📚 International AS Chemistry Unit 1 (CH01) Core Principles | 国际AS化学第一单元 (CH01) 核心原理
Unit 1 of the International AS Chemistry course (CH01) lays the groundwork for all advanced chemical study. It covers atomic structure, the mole concept, chemical bonding, introductory organic chemistry, and energetics. A firm grasp of these core principles enables you to interpret exam command words, perform multi‑step calculations, and construct the kind of concise, accurate responses expected in high‑stakes assessments. This article revisits each major area, linking theory to example‑response thinking.
国际AS化学第一单元（CH01）为整个高等化学学习奠定基础，覆盖面包括原子结构、摩尔概念、化学键、有机化学入门以及能量学。牢固掌握这些核心原理，能帮助你理解考题指令词、完成多步计算，并构建出在重要考试中出题人所期望的那种简洁而准确的答案。下文逐一重温各主要板块，将理论与答题思维相互衔接。
1. The Mole and Stoichiometry | 摩尔与化学计量
The mole is the SI unit for amount of substance, containing exactly 6.02214076×10²³ elementary entities (Avogadro’s constant, NA). Stoichiometry uses the ratios from balanced chemical equations to convert between masses, volumes of gases, and solution concentrations. For the reaction 2H₂ + O₂ → 2H₂O, two moles of hydrogen molecules react with one mole of oxygen to yield two moles of water. This fixed ratio underpins every quantitative prediction in chemistry.
摩尔是物质的量的国际单位，含有恰好 6.02214076×10²³ 个基本单元（阿伏伽德罗常数 NA）。化学计量运用配平化学方程式中的系数之比，在质量、气体体积和溶液浓度之间进行换算。对于反应 2H₂ + O₂ → 2H₂O，2 摩尔氢分子与 1 摩尔氧反应生成 2 摩尔水。这个固定比率为化学中任何定量预测奠定了基础。
Calculations typically follow the pathway mass → moles → moles of target substance → mass/volume/concentration. Using the formula n = m/M (moles = mass ÷ molar mass) and n = V/24.0 dm³ (at RTP for gases) allows you to solve titration, combustion, and yield problems. Always check that the equation is balanced before applying the mole ratio.
计算通常遵循“质量→摩尔→目标物质摩尔→质量/体积/浓度”的路径。运用公式 n = m/M（摩尔 = 质量 ÷ 摩尔质量）以及室温常压下气体的 n = V/24.0 dm³，便可解决滴定、燃烧和产率等问题。在应用摩尔比之前，务必确认方程式已配平。
2. Atomic Structure and Isotopes | 原子结构与同位素
Atoms consist of a central nucleus containing protons and neutrons, surrounded by electrons arranged in shells. Protons have a relative mass of 1 and charge of +1, neutrons are uncharged with mass 1, and electrons have negligible mass but a –1 charge. The atomic number (Z) is the number of protons, defining the element, while the mass number (A) is the total of protons and neutrons.
原子由一个包含质子和中子的中心原子核以及分层排布的核外电子构成。质子相对质量约为 1，带 1 个单位正电荷；中子不带电，质量也约为 1；电子质量极小但带 1 个单位负电荷。原子序数（Z）就是质子数，决定了元素种类；质量数（A）则是质子与中子数目之和。
Isotopes are atoms of the same element with different numbers of neutrons. They share identical chemical properties because they have the same electron configuration, but their physical properties such as mass differ. A common exam task is to deduce the number of subatomic particles for a specific isotope, e.g. ³⁷Cl has 17 protons, 20 neutrons, and 17 electrons.
同位素是同一种元素的中子数不同的原子。由于电子排布相同，它们具有相同的化学性质，但质量等物理性质存在差异。常见的考题是推断特定同位素中的亚原子粒子数目，例如 ³⁷Cl 中有 17 个质子、20 个中子和 17 个电子。
3. Relative Masses and Mass Spectrometry | 相对质量与质谱分析
Relative atomic mass (Aᵣ) is the weighted average mass of an atom compared to 1/12th the mass of a carbon‑12 atom. For an element with several isotopes, Aᵣ = Σ (isotopic mass × % abundance)/100. Mass spectrometry provides the isotopic masses and relative abundances. A vaporised sample is ionised, accelerated, deflected in a magnetic field, and detected; the resulting mass spectrum shows peaks whose positions give isotopic masses and heights reflect abundance.
