📚 IGCSE Chemistry Revision Notes: Complete Topic Checklist and Study Plan | IGCSE 化学复习笔记：完整考点清单与备考计划
These IGCSE Chemistry notes are a complete, exam-focused revision pack covering every topic that appears on the Cambridge 0620 / 0971 and Edexcel 4CH1 specifications. Each topic is written as a condensed set of definitions, equations, diagrams and mark-scheme phrases, so you can revise a whole unit in one sitting and still hit the exact wording examiners reward. The notes are supplied as a digital PDF, work on phone, tablet and computer, and can be printed for annotation. This guide explains what the pack contains, how it maps to the papers, and how to use it in a realistic revision timetable.
这套 IGCSE 化学笔记是一份完整的应试型复习资料，覆盖 Cambridge 0620 / 0971 与 Edexcel 4CH1 考纲中的全部考点。每个专题都以精简的定义、方程式、图示和得分点句式呈现，让你能一次复习完一个单元，并且写出考官真正给分的表述。资料为电子版 PDF，手机、平板、电脑均可打开，也可以打印出来做批注。本文将说明资料的内容结构、它与各份试卷的对应关系，以及如何在真实的复习时间表中使用它。
1. What These Notes Cover | 笔记涵盖范围
The pack is organised into 14 topic units: states of matter, atomic structure, the periodic table, chemical bonding, stoichiometry and the mole, electrochemistry, energetics, rates of reaction, equilibrium, acids bases and salts, metals and the reactivity series, air water and the environment, sulfur and carbonates, and organic chemistry. Every unit opens with a one-page summary sheet, followed by detailed notes, worked calculations, and a short self-test with answers. Practical skills and the Paper 6 / alternative-to-practical techniques are gathered in a separate unit, because these marks are the easiest to lose and the cheapest to secure.
资料按 14 个专题单元编排：物质状态、原子结构、元素周期表、化学键、化学计量与摩尔、电化学、能量学、反应速率、化学平衡、酸碱与盐、金属与活动性顺序、空气水与环境、硫与碳酸盐、有机化学。每个单元以单页速记表开头，随后是详细笔记、计算例题和一组带答案的自测题。实验技能与 Paper 6（实验操作替代卷）相关技巧单独成章，因为这部分分数最容易丢，也最容易稳拿。
Inside each unit you will find the standard exam ingredients: keyword definitions, labelled diagrams, ionic and symbol equations, colour changes, and the exact command-word responses that appear in mark schemes. Where a topic is commonly examined as a structured question, the notes show a model answer layout with the mark allocation, so you learn how much detail is enough. Long-form explanations are deliberately kept short: revision notes should remind you of knowledge, not teach it from zero.
每个单元都包含标准的应试要素：关键词定义、标注图示、离子方程式与化学方程式、颜色变化，以及评分标准中反复出现的"指令词"作答模板。对于常以结构化作答题形式考查的专题，笔记会给出带分值的示范答题框架，让你清楚"写多少才够"。长篇讲解被刻意压缩：复习笔记的作用是唤醒记忆，而不是从零讲课。
2. Syllabus Map and Paper Structure | 大纲地图与试卷结构
Before revising, know what you are being tested on. Cambridge IGCSE Chemistry has a Core and an Extended route, and the content differs between them: the Extended tier adds mole calculations with equations, equilibrium, and more demanding organic chemistry. A common mistake is to revise from a Core-only source and then sit the Extended paper. Check which tier your school has entered you for, then match your notes to that tier.
