📚 1.3.4 Shapes of Molecules and Ions | 分子和离子的形状
The three-dimensional arrangement of atoms in a molecule or ion is fundamental to understanding its physical and chemical properties. In the A‑Level Edexcel Chemistry specification, topic 1.3.4 focuses on predicting and explaining these shapes using the Valence Shell Electron Pair Repulsion (VSEPR) model. By considering the number of electron pairs – both bonding and non‑bonding – around a central atom, you can determine whether a species is linear, trigonal planar, tetrahedral, or adopts one of the other essential geometries. This article walks through the logic of VSEPR, the key shapes required for the exam, and how lone pairs compress bond angles. You will also learn to handle ions like ammonium and nitrate, as well as molecules with expanded octets such as phosphorus pentachloride and sulfur hexafluoride.
分子或离子中原子的三维排布是理解其物理和化学性质的基础。在 A‑Level Edexcel 化学大纲中，主题 1.3.4 着重于运用价层电子对互斥（VSEPR）模型预测和解释这些形状。通过考虑中心原子周围的电子对数量——包括成键电子对和非键电子对——你可以判断一个粒子是直线形、平面三角形、四面体形，还是属于其他重要几何构型。本文带你梳理 VSEPR 的逻辑、考试要求的关键形状以及孤对电子如何压缩键角。你还将学习如何处理铵根、硝酸根等离子，以及五氯化磷、六氟化硫等扩展八隅体分子。
1. Fundamentals of VSEPR Theory | VSEPR 理论基础
VSEPR theory states that electron pairs around a central atom arrange themselves as far apart as possible to minimise electrostatic repulsion. Both bonding pairs (shared with another atom) and lone (non‑bonding) pairs occupy space in the valence shell and repel one another. The shape adopted by the molecule or ion is determined by the number of these electron pairs and how many of them are bonding pairs. Lone pairs exert a stronger repulsive effect than bonding pairs because they are on average closer to the nucleus and spread out more. This means the presence of a lone pair will squeeze the bond angles slightly compared to the ideal geometry predicted by the total number of electron pairs.
VSEPR 理论指出，中心原子周围的电子对会尽可能远离彼此，以最大程度地减小静电排斥。成键电子对（与另一个原子共享）和孤对（非键）电子对都在价层中占据空间并相互排斥。分子或离子采用的形状取决于这些电子对的数量以及其中有多少是成键电子对。孤对电子对产生的排斥作用比成键电子对更强，因为它们平均更靠近原子核且分布更弥散。这意味着，与根据电子对总数预测的理想几何构型相比，孤对电子的存在会略微压缩键角。
2. Counting Electron Pairs | 电子对计数方法
To use VSEPR, you must first identify the central atom – usually the least electronegative element that is not hydrogen – and draw a dot‑and‑cross diagram or Lewis structure. Count all the valence electrons provided by the atoms and adjust for any charge if the species is an ion. Then work out the number of electron pairs around the central atom by dividing the total number of valence electrons used in bonds and lone pairs by two. It is useful to distinguish between the total number of electron pairs (bonding + lone) and the number of bonding pairs only, because the shape name often refers to the arrangement of atoms, not all electron pairs.
要运用 VSEPR，你必须首先确定中心原子——通常是除氢以外电负性最小的元素——并画出点叉图或路易斯结构。计算出所有原子提供的价电子总数，如果是离子还要调整电荷数。然后将中心原子用于成键和孤对的价电子总数除以 2，得到该原子周围的电子对数。区分电子对总数（成键对 + 孤对）和纯成键对的数量很重要，因为形状名称通常指的是原子的排布方式，而非所有电子对的排布。
3. Two Electron Pairs – Linear | 两对电子对——直线形
If the central atom possesses exactly two electron pairs and both are bonding pairs, the pairs will position themselves 180° apart to minimise repulsion. This gives a linear shape with a bond angle of 180°. Key examples are beryllium chloride (BeCl₂) and carbon dioxide (CO₂). In CO₂, carbon forms two double bonds to oxygen; each double bond counts as one region of electron density, so there are two bonding pairs and no lone pairs around carbon. Thus the O–C–O angle is exactly 180°.
