PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 26 marksAP 风格选择题 + 安/卑省考短答 · 共 26 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. For short-answer items, state units where applicable. No calculator needed on Q1-Q5.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写下足以让阅卷人核对的过程。短答题需要时写出单位。Q1-Q5 无需计算器。
How many valence electrons does a sulfur atom (Group 16) have, and how many additional electrons does it need to satisfy the octet rule?硫原子（第 16 族）有多少个价电子？它还需要额外得到多少个电子才能满足八隅规则？
Calcium chloride ($\text{CaCl}_2$) is an ionic compound formed from calcium (Group 2) and chlorine (Group 17).氯化钙（$\text{CaCl}_2$）是由钙（第 2 族）和氯（第 17 族）形成的离子化合物。
(a)State the charge on the calcium ion and the chloride ion.写出钙离子和氯离子的电荷。[2]
(b)Explain in terms of electron transfer why two chloride ions are needed per calcium ion.从电子转移的角度解释为什么每个钙离子需要两个氯离子。[2]
Nitrogen gas ($\text{N}_2$) and water ($\text{H}_2\text{O}$) are both covalent molecules.氮气（$\text{N}_2$）和水（$\text{H}_2\text{O}$）都是共价分子。
(a)State the bond order in $\text{N}_2$ and the number of shared electron pairs in that bond.写出 $\text{N}_2$ 中的键级以及该键中共用电子对的数目。[2]
(b)Identify the type of bond in $\text{H}_2\text{O}$ (single, double, or triple) and state one property of water that this bond type is consistent with.指出 $\text{H}_2\text{O}$ 中键的类型（单键、双键或三键），并写出水的一个与该键型一致的性质。[2]
Q4MEDIUM中🇺🇸 US美AP-style MCQAP 风格选择题§4 Lewis structures路易斯结构 · HS-PS1-2[3 marks][3 分]
In the Lewis structure of $\text{CO}_2$, what is the total number of lone pairs on all atoms?在 $\text{CO}_2$ 的路易斯结构中，所有原子上孤对电子的总数是多少？
The electronegativity values of hydrogen (H), chlorine (Cl), and fluorine (F) are approximately 2.1, 3.0, and 4.0 respectively. Consider the molecules $\text{HCl}$ and $\text{HF}$.氢（H）、氯（Cl）和氟（F）的电负性值分别约为 2.1、3.0 和 4.0。考察分子 $\text{HCl}$ 和 $\text{HF}$。
(a)Calculate the electronegativity difference for each bond and classify each bond (nonpolar covalent, polar covalent, or ionic).计算每种键的电负性差值，并对每种键进行分类（非极性共价键、极性共价键或离子键）。[2]
(b)Identify the partial negative charge ($\delta^-$) end of each molecule and explain your reasoning.指出每个分子中带部分负电荷（$\delta^-$）的一端，并解释你的理由。[2]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 30 marksAP 衔接简答题 + 荣誉级 · 共 30 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. State the chemical principle before applying it. Answer in complete sentences for explain/describe parts. Calculator permitted on Q6-Q9.每一步推理都要写出。在应用前先说明所用化学原理。"解释"和"描述"类问题须用完整句子作答。Q6-Q9 可用计算器。
Sodium chloride ($\text{NaCl}$) and hydrogen chloride ($\text{HCl}$) are both chloride-containing compounds, but one is ionic and one is covalent.氯化钠（$\text{NaCl}$）和氯化氢（$\text{HCl}$）都含有氯，但一个是离子化合物，一个是共价化合物。
(a)Identify which compound is ionic and which is covalent. Justify your answer using electronegativity difference.指出哪个化合物是离子化合物，哪个是共价化合物。用电负性差来支持你的答案。（Na 电负性 0.9，Cl 电负性 3.0，H 电负性 2.1）[2]
(b)Compare the melting points of $\text{NaCl}$ and $\text{HCl}$ and explain the difference in terms of bonding and structure.比较 $\text{NaCl}$ 和 $\text{HCl}$ 的熔点，并从键和结构的角度解释差异。[3]
(c)Explain why $\text{NaCl}$ conducts electricity when dissolved in water but $\text{HCl}$(g) does not conduct electricity in its pure gaseous form. (Note: $\text{HCl}$ dissolved in water does conduct, but that involves a reaction; focus on the pure substances here.)解释为什么 $\text{NaCl}$ 溶于水后能导电，而纯气态 $\text{HCl}$(g) 不能导电。（注：$\text{HCl}$ 溶于水后也能导电，但那涉及反应；此处聚焦于纯物质。）[3]
Draw the Lewis structure for each of the following molecules. Show all bonding pairs and lone pairs. Atom counts: $\text{NH}_3$ (N has 5 valence e$^-$, H has 1 each); $\text{O}_2$ (O has 6 valence e$^-$ each); $\text{CH}_4$ (C has 4 valence e$^-$, H has 1 each).为以下每个分子绘制路易斯结构，显示所有成键电子对和孤对电子。原子价电子数：$\text{NH}_3$（N 有 5 个价电子，每个 H 有 1 个）；$\text{O}_2$（每个 O 有 6 个价电子）；$\text{CH}_4$（C 有 4 个价电子，每个 H 有 1 个）。
