High School Chemistry - Chemical Bonding - Solutions

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Octet rule八隅体规则 [3 marks][3 分]

How many electrons does a sulfur atom need to gain to satisfy the octet rule?硫原子需要获得多少个电子才能满足八隅体规则？

Answer:答案： (A) 2

(a) Count valence electrons and apply the octet rule数价电子并应用八隅体规则 M1·A1·A1

Sulfur is in Group VIA (Group 16). It has 6 valence electrons. The octet rule requires 8 electrons in the valence shell. Therefore sulfur needs to gain:硫位于第 VIA 族（第 16 族），有 6 个价电子。八隅体规则要求价层有 8 个电子，因此硫需要获得： $$ 8 - 6 \;=\; 2 \;\text{ electrons.} $$ Answer is (A) 2. Sulfur forms the sulfide ion $\text{S}^{2-}$ in ionic compounds, consistent with a gain of 2 electrons.答案为 (A) 2。硫在离子化合物中形成硫离子 $\text{S}^{2-}$，与获得 2 个电子一致。

Why the distractors fail.干扰项分析。

(B) 6: confuses the number of valence electrons sulfur already has with the number it needs to gain.混淆了硫已有的价电子数与需要获得的电子数。

(D) 4: applies to carbon (Group IVA), which has 4 valence electrons and needs 4 more.适用于碳（第 IVA 族），碳有 4 个价电子，需再获得 4 个。

The group number tells you how many valence electrons an element has; 8 minus that number is how many it gains (for nonmetals).族序数等于元素的价电子数；8 减去该数即为非金属元素需要获得的电子数。 For main-group nonmetals, the octet rule shortcut is: electrons to gain = 8 - (group number). Group IA gains 7 but loses 1 instead; Groups VIA, VIIA, VA gain 2, 1, and 3 respectively. Sulfur (Group VIA) gains 2, forming $\text{S}^{2-}$. This same logic predicts the charge on the ion formed in ionic bonding and the number of covalent bonds formed in covalent bonding.对于主族非金属元素，八隅体快捷法则为：需获得的电子数 = 8 - 族序数。第 VIA、VIIA、VA 族分别获得 2、1、3 个电子。硫（第 VIA 族）获得 2 个电子，形成 $\text{S}^{2-}$。这一逻辑同样预测了离子键中形成的离子电荷以及共价键中形成的共价键数目。

Q2EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §2 Ionic bonding离子键 [4 marks][4 分]

CaCl₂: (a) Identify the ions formed. (b) Explain why two chloride ions are needed for every calcium ion.氯化钙 CaCl₂：(a) 确定所形成的离子；(b) 解释为何每个钙离子需要两个氯离子。

Answer:答案： (a) Ca²⁺ and Cl⁻Ca²⁺ 和 Cl⁻ · (b) Ca loses 2e⁻; each Cl gains only 1e⁻, so 2 Cl⁻ are needed for charge balanceCa 失去 2 个电子；每个 Cl 只获得 1 个电子，故需 2 个 Cl⁻ 以保持电荷守恒

(a) Identify the ions确定离子 A1·A1

Calcium is in Group IIA (Group 2); it has 2 valence electrons and loses them both to achieve the noble-gas configuration of argon, forming $\text{Ca}^{2+}$. Chlorine is in Group VIIA (Group 17); it has 7 valence electrons and gains 1 to reach the argon configuration, forming $\text{Cl}^{-}$.钙位于第 IIA 族（第 2 族），有 2 个价电子，失去全部 2 个电子后达到氩的稀有气体构型，形成 $\text{Ca}^{2+}$。氯位于第 VIIA 族（第 17 族），有 7 个价电子，获得 1 个电子后达到氩的构型，形成 $\text{Cl}^{-}$。

(b) Why two chloride ions?为何需要两个氯离子？ A1·A1

Each calcium atom loses 2 electrons. Each chlorine atom can accept only 1 electron (it only needs 1 to complete its octet). To account for the 2 electrons released by one Ca, exactly 2 Cl atoms must each accept 1 electron. The overall ionic compound is electrically neutral: $(+2) + 2(-1) = 0$.每个钙原子失去 2 个电子。每个氯原子只能接受 1 个电子（它只需 1 个即可完成八隅体）。1 个 Ca 释放 2 个电子，须由 2 个 Cl 各接受 1 个来承接。整体离子化合物电中性：$(+2) + 2(-1) = 0$。

Ionic formula ratios are set by the charge balance condition: total positive charge must equal total negative charge.离子式的比例由电荷守恒条件决定：正电荷总量必须等于负电荷总量。 A reliable method: write the charge on each ion, then cross the magnitudes to get the subscripts. $\text{Ca}^{2+}$ and $\text{Cl}^{-}$: cross gives 1 Ca and 2 Cl, so $\text{CaCl}_2$. This works for any binary ionic compound. The 2:1 ratio of $\text{Cl}^{-}$ to $\text{Ca}^{2+}$ also explains why dissolving $\text{CaCl}_2$ in water releases three ions per formula unit, giving it a higher conductivity than 1:1 ionic compounds at the same concentration.一个可靠的方法：写出每种离子的电荷，然后交叉电荷大小得到下标。$\text{Ca}^{2+}$ 和 $\text{Cl}^{-}$：交叉得 1 个 Ca 和 2 个 Cl，即 $\text{CaCl}_2$。此法适用于任何二元离子化合物。$\text{Cl}^{-}$ 与 $\text{Ca}^{2+}$ 的 2:1 比例也解释了为何溶解 $\text{CaCl}_2$ 时每个化学式单元释放 3 个离子，在相同浓度下导电性高于 1:1 离子化合物。

Q3MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §3 Covalent bonding共价键 [4 marks][4 分]

Compare bonding in N₂ and H₂O: (a) Describe the bond type and bond order in N₂. (b) Describe the bond in H₂O and explain how it is consistent with water being a liquid at room temperature.比较 N₂ 和 H₂O 中的化学键：(a) 描述 N₂ 中的键型和键级；(b) 描述 H₂O 中的化学键，并解释这与水在常温下为液态的联系。

Answer:答案： (a) triple bond, bond order 3, 3 shared pairs三键，键级 3，3 对共用电子 · (b) single O-H bonds; O lone pairs enable hydrogen bonding, raising the boiling point so water is liquid at room temperature单 O-H 键；O 上的孤对电子使水分子间形成氢键，沸点升高，常温下为液态

(a) Bond type and bond order in N₂N₂ 的键型与键级 A1·A1

Nitrogen has 5 valence electrons. Each N atom needs 3 more to complete its octet. The two N atoms share 3 pairs of electrons, forming a triple bond (bond order = 3). The Lewis structure is $\text{:N:::N:}$ with 1 lone pair on each N and 3 bonding pairs between them.氮有 5 个价电子，每个 N 原子还需 3 个才能满足八隅体。两个 N 原子共用 3 对电子，形成三键（键级 = 3）。路易斯结构为 $\text{:N:::N:}$，每个 N 上有 1 对孤对电子，N 之间有 3 对共用电子。

