High School Chemistry

Chemical Bonding化学键

Why do atoms stick together at all — and why do the resulting substances have such different properties? The answer lies in the electrons: specifically, in how atoms gain, lose, or share electrons to reach a lower-energy, more stable configuration. This guide moves from the driving force for bonding (the octet rule), through the two primary bond types (ionic and covalent), to the tools chemists use to represent and predict bonding (Lewis structures, VSEPR geometry, electronegativity and polarity), and finally to the weaker forces that act between molecules (intermolecular forces) and through metals (metallic bonding). Worked examples and KaTeX formulas are used throughout.原子为什么会结合在一起——由此形成的物质为何拥有如此不同的性质？答案在于电子：具体来说，在于原子如何得失或共享电子以达到能量更低、更稳定的状态。本指南从成键的驱动力（八隅规则）出发，经过两种主要键型（离子键与共价键），到化学家用来表示和预测成键的工具（路易斯结构、VSEPR 分子几何、电负性与极性），最终落脚于分子间的较弱作用力（分子间作用力）和金属中的结合力（金属键）。全程使用例题与公式。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB Honors block: VSEPR depth and named IMFs (ON SCH4U / AB Chem 20)荣誉级：VSEPR 深度与具名分子间作用力（ON SCH4U / AB Chem 20）
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How to use this guide如何使用本指南

Chemical bonding is the unit where all four curricula overlap most strongly. The core scope — ionic bonding, covalent bonding, Lewis structures, electronegativity, and molecular polarity — is assessed in US NGSS (HS-PS1-2), Ontario SCH3U (B2.4–B2.6, B3.4–B3.5), BC Chemistry 11 (bonding + Lewis structures + polarity), and Alberta Chemistry 20 (Unit A GO1/GO2). The divergence is in depth: NGSS HS-PS1-3 deliberately avoids naming specific IMF types, while Alberta Chemistry 20 GO2 and Ontario SCH4U C3.4 explicitly name London/dispersion forces, dipole-dipole forces, and hydrogen bonding. VSEPR geometry is BC Chemistry 11 core content; it enters Ontario at SCH4U C2.3. The table below tells you which sections are core for you.化学键是四套大纲重叠最多的单元。核心范围——离子键、共价键、路易斯结构、电负性与分子极性——在 US NGSS（HS-PS1-2）、安大略 SCH3U（B2.4–B2.6、B3.4–B3.5）、BC Chemistry 11（成键、路易斯结构、极性）和阿尔伯塔 Chemistry 20（A 单元 GO1/GO2）中均被评估。分歧在于深度：NGSS HS-PS1-3 故意不要求命名具体的分子间作用力类型，而阿尔伯塔 Chemistry 20 GO2 和安大略 SCH4U C3.4 明确命名伦敦/色散力、偶极-偶极力和氢键。VSEPR 几何是 BC Chemistry 11 的核心内容；它在安大略的 SCH4U C2.3 才引入。下表告诉你哪些节是你的核心内容。

If you are in…如果你在… Focus on these sections重点学习 Defer / lighter可推迟 / 减负 Source依据
🇺🇸 US NGSS HS Chemistry美国 NGSS 高中化学 §1–§4 and §6 (octet rule, ionic bonding, covalent bonding, Lewis structures, polarity) — HS-PS1-2 and HS-PS2-6. HS-PS1-3 requires understanding force strength from bulk properties but not naming IMF types.§1–§4 和 §6（八隅规则、离子键、共价键、路易斯结构、极性）—— HS-PS1-2 与 HS-PS2-6。HS-PS1-3 要求从宏观性质理解力的强度，但不要求命名分子间作用力类型。 §5 VSEPR geometry (above NGSS assessed floor for most states) and named IMFs in §7 (London, dipole-dipole, hydrogen bonding) — valuable context, not NGSS-assessed.§5 VSEPR 几何（高于大多数州的 NGSS 评估下限）和 §7 中具名的分子间作用力（伦敦力、偶极-偶极力、氢键）——有价值的背景知识，但 NGSS 未评估。 NGSS HS-PS1 (Chemistry) — HS-PS1-2 PE + Clarification + Assessment Boundary; HS-PS1-3 PE + emphasis note; HS-PS2-6 PE + molecular-structure link— HS-PS1-2 表现期望 + 澄清 + 评估边界；HS-PS1-3 表现期望 + 强调说明；HS-PS2-6 表现期望 + 分子结构链接
🇨🇦 ON SCH3U (Gr 11)安大略 SCH3U（11 年级） §1–§4 and §6 in full (ionic bonding B3.4; covalent bonding B3.4; Lewis structures B2.4, B2.6; electronegativity and bond prediction B2.5; comparison of ionic vs molecular properties B3.5).§1–§4 和 §6 完整学习（离子键 B3.4；共价键 B3.4；路易斯结构 B2.4、B2.6；电负性与键的预测 B2.5；离子化合物与分子化合物性质比较 B3.5）。 §5 VSEPR (Grade 12 SCH4U C2.3) and named IMFs in §7 (SCH4U C3.4) — flag with Honors for SCH3U students.§5 VSEPR（12 年级 SCH4U C2.3）和 §7 中具名的分子间作用力（SCH4U C3.4）—— SCH3U 学生标 Honors。 Ontario SCH3U/4U Chemistry — SCH3U B2.4–B2.6, B3.4–B3.5; SCH4U C2.3–C2.5, C3.4— SCH3U B2.4–B2.6、B3.4–B3.5；SCH4U C2.3–C2.5、C3.4
🇨🇦 BC Chemistry 11BC Chemistry 11 §1–§7 in full. BC Chemistry 11 makes VSEPR and all three IMF types core content (not honors): "electron configuration: molecular geometry, VSEPR theory"; "bonds/forces: covalent bond, hydrogen bond, intra- and intermolecular forces."§1–§7 完整学习。BC Chemistry 11 把 VSEPR 和全部三种分子间作用力列为核心内容（非荣誉）："电子排布：分子几何、VSEPR 理论"；"键/力：共价键、氢键、分子内和分子间作用力"。 Nothing — BC Chemistry 11 assesses the full bonding scope including VSEPR geometry and named IMFs at Grade 11.无 — BC Chemistry 11 在 11 年级评估完整成键范围，包括 VSEPR 几何与具名分子间作用力。 BC Chemistry 11/12 — Chemistry 11 Content "chemical bonding based on electronegativity"; "valence electrons and Lewis structures"; "bonds/forces"; Elaborations: "chemical bonding: Lewis structures, polarity"; "bonds/forces: covalent bond, hydrogen bond, IMFs, impact on properties"; "electron configuration: molecular geometry, VSEPR theory"— Chemistry 11 内容"基于电负性的化学键"；"价电子与路易斯结构"；"键/力"；细化："化学键：路易斯结构、极性"；"键/力：共价键、氢键、分子间作用力、对性质的影响"；"电子排布：分子几何、VSEPR 理论"
🇨🇦 AB Chemistry 20阿尔伯塔 Chemistry 20 §1–§7 in full. Chemistry 20 Unit A (GO1, GO2) names all three IMF types (London, dipole-dipole, hydrogen bonding) and VSEPR explicitly in the Knowledge outcomes; these are core, not honors, for AB students.§1–§7 完整学习。Chemistry 20 A 单元（GO1、GO2）在知识结果中明确命名全部三种分子间作用力类型（伦敦力、偶极-偶极力、氢键）和 VSEPR；对 AB 学生而言，这些是核心而非荣誉内容。 Nothing — AB Chemistry 20 expects Lewis structures, VSEPR, electronegativity, polarity, and named IMFs all in Grade 11. Diploma exam problems cover the full scope.无 — AB Chemistry 20 要求在 11 年级掌握路易斯结构、VSEPR、电负性、极性和具名分子间作用力。文凭考试覆盖完整范围。 Alberta Chemistry 20/30 — Chemistry 20 Unit A GO1/GO2 Knowledge outcomes and Key Concepts (verbatim from the Open Alberta program of studies)— Chemistry 20 A 单元 GO1/GO2 知识结果及关键概念（来自阿尔伯塔开放学习课程大纲原文）
🇺🇸 IB / AP feeder trackIB / AP 衔接轨道 All seven sections plus every going-deeper derivation. IB Chemistry HL Structure 2 (Bonding and Structure) and AP Chemistry Unit 2 (Molecular and Ionic Compound Structure and Properties) both assume Lewis structures, VSEPR, polarity, and IMFs from day one.全部 7 节，并完成每个"深入"推导。IB Chemistry HL Structure 2（成键与结构）和 AP Chemistry Unit 2（分子与离子化合物结构和性质）第一天就默认你掌握路易斯结构、VSEPR、极性和分子间作用力。 Nothing — chemical bonding is the conceptual backbone of IB and AP chemistry. Lewis structures appear in every subsequent unit.无 — 化学键是 IB 和 AP 化学的概念骨架。路易斯结构出现在后续每个单元中。 NGSS HS-PS1 (Chemistry) — the IB/AP feeder reads the full VSEPR + IMF depth; see the IB Chemistry HL link in "What This Feeds Into"— IB/AP 衔接读到完整 VSEPR + 分子间作用力深度；见"本单元的去向"中的 IB Chemistry HL 链接

Once you have located your row, use the two cards below for the pace at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

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If you are cramming the night before如果你在临阵磨枪

Memorise four things: metals transfer electrons to form ions (ionic bond); non-metals share electrons (covalent bond); Lewis structures count valence electrons and satisfy the octet; and electronegativity differences predict ionic vs polar covalent vs non-polar covalent character (ΔEN > 1.7 ionic, 0.4–1.7 polar covalent, < 0.4 non-polar covalent). Read every cram-cheat box. Skip the going-deeper VSEPR derivations if time is short.背熟四件事：金属转移电子形成离子（离子键）；非金属共享电子（共价键）；路易斯结构计算价电子并满足八隅；电负性差值预测离子型/极性共价型/非极性共价型（ΔEN > 1.7 离子键，0.4–1.7 极性共价键，< 0.4 非极性共价键）。读每个速记框。若时间紧，可跳过 VSEPR 的深入推导。

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If you are going for the top mark如果你目标顶分

Draw Lewis structures including resonance structures and formal charges; apply VSEPR to all five electron-domain geometries and convert to molecular geometry; distinguish bond polarity from molecular polarity using vector addition; name and rank all three IMF types by strength; and explain melting/boiling-point data using the dominant IMF in a given substance. ON SCH4U C2.3 and AB Chemistry 20 GO2 expect you to apply VSEPR and named IMFs, not just define them.绘制包括共振结构和形式电荷的路易斯结构；将 VSEPR 应用于全部五种电子域几何并转化为分子几何；用向量加法区分键的极性和分子极性；命名并按强度排列全部三种分子间作用力类型；并用给定物质中的主要分子间作用力解释熔沸点数据。ON SCH4U C2.3 与 AB Chemistry 20 GO2 要求你应用 VSEPR 和具名分子间作用力，而非仅仅定义它们。

