IGCSE Chemistry (0620) – Complete Revision Notes
Learning Objectives
- Recall the fundamental concepts of matter, atomic structure, isotopes, the periodic table and bonding (AO1).
- Apply the mole‑concept, Avogadro’s constant, relative formula masses and gas‑volume relationships to calculate quantities (AO2).
- Identify oxidation and reduction in redox reactions, write and balance half‑equations in acidic or basic media, and recognise oxidising/reducing agents (AO3).
- Explain rates of reaction, dynamic equilibrium, energetics, acid‑base behaviour, electrochemistry and the environmental impact of chemical processes (AO3).
1. States of Matter
1.1 Core Concepts
- Solids – particles vibrate in fixed positions; definite shape and volume.
- Liquids – particles slide past one another; definite volume, no fixed shape.
- Gases – particles move freely; no fixed shape or volume.
- Kinetic‑Particle Theory (KPT)
- Temperature = average kinetic energy of particles.
- Pressure = collisions of particles with container walls.
- Volume, temperature and pressure are related by pV = nRT.
1.2 Supplementary Material
- Diffusion & Effusion – movement of particles from high to low concentration.
- Graham’s law: rate ∝ 1/√M (M = molar mass).
- Heating / Cooling Curves – interpret plateaus (melting, boiling) and slopes (specific heat).
- Pressure‑Volume Work – ΔE = –pΔV for gases (useful for electrochemical cells).
2. Atoms, Elements & Compounds
2.1 Atomic Structure & Isotopes (Core)
- Protons (+), neutrons (0), electrons (–). Atomic number Z = # of protons.
- Isotopes: same Z, different neutron number; atomic mass = weighted average of isotopic masses.
- Relative atomic mass (Ar) – dimensionless; molar mass (M) = Ar g mol⁻¹.
2.2 Periodic Trends (Core)
| Trend | Across a Period (left → right) | Down a Group |
|---|
| Atomic radius | decreases | increases |
| Ion‑radius | decreases | increases |
| Ionisation energy | increases | decreases |
| Electronegativity | increases | decreases |
| Metallic character | decreases | increases |
2.3 Bonding (Core)
- Ionic bonding – transfer of electrons; formation of cations & anions; lattice structure.
- Covalent bonding – sharing of electrons; single, double, triple bonds; VSEPR basics for molecular shape.
- Metallic bonding – delocalised electrons give conductivity, malleability, ductility.
- Giant covalent structures – diamond, silicon, SiO₂; very high melting points.
2.4 Ions & Dot‑and‑Cross Diagrams (Core)
Write electron‑dot diagrams for any ion, showing gain or loss of electrons to achieve a noble‑gas configuration.
2.5 Formulae, Empirical & Molecular Formulas (Core)
- Relative formula mass (Mᵣ) = Σ(atomic masses of atoms in the formula).
- Molar mass (M) = Mᵣ g mol⁻¹.
- Empirical formula = simplest whole‑number ratio of atoms.
- Molecular formula = (n × empirical formula) where n = M / Mᵣ(empirical).
2.6 The Mole Concept (Core)
- 1 mol = 6.022 × 10²³ particles (Avogadro’s constant, Nₐ).
- Mass (g) = moles × molar mass.
- Molar volume of a gas at STP (0 °C, 1 atm) = 24 dm³ mol⁻¹ (IGCSE convention).
2.7 Supplementary Topics
- Percentage purity and percentage yield calculations.
- Fuel‑cell operation, advantages and disadvantages.
- Full method for writing and balancing half‑equations (see Section 5).
3. Stoichiometry
3.1 Calculations with Relative Masses (Core)
| Quantity | How to Find |
|---|
| Relative formula mass (Mᵣ) | Sum of atomic masses of atoms in the formula. |
| Molar mass (M) | Numerical value of Mᵣ in g mol⁻¹. |
| Number of moles | n = mass (g) ÷ M (g mol⁻¹). |
| Mass of gas at STP | V (dm³) ÷ 24 dm³ mol⁻¹ = moles → mass. |
3.2 Worked Example – Simple Stoichiometry
Question: 2 g of Na reacts with excess Cl₂. How many grams of NaCl are formed?
