinguished

Answer: C) Point Z

Point Z is the critical point of water (374°C, 217.75 atm). Above this temperature and pressure, water exists as a supercritical fluid — a state where distinct liquid and gas phases no longer exist. The two phases become indistinguishable.

ii) Point where all three phases coexist

Answer: B) Point V

Point V is the triple point of water (0.01°C, 0.006 atm). It is the unique combination of temperature and pressure at which solid ice, liquid water, and water vapour coexist simultaneously in equilibrium.

iii) Phase equilibrium represented by line XZ

Answer: B) H₂O(l) ⇌ H₂O(g)

Line XZ runs from the triple point (X/V area) to the critical point (Z), passing through the point at 100°C and 1 atm. This line represents the liquid-vapour equilibrium — the boundary between liquid water and water vapour (the boiling curve).

iv) What does the negative slope of the solid-liquid line indicate?

Answer: B) The melting point of water decreases with increasing pressure.

For most substances, the solid-liquid line slopes to the right (positive slope), meaning increasing pressure raises the melting point. Water is anomalous — ice is less dense than liquid water, so increasing pressure favours the liquid phase. The melting point of ice decreases as pressure increases (the line slopes left/negative). This is why ice skating works — pressure under the blade slightly lowers the melting point.

Question 16: Solubility Products (4 marks)

i) Correct Ksp expression for lead(II) chloride PbCl₂

Answer: D) Ksp = [Pb²⁺][Cl⁻]²

PbCl₂ dissociates as: ${\text{PbCl}}_2(s)\rightleftharpoons {\text{Pb}}^{2+}(aq)+2{\text{Cl}}^{-}(aq)$. The Ksp expression includes the concentrations raised to the power of their stoichiometric coefficients: $K_{sp}=[{\text{Pb}}^{2+}][{\text{Cl}}^{-}{]}^2$. Solid PbCl₂ is not included in the expression.

ii) In which solution does solubility of Ag₂CO₃ increase?

Answer: A) HNO₃(aq)

Ag₂CO₃ is sparingly soluble. In HNO₃, the H⁺ ions react with CO₃²⁻ ions: ${\text{CO}}_3^{2-}+2{\text{H}}^{+}\to {\text{H}}_2\text{O}+{\text{CO}}_2$. This removes CO₃²⁻ from solution, shifting the dissolution equilibrium to the right (Le Chatelier's principle), increasing solubility. In Na₂CO₃, AgNO₃, or K₂CO₃, the common ion effect (extra CO₃²⁻ or Ag⁺) decreases solubility.

iii) Molar solubility of PbSO₄ (Ksp = 1.7×10⁻⁸)

Answer: B) 1.3 × 10⁻⁴ mol dm⁻³

PbSO₄ dissociates as: ${\text{PbSO}}_4(s)\rightleftharpoons {\text{Pb}}^{2+}(aq)+{\text{SO}}_4^{2-}(aq)$

Let molar solubility = S. Then [Pb²⁺] = S and [SO₄²⁻] = S.

$$K_{sp}=[{\text{Pb}}^{2+}][{\text{SO}}_4^{2-}]=S\times S=S^2$$ $$S^2=1.7\times {10}^{-8}$$ $$S=\sqrt{1.7\times {10}^{-8}}=1.30\times {10}^{-4}{\text{ mol dm}}^{-3}$$

Question 17: Hess's Law and Enthalpy (4 marks)

i) Correct statement of Hess's Law

Answer: B) The heat absorbed or released in a reaction is the same regardless of the pathway taken.

Hess's Law states that the total enthalpy change for a reaction is independent of the route — only the initial and final states matter. This makes enthalpy a state function.

ii) Correct balanced equation for combustion of ethanamide (CH₃CONH₂)

Answer: A) 2CH₃CONH₂ + $\frac{11}{2}$O₂ → 4CO₂ + 5H₂O + N₂

Balancing: each CH₃CONH₂ has 2C, 5H, 1O, 1N.

