📚 AS Chemistry Unit 1 Core Principles: Key Concepts from the January 2021 Exam | AS化学单元1核心原理：2021年1月考试重点概念
AS Chemistry Unit 1 forms the foundation for all further study in the subject, covering atomic structure, bonding, the mole, and an introduction to organic chemistry. The January 2021 examination paper tested these core areas with a balance of recall, calculation, and explanation questions. This article revisits the most important principles examined, clarifying key definitions, trends, and reaction mechanisms with worked examples and clear diagrams of thought. Whether you are preparing for a retake or tackling the unit for the first time, mastering these concepts will help you achieve a high score.
AS化学单元1是化学学习的基础，涵盖原子结构、化学键、摩尔概念以及有机化学入门。2021年1月的试卷通过记忆、计算和解释题均衡地考查了这些核心领域。本文重新梳理其中最重要的原理，厘清关键定义、变化趋势和反应机理，并配有示例和清晰的思路图解。不论你是准备重考还是首次学习该单元，掌握这些概念都将帮助你取得高分。
1. Atomic Structure and Electron Configuration | 原子结构与电子排布
The atom consists of a nucleus containing protons and neutrons, orbited by electrons in shells. The atomic number Z equals the number of protons, while the mass number A is the sum of protons and neutrons. Isotopes are atoms of the same element with different numbers of neutrons. Electrons fill orbitals in a specific order: 1s, 2s, 2p, 3s, 3p, 4s, 3d. The Aufbau principle states that lowest-energy orbitals fill first, and Hund’s rule requires electrons to occupy separate orbitals of a subshell before pairing. The electronic configuration of sulfur (Z=16) is 1s² 2s² 2p⁶ 3s² 3p⁴. Understanding this arrangement is essential for explaining trends in ionisation energy and chemical bonding.
原子由包含质子和中子的原子核及核外电子组成。原子序数Z等于质子数，质量数A是质子数与中子数之和。同位素是质子数相同而中子数不同的原子。电子按特定顺序填充轨道：1s、2s、2p、3s、3p、4s、3d。构造原理指出电子优先占据能量最低的轨道，洪特规则要求电子在简并轨道上先分占再成对。硫（Z=16）的电子排布为1s² 2s² 2p⁶ 3s² 3p⁴。理解电子排布是解释电离能趋势和化学键的基础。
2. Ionisation Energy and Mass Spectrometry | 电离能与质谱
First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions. Across a period, ionisation energy generally increases due to greater nuclear charge and similar shielding, resulting in a stronger attraction for the outer electrons. Down a group, it decreases because the outer electron is further from the nucleus and more shielded, despite the increased nuclear charge. Mass spectrometry provides evidence for electron shells and isotopes. A sample is vaporised, ionised (often by electron impact), accelerated, and deflected by a magnetic field; ions are detected to produce a mass spectrum. The relative atomic mass can be calculated using the formula Σ (isotopic mass × relative abundance) / Σ relative abundance.
第一电离能是指从气态原子中移除一摩尔电子以形成一摩尔气态1+离子所需的能量。同一周期从左到右，电离能总体上升，因为核电荷增大而屏蔽作用相近，最外层电子受到的吸引力增强。同一族从上到下，电离能减小，因为最外层电子离核更远、屏蔽作用增强，即使核电荷增加。质谱技术为电子层和同位素的存在提供了证据。样品经气化、电离（通常电子轰击）、加速后，在磁场中偏转，离子被检测形成质谱图。相对原子质量可按公式Σ（同位素质量 × 相对丰度）/ Σ相对丰度计算。
3. Bonding: Ionic, Covalent and Metallic | 化学键：离子键、共价键与金属键
Ionic bonding involves the transfer of electrons from a metal to a non-metal, forming oppositely charged ions held in a giant lattice by electrostatic forces. Covalent bonding occurs between non-metal atoms that share pairs of electrons; a single bond is one shared pair, a double bond two pairs, and a triple bond three pairs. Polar covalent bonds arise when atoms of different electronegativities share electrons unequally, creating a dipole. Metallic bonding consists of a regular array of positive metal ions surrounded by a sea of delocalised electrons; this explains high electrical and thermal conductivity as well as malleability. January 2021 questions often asked candidates to relate bonding type to physical properties such as melting point and solubility.