- M(Na) = 23 g mol⁻¹ → n(Na) = 2 g ÷ 23 = 0.087 mol.
- Balanced equation: 2 Na + Cl₂ → 2 NaCl (1 mol Na → 1 mol NaCl).
- M(NaCl) = 58.5 g mol⁻¹ → mass NaCl = 0.087 mol × 58.5 = 5.1 g.
3.3 Limiting Reactant & Percent Yield (Supplement)
- Calculate moles of each reactant, compare using the stoichiometric coefficients; the reactant that would produce the fewest moles of product is the limiting reactant.
- Percent yield = (actual yield ÷ theoretical yield) × 100 %.
3.4 Empirical & Molecular Formula Determination (Supplement)
Example: A compound contains 40 % C, 6.7 % H, 53.3 % O by mass. Find its empirical formula.
- Assume 100 g sample → 40 g C, 6.7 g H, 53.3 g O.
- Convert to moles: C = 40 ÷ 12.01 = 3.33 mol; H = 6.7 ÷ 1.008 = 6.65 mol; O = 53.3 ÷ 16.00 = 3.33 mol.
- Divide by the smallest (3.33) → C = 1, H ≈ 2, O = 1 → empirical formula CH₂O.
- If the molar mass is 60 g mol⁻¹, n = 60 ÷ (12 + 2 + 16) = 2 → molecular formula C₂H₄O₂.
4. Electrochemistry
4.1 Electrolysis (Core)
- Electrolytic cell – external electricity forces a non‑spontaneous redox reaction.
- Cathode (reduction): cations gain electrons.
- Anode (oxidation): anions lose electrons.
- Inert electrodes (Pt, C) are used when the reactant itself is not a metal.
Typical Products
| Electrolyte | Cathode (Reduction) | Anode (Oxidation) |
|---|
| Molten NaCl | Na⁺ + e⁻ → Na (l) | 2Cl⁻ → Cl₂(g) + 2e⁻ |
| Aqueous CuSO₄ | Cu²⁺ + 2e⁻ → Cu(s) | 2H₂O → O₂(g) + 4H⁺ + 4e⁻ (water oxidation dominates) |
| Aqueous NaCl | 2H₂O + 2e⁻ → H₂(g) + 2OH⁻ (cathode) | 2Cl⁻ → Cl₂(g) + 2e⁻ (anode) |
4.2 Electroplating (Supplement)
Metal ions in solution are reduced onto a conductive object (cathode). Example: Cu²⁺ + 2e⁻ → Cu(s) deposits copper onto a steel nail.
4.3 Fuel Cells (Core)
Spontaneous redox reaction that generates electricity.
- Overall reaction (hydrogen‑oxygen cell): 2 H₂ + O₂ → 2 H₂O.
- Anode (oxidation): H₂ → 2H⁺ + 2e⁻.
- Cathode (reduction): ½O₂ + 2H⁺ + 2e⁻ → H₂O.
- Advantages: high efficiency, low emissions; Disadvantages: cost of catalysts, storage of H₂.
4.4 Half‑Equations – Method (Supplement)
- Separate the overall reaction into oxidation and reduction parts.
- Balance each half‑reaction for all atoms except O and H.
- Balance O by adding H₂O.
- Balance H by adding H⁺ (acidic) or H₂O/OH⁻ (basic).
- Balance charge by adding electrons.
- Multiply half‑reactions to equalise electrons, then add and cancel.
5. Redox Reactions
5.1 Key Definitions
- Oxidation – loss of electrons; oxidation number increases.
- Reduction – gain of electrons; oxidation number decreases.
- Oxidising agent – substance that is reduced (accepts electrons).
- Reducing agent – substance that is oxidised (donates electrons).