如果中心原子恰好有两对电子且均为成键电子对，这两对电子会以 180° 的夹角排布以使排斥最小，从而得到键角为 180° 的直线形。典型的例子是氯化铍 (BeCl₂) 和二氧化碳 (CO₂)。在 CO₂ 中，碳与氧形成两个双键；每个双键被视为一个电子密度区域，因此碳原子周围有两对成键电子且无孤对电子。所以 O–C–O 角精确为 180°。
4. Three Electron Pairs – Trigonal Planar and Bent | 三对电子对——平面三角形与弯曲形
With three electron pairs around the central atom, the ideal arrangement is trigonal planar, with bond angles of 120°. If all three pairs are bonding, as in boron trifluoride (BF₃), the molecule is perfectly flat and the F–B–F angles are 120°. If one of the three positions is occupied by a lone pair, only two atoms remain bonded to the central atom, producing a bent or V‑shaped appearance. An example is tin(II) chloride (SnCl₂) or sulfur dioxide (SO₂) if considered with one lone pair on sulfur. The bond angle here is compressed by the lone pair and falls below 120°; for SO₂ the O–S–O angle is about 119°, but in many exam contexts the angle for a bent species with three electron pairs is quoted as slightly less than 120°, such as 117°–119°.
若中心原子周围有三对电子，理想排布是平面三角形，键角为 120°。若三对均为成键电子对，如三氟化硼 (BF₃)，分子完全呈平面状，F–B–F 角为 120°。如果三个位置中有一个被孤对电子占据，则只剩两个原子与中心原子成键，因而呈现弯曲形或 V 形。典型的例子有氯化亚锡 (SnCl₂) 或二氧化硫 (SO₂)，当认为硫有一对孤对时。这里的键角因孤对电子压缩而小于 120°；对 SO₂ 而言 O–S–O 角约 119°，但在许多考试场合中，具有三对电子对的弯曲形粒子，其键角常被引用为略小于 120°，如 117°–119°。
5. Four Electron Pairs – Tetrahedral, Trigonal Pyramidal and Bent | 四对电子对——四面体形、三角锥形与弯曲形
The tetrahedral arrangement is one of the most common in A‑Level chemistry. With four electron pairs, the ideal geometry is tetrahedral and the bond angle is 109.5°. When all four are bonding pairs, as in methane (CH₄) or the ammonium ion (NH₄⁺), the molecule is a perfect tetrahedron. If one of the four electron pairs is a lone pair, the molecule has three bonding pairs and one lone pair, resulting in a trigonal pyramidal shape. Ammonia (NH₃) is the classic example; the lone pair on nitrogen pushes the three N–H bonds closer together, reducing the H–N–H angle to approximately 107°. If the central atom has two bonding pairs and two lone pairs, as in water (H₂O), the shape is bent or V‑shaped. The two lone pairs compress the bond angle even further, giving an H–O–H angle of about 104.5°.
四面体排布是 A‑Level 化学中最常见的构型之一。当存在四对电子时，理想几何是四面体，键角为 109.5°。若四对均为成键电子对，如甲烷 (CH₄) 或铵根离子 (NH₄⁺)，分子即呈现完美的四面体。若四对电子对中有一对是孤对，分子便有三对成键对和一对孤对，从而形成三角锥形。氨 (NH₃) 是经典例子；氮上的孤对电子把三个 N–H 键推得彼此靠拢，使 H–N–H 角减小至约 107°。若中心原子有两对成键对和两对孤对，如水 (H₂O)，形状为弯曲形或 V 形。两对孤对电子进一步压缩键角，使 H–O–H 角约为 104.5°。
6. Five Electron Pairs – Trigonal Bipyramidal and Related Shapes | 五对电子对——三角双锥形及其相关形状
For species in which the central atom can expand its octet, five electron pairs lead to a trigonal bipyramidal geometry. The electron pairs are not all equivalent: three occupy equatorial positions (120° apart in a plane) and two occupy axial positions (90° to the equatorial plane). An example is phosphorus pentachloride (PCl₅), which has five bonding pairs and no lone pairs; the Cl–P–Cl equatorial angles are 120° and the axial–equatorial angles are 90°. If one or more lone pairs are present, they preferentially occupy equatorial positions to minimise repulsion. A single lone pair gives a see‑saw shape, seen in sulfur tetrafluoride (SF₄); two equatorial lone pairs give a T‑shape, as in chlorine trifluoride (ClF₃); and three equatorial lone pairs produce a linear molecule, such as xenon difluoride (XeF₂).