Honors flag.荣誉级标记。 Section 5 (VSEPR molecular geometry) carries the Honors chip for US NGSS and Ontario SCH3U, where VSEPR is not an assessed expectation at Grade 11 depth. Named IMFs in §7 (London/dispersion, dipole-dipole, hydrogen bonding) carry the Honors chip for US NGSS (HS-PS1-3 explicitly avoids naming specific IMF types) and Ontario SCH3U (named IMFs enter at SCH4U C3.4). Both sections are core, not honors, in BC Chemistry 11 and Alberta Chemistry 20. If your row sends you to §5 and §7, treat them as required content.§5（VSEPR 分子几何）在 US NGSS 和安大略 SCH3U 轨道上标 Honors，VSEPR 在 11 年级深度不是被评估的期望。§7 中具名的分子间作用力（伦敦/色散力、偶极-偶极力、氢键）在 US NGSS（HS-PS1-3 明确避免命名具体分子间作用力类型）和安大略 SCH3U（具名分子间作用力在 SCH4U C3.4 才引入）轨道上标 Honors。这两节在 BC Chemistry 11 和阿尔伯塔 Chemistry 20 中是核心而非荣誉内容。如果你的行指向 §5 和 §7，就把它们视为必学内容。
Section 1 · Why atoms bond and the octet rule第 1 节 · 原子为何成键与八隅规则

Why Atoms Bond: The Octet Rule原子为何成键：八隅规则

Atoms bond to reach a lower-energy, more stable electron configuration.原子成键是为了达到能量更低、更稳定的电子构型。
  • Driving force驱动力 — isolated atoms are higher in energy than bonded atoms. Bond formation releases energy; bond breaking absorbs energy. Atoms bond because the product has a lower potential energy than the separated reactants.— 孤立原子的能量高于成键原子。成键释放能量；断键吸收能量。原子成键是因为生成物的势能低于分离反应物的势能。
  • Octet rule八隅规则 — main-group atoms tend to gain, lose, or share valence electrons until they have 8 valence electrons (a full outer shell, like a noble gas). Hydrogen is the key exception: it aims for 2 electrons (helium configuration). Exceptions beyond hydrogen include: atoms in Period 3 and below can expand their octet (more than 8 e$^-$ around the central atom) using d-orbital capacity; some molecules are electron-deficient (e.g. BF$_3$, with only 6 e$^-$ around B).— 主族原子倾向于得失或共享价电子，直到拥有 8 个价电子（满外层，类似稀有气体）。氢是关键例外：它的目标是 2 个电子（氦的构型）。氢以外的例外包括：第 3 周期及以下的原子可以利用 d 轨道容量扩展八隅（中心原子周围超过 8 个电子）；一些分子电子不足（如 BF$_3$，B 周围只有 6 个电子）。
  • Valence electrons价电子 — the electrons in the outermost shell; they are the ones involved in bonding. The group number of a main-group element equals its valence-electron count (Group 1 has 1, Group 17 has 7, Group 18 has 8).— 最外层壳层中的电子；这些电子参与成键。主族元素的族数等于其价电子数（第 1 族有 1 个，第 17 族有 7 个，第 18 族有 8 个）。
NGSS HS-PS1-2 grounds bonding in "outermost electron states of atoms." AB Chemistry 20 GO1 defines "valence electron" as a key concept. SCH3U B3.4 names the octet as the conceptual basis for both ionic and covalent bond formation.NGSS HS-PS1-2 把成键建立在"原子最外层电子状态"的基础上。AB Chemistry 20 GO1 将"价电子"列为关键概念。SCH3U B3.4 把八隅规则列为离子键和共价键形成的概念基础。
Two ways to reach an octet.达到八隅的两种方式。
  • Transfer electrons转移电子 — one atom gives electron(s) to another. This creates oppositely charged ions that attract each other: an ionic bond. Typical between metals and non-metals.— 一个原子将电子给予另一个原子。这产生相互吸引的异号带电离子：离子键。典型地发生在金属与非金属之间。
  • Share electrons共享电子 — two atoms each contribute electrons to a shared pair that both nuclei attract: a covalent bond. Typical between non-metals.— 两个原子各贡献电子形成两个原子核共同吸引的共用电子对：共价键。典型地发生在非金属之间。
How many valence electrons does a chlorine atom (Group 17) have?氯原子（第 17 族）有多少个价电子？
§1 · Q1
11
55
77
88
For main-group elements, group number = valence electron count. Chlorine is in Group 17, so it has 7 valence electrons. It needs 1 more to complete its octet.对主族元素，族数 = 价电子数。氯在第 17 族，故有 7 个价电子。它还需要 1 个才能完成八隅。
Group number = valence electron count for main-group elements. Group 17 means 7 valence electrons, not 1 or 8. Noble gases (Group 18) already have a full octet of 8.主族元素：族数 = 价电子数。第 17 族意味着 7 个价电子，而非 1 或 8。稀有气体（第 18 族）已有满的 8 个价电子。
According to the octet rule, why is hydrogen an exception with a target of only 2 valence electrons?根据八隅规则，为什么氢是例外，其目标只有 2 个价电子？
§1 · Q2
Hydrogen's outermost shell is n=1, which can hold only 2 electrons maximum氢的最外层壳层是 n=1，最多只能容纳 2 个电子
Hydrogen has no protons to attract more electrons氢没有质子来吸引更多电子
Hydrogen is a noble gas and already has a full octet氢是稀有气体，已有满的八隅
The octet rule does not apply to Period 1 elements at all八隅规则完全不适用于第 1 周期元素
The $n = 1$ shell contains only the 1s subshell, which holds a maximum of 2 electrons. The nearest noble-gas configuration for hydrogen is helium (2 electrons), so hydrogen follows a "duet rule" rather than an octet rule.$n = 1$ 壳层只包含 1s 亚壳层，最多容纳 2 个电子。氢最近的稀有气体构型是氦（2 个电子），因此氢遵循"二隅规则"而非八隅规则。
The octet rule does apply to Period 1, but the first shell ($n=1$) only has capacity for 2 electrons. Hydrogen aims for the helium configuration (2 e$^-$), not 8. It has 1 proton and forms 1 bond.八隅规则确实适用于第 1 周期，但第一壳层（$n=1$）只有 2 个电子的容量。氢的目标是氦构型（2 个电子），而非 8 个。氢有 1 个质子，形成 1 个键。
Section 2 · Ionic bonding第 2 节 · 离子键

Ionic Bonding离子键

Ionic bond: metal transfers electron(s) to non-metal; opposites attract.离子键：金属将电子转移给非金属；异号电荷相互吸引。
  • Formation形成 — a metal atom loses one or more valence electrons to become a cation (positive ion); a non-metal atom gains electrons to become an anion (negative ion). The electrostatic attraction between cation and anion is the ionic bond.— 金属原子失去一个或多个价电子成为阳离子（正离子）；非金属原子获得电子成为阴离子（负离子）。阳离子与阴离子之间的静电吸引力就是离子键。
  • Lattice structure晶格结构 — ionic compounds form regular 3-D lattices (not discrete molecules). The formula unit (e.g. NaCl) gives the simplest whole-number ratio of ions that gives a net charge of zero.— 离子化合物形成规则的三维晶格（而非离散分子）。化学式单元（如 NaCl）给出净电荷为零的最简整数比。
  • Properties of ionic compounds离子化合物的性质 — high melting/boiling points (strong lattice forces); brittle (shifting layers misalign like-charge rows); conduct electricity when dissolved or molten (free ions); soluble in polar solvents like water.— 高熔沸点（强晶格力）；脆性（层错位时同性电荷排列相对）；溶解或熔融时导电（自由离子）；溶于极性溶剂（如水）。
NGSS HS-PS1-2 uses Na + Cl as an example reaction driven by "outermost electron states." SCH3U B3.4 explicitly assesses "differences between the formation of ionic bonds and the formation of covalent bonds." AB Chemistry 20 GO1 knowledge outcomes: "explain how an ionic bond results from the simultaneous attraction of oppositely charged ions" and "explain that ionic compounds form lattices."NGSS HS-PS1-2 以 Na + Cl 反应为例，由"最外层电子状态"驱动。SCH3U B3.4 明确评估"离子键与共价键形成方式的区别"。AB Chemistry 20 GO1 知识结果："解释离子键如何由异号离子的同时吸引产生"和"解释离子化合物形成晶格"。
Worked Example 2 · Ionic bond formation: sodium chloride例题 2 · 离子键形成：氯化钠

Explain, using valence electrons and the octet rule, how sodium (Group 1) and chlorine (Group 17) form an ionic bond. Identify the resulting ions and their electron configurations.用价电子和八隅规则解释钠（第 1 族）和氯（第 17 族）如何形成离子键。确定生成的离子及其电子排布。

Sodium's situation.钠的情况。 Na has electron configuration $1s^2\, 2s^2\, 2p^6\, 3s^1$ — 1 valence electron. Losing it gives the neon configuration ($1s^2\, 2s^2\, 2p^6$, full outer shell) and creates Na$^+$ (cation, charge $+1$).Na 的电子排布为 $1s^2\, 2s^2\, 2p^6\, 3s^1$——1 个价电子。失去它得到氖的构型（$1s^2\, 2s^2\, 2p^6$，满外层），形成 Na$^+$（阳离子，电荷 $+1$）。

Chlorine's situation.氯的情况。 Cl has configuration $[\text{Ne}]\, 3s^2\, 3p^5$ — 7 valence electrons. Gaining 1 electron gives the argon configuration ($[\text{Ne}]\, 3s^2\, 3p^6$, full outer shell) and creates Cl$^-$ (anion, charge $-1$).Cl 的构型为 $[\text{Ne}]\, 3s^2\, 3p^5$——7 个价电子。获得 1 个电子得到氩的构型（$[\text{Ne}]\, 3s^2\, 3p^6$，满外层），形成 Cl$^-$（阴离子，电荷 $-1$）。

The bond.化学键。 Na$^+$ and Cl$^-$ attract electrostatically. In the solid, each Na$^+$ is surrounded by 6 Cl$^-$ and vice versa — a face-centred cubic lattice. The formula unit NaCl represents the 1:1 ratio that gives net charge zero. ✓Na$^+$ 与 Cl$^-$ 静电吸引。在固体中，每个 Na$^+$ 被 6 个 Cl$^-$ 包围，反之亦然——面心立方晶格。化学式 NaCl 表示净电荷为零的 1:1 比值。✓

