Atomic Structure in A-Level Chemistry | A-Level 化学：原子结构考点精讲

📚 Atomic Structure in A-Level Chemistry | A-Level 化学：原子结构考点精讲

Understanding atomic structure is the foundation for mastering A-Level Chemistry. From the arrangement of subatomic particles to electron configurations and ionisation energies, this topic connects directly to bonding, periodicity, and reactivity. This revision guide breaks down every key concept you need to excel in your exams, with clear explanations in both English and Chinese.

掌握原子结构是学好 A-Level 化学的基础。从亚原子粒子的排布到电子构型、电离能，这个主题直接关系到化学键、周期律和反应活性。本篇复习指南用中英双语为你拆解每个必须掌握的核心概念，助你轻松应对考试。

1. Fundamental Particles: Protons, Neutrons, Electrons | 基本粒子：质子、中子、电子

Atoms consist of three types of subatomic particles. Protons carry a positive charge and are found in the nucleus; neutrons are neutral and also reside in the nucleus; electrons are negatively charged and orbit the nucleus in energy levels. The relative masses of these particles are 1 for a proton, 1 for a neutron, and approximately 1/1836 for an electron.

原子由三种亚原子粒子组成。质子带正电，位于原子核内；中子不带电，同样存在于原子核中；电子带负电，在原子核外的能级上运动。它们的相对质量分别为：质子 1、中子 1、电子约为 1/1836。

Properties of subatomic particles can be summarised:

亚原子粒子的性质可总结如下：

Proton: charge +1, mass ~1 u, located in nucleus | 质子：电量 +1，质量约 1 u，位于原子核内
Neutron: charge 0, mass ~1 u, located in nucleus | 中子：电量 0，质量约 1 u，位于原子核内
Electron: charge -1, mass ~1/1836 u, located in orbitals outside nucleus | 电子：电量 -1，质量约 1/1836 u，位于核外轨道

The number of protons defines the element (atomic number, Z); the sum of protons and neutrons gives the mass number (A).

质子数决定了元素的种类（原子序数 Z）；质子数与中子数之和为质量数（A）。

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. In a neutral atom, the number of electrons equals Z. The mass number (A) is the total number of protons and neutrons. Isotopes of an element have the same Z but different A due to varying numbers of neutrons.

原子序数（Z）是原子核内的质子数。在中性原子中，电子数等于 Z。质量数（A）是质子数与中子数的总和。同一元素的同位素具有相同的 Z，但因中子数不同而导致 A 不同。

Chemical properties are determined by the electron configuration, so isotopes exhibit nearly identical chemical behaviour.

化学性质由电子排布决定，因此同位素的化学行为几乎完全相同。

3. Isotopes and Relative Atomic Mass | 同位素与相对原子质量

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. Because the chemical properties are governed by the electron arrangement, isotopes react in the same way. The relative atomic mass (Aᵣ) of an element takes into account the mass and abundance of each isotope.

同位素是指质子数相同而中子数不同的同种元素的原子。由于化学性质由电子排布决定，同位素的反应方式相同。元素的相对原子质量（Aᵣ）综合考虑了每种同位素的质量和丰度。

Relative atomic mass is calculated using the formula:

相对原子质量的计算公式为：

Aᵣ = Σ (isotopic mass × % abundance) / 100

or, when abundances are given as fractions: Aᵣ = Σ (isotopic mass × fractional abundance). The mass spectrometer is used to obtain data on isotopic masses and their relative abundances.

或者，当丰度以分数给出时：Aᵣ = Σ (同位素质量 × 丰度分数)。质谱仪可用于获取同位素质量及其相对丰度的数据。

4. Mass Spectrometry: Determining Isotopic Abundance | 质谱法：测定同位素丰度

A mass spectrometer can determine the relative isotopic masses and the percentage abundances of isotopes. The sample is vaporised, ionised (usually by electron impact or electrospray), accelerated, deflected by a magnetic field, and detected. Ions are separated according to their mass-to-charge ratio (m/z).

质谱仪可以测定相对同位素质量以及同位素的丰度百分比。样品经过气化、电离（通常采用电子轰击或电喷雾）、加速、通过磁场偏转后被检测。离子根据其质荷比（m/z）被分离。

For an element, the mass spectrum shows peaks for each isotope. The height of each peak is proportional to the abundance. Using the data, you can calculate Aᵣ. For example, chlorine has two main isotopes: ³⁵Cl (75%) and ³⁷Cl (25%). Aᵣ = (35 × 75 + 37 × 25) / 100 = 35.5.

对于元素，质谱图上会显示每种同位素的峰，峰高与丰度成正比。利用这些数据可以计算 Aᵣ。例如，氯有两种主要同位素：³⁵Cl（75%）和 ³⁷Cl（25%）。Aᵣ = (35 × 75 + 37 × 25) / 100 = 35.5。

5. Electron Configuration: Shells, Subshells, and Orbitals | 电子排布：电子层、亚层与轨道

Electrons occupy energy levels known as shells, labelled with principal quantum numbers n = 1, 2, 3, etc. Each shell contains subshells: s, p, d, f. An orbital is a region where there is a high probability of finding an electron. Each orbital can hold a maximum of two electrons with opposite spins.