对于中心原子可以扩展八隅体的粒子，五对电子会导致三角双锥几何构型。这些电子对并不全都等价：三对位于赤道位（平面上彼此相隔 120°），两对位于轴向位（与赤道平面呈 90°）。一个例子是五氯化磷 (PCl₅)，它有五对成键对且无孤对；Cl–P–Cl 赤道角为 120°，轴向-赤道角为 90°。若存在一对或多对孤对电子，它们优先占据赤道位置以最小化排斥。一对孤对形成翘翘板形，见于四氟化硫 (SF₄)；两对赤道孤对形成 T 形，如三氟化氯 (ClF₃)；三对赤道孤对则产生直线形分子，例如二氟化氙 (XeF₂)。
7. Six Electron Pairs – Octahedral and Square Planar | 六对电子对——八面体形与平面四边形
When a central atom accommodates six electron pairs, the shape that minimises repulsion is octahedral. With six bonding pairs, as in sulfur hexafluoride (SF₆), all bond angles are 90°. If two lone pairs replace two bonding pairs, they position themselves opposite each other (trans arrangement) to keep the lone–lone repulsion low, leaving four bonding pairs in a square planar arrangement. Xenon tetrafluoride (XeF₄) exhibits exactly this shape, with bond angles of 90°. A species with one lone pair and five bonding pairs is square pyramidal, but this is less commonly examined at A‑Level.
当中心原子容纳六对电子时，能使排斥最小的形状为八面体形。若六对均为成键对，如六氟化硫 (SF₆)，所有键角均为 90°。如果有两对孤对替代了两对成键对，它们会位于彼此相对的位置（反式排布）以降低孤对-孤对排斥，剩余的四个成键对则呈平面四边形。四氟化氙 (XeF₄) 正好展现这种形状，键角为 90°。具有一对孤对和五对成键对的粒子为四方锥形，但在 A‑Level 考试中较少涉及。
8. Bond Angle Trends and Lone Pair Compression | 键角趋势与孤对压缩效应
A crucial skill is predicting how bond angles depart from ideal values when lone pairs are present. The order of repulsion strength is: lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair. Thus, in the series CH₄ (tetrahedral, 109.5°), NH₃ (pyramidal, ~107°) and H₂O (bent, ~104.5°), the angle shrinks progressively as the number of lone pairs on the central atom increases. Similarly, for a trigonal planar system, replacing a bonding pair with a lone pair reduces the angle from 120° to between 117° and 119°. Recognising this trend allows you to rationalise the slight differences in bond angles and to make accurate predictions for unfamiliar molecules.
一项关键技能是预测当存在孤对电子时，键角如何偏离理想值。排斥强度顺序为：孤对–孤对 > 孤对–成键对 > 成键对–成键对。因此，在 CH₄（四面体，109.5°）、NH₃（三角锥，约 107°）和 H₂O（弯曲形，约 104.5°）这个序列中，随着中心原子上孤对电子的数量增加，键角逐渐缩小。类似地，在平面三角形体系中，用孤对电子取代一对成键对会将角度从 120° 减小到 117°–119° 之间。认识到这一趋势，你就能合理解释键角的细微差别，并对陌生分子做出准确预测。
9. Shapes of Common Ions | 常见离子的形状
VSEPR theory applies equally well to polyatomic ions. The ammonium ion (NH₄⁺) has four bonding pairs and no lone pair, so it is tetrahedral. The nitrate ion (NO₃⁻) contains three regions of electron density around nitrogen, all bonding, giving a trigonal planar shape with 120° bond angles. The sulfate ion (SO₄²⁻) is tetrahedral around sulfur. In the same way, the carbonate ion (CO₃²⁻) is trigonal planar. When counting electrons for ions, remember to add one electron for each negative charge and remove one for each positive charge before determining the number of electron pairs.
