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Module 1 · L21 of 21 45 min ⚡ +50 XP in Learn · +25 to complete Year 11 · Module 1 · IQ1–IQ4

Module 1 Synthesis and Review

Today's hook, Every question in this module comes back to one idea: a substance behaves the way it does because of its structure and its bonding. Here is how it all connects.
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Warm up and recall

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions
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Recall, your gut answer first
+5 XP warm-up

Write your initial answer before reading on.

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What you'll master, and the words for it

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What you'll master
Know

Key facts

  • Module 1 has four Inquiry Questions: IQ1 (properties, classification and separation), IQ2 (atomic structure and mass), IQ3 (periodicity), and IQ4 (bonding and structure → property links).
  • The structure → bonding → properties → separation master chain connects all Module 1 concepts into one logical framework.
Understand

Concepts

  • Atomic structure (proton number, electron configuration) determines bonding type, which dictates physical properties and appropriate separation techniques.
  • Periodic trends (radius, IE, EN) arise from effective nuclear charge and atomic radius; elements in the same group have identical valence electron counts.
Can do

Skills

  • Predict properties of unknown substances from given data and classify them as ionic, covalent molecular, covalent network, or metallic.
  • Construct coherent short-answer responses using the classification-first approach, linking structure to bonding to properties.
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Key terms
Pure substance
A single chemical substance (element or compound) with a uniform, fixed composition; its particles may be atoms, molecules, ions or a network.
Homogeneous mixture
A mixture with uniform composition throughout, appearing as a single phase.
Heterogeneous mixture
A mixture with non-uniform composition and visibly distinct regions.
Element
A pure substance that cannot be broken down into simpler substances chemically.
Compound
A pure substance formed from two or more elements chemically bonded in fixed ratio.
Filtration
A separation technique using a porous barrier to separate insoluble solids from liquids.
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Structure → Bonding → Properties: The Master Chain

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Structure → Bonding → Properties: The Master Chain
core concept

Every topic in Module 1 belongs to this chain. This is the most important thing to understand as you head into exams:

How atomic structure leads through electron configuration, bond type, physical properties and finally separation and purification techniques.
The Central Idea: You can predict almost any property of a substance if you know its atomic structure and how its atoms are bonded. And you can work backward, from observed properties, determine what the structure and bonding must be. This bidirectional reasoning is what chemistry exam questions test.

Module 1 follows a single chain: atomic structure → electron configuration → bond type → physical properties → separation technique. Knowing the structure predicts all properties and the right purification method. This chain connects IQ1 (classification and separation), IQ2 (bonding and properties), and IQ3 (atomic structure and trends) into one unified framework.

Pause, copy the highlighted master chain into your book before moving on.

Cloze: drag the terms to complete the Module 1 master chain.

atomic structure bond type physical properties separation technique

In Module 1, ___ determines the electron configuration and electronegativity of an atom. Together these decide the ___ a substance forms (ionic, covalent or metallic), which in turn fixes its ___ such as melting point, conductivity and solubility. Those properties are then used to choose the most appropriate ___ for purifying or analysing the substance.

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Master Example: Diamond vs Graphite, Same Element, Different Everything

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Master Example: Diamond vs Graphite, Same Element, Different Everything
core concept
PropertyDiamondGraphiteStructural reason
HardnessHardest natural substanceSoft, slipperyDiamond: each C covalently bonded to 4 others in 3D tetrahedral network, must break covalent bonds to deform. Graphite: layers of hexagonal sheets, only weak dispersion forces between layers → layers slide past each other easily.
Electrical conductivityInsulatorGood conductor (in layer direction)Diamond: all 4 valence electrons bonded → no delocalised electrons. Graphite: each C uses 3 bonds in plane, leaving 1 electron per C delocalised across the entire layer → mobile π-electron system → conducts electricity.
Melting point>3500°C (sublimes)~3600°C (also very high)Both require breaking C–C covalent bonds to melt (very strong). Despite different structures, both have extremely high MP because covalent network bonding must be overcome.
Thermal conductivityExceptionally highAnisotropic (high in plane, low between layers)Diamond: rigid 3D covalent network conducts heat through phonons efficiently in all directions. Graphite: efficient in-plane but poor between layers (only weak IMFs).
AppearanceTransparent, clearBlack, opaqueDiamond: large HOMO-LUMO gap (no visible light absorption). Graphite: delocalised electrons absorb all visible wavelengths → black.
Density3.51 g/cm³2.09–2.23 g/cm³Diamond: denser 3D packing of C atoms. Graphite: layered structure with spacing between layers → less dense.
Why this example is powerful: Diamond and graphite have the same Z (6), same atomic mass, and same electron configuration ([He]2s²2p²). The only difference is how the atoms are arranged, the macroscopic properties differ completely because the structural arrangement of atoms is different. This is the purest demonstration that structure determines properties.