When magnesium (Group 2) reacts with oxygen (Group 16) to form MgO, what ions are produced?当镁（第 2 族）与氧（第 16 族）反应形成 MgO 时，会产生哪些离子？
§2 · Q1
Mg$^+$ and O$^-$Mg$^+$ 和 O$^-$
Mg$^{2+}$ and O$^{2-}$Mg$^{2+}$ 和 O$^{2-}$
Mg$^{2-}$ and O$^{2+}$Mg$^{2-}$ 和 O$^{2+}$
Mg$^+$ and O$^{3-}$Mg$^+$ 和 O$^{3-}$
Mg (Group 2) loses 2 valence electrons to reach the neon configuration: Mg$^{2+}$. O (Group 16) needs 2 more electrons to reach a full octet (neon configuration): O$^{2-}$. The 2+ and 2$-$ charges balance to give the neutral formula MgO.Mg（第 2 族）失去 2 个价电子达到氖的构型：Mg$^{2+}$。O（第 16 族）需要再获得 2 个电子才能达到满八隅（氖的构型）：O$^{2-}$。2+ 和 2$-$ 电荷平衡，给出中性化学式 MgO。
Group 2 metals always lose 2 electrons (2+ charge) to reach the previous noble gas configuration. Group 16 non-metals always gain 2 electrons (2$-$ charge). The charges must balance to give a neutral compound.第 2 族金属总是失去 2 个电子（2+ 电荷）以达到前一个稀有气体构型。第 16 族非金属总是获得 2 个电子（2$-$ 电荷）。电荷必须平衡以给出中性化合物。
Which property is characteristic of ionic compounds in the solid state?下列哪种性质是离子化合物固态时的特征？
§2 · Q2
They conduct electricity in the solid state固态时导电
They have low melting points due to weak bonds由于键弱而熔点低
They exist as discrete molecules以离散分子形式存在
They form a regular lattice and have high melting points形成规则晶格，熔点高
Ionic compounds form 3-D lattices with strong electrostatic forces in all directions. Overcoming these forces requires a large amount of energy, giving high melting points. In the solid state, ions are locked in place and cannot carry current; conductivity requires the liquid (molten) or dissolved state.离子化合物形成三维晶格，各方向均有强静电力。克服这些力需要大量能量，赋予高熔点。固态时，离子被锁定在原位，无法传递电流；导电需要液态（熔融）或溶解状态。
Solid ionic compounds do NOT conduct electricity (ions are fixed). They have HIGH melting points (strong lattice). They exist as lattices (not discrete molecules). SCH3U B3.5 contrasts these properties with molecular compounds.固态离子化合物不导电（离子固定）。它们有高熔点（强晶格）。它们以晶格形式存在（不是离散分子）。SCH3U B3.5 将这些性质与分子化合物进行对比。
Section 3 · Covalent bonding第 3 节 · 共价键

Covalent Bonding共价键

Covalent bond: two atoms share electrons; both nuclei attract the shared pair.共价键：两个原子共享电子；两个原子核共同吸引共用电子对。
  • Single bond单键 — one shared pair of electrons (2 e$^-$). Example: H–H in H$_2$; H–Cl in HCl.— 一个共用电子对（2 个电子）。示例：H$_2$ 中的 H–H；HCl 中的 H–Cl。
  • Double bond双键 — two shared pairs (4 e$^-$). Shorter and stronger than a single bond. Example: O=O in O$_2$; C=O in CO$_2$.— 两个共用电子对（4 个电子）。比单键短且强。示例：O$_2$ 中的 O=O；CO$_2$ 中的 C=O。
  • Triple bond三键 — three shared pairs (6 e$^-$). Even shorter and stronger. Example: N$\equiv$N in N$_2$; C$\equiv$O in CO.— 三个共用电子对（6 个电子）。更短更强。示例：N$_2$ 中的 N$\equiv$N；CO 中的 C$\equiv$O。
  • Properties of molecular (covalent) compounds分子（共价）化合物的性质 — generally low melting/boiling points (only IMFs between molecules, not covalent bonds, must be overcome); poor conductors of electricity; can be gases, liquids, or low-melting solids at room temperature.— 通常熔沸点低（只需克服分子间的分子间作用力，而非共价键）；导电性差；在室温下可以是气体、液体或低熔点固体。
NGSS HS-PS1-2 explicitly names carbon + hydrogen reactions (covalent) alongside Na + Cl (ionic). SCH3U B2.6 requires students to "build molecular models and write structural formulae for molecular compounds containing single and multiple bonds." AB Chemistry 20 GO2: "relate electron pairing to multiple and covalent bonds."NGSS HS-PS1-2 明确将碳 + 氢反应（共价键）与 Na + Cl（离子键）并列举例。SCH3U B2.6 要求学生"构建含单键和多键的分子化合物的分子模型并书写结构式"。AB Chemistry 20 GO2："将电子配对与多键和共价键联系起来"。
Bond type键型Shared pairs共用对数Electrons shared共享电子数Relative strength相对强度Example示例
Single单键12weakest最弱H–H, H–Cl, C–H
Double双键24intermediate中等O=O, C=O, C=C
Triple三键36strongest最强N$\equiv$N, C$\equiv$O
A nitrogen molecule (N$_2$) contains a triple bond. How many electrons are shared between the two nitrogen atoms?氮分子（N$_2$）含有一个三键。两个氮原子之间共享多少个电子？
§3 · Q1
66
22
33
44
A triple bond consists of 3 shared pairs of electrons = 6 electrons total. Each nitrogen contributes 3 electrons to the bond (3 of its 5 valence electrons), giving each atom an octet: 3 bonding pairs + 1 lone pair.三键由 3 个共用电子对组成 = 共 6 个电子。每个氮为键贡献 3 个电子（其 5 个价电子中的 3 个），使每个原子达到八隅：3 个成键对 + 1 个孤电子对。
Single bond = 1 pair = 2 electrons; double bond = 2 pairs = 4 electrons; triple bond = 3 pairs = 6 electrons. Multiply the number of shared pairs by 2.单键 = 1 对 = 2 个电子；双键 = 2 对 = 4 个电子；三键 = 3 对 = 6 个电子。将共用对数乘以 2。
Compared to a C–C single bond, a C=C double bond is:与 C–C 单键相比，C=C 双键：
§3 · Q2
Longer and weaker更长且更弱
Longer and stronger更长且更强
Shorter and stronger更短且更强
Shorter and weaker更短且更弱
More shared electrons pull the nuclei closer together (shorter bond length) and hold them more tightly (greater bond energy = stronger bond). Every additional bond order makes bonds shorter and stronger: triple > double > single in both strength and shortness.更多的共享电子将原子核拉得更近（键长更短），并更牢固地结合（键能更大 = 键更强）。每增加一个键级，键就更短更强：三键 > 双键 > 单键，无论在强度还是短度上。
More electron pairs between nuclei → stronger electrostatic attraction → shorter bond AND higher bond energy (stronger). The trend is: single < double < triple in strength; single > double > triple in bond length.原子核间的电子对越多 → 静电吸引力越强 → 键越短且键能越高（越强）。趋势是：强度：单键 < 双键 < 三键；键长：单键 > 双键 > 三键。
Section 4 · Lewis (electron-dot) structures第 4 节 · 路易斯（电子点）结构

Lewis Structures路易斯结构

Lewis structures show all valence electrons — bonding pairs as lines, lone pairs as dots.路易斯结构显示所有价电子——成键对用线表示，孤电子对用点表示。
  • Step 1 — Count total valence electrons.第 1 步——计算总价电子数。 Add valence electrons from all atoms. For ions: subtract 1 for each positive charge; add 1 for each negative charge.将所有原子的价电子数相加。对离子：每个正电荷减 1；每个负电荷加 1。
  • Step 2 — Arrange atoms and connect with single bonds.第 2 步——排列原子并用单键连接。 The least electronegative atom is usually the central atom (except H, which is always terminal). Each single bond uses 2 electrons.电负性最小的原子通常是中心原子（氢除外，氢始终是末端原子）。每个单键使用 2 个电子。
  • Step 3 — Complete octets on terminal atoms first, then on the central atom.第 3 步——先完成末端原子的八隅，再完成中心原子的八隅。 Place remaining electrons as lone pairs. If the central atom still lacks an octet, convert lone pairs on adjacent atoms to multiple bonds.将剩余电子作为孤电子对放置。若中心原子仍不满八隅，将相邻原子上的孤电子对转变为多键。
  • Step 4 — Verify.第 4 步——验证。 Total electrons in the structure must equal the count from Step 1. Every atom (except H) should have 8 electrons around it (unless it is an expanded-octet or electron-deficient exception).结构中的总电子数必须等于第 1 步的计数。每个原子（氢除外）周围应有 8 个电子（除非是扩展八隅或电子不足的例外）。
SCH3U B2.4 requires drawing Lewis structures for ionic and molecular compounds. AB Chemistry 20 GO2: "draw electron dot diagrams of atoms and molecules, writing structural formulas for molecular substances and using Lewis structures to predict bonding in simple molecules." BC Chemistry 11 Content: "valence electrons and Lewis structures"; elaboration: "chemical bonding: Lewis structures of compounds, polarity."SCH3U B2.4 要求绘制离子和分子化合物的路易斯结构。AB Chemistry 20 GO2："绘制原子和分子的电子点图，为分子物质书写结构式，并用路易斯结构预测简单分子中的成键"。BC Chemistry 11 内容："价电子与路易斯结构"；细化："化学键：化合物的路易斯结构、极性"。
Worked Example 4 · Lewis structure of water (H$_2$O)例题 4 · 水（H$_2$O）的路易斯结构

Draw the Lewis structure of water. Count valence electrons, arrange atoms, and verify the octet rule is satisfied.绘制水的路易斯结构。计算价电子数，排列原子，并验证八隅规则是否满足。

Step 1 — Count valence electrons.第 1 步——计算价电子数。 O (Group 16) contributes 6; each H (Group 1) contributes 1. Total: $6 + 1 + 1 = 8$ valence electrons.O（第 16 族）贡献 6 个；每个 H（第 1 族）贡献 1 个。总计：$6 + 1 + 1 = 8$ 个价电子。

Step 2 — Arrange and connect.第 2 步——排列并连接。 O is the central atom (H is always terminal). Place H–O–H. Two single bonds use $2 \times 2 = 4$ electrons; $8 - 4 = 4$ electrons remain.O 是中心原子（H 始终是末端原子）。放置 H–O–H。两个单键使用 $2 \times 2 = 4$ 个电子；剩余 $8 - 4 = 4$ 个电子。

Step 3 — Complete octets.第 3 步——完成八隅。 H already has its duet (1 bond = 2 electrons). Place the remaining 4 electrons (2 lone pairs) on O. O now has: 2 bonds + 2 lone pairs = $4 + 4 = 8$ electrons. ✓H 已有二隅（1 键 = 2 个电子）。将剩余 4 个电子（2 个孤电子对）放在 O 上。O 现在有：2 键 + 2 孤电子对 = $4 + 4 = 8$ 个电子。✓