(b) Bonding in H₂O and its link to liquid stateH₂O 中的化学键及其与液态的关系 A1·A1

Water has two O-H single bonds (bond order = 1). Oxygen is highly electronegative, so the O-H bonds are strongly polar. Oxygen also carries 2 lone pairs. These lone pairs act as hydrogen-bond acceptors, allowing each water molecule to form up to 4 hydrogen bonds with neighbors. Hydrogen bonding is a strong intermolecular force, requiring significant energy to overcome. Therefore water has a much higher boiling point (100 °C) than expected for a molecule of its size, making it a liquid at room temperature.水有两条 O-H 单键（键级 = 1）。氧的电负性很强，使 O-H 键极性显著。氧上还有 2 对孤对电子，可作为氢键受体，使每个水分子与邻近分子最多形成 4 条氢键。氢键是较强的分子间作用力，克服它需要较多能量。因此水的沸点（100 °C）远高于同等大小分子的预期值，在室温下呈液态。

Bond order determines bond strength and length; intermolecular forces (not bond type) determine physical state.键级决定键强度和键长；分子间作用力（而非键型）决定物理状态。 N₂ has a very strong, short triple bond (bond energy ~945 kJ/mol), but it is a gas at room temperature because N₂ molecules are nonpolar and only experience weak London dispersion forces between them. The intramolecular bond strength and the intermolecular forces are completely separate concepts. H₂O has weaker O-H bonds, but the hydrogen bonding between molecules is strong enough to keep water liquid. Always distinguish between bonds within a molecule (intramolecular) and forces between molecules (intermolecular) when explaining physical properties.N₂ 有很强的短三键（键能约 945 kJ/mol），但在室温下是气体，因为 N₂ 分子是非极性的，分子间只有弱的伦敦色散力。分子内键强度与分子间作用力是完全不同的概念。H₂O 的 O-H 键相对较弱，但分子间的氢键足够强，使水保持液态。解释物理性质时，始终要区分分子内化学键（intramolecular）和分子间作用力（intermolecular）。

Q4MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §4 Lewis structures路易斯结构 [3 marks][3 分]

How many lone pairs are on the oxygen atoms in a correct Lewis structure of CO₂?在 CO₂ 的正确路易斯结构中，氧原子上共有多少对孤对电子？

Answer:答案： (B) 4 lone pairs(B) 4 对孤对电子

(a) Draw the Lewis structure of CO₂ and count lone pairs画出 CO₂ 的路易斯结构并数孤对电子 M1·A1·A1

Total valence electrons: $\text{C} = 4,\ \text{O} \times 2 = 12$, total $= 16$. Carbon is the central atom. Connecting with single bonds uses $4$ electrons, leaving $12$ for lone pairs on the oxygens ($3$ pairs each). Checking octets: each O has $6$ electrons from lone pairs + $2$ from the C-O bond = $8$ (satisfied), but C has only $4$ from two single bonds, so C's octet is not satisfied. Promote to double bonds: $\text{O=C=O}$. In the final structure each O has $2$ lone pairs (4 electrons) and is double-bonded to C. Carbon has no lone pairs and its octet is satisfied by the two double bonds (4 electrons each side = 4 + 4 = 8).总价电子数：$\text{C} = 4,\ \text{O} \times 2 = 12$，共 $16$ 个。碳为中心原子。用单键连接消耗 $4$ 个电子，剩余 $12$ 个作为 O 上的孤对电子（每个 O 各 3 对）。核查八隅体：每个 O 有 6 个孤对电子 + 2 个 C-O 键电子 = 8（满足），但 C 仅有 2 条单键提供的 4 个电子，未满足八隅体，故需升级为双键：$\text{O=C=O}$。最终结构中每个 O 有 2 对孤对电子（4 个电子）并与 C 形成双键。碳没有孤对电子，两个双键各提供 4 个电子使碳的八隅体满足。

Total lone pairs on O atoms: $2 + 2 = \mathbf{4}$ lone pairs. Answer: (B).O 原子上孤对电子总对数：$2 + 2 = \mathbf{4}$ 对。答案：(B)。

Why the distractors fail.干扰项分析。

(A) 2 lone pairs: counts only the pairs on one oxygen atom instead of both.只计算了一个氧原子上的孤对电子，而非两个氧。

(C) 6 lone pairs: uses the incorrect single-bond structure where each O has 3 lone pairs, failing to recognize that carbon's octet requires double bonds.使用了错误的单键结构（每个 O 有 3 对孤对电子），未意识到碳的八隅体需要双键。

(D) 8 lone pairs: confuses lone pairs on C (zero) with lone pairs on O, or adds incorrectly.将 C 上的孤对电子（零对）与 O 上的混淆，或计算有误。

Always check all octets after placing lone pairs; if the central atom is unsatisfied, convert lone pairs on terminal atoms to bonding pairs (multiple bonds).分配孤对电子后，务必核查所有原子的八隅体；若中心原子不满足，则将端基原子上的孤对电子转化为成键对（形成多重键）。 The systematic method for Lewis structures: (1) count total valence electrons; (2) connect all atoms with single bonds; (3) distribute remaining electrons as lone pairs on terminal atoms first; (4) check the central atom's octet; (5) if the central atom is short, convert terminal lone pairs to double/triple bonds until all octets are satisfied. In CO₂, each O donates one lone pair to form a double bond with C, reducing the lone pairs on each O from 3 to 2, and satisfying C's octet.路易斯结构的系统方法：(1) 计算总价电子数；(2) 用单键连接所有原子；(3) 优先将剩余电子以孤对电子形式分配给端基原子；(4) 核查中心原子的八隅体；(5) 若中心原子不满足，则将端基原子的孤对电子转化为双键/三键，直至所有原子的八隅体满足。在 CO₂ 中，每个 O 贡献 1 对孤对电子与 C 形成双键，使每个 O 的孤对电子从 3 对减为 2 对，同时满足 C 的八隅体。

Q5MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §6 Bond polarity键的极性 [4 marks][4 分]

(a) Classify the bonds in HCl (ΔEN = 0.9) and HF (ΔEN = 1.9) as ionic, polar covalent, or nonpolar covalent. (b) In each molecule, identify which atom carries the partial negative charge (δ⁻).(a) 将 HCl（ΔEN = 0.9）和 HF（ΔEN = 1.9）中的化学键分类为离子键、极性共价键或非极性共价键。(b) 在每个分子中，指出哪个原子带部分负电荷（δ⁻）。

Answer:答案： (a) both polar covalent均为极性共价键 · (b) δ⁻ on Cl in HCl; δ⁻ on F in HFHCl 中 Cl 带 δ⁻；HF 中 F 带 δ⁻

(a) Classify by electronegativity difference按电负性差分类 A1·A1

The standard thresholds are: $\Delta\text{EN} < 0.4$ = nonpolar covalent; $0.4 \le \Delta\text{EN} < 1.7$ = polar covalent; $\Delta\text{EN} \ge 1.7$ = ionic.标准阈值为：$\Delta\text{EN} < 0.4$ = 非极性共价键；$0.4 \le \Delta\text{EN} < 1.7$ = 极性共价键；$\Delta\text{EN} \ge 1.7$ = 离子键。

HCl: $\Delta\text{EN} = 0.9$ (between 0.4 and 1.7) → polar covalent.HCl：$\Delta\text{EN} = 0.9$（介于 0.4 和 1.7 之间）→ 极性共价键。
HF: $\Delta\text{EN} = 1.9$ (above 1.7) → strictly ionic by the threshold, but HF is commonly classified as polar covalent in high-school curricula because it exists as discrete molecules. At the HS level, accept "polar covalent" for $\Delta\text{EN} = 1.9$.HF：$\Delta\text{EN} = 1.9$（超过 1.7）→ 按阈值严格分类为离子键，但高中课程中因 HF 以独立分子形式存在，通常归类为极性共价键。高中阶段接受 $\Delta\text{EN} = 1.9$ 被归为极性共价键。

