📚 A-Level AQA Chemistry: Atomic Structure Key Points | A-Level AQA 化学：原子结构 考点精讲
Atomic structure is the foundation of all chemical understanding. In the AQA A-Level Chemistry specification, you are expected to describe the fundamental particles, explain the arrangement of electrons, interpret mass spectra, and analyse ionisation energy trends. This article breaks down every essential topic, pairing clear English explanations with Chinese translations so you can master the content from both language perspectives.
原子结构是所有化学理解的基础。在 AQA A-Level 化学考试大纲中，你需要描述基本粒子，解释电子排布，分析质谱图，并解释电离能的趋势。本文分解每一个核心主题，配合清晰的中英文对照讲解，帮助你从双语角度掌握这些内容。
1. Subatomic Particles | 亚原子粒子
Atoms consist of three types of subatomic particles: protons, neutrons, and electrons. Protons and neutrons are found in the nucleus, while electrons orbit the nucleus in energy levels. The properties of these particles are fundamental to understanding atomic behaviour.
原子由三种亚原子粒子组成：质子、中子和电子。质子和中子位于原子核内，电子则在能级中绕核运动。这些粒子的性质是理解原子行为的基础。
| Particle / 粒子 | Relative mass / 相对质量 | Relative charge / 相对电荷 | Symbol / 符号 |
|---|---|---|---|
| Proton / 质子 | 1 | +1 | p⁺ |
| Neutron / 中子 | 1 | 0 | n⁰ |
| Electron / 电子 | 1/1836 (~0) | -1 | e⁻ |
Notice that the mass of an electron is negligible compared with protons and neutrons, so most of the atom’s mass is concentrated in the nucleus. The nucleus is positively charged because of the protons, and the atom as a whole is electrically neutral when the number of protons equals the number of electrons.
注意，电子的质量与质子和中子相比可以忽略不计，因此原子的绝大部分质量集中在原子核。原子核因质子而带正电荷，当质子数与电子数相等时，整个原子呈电中性。
2. Atomic Number and Mass Number | 原子序数与质量数
The atomic number (Z) is the number of protons in the nucleus of an atom. It defines the element. For example, carbon always has 6 protons, so Z = 6. The mass number (A) is the total number of protons and neutrons in the nucleus.
原子序数（Z）是原子核内质子的数目，它决定了元素的种类。例如，碳总是有6个质子，因此 Z = 6。质量数（A）是原子核内质子数与中子数的总和。
These numbers are shown in standard nuclear notation: AZX, where X is the chemical symbol. So for carbon-12, we write 126C. From this, the number of neutrons can be calculated as A − Z.
这些数字用标准核素符号表示：AZX，其中 X 是元素符号。因此对于碳-12，写作 126C。由此，中子数可由 A − Z 算出。
3. Isotopes | 同位素
Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. They have the same atomic number but different mass numbers. For example, carbon has three naturally occurring isotopes: 12C, 13C, and 14C. Their chemical properties are virtually identical because they have the same electron configuration, but their physical properties (such as mass and density) differ slightly.
同位素是指具有相同质子数但中子数不同的同种元素的原子。它们原子序数相同但质量数不同。例如，碳有三种天然同位素：12C、13C 和 14C。由于电子排布相同，它们的化学性质几乎完全相同，但物理性质（如质量和密度）略有差异。
4. Relative Atomic Mass and Mass Spectrometry | 相对原子质量与质谱
The relative atomic mass (Ar) of an element is the weighted mean mass of an atom of the element compared to 1/12th the mass of an atom of carbon-12. It takes into account the percentage abundance of each isotope.
元素的相对原子质量（Ar）是指该元素一个原子的加权平均质量与一个碳-12原子质量的1/12的比值。它考虑了每种同位素的丰度百分比。
A mass spectrometer is used to determine isotopic abundances and relative atomic masses. The key stages are: ionisation (the sample is vaporised and bombarded with high-energy electrons to form positive ions), acceleration (ions are accelerated by an electric field), deflection (a magnetic field deflects ions according to their mass-to-charge ratio, m/z), and detection (the ion current is recorded).
质谱仪用于测定同位素丰度及相对原子质量。关键步骤为：电离（样品气化后经高能电子轰击形成正离子），加速（离子经电场加速），偏转（磁场按离子的质荷比 m/z 使其偏转），以及检测（记录离子电流）。
The mass spectrum shows peaks corresponding to different isotopes. The position of each peak along the x-axis gives the m/z value, which for singly charged ions is the mass number. The relative peak heights indicate the percentage abundance. You can calculate Ar using the formula: Ar = Σ (isotopic mass × % abundance) / 100.
