📚 A-Level OCR Chemistry Last-Minute Revision Notes | A-Level OCR 化学考前冲刺笔记
As your OCR A-Level Chemistry exams approach, this concise set of last-minute revision notes highlights the essential concepts, equations, and common pitfalls to boost your confidence. Review these key topics methodically to ensure you are fully prepared.
随着OCR A-Level化学考试的临近，这套简洁的考前冲刺笔记浓缩了核心概念、公式和常见陷阱，助你信心满满地迎考。请有条理地复习这些关键主题，确保万无一失。
1. Atomic Structure and Ionisation Energies | 原子结构及电离能
Electronic configurations follow the Aufbau principle: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p… Remember that 4s fills before 3d, but 4s electrons are lost first when forming transition metal ions (e.g., Fe atom is [Ar] 3d⁶4s², Fe²⁺ is [Ar] 3d⁶).
电子排布遵循构造原理：1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p……注意4s先于3d填充，但当形成过渡金属离子时，4s电子优先失去（例如Fe原子：[Ar] 3d⁶4s²，Fe²⁺：[Ar] 3d⁶）。
First ionisation energy increases across a period due to increasing nuclear charge and similar shielding; it drops sharply between groups 2–3 and 5–6, because of p-orbital shielding and electron repulsion in paired orbitals.
第一电离能在同一周期中随核电荷增大和屏蔽相近而升高；但在第2–3族和5–6族之间急剧下降，这是因为p轨道屏蔽以及成对电子的排斥。
2. Bonding and Structure | 化学键与结构
Ionic bonding occurs between metals and non-metals; giant ionic lattices are held by strong electrostatic attractions. Covalent bonding involves shared pairs of electrons. Metallic bonding is a lattice of positive ions in a sea of delocalised electrons.
离子键存在于金属与非金属之间；离子巨晶体由强静电吸引力维系。共价键涉及共享电子对。金属键是阳离子在离域电子海中的晶格。
Dative covalent bonds form when an atom with a lone pair donates both electrons to an electron-deficient species. Examples: NH₃ → BF₃, or H₃O⁺ formation from H₂O and H⁺.
配位共价键（配位键）形成于含孤对电子的原子将两个电子都提供给缺电子物种时。例如：NH₃→BF₃，或 H₂O 与 H⁺形成 H₃O⁺。
3. Shapes of Molecules and Polarity | 分子形状与极性
Use VSEPR theory: 2 electron pairs → linear (180°), 3 pairs → trigonal planar (120°), 4 pairs → tetrahedral (109.5°), 5 pairs → trigonal bipyramidal (120°/90°), 6 pairs → octahedral (90°). Lone pairs repel more, reducing bond angles (e.g., NH₃: 107°, H₂O: 104.5°).
运用价层电子对互斥理论：2对电子→直线形(180°)，3对→平面三角形(120°)，4对→四面体形(109.5°)，5对→三角双锥形(120°/90°)，6对→八面体形(90°)。孤对电子排斥力更强，使键角减小（如NH₃：107°，H₂O：104.5°）。
A molecule is polar if it has polar bonds and a non-symmetrical shape, so bond dipoles do not cancel. CO₂ is non-polar (linear), whereas SO₂ is polar (bent).
如果分子具有极性键且形状不对称，键偶极无法抵消，则分子为极性分子。CO₂为非极性（直线形），而SO₂为极性（V形）。
4. The Mole and Gas Equations | 摩尔与气体方程
n = m / M, concentration c = n / V (in mol dm⁻³). At room temperature and pressure (RTP, 298 K, 101 kPa), molar volume Vₘ ≈ 24 dm³ mol⁻¹. Use the ideal gas equation pV = nRT, where R = 8.31 J mol⁻¹ K⁻¹, p in Pa, V in m³, T in K.
n = m / M，浓度 c = n / V（单位mol dm⁻³）。在室温常压下（298 K，101 kPa），气体摩尔体积约为24 dm³ mol⁻¹。使用理想气体方程pV = nRT，其中R = 8.31 J mol⁻¹ K⁻¹，p的单位为Pa，V的单位为m³，T的单位为K。
pV = nRT
pV = nRT
For reacting masses, convert masses to moles, use the stoichiometric ratio, then convert back to mass or volume.
针对反应的质量，先将质量转化为摩尔，根据化学计量比换算，再转换为质量或体积。
5. Energetics and Hess’s Law | 能量学与赫斯定律
Enthalpy change ΔH is measured under standard conditions (100 kPa, 298 K). q = mcΔT, then ΔH = –q / n (exothermic if negative). Calorimetry experiments often underestimate ΔH because of heat loss.
焓变ΔH在标准条件下（100 kPa、298 K）测定。q = mcΔT，后续ΔH = –q/n（放热时为负值）。量热实验常因热损失而低估ΔH。
Hess’s Law: ΔH for a reaction is the same regardless of route. Use enthalpy of formation (ΔHf) or combustion (ΔHc) cycles. ΔH = ΣΔHf(products) − ΣΔHf(reactants).
赫斯定律：反应焓变与路径无关。利用生成焓(ΔHf)或燃烧焓(ΔHc)循环计算。ΔH = ΣΔHf(产物) − ΣΔHf(反应物)。
ΔH = ΣΔHf(products) − ΣΔHf(reactants)
ΔH = ΣΔHf(产物)
Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导