开始复习前，先弄清自己要考哪条路线。Cambridge IGCSE 化学分为 Core（核心）与 Extended（拓展）两级，内容并不相同：拓展级会增加带方程式的摩尔计算、化学平衡以及更深入的有机化学。一个常见错误是用只覆盖核心级的内容复习，却去考拓展卷。先向学校确认你被报考的层级，再让笔记与之对应。
| Paper | Content | Typical marks | Best revision source |
| Paper 1 / 2 | Multiple choice, Core or Extended | 40 | One-page summary sheets |
| Paper 3 / 4 | Theory, Core or Extended | 80 / 80 | Detailed notes + worked examples |
| Paper 5 | Practical test | 40 | Practical unit + apparatus diagrams |
| Paper 6 | Alternative to practical | 40 | Practical unit + past-paper drills |
Edexcel IGCSE Chemistry follows a similar split but with different paper codes: Paper 1 covers the full content in 2 hours, while Paper 2 tests practical-based reasoning in 1 hour 15 minutes. Whatever the board, the command words are the same, and they carry marks: 'state' needs one line, 'describe' needs a sequence, 'explain' needs a cause and its consequence, and 'suggest' means applying knowledge to an unfamiliar context.
Edexcel IGCSE 化学的结构类似，但卷号不同：Paper 1 用 2 小时考查全部内容，Paper 2 用 1 小时 15 分钟考查基于实验的推理。无论哪个考试局，指令词都一样，而且直接对应分值："state（写出）"一行即可，"describe（描述）"需要过程顺序，"explain（解释）"需要原因加结果，而"suggest（推测）"意味着把知识迁移到陌生情境中。
3. Atomic Structure and the Periodic Table | 原子结构与元素周期表
Define the basics precisely, because definitions score marks with no calculation needed. An atom is the smallest particle of an element that still has the element's chemical properties. The proton number (atomic number) Z is the number of protons in the nucleus and uniquely identifies the element. The nucleon number (mass number) A is the total number of protons and neutrons. In a neutral atom, the number of electrons equals Z.
先把基础定义写准，因为定义题不需要计算就能得分。原子是保持该元素化学性质的最小粒子。质子数（原子序数）Z 是原子核中质子的数量，唯一确定元素种类。核子数（质量数）A 是质子与中子的总数。电中性原子中，电子数等于 Z。
number of neutrons = A − Z
Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. They have identical chemical properties because chemical behaviour depends on electrons, but slightly different physical properties such as density and rate of diffusion. Relative atomic mass is the weighted average of isotope masses, so a sample with more of the heavier isotope has a higher Aᵣ. Watch out for the classic trap: the mass number of the most abundant isotope is not the same as the relative atomic mass.
同位素是质子数相同、中子数不同的同种元素的原子。由于化学性质取决于电子，它们的化学性质完全相同，但密度、扩散速率等物理性质略有差异。相对原子质量是各同位素质量的加权平均，因此重同位素比例更高的样品，其 Aᵣ 更大。注意经典陷阱：丰度最高的同位素的质量数，并不等于相对原子质量。
Electronic configuration is written in shells, for example sodium is 2,8,1 and chlorine is 2,8,7. Electrons fill the lowest available shell first, and the outer-shell electron count determines the group number, while the number of occupied shells determines the period. Group I metals lose one electron to form 1+ ions, Group VII non-metals gain one electron to form 1− ions, and noble gases have full outer shells so they are unreactive.
电子排布按电子层书写，例如钠为 2,8,1，氯为 2,8,7。电子先填入能量最低的层，最外层电子数决定族序数，占据的电子层数决定周期。第 I 族金属失去 1 个电子形成 1+ 离子，第 VII 族非金属得到 1 个电子形成 1− 离子，稀有气体最外层已满，因此化学性质极不活泼。
4. Bonding, Structure and Properties | 化学键、结构与性质
Ionic bonding happens between a metal and a non-metal, and involves transfer of electrons from the metal to the non-metal. The resulting oppositely charged ions are held in a giant lattice by strong electrostatic attraction, so ionic compounds have high melting points, are usually soluble in water, and conduct electricity only when molten or in aqueous solution, because the ions must be free to move.
离子键存在于金属与非金属之间，本质是金属原子把电子转移给非金属原子。生成的阴阳离子靠强静电作用构成巨型晶格，因此离子化合物熔点高、通常溶于水，并且只在熔融或水溶液中导电，因为必须让离子能够自由移动。
Covalent bonding happens between non-metals, and involves sharing pairs of electrons. Simple molecular substances have weak intermolecular forces, so they have low melting and boiling points and do not conduct electricity, even though the covalent bonds inside each molecule are strong. Giant covalent structures such as diamond, graphite and silicon dioxide have very high melting points because every atom is joined by strong covalent bonds that must be broken before melting.