We just saw the master structure → property → separation chain that unifies Module 1. That raises a question: what is the most striking example that illustrates how the same element can have completely opposite properties depending on atomic arrangement? This card answers it → diamond and graphite, both pure carbon, are the canonical example of structure determining properties.

Diamond vs graphite, same element (C), different structure, opposite properties. Diamond: 3D covalent network (4 bonds per C) → hardest natural substance, non-conducting, transparent, MP 3550°C. Graphite: layered sheets (3 bonds per C + 1 delocalised electron) → soft (weak dispersion between layers), conducts in-plane, opaque. Allotropism: same element + different atomic arrangement → completely different properties.

Add the highlighted diamond vs graphite comparison to your notes before the check below.

True or false: diamond and graphite differ in hardness and conductivity even though both are pure carbon, because their atoms are arranged in different bonded structures.

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Full Module 1 Concept Map

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Full Module 1 Concept Map
core concept
Each idea determines the next — from matter's classification through structure and bonding to properties and separation.

We just saw that diamond and graphite demonstrate structure determining properties. That raises a question: stepping back across the whole module, how do the four inquiry questions connect and what are the key ideas within each? This card answers it → IQ1 is about classifying and separating; IQ2 is about atomic structure and mass; IQ3 is about periodicity (patterns across the table); IQ4 is about bonding and how structure gives rise to properties; all four connect through the structure → property chain.

How structure determines bonding, which determines properties, which determines the separation technique.

IQ1, Classification and Separation

Matter → pure substances (elements/compounds) or mixtures (homogeneous/heterogeneous). Separation techniques match to properties: filtration (particle size), distillation (BP difference), crystallisation (solubility/temperature), chromatography (differential distribution), gravimetric analysis (mass of precipitate). Technique choice depends on the physical property difference being exploited.

IQ2, Atomic Structure and Atomic Mass

Atomic models: Dalton → Thomson → Rutherford (Geiger–Marsden gold foil) → Bohr (emission spectra) → quantum mechanical. Subatomic particles: proton, neutron, electron. Isotopes: same Z, different A, very similar chemistry. Ar = Σ(mass × fraction). Electron configuration: Aufbau, Pauli, Hund's. Subshells: 1s 2s 2p 3s 3p 4s 3d 4p.

IQ3, Periodicity

The periodic table is organised by atomic number into periods and groups. Periodic trends: across a period → ↑Z_eff → ↓radius, ↑IE, ↑EN; down a group → ↑shells → ↑radius, ↓IE, ↓EN. Valence electrons help determine ion charge and reactivity. Trends are patterns with exceptions, not exact laws.

IQ4, Bonding, Structure and Properties

Bonding type largely determines properties. Ionic: strong electrostatic forces, often high MP/BP, conducts when molten/dissolved, brittle. Covalent molecular: weak IMFs (dispersion, dipole-dipole, H-bonds), low MP/BP, poor conductor, solubility depends on polarity. Covalent network: covalent bonds throughout, very high MP, insoluble, hard. Metallic: delocalised electron sea, generally conductive, malleable/ductile, variable MP. IMF hierarchy (per contact): H-bond > dipole-dipole > dispersion. Polymers: addition vs condensation; thermoplastic vs thermosetting. Solubility: like dissolves like is a guide, polarity compatibility and entropy determine dissolution.