相对原子质量（Aᵣ）是原子质量与碳‑12 原子质量的 1/12 相比较得出的加权平均值。对于含有多种同位素的元素，Aᵣ = Σ（同位素质量 × 百分丰度）/100。质谱分析法可提供同位素质量和相对丰度。气态样品经电离、加速、在磁场中偏转后被检测；得到的质谱图中，峰的位置给出同位素质量，峰的高度反映丰度。
Relative molecular mass (Mᵣ) is the sum of the Aᵣ values of all atoms in a molecule. When using mass spectra, beware of fragmentation patterns: the peak at the highest m/z is often the molecular ion peak (M⁺), which gives the Mᵣ of the compound. Practice deducing relative atomic mass from a given mass spectrum is essential for achieving full marks on data‑response questions.
相对分子质量（Mᵣ）是分子中所有原子的 Aᵣ 值之和。在使用质谱图时，要注意碎片化图像：质荷比（m/z）最大的峰往往是分子离子峰（M⁺），它提供了化合物的 Mᵣ。通过给定的质谱图推算相对原子质量，是数据应答类题目取得满分所必需的训练。
4. Empirical and Molecular Formulae | 实验式与分子式
The empirical formula shows the simplest whole‑number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms of each element in a molecule. Combustion analysis or mass composition data lead first to the empirical formula. If the sample contains C, H, and O, you convert masses to moles by dividing by the Aᵣ of each element, then find the simplest ratio by dividing by the smallest number of moles.
实验式表示化合物中原子的最简整数比，分子式则给出分子中每种原子的真实数目。燃烧分析或质量组成数据首先得到的是实验式。如果样品含有碳、氢和氧，就需要把各元素质量分别除以其 Aᵣ，转化为摩尔数，然后除以最小摩尔数以得出最简整数比。
Molecular formula = n × empirical formula, where n = Mᵣ(compound) / Mᵣ(empirical formula). The Mᵣ of the compound is often obtained from the mass spectrum or the ideal gas equation. In exam responses, clearly show the division steps and explicitly state the integer multiplier n to secure the final mark.
分子式 = n × 实验式，其中 n = 化合物的 Mᵣ / 实验式的 Mᵣ。化合物的 Mᵣ 通常通过质谱或理想气体状态方程获得。在作答时，要清晰展示除法步骤，并明确写出整数乘数 n，以确保得到最终分数。
5. Chemical Bonding: Ionic, Covalent and Metallic | 化学键：离子键、共价键与金属键
Ionic bonding occurs between metals and non‑metals by the transfer of electrons, forming a giant ionic lattice of oppositely charged ions held together by strong electrostatic attractions. Covalent bonding results from the sharing of electron pairs between non‑metal atoms, forming either simple molecular or giant covalent structures. Metallic bonding consists of a lattice of positive metal ions surrounded by a sea of delocalised electrons, which explains electrical conductivity and malleability.
离子键发生在金属和非金属之间，通过电子转移形成，构成由相反电荷的离子通过强静电吸引力结合而成的巨型离子晶格。共价键是由非金属原子之间共享电子对而形成的，构成简单分子结构或巨型共价结构。金属键则由正电荷金属离子规则排列、沉浸在离域电子的“海洋”中所构成，这一模型解释了导电性和延展性。
Understanding bond polarity is vital. In a covalent bond between different atoms, the bonding electrons are attracted more strongly to the more electronegative atom, creating partial charges δ⁺ and δ⁻. The resulting dipole influences physical properties and chemical reactivity. Use electronegativity values to predict whether a bond is non‑polar, polar, or ionic.