离子键涉及电子从金属转移到非金属，形成带相反电荷的离子，通过静电作用力构成巨型晶格。共价键发生在非金属原子之间，通过共用电子对形成；单键共用一对电子，双键共用两对，三键共用三对。当电负性不同的原子共用电子时，产生极性共价键，形成偶极。金属键由规则排列的正金属离子和离域电子海构成，这解释了金属具有高导电性、高导热性和延展性。2021年1月的试题常要求考生将键合类型与熔点和溶解性等物理性质联系起来。
4. Molecular Shapes and Polarity | 分子形状与极性
The shape of a molecule is determined by the number of electron pairs (bonding and lone pairs) around the central atom, according to VSEPR theory. Electron pairs repel each other and arrange themselves as far apart as possible. Methane (CH₄) has 4 bonding pairs and no lone pairs → tetrahedral, bond angle 109.5°. Ammonia (NH₃) has 3 bonding pairs and 1 lone pair → trigonal pyramidal, angle about 107°. Water (H₂O) has 2 bonding pairs and 2 lone pairs → bent, angle approximately 104.5°. Carbon dioxide (CO₂) is linear with 180° angles because of two double-bond regions. The overall polarity of a molecule depends on both bond polarity and molecular symmetry; CO₂ is non-polar despite polar C=O bonds due to its linear shape, while H₂O is polar because of its bent geometry.
根据价层电子对互斥理论，分子的形状由中心原子周围的电子对数（成键电子对和孤对电子）决定。电子对互相排斥，尽可能远离。甲烷（CH₄）有4个成键电子对、无孤对电子 → 四面体形，键角109.5°。氨（NH₃）有3个成键电子对和1个孤对电子 → 三角锥形，键角约107°。水（H₂O）有2个成键电子对和2个孤对电子 → 角形，键角约104.5°。二氧化碳（CO₂）有两个双键区域，呈直线形，键角180°。分子的整体极性取决于键的极性和分子对称性；CO₂尽管C=O键为极性键，但因直线形对称而成为非极性分子，而水分子因角形结构呈现极性。
5. Intermolecular Forces | 分子间作用力
Intermolecular forces are attractions between molecules that affect physical properties such as boiling and melting points. London dispersion forces exist between all molecules and arise from temporary fluctuations in electron density, inducing temporary dipoles; their strength increases with the number of electrons and molecular surface area. Permanent dipole–dipole interactions occur between polar molecules, adding to the overall attraction. Hydrogen bonding is a particularly strong dipole–dipole interaction when hydrogen is covalently bonded to highly electronegative nitrogen, oxygen, or fluorine and is attracted to a lone pair on another such atom. Water’s unexpectedly high boiling point and ice’s low density are direct consequences of hydrogen bonding. In the January 2021 paper, candidates had to explain trends in boiling points of hydrides or organic compounds by referring to the dominant intermolecular forces.
分子间作用力是分子之间的吸引力，影响着沸点、熔点等物理性质。伦敦色散力存在于所有分子之间，由瞬时电子密度涨落引起的瞬时偶极诱发，其强度随电子数和分子表面积的增加而增大。永久偶极-偶极作用力存在于极性分子之间，进一步增强了分子间的吸引力。氢键是一种特别强的偶极-偶极作用，当氢原子与电负性极强的氮、氧或氟原子共价结合后，被另一个这类原子上的孤对电子所吸引。水的沸点异常高以及冰的密度较低，都是氢键的直接结果。2021年1月的试卷中，考生需通过指认主要分子间力来解释氢化物或有机物沸点的变化趋势。
6. The Mole and Stoichiometry | 摩尔与化学计量
The mole is the SI unit for amount of substance; one mole contains exactly 6.02214076 × 10²³ elementary entities (Avogadro’s constant). The molar mass (g mol⁻¹) of a substance is the mass per mole of its formula units. In a balanced chemical equation, coefficients represent the mole ratio of reactants and products. Stoichiometry uses these ratios to calculate masses, volumes, or concentrations of substances involved in a reaction. For example, the decomposition of calcium carbonate: CaCO₃ → CaO + CO₂ shows a 1:1:1 mole ratio. If 10.0 g of CaCO₃ (M = 100.1 g mol⁻¹) is heated, the amount is 10.0/100.1 = 0.0999 mol, which produces 0.0999 mol of CO₂, occupying about 2.24 dm³ at room temperature and pressure (24 dm³ mol⁻¹). The limiting reagent concept is essential when reactants are not in stoichiometric proportion; the reactant that is completely consumed determines the maximum yield.