IQ1: classification of matter + separation techniques (filtration, distillation, crystallisation, chromatography, gravimetric). IQ2: atomic structure, subatomic particles, isotopes and relative atomic mass, electron configuration (Aufbau/Pauli/Hund's). IQ3: periodicity, the periodic table and its trends (radius, IE, electronegativity, reactivity). IQ4: bonding types (ionic, covalent molecular, covalent network, metallic), intermolecular forces, polymers and structure → property links. Start a property prediction by identifying the bonding type (IQ4), and remember that atomic structure (IQ2) and periodicity (IQ3) explain why those bonds form.

Pause, write the highlighted IQ summary into your book.

Lock-in task: in one sentence, explain how an element's position on the periodic table (group + period) lets you predict (a) its bonding behaviour and (b) the relative strength of its bonds.

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Module 1 Exam Strategy

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Module 1 Exam Strategy
core concept

The most common Module 1 question types and how to approach them:

Key approach
Identify bond/IMF type → state its relative strength → link to energy needed to overcome it
Identify whether each has mobile charge carriers (delocalised electrons or free ions)
Compare atomic radius and Z_eff → link to ease of electron removal
State the direction of change → explain Z_eff/shielding/radius → link to the property
Check for C=C (addition) or ester/amide links + byproduct (condensation)
Critical phrases to include
"Strong/weak [bond type]... requires large/small energy to overcome... therefore high/low MP"
"[Substance] has/has no delocalised electrons/free ions... therefore conducts/does not conduct"
"X has a larger atomic radius / lower Z_eff... valence electron is further from nucleus / less attracted... lower energy required"
Across period: "Z_eff increases, shielding constant..." Down group: "New shell added, radius increases, more shielded..."
Addition: "C=C opens, no byproduct." Condensation: "ester/amide linkage formed, H₂O released"

We just saw how the three IQs form a unified framework. That raises a question: faced with any Module 1 exam question, what systematic approach guarantees you cover the examiners' marking criteria? This card answers it → five-step module strategy: classify → link bond type to energy → name charge carrier → explain trends with Z_eff → handle allotrope/isomer exceptions.

Module 1 exam strategy: (1) Always start a property-prediction SAQ with a one-line bonding classification. (2) Link bond type to the energy required to overcome it (explains MP and BP). (3) For conductivity, explicitly name the mobile charge carrier and whether it is free. (4) For periodic trends, explain using Z_eff, shielding, and shell count, never just state the trend. (5) For C₆₀ (buckminsterfullerene): covalent molecular (discrete C₆₀ units, weak dispersion forces between them) → low MP, soluble in non-polar solvents.

Add the highlighted exam strategy to your notes before the check below.

Fill the blanks: drag each token into the matching blank, the standard structure for a periodic-trend SAQ.

Zeff atomic radius shielding direction of change

Across a period, state the ___, then explain using ___ and ___, and finally link to the ___ and the property in question.

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Short Answer, Module Integration Questions

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Short Answer, Module Integration Questions
core concept

6. Buckminsterfullerene (C₆₀, "buckyballs") consists of carbon atoms arranged in a hollow spherical cage of 60 carbon atoms (hexagons and pentagons, like a soccer ball). Each C is bonded to 3 others. (a) Classify C₆₀ as ionic, covalent molecular, covalent network, or metallic. Justify. (b) Compare the likely melting point of C₆₀ to diamond and graphite. (c) Would C₆₀ be expected to dissolve in water or in a non-polar solvent like toluene? Justify using IQ2 principles. 5 MARKS

✏️ Answer in your book

7. Extended response: A student claims: "The periodic table is just a list of elements arranged in alphabetical order of their names." Using evidence and reasoning from IQ3, evaluate this claim. In your answer: (a) identify the actual organising principle, (b) explain why this principle works to group elements with similar chemical behaviour together, and (c) give two specific examples from the periodic table that demonstrate the table's predictive power. 6 MARKS

✏️ Answer in your book

We just saw the five-step module exam strategy. That raises a question: in an integration question that spans IQ1, IQ2, and IQ3, what is the logical order to structure your written response? This card answers it → always start from structure (configuration → bonding) and move outward to properties, then to techniques or trends.