Final structure:最终结构： H–O–H with 2 lone pairs on O. Total electrons in structure = 8. ✓H–O–H，O 上有 2 个孤电子对。结构中总电子数 = 8。✓

How many total valence electrons must be shown in the Lewis structure of CO$_2$?CO$_2$ 的路易斯结构中必须显示多少个总价电子？
§4 · Q1
1212
88
1818
1616
C (Group 14) contributes 4; each O (Group 16) contributes 6. Total: $4 + 6 + 6 = 16$ valence electrons. The Lewis structure is O=C=O, with 2 lone pairs on each O (4 lone pairs total = 8 e$^-$) and 2 double bonds (8 e$^-$) = 16 e$^-$. ✓C（第 14 族）贡献 4 个；每个 O（第 16 族）贡献 6 个。总计：$4 + 6 + 6 = 16$ 个价电子。路易斯结构为 O=C=O，每个 O 上有 2 个孤电子对（共 4 个孤电子对 = 8 个电子），2 个双键（8 个电子）= 16 个电子。✓
Add valence electrons from all atoms: C has 4 (Group 14), each O has 6 (Group 16). $4 + 6 + 6 = 16$. The structure has two C=O double bonds plus two lone pairs on each oxygen.将所有原子的价电子数相加：C 有 4 个（第 14 族），每个 O 有 6 个（第 16 族）。$4 + 6 + 6 = 16$。结构有两个 C=O 双键加上每个氧上的两个孤电子对。
In the Lewis structure of NH$_3$, how many lone pairs does the nitrogen atom have?NH$_3$ 的路易斯结构中，氮原子有多少个孤电子对？
§4 · Q2
00
11
22
33
N (Group 15) has 5 valence electrons. In NH$_3$, it forms 3 bonds to H (using 3 electrons). The remaining $5 - 3 = 2$ electrons form 1 lone pair. N has 3 bonding pairs + 1 lone pair = 8 electrons total. ✓N（第 15 族）有 5 个价电子。在 NH$_3$ 中，它与 H 形成 3 个键（使用 3 个电子）。剩余的 $5 - 3 = 2$ 个电子形成 1 个孤电子对。N 有 3 个成键对 + 1 个孤电子对 = 共 8 个电子。✓
N has 5 valence electrons. It uses 3 to bond to the three H atoms, leaving $5 - 3 = 2$ electrons = 1 lone pair. The lone pair on N is crucial — it gives NH$_3$ its pyramidal shape (VSEPR, §5).N 有 5 个价电子。它用 3 个与三个 H 原子成键，剩下 $5 - 3 = 2$ 个电子 = 1 个孤电子对。N 上的孤电子对至关重要——它赋予 NH$_3$ 三角锥形（VSEPR，§5）。
Going deeper — resonance structures and formal charges深入 — 共振结构与形式电荷

Some molecules cannot be described by a single Lewis structure. Ozone (O$_3$) and carbonate (CO$_3^{2-}$) are classic examples. In ozone, two equivalent Lewis structures can be drawn that differ only in which O–O bond is single and which is double. The actual molecule is an average (resonance hybrid) of both: both bonds are identical, with a bond order of 1.5. Resonance structures are drawn with a double-headed arrow between them ($\leftrightarrow$). Formal charge helps identify the best Lewis structure: Formal charge $= $ (valence electrons) $-$ (lone-pair electrons) $-$ $\frac{1}{2}$(bonding electrons). The best structure minimises formal charges and places any negative formal charge on the most electronegative atom. SCH4U C3.4 and AB Chemistry 20 GO2 both implicitly require resonance awareness for explaining equivalent bond lengths in ozone and benzene.某些分子不能用单一路易斯结构描述。臭氧（O$_3$）和碳酸根（CO$_3^{2-}$）是经典例子。在臭氧中，可以画出两个等价的路易斯结构，只有 O–O 单键和双键的位置不同。实际分子是两者的平均（共振杂化体）：两个键完全相同，键级为 1.5。共振结构之间用双向箭头（$\leftrightarrow$）连接。形式电荷有助于识别最佳路易斯结构：形式电荷 $= $（价电子数）$-$（孤电子对数）$-$ $\frac{1}{2}$（成键电子数）。最佳结构使形式电荷最小化，并将任何负形式电荷放在电负性最强的原子上。SCH4U C3.4 和 AB Chemistry 20 GO2 均隐含要求对共振的理解，以解释臭氧和苯中等价键长的问题。

Section 5 · Molecular geometry (VSEPR) Honors — US NGSS / ON SCH3U第 5 节 · 分子几何（VSEPR）荣誉 — US NGSS / ON SCH3U

Molecular Geometry: VSEPR Theory分子几何：价层电子对互斥理论

Curriculum note.课纲提示。 VSEPR is core content in BC Chemistry 11 (Elaboration: "electron configuration: molecular geometry, valence shell electron pair repulsion (VSEPR) theory") and Alberta Chemistry 20 (GO2 Knowledge outcome: "apply VSEPR theory to predict molecular shapes for linear, angular, tetrahedral, trigonal pyramidal and trigonal planar molecules"). It enters Ontario at SCH4U C2.3 (Grade 12). NGSS does not name VSEPR as an assessed expectation, though HS-PS2-6 requires communicating how molecular-level structure affects function. If your curriculum is NGSS-only or SCH3U, read this section for conceptual depth but check your exam scope.VSEPR 是 BC Chemistry 11 的核心内容（细化："电子排布：分子几何、VSEPR 理论"）和阿尔伯塔 Chemistry 20（GO2 知识结果："应用 VSEPR 理论预测线型、角型、四面体形、三角锥形和三角平面形分子的形状"）。它在安大略 SCH4U C2.3（12 年级）才引入。NGSS 未将 VSEPR 列为评估期望，尽管 HS-PS2-6 要求说明分子级别结构如何影响功能。如果你的课程仅为 NGSS 或 SCH3U，请阅读本节以加深概念理解，但需核实考试范围。
Electron domains (bonding + lone pairs) repel each other; the molecule adopts the shape that minimises repulsion.电子域（成键对 + 孤电子对）相互排斥；分子采取使排斥最小化的形状。
  • Electron-domain geometry电子域几何 describes where all electron domains (bonding pairs + lone pairs) point. Molecular geometry describes only where atoms sit (ignoring lone pairs).描述所有电子域（成键对 + 孤电子对）的指向。分子几何只描述原子的位置（忽略孤电子对）。
  • Lone pairs occupy more space孤电子对占据更多空间 than bonding pairs (closer to the central nucleus, not shared) and compress the bond angles below the ideal values.比成键对（更靠近中心原子核，不共享）占据更多空间，并将键角压缩至低于理想值。
Domains电子域数Lone pairs孤对数Molecular geometry分子几何Bond angle键角Example示例
20Linear线型180°CO$_2$, BeCl$_2$
30Trigonal planar三角平面形120°BF$_3$, SO$_3$
31Bent / angular角型 / 弯曲形<120°SO$_2$, O$_3$
40Tetrahedral四面体形109.5°CH$_4$, CCl$_4$
41Trigonal pyramidal三角锥形<109.5°NH$_3$, PCl$_3$
42Bent / angular角型 / 弯曲形<109.5°H$_2$O, H$_2$S
Worked Example 5 · VSEPR geometry of NH$_3$例题 5 · NH$_3$ 的 VSEPR 几何

Use VSEPR theory to predict the molecular geometry and approximate bond angles of ammonia (NH$_3$). Compare to the electron-domain geometry.用 VSEPR 理论预测氨（NH$_3$）的分子几何和近似键角。与电子域几何进行比较。

Lewis structure from §4.来自 §4 的路易斯结构。 N has 3 bonding pairs (to each H) and 1 lone pair. Total electron domains = 4.N 有 3 个成键对（连接每个 H）和 1 个孤电子对。总电子域数 = 4。

Electron-domain geometry.电子域几何。 4 domains → tetrahedral electron-domain geometry (ideal 109.5°).4 个域 → 四面体形电子域几何（理想 109.5°）。

Molecular geometry.分子几何。 One domain is a lone pair (invisible in molecular geometry). The three N–H bonds point to three corners of a tetrahedron → trigonal pyramidal molecular geometry.一个域是孤电子对（在分子几何中不可见）。三个 N–H 键指向四面体的三个顶点 → 三角锥形分子几何。

Bond angle.键角。 The lone pair compresses the H–N–H angle below 109.5°. Actual H–N–H angle ≈ 107°. ✓孤电子对将 H–N–H 角压缩至低于 109.5°。实际 H–N–H 角 ≈ 107°。✓

Water (H$_2$O) has 2 bonding pairs and 2 lone pairs on oxygen. What is its molecular geometry?水（H$_2$O）的氧上有 2 个成键对和 2 个孤电子对。它的分子几何是什么？
§5 · Q1
Linear线型
Trigonal planar三角平面形
Bent角型
Tetrahedral四面体形
H$_2$O has 4 electron domains (2 bonding + 2 lone pairs) → tetrahedral electron-domain geometry. But the molecular geometry (atoms only) is bent. The two lone pairs compress the H–O–H angle to about 104.5° (below the tetrahedral ideal of 109.5°). This bent shape makes water a polar molecule with unique properties.H$_2$O 有 4 个电子域（2 个成键对 + 2 个孤电子对）→ 四面体形电子域几何。但分子几何（仅原子）是角型。两个孤电子对将 H–O–H 角压缩至约 104.5°（低于四面体理想值 109.5°）。这种角型使水成为具有独特性质的极性分子。
Tetrahedral is the electron-domain geometry (4 domains), not the molecular geometry. Molecular geometry ignores lone pairs and looks at atom arrangement only: H–O–H gives a bent (V-shaped) molecule, not tetrahedral.四面体形是电子域几何（4 个域），而非分子几何。分子几何忽略孤电子对，只看原子排列：H–O–H 形成角型（V 形）分子，而非四面体形。
Which molecule has a linear molecular geometry?哪种分子具有线型分子几何？
§5 · Q2
CO$_2$CO$_2$
H$_2$OH$_2$O
NH$_3$NH$_3$
SO$_2$SO$_2$
CO$_2$ has 2 electron domains on the central C (2 double bonds, no lone pairs) → linear electron-domain AND molecular geometry (180° bond angle). H$_2$O is bent (2 lone pairs); NH$_3$ is trigonal pyramidal (1 lone pair); SO$_2$ is bent (1 lone pair on S).CO$_2$ 中心 C 有 2 个电子域（2 个双键，无孤电子对）→ 线型电子域几何和分子几何（键角 180°）。H$_2$O 是角型（2 个孤电子对）；NH$_3$ 是三角锥形（1 个孤电子对）；SO$_2$ 是角型（S 上 1 个孤电子对）。
Linear requires 2 electron domains and 0 lone pairs on the central atom. CO$_2$: C has 2 double bonds and 0 lone pairs → linear. H$_2$O, NH$_3$, and SO$_2$ all have lone pairs on the central atom, giving non-linear geometries.线型需要中心原子有 2 个电子域且无孤电子对。CO$_2$：C 有 2 个双键且无孤电子对 → 线型。H$_2$O、NH$_3$ 和 SO$_2$ 的中心原子上都有孤电子对，导致非线型几何。
Section 6 · Bond polarity and molecular polarity第 6 节 · 键的极性与分子极性