摩尔是物质数量的SI单位，1摩尔恰好含有6.02214076 × 10²³个基本单元（阿伏伽德罗常数）。物质的摩尔质量（g mol⁻¹）是每摩尔式量单位的质量。在配平的化学方程式中，系数表示反应物和生成物的摩尔比。化学计量学利用这些比例来计算反应中物质的质量、体积或浓度。例如碳酸钙的分解：CaCO₃ → CaO + CO₂，摩尔比为1:1:1。若加热10.0 g CaCO₃（M = 100.1 g mol⁻¹），物质的量为0.0999 mol，产生0.0999 mol CO₂，在室温常压下（气体摩尔体积24 dm³ mol⁻¹）约占有2.24 dm³。当反应物并非按化学计量比例投料时，必须考虑限量试剂；完全消耗掉的反应物决定了最大产率。
7. Empirical and Molecular Formulae | 经验式与分子式
The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound, whereas the molecular formula shows the actual number of atoms of each element in a molecule. Combustion analysis or elemental composition data can be used to determine the empirical formula. The steps are: convert masses (or percentages) to moles by dividing by atomic masses, then divide each mole value by the smallest to obtain the simplest ratio. The molecular formula is found by comparing the empirical formula mass to the relative molecular mass (Mᵣ). For instance, a compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Moles: C = 40.0/12.0 = 3.33, H = 6.7/1.0 = 6.7, O = 53.3/16.0 = 3.33. Dividing by 3.33 gives CH₂O as the empirical formula. If the Mᵣ is 180, then the molecular formula is C₆H₁₂O₆ (n = 180/30 = 6).
经验式表示化合物中各元素原子的最简整数比，而分子式表示分子中各元素原子的实际数目。可以通过燃烧分析或元素组成数据求算经验式。步骤为：将质量（或百分比）除以相对原子质量转化为摩尔数，再将各摩尔值除以最小值以获得最简整数比。通过比较经验式质量与相对分子质量（Mᵣ）可确定分子式。例如，某化合物含碳40.0%、氢6.7%、氧53.3%。摩尔数：C = 40.0/12.0 = 3.33，H = 6.7/1.0 = 6.7，O = 53.3/16.0 = 3.33。同时除以3.33得到经验式CH₂O。若Mᵣ为180，则分子式为C₆H₁₂O₆（n = 180/30 = 6）。
8. Alkanes and Free-Radical Substitution | 烷烃与自由基取代
Alkanes are saturated hydrocarbons with the general formula CₙH₂ₙ₊₂. They are relatively unreactive due to the strength of C–C and C–H bonds, but undergo free-radical substitution with halogens in the presence of ultraviolet light. The mechanism proceeds in three stages: initiation, propagation, and termination. Initiation: Cl₂ → 2Cl• (homolytic fission by UV). Propagation: Cl• + CH₄ → •CH₃ + HCl, then •CH₃ + Cl₂ → CH₃Cl + Cl•. Termination: two radicals combine, e.g. Cl• + Cl• → Cl₂. The product is a mixture of halogenoalkanes, as further substitution can occur. In the exam, you may be asked to write equations for the propagation steps using skeletal or displayed formulae. Safety note: these reactions must be carried out with care due to the poisonous nature of chlorine.