For any module integration question: (1) identify the substance's bonding type from its formula or data; (2) predict properties using the bonding model; (3) choose the separation technique that exploits the key property difference; (4) cite the relevant periodic trend if ion charge or atomic size is asked. The valence electron count determines chemical behaviour, elements in the same group show similar reactivity because their valence electron configuration is identical.

Pause, copy the highlighted integration approach into your book before moving on.

Did you get this? True or false: a strong Module 1 SAQ answer starts with a one-line classification (e.g. "ionic / covalent network / molecular") before any explanation.

Worked examples

Worked examples · reveal as you go

Worked example +5 XP on full reveal

Silicon dioxide (SiO₂, quartz) is used in glass, semiconductors, and optical fibres. Predict its: melting point, electrical conductivity, solubility in water, and hardness. Justify each from structure and bonding.

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Identify substance type: Si (Z=14, Group 14, [Ne]3s²3p²) forms 4 covalent bonds; O (Z=8, Group 16) forms 2 covalent bonds. In SiO₂: each Si bonds to 4 O atoms (tetrahedral); each O bridges 2 Si atoms. Result: 3D covalent network lattice.
This identifies the bonding structure, which determines all properties.
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Melting point: VERY HIGH (~1710°C). Why: Must break strong covalent Si–O bonds throughout the 3D network. Electrical conductivity: POOR (insulator). Why: All valence electrons locked in Si–O bonds; no delocalised electrons. Solubility: INSOLUBLE. Why: Water cannot overcome strong covalent bonds throughout lattice. Hardness: VERY HARD. Why: 3D network resists deformation in all directions.
Each property flows directly from the covalent network structure.
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Final answer: SiO₂ is a covalent network solid. MP very high (~1710°C), must break covalent Si–O bonds throughout lattice. Electrical conductivity poor, no delocalised electrons. Insoluble in water, covalent bonds cannot be disrupted by dissolution. Very hard, 3D covalent network resists all deformation.
Summarise: structure determines all properties.
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Worked example +5 XP on full reveal

An unknown substance X has these properties: MP = 801°C, dissolves in water giving a conducting solution, does not conduct electricity as a solid, brittle when struck. Identify the substance type, suggest a likely structure, and name a real example.

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Analyse each property: MP = 801°C: too high for molecular (<300°C) but lower than covalent network (>1000°C). Could be ionic. Dissolves in water with conducting solution: ionic compound (ions free to move). Does not conduct as solid: ions fixed in lattice. Brittle when struck: consistent with ionic lattice fracture (layers shift → like charges repel → shatters).
Each clue points towards ionic bonding.
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Identify substance type: All properties consistent with IONIC solid. High but not extreme MP (ionic bonding, not as strong as covalent network) → conducts when molten/dissolved (free ions). Insulates as solid (fixed ions). Brittle (ionic crystal fracture). Dissolves in polar solvent like water (ion-dipole forces compatible).
The pattern of properties matches only ionic compounds.
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Suggest real example: MP of 801°C matches NaCl (sodium chloride), which melts at 801°C. Structure: face-centred cubic lattice of Na⁺ and Cl⁻ ions alternating, held by strong electrostatic forces. Alternative examples: NaBr (748°C), LiCl (605°C).
NaCl is the most common ionic salt with this melting point.
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Misconception to fix

Common errors · the 3 traps that cost marks

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Misconception to fix

Wrong: Synthesis reactions always produce a single product from two elements.

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Misconception to fix

Right: Synthesis reactions combine two or more reactants into a single product, but the reactants need not be elements. Compounds can also combine in synthesis reactions (e.g., SO₃ + H₂O → H₂SO₄). The defining feature is one product forming from multiple reactants.

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Covalent molecular substances have low MP because covalent bonds are weak

Wrong: Covalent molecular substances like iodine or water have low melting points because the covalent bonds inside the molecules are weak.

Fix: The covalent bonds within molecules are strong. The low melting points are due to weak intermolecular forces (dispersion, dipole-dipole, hydrogen bonds) between molecules. Do not confuse intramolecular covalent bonds with intermolecular forces, they operate on completely different energy scales.

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Drill, then revisit

Quick-fire practice · 5 reps +2 XP per reveal

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State the Module 1 master chain in order from atomic structure to separation technique.