理解键的极性非常重要。在不同原子之间形成的共价键中，成键电子会被电负性更强的原子强烈吸引，产生部分正电 δ⁺ 和部分负电 δ⁻。由此形成的偶极会影响物理性质和化学反应活性。利用电负性数值可以预测某个键是非极性、极性还是离子键。
6. Shapes of Molecules and VSEPR Theory | 分子形状与价层电子对互斥理论
VSEPR theory states that electron pairs around a central atom repel each other and adopt an arrangement that minimises repulsion. The shape depends on the number of bonding pairs and lone pairs in the valence shell. For example, 4 bonding pairs with no lone pairs give a tetrahedral shape (bond angle 109.5°, e.g. CH₄), while 3 bonding pairs and 1 lone pair produce a trigonal pyramidal shape (bond angle ≈107°, e.g. NH₃).
VSEPR 理论认为，中心原子周围的电子对互相排斥，并采取排斥最小的空间排布方式。分子形状取决于价层中成键电子对和孤电子对的数目。例如，4 对成键电子、无孤对电子时，分子呈四面体形（键角 109.5°，如 CH₄）；而 3 对成键电子加 1 对孤对电子时，则形成三角锥形（键角约 107°，如 NH₃）。
Two bonding pairs and two lone pairs lead to a bent or V‑shaped molecule (104.5°, e.g. H₂O). Linear molecules occur with 2 bonding pairs (180°, e.g. BeCl₂) or 1 bonding pair and 3 lone pairs (HCl, which is still linear). Always draw a dot‑and‑cross diagram first, count the total number of electron pairs, and then deduce the shape. Exam responses must state the shape, the bond angle, and the number of bonding and lone pairs.
当有 2 对成键电子和 2 对孤对电子时，分子为角形或 V 形（104.5°，如 H₂O）。直线形分子出现在有 2 对成键电子（180°，如 BeCl₂），或 1 对成键电子配 3 对孤对电子的情形（如 HCl，虽然仍为直线）。解题时，应先画出电子点叉图，数出总电子对数，再推断分子形状。答案必须写明形状、键角以及成键和孤对电子对数。
7. Energetics: Enthalpy Changes and Hess’s Law | 能量学：焓变与赫斯定律
Enthalpy change (ΔH) is the heat energy transferred in a reaction at constant pressure. Standard conditions are 100 kPa and a stated temperature, usually 298 K. Exothermic reactions release energy (ΔH negative), while endothermic reactions absorb energy (ΔH positive). Bond breaking is endothermic, bond making is exothermic.
焓变（ΔH）是反应在恒压条件下传递的热量。标准条件为 100 kPa 和指定的温度，通常为 298 K。放热反应释放能量（ΔH 为负值），吸热反应吸收能量（ΔH 为正值）。断裂化学键需要吸收能量，形成化学键则放出能量。
Hess’s Law states that the total enthalpy change for a reaction is independent of the route taken, as long as the initial and final conditions are the same. This allows calculation of ΔH using enthalpy cycles, such as those involving enthalpies of formation, combustion, or bond enthalpies. A typical calculation pattern is:
ΔH = ΣΔHf°(products) – ΣΔHf°(reactants)
赫斯定律指出，在始态和终态相同的条件下，反应的总焓变与所经路径无关。这就允许我们利用生成焓、燃烧焓或键焓构建焓循环来计算 ΔH。典型的计算模式为：
ΔH = ΣΔHf°（产物） – ΣΔHf°（反应物）
Always construct a clear cycle, label each arrow with the correct ΔH, and show the algebraic summation. Marks are awarded for the cycle itself, so do not skip this step.
作答时务必画出清晰的循环图，用正确的 ΔH 标注每一箭头，并展示代数求和过程。循环图本身就能得分，因此切勿省略这一步。
8. Chemical Equilibria and Le Chatelier’s Principle | 化学平衡与勒夏特列原理
A reversible reaction reaches dynamic equilibrium when the rates of the forward and reverse reactions become equal, and the concentrations of reactants and products remain constant. The equilibrium constant Kc is expressed in terms of concentrations: for aA + bB ⇌ cC + dD, Kc = [C]ᶜ [D]ᵈ / [A]ᵃ [B]ᵇ. Kc is temperature‑dependent only.
