📚 Unit 2 pp024–038 Homework 9C: Bonding, Polarity & Intermolecular Forces | 第2单元第24–38页9C作业：化学键、极性与分子间力
This revision block covers the ideas that usually appear in Unit 2 pages 24–38 and drive Homework 9C: how bonds become polar, how molecular shape controls polarity, and how the three types of intermolecular forces determine physical properties such as boiling point and solubility.
本复习模块涵盖第2单元第24–38页以及9C作业中的核心思路：化学键如何产生极性、分子形状如何控制分子极性，以及三种分子间作用力如何决定沸点和溶解性等物理性质。
1. Why pp024–038 Matters | 为什么第24–38页很重要
This page range links the bonding model to observable behaviour. You cannot explain why ice floats or why iodine sublimes without first asking whether a molecule is polar and which intermolecular forces dominate.
这一页数范围将化学键模型与可观察到的行为联系起来。如果不先判断分子是否具有极性以及哪种分子间作用力占主导，就无法解释为什么冰会浮在水面上，或者碘为什么会升华。
The sequence is logical: electronegativity first, then VSEPR shapes, then molecular dipoles, then London, permanent dipole and hydrogen bonding. Homework 9C usually tests this whole chain in one structured question.
知识顺序非常清晰：先看电负性，再看 VSEPR 形状，再看分子偶极，最后看伦敦色散力、永久偶极-偶极力和氢键。9C作业通常会用一道结构化大题考查这条完整逻辑链。
2. Key Definitions from the Spread | 该范围中的关键定义
Start with precise language. Many marks in 9C are lost because candidates say ‘a dipole is a charge’ instead of separating a bond dipole from a molecular dipole.
先要使用准确的语言。9C作业中很多失分是因为考生把“偶极”说成了“电荷”，而没有区分键偶极和分子偶极。
- Electronegativity is the ability of an atom in a covalent bond to attract the bonding pair of electrons.
- Bond dipole is a separation of charge across a bond, shown as δ⁺ and δ⁻.
- Molecular dipole is the vector sum of all individual bond dipoles in a molecule.
- Intermolecular forces are attractions between neighbouring molecules, not chemical bonds inside a molecule.
- 电负性是共价键中一个原子吸引成键电子对的能力。
- 键偶极是化学键两端电荷的分离，用 δ⁺ 和 δ⁻ 表示。
- 分子偶极是分子中所有单个键偶极的矢量和。
- 分子间作用力是相邻分子之间的吸引力，不是分子内部的化学键。
3. Electronegativity and Bond Polarity | 电负性与键的极性
A small electronegativity difference gives a non-polar covalent bond; a large difference gives a polar covalent bond or, at the extreme, an ionic bond.
电负性差异很小会形成非极性共价键；差异较大会形成极性共价键，甚至形成离子键。
A rough boundary used in exam answers is ΔEN > 1.7 for ionic character, but the bonding is best treated as a continuum rather than three rigid boxes.
考试答案中常用的粗略界限是 ΔEN > 1.7 时具有离子性，但最好把化学键看作一种连续变化，而不是三个固定不变的分类。
| Bond type | ΔEN range | Electron pair |
| Non-polar covalent | 0–0.4 | Shared evenly |
| Polar covalent | 0.4–1.7 | Shared unevenly |
| Ionic | > 1.7 | Transferred |
For 9C, always quote the bond polarity first, then check whether the shape cancels the dipoles. A molecule with polar bonds can be completely non-polar overall.
在9C作业中，一定要先指出键的极性，再检查分子形状是否将各偶极抵消。含有极性键的分子整体也可能是完全非极性的。
4. VSEPR and Molecular Shapes | VSEPR与分子形状
VSEPR theory treats electron pairs around a central atom as repelling charge clouds. The molecule adopts the shape that keeps the bonded and non-bonded pairs as far apart as possible.