VSEPR 理论同样适用于多原子离子。铵根离子 (NH₄⁺) 有四对成键对，无孤对，故为四面体形。硝酸根离子 (NO₃⁻) 在氮原子周围有三个电子密度区域，全是成键区域，因此呈现平面三角形，键角 120°。硫酸根离子 (SO₄²⁻) 在硫周围为四面体。类似地，碳酸根离子 (CO₃²⁻) 为平面三角形。在计算离子的电子数时，请记住每个负电荷需加一个电子，每个正电荷需减一个电子，然后再确定电子对数。
10. Molecules with Multiple Central Atoms | 多中心原子分子的局部形状
For larger molecules, you should identify the shape around each central atom separately. Ethane (C₂H₆) has two carbon atoms, each surrounded by four bonding pairs, so both carbon centres are tetrahedral. Ethene (C₂H₄) has each carbon attached to three bonding regions (two single bonds and one double bond); the shape around each carbon is trigonal planar with an approximate bond angle of 120°. In ethanol (C₂H₅OH), one carbon is tetrahedral and the oxygen adopts a bent arrangement (two bonding pairs, two lone pairs, ~104.5°). Describing local geometry in this way is often required when analysing organic mechanisms.
对于较大的分子，你应当分别识别每个中心原子周围的形状。乙烷 (C₂H₆) 有两个碳原子，每个碳周围均有四对成键对，因此两个碳中心均为四面体形。乙烯 (C₂H₄) 中每个碳连有三个成键区域（两个单键和一个双键）；每个碳周围的形状为平面三角形，键角约 120°。在乙醇 (C₂H₅OH) 中，一个碳呈四面体形，氧则采取弯曲排布（两对成键对、两对孤对，约 104.5°）。在分析有机机理时，常需要这样描述局部几何结构。
11. Common Misconceptions and Exam Tips | 常见误区与考试技巧
One frequent mistake is to confuse the arrangement of electron pairs with the molecular shape. Always state the shape based on the positions of the atoms, not the electron pairs. For instance, ammonia has a tetrahedral arrangement of four electron pairs, but its molecular shape is trigonal pyramidal. Another pitfall is forgetting that double and triple bonds each count as one set of electron density. Remind yourself that the lone pair on nitrogen is responsible for ammonia's lower bond angle; in an exam, you may be asked to explain this using the VSEPR model. It is also useful to practise drawing dot‑and‑cross diagrams quickly and accurately, as they form the foundation for all shape predictions.
一个常见错误是混淆电子对排布与分子形状。务必根据原子的位置而非电子对来描述形状。例如，氨有四对电子对的四面体排布，但其分子形状为三角锥形。另一个误区是忘记双键和三键各被当作一个电子密度集合。提醒自己氮上的孤对电子是造成氨键角低于理想值的原因；考试中可能会要求你用 VSEPR 模型解释这一点。练习快速准确地绘制点叉图也很有用，因为它们是所有形状预测的基础。
12. Summary of Key Shapes and Bond Angles | 关键形状与键角总结
The table below summarises the shapes, bond angles, and essential examples required for the Edexcel A‑Level specification.
下表演练了 Edexcel A‑Level 大纲要求的关键形状、键角和典型例子。
| Total electron pairs | Bonding pairs / Lone pairs | Shape name | Bond angle | Examples |
|---|---|---|---|---|
| 2 | 2 / 0 | Linear | 180° | BeCl₂, CO₂ |
| 3 | 3 / 0 | Trigonal planar | 120° | BF₃, NO₃⁻ |
| 3 | 2 / 1 | Bent / V‑shaped | slightly < 120° | SO₂, SnCl₂ |
| 4 | 4 / 0 | Tetrahedral | 109.5° | CH₄, NH₄⁺ |
| 4 | 3 / 1 | Trigonal pyramidal | ~107° | NH₃ |
| 4 | 2 / 2 | Bent / V‑shaped | ~104.5° | H₂O |
| 5 | 5 / 0 | Trigonal bipyramidal | 90°, 120° | PCl₅ |
| 6 | 6 / 0 | Octahedral | 90° | SF₆ |
| 6 | 4 / 2 | Square planar | 90° | XeF₄ |
By systematically applying the VSEPR steps—draw the Lewis structure, count electron pairs, decide their arrangement, and then report the shape based on atom positions—you can confidently tackle any shape prediction question on the Edexcel examination.
通过系统地运用 VSEPR 步骤——绘制路易斯结构、计数电子对、确定其排布方式，然后根据原子位置报告形状——你就能从容应对 Edexcel 考试中任何形状预测题。
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