(b) Identify the δ⁻ atom in each molecule确定每个分子中的 δ⁻ 原子 A1·A1

The partial negative charge resides on the more electronegative atom (the one that pulls electron density toward itself).部分负电荷位于电负性较大的原子上（即吸引电子密度更强的一方）。

HCl: Cl (EN = 3.0) is more electronegative than H (EN = 2.1), so Cl carries δ⁻.HCl：Cl（EN = 3.0）比 H（EN = 2.1）电负性更大，故 Cl 带 δ⁻。
HF: F (EN = 4.0) is the most electronegative element, more electronegative than H (EN = 2.1), so F carries δ⁻.HF：F（EN = 4.0）是电负性最强的元素，比 H（EN = 2.1）强，故 F 带 δ⁻。

Electronegativity difference predicts bond polarity; the more electronegative atom always carries δ⁻.电负性差预测键的极性；电负性较大的原子始终带 δ⁻。 Electronegativity (EN) measures an atom's pull on shared electrons. In a polar covalent bond, electrons are not shared equally; they spend more time near the more electronegative atom, creating a permanent dipole ($\delta^-$ on the EN-rich end, $\delta^+$ on the EN-poor end). HF has a larger $\Delta\text{EN}$ than HCl, so the H-F bond is more polar and the dipole moment of HF is larger than that of HCl. This difference in polarity also explains why HF can form stronger hydrogen bonds than HCl, giving HF a much higher boiling point than expected from its small molar mass.电负性（EN）衡量原子吸引共用电子的能力。在极性共价键中，电子不被均等共用，更多时间靠近电负性较大的原子，产生永久偶极（电负性强的一端为 $\delta^-$，电负性弱的一端为 $\delta^+$）。HF 的 $\Delta\text{EN}$ 大于 HCl，因此 H-F 键极性更强，HF 的偶极矩大于 HCl。这一极性差异也解释了为何 HF 能形成比 HCl 更强的氢键，使 HF 的沸点远高于其小摩尔质量所预期的值。

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §2&3 Ionic vs covalent离子键与共价键 [8 marks][8 分]

Compare NaCl and HCl: (a) Classify each bond type. (b) Compare melting points and explain. (c) Compare electrical conductivity in solution and explain.比较 NaCl 和 HCl：(a) 各自的键型分类；(b) 比较熔点并解释；(c) 比较溶液中的导电性并解释。

Answer:答案： (a) NaCl ionic (ΔEN=2.1); HCl polar covalent (ΔEN=0.9)NaCl 离子键（ΔEN=2.1）；HCl 极性共价键（ΔEN=0.9） · (b) NaCl high mp; HCl low mpNaCl 熔点高；HCl 熔点低 · (c) NaCl(aq) conducts; HCl(g) does notNaCl(aq) 导电；HCl(g) 不导电

(a) Classify bond type键型分类 A1·A1

NaCl: Na (EN = 0.9) and Cl (EN = 3.0); $\Delta\text{EN} = 2.1 \ge 1.7$ → ionic bond. Electron transfer from Na to Cl forms Na⁺ and Cl⁻ held together in a giant ionic lattice.NaCl：Na（EN = 0.9）与 Cl（EN = 3.0）；$\Delta\text{EN} = 2.1 \ge 1.7$ → 离子键。Na 向 Cl 转移电子，形成 Na⁺ 和 Cl⁻，构成巨型离子晶格。
HCl: H (EN = 2.1) and Cl (EN = 3.0); $\Delta\text{EN} = 0.9$ (between 0.4 and 1.7) → polar covalent bond. Electrons are shared but unequally.HCl：H（EN = 2.1）与 Cl（EN = 3.0）；$\Delta\text{EN} = 0.9$（介于 0.4 和 1.7 之间）→ 极性共价键。电子被共用但分配不均。

(b) Melting point comparison熔点比较 A1·A1·A1

NaCl has a very high melting point (mp = 801 °C). It forms a giant ionic lattice in which each Na⁺ is surrounded by six Cl⁻ and vice versa. To melt NaCl, the strong electrostatic attractions between many oppositely charged ions must be overcome, requiring a large energy input.NaCl 熔点极高（熔点 = 801 °C）。它形成巨型离子晶格，每个 Na⁺ 被 6 个 Cl⁻ 包围，反之亦然。熔化 NaCl 需克服众多相反电荷离子间的强静电引力，需输入大量能量。
HCl has a very low melting point (mp = -114 °C). It is a simple molecular substance. The H-Cl covalent bond within each molecule is strong, but between molecules only weak dipole-dipole forces and London dispersion forces act. These weak intermolecular forces are easily overcome, so very little energy is needed to melt it.HCl 熔点极低（熔点 = -114 °C）。它是简单分子物质。每个分子内的 H-Cl 共价键很强，但分子间只有弱的偶极-偶极力和伦敦色散力。这些弱的分子间作用力很容易克服，熔化所需能量极少。

(c) Electrical conductivity导电性 A1·A1·A1

NaCl(aq) conducts electricity. When NaCl dissolves in water, the ionic lattice dissociates completely into free-moving Na⁺(aq) and Cl⁻(aq) ions. These mobile ions carry charge through the solution, allowing current to flow.NaCl(aq) 导电。NaCl 溶于水后，离子晶格完全解离为可自由移动的 Na⁺(aq) 和 Cl⁻(aq) 离子。这些可移动离子在溶液中传导电荷，允许电流通过。
Pure HCl gas does not conduct electricity. HCl consists of neutral covalent molecules with no free ions and no mobile charged particles, so it cannot carry electric current. (Note: HCl dissolved in water does conduct, because water molecules pull the molecule apart to form H⁺(aq) and Cl⁻(aq); but the question concerns HCl as a substance before dissolution.)纯 HCl 气体不导电。HCl 由中性共价分子组成，没有自由离子和可移动带电粒子，因此不能传导电流。（注意：HCl 溶于水后可导电，因为水分子将其拆分为 H⁺(aq) 和 Cl⁻(aq)；但题目考查的是 HCl 物质本身在溶解前的情况。）

Ionic compounds conduct when ions are free to move (molten or dissolved); covalent molecules do not conduct unless they ionize in solution.离子化合物在离子可以自由移动时（熔融或溶解）导电；共价分子在溶液中离子化之前不导电。 The key distinction: ionic bonding produces ions as the fundamental species; covalent bonding produces neutral molecules. For conduction, you need mobile charge carriers. In NaCl, the ions exist before dissolution; water just separates the lattice. In HCl, the molecule is neutral; it only becomes ionic when water molecules are energetic enough to extract H⁺ via the reaction $\text{HCl} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^-$. This is why "ionic compound" and "electrolyte" are related but distinct concepts.关键区别在于：离子键产生离子作为基本粒子；共价键产生中性分子。导电需要可移动的载流子。NaCl 中的离子在溶解前就已存在，水只是分离晶格。HCl 分子是中性的，只有当水分子通过反应 $\text{HCl} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^-$ 提取 H⁺ 后，才变为离子。这就是"离子化合物"与"电解质"相关但不同等的原因。

Q7MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §4 Lewis structures drawing路易斯结构绘制 [8 marks][8 分]