共价键存在于非金属之间，本质是共用电子对。简单分子物质分子间作用力弱，因此熔点、沸点低，也不导电，尽管分子内部的共价键很强。金刚石、石墨、二氧化硅等巨型共价结构熔点很高，因为熔化前必须破坏每个原子之间的强共价键。
| Structure | Bonding | Melting point | Conducts? |
| Sodium chloride | Ionic lattice | High | Molten / aqueous only |
| Iodine | Simple molecular | Low | No |
| Diamond | Giant covalent | Very high | No |
| Graphite | Giant covalent, layered | Very high | Yes, delocalised electrons |
| Copper | Metallic | High | Yes |
ns the high melting points, and the mobile electrons explain why metals conduct heat and electricity and why they are malleable. When writing ion formation, remember the charges: Na → Na⁺ + e⁻, Cl + e⁻ → Cl⁻, Mg → Mg²⁺ + 2e⁻, and O + 2e⁻ → O²⁻.
金属键是正离子晶格浸在自由电子海洋中。两者之间的引力很强，这解释了金属的高熔点；可移动的电子解释了金属为何导热导电、以及为什么具有延展性。书写离子生成时记住电荷：Na → Na⁺ + e⁻、Cl + e⁻ → Cl⁻、Mg → Mg²⁺ + 2e⁻、O + 2e⁻ → O²⁻。
5. Stoichiometry and the Mole | 化学计量与摩尔
Mole calculations separate the Extended grades from the rest, and almost all of them use three equations. Learn the equations, learn the units, and use a clear working format: write the balanced equation, convert everything you are given into moles, use the ratio, convert back. Most lost marks in calculations come from unbalanced equations, not from arithmetic.
摩尔计算是拓展级分数的分界线，而其中几乎所有题目只用到三个公式。记住公式、记住单位，并采用清晰的书写格式：先写配平方程式，把已知量全部换算成摩尔，再按系数比换算回去，最后换算成题目要求的量。计算题丢分大多源于方程式没配平，而不是算错数。
n = m ÷ Mᵣ c = n ÷ V V = n × 24 dm³ (r.t.p.)
= 1000 cm³). Molar gas volume is 24 dm³ at room temperature and pressure, so 0.5 mol of any gas occupies 12 dm³. Percentage yield and atom economy are also frequently examined and reward careful definitions.
其中 n 为物质的量（mol），m 为质量（g），Mᵣ 为相对式量，c 为浓度（mol/dm³），V 为换算成 dm³ 的体积（记住 1 dm³ = 1000 cm³）。室温常压下气体摩尔体积为 24 dm³，因此 0.5 mol 任何气体都占 12 dm³。产率与原子经济性也常考，写清定义同样能得分。
% yield = (actual yield ÷ theoretical yield) × 100
% atom economy = (Mᵣ of desired product ÷ sum of Mᵣ of all reactants) × 100
whole-number ratio. If the ratio comes out as 1 : 1.5, multiply everything by 2. For titration or solution calculations, always convert the volume to dm³ before using c = n ÷ V, and remember that a limiting reactant is the one that is used up first and therefore determines the maximum amount of product.
求实验式的方法是：把各元素的质量或百分含量除以各自的 Aᵣ，再把所有结果除以最小值，得到最简整数比。若比值出现 1 : 1.5，就把所有数乘 2。做滴定或溶液计算时，使用 c = n ÷ V 前务必把体积换算成 dm³；另外要记住，限量反应物是最先用完、从而决定产物最大量的那种反应物。
6. Electrochemistry and the Reactivity Series | 电化学与金属活动性顺序
Electrolysis is the breakdown of a compound using a direct current. The electrolyte must be molten or aqueous so that ions can move. In every cell, positive ions (cations) move to the cathode and are reduced, while negative ions (anions) move to the anode and are oxidised. Remember the mnemonic: reduction happens at the cathode, oxidation at the anode.