当可逆反应的正、逆反应速率相等，且各物质浓度保持不变时，体系达到动态平衡。平衡常数 Kc 以浓度表示：对于 aA + bB ⇌ cC + dD，Kc = [C]ᶜ [D]ᵈ / [A]ᵃ [B]ᵇ。Kc 仅取决于温度。
Le Chatelier’s principle states that if a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the equilibrium position shifts to counteract the imposed change. Increasing the concentration of a reactant shifts equilibrium to the right, producing more product. For gaseous systems, increasing pressure favours the side with fewer moles of gas. Temperature changes depend on whether the forward reaction is exothermic or endothermic.
勒夏特列原理指出，若一个处于平衡状态的体系受到浓度、压强或温度变化的干扰，平衡位置会向削弱这种变化的方向移动。增加某反应物浓度会使平衡向右移动，生成更多产物。对于气体反应，增大压强有利于气体总物质的量较小的一侧。温度变化的效果则取决于正反应是放热还是吸热。
In an exam response, clearly state the direction of shift and the underlying reason. Do not say equilibrium “shifts to the side of fewer moles” without linking it to the balanced equation. Use terms such as “shifts to the right” or “favours the forward reaction” precisely.
在考试作答中，要明确说出移动方向及其根本原因。不要只是含糊地说“平衡向物质的量少的一侧移动”，而不联系配平方程式。要准确使用“向右移动”或“有利于正向反应”等术语。
9. Organic Chemistry: Alkanes and Alkenes | 有机化学：烷烃与烯烃
Alkanes are saturated hydrocarbons with the general formula CₙH₂ₙ₊₂. They are relatively unreactive because of their strong C–C and C–H bonds, but they undergo combustion and substitution reactions with halogens in the presence of UV light (free‑radical substitution). The mechanism involves initiation, propagation, and termination steps that produce a mixture of halogenoalkane products.
烷烃是通式为 CₙH₂ₙ₊₂ 的饱和烃。由于 C–C 键和 C–H 键较强，它们的反应活性相对较低，但能发生燃烧反应，以及在紫外光作用下与卤素发生取代反应（自由基取代）。其机理包含链引发、链增长和链终止步骤，产物是卤代烷的混合物。
Alkenes contain a carbon‑carbon double bond (C=C) and have the general formula CₙH₂ₙ. The double bond is an area of high electron density, making alkenes susceptible to electrophilic addition. Common reactions include addition of hydrogen, halogens, hydrogen halides, and steam, as well as oxidation with acidified KMnO₄. The major product of addition to unsymmetrical alkenes often follows Markovnikov’s rule.
烯烃含有碳‑碳双键（C=C），通式为 CₙH₂ₙ。双键区域电子密度高，使得烯烃易于发生亲电加成反应。常见反应包括与氢气、卤素、卤化氢和水蒸气的加成，以及用酸化高锰酸钾进行的氧化反应。不对称烯烃发生加成反应时，主要产物往往遵循马氏规则。
When writing mechanisms, use curly arrows to show electron movement. Clearly indicate the electrophile, the intermediate carbocation, and the final product. In extended‑response questions, always name the organic product and give the displayed formula where required.
书写反应机理时，要用弯箭头表示电子移动。清楚标出亲电试剂、中间体碳正离子和最终产物。在拓展性作答中，要根据要求命名有机产物并写出结构式。
10. Oxidation and Reduction: Redox Reactions | 氧化与还原：氧化还原反应
Oxidation is the loss of electrons, and reduction is the gain of electrons; they always occur simultaneously. Oxidation numbers (or states) help track electron transfer. The oxidation number of an element in its standard state is 0; for a monatomic ion it equals the ion’s charge. In compounds, H is usually +1, O is –2, and the sum of oxidation numbers equals the overall charge.