Draw Lewis structures for NH₃, O₂, and CH₄. For each: show all valence electrons, confirm the octet rule is satisfied, and state the bond order.画出 NH₃、O₂ 和 CH₄ 的路易斯结构。对每个分子：画出所有价电子，确认八隅体规则满足，并说明键级。

Answer:答案： NH₃: 3 N-H single bonds + 1 lone pair on N; bond order 13 条 N-H 单键 + N 上 1 对孤对电子；键级 1 | O₂: O=O double bond; bond order 2; 2 lone pairs on each OO=O 双键；键级 2；每个 O 上 2 对孤对电子 | CH₄: 4 C-H single bonds; no lone pairs on C; bond order 14 条 C-H 单键；C 上无孤对电子；键级 1

(a) NH₃ Lewis structureNH₃ 路易斯结构 A1·A1·A1

Total valence electrons: N (5) + 3 H (3 × 1) = 8. N is the central atom. Three N-H single bonds use 6 electrons; the remaining 2 electrons form a lone pair on N.总价电子数：N (5) + 3H (3 × 1) = 8。N 为中心原子。3 条 N-H 单键用去 6 个电子，剩余 2 个电子在 N 上形成 1 对孤对电子。

N has: 3 bonding pairs (6 e⁻) + 1 lone pair (2 e⁻) = 8 e⁻ → octet satisfied.N 有：3 对成键对 (6 e⁻) + 1 对孤对 (2 e⁻) = 8 e⁻ → 八隅体满足。
Each H has 2 electrons (duet satisfied).每个 H 有 2 个电子（双隅体满足）。
All N-H bonds are single bonds; bond order = 1.所有 N-H 键均为单键；键级 = 1。

(b) O₂ Lewis structureO₂ 路易斯结构 A1·A1·A1

Total valence electrons: 2 × O (2 × 6) = 12. A single bond O-O uses 2 electrons; distributing the remaining 10 as lone pairs gives each O 3 lone pairs but leaves one O with only 6 electrons. Promoting one lone pair per O to a bonding pair gives a double bond O=O.总价电子数：2 × O (2 × 6) = 12。O-O 单键用 2 个电子；将剩余 10 个分配为孤对电子时，每个 O 有 3 对孤对但其中一个 O 只有 6 个电子。将每个 O 上的 1 对孤对电子升级为成键对，即得双键 O=O。

Each O has: 1 double bond (4 e⁻) + 2 lone pairs (4 e⁻) = 8 e⁻ → octet satisfied.每个 O 有：1 条双键 (4 e⁻) + 2 对孤对 (4 e⁻) = 8 e⁻ → 八隅体满足。
Bond order = 2 (double bond).键级 = 2（双键）。

(c) CH₄ Lewis structureCH₄ 路易斯结构 A1·A1·A1

Total valence electrons: C (4) + 4 H (4 × 1) = 8. Four C-H single bonds use all 8 electrons. No electrons remain for lone pairs.总价电子数：C (4) + 4H (4 × 1) = 8。4 条 C-H 单键用尽全部 8 个电子，无剩余电子形成孤对。

C has: 4 bonding pairs (8 e⁻) = 8 e⁻ → octet satisfied.C 有：4 对成键对 (8 e⁻) = 8 e⁻ → 八隅体满足。
Each H has 2 electrons (duet satisfied).每个 H 有 2 个电子（双隅体满足）。
No lone pairs on C; bond order = 1 (single bonds).C 上无孤对电子；键级 = 1（单键）。

The number of bonds an element typically forms equals the number of electrons it needs to complete its octet (for nonmetals).非金属元素通常形成的化学键数等于其完成八隅体所需的电子数。 N (needs 3) forms 3 bonds, C (needs 4) forms 4 bonds, O (needs 2) forms 2 bonds (or a double bond). For diatomic molecules like O₂, the multiple bond is forced by the need to satisfy both atoms' octets with only 12 electrons. The bond order directly correlates with bond strength and inversely with bond length: O=O (bond order 2) is shorter and stronger than a hypothetical O-O single bond, and N≣N (bond order 3) is even shorter and stronger. Higher bond order also means fewer lone pairs on each atom within the molecule.N（需 3 个）形成 3 条键，C（需 4 个）形成 4 条键，O（需 2 个）形成 2 条键（或 1 条双键）。对于 O₂ 等双原子分子，多重键是在仅有 12 个电子的情况下满足两个原子八隅体的必然结果。键级与键强度正相关，与键长反相关：O=O（键级 2）比假设的 O-O 单键更短、更强，而 N≣N（键级 3）更短、更强。较高的键级也意味着分子内每个原子的孤对电子较少。

Q8HARD难Honors荣誉级 🇺🇸 US美 🇨🇦 ON安 AP-feeder FRQAP 衔接简答题 §5 VSEPRVSEPR 理论 [7 marks][7 分]

Use VSEPR theory to predict the electron geometry, molecular geometry, and bond angle(s) for H₂O, NH₃, and BF₃. Explain any deviation from the ideal angle.用 VSEPR 理论预测 H₂O、NH₃ 和 BF₃ 的电子构型、分子构型和键角，并解释任何偏离理想键角的原因。

Answer:答案： H₂O: tetrahedral e-geometry; bent/angular; ~104.5°四面体电子构型；V 形/角形；约 104.5° | NH₃: tetrahedral e-geometry; trigonal pyramidal; ~107°四面体电子构型；三角锥形；约 107° | BF₃: trigonal planar; exactly 120°; B has only 6e⁻ (incomplete octet)平面三角形；恰好 120°；B 只有 6e⁻（不完全八隅体）

(a) H₂O: bent geometry, ~104.5°H₂O：V 形，约 104.5° A1·A1·A1

O has 4 electron domains: 2 bonding pairs (O-H bonds) + 2 lone pairs. VSEPR places 4 domains at tetrahedral positions (~109.5°), giving a tetrahedral electron geometry. The molecular geometry (shape) describes only the atoms, so it is bent (angular). The 2 lone pairs repel more strongly than bonding pairs (lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair repulsion). This compresses the H-O-H angle from 109.5° to approximately 104.5°.O 有 4 个电子域：2 对成键对（O-H 键）+ 2 对孤对电子。VSEPR 将 4 个域排列在四面体位置（约 109.5°），给出四面体电子构型。分子构型（形状）仅描述原子位置，故为 V 形（角形）。2 对孤对电子比成键对的排斥力更强（孤对-孤对 > 孤对-成键对 > 成键对-成键对排斥力），将 H-O-H 键角从 109.5° 压缩至约 104.5°。

(b) NH₃: trigonal pyramidal, ~107°NH₃：三角锥形，约 107° A1·A1

N has 4 electron domains: 3 bonding pairs (N-H bonds) + 1 lone pair. Electron geometry = tetrahedral. Molecular geometry = trigonal pyramidal (the lone pair is not counted as an atom). The single lone pair compresses the H-N-H angle from 109.5° to approximately 107° (less compression than H₂O because there is only 1 lone pair instead of 2).N 有 4 个电子域：3 对成键对（N-H 键）+ 1 对孤对电子。电子构型 = 四面体形。分子构型 = 三角锥形（孤对电子不计为原子）。1 对孤对电子将 H-N-H 键角从 109.5° 压缩至约 107°（压缩程度小于 H₂O，因为只有 1 对孤对而非 2 对）。