电子占据称为电子层的能级，主量子数以 n = 1, 2, 3 等标记。每层包含亚层：s、p、d、f。轨道是电子出现概率很高的区域。每个轨道最多可容纳两个自旋相反的电子。

The s subshell has 1 orbital (holds 2 electrons), p has 3 orbitals (6 electrons), d has 5 orbitals (10 electrons), and f has 7 orbitals (14 electrons). The order of filling follows the Aufbau principle, but for A-Level, note the 4s orbital fills before 3d, yet 4s electrons are lost before 3d when forming transition metal ions.

s 亚层有 1 个轨道（容纳 2 个电子），p 有 3 个轨道（6 个电子），d 有 5 个轨道（10 个电子），f 有 7 个轨道（14 个电子）。填充顺序遵循构造原理，但在 A-Level 中需注意 4s 轨道先于 3d 被填充，而过渡金属形成离子时 4s 电子先于 3d 电子失去。

6. Quantum Numbers and Orbital Shapes (s, p, d) | 量子数与轨道形状 (s, p, d)

Each electron in an atom is described by four quantum numbers: principal (n), azimuthal/angular momentum (l), magnetic (mₗ), and spin (mₛ). However, at A-Level, you mainly focus on the shapes and relative energies of s and p orbitals, with an introduction to d orbitals.

原子中每个电子由四个量子数描述：主量子数（n）、角量子数（l）、磁量子数（mₗ）和自旋量子数（mₛ）。但在 A-Level 中，主要关注的是 s 和 p 轨道的形状及相对能量，并初步了解 d 轨道。

s orbitals are spherical. p orbitals have a dumbbell shape and are oriented along the x, y, z axes (pₓ, pᵧ, p₂). d orbitals have more complex cloverleaf shapes (with one having a doughnut shape). Understanding orbital shapes helps explain molecular geometry and bonding.

s 轨道呈球形；p 轨道为哑铃形，沿 x、y、z 轴定向（pₓ、pᵧ、p₂）；d 轨道具有更复杂的四叶草形状（其中有一个轨道呈甜甜圈形）。了解轨道形状有助于解释分子的几何构型和化学键。

7. Filling Orbitals: Aufbau Principle, Hund’s Rule, Pauli Exclusion | 轨道填充：构造原理、洪特规则、泡利不相容

To write electron configurations correctly, three rules must be followed. The Aufbau principle states that electrons fill the lowest energy orbitals first. Hund’s rule says that electrons fill degenerate orbitals (orbitals of the same energy) singly with parallel spins before pairing up. Pauli exclusion principle: no two electrons in an atom can have the same set of four quantum numbers; an orbital can hold at most two electrons with opposite spins.

要正确书写电子排布，必须遵循三条规则。构造原理：电子优先填充能量最低的轨道。洪特规则：电子在填充简并轨道（能量相同的轨道）时，会先以相同的自旋方向单独占据，然后再配对。泡利不相容原理：原子内没有两个电子具有完全相同的四个量子数；每个轨道最多容纳两个自旋相反的电子。

For example, the electron configuration of nitrogen (Z=7) is 1s² 2s² 2p³. The three 2p electrons occupy three separate p orbitals with parallel spins. Oxygen (Z=8) adds one more electron, causing one 2p orbital to have a pair: 1s² 2s² 2p⁴. Using box-and-arrow diagrams can help visualise spin.

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 p 轨道，且自旋平行。氧（Z=8）增加一个电子，导致一个 2p 轨道出现配对：1s² 2s² 2p⁴。使用轨道方框图有助于直观表示自旋。

8. Electron Configurations of Ions: Transition Metals | 离子的电子排布：过渡金属

When atoms form positive ions, electrons are removed from the highest energy level (outermost shell) first. For the first 20 elements, this is straightforward: Na: 1s² 2s² 2p⁶ 3s¹ becomes Na⁺: 1s² 2s² 2p⁶. However, transition metals present an important exception: when forming cations, the 4s electrons are lost before the 3d electrons.

当原子形成阳离子时，电子首先从最高能级（最外层）失去。对前 20 号元素这很直接：Na: 1s² 2s² 2p⁶ 3s¹ → Na⁺: 1s² 2s² 2p⁶。然而，过渡金属是一个重要的例外：形成阳离子时，4s 电子先于 3d 电子失去。

For instance, iron (Fe, Z=26) has electron configuration [Ar] 3d⁶ 4s². Fe²⁺ loses the two 4s electrons giving [Ar] 3d⁶, not [Ar] 3d⁴ 4s². Fe³⁺ loses one 3d electron plus the two 4s: [Ar] 3d⁵. This is because once the 3d orbitals are occupied, they become lower in energy than 4s; thus 4s electrons are removed first.

