NJC J2 H2 Chemistry, 2024 A-Level Paper 2, Questions 1–3
Difficulty: Intermediate to Advanced | Prerequisites: Atomic structure, basic thermodynamics, organic functional groups.
Big picture: These three questions span physical and organic chemistry. Question 1 tests energetics through the lens of SrCl₂ (lattice energy, hydration, entropy, Born-Haber cycles). Question 2 covers the gaseous state, focusing on ideal vs. non-ideal gas behaviour and intermolecular forces. Question 3 bridges organic chemistry with transition metal chemistry, asking you to determine molecular formulae from gas data, classify alcohols via oxidation reactions, and explain d-d electronic transitions in chromium ions. If you are comfortable with Hess's law cycles and can draw mechanisms for alcohol reactions, you are in good shape.
Paper 2 Questions 1–3 test your ability to calculate lattice energies and hydration enthalpies from Born-Haber data, explain entropy changes for dissolving, apply the ideal gas equation to real problems (including deviations at high pressure), and identify alcohols through their oxidation products. You also need to explain why transition metal ions are coloured, using crystal field theory.
Lattice energy (LE)
The energy released when gaseous ions combine to form one mole of an ionic solid, under standard conditions. Think of it as a measure of how strongly the ions grip each other in the crystal.
Enthalpy change of hydration (ΔH_hyd)
The enthalpy change when one mole of gaseous ions is surrounded by water molecules to form an aqueous solution. In simple terms, it is the energy released when water molecules cluster around an ion.
Ion-dipole interaction
The electrostatic attraction between a charged ion and the partial charge on a polar molecule (here, water). This is the force responsible for dissolving ionic compounds in water.
Entropy (ΔS)
A measure of the disorder or number of ways energy can be distributed in a system. When a solid dissolves, particles disperse into solution, typically increasing entropy.
Thermal decomposition
The breakdown of a compound by heating. For Group 2 carbonates, smaller cations (higher charge density) polarise the carbonate ion more, making decomposition easier.
Ideal gas
A theoretical gas whose particles have negligible volume and no intermolecular forces. It obeys pV = nRT perfectly at all conditions.
Non-ideal (real) gas behaviour
Deviations from the ideal gas law that occur at high pressures (particle volume matters) and low temperatures (intermolecular attractions matter).
pV/RT ratio
A dimensionless quantity used to assess deviation from ideal behaviour. For an ideal gas, pV/nRT = 1 at all pressures.
Molecular formula
The actual number of each type of atom in one molecule of a compound, determined here by applying pV = nRT to gas-phase data.
Iodoform reaction (CHI₃ test)
A test for methyl ketones and secondary alcohols of the form CH₃CH(OH)R. Produces a pale yellow precipitate of triiodomethane (CHI₃).
Dichromate oxidation (Cr₂O₇²⁻ / H⁺)
An oxidising agent that turns from orange to green (Cr³⁺) when it oxidises primary and secondary alcohols. Tertiary alcohols show no reaction.
d-d electronic transition
The absorption of visible light that promotes an electron from a lower-energy d orbital to a higher-energy d orbital in a transition metal complex, producing colour.
Group 2 carbonates (CaCO₃, SrCO₃) decompose on heating: MCO₃ → MO + CO₂
Thermal stability increases down the group (BeCO₃ < MgCO₃ < CaCO₃ < SrCO₃ < BaCO₃).
Explanation: smaller cations have higher charge density, which polarises the carbonate anion more strongly, weakening the C–O bond and making decomposition easier.
CaCO₃ decomposes at a lower temperature than SrCO₃ because Ca²⁺ is smaller and has higher charge density than Sr²⁺.
When SrCl₂ dissolves, Sr²⁺ and Cl⁻ ions interact with water molecules via ion-dipole forces.
Sr²⁺ attracts the δ⁻ oxygen end of water; Cl⁻ attracts the δ⁺ hydrogen end.
Diagram should show: the ion at the centre, water molecules oriented with the correct partial charge facing the ion, and arrows or labels for the electrostatic attraction.
Dissolving SrCl₂(s) → Sr²⁺(aq) + 2Cl⁻(aq)
Overall entropy change has two competing contributions:
Positive: the solid lattice breaks apart, increasing disorder as ions disperse into solution.
Negative: water molecules become ordered around each ion (hydration shells), reducing their freedom.
Net effect for SrCl₂ dissolving: the increase in disorder from breaking the lattice outweighs the ordering of water, so ΔS is positive overall.
ΔH_hyd is the enthalpy change when 1 mol of gaseous ions becomes hydrated.
Always exothermic (negative) because ion-dipole bonds form.