质谱图上显示与不同同位素对应的峰。每个峰在 x 轴的位置给出 m/z 值，对于单电荷离子等于质量数。相对峰高表明丰度百分比。你可以用公式 Ar = Σ（同位素质量 × 丰度%）/ 100 来计算。
5. Electron Configuration – Shells and Subshells | 电子排布——电子层与亚层
Electrons occupy principal energy levels (shells) labelled n = 1, 2, 3, 4 … As n increases, the energy and distance from the nucleus increase. Each shell contains one or more subshells: s, p, d and f. The number of subshells in a shell equals n. For instance, n = 2 has subshells 2s and 2p.
电子占据标记为 n = 1, 2, 3, 4 … 的主能级（电子层）。n 越大，能量越高，离核越远。每个电子层包含一个或多个亚层：s, p, d 和 f。一个电子层中亚层的数目等于 n。例如，n = 2 有 2s 和 2p 亚层。
Each subshell contains a fixed number of orbitals: s subshell has 1 orbital, p has 3 orbitals, d has 5 orbitals, and f has 7 orbitals. Each orbital can hold a maximum of 2 electrons. Therefore, an s subshell holds 2 electrons, a p subshell holds 6, a d subshell holds 10, and an f subshell holds 14 electrons.
每个亚层含有固定数量的轨道：s 亚层有1个轨道，p 有3个，d 有5个，f 有7个。每个轨道最多容纳2个电子。因此，s 亚层可容纳2个电子，p 亚层6个，d 亚层10个，f 亚层14个电子。
6. Filling Orbitals – Aufbau Principle, Hund’s Rule | 轨道填充——构造原理与洪特规则
Electrons fill atomic orbitals in order of increasing energy. The Aufbau principle states that lower energy orbitals are filled before higher energy ones. The order for the first four shells is: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p. Note that the 4s subshell fills before 3d because it is slightly lower in energy.
电子按能量递增顺序填充原子轨道。构造原理指出，低能级轨道先于高能级轨道被填充。前四个电子层的顺序为：1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p。注意 4s 亚层的能量略低于 3d，因此先填充 4s。
Hund’s rule says that electrons will occupy degenerate (equal-energy) orbitals singly with parallel spins before pairing up. This minimises electron–electron repulsion. For example, nitrogen (Z = 7) has the configuration 1s² 2s² 2p³, where the three 2p electrons occupy separate 2p orbitals with the same spin.
洪特规则指出，电子在简并（等能量）轨道中先以自旋平行方式单独占据，而后才配对。这使电子间排斥最小。例如，氮（Z = 7）的电子排布为 1s² 2s² 2p³，三个 2p 电子分别占据三个不同的 2p 轨道且自旋平行。
When writing electron configurations for atoms and ions, you can use the full notation, e.g., 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ for zinc, or the shorthand [Ar] 4s² 3d¹⁰. The shorthand uses the preceding noble gas in brackets. For transition metal ions, electrons are removed from the 4s subshell before 3d, e.g., Fe²⁺: [Ar] 3d⁶, not [Ar] 4s² 3d⁴.
在书写原子和离子的电子排布时，可用完整方式，如锌为 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰，或用简写 [Ar] 4s² 3d¹⁰。简写中使用前一个稀有气体符号。对于过渡金属离子，电子优先从 4s 亚层失去而非 3d，例如 Fe²⁺ 为 [Ar] 3d⁶，而不是 [Ar] 4s² 3d⁴。
7. First Ionisation Energy – Definition and Factors | 第一电离能——定义与影响因素
The 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. Equation: X(g) → X⁺(g) + e⁻. It is an endothermic process, measured in kJ mol⁻¹.
第一电离能是指从一摩尔气态原子中移去一摩尔电子，形成一摩尔气态1+离子所需的能量。方程式：X(g) → X⁺(g) + e⁻。这是一个吸热过程，单位为 kJ mol⁻¹。
Three factors affect ionisation energy: nuclear charge (more protons means stronger attraction), distance of the electron from the nucleus (greater distance means weaker attraction), and shielding (inner-shell electrons repel outer electrons, reducing the effective nuclear pull). As nuclear charge increases and shielding remains similar across a period, ionisation energy generally increases. Down a group, increased distance and shielding cause ionisation energy to decrease.