VSEPR理论将中心原子周围的电子对看作相互排斥的电荷云。分子会采取使成键电子对和孤电子对彼此尽量远离的形状。
- 4 electron pairs, 0 lone pairs → tetrahedral, bond angle 109.5°
- 4 electron pairs, 1 lone pair → pyramidal, bond angle about 107°
- 4 electron pairs, 2 lone pairs → bent, bond angle about 104.5°
- 3 electron pairs, 0 lone pairs → trigonal planar, bond angle 120°
- 2 electron pairs, 0 lone pairs → linear, bond angle 180°
- 4 对电子，0 孤对电子 → 四面体，键角 109.5°
- 4 对电子，1 孤对电子 → 三角锥形，键角约 107°
- 4 对电子，2 孤对电子 → 角形，键角约 104.5°
- 3 对电子，0 孤对电子 → 平面三角形，键角 120°
- 2 对电子，0 孤对电子 → 直线形，键角 180°
Lone pairs repel more strongly than bonding pairs, so the bond angle is compressed. For example, CH₄ is 109.5°, NH₃ is 107°, and H₂O is 104.5°.
孤对电子的排斥力比成键电子对更强，因此键角会被压缩。例如 CH₄ 的键角为 109.5°，NH₃ 为 107°，H₂O 为 104.5°。
5. Dipole Moments: Bond vs Molecular Polarity | 偶极矩：键极性与分子极性
A molecule is polar only if it has polar bonds and the geometry does not cancel them. Symmetry is the key test: a highly symmetric molecule is often non-polar.
分子具有极性时必须同时具备两个条件：存在极性键，且几何形状没有将键偶极抵消。对称性是关键判断依据：高度对称的分子通常是非极性的。
μ = Q × r
The dipole moment μ depends on the magnitude of the partial charge Q and the distance r between the charge centres. In Practice 9C, you may be given μ values and asked to infer geometry.
偶极矩 μ 取决于部分电荷的大小 Q 和电荷中心之间的距离 r。在9C练习中，可能会给出 μ 数值并要求推断分子几何构型。
| Molecule | Polar bonds? | Shape | Polar overall? |
| CO₂ | Yes | Linear | No |
| H₂O | Yes | Bent | Yes |
| CH₄ | Weakly polar | Tetrahedral | No |
| NH₃ | Yes | Pyramidal | Yes |
6. Intermolecular Forces Overview | 分子间作用力概览
Intermolecular forces are much weaker than covalent, ionic or metallic bonds, but they control physical state, boiling point, melting point, viscosity and solubility.
分子间作用力比共价键、离子键或金属键弱得多，但它们决定物质的物理状态、沸点、熔点、黏度和溶解性。
There are three main types in this unit: London dispersion forces, permanent dipole–dipole forces, and hydrogen bonding. You must be able to name the strongest force for any given molecule.
本单元主要有三种类型：伦敦色散力、永久偶极-偶极作用力和氢键。你必须能够判断任意给定分子中最强的分子间作用力。
7. London Dispersion Forces | 伦敦色散力
London forces exist between all molecules and atoms. They arise from instantaneous fluctuations in electron density that create temporary dipoles, which then induce dipoles in neighbouring particles.
伦敦色散力存在于所有分子和原子之间。它源于电子密度的瞬时波动形成瞬时偶极，进而使邻近粒子产生诱导偶极。
Larger electron clouds are more polarisable, so London forces increase with molecular size and surface area. This explains why boiling points of the noble gases increase down the group.
电子云越大越容易极化，因此伦敦色散力随分子大小和表面积的增加而增强。这就解释了为什么稀有气体的沸点沿族向下逐渐升高。
- He < Ne < Ar < Kr < Xe in boiling point
- Straight-chain alkanes have stronger London forces than branched isomers
- 沸点大小：He < Ne < Ar < Kr < Xe
- 直链烷烃的伦敦色散力比支链异构体更强
8. Permanent Dipole–Dipole Forces | 永久偶极-偶极作用力
Permanent dipole–dipole forces act between polar molecules. The δ⁺ end of one molecule attracts the δ⁻ end of a neighbouring molecule.
永久偶极-偶极作用力存在于极性分子之间。一个分子的 δ⁺ 端会吸引相邻分子的 δ⁻ 端。
These forces are stronger than London forces for similarly sized molecules. Therefore polar molecules often have higher boiling points than non-polar molecules with comparable molar mass.
对于大小相近的分子，这种作用力比伦敦色散力更强。因此极性分子的沸点通常高于摩尔质量相近的非极性分子。
Example: CH₃Cl is polar and has a higher boiling point than non-polar CH₄, even though its molecule is larger and heavier.