Magnitude increases with higher charge density (smaller ionic radius and/or higher charge).
Use a Hess's law cycle to calculate lattice energy of SrCl₂ from:
Enthalpy of formation of SrCl₂
Enthalpy of atomisation of Sr
First and second ionisation energies of Sr
Enthalpy of atomisation of Cl₂ (bond energy of Cl–Cl, halved per Cl)
Electron affinity of Cl (×2 for two Cl⁻ ions)
Remember: lattice energy is exothermic (negative by the thermodynamic convention, or a positive magnitude if defined as energy to break the lattice apart).
Enthalpy change of hydration can also be derived from the cycle: ΔH_hyd = ΔH_solution + LE (signs depend on convention used).
For a salt to dissolve, the hydration enthalpy must be sufficient to overcome the lattice energy.
CaF₂ has a very high lattice energy because both Ca²⁺ and F⁻ are small ions with high charge density, giving strong electrostatic attraction in the lattice.
The hydration enthalpies of Ca²⁺ and 2F⁻ are not large enough to compensate, so dissolution is not energetically favourable.
Particles have negligible volume compared to the volume of the container.
No intermolecular forces between particles (no attraction or repulsion).
Collisions between particles (and with the container walls) are perfectly elastic: kinetic energy is conserved.
Duration of collisions is negligible compared to time between collisions.
For one mole of an ideal gas, pV/RT = 1 at all pressures.
The graph is a horizontal line at y = 1, regardless of pressure.
At very high pressures, the volume of N₂ molecules is no longer negligible relative to the container volume.
The actual volume available for molecular motion is less than the measured container volume.
This means the effective molar volume is larger than predicted, so pV/nRT > 1.
At moderate pressures, intermolecular attractions cause pV/nRT to dip below 1 first, but at very high pressures the volume effect dominates and the ratio rises well above 1.
N₂ is non-polar. The only intermolecular forces are instantaneous dipole-induced dipole (id-id) interactions, also called London dispersion forces.
CO is polar (small permanent dipole due to the difference in electronegativity between C and O). CO exhibits:
Permanent dipole-permanent dipole (pd-pd) interactions
Instantaneous dipole-induced dipole (id-id) interactions
N₂ and CO have very similar electron counts (both 14 electrons) and similar molecular sizes, so their id-id forces are comparable.
CO has additional pd-pd forces, so overall intermolecular forces in CO are slightly stronger than in N₂.
This means CO deviates from ideal gas behaviour slightly more than N₂ under the same conditions.
You are given a volatile compound A (an alcohol or phenol) and gas-phase data (pressure, volume, temperature, mass).
Steps:
Use pV = nRT to find n (moles of gaseous A).
Calculate molar mass: M = mass / n.
From the molar mass and given empirical formula, deduce the molecular formula.
Common pitfall: make sure temperature is in kelvin and pressure is in the correct units (Pa, atm, etc.) matching R.
B is a methyl alcohol (primary, with the structure CH₃OH or similar).
C is a tertiary alcohol.
Distinguishing tests:
Oxidation with acidified K₂Cr₂O₇: primary and secondary alcohols are oxidised (orange Cr₂O₇²⁻ turns green Cr³⁺). Tertiary alcohols show no colour change.
Iodoform test (I₂/NaOH): positive for methyl alcohols of the form CH₃CH(OH)R or methyl ketones. Gives a pale yellow precipitate of CHI₃ (triiodomethane).
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
This is the reduction half-equation. The oxidation half-equation depends on the alcohol being oxidised.
Reagent: I₂ in NaOH (or NaOI).
Observation: pale yellow precipitate of CHI₃.
Indicates the presence of a CH₃CO– group (methyl ketone) or a CH₃CH(OH)– group (methyl secondary alcohol, which is first oxidised to the methyl ketone in situ).
Cr³⁺ has a partially filled 3d subshell.
In an aqueous complex, the d orbitals split into two energy levels due to the electric field of the ligands.
Electrons absorb visible light of a specific frequency to be promoted from the lower to the upper set of d orbitals (d-d transition).
The colour observed is the complementary colour of the light absorbed.
Key conditions for colour: partially filled d orbitals and d orbital splitting by ligands.
Students often think lattice energy and enthalpy of formation are the same thing. They are not. Lattice energy refers specifically to the energy change when gaseous ions form a solid lattice, while enthalpy of formation is the energy change when one mole of a compound forms from its elements in their standard states.
Students sometimes claim that entropy always increases when a solid dissolves. In fact, the ordering of solvent molecules around ions can make the entropy change less positive (or even negative for highly charged, small ions).