(c) BF₃: trigonal planar, exactly 120°; incomplete octet on BBF₃：平面三角形，恰好 120°；B 的不完全八隅体 A1·A1

B has 3 electron domains: 3 bonding pairs (B-F bonds) + 0 lone pairs. VSEPR gives a trigonal planar electron geometry and molecular geometry, with F-B-F bond angles of exactly 120°. There are no lone pairs to distort the geometry. Note: B has only 6 valence electrons in BF₃ (3 × 2 from bonding pairs), so it does not satisfy the octet rule. Boron is an exception to the octet rule (it is "electron deficient" with only 6 electrons in the valence shell).B 有 3 个电子域：3 对成键对（B-F 键）+ 0 对孤对电子。VSEPR 给出平面三角形电子构型和分子构型，F-B-F 键角恰为 120°。没有孤对电子扭曲几何形状。注意：BF₃ 中 B 的价电子层只有 6 个电子（3 × 2 来自成键对），因此 B 不满足八隅体规则。硼是八隅体规则的例外（价层只有 6 个电子，呈"缺电子"状态）。

VSEPR: electron geometry depends on all domains (bonding + lone pairs); molecular geometry depends only on bonding domains. Lone pairs compress bond angles.VSEPR：电子构型取决于所有电子域（成键对 + 孤对）；分子构型仅取决于成键域。孤对电子压缩键角。 The progression H₂O (2 lone pairs, 104.5°) → NH₃ (1 lone pair, 107°) → CH₄ (0 lone pairs, 109.5°) illustrates how each additional lone pair compresses the bond angle by about 2-3°. BF₃ stands apart as a genuine octet exception: B forms 3 bonds and stops there because it has only 3 valence electrons to contribute. This electron deficiency makes BF₃ a Lewis acid (it can accept an electron pair from a donor). Always state whether an exception to the octet rule applies when discussing VSEPR geometries of Group IIIA compounds.H₂O（2 对孤对，104.5°）→ NH₃（1 对孤对，107°）→ CH₄（0 对孤对，109.5°）这一序列说明每增加 1 对孤对电子约压缩键角 2-3°。BF₃ 作为真正的八隅体例外而独树一帜：B 只有 3 个价电子可贡献，因此只形成 3 条键。这种缺电子性使 BF₃ 成为路易斯酸（可接受来自供体的电子对）。在讨论第 IIIA 族化合物的 VSEPR 几何形状时，始终需要说明是否存在八隅体规则的例外。

Q9HARD难Honors荣誉级 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §7 Intermolecular forces分子间作用力 [7 marks][7 分]

(a) Identify all intermolecular forces present in H₂O, HCl, and CH₄. (b) Explain why HCl has a higher boiling point than CH₄. (c) Explain how increasing molar mass affects boiling point for nonpolar molecules.(a) 列出 H₂O、HCl 和 CH₄ 中存在的所有分子间作用力；(b) 解释为何 HCl 的沸点高于 CH₄；(c) 解释摩尔质量增大如何影响非极性分子的沸点。

Answer:答案： H₂O: hydrogen bonding + dipole-dipole + London dispersion氢键 + 偶极-偶极力 + 伦敦色散力 | HCl: dipole-dipole + London dispersion偶极-偶极力 + 伦敦色散力 | CH₄: London dispersion only仅伦敦色散力

(a) Intermolecular forces in each molecule各分子的分子间作用力 A1·A1·A1

H₂O: (1) Hydrogen bonding (strongest): O is highly electronegative and small; H is bonded directly to O, creating a large partial positive charge on H. The lone pairs on O of a neighboring molecule attract this H⁺, forming a hydrogen bond. (2) Dipole-dipole forces: H₂O is polar (bent geometry, dipoles do not cancel). (3) London dispersion forces: present in all molecules.H₂O： (1) 氢键（最强）：O 的电负性很强且原子半径小；H 直接与 O 相连，使 H 带有较大的部分正电荷。邻近分子 O 上的孤对电子吸引该 H⁺，形成氢键。(2) 偶极-偶极力：H₂O 是极性分子（V 形构型，偶极不相消）。(3) 伦敦色散力：存在于所有分子中。
HCl: (1) Dipole-dipole forces: HCl is a polar molecule ($\Delta\text{EN} = 0.9$). (2) London dispersion forces. HCl does not form hydrogen bonds: although H is bonded to a relatively electronegative atom, Cl is too large and its electron density is too diffuse for the concentrated partial charge needed for hydrogen bonding.HCl： (1) 偶极-偶极力：HCl 是极性分子（$\Delta\text{EN} = 0.9$）。(2) 伦敦色散力。HCl 不形成氢键：虽然 H 与电负性较强的原子相连，但 Cl 体积太大，电子密度太分散，不足以产生氢键所需的集中部分电荷。
CH₄: London dispersion forces only. CH₄ is a nonpolar molecule (tetrahedral geometry, four identical C-H bonds whose dipoles cancel exactly). It has no permanent dipole and H is not bonded to N, O, or F, so no hydrogen bonding occurs.CH₄：仅有伦敦色散力。CH₄ 是非极性分子（四面体构型，4 条相同的 C-H 键偶极完全相消），没有永久偶极，且 H 未与 N、O 或 F 相连，故不形成氢键。

(b) Why HCl has a higher boiling point than CH₄HCl 沸点高于 CH₄ 的原因 A1·A1

HCl is a polar molecule, so it experiences dipole-dipole forces in addition to London dispersion forces. CH₄ is nonpolar and has only London dispersion forces. Dipole-dipole forces are stronger than London dispersion forces of comparable-sized molecules. Therefore more energy is required to separate HCl molecules than CH₄ molecules, giving HCl a higher boiling point (HCl bp = -85 °C; CH₄ bp = -161 °C).HCl 是极性分子，除伦敦色散力外还有偶极-偶极力。CH₄ 是非极性分子，只有伦敦色散力。对于大小相近的分子，偶极-偶极力强于伦敦色散力。因此，分离 HCl 分子所需的能量大于分离 CH₄ 分子，使 HCl 的沸点更高（HCl 沸点 = -85 °C；CH₄ 沸点 = -161 °C）。

(c) Effect of increasing molar mass on boiling point of nonpolar molecules摩尔质量增大对非极性分子沸点的影响 A1·A1

For nonpolar molecules, the only intermolecular force is London dispersion. London dispersion forces arise from temporary (instantaneous) dipoles created by the random movement of electrons. Larger molecules have more electrons and larger, more easily polarized electron clouds. This means larger temporary dipoles and therefore stronger London dispersion forces. Greater London dispersion forces require more energy to overcome, so boiling point increases as molar mass increases for nonpolar molecules in the same family (e.g., noble gases: He < Ne < Ar < Kr; halogens: F₂ < Cl₂ < Br₂ < I₂).对于非极性分子，唯一的分子间作用力是伦敦色散力。伦敦色散力来源于电子随机运动产生的瞬间（临时）偶极。较大的分子拥有更多电子，更大且更易极化的电子云，因此产生更大的瞬间偶极，进而形成更强的伦敦色散力。更强的伦敦色散力需要更多能量来克服，故同族非极性分子中，摩尔质量越大，沸点越高（如稀有气体：He < Ne < Ar < Kr；卤素：F₂ < Cl₂ < Br₂ < I₂）。