烷烃是通式为CₙH₂ₙ₊₂的饱和烃。由于C–C和C–H键键能较高，烷烃较为稳定，但在紫外光下能与卤素发生自由基取代反应。反应机理分为链引发、链增长和链终止三个阶段。引发：Cl₂ → 2Cl•（紫外光导致键均裂）。增长：Cl• + CH₄ → •CH₃ + HCl，接着•CH₃ + Cl₂ → CH₃Cl + Cl•。终止：两个自由基结合，如Cl• + Cl• → Cl₂。产物通常是卤代烷的混合物，因还会发生进一步取代。考试中可能要求用骨架式或结构式写出链增长步骤的方程式。安全须知：此类反应因氯气有毒而须谨慎进行。
9. Alkenes and Electrophilic Addition | 烯烃与亲电加成
Alkenes contain at least one carbon–carbon double bond (C=C) and have the general formula CₙH₂ₙ. The π‑bond in the double bond is an area of high electron density, making alkenes susceptible to attack by electrophiles. The characteristic reaction is electrophilic addition. For example, ethene reacts with bromine (Br₂) at room temperature, decolourising the orange-brown bromine water. The mechanism involves the polarisation of the Br–Br bond as it approaches the electron-rich double bond; a cyclic bromonium ion intermediate may form, followed by attack of a bromide ion to give 1,2-dibromoethane. With unsymmetrical alkenes and polar reagents such as HBr, Markovnikov’s rule applies: the hydrogen atom of the reagent attaches to the carbon with more hydrogen atoms already, leading to the more stable carbocation intermediate. Alkenes also exhibit geometric (E/Z) isomerism due to restricted rotation about the C=C bond.
烯烃含有至少一个碳碳双键（C=C），通式为CₙH₂ₙ。双键中的π键是电子密度较高的区域，使烯烃容易受到亲电试剂的进攻。其特征反应为亲电加成。例如，乙烯与溴在室温下反应，能使橙黄色的溴水褪色。机理为：Br–Br键在接近富电子的双键时被极化，可能形成环状溴鎓离子中间体，随后溴离子进攻生成1,2-二溴乙烷。对于不对称烯烃与极性试剂（如HBr）的反应，遵循马尔科夫尼科夫规则：试剂中的氢原子加在原本氢原子较多的碳上，从而形成更稳定的碳正离子中间体。由于C=C双键不能自由旋转，烯烃还能表现出几何异构（E/Z异构）。
10. Isomerism and Oxidation Numbers | 同分异构与氧化数
Structural isomers have the same molecular formula but different structural formulae. In AS Unit 1, you encounter chain isomerism (different carbon skeleton), position isomerism (functional group at different position), and functional group isomerism (different functional group, e.g. alcohols and ethers). Stereoisomerism, specifically E/Z isomerism, occurs in alkenes where each carbon of the double bond is attached to two different groups; the Z isomer has the higher-priority groups on the same side, while the E isomer has them on opposite sides. Separately, oxidation numbers are useful for identifying redox reactions. The rules include: free elements have oxidation number 0; hydrogen is usually +1; oxygen is usually –2; the sum of oxidation numbers in a neutral compound is zero. A substance is oxidised if its oxidation number increases, and reduced if it decreases. For example, in the reaction 2Mg + O₂ → 2MgO, Mg goes from 0 to +2 (oxidation) and O from 0 to –2 (reduction).
结构异构体具有相同的分子式但结构式不同。在AS单元1中，你会遇到碳链异构（碳骨架不同）、位置异构（官能团位置不同）和官能团异构（官能团不同，如醇和醚）。立体异构，特别是E/Z异构，发生在双键碳上各连有两个不同基团的烯烃中；Z异构体中优先基团在双键同侧，E异构体中优先基团在异侧。另外，氧化数规则有助于识别氧化还原反应。主要规则：游离态单质的氧化数为0；氢通常为+1；氧通常为–2；中性化合物中氧化数总和为0。氧化数升高则为被氧化，降低则为被还原。例如反应2Mg + O₂ → 2MgO中，Mg由0升至+2（被氧化），O由0降至–2（被还原）。
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