• 2 moles → 4C, 10H, 2O, 2N
• Products: 4CO₂ (8 O), 5H₂O (5 O), N₂
• Total O needed on right = 8 + 5 = 13. Already have 2 from reactant, so need 11 from O₂ → $\frac{11}{2}$ O₂ ✓

iii) Enthalpy of combustion of one mole of ethanamide

Answer: A) −1823 kJ mol⁻¹

Using: $\mathrm{\Delta }{H}_c=\sum \mathrm{\Delta }{H}_f(\text{products})-\sum \mathrm{\Delta }{H}_f(\text{reactants})$

For the equation with 2 moles of ethanamide:

$$\mathrm{\Delta }H=[4(-394)+5(-286)+0]-[2(-320)+0]$$ $$=[-1576+(-1430)]-[-640]$$ $$=-3006-(-640)=-3006+640=-2366\text{ kJ}$$

This is for 2 moles. For 1 mole:

$$\mathrm{\Delta }{H}_c=\frac{-2366}{2}=\mathbf{-}\mathbf{1183}{\text{ kJ mol}}^{-1}$$

⚠️ Correction: The calculation gives −1183 kJ mol⁻¹, which matches option D. Option A (−1823) does not match the data provided. The correct answer based on the given ΔHf values is D) −1183 kJ mol⁻¹. Always follow the calculation, not the expected answer.

Question 18: Dilution and Titration (4 marks)

i) Volume of 18.4 M H₂SO₄ needed to prepare 1000 cm³ of 0.5 M solution

Using the dilution formula: $C_1{V}_1=C_2{V}_2$

$$18.4\times V_1=0.5\times 1000$$ $$V_1=\frac{0.5\times 1000}{18.4}=\frac{500}{18.4}$$ $$\boxed{{\displaystyle V_1=27.2{\text{ cm}}^3}}$$

Procedure: Carefully add 27.2 cm³ of concentrated H₂SO₄ into approximately 500 cm³ of distilled water in a volumetric flask (always add acid to water, never water to acid). Allow to cool, then make up to exactly 1000 cm³ with distilled water and mix thoroughly.

ii) Acid-base titration vs redox titration

Question 19: Energy Profile Diagram (4 marks)

The diagram shows reaction AB + C → CB + A with a transition state (ABC) at the energy peak, reactants (AB + C) at a higher energy level than products (CB + A).

i) Energy changes F and G

F = Activation energy (Ea) of the reverse reaction
F is the energy difference between the transition state (ABC) and the products (CB + A). It is the minimum energy needed for the reverse reaction to proceed.

G = Enthalpy change (ΔH) of the reaction
G is the energy difference between reactants and products. Since products are at higher energy than reactants... wait — re-reading the diagram: reactants (AB + C) are at a lower level than products (CB + A), making G the positive enthalpy difference, which represents the overall energy change ΔH.

ii) Chemical species ABC

ABC is the activated complex (transition state).
It is the unstable, high-energy intermediate formed momentarily when AB and C collide with sufficient energy. It sits at the peak of the energy profile and cannot be isolated.

iii) Exothermic or endothermic?

From the diagram, the products (CB + A) are at a higher energy level than the reactants (AB + C). Energy is absorbed during the reaction.

The reaction is endothermic (ΔH is positive).

Section B: Worked Solutions (Attempt any THREE)

Question 20:Reaction Kinetics (10 marks)

Rate equation: $\text{rate}=k[\text{A}][\text{B}{]}^2$

i) Finding M, N, and P

Finding M (Experiment 2 vs Experiment 1: [A] same, [B] halved):

$$\frac{{\text{rate}}_1}{{\text{rate}}_2}=\frac{k[A][B_1{]}^2}{k[A][B_2{]}^2}={\left(\frac{6.60}{3.30}\right)}^2=4$$ $$M=\frac{10.4\times {10}^{-3}}{4}=\mathbf{2.6}\times {\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{\text{ mol dm}}}^{\mathbf{-}\mathbf{3}}{\mathbf{\text{s}}}^{\mathbf{-}\mathbf{1}}$$

