Grade 12 — Term 1
3 study notes · 33 practice questions · Aligned with NIE syllabus
Counted in the Idasara Academy app, which you open with a free account. The free notes, revision, practice and library pages for each lesson are linked below.
This term in the Idasara Academy app has 3 lessons (NIE Units 1, 2, 3). For each lesson, this page gives you the points you must know and one exam-style question with its answer. None of it needs an account.
Lessons in This Term
Atomic Structure
What you must know
- Anomalous configurations: Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹ due to half-filled and fully-filled 3d orbital stability.
- Where do the first ionization energy drops fall across Periods 2 and 3, and why?: Period 2 & 3 first ionization energy drops occur at Group 2 to 13 (s² to s²p¹) and Group 15 to 16 (stable p³ to paired p⁴ repulsion).
- Photon energy transition relationship: ΔE = hν = hc/λ.
Try one question
The first five successive ionization energies of an element X are 738, 1450, 7730, 10540, and 13630 kJ mol⁻¹. Deduce the periodic group of element X and the chemical formula of its stable chloride.
- A. Group 13, with formula XCl₃
- B. Group 2, with formula XCl₂
- C. Group 1, with formula XCl
- D. Group 14, with formula XCl₄
Show the answer
Answer: B. Group 2, with formula XCl₂. A sharp jump in ionization energy occurs between IE₂ (1450 kJ/mol) and IE₃ (7730 kJ/mol), indicating that the first two electrons are in the valence shell and the third electron is removed from a stable inner noble gas core. Thus, element X belongs to Group 2 (alkaline earth metals) and forms a divalent cation X²⁺, producing the chloride XCl₂.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Structure and Bonding
What you must know
- Common molecular geometries: BeCl₂ (linear, 180°), BF₃ (trigonal planar, 120°), CH₄ (tetrahedral, 109.5°), NH₃ (trigonal pyramidal, 107°), H₂O (bent, 104.5°), PCl₅ (trigonal bipyramidal), SF₆ (octahedral, 90°).
- Dative (coordinate) covalent bonding: shared pair of electrons provided entirely by one donor atom with a lone pair (e.g., NH₄⁺, H₃O⁺).
- Why does graphite conduct electricity when diamond does not?: Graphite conducts electricity parallel to layers due to delocalized π electrons, whereas diamond is an electrical insulator.
Try one question
According to VSEPR theory, what is the molecular shape and the approximate bond angles of sulfur tetrafluoride (SF₄)?
- A. Tetrahedral shape with bond angles of 109.5°
- B. Square planar shape with bond angles of 90°
- C. See-saw shape with bond angles < 120° and < 90°
- D. Trigonal bipyramidal shape with bond angles of 120° and 90°
Show the answer
Answer: C. See-saw shape with bond angles < 120° and < 90°. Sulfur has 6 valence electrons; in SF₄ it forms 4 single bonds with fluorine and retains 1 lone pair (total 5 electron pairs). The electron pair geometry is trigonal bipyramidal. The lone pair occupies an equatorial position to minimize repulsion, resulting in a see-saw molecular shape with bond angles slightly compressed to <120° and <90°.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Chemical Calculations
What you must know
- Ideal gas equation: PV = nRT with consistent SI units (P in Pa, V in m³, T in K, R = 8.314 J K⁻¹ mol⁻¹).
- Standard conditions: STP is 0 °C (273.15 K) and 100 kPa; molar volume of an ideal gas at STP = 22.7 dm³ mol⁻¹ (or 22.4 dm³ mol⁻¹ at 1 atm).
- How is percentage purity calculated?: Percentage purity = (mass of pure substance / total mass of impure sample) × 100%.
Try one question
A 25.00 cm³ sample of 0.100 mol dm⁻³ ethanedioic acid ((COOH)₂) requires 20.00 cm³ of acidified KMnO₄ solution for complete reaction. What is the molar concentration of the KMnO₄ solution?
- A. 0.0400 mol dm⁻³
- B. 0.0500 mol dm⁻³
- C. 0.0200 mol dm⁻³
- D. 0.1250 mol dm⁻³
Show the answer
Answer: B. 0.0500 mol dm⁻³. The balanced redox equation is 2MnO₄⁻ + 5(COOH)₂ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. The mole ratio is 2 moles of KMnO₄ to 5 moles of oxalic acid. Moles of oxalic acid = 0.02500 × 0.100 = 0.002500 mol. Moles of KMnO₄ = (2/5) × 0.002500 = 0.001000 mol. Concentration = 0.001000 / 0.02000 = 0.0500 mol dm⁻³.
A model question written by Idasara on the exam pattern. It is not from a past paper.
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