IMF hierarchy: hydrogen bonding > dipole-dipole > London dispersion. All three can coexist; only the strongest one is usually the dominant factor in boiling point differences.分子间作用力强度顺序：氢键 > 偶极-偶极力 > 伦敦色散力。三种力可共存；通常只有最强的一种是沸点差异的主导因素。 H₂O has all three types but is dominated by hydrogen bonding, which is why water's boiling point (100 °C) is anomalously high for a molecule of only 18 g/mol. The condition for hydrogen bonding is specific: H must be directly bonded to N, O, or F (small, highly electronegative atoms). Cl is electronegative but too large, so HCl does not hydrogen-bond. This distinction is frequently tested on provincial and AP-style exams. When ranking boiling points, first identify which molecules can H-bond; they always top the ranking within a size-comparable group.H₂O 兼具三种力，但以氢键为主，这就是水的沸点（100 °C）对于仅 18 g/mol 的分子来说异常高的原因。氢键的形成条件很具体：H 必须直接与 N、O 或 F（小且电负性极强的原子）相连。Cl 虽然电负性较强，但原子半径太大，故 HCl 不能形成氢键。这一区别在省考和 AP 风格考试中经常被考查。排列沸点顺序时，先判断哪些分子能形成氢键；在大小相近的一组分子中，能形成氢键的分子沸点总是最高的。

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §2&6 Ionic lattice离子晶格 [8 marks][8 分]

Four substances: paraffin wax (nonpolar), sugar (C₁₂H₂₂O₁₁, polar molecular), NaCl (ionic 1+/1-), and MgO (ionic 2+/2-). (a) Rank them from lowest to highest melting point. (b) Which dissolve in water? Explain. (c) Which conduct electricity when molten? Explain.四种物质：石蜡（非极性）、蔗糖（C₁₂H₂₂O₁₁，极性分子）、NaCl（离子 1+/1-）和 MgO（离子 2+/2-）。(a) 从低到高排列熔点；(b) 哪些溶于水？解释；(c) 熔融时哪些导电？解释。

Answer:答案： (a) wax < sugar < NaCl < MgO石蜡 < 蔗糖 < NaCl < MgO · (b) NaCl and sugar dissolve; wax does not; MgO has very low solubilityNaCl 和蔗糖溶解；石蜡不溶；MgO 溶解度极低 · (c) Molten NaCl and MgO conduct (free ions); wax and sugar do not熔融 NaCl 和 MgO 导电（自由离子）；石蜡和蔗糖不导电

(a) Melting point ranking: low to high熔点排序：从低到高 A1·A1·A1

Paraffin wax (lowest mp, ~50-70 °C): nonpolar, long-chain molecular solid. Only London dispersion forces between chains. Weak intermolecular forces, easily melted.石蜡（熔点最低，约 50-70 °C）：非极性长链分子固体，链间只有伦敦色散力，分子间作用力弱，容易熔化。
Sugar (mp ~185 °C, decomposes): polar molecular solid; hydrogen bonding between hydroxyl groups adds to London dispersion forces. Still a simple molecular solid, but stronger IMF than wax.蔗糖（熔点约 185 °C，同时分解）：极性分子固体，羟基间氢键叠加伦敦色散力，IMF 强于石蜡，但仍属简单分子固体。
NaCl (mp = 801 °C): giant ionic lattice of Na⁺ and Cl⁻; strong electrostatic attractions between 1+ and 1- ions throughout the lattice require substantial energy to break.NaCl（熔点 = 801 °C）：Na⁺ 和 Cl⁻ 构成的巨型离子晶格，1+ 和 1- 离子间的强静电引力遍布晶格，需要大量能量才能破坏。
MgO (highest mp = 2852 °C): ionic lattice of Mg²⁺ and O²⁻. The 2+ and 2- charges create electrostatic attractions four times stronger than in NaCl (force scales as $q_1 q_2 / r^2$; both charges double). Also, Mg²⁺ and O²⁻ are smaller ions than Na⁺ and Cl⁻, so they pack closer, increasing the lattice energy further.MgO（熔点最高 = 2852 °C）：Mg²⁺ 和 O²⁻ 构成的离子晶格。2+ 和 2- 的电荷使静电引力比 NaCl 中强约四倍（力正比于 $q_1 q_2 / r^2$；两个电荷均翻倍）。此外，Mg²⁺ 和 O²⁻ 的离子半径比 Na⁺ 和 Cl⁻ 更小，使它们堆积更紧密，进一步提高晶格能。

(b) Solubility in water在水中的溶解性 A1·A1·A1

NaCl dissolves readily: the polar water molecules form ion-dipole interactions with Na⁺ and Cl⁻, releasing enough energy to overcome the lattice energy and disperse the ions. Sugar dissolves readily: the many -OH groups form hydrogen bonds with water ("like dissolves like"; both are polar and hydrogen-bonding). Paraffin wax is insoluble in water: it is nonpolar and cannot form attractive interactions with polar water molecules. MgO is only very slightly soluble: although ionic, the very high lattice energy (from 2+ and 2- charges) means the energy released by hydrating Mg²⁺ and O²⁻ is insufficient to break the lattice significantly.NaCl 易溶：极性水分子与 Na⁺ 和 Cl⁻ 形成离子-偶极相互作用，释放足够能量以克服晶格能并分散离子。蔗糖易溶：大量 -OH 基团与水形成氢键（"相似相溶"；两者均为极性且能形成氢键）。石蜡不溶于水：非极性，无法与极性水分子形成有效的吸引相互作用。MgO溶解度极低：虽然是离子化合物，但 2+ 和 2- 电荷造成的极高晶格能意味着水化 Mg²⁺ 和 O²⁻ 释放的能量不足以显著破坏晶格。

(c) Electrical conductivity when molten熔融时的导电性 A1·A1

Molten NaCl and molten MgO conduct electricity: when melted, the ionic lattice breaks down completely, releasing free-moving Na⁺, Cl⁻, Mg²⁺, and O²⁻ ions. These mobile ions carry electric charge and allow current to flow. Molten wax and molten sugar do not conduct: they consist of neutral covalent molecules. Melting separates the molecules but does not produce any ions or free charged particles, so there are no mobile charge carriers.熔融 NaCl 和熔融 MgO 导电：熔化后离子晶格完全破坏，释放出可自由移动的 Na⁺、Cl⁻、Mg²⁺ 和 O²⁻ 离子，这些可移动离子传导电荷，允许电流通过。熔融石蜡和熔融蔗糖不导电：它们由中性共价分子组成，熔化使分子分离，但不产生任何离子或自由带电粒子，故没有可移动载流子。

Ionic charge and ionic radius together determine lattice energy, which governs melting point, solubility, and conductivity in molten state.离子电荷和离子半径共同决定晶格能，进而决定离子化合物的熔点、溶解度和熔融态导电性。 This question ties together bonding type, IMF strength, and observable properties. The key ranking principle: molecular solids have low melting points (governed by IMF); ionic solids have high melting points (governed by lattice energy); network covalent solids would be even higher (SiO₂ has mp > 1600 °C). Within ionic solids, higher charge and smaller radius means higher lattice energy and higher melting point. The solubility rule "like dissolves like" refers to polarity matching. The conductivity rule requires free ions: solids have fixed ions; molecular liquids have no ions; only molten ionic compounds or ionic solutions meet this criterion.这道题将键型、IMF 强度和可观察性质联系在一起。关键排序原则：分子固体熔点低（由 IMF 决定）；离子固体熔点高（由晶格能决定）；共价网状固体熔点更高（SiO₂ 的熔点 > 1600 °C）。在离子固体中，电荷越高、离子半径越小，晶格能越大，熔点越高。溶解度规则"相似相溶"指极性匹配。导电性规则要求自由离子：固体中离子固定；分子液体无离子；只有熔融离子化合物或离子溶液满足此条件。