Finding N (Experiment 3 vs Experiment 1: [B] doubled, rate halved):

$$\frac{{\text{rate}}_1}{{\text{rate}}_3}=\frac{[A_1][B_1{]}^2}{[N][B_3{]}^2}$$ $$\frac{10.4\times {10}^{-3}}{5.20\times {10}^{-3}}=\frac{4.80\times {10}^{-2}\times (6.60{)}^2}{N\times (13.2{)}^2}$$ $$2=\frac{4.80\times {10}^{-2}\times 43.56}{N\times 174.24}$$ $$N=\frac{4.80\times {10}^{-2}\times 43.56}{2\times 174.24}=\frac{2.09}{348.48}=\mathbf{6.0}\times {\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{\text{ mol dm}}}^{\mathbf{-}\mathbf{3}}$$

Finding P (Experiment 4 vs Experiment 1: rate same, [A] one-third):

If rate is the same but [A] is reduced to 1.60×10⁻² (one-third of 4.80×10⁻²), then [B]² must be 3 times larger to compensate:

$$[B_4{]}^2=3\times [B_1{]}^2=3\times (6.60\times {10}^{-2}{)}^2$$ $$P=6.60\times {10}^{-2}\times \sqrt{3}=6.60\times {10}^{-2}\times 1.732=\mathbf{1.14}\times {\mathbf{10}}^{\mathbf{-}\mathbf{1}}{\mathbf{\text{ mol dm}}}^{\mathbf{-}\mathbf{3}}$$

ii) Rate constant k from Experiment 1

Answer: B) 49.73 mol⁻² dm⁶ s⁻¹

$$k=\frac{\text{rate}}{[\text{A}][\text{B}{]}^2}=\frac{10.4\times {10}^{-3}}{(4.80\times {10}^{-2})(6.60\times {10}^{-2}{)}^2}$$ $$=\frac{10.4\times {10}^{-3}}{4.80\times {10}^{-2}\times 4.356\times {10}^{-3}}$$ $$=\frac{10.4\times {10}^{-3}}{2.091\times {10}^{-4}}=\mathbf{49.73}{\mathbf{\text{ mol}}}^{\mathbf{-}\mathbf{2}}{\mathbf{\text{ dm}}}^{\mathbf{6}}{\mathbf{\text{ s}}}^{\mathbf{-}\mathbf{1}}$$

iii) Overall order of reaction

Answer: C) Third order (3)

From rate = k[A]¹[B]²: order with respect to A = 1, order with respect to B = 2. Overall order = 1 + 2 = 3.

iv) Factor that does NOT affect reaction rate

Answer: A) Colour of reactants

Colour is a physical property unrelated to reaction kinetics. Surface area, concentration, and temperature all affect how frequently and energetically reactant particles collide.

Question 21 : Transition Metals (10 marks)

i) Definition of transition metals

Answer: B) Partially filled d-orbitals

Transition metals are elements that have at least one ion with a partially filled d-subshell. This gives them properties like variable oxidation states, coloured compounds, and catalytic activity.

ii) Oxidation state of Pt in [PtCl₂F₂]²⁻

Answer: C) +2

Let Pt oxidation state = x. Cl is −1, F is −1, overall charge is −2:

$$x+2(-1)+2(-1)=-2$$ $$x-4=-2$$ $$x=+2$$

iii) Complex compound naming: matching

iv a) Fe³⁺ more paramagnetic than Fe²⁺ — True or False?

TRUE

Justification: Fe (Z=26) has electron configuration [Ar]3d⁶4s². Fe²⁺ loses 2 electrons → [Ar]3d⁶ 4 unpaired electrons. Fe³⁺ loses 3 electrons → [Ar]3d⁵ — 5 unpaired electrons (half-filled d-subshell, one electron in each orbital). Paramagnetism increases with the number of unpaired electrons. Fe³⁺ has more unpaired electrons (5 > 4), so it is more paramagnetic.

iv b) Correct electron configuration for Cr³⁺ is 1s²2s²2p⁶3s²3p⁶3d⁴ — True or False?