例如，铁（Fe，Z=26）的电子排布为 [Ar] 3d⁶ 4s²。Fe²⁺ 先失去两个 4s 电子，形成 [Ar] 3d⁶，而不是 [Ar] 3d⁴ 4s²。Fe³⁺ 再失去一个 3d 电子以及两个 4s 电子，得到 [Ar] 3d⁵。这是因为一旦 3d 轨道被占据，其能量会低于 4s，因此 4s 电子优先被移走。

9. First Ionisation Energy: Definition and Trends | 第一电离能：定义与趋势

The first ionisation energy (IE₁) is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous ions with a single positive charge:

第一电离能（IE₁）是指从一摩尔气态原子中移除一摩尔电子，形成一摩尔带一个正电荷的气态离子所需的能量：

X(g) → X⁺(g) + e⁻

Across a period, IE₁ generally increases because nuclear charge increases while shielding remains similar, pulling the outer electrons more strongly. However, there are two notable drops in the trend: from Group 2 to Group 3 (e.g., Be to B) and from Group 5 to Group 6 (e.g., N to O).

在同一周期中，由于核电荷增加而屏蔽效应几乎不变，对外层电子的吸引力增强，因此 IE₁ 总体呈上升趋势。但趋势中存在两次明显的下降：从第 2 族到第 3 族（如 Be 到 B）以及从第 5 族到第 6 族（如 N 到 O）。

Going down a group, IE₁ decreases. This is because the outer electron is in a higher energy level, farther from the nucleus, and experiences greater shielding by inner electrons, making it easier to remove.

向下沿同一族，IE₁ 下降。这是因为外层电子处于更高的能级，离核更远，且受到内层电子更强的屏蔽，更容易被移除。

10. Factors Affecting Ionisation Energy | 影响电离能的要素

Three main factors influence ionisation energy:

影响电离能大小的三个主要因素：

Atomic radius: the greater the distance between the nucleus and the outer electrons, the weaker the attraction, lowering IE. | 原子半径：原子核与外层电子距离越大，吸引力越弱，电离能越低。
Nuclear charge: a higher number of protons increases the attractive force on electrons, raising IE. | 核电荷：质子数越多，对电子的吸引力越大，电离能越高。
Shielding: inner electrons repel outer electrons, reducing the effective nuclear attraction; more shielding lowers IE. | 屏蔽效应：内层电子排斥外层电子，削弱有效核引力；屏蔽越多，电离能越低。

The drop from Be to B occurs because the 2p electron in B is higher in energy and better shielded than the 2s electrons in Be. The drop from N to O occurs because in O, one 2p orbital contains a pair of electrons that experience repulsion, making it easier to remove one electron.

从 Be 到 B 的下降是由于 B 的 2p 电子比 Be 的 2s 电子能量更高、屏蔽更强。从 N 到 O 的下降是由于 O 中某一个 2p 轨道含有一对电子，电子间排斥作用使其更容易失去一个电子。

11. Successive Ionisation Energies and Evidence for Shells | 连续电离能与电子层证据

Successive ionisation energies are the energies needed to remove electrons one after another from a gaseous ion. For an element, these values provide strong evidence for the existence of electron shells. A large jump in successive ionisation energies indicates the removal of an electron from a new, inner shell that is much closer to the nucleus and experiences less shielding.

连续电离能是指从气态离子中逐个移除电子所需的能量。对于某元素，这些数据为电子层的存在提供了有力证据。连续电离能的大幅跃升表明正在移走的电子来自一个新的、更靠近原子核且屏蔽更少的内层。

For example, sodium (1s² 2s² 2p⁶ 3s¹) has a relatively low first ionisation energy, but the second IE is enormous because it involves removing an electron from the stable neon-like 2p subshell. The pattern of jumps helps deduce an element’s group and electron arrangement.

例如，钠（1s² 2s² 2p⁶ 3s¹）的第一电离能相对较低，但第二电离能急剧升高，因为这次需要从类似氖的稳定 2p 亚层移除电子。这种跃升模式有助于推断元素所在的族及其电子排布。

12. Summary and Exam Tips | 总结与考试技巧

Master atomic structure by practising electron configurations, ionisation energy graphs, and mass spectrum calculations. When explaining trends, always link your answer to nuclear charge, shielding, and distance. Remember the exceptions for transition metal ions (4s removed first) and the ionisation energy drops across periods. Use precise terminology like “effective nuclear charge” and “electron–electron repulsion” to gain full marks.

通过练习电子排布、电离能图示与质谱计算掌握原子结构。在解释趋势时，始终把答案与核电荷、屏蔽效应及距离联系起来。牢记过渡金属离子的例外（先失去 4s 电子）和周期内电离能的下降点。运用“有效核电荷”“电子-电子排斥”等精确术语以获得满分。

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Atomic Structure in A-Level Chemistry | A-Level 化学：原子结构 考点精讲