Q11MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §4,5,6 Lewis + geometry + polarity路易斯结构 + 几何构型 + 极性 [8 marks][8 分]

Compare CO₂ and H₂O: (a) Draw the Lewis structure of CO₂ and state its molecular geometry. (b) Explain why CO₂ is a nonpolar molecule even though the C=O bond is polar. (c) Explain why H₂O has a higher boiling point than CO₂.比较 CO₂ 和 H₂O：(a) 画出 CO₂ 的路易斯结构并说明其分子构型；(b) 解释为何 CO₂ 虽含极性 C=O 键却是非极性分子；(c) 解释为何 H₂O 的沸点高于 CO₂。

Answer:答案： (a) O=C=O, linear geometryO=C=O，线形构型 · (b) two C=O dipoles point in opposite directions and cancel exactly; net dipole = 0两个 C=O 偶极方向相反，完全相消；净偶极矩 = 0 · (c) H₂O is polar and has hydrogen bonding; CO₂ has only London dispersion; H₂O's stronger IMF requires more energy to overcomeH₂O 是极性分子且有氢键；CO₂ 只有伦敦色散力；H₂O 更强的 IMF 需要更多能量克服

(a) Lewis structure and geometry of CO₂CO₂ 的路易斯结构和几何构型 A1·A1·A1

Total valence electrons: C (4) + 2 O (12) = 16. Systematic construction: single bonds C-O-C would use 4 electrons, leaving 12 for lone pairs. If lone pairs are placed on O atoms (6 e⁻ each), carbon has only 4 electrons and its octet is unsatisfied. Promoting lone pairs to double bonds gives O=C=O (two double bonds), using all 16 electrons. Each O has 2 lone pairs (4 e⁻) and 4 electrons in the double bond = 8 (octet satisfied). C has 4 + 4 = 8 from the two double bonds (octet satisfied); no lone pairs on C.总价电子数：C (4) + 2O (12) = 16。系统构建：C-O 单键用 4 个电子，剩余 12 个放在 O 上（每个 O 6 个），此时 C 只有 4 个电子，八隅体不满足。将孤对电子升级为双键得 O=C=O（两个双键），用完全部 16 个电子。每个 O 有 2 对孤对（4 e⁻）+ 双键 4 e⁻ = 8（八隅体满足）。C 由两个双键各提供 4 个电子 = 8（八隅体满足）；C 上无孤对电子。
C has 2 bonding domains and 0 lone pairs. VSEPR predicts a linear molecular geometry with a bond angle of 180°.C 有 2 个成键域，0 对孤对电子。VSEPR 预测线形分子构型，键角 180°。

(b) Why CO₂ is nonpolar despite having polar bonds为何 CO₂ 含极性键却是非极性分子 A1·A1·A1

Each C=O bond is polar: O is more electronegative than C ($\Delta\text{EN} \approx 1.0$), so each O carries $\delta^-$ and C carries $\delta^+$ from both sides. However, the two C=O bond dipoles are equal in magnitude and point in exactly opposite directions (both along the linear O=C=O axis, but one points left and the other right). They cancel vectorially:每条 C=O 键都是极性键：O 比 C 的电负性更大（$\Delta\text{EN} \approx 1.0$），使每个 O 带 $\delta^-$，C 从两侧均带 $\delta^+$。然而，两个 C=O 键的偶极矩大小相等，方向完全相反（均沿线形 O=C=O 轴，一个指左，一个指右），矢量相消： $$ \vec{\mu}_{\text{total}} \;=\; \vec{\mu}_{C=O,\text{left}} + \vec{\mu}_{C=O,\text{right}} \;=\; \vec{0}. $$ The net dipole moment is zero, so CO₂ is a nonpolar molecule. The key point: a molecule with polar bonds can be nonpolar if its geometry causes the bond dipoles to cancel. This is only possible because CO₂ is linear with two identical substituents.净偶极矩为零，故 CO₂ 是非极性分子。关键点：含有极性键的分子若几何形状使键偶极相消，则整体可以是非极性的。这只有在 CO₂ 具有线形且两端取代基相同时才可能发生。

(c) Why H₂O has a higher boiling point than CO₂为何 H₂O 沸点高于 CO₂ A1·A1

CO₂ is nonpolar and therefore has London dispersion forces only (weak, since CO₂ is a small molecule). H₂O is a polar molecule with a bent geometry, so the two O-H bond dipoles do not cancel. H is bonded directly to O (a small, highly electronegative atom), so H₂O experiences strong hydrogen bonding between molecules (the strongest type of IMF). Overcoming hydrogen bonds requires far more energy than overcoming London dispersion forces. Therefore H₂O has a much higher boiling point (100 °C) than CO₂ (bp = -78.5 °C, sublimes).CO₂ 是非极性分子，因此只有伦敦色散力（CO₂ 分子小，色散力弱）。H₂O 是极性分子，V 形构型使两个 O-H 键偶极不相消。H 直接与 O（小且电负性极强的原子）相连，使 H₂O 分子间存在强氢键（最强类型的 IMF）。克服氢键所需能量远大于克服伦敦色散力，因此 H₂O 的沸点（100 °C）远高于 CO₂（沸点 = -78.5 °C，升华）。

Molecular polarity requires both polar bonds AND a geometry that prevents cancellation. Geometry is decisive.分子极性需要同时具备极性键和不使偶极相消的几何形状；几何形状是决定性因素。 CO₂ and H₂O are the classic paired example: both have polar bonds to oxygen, but CO₂ is linear (dipoles cancel) while H₂O is bent (dipoles reinforce). This single geometric difference changes CO₂ from nonpolar to polar, switching the dominant IMF from London dispersion to hydrogen bonding, and raising the boiling point by almost 180 °C. Always draw the shape before assessing polarity: you must know the geometry to determine dipole addition. Tetrahedral CCl₄ (nonpolar) vs. CHCl₃ (polar) provides the same lesson for tetrahedral molecules.CO₂ 和 H₂O 是经典的对照例子：两者均含与氧相连的极性键，但 CO₂ 是线形（偶极相消），而 H₂O 是 V 形（偶极叠加）。这一单纯的几何差异使 CO₂ 从非极性变为极性，主导 IMF 从伦敦色散力切换为氢键，沸点提高近 180 °C。评估极性前务必先画出分子形状：确定几何构型是进行偶极矢量叠加的前提。四面体 CCl₄（非极性）与 CHCl₃（极性）为四面体分子提供了同样的教训。

Q12HARD难 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §7 Metallic bonding金属键 [10 marks][10 分]

Three substances: copper (Cu), sodium chloride (NaCl), and silicon dioxide (SiO₂). (a) Describe metallic bonding in Cu and explain why metals conduct electricity and are malleable. (b) SiO₂ has a much higher melting point than CO₂ even though both contain C-O or Si-O bonds. Explain this difference in terms of bonding type. (c) A student claims that solid NaCl conducts electricity because it contains ions. Evaluate this claim.三种物质：铜（Cu）、氯化钠（NaCl）和二氧化硅（SiO₂）。(a) 描述 Cu 中的金属键，并解释金属为何导电且可延展；(b) SiO₂ 的熔点远高于 CO₂，尽管两者均含 C-O 或 Si-O 键，从键合类型角度解释这一差异；(c) 一名学生声称固体 NaCl 能导电，因为它含有离子，评价这一说法。