Bond and Molecular Polarity键的极性与分子极性

Electronegativity difference determines bond polarity; geometry determines whether the whole molecule is polar.电负性差决定键的极性；分子几何决定整个分子是否有极性。
  • Electronegativity (χ)电负性（χ） — an atom's ability to attract shared electrons toward itself. Increases left to right across a period and bottom to top within a group. F is the most electronegative element (χ = 4.0 on the Pauling scale).— 原子将共享电子向自身吸引的能力。在同一周期从左到右增大，在同一族从下到上增大。F 是电负性最强的元素（泡利标度 χ = 4.0）。
  • Bond polarity: Δχ determines type键的极性：Δχ 决定类型
Δχ rangeΔχ 范围Bond type键型Example示例
< 0.4Non-polar covalent非极性共价键H–H, C–C
0.4 – 1.7Polar covalent极性共价键H–Cl, H–O
> 1.7Ionic character dominates离子性为主Na–Cl, K–F
  • Molecular polarity分子极性 — a molecule is polar if (1) it has polar bonds AND (2) the bond dipoles do not cancel due to molecular geometry. Dipoles are vectors; they cancel only if the molecule is symmetric. CO$_2$ has two polar C=O bonds but is linear and non-polar (dipoles cancel 180°). H$_2$O has two polar O–H bonds and a bent shape, so the dipoles do not cancel — H$_2$O is polar.— 如果分子 (1) 有极性键，且 (2) 键偶极矩因分子几何不相消，则该分子有极性。偶极矩是矢量；只有在分子对称时才相消。CO$_2$ 有两个极性 C=O 键，但为线型且无极性（偶极矩 180° 相消）。H$_2$O 有两个极性 O–H 键且为角型，偶极矩不相消——H$_2$O 是极性分子。
SCH3U B2.5: "predict the nature of a bond using electronegativity values." SCH4U C2.4: "predict the polarity of various chemical compounds, based on their molecular shapes and electronegativity." AB Chemistry 20 GO2: "determine the polarity of a molecule based on simple structural shapes and unequal charge distribution." BC Chemistry 11 Elaboration: "chemical bonding: Lewis structures of compounds, polarity."SCH3U B2.5："用电负性值预测化学键的性质"。SCH4U C2.4："根据分子形状和电负性预测各种化合物的极性"。AB Chemistry 20 GO2："根据简单结构形状和不均等电荷分布确定分子的极性"。BC Chemistry 11 细化："化学键：化合物的路易斯结构、极性"。
Worked Example 6 · Is CCl$_4$ a polar molecule?例题 6 · CCl$_4$ 是极性分子吗？

Carbon tetrachloride (CCl$_4$) has four C–Cl bonds. Electronegativity: C = 2.5, Cl = 3.0. Determine (a) whether each C–Cl bond is polar, and (b) whether the overall molecule is polar.四氯化碳（CCl$_4$）有四个 C–Cl 键。电负性：C = 2.5，Cl = 3.0。判断 (a) 每个 C–Cl 键是否有极性，(b) 整体分子是否有极性。

(a) Bond polarity.(a) 键的极性。 Δχ = 3.0 − 2.5 = 0.5. Since 0.4 < 0.5 < 1.7, each C–Cl bond is polar covalent. Cl is more electronegative, so it carries the partial negative charge (δ−); C carries partial positive (δ+).Δχ = 3.0 − 2.5 = 0.5。由于 0.4 < 0.5 < 1.7，每个 C–Cl 键是极性共价键。Cl 电负性更强，携带部分负电荷（δ−）；C 携带部分正电荷（δ+）。

(b) Molecular polarity.(b) 分子极性。 CCl$_4$ is tetrahedral (4 bonding pairs, 0 lone pairs on C). The four C–Cl dipoles point symmetrically toward the four corners of a tetrahedron and cancel exactly. Net dipole moment = 0. CCl$_4$ is non-polar overall, despite having polar bonds. ✓CCl$_4$ 是四面体形（C 上 4 个成键对，0 个孤电子对）。四个 C–Cl 偶极矩对称地指向四面体的四个顶点，完全相消。净偶极矩 = 0。尽管有极性键，CCl$_4$ 整体上是非极性的。✓

The electronegativity of H is 2.1 and of O is 3.5. The H–O bond in water is best described as:H 的电负性为 2.1，O 的电负性为 3.5。水中 H–O 键最好描述为：
§6 · Q1
Non-polar covalent非极性共价键
Polar covalent极性共价键
Ionic离子键
Metallic金属键
Δχ = 3.5 − 2.1 = 1.4. Since 0.4 < 1.4 < 1.7, the bond is polar covalent. The electrons are shared but unequally — O pulls the shared pair closer, acquiring a partial negative charge (δ−); H becomes slightly positive (δ+). This polarity makes water a polar molecule and gives it its unique solvent properties.Δχ = 3.5 − 2.1 = 1.4。由于 0.4 < 1.4 < 1.7，该键是极性共价键。电子被共享但不均等——O 将共用电子对拉得更近，获得部分负电荷（δ−）；H 变得略带正电荷（δ+）。这种极性使水成为极性分子，赋予其独特的溶剂性质。
Non-polar covalent requires Δχ < 0.4; ionic dominates when Δχ > 1.7. Here Δχ = 1.4, which falls in the polar covalent range (0.4–1.7). Metallic bonding only occurs in metal lattices, not molecular bonds.非极性共价键要求 Δχ < 0.4；离子性在 Δχ > 1.7 时占主导。这里 Δχ = 1.4，属于极性共价键范围（0.4–1.7）。金属键只发生在金属晶格中，不发生在分子键中。
CO$_2$ has two polar C=O bonds. Why is the overall CO$_2$ molecule non-polar?CO$_2$ 有两个极性 C=O 键。为什么整体 CO$_2$ 分子是非极性的？
§6 · Q2
C and O have the same electronegativity, so no dipole existsC 和 O 的电负性相同，因此不存在偶极矩
CO$_2$ is an ionic compound, not a molecular oneCO$_2$ 是离子化合物，而非分子化合物
The C=O bonds cancel because they are double bondsC=O 键因为是双键而相消
The linear geometry makes the two C=O dipoles point in opposite directions and cancel线型几何使两个 C=O 偶极矩方向相反而相消
CO$_2$ is linear (O=C=O). The two C=O bond dipoles both point from C toward O, but in exactly opposite directions (180° apart). Vector addition gives a net dipole of zero — the molecule is non-polar. This is a key example of the rule: polar bonds + symmetric geometry = non-polar molecule.CO$_2$ 是线型（O=C=O）。两个 C=O 键偶极矩都从 C 指向 O，但方向恰好相反（相距 180°）。向量相加得到净偶极矩为零——分子是非极性的。这是一个关键例子：极性键 + 对称几何 = 非极性分子。
C (χ=2.5) and O (χ=3.5) have different electronegativities, so each C=O bond IS polar. Bond order (single vs double) does not determine whether dipoles cancel — geometry does. CO$_2$'s linear shape makes the two equal, opposite dipoles sum to zero.C（χ=2.5）和 O（χ=3.5）的电负性不同，所以每个 C=O 键确实是极性的。键级（单键与双键）不决定偶极矩是否相消——几何结构才决定。CO$_2$ 的线型使两个大小相等、方向相反的偶极矩之和为零。
Section 7 · Intermolecular forces and metallic bonding Named IMFs: Honors — US NGSS第 7 节 · 分子间作用力与金属键 具名分子间作用力：荣誉 — US NGSS

Intermolecular Forces and Metallic Bonding分子间作用力与金属键

IMFs are the forces between molecules (not inside them). They are weaker than covalent and ionic bonds but govern physical properties.分子间作用力是分子之间（而非分子内部）的力。它们比共价键和离子键弱，但决定物质的物理性质。
  • London (dispersion) forces伦敦（色散）力 — present in ALL molecules (polar or not). Arise from instantaneous dipoles caused by random electron motion. Strength increases with molar mass and surface area. Example: CH$_4$ (bp −161°C) vs C$_4$H$_{10}$ (bp −1°C) — larger molecule, stronger London forces.— 存在于所有分子中（无论极性与否）。由电子随机运动产生的瞬间偶极矩引起。强度随摩尔质量和表面积增大而增强。示例：CH$_4$（沸点 −161°C）vs C$_4$H$_{10}$（沸点 −1°C）——分子更大，伦敦力更强。
  • Dipole-dipole forces偶极-偶极力 — between polar molecules. The positive end of one molecule attracts the negative end of a neighbour. Stronger than London forces for molecules of similar size. Example: HCl (polar) has higher boiling point than Ar (non-polar) of similar molar mass.— 存在于极性分子之间。一个分子的正端吸引邻近分子的负端。对于大小相似的分子，比伦敦力更强。示例：HCl（极性）的沸点高于摩尔质量相近的 Ar（非极性）。
  • Hydrogen bonding氢键 — a special, unusually strong dipole-dipole attraction between an H bonded to F, O, or N and a lone pair on an adjacent F, O, or N. Much stronger than ordinary dipole-dipole forces. Responsible for water's anomalously high boiling point (100°C vs −60°C predicted from mass alone), ice being less dense than liquid water, and the double-helix structure of DNA.— 一种特殊的、异常强的偶极-偶极吸引力，存在于与 F、O 或 N 相连的 H 和相邻 F、O 或 N 的孤电子对之间。比普通偶极-偶极力强得多。负责水的异常高沸点（100°C vs 仅凭质量预测的 −60°C）、冰的密度比液态水小，以及 DNA 双螺旋结构。
Ranking: hydrogen bonding > dipole-dipole > London (dispersion). Metallic bonding: in metals, valence electrons are delocalised in a "sea" of electrons throughout the lattice. This "sea" of mobile electrons explains electrical and thermal conductivity, malleability, and ductility of metals. SCH4U C3.4 and AB Chemistry 20 GO2 both explicitly require naming all three IMF types and distinguishing them from intramolecular bonds.排名：氢键 > 偶极-偶极力 > 伦敦（色散）力。金属键：在金属中，价电子在整个晶格的电子"海洋"中离域。这片移动电子的"海洋"解释了金属的导电性、导热性、延展性和可锻性。SCH4U C3.4 和 AB Chemistry 20 GO2 均明确要求命名全部三种分子间作用力类型，并将其与分子内键区分开来。
Worked Example 7 · Predicting boiling point order using IMFs例题 7 · 用分子间作用力预测沸点顺序

