Electron Configuration for A-Level AQA Chemistry | A-Level AQA 化学：电子排布考点精讲

📚 Electron Configuration for A-Level AQA Chemistry | A-Level AQA 化学：电子排布考点精讲

Electron configuration is the foundation for understanding chemical bonding, reactivity, and periodic trends. Mastering how electrons are arranged in atoms will help you tackle questions on ionisation energies, transition metal chemistry, and oxidation states with confidence. This guide covers everything you need to know for the AQA A-Level Chemistry specification, from the basics of subshells to tricky exceptions like chromium and copper.

电子排布是理解化学键、反应活性以及元素周期律的基础。掌握原子中电子的排布方式，能帮助你自信地应对电离能、过渡金属化学以及氧化数等考题。本指南涵盖 AQA A-Level 化学考纲中你需要掌握的全部内容，从亚层的基础知识到铬和铜等棘手的例外情况，一一为你精讲。

1. Energy Levels and Subshells | 能层与亚层

Electrons in atoms are organised into principal energy levels (shells) labelled n = 1, 2, 3, 4, and so on. As n increases, the energy of the electron increases and the average distance from the nucleus becomes larger. Each principal energy level is split into subshells: s, p, d, and f. The first shell (n=1) contains only a 1s subshell; n=2 has 2s and 2p subshells; n=3 has 3s, 3p, and 3d subshells; and from n=4 you also find 4f subshells. The maximum number of electrons a subshell can hold depends on the number of orbitals it contains: an s subshell holds up to 2 electrons, a p subshell holds up to 6, a d subshell holds up to 10, and an f subshell holds up to 14.

原子中的电子排布在不同的主能层（壳层）中，主能层用 n = 1, 2, 3, 4… 标记。n 越大，电子能量越高，离原子核的平均距离也越远。每个主能层又分为若干亚层：s、p、d、f。第一层 (n=1) 只有一个 1s 亚层；n=2 有 2s 和 2p 亚层；n=3 有 3s、3p 和 3d 亚层；从 n=4 开始还会出现 4f 亚层。亚层最多能容纳的电子数取决于其包含的轨道数：s 亚层最多容纳 2 个电子，p 亚层最多容纳 6 个电子，d 亚层最多容纳 10 个电子，f 亚层最多容纳 14 个电子。

2. Orbitals and Their Shapes | 轨道及其形状

An orbital is a region of space around the nucleus where there is a high probability (typically > 90%) of finding an electron. Each orbital can hold a maximum of two electrons. The s orbitals are spherical in shape; the p orbitals are dumbbell-shaped and come in three orientations (p_x, p_y, p_z), each at right angles to the others. For AQA, you are not required to recall the shapes of d orbitals, but it is useful to know that there are five d orbitals in a d subshell. The concept of orbitals helps explain why electrons do not simply collapse into the nucleus and why chemical bonds have directional properties.

轨道是原子核周围空间中的一个区域，在那里找到电子的概率很高（通常大于 90%）。每个轨道最多可容纳两个电子。s 轨道呈球形；p 轨道呈哑铃形，分为三种取向（p_x、p_y、p_z），彼此相互垂直。AQA 考纲不要求记忆 d 轨道的形状，但了解一个 d 亚层含有五个 d 轨道是有益的。轨道的概念有助于解释为什么电子不会直接坠落到原子核里，以及为什么化学键具有方向性的特征。

3. The Aufbau Principle | 构造原理

The Aufbau Principle states that electrons fill the lowest energy orbitals first before occupying higher energy levels. The order of filling is not simply 1s → 2s → 2p → 3s → 3p → 3d → 4s, because the 4s subshell has a slightly lower energy than the 3d subshell in neutral atoms. The complete sequence can be remembered using the diagonal rule or by following the pattern: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p. You must be able to use this sequence to predict electron configurations for elements up to krypton (Z=36) and for any element when given the appropriate filling order.