FALSE

Justification: Cr (Z=24) has the anomalous configuration [Ar]3d⁵4s¹ (not [Ar]3d⁴4s² as expected) because a half-filled 3d subshell is extra stable. Cr³⁺ loses 3 electrons — first the 4s electron, then two 3d electrons: [Ar]3d³, which expands to 1s²2s²2p⁶3s²3p⁶3d³. The configuration given in the question (3d⁴) is incorrect — it would require losing only 2 electrons from a 3d⁵4s¹ configuration.

Question 22: Electrochemistry (10 marks)

i) Equation when iron filings are added to copper(II) solution

Answer: B) Fe(s) + Cu²⁺(aq) → Cu(s) + Fe²⁺(aq)

From the table: E°(Cu²⁺/Cu) = +0.34 V; E°(Fe²⁺/Fe) = −0.44 V. A more positive reduction potential means Cu²⁺ is more readily reduced. Iron has the more negative potential — it is the better reducing agent. Iron is oxidised (Fe → Fe²⁺ + 2e⁻) while copper ions are reduced (Cu²⁺ + 2e⁻ → Cu). Iron displaces copper from solution.

ii) Most oxidising reaction

Answer: A) NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ → NO(g) + 2H₂O(l)

The most oxidising agent is the one most easily reduced — the half-reaction with the most positive standard reduction potential. NO₃⁻/NO has E° = +0.96 V, the highest in the table. A strong oxidising agent accepts electrons readily.

iii) Most reducing electrode

Answer: D) Zn²⁺(aq) + 2e⁻ ⇌ Zn(s)

The strongest reducing agent is the species most easily oxidised — the half-reaction with the most negative standard reduction potential. Zn has E° = −0.76 V, the most negative in the table. Zn metal gives up electrons most readily.

iv) EMF of the iron-zinc cell (Fe-Zn)

Answer: D) +0.32 V

In a Fe-Zn cell, zinc is the anode (oxidised, E° = −0.76 V) and iron is the cathode (reduced, E° = −0.44 V):

$${\text{E°}}_{cell}={\text{E°}}_{cathode}-{\text{E°}}_{anode}=(-0.44)-(-0.76)=+0.32\text{ V}$$

v a) Can copper protect iron from corrosion?

FALSE

Justification: For cathodic protection, the protective metal must be more reactive (lower reduction potential) than iron, so it acts as a sacrificial anode and preferentially corrodes instead. Copper has a higher reduction potential (+0.34 V) than iron (−0.44 V), meaning copper is less reactive. If copper is in contact with iron, iron will corrode faster — copper would actually accelerate corrosion of iron by acting as a cathode in a galvanic cell. Metals like zinc or magnesium protect iron; copper does not.

v b) Can the SHE measure standard reduction potentials of other substances?

TRUE

Justification: The Standard Hydrogen Electrode (SHE) is assigned E° = 0.00 V by convention. To measure the standard reduction potential of another electrode, it is connected to the SHE to form a cell. The measured cell voltage equals the E° of the test electrode directly (since SHE = 0). If the test electrode acts as cathode, its E° is positive; if it acts as anode, its E° is negative. This is precisely how all standard reduction potentials in electrochemistry tables are determined.

Question 23: Group 17 Elements (10 marks)

i) What is a diatomic molecule?

A diatomic molecule is a molecule consisting of exactly two atoms bonded together. These atoms may be of the same element (homodiatomic, e.g. F₂, Cl₂, Br₂, I₂) or different elements (heterodiatomic, e.g. HCl, HF). All Group 17 elements exist as diatomic molecules $X_2$ under standard conditions.

ii) Why does volatility decrease down Group 17?

Volatility is the tendency of a substance to vaporise — lower boiling point means higher volatility. Down Group 17, molecular mass increases: F₂ (38) → Cl₂ (71) → Br₂ (160) → I₂ (254). Larger molecules have more electrons, leading to stronger London dispersion (van der Waals) intermolecular forces. More energy is needed to overcome these forces during vaporisation, so boiling points increase and volatility decreases down the group.

iii) Why does reducing power of hydrogen halides decrease: HI > HBr > HCl > HF?