氧化是指失去电子，还原是指得到电子；两者必然同时发生。氧化数（或氧化态）有助于追踪电子转移。单质的氧化数为 0；单原子离子的氧化数等于其所带电荷。在化合物中，H 通常为 +1，O 为 –2，氧化数的总和等于物质所带的总体电荷。
A chemical change is redox if any atom’s oxidation number changes. Disproportionation is a special type of redox where the same element is simultaneously oxidised and reduced. In exam answers, annotate the oxidation number changes above each species, and explicitly state which species is oxidised and which is reduced.
如果反应中任一原子的氧化数发生变化，则该变化为氧化还原反应。歧化反应是一种特殊的氧化还原反应，其特点是同一元素同时被氧化和被还原。在答题时，应在每一种物质上方标出氧化数的变化，并明确指出何种物质被氧化、何种物质被还原。
Identifying the oxidising agent (the species that is reduced) and the reducing agent (the species that is oxidised) is a standard requirement. Combine this with half‑equations to build full ionic equations. Balance oxygen atoms with H₂O and hydrogen atoms with H⁺ (for acidic conditions) and add electrons to equalise the charge.
辨认氧化剂（自身被还原的物质）和还原剂（自身被氧化的物质）是标准考点。将其与半反应方程式结合，进而构建完整的离子方程式。在酸性条件下，用 H₂O 配平氧原子，用 H⁺ 配平氢原子，并通过添加电子使电荷相等。
11. Bond Polarity and Intermolecular Forces | 键极性与分子间作用力
Electronegativity is the ability of an atom to attract the bonding electrons in a covalent bond. When atoms of different electronegativity are bonded, the bond is polar. If polar bonds are arranged asymmetrically in a molecule, the molecule as a whole has a permanent dipole – it is polar. Symmetrical molecules such as CCl₄ have polar bonds but no overall dipole because the bond dipoles cancel.
电负性是原子在共价键中吸引成键电子的能力。电负性不同的原子成键时，形成极性键。如果极性键在分子中不对称排列，分子整体就会具有永久偶极，即为极性分子。对称型分子如 CCl₄ 虽含有极性键，但因键偶极互相抵消，无整体偶极。
Intermolecular forces include London (dispersion) forces, permanent dipole‑dipole attractions, and hydrogen bonding. London forces exist between all molecules and increase with molecular size. Hydrogen bonding, which occurs between molecules where H is bonded to N, O, or F, is the strongest type of intermolecular force and profoundly affects boiling points and solubility.
分子间作用力包括伦敦（色散）力、永久偶极‑偶极吸引力和氢键。伦敦力存在于所有分子之间，并随分子体积的增大而增强。氢键则存在于含有与 N、O 或 F 键合的 H 的分子之间，属于最强的一类分子间作用力，对物质的沸点和溶解度有显著影响。
In “explain the trend in boiling points” questions, first identify the type of intermolecular force present, then explain how the strength of those forces varies with chain length or branching. Use comparative language such as “more surface contact → stronger London forces → more energy required to separate molecules”.
在“解释沸点变化趋势”类题目中，首先要识别存在的分子间作用力类型，然后解释这些作用力强度如何随链长或支链程度而变化。应使用比较性语言，如“表面接触越多 → 伦敦力越强 → 分离分子所需能量越高”。
12. Applying Core Principles to Example Responses | 将核心原理应用于典型答题
Examiners expect you to select the appropriate principle and communicate it precisely. When you see a command word such as “explain”, “calculate”, or “predict”, pause to identify the underlying concept. For example, a question about the shape of PCl₅ requires VSEPR theory; a question about the electrical conductivity of magnesium needs metallic bonding knowledge.
阅卷人期望你能够选择合适的原理并精准表述。当你看到“解释”“计算”或“预测”等指令词时，应先停下来找出其背后的核心概念。例如，关于 PCl₅ 形状的题目需要运用
Published by TutorHao | 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