电解是利用直流电使化合物分解。电解质必须处于熔融或溶液状态，离子才能移动。任何电解池中，阳离子移向阴极并被还原，阴离子移向阳极并被氧化。记住口诀：阴极还原，阳极氧化。
cathode: Pb²⁺ + 2e⁻ → Pb anode: 2Br⁻ → Br₂ + 2e⁻
In the electrolysis of concentrated sodium chloride solution, the products are hydrogen at the cathode, chlorine at the anode, and sodium hydroxide left in solution. In the electrolysis of copper sulfate solution with copper electrodes, copper is deposited at the cathode and dissolves from the anode, so the concentration of the electrolyte stays constant - which is why this system is used in electroplating and in refining copper.
电解浓氯化钠溶液时，阴极得到氢气，阳极得到氯气，溶液中剩余氢氧化钠。用铜电极电解硫酸铜溶液时，阴极析出铜、阳极铜溶解，因此电解质浓度基本不变——这正是该体系用于电镀与精炼铜的原因。
The reactivity series orders metals from potassium, sodium, calcium, magnesium, aluminium, zinc, iron, lead, copper, silver to gold. Metals above carbon in the series are extracted by electrolysis, metals below carbon are extracted by reduction with carbon, and metals far below, such as gold, occur native. A more reactive metal displaces a less reactive metal from its salt solution, and this is a standard test question: zinc will displace copper from copper sulfate, but copper will not displace zinc.
金属活动性顺序由钾、钠、钙、镁、铝、锌、铁、铅、铜、银到金排列。排在碳之前的金属用电解法冶炼，排在碳之后的用碳还原法冶炼，而金等极不活泼金属以单质形式存在。较活泼金属能把较不活泼金属从其盐溶液中置换出来，这是标准考点：锌能置换硫酸铜中的铜，铜却不能置换锌。
7. Energetics, Rates and Equilibrium | 能量学、反应速率与平衡
An exothermic reaction releases heat to the surroundings and shows a temperature rise; an endothermic reaction absorbs heat and shows a temperature fall. In bond terms, breaking bonds absorbs energy and making bonds releases energy, so the sign of the overall energy change follows directly from the totals.
放热反应向环境释放热量，表现为温度升高；吸热反应吸收热量，表现为温度降低。从键的角度看，断键吸能、成键放能，因此总能量变化的符号直接由两者总量决定。
ΔH = Σ (bonds broken) − Σ (bonds formed)
Reaction rate is explained by collision theory: particles must collide with energy greater than or equal to the activation energy. Increasing temperature, increasing concentration or pressure, increasing surface area of a solid, and adding a catalyst all increase the rate. Explain each with the same chain of reasoning: more frequent or more energetic collisions, more particles with energy greater than the activation energy, therefore more successful collisions per second.
反应速率用碰撞理论解释：粒子碰撞时的能量必须大于或等于活化能。升高温度、增大浓度或压强、增大固体表面积、加入催化剂都会加快反应。解释时要使用同一条推理链：碰撞更频繁或能量更高，能量超过活化能的粒子更多，因此每秒有效碰撞次数增加。
A catalyst increases the rate by providing an alternative pathway with a lower activation energy, and it is not used up. In a reversible reaction at equilibrium, the forward and reverse rates are equal and the concentrations stay constant, but the reaction has not stopped. If conditions change, the position of equilibrium shifts to oppose the change, which is the qualitative reasoning expected at IGCSE level.
催化剂通过提供活化能更低的反应途径来加快反应，本身不被消耗。可逆反应达到平衡时，正逆反应速率相等、各物质浓度保持不变，但反应并未停止。若条件改变，平衡会向抵消该改变的方向移动——这就是 IGCSE 阶段要求的定性推理。
8. Acids, Bases and Salt Preparation | 酸碱与盐的制备
An acid is a proton donor and produces H⁺ ions in aqueous solution; a base is a proton acceptor, and an alkali is a soluble base that produces OH⁻ ions. The pH scale runs
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