A reducing agent donates electrons — this requires breaking the H–X bond. Reducing power depends on how easily the H–X bond breaks (bond enthalpy). Down the group, the halide ion X⁻ becomes larger, and the H–X bond length increases, making the bond weaker and easier to break:

• HF: very strong, short bond (568 kJ/mol) — very poor reducing agent
• HCl: 432 kJ/mol
• HBr: 366 kJ/mol
• HI: weakest bond (297 kJ/mol) — easiest to break — strongest reducing agent

Therefore reducing power follows: HI > HBr > HCl > HF.

iv) Chemical equation for bleach formation — Cl₂ with water

$${\text{Cl}}_2(g)+{\text{H}}_2\text{O}(l)\rightleftharpoons \text{HCl}(aq)+\text{HClO}(aq)$$

The bleaching agent formed is hypochlorous acid (HClO / HOCl). It releases nascent oxygen [O] which oxidises and destroys coloured compounds in fabrics.

v) Showing Cl₂ + 2NaOH → NaCl + NaClO + H₂O is a disproportionation reaction

In disproportionation, the same element is simultaneously oxidised and reduced. Assign oxidation states to chlorine:

• Cl₂: oxidation state of Cl = 0
• NaCl: oxidation state of Cl = −1 (reduced)
• NaClO: oxidation state of Cl = +1 (oxidised)

Chlorine starts at 0 and goes to both −1 and +1 in the same reaction. One Cl atom is reduced (0 → −1) and another is oxidised (0 → +1). Since the same element undergoes both oxidation and reduction simultaneously, this is a disproportionation reaction. 

Question 24: Benzene Reactions (10 marks)

i a) Bromination of benzene to give bromobenzene + HBr

Reagents and conditions:

• Reagent: Br₂ (liquid bromine)
• Catalyst: anhydrous FeBr₃ or anhydrous AlBr₃ (halogen carrier / Lewis acid catalyst)
• Conditions: room temperature, dry conditions (no water/UV light)
• Mechanism: Electrophilic aromatic substitution (EAS)

$${\text{C}}_6{\text{H}}_6+{\text{Br}}_2\stackrel{{\text{FeBr}}_3}{\to }{\text{C}}_6{\text{H}}_5\text{Br}+\text{HBr}$$

i b) Friedel-Crafts alkylation of benzene: benzene → methylbenzene + CH₂Cl side chain giving ethylbenzene product + HCl

Reagents and conditions:

• Reagent: chloromethane (CH₃Cl) or chloroethane (CH₃CH₂Cl) depending on the required alkyl group
• Catalyst: anhydrous AlCl₃ (Lewis acid)
• Conditions: dry conditions, room temperature or gentle warming
• Mechanism: Friedel-Crafts alkylation (electrophilic aromatic substitution)

$${\text{C}}_6{\text{H}}_6+{\text{CH}}_3{\text{CH}}_2\text{Cl}\stackrel{{\text{AlCl}}_3}{\to }{\text{C}}_6{\text{H}}_5{\text{CH}}_2{\text{CH}}_3+\text{HCl}$$

ii a) Converting ethylbenzene to acetophenone (phenyl methyl ketone)

Target: C₆H₅CH₂CH₃ → C₆H₅COCH₃

Step 1: Side-chain oxidation using acidified KMnO₄ (conc.) or CrO₃/H₂SO₄ would over-oxidise. Better route:

Step 1: Halogenation of side chain: C₆H₅CH₂CH₃ + Cl₂ (UV light, no catalyst) → C₆H₅CHClCH₃ (1-chloro-1-phenylethane)

Step 2: Hydrolysis: C₆H₅CHClCH₃ + NaOH(aq)/heat → C₆H₅CH(OH)CH₃ (1-phenylethan-1-ol)

Step 3: Oxidation: C₆H₅CH(OH)CH₃ + acidified K₂Cr₂O₇ (or KMnO₄) → C₆H₅COCH₃ (acetophenone)

ii b) Converting benzene to cumene (isopropylbenzene — C₆H₅CH(CH₃)₂)