Rank the following in order of increasing boiling point and explain using IMF types: CH$_4$ (molar mass 16 g/mol, non-polar), HCl (molar mass 36.5 g/mol, polar, Δχ=0.9), H$_2$O (molar mass 18 g/mol, polar with H-bonding).按升高的沸点顺序排列以下物质，并用分子间作用力类型解释：CH$_4$（摩尔质量 16 g/mol，非极性）、HCl（摩尔质量 36.5 g/mol，极性，Δχ=0.9）、H$_2$O（摩尔质量 18 g/mol，极性且有氢键）。

CH$_4$.CH$_4$。 Non-polar; only London forces. Smallest and lightest; weakest IMFs. Lowest boiling point (−161°C).非极性；只有伦敦力。最小最轻；最弱的分子间作用力。最低沸点（−161°C）。

HCl.HCl。 Polar (dipole-dipole + London). Heavier than CH$_4$, stronger London forces; also has dipole-dipole. But H is bonded to Cl, not F/O/N, so no hydrogen bonding. Intermediate boiling point (−85°C).极性（偶极-偶极力 + 伦敦力）。比 CH$_4$ 重，伦敦力更强；还有偶极-偶极力。但 H 与 Cl 相连，而非 F/O/N，因此无氢键。中间沸点（−85°C）。

H$_2$O.H$_2$O。 Polar; H bonded to O → hydrogen bonding (the strongest IMF) plus dipole-dipole plus London. Despite being lighter than HCl, its hydrogen bonding is so strong that it has a much higher boiling point (100°C). ✓极性；H 与 O 相连 → 氢键（最强的分子间作用力）加上偶极-偶极力和伦敦力。尽管比 HCl 轻，但其氢键如此之强，沸点远高于 HCl（100°C）。✓

Ranking (lowest to highest bp):排名（从低到高沸点）： CH$_4$ < HCl < H$_2$O.CH$_4$ < HCl < H$_2$O。

Which type of intermolecular force is present in ALL substances, including non-polar molecules like N$_2$?哪种分子间作用力存在于所有物质中，包括 N$_2$ 等非极性分子？
§7 · Q1
Hydrogen bonding氢键
Dipole-dipole forces偶极-偶极力
London (dispersion) forces伦敦（色散）力
Ionic forces离子力
London (dispersion) forces arise from instantaneous, temporary dipoles that form in ANY electron cloud — polar or not. They are present in all substances. Dipole-dipole only occurs in polar molecules; hydrogen bonding only when H is bonded to F, O, or N. Ionic forces operate between ions, not molecules.伦敦（色散）力来自任何电子云（无论极性与否）中形成的瞬间临时偶极矩。它们存在于所有物质中。偶极-偶极力只存在于极性分子中；氢键只在 H 与 F、O 或 N 相连时存在。离子力作用于离子之间，而非分子之间。
Hydrogen bonding requires H bonded to F, O, or N. Dipole-dipole requires a permanent dipole (polar molecule). Only London forces are universal — they arise from temporary fluctuations in electron density that exist in every atom and molecule.氢键要求 H 与 F、O 或 N 相连。偶极-偶极力要求有永久偶极矩（极性分子）。只有伦敦力是普遍存在的——它来自每个原子和分子中电子密度的临时涨落。
Why does water have an anomalously high boiling point (100°C) for such a small molecule?为什么水这么小的分子却有异常高的沸点（100°C）？
§7 · Q2
Extensive hydrogen bonding between water molecules requires more energy to break水分子间广泛的氢键需要更多能量才能断裂
Water molecules are very large and have strong London forces水分子非常大，具有强伦敦力
Water is an ionic compound with a crystal lattice水是具有晶格结构的离子化合物
Water has three O–H bonds, all of which must break to boil水有三个 O–H 键，沸腾时所有键都必须断裂
Each water molecule can form up to 4 hydrogen bonds (2 as H-bond donor through its O–H groups; 2 as acceptor through its lone pairs on O). This extensive hydrogen-bond network requires large amounts of energy to disrupt during boiling — far more than dipole-dipole or London forces alone would require for a molecule of this size. H$_2$S (same Group 16, larger) boils at −60°C because S is not electronegative enough to form hydrogen bonds.每个水分子最多可形成 4 个氢键（2 个作为 H 键供体通过 O–H 基团；2 个作为受体通过 O 上的孤电子对）。沸腾时破坏这个广泛的氢键网络需要大量能量——远远超过单纯的偶极-偶极力或伦敦力对这个大小分子的要求。H$_2$S（同第 16 族，更大）的沸点为 −60°C，因为 S 的电负性不够强，无法形成氢键。
Water (MW 18) is actually a small molecule — its high boiling point is not due to size or London forces. When water boils, covalent O–H bonds do NOT break; the hydrogen bonds between molecules break. The answer is hydrogen bonding, not bond strength, lattice energy, or molecular size.水（摩尔质量 18）实际上是一个小分子——其高沸点不是由于大小或伦敦力。当水沸腾时，共价 O–H 键不会断裂；分子间的氢键断裂。答案是氢键，而非键强度、晶格能或分子大小。
Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Setup discipline for every bonding question每道成键题的解题纪律
  • Count valence electrons before drawing anything.绘图前先计算价电子数。 Add up valence electrons from all atoms. Adjust for charge (subtract 1 per positive charge; add 1 per negative). This total must equal the number of electrons in your Lewis structure — check at the end.将所有原子的价电子数相加。根据电荷调整（每个正电荷减 1；每个负电荷加 1）。这个总数必须等于路易斯结构中的电子数——最后核验。
  • Identify whether the bond is ionic or covalent first.先判断键是离子键还是共价键。 Metal + non-metal → likely ionic. Non-metal + non-metal → covalent. Δχ > 1.7 → ionic character dominates. This determines which type of structure to draw.金属 + 非金属 → 可能是离子键。非金属 + 非金属 → 共价键。Δχ > 1.7 → 离子性为主。这决定应绘制哪种类型的结构。
  • Don't confuse intramolecular bonds with IMFs.不要将分子内键与分子间作用力混淆。 Covalent and ionic bonds are intramolecular (inside a formula unit). IMFs (London, dipole-dipole, H-bonding) are intermolecular (between molecules). Boiling/melting overcomes IMFs, not covalent bonds.共价键和离子键是分子内键（在化学式单元内部）。分子间作用力（伦敦力、偶极-偶极力、氢键）是分子间的（分子之间）。沸腾/熔化克服的是分子间作用力，而非共价键。
Lewis structures and VSEPR (§4–§5)路易斯结构与 VSEPR（§4–§5）
  • Electron-domain geometry ≠ molecular geometry.电子域几何 ≠ 分子几何。 Count ALL electron domains (bonds + lone pairs) for geometry; use only atom positions for molecular shape. A common mistake: calling H$_2$O "tetrahedral" (that's the electron-domain geometry) rather than "bent" (the molecular geometry).计算所有电子域（键 + 孤电子对）以确定几何；仅用原子位置确定分子形状。常见错误：将 H$_2$O 称为"四面体形"（那是电子域几何）而非"角型"（分子几何）。
  • Multiple bonds = one domain.多重键 = 一个域。 A double or triple bond counts as one electron domain for VSEPR purposes. CO$_2$ has 2 domains (two double bonds) → linear, not bent.双键或三键在 VSEPR 中计为一个电子域。CO$_2$ 有 2 个域（两个双键）→ 线型，而非角型。
Polarity and IMFs (§6–§7) Honors — US NGSS (named IMFs)极性与分子间作用力（§6–§7）荣誉 — US NGSS（具名分子间作用力）
  • Polar bonds + symmetric shape = non-polar molecule.极性键 + 对称形状 = 非极性分子。 CO$_2$ and CCl$_4$ are the classic examples. Always check both: (1) is there a Δχ creating polar bonds? (2) does the geometry make the dipoles cancel?CO$_2$ 和 CCl$_4$ 是经典例子。始终同时检查两点：(1) 是否存在 Δχ 产生极性键？(2) 几何结构是否使偶极矩相消？
  • Hydrogen bonding checklist.氢键清单。 H must be bonded to F, O, or N (not Cl, S, etc.) for H-bonding to occur. HF, H$_2$O, NH$_3$, alcohols, and amines — yes. HCl, H$_2$S — no.H 必须与 F、O 或 N（而非 Cl、S 等）相连才能形成氢键。HF、H$_2$O、NH$_3$、醇类和胺类——是。HCl、H$_2$S——否。
  • Explain property differences using the strongest IMF present.用存在的最强分子间作用力解释性质差异。 Higher boiling point, melting point, or viscosity → stronger IMFs. Rank IMFs: H-bonding > dipole-dipole > London (dispersion). When comparing substances of the same IMF type, larger/heavier molecules have stronger London forces.更高的沸点、熔点或粘度 → 更强的分子间作用力。分子间作用力排名：氢键 > 偶极-偶极力 > 伦敦（色散）力。当比较相同分子间作用力类型的物质时，更大/更重的分子具有更强的伦敦力。
Ionic vs covalent compounds (§2–§3) and answer hygiene离子化合物与共价化合物（§2–§3）与作答规范
  • Never say an ionic compound "has molecules."永远不要说离子化合物"有分子"。 Ionic compounds have formula units and lattices; covalent compounds have molecules. SCH3U B3.5 tests this distinction explicitly.离子化合物有化学式单元和晶格；共价化合物有分子。SCH3U B3.5 明确测试这一区别。
  • Link bond type to physical properties.将键型与物理性质联系起来。 Ionic: high mp/bp, brittle, conducts when dissolved/molten. Polar covalent molecular: moderate mp/bp (IMF-dependent), non-conductor. Non-polar covalent molecular: low mp/bp (London only). Metallic: conductor, malleable, ductile.离子型：高熔沸点、脆性、溶解/熔融时导电。极性共价分子型：中等熔沸点（取决于分子间作用力）、不导电。非极性共价分子型：低熔沸点（仅伦敦力）。金属型：导电、可锻、可延。
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Octet rule?八隅规则？
Main-group atoms gain, lose, or share valence electrons to reach 8 valence electrons (noble gas configuration). H exception: targets 2 electrons.主族原子得失或共享价电子以达到 8 个价电子（稀有气体构型）。氢例外：目标为 2 个电子。
Ionic bond: definition?离子键：定义？
Electrostatic attraction between a cation (+) and anion (−) formed by electron transfer from metal to non-metal. Ions arrange in a 3-D lattice.由金属向非金属转移电子形成的阳离子（+）和阴离子（−）之间的静电吸引力。离子排列在三维晶格中。
Covalent bond: definition?共价键：定义？
Shared pair of electrons between two non-metal atoms; both nuclei attract the shared electrons. Single (2e⁻), double (4e⁻), triple (6e⁻).两个非金属原子之间的共用电子对；两个原子核共同吸引共享电子。单键（2e⁻）、双键（4e⁻）、三键（6e⁻）。
Lewis structure — step 1?路易斯结构——第 1 步？
Count total valence electrons: sum all atoms' valence e⁻; subtract 1 per + charge; add 1 per − charge.计算总价电子数：将所有原子的价电子数相加；每个正电荷减 1；每个负电荷加 1。
Bond order vs bond length and strength?键级与键长、键能的关系？
Higher bond order → shorter bond AND stronger bond. Triple > double > single in strength and shortness.键级越高 → 键越短且键越强。强度和短度：三键 > 双键 > 单键。
VSEPR: 4 domains, 0 lone pairs?VSEPR：4 个域，0 个孤电子对？
Tetrahedral molecular geometry. Bond angle: 109.5°. Example: CH₄.四面体形分子几何。键角：109.5°。示例：CH₄。
VSEPR: 4 domains, 1 lone pair?VSEPR：4 个域，1 个孤电子对？
Trigonal pyramidal molecular geometry. Bond angle: <109.5° (≈107°). Example: NH₃.三角锥形分子几何。键角：<109.5°（约 107°）。示例：NH₃。
VSEPR: 4 domains, 2 lone pairs?VSEPR：4 个域，2 个孤电子对？
Bent (angular) molecular geometry. Bond angle: <109.5° (≈104.5°). Example: H₂O.角型（弯曲形）分子几何。键角：<109.5°（约 104.5°）。示例：H₂O。
Electronegativity difference → bond type?电负性差 → 键型？
Δχ <0.4: non-polar covalent. Δχ 0.4–1.7: polar covalent. Δχ >1.7: ionic character dominates.Δχ <0.4：非极性共价键。Δχ 0.4–1.7：极性共价键。Δχ >1.7：离子性为主。
Polar bonds but non-polar molecule — example?有极性键但分子无极性——示例？
CO₂ (linear) and CCl₄ (tetrahedral). Symmetric geometry makes bond dipoles cancel exactly. Net dipole = 0.CO₂（线型）和 CCl₄（四面体形）。对称几何使键偶极矩完全相消。净偶极矩 = 0。
London (dispersion) forces?伦敦（色散）力？
Present in ALL substances. Arise from instantaneous dipoles. Strength increases with molar mass and surface area.存在于所有物质中。由瞬间偶极矩产生。强度随摩尔质量和表面积增大而增强。
Hydrogen bonding: conditions?氢键：条件？
H must be bonded to F, O, or N. The H then attracts a lone pair on a neighbouring F, O, or N. Strongest IMF. Examples: H₂O, NH₃, HF.H 必须与 F、O 或 N 相连。然后 H 吸引邻近 F、O 或 N 上的孤电子对。最强的分子间作用力。示例：H₂O、NH₃、HF。
IMF strength ranking?分子间作用力强度排名？
Hydrogen bonding > dipole-dipole > London (dispersion). Stronger IMF → higher bp/mp.氢键 > 偶极-偶极力 > 伦敦（色散）力。更强的分子间作用力 → 更高的沸点/熔点。
Metallic bonding?金属键？
Delocalised valence electrons form a "sea" flowing through a lattice of cations. Explains electrical conductivity, thermal conductivity, malleability, and ductility.离域的价电子形成流经阳离子晶格的"海洋"。解释导电性、导热性、可锻性和延展性。
Mixed Practice综合练习

