OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2025: Question 16
17 marks · Medium difficulty · Structured Questions
Explore Period 3 trends including melting points, first ionisation energy, halogen displacement reactions, and molecular shapes of hydrides.
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Period 3 Trends, Halogen Displacements & Molecular Shapes
- Period 3 Physical Properties: Melting point trends governed by giant covalent vs simple molecular vs giant metallic lattices (P₄, S₈, Cl₂, Ar).
- Bonding Comparison: Energetics of breaking strong covalent bonds in giant macromolecular lattices versus metallic bonds.
- First Ionisation Energy: Writing standard formation/ionisation equations with gas phase state symbols; explaining sub-shell anomalies (P to S spin-pair repulsion).
- Group 7 Halogen Reactivity: Redox displacement reactions, associated colour changes in aqueous solution, and writing ionic equations.
- VSEPR & Molecular Geometry: Level-of-Response 6-marker on dot-and-cross drawings, electron-pair repulsion theory, shapes, and relative bond angles (SiH₄, PH₃, H₂S).
Part (a) Periodic Trends in Melting Points
Part (a)(i): Completing the Melting Point Graph for P, S, Cl, and Ar [2 marks]
✅ Mark Scheme Requirements
- Mark 1: Relative heights: S > P > Cl > Ar .
- Mark 2: All four plotted points (or connecting lines) must be distinctly below the level of Mg (< 923 K, below ~900 K on the grid).
💡 Key Knowledge
These elements exist as simple non-metal molecules:
- Phosphorus: exists as P₄
- Sulfur: exists as S₈ (largest molecule, most electrons, strongest London dispersion forces → highest melting point of the four)
- Chlorine: exists as Cl₂
- Argon: monoatomic Ar (weakest London forces → lowest melting point)
❌ Common Errors
- Plotting sulfur lower than phosphorus by confusing the melting point trend with the ionisation energy dip.
- Drawing sulfur higher than Mg or Al. P, S, Cl, and Ar are simple molecular structures with weak London forces, so their melting points are all vastly lower than giant structures.
🧠 Exam Technique
Always write down molecular formulae (P₄, S₈, Cl₂, Ar) on the paper when deciding relative melting points. The number of electrons determines the magnitude of induced dipole-dipole interactions!
Part (a)(ii): Why Silicon has a Higher Melting Point than Aluminium [3 marks]
✅ Model Answer
Aluminium has a giant metallic lattice, whereas silicon has a giant covalent (macromolecular) lattice.
Aluminium contains metallic bonding (electrostatic attraction between Al³⁺ cations and delocalised electrons), while silicon contains covalent bonds.
The covalent bonds in Si are stronger than the metallic bonds in Al, so more energy is needed to break the covalent bonds.
🧠 Structure & Bonding Comparison Strategy
- Step 1 (Structure): Identify both lattice types as giant (giant metallic vs giant covalent).
- Step 2 (Bonding): Explicitly name the bonds broken during melting (metallic bonds vs covalent bonds).
- Step 3 (Energy Comparison): State that the covalent bonds are stronger and require more energy to break than the metallic bonds.
❌ Examiner Contradiction Warning
Fatal error: Mentioning "intermolecular forces" or "van der Waals forces" in relation to either silicon or aluminium is a direct contradiction (CON on the mark scheme) and will cost you marks. Neither substance is simple molecular!
Part (b) First Ionisation Energy
Part (b)(i): Equation for First Ionisation Energy of Sulfur [1 mark]
✅ Correct Equation
S(g) → S⁺(g) + e⁻
🧠 Exam Technique: State Symbols
State symbols on S(g) and S⁺(g) are mandatory. Ionisation energy is always defined for one mole of gaseous atoms to form one mole of gaseous 1+ ions.
Part (b)(ii): Why First IE of S is Lower than P [2 marks]
✅ Model Answer
- Sulfur has a pair of electrons in a (3p) orbital (whereas phosphorus has singly occupied 3p orbitals).
- Spin-pair repulsion between the paired electrons in sulfur means less energy is required to remove one electron.
💡 Electronic Configurations
- P: 1s² 2s² 2p⁶ 3s² 3p³ → [↑ ][↑ ][↑ ] (half-filled sub-shell, stable, no paired 3p electrons)
- S: 1s² 2s² 2p⁶ 3s² 3p⁴ → [↑↓][↑ ][↑ ] (one paired 3p orbital)
❌ Common Misconceptions
- Vague terminology: Saying "in the shell" or "outer shell" loses Mark 1. You must refer to an orbital or sub-shell.