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Name the four main bonding types and give one key property of each.

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Diamond and graphite are both pure carbon. Why does diamond not conduct electricity while graphite does?

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A white solid melts at 801°C and its aqueous solution conducts electricity. Classify the bonding type and name a real example.

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Explain why atomic radius decreases across a period but increases down a group.

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Revisit your thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

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Interactive Tool, Atomic Structure Builder Open fullscreen ↗
The Atomic Structure tool shows that the atomic number of an element equals…
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Multiple choice

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next.

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Short answer
ApplyApply5 MARKS

Q1. 6. Buckminsterfullerene (C₆₀, "buckyballs") consists of carbon atoms arranged in a hollow spherical cage of 60 carbon atoms (hexagons and pentagons, like a soccer ball). Each C is bonded to 3 others. (a) Classify C₆₀ as ionic, covalent molecular, covalent network, or metallic. Justify. (b) Compare the likely melting point of C₆₀ to diamond and graphite. (c) Would C₆₀ be expected to dissolve in water or in a non-polar solvent like toluene? Justify using IQ2 principles.

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ApplyApply6 MARKS

Q2. 7. Extended response: A student claims: "The periodic table is just a list of elements arranged in alphabetical order of their names." Using evidence and reasoning from IQ3, evaluate this claim. In your answer: (a) identify the actual organising principle, (b) explain why this principle works to group elements with similar chemical behaviour together, and (c) give two specific examples from the periodic table that demonstrate the table's predictive power.

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📖 Comprehensive answers (click to reveal)

Synthesis Activity

1. Si: element, covalent network solid (each Si bonded to 4 others tetrahedrally throughout 3D lattice), very high MP (~1414°C) because must break strong Si–Si covalent bonds, poor electrical conductor (semiconductor; valence band full, small band gap), insoluble in water. NaCl: ionic compound, ionic lattice (alternating Na⁺ and Cl⁻ in face-centred cubic array), high but not extreme MP (~801°C) because strong electrostatic forces but weaker than covalent bonds, solid insulates (fixed ions), liquid/aqueous conducts (free ions), soluble in water (ion-dipole forces).

2. SiO₂ ceramic: (a) thermal conductivity, moderate (phonon conduction through rigid covalent network, lower than Cu); (b) electrical insulator ✓ (no free electrons); (c) very high MP ✓ (~1710°C); (d) very hard ✓ (3D covalent network). Cu: (a) excellent thermal conductor ✓ (delocalised electrons); (b) excellent electrical conductor ✗ (fails requirement b); (c) MP 1085°C ✓; (d) moderately hard. Best material for all 4 requirements: SiO₂ ceramic satisfies b, c, d better, though with lower thermal conductivity than Cu. Copper fails requirement (b). A ceramic (SiO₂-based) would be the better choice if all four properties are equally required.

3. C (Z=6): config [He]2s²2p² with 4 valence electrons (Group 14, Period 2). To form C⁴⁺ (ionic): would require removing all 4 valence electrons; cumulative IE₁+IE₂+IE₃+IE₄ is enormous (IE₁=1086, IE₂=2353, IE₃=4621, IE₄=6223 kJ/mol), total ~14,000 kJ/mol. No chemical reaction releases enough energy to compensate. To form C⁴⁻: gaining 4 electrons would require enormous electron-electron repulsion in a small C atom. Energetically, covalent sharing is overwhelmingly preferred. In diamond: C uses all 4 valence electrons in sp³ hybridised C–C bonds, forming a 3D tetrahedral network, each C bonded to 4 others. The result is an extremely rigid, non-conducting, transparent lattice. In graphite: C uses 3 valence electrons in sp² hybridised C–C bonds within planar hexagonal layers; the remaining 1 electron per C forms a delocalised π system across the entire layer. This explains conductivity and softness (layers slide easily). Same element, same valence electron count (4), different structural arrangements.

❓ Multiple Choice

~2800°C MP: too high for ionic (mostly 600–1000°C) or molecular (< 300°C), but consistent with covalent network. Does not conduct in any state: rules out metals (conduct) and ionic (conducts when molten). Extremely hard + insoluble: covalent network. Example: SiC (silicon carbide), Al₂O₃ (corundum).