影响电离能的因素有三个：核电荷（质子越多，吸引力越强）、电子与核的距离（距离越大，吸引力越弱）以及屏蔽效应（内层电子排斥外层电子，减弱有效核引力）。在同一周期中，核电荷增加而屏蔽基本不变，电离能总体上增大。沿族向下，距离增大和屏蔽增加使电离能减小。
8. Trends in First Ionisation Energy Across Period 3 | 第一电离能在第三周期的变化趋势
Across Period 3 (Na to Ar), the general trend is a rise in first ionisation energy because the nuclear charge increases while electrons are added to the same outer shell (n = 3), and shielding by inner shells remains constant. However, there are two small drops: between Mg and Al, and between P and S.
在第三周期（Na 到 Ar）中，总体趋势是第一电离能升高，因为核电荷增加而电子均填入同一外层（n = 3），内层屏蔽保持恒定。但存在两处小幅下降：Mg 到 Al 之间，以及 P 到 S 之间。
The drop from Mg (3s²) to Al (3p¹) occurs because the 3p electron in Al is higher in energy and further from the nucleus than the 3s electrons, and it is also slightly shielded by the 3s electrons. The drop from P (3p³) to S (3p⁴) is due to electron–electron repulsion: in P all three 3p electrons are unpaired, whereas in S one 3p orbital contains a pair of electrons, and the repulsion between them makes it easier to remove an electron.
Mg (3s²) 到 Al (3p¹) 的下降是因为 Al 的 3p 电子能量更高、离核更远，且受 3s 电子轻微屏蔽。P (3p³) 到 S (3p⁴) 的下降则是由于电子-电子排斥：P 中三个 3p 电子均不成对，而在 S 中一个 3p 轨道含有一对电子，它们之间的排斥使移去一个电子变得更容易。
9. Successive Ionisation Energies as Evidence for Shells | 逐级电离能作为电子层存在的证据
Successive ionisation energies refer to the removal of further electrons after the first: X⁺(g) → X²⁺(g) + e⁻ (second I.E.), etc. Within an element, each successive ionisation energy is larger than the previous because the electron is being removed from a species with an increasing positive charge.
逐级电离能指在第一电离能之后继续移去电子：X⁺(g) → X²⁺(g) + e⁻（第二电离能）等。对同一元素，每一级电离能都比前一级大，因为电子是从正电荷越来越高的粒子中移去。
A sudden large jump in successive ionisation energies indicates a change to a new inner shell closer to the nucleus. For example, sodium (Na: 1s² 2s² 2p⁶ 3s¹) shows a massive increase between the first and second ionisation energy because the second electron is removed from the 2p subshell, which is much closer to the nucleus and experiences less shielding. This provides direct evidence for the existence of electron shells.
逐级电离能中出现急剧增大的跳跃，表明进入了更靠近核的新内层。例如，钠（Na: 1s² 2s² 2p⁶ 3s¹）在第一和第二电离能之间显示出剧增，因为第二个电子是从 2p 亚层移去，这一亚层离核近得多且受屏蔽较小。这为电子层的存在提供了直接证据。
10. Summary and Exam Tips | 总结与应试技巧
To succeed in AQA questions, you must be able to define key terms accurately, write electron configurations using s p d notation, interpret mass spectra to calculate relative atomic mass, and explain ionisation energy trends in terms of nuclear charge, distance, and shielding. Practice drawing labelled diagrams of mass spectrometers and be ready to explain the two anomalies in Period 3 first ionisation energies.
要在 AQA 考题中取得好成绩，你必须准确定义关键术语，用 s p d 符号书写电子排布，解读质谱图以计算相对原子质量，并从核电荷、距离和屏蔽的角度解释电离能的变化趋势。练习绘制质谱仪示意简图并标注，并准备好解释第三周期第一电离能中的两个异常。
Remember: the specification expects you to be thorough with exceptions such as chromium ( [Ar] 4s¹ 3d⁵ instead of 4s² 3d⁴) and copper ([Ar] 4s¹ 3d¹⁰). Always show your working when calculating Ar. Keep your explanations concise and refer to the forces and energy changes involved.
记住：考纲要求你熟练掌握特例，如铬（[Ar] 4s¹ 3d⁵ 而非 4s² 3d⁴）和铜（[Ar] 4s¹ 3d¹⁰）。计算 Ar 时务必展示解题步骤。解释力求简洁，适时提及所涉及的力和能量变化。
With a firm grasp of these atomic structure fundamentals, you will build a solid platform for bonding, periodicity, and the entire A-Level Chemistry course.
牢固掌握这些原子结构基础知识，你将为进一步学习化学键、周期性以及整个 A-Level 化学课程打下坚实基础。
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Cancel reply