构造原理指出，电子会优先填充能量最低的轨道，然后再占据能量更高的轨道。填充顺序并非简单的 1s → 2s → 2p → 3s → 3p → 3d → 4s，因为在中性原子中，4s 亚层的能量略低于 3d 亚层。完整的填充顺序可以通过对角规则来记忆，或按以下模式排列：1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p。你必须能够运用该顺序来预测氪（Z=36）之前元素的电子排布，并在给出适当填充顺序的条件下预测任何元素的电子排布。

4. Pauli Exclusion Principle and Hund’s Rule | 泡利不相容原理与洪特规则

The Pauli Exclusion Principle tells us that no two electrons in the same atom can have the same set of four quantum numbers. In practical terms, this means an orbital can hold a maximum of two electrons, and these two electrons must have opposite spins (often represented as ↑ and ↓). Hund’s Rule states that when filling degenerate orbitals (orbitals of the same energy, such as the three p orbitals or the five d orbitals), electrons will occupy separate orbitals singly and with parallel spins before any pairing occurs. This minimises electron‑electron repulsion and leads to half‑filled or fully‑filled subshells being especially stable.

泡利不相容原理告诉我们，同一个原子中没有两个电子可以具有完全相同的四个量子数。在实际应用中，这意味着一个轨道最多只能容纳两个电子，并且这两个电子的自旋方向必须相反（通常用 ↑ 和 ↓ 来表示）。洪特规则指出，在填充简并轨道（能量相同的轨道，例如三个 p 轨道或五个 d 轨道）时，电子会先以自旋平行方式单独占据不同的轨道，然后再进行配对。这样可以最大限度地减少电子间的排斥，并使半充满或全充满的亚层具有特殊的稳定性。

5. Writing Electron Configurations | 书写电子排布

You must be able to write electron configurations using the standard notation, for example: carbon (6 electrons) → 1s² 2s² 2p². You may also be asked to present the configuration in terms of sub‑shells using ‘boxes’ to represent orbitals and arrows to represent electrons (electrons‑in‑box diagrams). For example, nitrogen would show three unpaired electrons in the 2p orbitals. Additionally, you should be able to write the shorthand or ‘noble gas’ configuration, e.g. chlorine: [Ne] 3s² 3p⁵. Always count the total number of electrons to check that it matches the atomic number of the element.

你必须能够用标准符号书写电子排布，例如：碳（6 个电子）的排布为 1s² 2s² 2p²。你还可能需要用“方框”代表轨道、用箭头代表电子的方式绘制电子在亚层中的排布图（即电子填充盒图）。例如，氮的电子填充图会显示 2p 轨道中有三个未成对电子。此外，你还应能够书写简化的“稀有气体”排布式，例如氯： [Ne] 3s² 3p⁵。书写后务必核验电子总数是否与元素的原子序数一致。

6. Exceptions: Chromium and Copper | 例外情况：铬和铜

Two famous exceptions to the Aufbau principle in Period 4 are chromium (Cr, Z=24) and copper (Cu, Z=29). Instead of the expected configurations of [Ar] 4s² 3d⁴ and [Ar] 4s² 3d⁹, the actual ground‑state configurations are [Ar] 4s¹ 3d⁵ and [Ar] 4s¹ 3d¹⁰ respectively. The reason is that a precisely half‑filled d subshell (d⁵) or a fully‑filled d subshell (d¹⁰) provides extra stability, and the energy gap between 4s and 3d is very small. In the exam, you will often be asked to recall these exceptions or to explain why they occur in terms of stability.

第四周期中，构造原理有两个著名的例外情况：铬（Cr，Z=24）和铜（Cu，Z=29）。它们并非按预期的 [Ar] 4s² 3d⁴ 和 [Ar] 4s² 3d⁹ 排布，实际的基态电子排布分别是 [Ar] 4s¹ 3d⁵ 和 [Ar] 4s¹ 3d¹⁰。其原因在于 d 亚层达到精确的半充满（d⁵）或全充满（d¹⁰）状态会带来额外的稳定性，且 4s 与 3d 之间的能量差极小。在考试中，你常常会被要求回忆这些例外情况，或从稳定性的角度解释其产生的原因。

7. Electron Configurations of Ions | 离子的电子排布

When writing the electron configuration of an ion, you must apply the correct sequence for removing or adding electrons. For the first‑row transition metals, electrons are always removed from the 4s subshell before the 3d subshell because once the 3d orbitals are occupied, the 4s electrons are higher in energy than the 3d electrons. For example, the Fe atom is [Ar] 4s² 3d⁶, but Fe²⁺ is [Ar] 3d⁶ and Fe³⁺ is [Ar] 3d⁵. This contrasts with the filling order where 4s fills before 3d. For negative ions (anions), you simply add electrons to the next available orbitals following the Aufbau order, e.g. O²⁻: 1s² 2s² 2p⁶.