A common error is writing that N₂ has no intermolecular forces at all. N₂ does have instantaneous dipole-induced dipole forces. Saying "no intermolecular forces" is an ideal gas assumption, not a real-world description.
Many students forget that tertiary alcohols do not react with acidified dichromate. They write an oxidation product anyway, which costs marks.
⚠️ Born-Haber cycle calculations are high-value (4 marks). Draw the cycle clearly and label every step. Sign errors are the most common way to lose marks.
⚠️ The pV/RT graph is a classic 1-mark question. Know the shape: dip below 1 at moderate pressure (attractions dominate), then rise above 1 at high pressure (volume dominates).
⚠️ Explaining colour of transition metal ions requires three elements: partially filled d orbitals, d orbital splitting by ligands, and absorption of visible light for d-d transitions. Missing any one of these costs marks.
⚠️ When asked to compare intermolecular forces in two molecules, state every type present in each, then compare strengths. Simply naming forces without comparing is incomplete.
⚠️ The iodoform test is frequently tested. Know the reagent (I₂/NaOH), the observation (pale yellow ppt), and which structural features give a positive result.
True or false: The lattice energy of SrCl₂ is endothermic. (False – lattice energy is exothermic; energy is released when gaseous ions form the lattice.)
Fill in the blank: For an ideal gas, the pV/nRT ratio equals ______ at all pressures. (1)
True or false: CO has stronger overall intermolecular forces than N₂ because CO has permanent dipole-permanent dipole interactions in addition to London forces. (True)
Fill in the blank: The iodoform test produces a pale yellow precipitate of ______. (CHI₃, triiodomethane)
True or false: Tertiary alcohols can be oxidised by acidified potassium dichromate. (False – tertiary alcohols resist oxidation under these conditions.)
Q: Explain why SrCO₃ has a higher decomposition temperature than CaCO₃.
A: Sr²⁺ is larger than Ca²⁺ and has lower charge density. It polarises the carbonate ion less, so the C–O bonds in SrCO₃ are less weakened. More thermal energy is needed to break them, giving a higher decomposition temperature.
Q: Draw a labelled diagram showing the ion-dipole interactions when SrCl₂ dissolves in water.
A: Show Sr²⁺ surrounded by water molecules with the oxygen (δ⁻) end pointing toward the cation. Show Cl⁻ surrounded by water molecules with the hydrogen (δ⁺) end pointing toward the anion. Label the ion-dipole attraction in each case.
Q: At very high pressures, the pV/nRT value for nitrogen is greater than 1. Explain why.
A: At very high pressures, the volume of N₂ molecules is no longer negligible compared to the container volume. The effective free volume is less than the measured volume, so the gas occupies more space than the ideal gas equation predicts, and pV/nRT exceeds 1.
Q: An unknown compound A has the molecular formula C₇H₈O. It gives a positive iodoform test. Suggest a possible structure for A.
A: A could be a methyl secondary alcohol such as 1-phenylethan-1-ol (C₆H₅CH(OH)CH₃), which contains the CH₃CH(OH)– group required for a positive iodoform test.
Q: Explain why Cr³⁺(aq) is coloured but Zn²⁺(aq) is colourless.
A: Cr³⁺ has a partially filled 3d subshell (3d³). In an aqueous complex, d orbitals split, and electrons can absorb visible light for d-d transitions, producing colour. Zn²⁺ has a fully filled 3d subshell (3d¹⁰), so no d-d transition is possible, and no visible light is absorbed.
This material connects directly to Chemical Bonding (ion-dipole forces, electronegativity, polarity), which underpins both the dissolving of ionic compounds and the comparison of intermolecular forces in gases. The Born-Haber cycle links to broader Energetics (Hess's law, enthalpy definitions), and Gibbs Free Energy (ΔG = ΔH – TΔS) brings entropy and enthalpy together to predict spontaneity of dissolving. The organic section feeds into Transition Metals (d-d transitions, colour) and Stereoisomers (Q4 tests chirality and optical activity in alcohols).
SrCl₂, strontium chloride, lattice energy, Born-Haber cycle, enthalpy of hydration, ion-dipole, entropy of dissolving, Group 2 carbonates, thermal decomposition, charge density, ideal gas, pV = nRT, non-ideal gas, real gas, pV/RT graph, intermolecular forces, London dispersion forces, permanent dipole, N₂ vs CO, iodoform test, CHI₃, triiodomethane, dichromate oxidation, Cr₂O₇²⁻, tertiary alcohol, d-d transition, crystal field splitting, transition metal colour, H2 Chemistry, A-Level, NJC, 2024 Paper 2