Step 1: Friedel-Crafts alkylation:

$${\text{C}}_6{\text{H}}_6+({\text{CH}}_3{)}_2\text{CHCl}\stackrel{{\text{AlCl}}_3}{\to }{\text{C}}_6{\text{H}}_5\text{CH}({\text{CH}}_3{)}_2+\text{HCl}$$

Alternatively using propene + HF catalyst (industrial route), but for exam purposes, use chloro-2-methylpropane (2-chloropropane) with AlCl₃ in a single Friedel-Crafts step.

Need More Help?

Chemistry II covers a wide range of topics  organic reactions, electrochemistry, kinetics, thermodynamics, and transition metals all in one paper. If any section above is not yet fully clear, working through it with a tutor is the fastest way to close the gap.

Book a Chemistry tutor on Mathrone Academy  online or at-home in Kigali for personalised past paper practice or text us via Whatsapp  +250786684285 . Also check the 2026 national exams study guide to plan your full revision schedule, and the study techniques guide for Rwandan students for smarter exam preparation strategies.

Frequently Asked Questions

Which combinations sit S6 Chemistry II (014) in Rwanda?

Chemistry II paper 014 is sat by students in five combinations: BCG (Biology-Chemistry-Geography), MCB (Mathematics-Chemistry-Biology), PCB (Physics-Chemistry-Biology), PCM (Physics-Chemistry-Mathematics), and ANP (Associate Nursing Program). All combinations use the same paper.

What topics are covered in S6 Chemistry II?

The 2025 paper covered atomic structure and isotopes, Group 15 elements and fertilisers, organic reactions (elimination, addition, substitution, esterification), colligative properties, Group 17 hydrides, base strength and Kb values, electrolysis, nuclear equations, periodic trends, phase diagrams, solubility products, Hess's law and enthalpy calculations, reaction kinetics and rate equations, transition metals, electrochemistry (standard reduction potentials, EMF), and aromatic chemistry (benzene reactions).

What is Zaitsev's rule and when does it apply?

Zaitsev's rule states that in elimination reactions, the major product is the most substituted alkene — the one with more alkyl groups on the double bond carbons. It applies when an alcohol undergoes acid-catalysed dehydration or when a halogenoalkane reacts with a strong base (like KOH/alcohol) under heat. It does not apply to Hofmann elimination from ammonium salts.

How do you calculate molar solubility from Ksp?

Write the dissolution equation and define S as the molar solubility. Express ion concentrations in terms of S, substitute into the Ksp expression, and solve. For a 1:1 salt like PbSO₄: Ksp = S², so $S=\sqrt{K_{sp}}$. For a 1:2 salt like PbCl₂: Ksp = [Pb²⁺][Cl⁻]² = S(2S)² = 4S³, so $S=\sqrt[3]{K_{sp}/4}$.

What is the difference between SN1 and SN2 mechanisms?

SN1 (unimolecular) is a two-step mechanism: the leaving group departs first to form a carbocation, then the nucleophile attacks. It is favoured by tertiary halides and polar protic solvents. SN2 (bimolecular) is a one-step mechanism where the nucleophile attacks at the same time as the leaving group departs — it is favoured by primary halides and gives inversion of configuration (Walden inversion).

Why is HF anomalous in Group 17 boiling points?

HF has a much higher boiling point (+19.5°C) than expected from its low molecular mass. This is because fluorine is the most electronegative element, creating a very polar H–F bond. Strong intermolecular hydrogen bonds form between the δ+ H of one molecule and the lone pair on the δ− F of another. These hydrogen bonds require much more energy to break than the van der Waals forces in HCl, HBr, and HI.

Where can I find more S6 past papers with worked solutions?

Mathrone Academy publishes worked solutions for S6 past papers across all subjects. See the S6 Mathematics II 2025 worked solutions, the S6 Physics 2025 worked solutions, and the S6 Computer Science 2025 worked solutions. You can also book a tutor for live past paper sessions.

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