书写离子的电子排布时，必须遵循正确的得失电子顺序。对于第一行过渡金属，电子总是先从 4s 亚层中移去，再从 3d 亚层中移去，因为一旦 3d 轨道上有电子占据，4s 电子的能量就会高于 3d 电子。例如，Fe 原子的排布为 [Ar] 4s² 3d⁶，而 Fe²⁺ 为 [Ar] 3d⁶，Fe³⁺ 为 [Ar] 3d⁵。这与填充时 4s 先于 3d 的顺序恰好相反。对于负离子（阴离子），只需按构造原理的填充顺序将电子添加到下一个可用的轨道中即可，例如 O²⁻：1s² 2s² 2p⁶。

8. Electron Configuration and the Periodic Table | 电子排布与周期表

The periodic table is structured around electron configurations. The block an element belongs to (s‑block, p‑block, d‑block, f‑block) is determined by the type of subshell that receives the last electron. For AQA, you need to be able to deduce the position of an element in the periodic table from its electron configuration, and vice versa. For instance, an element ending in 3p⁴ is in Period 3 (n=3), Group 6 (p⁴ indicates 6 valence electrons when the s and p electrons are totalled). Transition metals are d‑block elements that form at least one ion with a partially filled d subshell, which is a key definition you must remember.

周期表的结构是以电子排布为基础的。元素所属的区（s 区、p 区、d 区、f 区）由其最后一个电子所填充的亚层类型决定。在 AQA 考试中，你需要能够从电子排布推断元素在周期表中的位置，反之亦然。例如，电子排布以 3p⁴ 结尾的元素位于第三周期（n=3）、第六族（将 s 和 p 电子总数相加，p⁴ 表明有 6 个价电子）。过渡金属是 d 区元素，它们能形成至少一种具有部分填充 d 亚层的离子，这是你必须牢记的关键定义。

9. Spin and Magnetic Properties | 电子自旋与磁性

Unpaired electrons give an atom or ion paramagnetic properties, meaning it is attracted by a magnetic field. When all electrons are paired, the substance is diamagnetic and is weakly repelled by a magnetic field. The number of unpaired electrons in a transition metal ion strongly influences its colour and catalytic activity. In AQA exams, you may be asked to predict whether a species is paramagnetic by looking at its electron configuration, remembering that Hund’s rule maximises unpaired spins in the ground state.

未成对电子使原子或离子具有顺磁性，即会被磁场吸引。如果所有电子均已配对，则该物质为抗磁性，会微弱地排斥磁场。过渡金属离子中未成对电子的数量会显著影响其颜色和催化活性。在 AQA 考试中，你可能会被要求通过观察电子排布来判断一种微粒是否具有顺磁性，并要记住根据洪特规则，基态下未成对自旋的数量是达到最大的。

10. Common Pitfalls and Exam Tips | 常见易错点与应试技巧

A common mistake is writing electron configurations in the wrong order, especially mixing up the sequence of 4s and 3d. Remember that when writing the configuration for a neutral atom, you list subshells in order of increasing principal quantum number (e.g. 3d before 4s for Sc: [Ar] 3d¹ 4s²), even though 4s fills first. Always double‑check your electron count. When drawing box diagrams, make sure electrons in degenerate orbitals are drawn with parallel spins before pairing. Finally, practise linking electron configurations to trends in ionisation energies, as this is a favourite topic in AQA paper 1.

一个常见的错误是书写电子排布时弄错顺序，尤其是混淆 4s 和 3d 的顺序。请记住，对于中性原子，书写时要按主量子数递增的顺序列出亚层（例如钪应写为 [Ar] 3d¹ 4s²，而非 [Ar] 4s² 3d¹），尽管填充时 4s 是先于 3d 的。每次都要复核电子总数。在绘制电子填充盒图时，务必确保在简并轨道中，电子先以自旋平行方式单独占据轨道，然后再配对。最后，多加练习将电子排布与电离能的变化趋势联系起来，这是 AQA 试卷一中最常见的考点之一。

Published by TutorHao | AQA Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Electron Configuration for A-Level AQA Chemistry | A-Level AQA 化学：电子排布 考点精讲