Both EN and IE₁ generally increase left to right due to increasing Z_eff with constant shielding. Both show anomalies (IE: dips at Group 13 and 16; EN is smoother). Both decrease going down a group.

Solvent extraction: I₂ non-polar → dissolves in hexane (like dissolves like). KCl ionic → remains in water (ion-dipole forces). After shaking with hexane, I₂ partitions into hexane layer; KCl stays in water. Separate layers (hexane floats on water). Filtration won't work (both dissolved). Distillation requires large BP differences and doesn't separate solutes well. Crystallisation doesn't separate I₂ from KCl by crystal shape.

H-bonding in HF (F is highly electronegative, N–H or O–H or F–H can H-bond). HCl: Cl not electronegative enough for H-bonding → only dipole-dipole (plus dispersion). H-bonds are stronger → higher energy to overcome → higher BP for HF despite lower molar mass.

Graphene: sp² C–C bonds in hexagonal lattice → strongest bonds → exceptional strength. 3 bonds per C leave 1 electron per C delocalised as π electrons across the layer → conductor. Single atom thick → thinnest possible material → absorbs ~2.3% of visible light (quantum mechanical opacity) → nearly transparent. All properties follow directly from the sp² covalent 2D structure with delocalised electrons.

Short Answer Model Answers

Q6 (5 marks): (a) C₆₀ is a covalent molecular substance (1 mark). It consists of discrete, closed cage molecules (60 C atoms), unlike an infinite covalent network lattice (diamond, graphite). Each C is covalently bonded to 3 others within the cage; the C₆₀ molecules are held to each other only by dispersion forces (IMFs between closed-shell molecules). (b) MP of C₆₀ is much lower than both diamond and graphite (~600°C sublimation) (1 mark). Diamond and graphite require breaking C–C covalent bonds throughout their entire structure to melt (>3500°C). C₆₀ only requires overcoming the relatively weak dispersion forces between whole C₆₀ molecules, much less energy (1 mark). (c) C₆₀ is insoluble in water, the non-polar C₆₀ molecule has only dispersion forces, which are incompatible with water's H-bond network; disrupting H-bonds to accommodate C₆₀ provides no compensating interaction energy (1 mark). C₆₀ dissolves in toluene (a non-polar solvent), dispersion forces between C₆₀ and toluene are compatible; "like dissolves like", both non-polar (1 mark).

Q7 (6 marks): (a) The student's claim is incorrect. The periodic table is NOT alphabetical, it is organised by increasing atomic number (Z), where Z = number of protons in the nucleus (1 mark). Elements are arranged in 7 periods (horizontal rows) and 18 groups (vertical columns). (b) This principle works because Z determines the electron configuration of an atom. The valence electron configuration (number and type of outermost electrons) determines how atoms bond, what ions they form, and how they react. Elements in the same group have the same number of valence electrons in the same subshell type (e.g. all Group 1 elements have ns¹ valence configuration) → same bonding tendencies → similar chemistry (1 mark). Additionally, the placement by Z resolves anomalies that Mendeleev's mass-based table could not: Te (Z=52) correctly precedes I (Z=53) by Z-ordering, consistent with their chemical group memberships (Group 16 and 17 respectively) (1 mark). (c) Example 1: All Group 1 alkali metals (Li, Na, K, Rb, Cs, Fr) react with water to form M⁺(aq) + OH⁻(aq) + H₂(g). The same reaction type occurs because all have ns¹ valence configuration and a single electron to lose. The vigour increases down the group as atomic radius increases and IE₁ decreases, but the reaction type is the same, a direct consequence of the shared valence configuration (1 mark). Example 2: Mendeleev used the periodic table's pattern to predict the existence and properties of undiscovered elements, eka-aluminium (predicted 1871, discovered as gallium 1875) and eka-silicon (predicted 1871, discovered as germanium 1886). His predictions (atomic mass, density, valence, oxide formula) were confirmed on discovery, demonstrating the table's genuine predictive power based on periodic patterns in electron configuration (1 mark). Overall: the periodic table is a powerful scientific model, not a list but a predictive framework grounded in the physics of electron configuration (1 mark).