5.2 Oxidation‑Number Rules (IGCSE level)
| Rule | Application |
|---|
| 1 | Element in its standard state has oxidation number 0 (e.g., Na, O₂, N₂). |
| 2 | For a mono‑atomic ion, oxidation number = ionic charge (Na⁺ = +1, Cl⁻ = –1). |
| 3 | Oxygen is –2, except in peroxides (–1) and OF₂ (+2). |
| 4 | Hydrogen is +1 when bonded to non‑metals, –1 when bonded to metals. |
| 5 | Sum of oxidation numbers = 0 for a neutral species; = overall charge for polyatomic ions. |
| 6 | Group‑1 metals are +1, Group‑2 metals are +2 in compounds. |
5.3 Step‑by‑Step Method to Identify Oxidation & Reduction
- Write the balanced chemical equation.
- Assign oxidation numbers to every atom using the rules above.
- Compare oxidation numbers on the reactant and product sides.
- The element whose oxidation number increases is oxidised (loses electrons).
- The element whose oxidation number decreases is reduced (gains electrons).
- Identify the oxidising and reducing agents accordingly.
5.4 Worked Example 1 – Single‑Displacement
Equation: Zn + CuSO₄ → ZnSO₄ + Cu
- Balanced as written.
- Oxidation numbers:
- Zn (reactant) = 0
- Cu in CuSO₄ = +2
- S in SO₄²⁻ = +6, O = –2
- Zn in ZnSO₄ = +2
- Cu (product) = 0
- Changes:
- Zn: 0 → +2 → oxidised.
- Cu: +2 → 0 → reduced.
- Oxidising agent = Cu²⁺ (it is reduced). Reducing agent = Zn (it is oxidised).
5.5 Worked Example 2 – Combination
Equation: 2 Fe + 3 Cl₂ → 2 FeCl₃
- Balanced as written.
- Oxidation numbers:
- Fe (reactant) = 0 → Fe in FeCl₃ = +3
- Cl₂ = 0 → Cl in FeCl₃ = –1
- Changes:
- Fe: 0 → +3 → oxidised.
- Cl: 0 → –1 → reduced.
- Oxidising agent = Cl₂. Reducing agent = Fe.
5.6 Disproportionation (Same Species Both Oxidised & Reduced)
Example: 2 ClO⁻ → Cl⁻ + ClO₃⁻
- Cl in ClO⁻ = +1.
- In Cl⁻ = –1 (reduction) and in ClO₃⁻ = +5 (oxidation).
- ClO⁻ acts as both oxidising and reducing agent.
5.7 Balancing Redox in Acidic Medium (Example)
Unbalanced reaction: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺
- Separate half‑reactions:
- Reduction: MnO₄⁻ → Mn²⁺
- Oxidation: Fe²⁺ → Fe³⁺
- Balance atoms (except O, H):
- Mn: already balanced.
- Fe: already balanced.
- Balance O by adding H₂O:
- Balance H by adding H⁺ (acidic):
- 4 H₂O → 8 H⁺ on the right, so add 8 H⁺ to the left.
- Balance charge by adding electrons:
- Reduction: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
- Oxidation: Fe²⁺ → Fe³⁺ + e⁻
- Equalise electrons (multiply oxidation half‑reaction by 5) and add:
5 Fe²⁺ → 5 Fe³⁺ + 5 e⁻
MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
------------------------------------------------
MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O
5.8 Common Redox Reactions in the Syllabus (Core)
- Metal + acid → salt + H₂ (e.g., Zn + 2 HCl → ZnCl₂ + H₂).
- Metal + salt solution → displaced metal + new salt (e.g., Cu + 2 AgNO₃ → Cu(NO₃)₂ + 2 Ag).
- Combustion of hydrocarbons (e.g., CH₄ + 2 O₂ → CO₂ + 2 H₂O).
- Disproportionation (e.g., 2 ClO⁻ → Cl⁻ + ClO₃⁻).