例如：CH₃Cl 是极性分子，其沸点高于非极性的 CH₄，尽管前者分子更大更重。
9. Hydrogen Bonding | 氢键
Hydrogen bonding is a special and stronger dipole–dipole attraction. It occurs when hydrogen is bonded to nitrogen, oxygen or fluorine, and a lone pair on one of these atoms in another molecule is available.
氢键是一种特殊且更强的偶极-偶极吸引力。当氢与氮、氧或氟成键，并且另一个分子中这些原子上有可供利用的孤对电子时，就会形成氢键。
Hydrogen bonding has a major effect on the properties of water, ammonia and alcohols. It explains the unusually high boiling point of H₂O compared with H₂S, H₂Se and H₂Te.
氢键对水、氨和醇的性质影响很大。它解释了为什么 H₂O 的沸点远高于 H₂S、H₂Se 和 H₂Te。
| Force | Present between | Relative strength |
| London | All particles | Weak |
| Dipole–dipole | Polar molecules only | Moderate |
| Hydrogen bonding | Molecules with N–H, O–H or F–H | Strongest of the three |
10. Homework 9C: Common Question Types | 9C作业：常见题型
Homework 9C usually combines a naming task, a shape-and-angle task, a polarity justification, and a comparison of boiling points. Use a standard answer structure: bond polarity, shape, molecular polarity, dominant intermolecular force, and physical property.
9C作业通常包含命名、形状和键角判断、极性解释以及沸点比较。答题时使用标准结构：键的极性、分子形状、分子极性、主要分子间作用力和物理性质。
- Draw the 3D shape and state the bond angle.
- Explain why a molecule is polar or non-polar.
- Identify the strongest intermolecular force.
- Predict which of two substances has the higher boiling point.
- 画出三维形状并说明键角。
- 解释分子为什么具有极性或非极性。
- 指出最强的分子间作用力。
- 预测两种物质中哪一种沸点更高。
11. Worked Example for 9C | 9C作业例题
Question: compare the boiling points of propane C₃H₈ and ethanol C₂H₅OH. Both molecules are of similar size, yet ethanol boils at a much higher temperature. Explain why.
题目：比较丙烷 C₃H₈ 和乙醇 C₂H₅OH 的沸点。两种分子大小相近，但乙醇的沸点高得多。请解释原因。
Propane is non-polar because its C–H bonds have very small polarity and the tetrahedral shape is symmetric. Its only intermolecular forces are London dispersion forces, which are weak.
丙烷是非极性分子，因为其 C–H 键极性很小，且四面体形状对称。它的分子间作用力只有较弱的伦敦色散力。
Ethanol contains a polar O–H bond. The molecule is bent around oxygen, so it is polar overall, and it can form hydrogen bonds between O–H groups in neighbouring molecules.
乙醇含有极性的 O–H 键。氧原子周围呈角形，因此乙醇整体为极性分子，并且相邻分子之间的 O–H 基团可以形成氢键。
Hydrogen bonding is much stronger than London forces, so more energy is required to separate ethanol molecules. Therefore ethanol has a much higher boiling point than propane.
氢键比伦敦色散力强得多，因此分离乙醇分子需要更多能量。所以乙醇的沸点远高于丙烷。
12. Exam Tips for Unit 2 pp024–038 | 第24–38页的考试技巧
Never write that hydrogen bonds break between oxygen and hydrogen atoms within a water molecule. Hydrogen bonds are intermolecular; the O–H covalent bond is intramolecular and much stronger.
不要写氢键是水分子内部氧原子和氢原子之间的化学键断裂。氢键是分子间作用力；O–H 共价键是分子内力，而且强得多。
When asked about a trend in boiling points, always name the force and describe how it changes with molecular size, surface area or polarity. A named force without a reason is only half an answer.
当被问到沸点变化规律时，一定要指出作用力类型，并描述它如何随分子大小、表面积或极性而变化。只写出作用力名称而不给理由只能拿一半分。
For shape questions, count electron pairs first, then lone pairs, then draw the shape. If a bond angle is not exactly 109.5°, explain that lone pairs repel more strongly.
关于形状的题目，先统计电子对总数，再统计孤对电子数，然后画出形状。如果键角不是精确的 109.5°，要解释孤对电子排斥力更强。
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