- Wrong sub-shell: Referring to the 3s orbital instead of 3p.
- Irrelevant points: Explaining that sulfur has more protons or greater shielding does not explain this dip. Nuclear charge actually increases, so referencing radius or distance from nucleus alone gains no credit.
Part (c) Halogen Displacement Reactions [3 marks]
✅ Table of Observations & Ionic Equation
| Solutions | Observations |
|---|---|
| Br₂(aq) + NaI(aq) | (Orange solution) turns brown (or yellow/brown) |
| Br₂(aq) + NaCl(aq) | Stays orange / no colour change (do NOT write "turns orange") |
Br₂ + 2I⁻ → 2Br⁻ + I₂
❌ Observation Traps
- Do NOT write "purple": Iodine in water is brown (or yellow/brown). Purple is only observed when an organic solvent such as cyclohexane is added!
- NaCl observation: The solution was orange at the start because Br₂(aq) is orange. Therefore, write "stays orange" or "no colour change". Writing "turns orange" implies a reaction produced bromine, which is incorrect.
🧠 Writing Ionic Equations
Spectator ions ( Na⁺ ) must be omitted. Always check both atom balance and charge balance: 2 negative charges on the left ( 2I⁻ ) balance 2 negative charges on the right ( 2Br⁻ ).
Part (d)* Extended Response: Molecular Shapes of SiH₄, PH₃, and H₂S [6 marks]
✅ Complete Indicative Content
| Molecule | Electron Pairs on Central Atom | Name of Shape | Bond Angle |
|---|---|---|---|
| SiH₄ | 4 bonding pairs, 0 lone pairs | Tetrahedral | 109.5° |
| PH₃ | 3 bonding pairs, 1 lone pair | Pyramidal (trigonal pyramidal) | < 109.5° (around 93°–107°) |
| H₂S | 2 bonding pairs, 2 lone pairs | Non-linear / bent / V-shaped | < PH₃ (around 92°–104.5°) |
- SiH₄: Central Si with 4 single bonds to 4 H atoms. Each bond has one dot and one cross ( Si•×H ). No lone pairs on Si (total 8 valence electrons).
- PH₃: Central P with 3 single covalent bonds to 3 H atoms ( P•×H ), plus one lone pair of electrons ( •• ) on P.
- H₂S: Central S with 2 single covalent bonds to 2 H atoms ( S•×H ), plus two lone pairs of electrons ( •• and •• ) on S.
💡 VSEPR Explanation & Bond Angle Trend
- Trend: SiH₄ > PH₃ > H₂S
- Core Principle: Electron pairs repel each other to get as far apart as possible to minimise repulsion.
- Repulsion Hierarchy: Lone pairs repel more strongly than bonding pairs.
- As the number of lone pairs increases from 0 (SiH₄) to 1 (PH₃) to 2 (H₂S), the lone pairs push the bonding pairs closer together, progressively decreasing the bond angle.
🧠 Securing Level 3 (5–6 Marks)
Examiners look for a fully structured answer covering all 4 components:
- All 3 dot-and-cross diagrams accurate.
- All 3 shape names correct (Tetrahedral, Pyramidal, Bent).
- Correct bond angle order explicitly stated: SiH₄ > PH₃ > H₂S .
- VSEPR theory explained comparing lone pair vs bonding pair repulsions.
❌ What Cost Students Marks
- Saying "atoms repel": NEVER say atoms repel; it is strictly electron pairs (or electron regions) that repel.
- Calling H₂S "planar" or "straight": Must be named non-linear, bent, or V-shaped.
- Forgetting the lone pairs in the dot-and-cross diagrams: Leaving off the lone pairs on P or S immediately caps your mark.
• Level 3 (5–6 marks): All 3 dot-and-cross diagrams correct + all 3 shapes named + correct trend of bond angles + complete explanation of LP > BP repulsion.
• Level 2 (3–4 marks): 2 molecules fully correct (diagram, shape, repulsion) OR all 3 molecules attempted across categories.
• Level 1 (1–2 marks): 1 molecule correct OR partial points across two molecules.
Topics
Module 2: Foundations in chemistry · Module 3: Periodic table and energy · 2.2 Electrons, bonding and structure · 3.1 The periodic table
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.