- Redox in electrolysis and fuel cells (see Sections 4 & 5).
5.9 Practice Questions (Redox)
- Mg + 2 HCl → MgCl₂ + H₂
- Assign oxidation numbers.
- State which element is oxidised and which is reduced.
- Fe₂O₃ + 2 Al → 2 Fe + Al₂O₃
- Identify the oxidising and reducing agents.
- Balance in acidic solution: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺
- Show the two half‑reactions.
- Give the final balanced equation.
6. Chemical Energetics
6.1 Enthalpy Change (ΔH) (Core)
- Exothermic: ΔH < 0 (heat released).
- Endothermic: ΔH > 0 (heat absorbed).
6.2 Activation Energy & Reaction Profiles (Core)
Diagram: Reactants → Activated complex → Products. Catalysts lower the activation energy without being consumed.
6.3 Bond‑Energy Calculations (Supplement)
ΔH ≈ Σ(bond energies broken) – Σ(bond energies formed).
7. Chemical Reactions – Rates & Equilibrium
7.1 Rates of Reaction (Core)
- Factors: concentration, temperature, surface area, catalyst.
- Collision theory – effective collisions must have sufficient energy and proper orientation.
7.2 Reversible Reactions & Dynamic Equilibrium (Core)
- At equilibrium, forward and reverse rates are equal; concentrations remain constant.
- Le Chatelier’s principle – predict the shift when concentration, pressure, temperature or catalyst changes.
8. Acids, Bases & Salts
8.1 Definitions (Core)
- Arrhenius: acid → H⁺ in water; base → OH⁻ in water.
- Bronsted‑Lowry: acid donates a proton, base accepts a proton.
8.2 Characteristic Reactions (Core)
- Acid + metal → salt + H₂.
- Acid + carbonate → salt + CO₂ + H₂O.
- Base + acid → salt + H₂O (neutralisation).
8.3 pH Scale & Indicators (Core)
pH = –log[H⁺]; pH < 7 acidic, pH = 7 neutral, pH > 7 basic. Indicators change colour at characteristic pH ranges.
8.4 Salt Preparation (Supplement)
- Acid‑base neutralisation.
- Metal‑acid reaction.
- Double‑displacement (metathesis) reactions.
9. The Periodic Table – Key Groups
9.1 Important Groups (Core)
- Group 1 – Alkali metals: very reactive, +1 oxidation state.
- Group 2 – Alkaline earth metals: +2 oxidation state.
- Group 17 – Halogens: –1 oxidation state (except in peroxides).
- Group 18 – Noble gases: inert under normal conditions.
9.2 Transition Metals (Supplement)
- Variable oxidation states; often act as catalysts and form coloured complexes.
- Typical examples in the syllabus: Fe, Cu, Zn, Mn.
10. Quick Revision Checklist
- Can you assign oxidation numbers to every atom in a given formula?
- Do you know how to identify the oxidising and reducing agents?
- Can you write and balance half‑equations in acidic and basic media?
- Are you comfortable with mole‑concept calculations, including limiting reactant and percent yield?
- Can you interpret heating/cooling curves and explain the effect of temperature on equilibrium?
- Do you remember the main products of electrolysis for molten vs. aqueous electrolytes?
- Can you name the acid‑base definitions (Arrhenius, Bronsted‑Lowry) and calculate pH?
Answers to Practice Questions (for self‑checking)
- Mg + 2 HCl → MgCl₂ + H₂
- Mg: 0 → +2 (oxidised)
- H in HCl: +1 → 0 (reduced)
- Oxidising agent: H⁺ (it is reduced). Reducing agent: Mg.
- Fe₂O₃ + 2 Al → 2 Fe + Al₂O₃
- Fe in Fe₂O₃: +3 → 0 (reduced)
- Al: 0 → +3 (oxidised)
- Oxidising agent: Fe₂O₃ (it is reduced). Reducing agent: Al.
- Balanced acidic equation (see Section 5.7):
MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O