Practice Quiz综合测验

Calcium (Group 2) reacts with fluorine (Group 17) to form CaF$_2$. What type of bond is present and what ions are formed?钙（第 2 族）与氟（第 17 族）反应形成 CaF$_2$。存在什么类型的键，形成了哪些离子？
Q1
Covalent bond; Ca$^-$ and F$^+$共价键；Ca$^-$ 和 F$^+$
Ionic bond; Ca$^{2+}$ and F$^-$离子键；Ca$^{2+}$ 和 F$^-$
Covalent bond; Ca$^{2+}$ and F$^{2-}$共价键；Ca$^{2+}$ 和 F$^{2-}$
Metallic bond; Ca$^+$ and F$^-$金属键；Ca$^+$ 和 F$^-$
Metal (Ca, Group 2) + non-metal (F, Group 17) → ionic bond. Ca loses its 2 valence electrons to reach the Ar configuration: Ca$^{2+}$. F needs 1 electron to complete its octet: F$^-$. Two F$^-$ balance one Ca$^{2+}$, giving CaF$_2$. Δχ = 4.0 − 1.0 = 3.0 > 1.7 → ionic.金属（Ca，第 2 族）+ 非金属（F，第 17 族）→ 离子键。Ca 失去其 2 个价电子达到 Ar 构型：Ca$^{2+}$。F 需要 1 个电子完成八隅：F$^-$。两个 F$^-$ 平衡一个 Ca$^{2+}$，给出 CaF$_2$。Δχ = 4.0 − 1.0 = 3.0 > 1.7 → 离子键。
Metal + non-metal pairs almost always form ionic bonds. Ca (Group 2) loses 2 electrons (not 1), forming Ca$^{2+}$. F (Group 17) gains 1 electron, forming F$^-$ (not F$^{2-}$). Two F$^-$ needed to balance Ca$^{2+}$.金属 + 非金属组合几乎总是形成离子键。Ca（第 2 族）失去 2 个电子（而非 1 个），形成 Ca$^{2+}$。F（第 17 族）获得 1 个电子，形成 F$^-$（而非 F$^{2-}$）。需要两个 F$^-$ 来平衡 Ca$^{2+}$。
How many total valence electrons must be shown in the Lewis structure of the nitrate ion (NO$_3^-$)?亚硝酸根离子（NO$_3^-$）的路易斯结构中必须显示多少个总价电子？
Q2
2424
2323
1818
2020
N (Group 15) contributes 5; each O (Group 16) contributes 6 × 3 = 18; total from atoms = 23. The 1− charge adds 1 electron. Grand total: 23 + 1 = 24 valence electrons. Always adjust for ionic charge before drawing.N（第 15 族）贡献 5；每个 O（第 16 族）贡献 6 × 3 = 18；来自原子的总计 = 23。1− 电荷增加 1 个电子。总计：23 + 1 = 24 个价电子。绘图前始终根据离子电荷调整。
Valence electrons: N = 5, O = 6 each (×3 = 18). Atom total = 5 + 18 = 23. But the ion has a 1− charge, meaning one extra electron has been added: 23 + 1 = 24. For anions, add 1 per negative charge; for cations, subtract 1 per positive charge.价电子：N = 5，每个 O = 6（×3 = 18）。原子总计 = 5 + 18 = 23。但离子有 1− 电荷，意味着增加了一个额外电子：23 + 1 = 24。对于阴离子，每个负电荷加 1；对于阳离子，每个正电荷减 1。
Which molecule has a tetrahedral molecular geometry? 🇨🇦 SCH4U C2.3 / AB Chem 20 GO2哪种分子具有四面体形分子几何？🇨🇦 SCH4U C2.3 / AB Chem 20 GO2
Q3
H$_2$O (4 domains, 2 lone pairs)H$_2$O（4 个域，2 个孤电子对）
NH$_3$ (4 domains, 1 lone pair)NH$_3$（4 个域，1 个孤电子对）
CO$_2$ (2 domains, 0 lone pairs)CO$_2$（2 个域，0 个孤电子对）
CH$_4$ (4 domains, 0 lone pairs)CH$_4$（4 个域，0 个孤电子对）
CH$_4$ has 4 bonding pairs and 0 lone pairs on C → 4 electron domains → tetrahedral electron-domain geometry → tetrahedral molecular geometry (all domains are bond pairs, so molecular geometry = electron-domain geometry). Bond angle: 109.5°.CH$_4$ 在 C 上有 4 个成键对和 0 个孤电子对 → 4 个电子域 → 四面体形电子域几何 → 四面体形分子几何（所有域均为成键对，故分子几何 = 电子域几何）。键角：109.5°。
H$_2$O: 4 domains but 2 lone pairs → electron-domain tetrahedral, molecular geometry bent. NH$_3$: 4 domains, 1 lone pair → trigonal pyramidal. CO$_2$: 2 domains → linear. Only CH$_4$ (4 domains, 0 lone pairs) is truly tetrahedral in molecular geometry.H$_2$O：4 个域但 2 个孤电子对 → 电子域四面体形，分子几何角型。NH$_3$：4 个域，1 个孤电子对 → 三角锥形。CO$_2$：2 个域 → 线型。只有 CH$_4$（4 个域，0 个孤电子对）在分子几何上真正是四面体形。
The electronegativity of Na is 0.9 and of Cl is 3.0. What type of bond does Na–Cl form, and why? 🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U B2.5Na 的电负性为 0.9，Cl 的电负性为 3.0。Na–Cl 形成什么类型的键，为什么？🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U B2.5
Q4
Non-polar covalent; Δχ = 2.1, which is just above 1.7非极性共价键；Δχ = 2.1，略高于 1.7
Polar covalent; Δχ = 2.1, which falls in the polar covalent range极性共价键；Δχ = 2.1，处于极性共价键范围内
Ionic; Δχ = 2.1 > 1.7, so ionic character dominates离子键；Δχ = 2.1 > 1.7，离子性为主
Metallic; Na is a metal金属键；Na 是金属
Δχ = 3.0 − 0.9 = 2.1 > 1.7 → ionic character dominates. Na transfers its lone valence electron to Cl, forming Na$^+$ and Cl$^-$. NGSS HS-PS1-2 explicitly uses Na + Cl as an example of bonding driven by outermost electron states.Δχ = 3.0 − 0.9 = 2.1 > 1.7 → 离子性为主。Na 将其唯一的价电子转移给 Cl，形成 Na$^+$ 和 Cl$^-$。NGSS HS-PS1-2 明确以 Na + Cl 为例，说明由最外层电子状态驱动的成键。
Δχ = 3.0 − 0.9 = 2.1, not 2.1 as non-polar. The threshold for ionic character is Δχ > 1.7. Since 2.1 > 1.7, the bond is ionic (electron transfer), not covalent (electron sharing). Metallic bonding only occurs within a metal lattice, not between Na and Cl.Δχ = 3.0 − 0.9 = 2.1。离子性的阈值为 Δχ > 1.7。由于 2.1 > 1.7，该键是离子键（电子转移），而非共价键（电子共享）。金属键只发生在金属晶格内部，而非 Na 和 Cl 之间。
Which substance has the highest boiling point, and why? 🇨🇦 SCH4U C3.4 / AB Chem 20 GO2哪种物质的沸点最高，为什么？🇨🇦 SCH4U C3.4 / AB Chem 20 GO2
Q5
H$_2$O (polar, H-bonding)H$_2$O（极性，有氢键）
H$_2$S (polar, dipole-dipole only)H$_2$S（极性，仅偶极-偶极力）
CH$_4$ (non-polar, London only)CH$_4$（非极性，仅伦敦力）
Ar (non-polar, London only)Ar（非极性，仅伦敦力）
H$_2$O has hydrogen bonding (H bonded to O) — the strongest IMF type. H$_2$S has only dipole-dipole + London forces (S is not electronegative enough for H-bonding). Despite H$_2$O being lighter than H$_2$S, its hydrogen bonding network gives it a boiling point of 100°C vs −60°C for H$_2$S. CH$_4$ and Ar have only weak London forces, giving the lowest boiling points.H$_2$O 有氢键（H 与 O 相连）——最强的分子间作用力类型。H$_2$S 只有偶极-偶极力 + 伦敦力（S 的电负性不足以形成氢键）。尽管 H$_2$O 比 H$_2$S 轻，但其氢键网络赋予它 100°C 的沸点，而 H$_2$S 仅为 −60°C。CH$_4$ 和 Ar 只有弱伦敦力，沸点最低。
The key is whether H is bonded to F, O, or N. H$_2$O: H bonded to O → hydrogen bonding (strongest IMF) → highest boiling point. H$_2$S: H bonded to S (not F/O/N) → no H-bonding → much lower bp despite larger size.关键在于 H 是否与 F、O 或 N 相连。H$_2$O：H 与 O 相连 → 氢键（最强分子间作用力）→ 最高沸点。H$_2$S：H 与 S 相连（非 F/O/N）→ 无氢键 → 尽管体积更大，沸点却低得多。
A student draws the Lewis structure of SO$_2$ and finds that S has 3 electron domains (1 lone pair + 2 bonds). What molecular geometry results? 🇨🇦 AB Chem 20 GO2 / SCH4U C2.3一名学生绘制了 SO$_2$ 的路易斯结构，发现 S 有 3 个电子域（1 个孤电子对 + 2 个键）。产生什么分子几何？🇨🇦 AB Chem 20 GO2 / SCH4U C2.3
Q6
Linear (180°)线型（180°）
Bent / angular (<120°)角型（<120°）
Trigonal planar (120°)三角平面形（120°）
Trigonal pyramidal (<109.5°)三角锥形（<109.5°）
3 electron domains (1 lone pair + 2 bonding domains) → trigonal planar electron-domain geometry. The lone pair occupies one of the three positions, leaving the two O atoms at a bent angle. Molecular geometry (atoms only) = bent/angular, with bond angle compressed below 120° by the lone pair.3 个电子域（1 个孤电子对 + 2 个成键域）→ 三角平面形电子域几何。孤电子对占据三个位置之一，使两个 O 原子呈角型排列。分子几何（仅原子）= 角型，键角被孤电子对压缩至低于 120°。
