QCE Chemistry study materials
QCE Chemistry study resources

QCE Chemistry: Chemical Fundamentals: Structure, Properties and Reactions

QCE Chemistry · Work through your exam, one topic at a time.

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Chemical Fundamentals: Structure, Properties and Reactions

What you must understand

Start with the periodic table as a predictive tool: effective nuclear charge, atomic radius, ionisation energy and electronegativity trends explain bonding type. Distinguish ionic (lattice of ions, electrostatic attraction), covalent molecular (discrete molecules, weak forces between them), covalent network (continuous lattice, e.g. diamond, SiO₂) and metallic bonding. Connect structure to properties: melting point, conductivity, solubility and malleability all follow from which particles move and which forces must be overcome. Master electron configuration, Lewis structures, VSEPR shapes and polarity, then stoichiometry: mole concept, empirical and molecular formulae, limiting reagents, percentage yield and solution concentration. Thermochemistry (ΔH, calorimetry, Hess's law) and reaction rates belong here as energy and collision ideas.

Traps in questions

  • Melting a molecular substance breaks intermolecular forces, not covalent bonds; boiling a metal does not break covalent bonds because none exist.
  • Ionic compounds conduct only when molten or dissolved, not as solids.
  • VSEPR counts lone pairs as electron domains but describes shape using atoms only: four domains with one lone pair is trigonal pyramidal, not tetrahedral.
  • Limiting reagent questions often hide a unit mismatch (grams versus moles) or a stoichiometric ratio that is not 1:1.
  • Percentage yield compares actual to theoretical moles, not masses, unless molar masses cancel.

Rule of thumb and worked example

Rule: before any calculation, convert every quantity to moles and write the balanced equation; then compare mole ratios to find the limiting reagent.

Example: 4.0 g of CaCO₃ (M = 100 g mol⁻¹) reacts with 50 mL of 1.0 mol L⁻¹ HCl. Moles CaCO₃ = 0.040 mol; moles HCl = 0.050 mol. The equation CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O needs 0.080 mol HCl for all the carbonate, so HCl is limiting. Moles CO₂ = 0.050/2 = 0.025 mol, giving 0.025 × 44 = 1.1 g CO₂.