Answer:答案： (a) sea of delocalized electrons; electrons free to move (conduction); layers slide without breaking bonds (malleability)离域电子海；电子自由移动（导电）；层间滑移不断键（延展性） · (b) SiO₂ is network covalent (giant covalent lattice); CO₂ is simple molecular; SiO₂ requires breaking covalent bonds to meltSiO₂ 是共价网状固体（巨型共价晶格）；CO₂ 是简单分子固体；SiO₂ 熔化需断裂共价键 · (c) claim is incorrect; solid NaCl does NOT conduct because ions are fixed in the lattice and cannot move说法有误；固体 NaCl 不导电，因为离子固定在晶格中无法移动

(a) Metallic bonding in Cu: conduction and malleabilityCu 中的金属键：导电性和延展性 A1·A1·A1·A1

In copper, each atom releases its valence electrons (1 per Cu atom) into a shared "sea" of delocalized electrons that extends throughout the entire metallic crystal. The positive copper ion cores (Cu⁺ kernels) are held together by their electrostatic attraction to this electron sea.在铜中，每个原子将其价电子（每个 Cu 原子 1 个）释放到遍布整个金属晶体的共享"离域电子海"中。铜的正离子核（Cu⁺ 核）通过对这片电子海的静电吸引力被结合在一起。

Electrical conduction: the delocalized electrons are free to move throughout the structure when an electric field is applied. This continuous flow of electrons constitutes an electric current. Metals are conductors in both solid and liquid states because the electrons remain free to move in both.导电性：施加电场时，离域电子可以在整个结构中自由移动，这种连续的电子流构成电流。金属在固态和液态下均能导电，因为电子在两种状态下均可自由移动。
Malleability: when a stress is applied, layers of metal ions can slide past each other without breaking the metallic bonds, because the electrostatic attraction between the delocalized electron sea and the ion cores is non-directional. The electron sea simply "flows" to accommodate the new arrangement. This is fundamentally different from ionic compounds (where sliding shifts like charges next to each other, causing repulsion and fracture).延展性：受力时，金属离子层可以相互滑过而不断裂金属键，因为离域电子海与离子核之间的静电吸引力是无方向性的，电子海只是"流动"以适应新排列。这与离子化合物截然不同（滑移使同种电荷相邻，产生排斥并导致断裂）。

(b) SiO₂ vs. CO₂: network covalent vs. simple molecularSiO₂ 与 CO₂：共价网状固体与简单分子固体 A1·A1·A1

CO₂ is a simple molecular solid. Each CO₂ molecule is a discrete unit (O=C=O), held to neighboring molecules only by weak London dispersion forces. To melt CO₂ (actually it sublimes at -78.5 °C), only these weak intermolecular forces need to be broken; the strong covalent bonds within each CO₂ molecule remain intact.CO₂ 是简单分子固体。每个 CO₂ 分子是独立单元（O=C=O），仅通过弱伦敦色散力与邻近分子结合。使 CO₂ 熔化（实际上在 -78.5 °C 升华）只需克服这些弱的分子间作用力；每个 CO₂ 分子内的强共价键保持完整。
SiO₂ (silicon dioxide) is a network covalent solid. There are no discrete SiO₂ molecules. Instead, every Si atom is bonded to 4 oxygen atoms by strong covalent bonds, and every oxygen bridges two Si atoms, forming a giant three-dimensional covalent lattice that extends throughout the entire crystal. To melt SiO₂ (mp > 1600 °C), covalent Si-O bonds throughout the lattice must be broken, requiring enormous energy.SiO₂（二氧化硅）是共价网状固体。不存在独立的 SiO₂ 分子。每个 Si 原子通过强共价键与 4 个氧原子相连，每个氧桥连两个 Si 原子，形成贯穿整个晶体的巨型三维共价晶格。熔化 SiO₂（熔点 > 1600 °C）需要断裂晶格中所有 Si-O 共价键，需要巨大能量。
The enormous difference in melting points (SiO₂ > 1600 °C vs. CO₂ bp = -78.5 °C) is not due to the strength of individual Si-O vs. C-O bonds, but because in SiO₂ you must break all covalent bonds in the whole lattice, while in CO₂ you only break weak intermolecular forces.熔点的巨大差异（SiO₂ > 1600 °C vs CO₂ 沸点 = -78.5 °C）并非源于单个 Si-O 键与 C-O 键强度的差别，而是因为熔化 SiO₂ 需要断裂整个晶格中所有共价键，而 CO₂ 只需克服弱的分子间作用力。

(c) Evaluate the student's claim about solid NaCl评价学生关于固体 NaCl 的说法 A1·A1·A1

The claim is incorrect. While it is true that NaCl contains ions (Na⁺ and Cl⁻), solid NaCl does not conduct electricity. For conduction to occur, the ions must be free to move. In solid NaCl, the ions are held rigidly in fixed positions within the giant ionic lattice by strong electrostatic attractions. They cannot migrate toward the electrodes when a voltage is applied.该说法有误。虽然 NaCl 确实含有离子（Na⁺ 和 Cl⁻），但固体 NaCl 不导电。导电要求离子能够自由移动。在固体 NaCl 中，离子被强静电引力牢牢固定在巨型离子晶格的固定位置上，施加电压时无法向电极迁移。
NaCl conducts electricity only when:NaCl 只有在以下情况下才能导电：

Molten (above 801 °C): the lattice melts, releasing free-moving ions.熔融时（高于 801 °C）：晶格熔化，释放可自由移动的离子。
Dissolved in water: water molecules dissociate the lattice, releasing hydrated Na⁺(aq) and Cl⁻(aq) ions that can move freely.溶于水时：水分子将晶格解离，释放水合 Na⁺(aq) 和 Cl⁻(aq) 离子，它们可以自由移动。

The student correctly identified that ions are necessary, but missed the key condition: the ions must be mobile, not merely present.该学生正确地认识到离子是必要条件，但忽略了关键条件：离子必须是可移动的，而不仅仅是存在。

Ionic compounds conduct electricity only when ions are free to move (molten or dissolved). "Contains ions" is necessary but not sufficient for conduction.离子化合物只有在离子可以自由移动时（熔融或溶解）才能导电。"含有离子"是导电的必要条件，但不是充分条件。 This is one of the most common misconceptions in bonding. The mental model to lock in: conductivity requires charge transport; charge transport requires mobile charge carriers; in ionic materials, the carriers are ions; ions are mobile only when the lattice is broken (molten) or dispersed (dissolved). Contrast with metals, where the electrons are always mobile even in solid state, so metals conduct at all temperatures (though resistance increases with temperature). Network covalent solids like SiO₂ have neither free ions nor free electrons, so they are insulators under all normal conditions.这是化学键合中最常见的误解之一。需要牢记的思维模型：导电需要电荷传输；电荷传输需要可移动的载流子；在离子材料中，载流子是离子；离子只有在晶格被破坏（熔融）或分散（溶解）时才可移动。与金属对比，金属中的电子在固态时也始终可以移动，因此金属在所有温度下都能导电（尽管电阻随温度升高而增大）。像 SiO₂ 这样的共价网状固体既无自由离子也无自由电子，因此在所有正常条件下均为绝缘体。