Trigonal planar is the electron-domain geometry (3 domains), not the molecular geometry. The lone pair "uses up" one of those three positions — so the O–S–O angle is bent, not linear or trigonal planar. 3 domains + 1 lone pair → bent molecular geometry.三角平面形是电子域几何（3 个域），而非分子几何。孤电子对"占用"了那三个位置之一——所以 O–S–O 角是角型的，而非线型或三角平面形。3 个域 + 1 个孤电子对 → 角型分子几何。
Why do metals conduct electricity? 🇨🇦 SCH4U C2.5, C3.4 / AB Chem 20为什么金属导电？🇨🇦 SCH4U C2.5, C3.4 / AB Chem 20
Q7
Metals are ionic compounds with mobile ions in the lattice金属是晶格中有移动离子的离子化合物
Metals have polar covalent bonds that allow electron sharing across the lattice金属有极性共价键，允许电子在晶格中共享
Delocalised valence electrons form a "sea" that moves freely in response to an electric field离域的价电子形成"海洋"，可在电场作用下自由移动
Metals have hydrogen bonding between atoms that transfers charge金属原子间有氢键传递电荷
In metallic bonding, valence electrons are delocalised — they are not attached to specific atoms but are free to move throughout the entire metal lattice. When a voltage is applied, these "sea" electrons flow toward the positive terminal, carrying charge. This is why all metals are conductors in the solid state (unlike ionic compounds, which only conduct when dissolved or molten).在金属键中，价电子是离域的——它们不附着于特定原子，而是可以在整个金属晶格中自由移动。施加电压时，这些"海洋"电子向正极流动，携带电荷。这就是为什么所有金属在固态下都是导体（不同于离子化合物，后者只有在溶解或熔融时才导电）。
Metals are NOT ionic compounds. Polar covalent bonding and hydrogen bonding do not explain conductivity in solid metals. The key is delocalised electrons — a "sea" of free electrons that can flow under an applied voltage.金属不是离子化合物。极性共价键和氢键不能解释固态金属的导电性。关键是离域电子——一片可在施加电压下流动的自由电子"海洋"。
Which statement correctly describes the difference between ionic and molecular compounds? 🇨🇦 SCH3U B3.5 / BC Chemistry 11下列哪句话正确描述了离子化合物和分子化合物之间的区别？🇨🇦 SCH3U B3.5 / BC Chemistry 11
Q8
Ionic compounds conduct electricity in the solid state; molecular compounds do not离子化合物在固态下导电；分子化合物不导电
Molecular compounds generally have higher melting points than ionic compounds分子化合物通常比离子化合物的熔点高
Ionic compounds are held together by shared electron pairs; molecular compounds by ion-ion forces离子化合物由共用电子对结合；分子化合物由离子-离子力结合
Ionic compounds form lattices with high melting points; molecular compounds have discrete molecules and generally lower melting points离子化合物形成具有高熔点的晶格；分子化合物有离散分子，通常熔点较低
Ionic compounds (e.g. NaCl, MgO) form 3-D lattices held by strong electrostatic forces — high melting and boiling points (NaCl melts at 801°C). Molecular compounds (e.g. H$_2$O, CO$_2$) consist of discrete molecules held together by IMFs, which are weaker — generally lower boiling/melting points. This is exactly the SCH3U B3.5 distinction.离子化合物（如 NaCl、MgO）形成由强静电力结合的三维晶格——高熔沸点（NaCl 熔点 801°C）。分子化合物（如 H$_2$O、CO$_2$）由分子间作用力结合的离散分子组成，分子间作用力较弱——通常沸沸点较低。这正是 SCH3U B3.5 的区别。
Ionic solids do NOT conduct electricity (ions are locked in the lattice). Molecular compounds generally have LOWER melting points. It is ionic compounds that are held by electrostatic ion-ion forces (not shared pairs), and molecular compounds that use shared electron pairs (covalent bonds).离子固体不导电（离子被锁定在晶格中）。分子化合物通常熔点更低。离子化合物由静电离子-离子力结合（而非共用对），分子化合物使用共用电子对（共价键）。
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Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

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Next Steps后续单元

What This Feeds Into本单元的去向

Chemical bonding is the link between atomic structure and every branch of chemistry that follows. The ability to draw Lewis structures, predict molecular geometry, and identify dominant IMFs is assumed in every subsequent unit — from reaction types (Unit 6) to solutions and solubility (Unit 8) to organic chemistry (Unit 14). The cross-references below point at the college-credit feeder and the next High School Chemistry units.化学键是原子结构与后续每个化学分支之间的桥梁。绘制路易斯结构、预测分子几何和识别主要分子间作用力的能力在后续每个单元中都被默认掌握——从反应类型（第 6 单元）到溶液与溶解度（第 8 单元）再到有机化学（第 14 单元）。以下链接指向大学学分衔接课程和后续高中化学单元。

Within High School Chemistry.在 HS Chemistry 内部。

Nomenclature and Chemical Formulae (Unit 4) builds directly on the ionic/covalent distinction from §2–§3. The Mole and Stoichiometry (Unit 5) and Chemical Reactions (Unit 6) require Lewis structures to predict reaction products from bond type. States of Matter (Unit 7) and Solutions (Unit 8) are governed by the IMFs you mastered in §7 — polarity determines solubility ("like dissolves like"), and IMF strength determines phase-change temperatures. Chemical Bonding is the most-used prior knowledge in the entire HS Chemistry course.命名与化学式（第 4 单元）直接建立在 §2–§3 的离子/共价区分基础上。摩尔与化学计量学（第 5 单元）和化学反应（第 6 单元）需要路易斯结构来根据键型预测反应产物。物质状态（第 7 单元）和溶液（第 8 单元）由你在 §7 中掌握的分子间作用力决定——极性决定溶解性（"相似相溶"），分子间作用力强度决定相变温度。化学键是整个 HS Chemistry 课程中用得最多的先修知识。

College-credit feeder and cross-subject links.大学学分衔接与跨学科链接。

IB Chemistry HL · Structure 2: Bonding and Structure (the college-credit feeder for chemical bonding, Lewis structures, VSEPR, molecular polarity, and solid-state structure at IB HL depth)IB Chemistry HL · Structure 2：成键与结构（化学键、路易斯结构、VSEPR、分子极性以及 IB HL 深度下固态结构的大学学分衔接）

If you are aiming for IB Chemistry HL or AP Chemistry, the Lewis structures, VSEPR geometry, polarity, and IMF material here is assumed from the first week of the college-credit course. IB Chemistry HL Structure 2 extends this with formal charges, resonance structures, hybridisation (sp, sp², sp³), sigma and pi bonds, and the band theory of metallic bonding. AP Chemistry Unit 2 (Molecular and Ionic Compound Structure and Properties) adds lattice energy, Born-Haber cycles, and VSEPR for 5- and 6-electron-domain geometries (trigonal bipyramidal, octahedral). Every advanced bonding topic at IB/AP depth traces back to the Lewis structure skills you practice in §4.如果你目标是 IB Chemistry HL 或 AP Chemistry，这里的路易斯结构、VSEPR 几何、极性和分子间作用力材料从大学学分课程的第一周就被默认掌握。IB Chemistry HL Structure 2 通过形式电荷、共振结构、杂化（sp、sp²、sp³）、σ 键和 π 键以及金属键的能带理论来延伸这部分内容。AP Chemistry Unit 2（分子与离子化合物结构和性质）增加了晶格能、玻恩-哈伯循环以及 5 和 6 电子域几何（三角双锥形、八面体形）的 VSEPR。IB/AP 深度下每个高级成键主题都可以追溯到你在 §4 中练习的路易斯结构技能。

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High School Chemistry · Chemical Bonding · v1高中化学 · 化学键 · v1