(a)Draw the Lewis structure of $\text{NH}_3$. Include the lone pair on nitrogen. State the total number of valence electrons.绘制 $\text{NH}_3$ 的路易斯结构，包括氮上的孤对电子，并写出价电子总数。[3]
(b)Draw the Lewis structure of $\text{O}_2$. Identify the bond order.绘制 $\text{O}_2$ 的路易斯结构，并标明键级。[2]
(c)Draw the Lewis structure of $\text{CH}_4$. Verify the octet rule is satisfied for carbon.绘制 $\text{CH}_4$ 的路易斯结构，并验证碳满足八隅规则。[3]
Use VSEPR theory to predict the molecular geometry and bond angles of $\text{H}_2\text{O}$, $\text{NH}_3$, and $\text{BF}_3$.用 VSEPR 理论预测 $\text{H}_2\text{O}$、$\text{NH}_3$ 和 $\text{BF}_3$ 的分子几何形状和键角。
(a)For $\text{H}_2\text{O}$: state the electron geometry, the molecular geometry, and the approximate bond angle. Explain why the bond angle is less than the ideal tetrahedral angle of $109.5^\circ$.对于 $\text{H}_2\text{O}$：写出电子几何形、分子几何形及近似键角。解释为什么键角小于理想正四面体角 $109.5^\circ$。[3]
(b)For $\text{NH}_3$: state the electron geometry, the molecular geometry, and the approximate bond angle.对于 $\text{NH}_3$：写出电子几何形、分子几何形及近似键角。[2]
(c)For $\text{BF}_3$: state the molecular geometry and the exact bond angle. Note that $\text{BF}_3$ is an exception to the octet rule. State what this exception is.对于 $\text{BF}_3$：写出分子几何形及精确键角。注意 $\text{BF}_3$ 是八隅规则的例外，写出该例外是什么。[2]
The boiling points of $\text{H}_2\text{O}$ ($100\ ^\circ\text{C}$), $\text{HF}$ ($19.5\ ^\circ\text{C}$), $\text{HCl}$ ($-85\ ^\circ\text{C}$), and $\text{CH}_4$ ($-161\ ^\circ\text{C}$) reflect differences in intermolecular forces.$\text{H}_2\text{O}$（$100\ ^\circ\text{C}$）、$\text{HF}$（$19.5\ ^\circ\text{C}$）、$\text{HCl}$（$-85\ ^\circ\text{C}$）和 $\text{CH}_4$（$-161\ ^\circ\text{C}$）的沸点差异反映了分子间作用力的不同。
(a)Identify the strongest type of intermolecular force present in $\text{H}_2\text{O}$ and explain why it is unusually strong compared to other molecules of similar molar mass.指出 $\text{H}_2\text{O}$ 中存在的最强类型的分子间作用力，并解释与类似摩尔质量的分子相比，它为什么异常强。[3]
(b)Explain why $\text{HCl}$ has a higher boiling point than $\text{CH}_4$ even though both lack hydrogen bonding. Name the type(s) of intermolecular force present in each.解释为什么 $\text{HCl}$ 的沸点高于 $\text{CH}_4$，即使两者都没有氢键。分别指出每种分子中存在的分子间作用力类型。[2]
(c)Predict and explain: if two substances have similar types of intermolecular forces, which one would have the higher boiling point, and why?预测并解释：如果两种物质具有相似类型的分子间作用力，哪种物质的沸点更高，为什么？[2]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 26 marks阿省毕业考 + 通用题型 · 共 26 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define all symbols and structures clearly. Cite the bonding principle before applying it. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.清晰定义所有符号和结构。在应用前先说明所用成键原理。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。
A student tests four substances: table salt ($\text{NaCl}$), sugar ($\text{C}_{12}\text{H}_{22}\text{O}_{11}$, a polar molecular solid), wax (a nonpolar molecular solid), and magnesium oxide ($\text{MgO}$, an ionic solid with $2+$ and $2-$ ions).一名学生测试四种物质：食盐（$\text{NaCl}$）、蔗糖（$\text{C}_{12}\text{H}_{22}\text{O}_{11}$，极性分子固体）、石蜡（非极性分子固体）和氧化镁（$\text{MgO}$，含 $2+$ 和 $2-$ 离子的离子固体）。
(a)Rank the four substances from lowest to highest melting point. Explain the ranking using bonding and structure.将四种物质按熔点从低到高排序，并用键和结构解释排序依据。[3]
(b)Predict which two substances are soluble in water and explain why, using the principle "like dissolves like" in terms of polarity.预测哪两种物质溶于水，并用极性角度的"相似相溶"原理解释。[3]
(c)Predict which substances conduct electricity in the molten (liquid) state and explain why.预测哪种物质在熔融（液态）状态下能导电，并解释原因。[2]
Consider $\text{CO}_2$ (carbon dioxide) and $\text{H}_2\text{O}$ (water). Both contain polar bonds, yet one molecule is nonpolar overall and the other is polar.考察 $\text{CO}_2$（二氧化碳）和 $\text{H}_2\text{O}$（水）。两者都含有极性键，但一个分子整体上是非极性的，另一个是极性的。
(a)Draw the Lewis structure for $\text{CO}_2$ and state its molecular geometry. Use VSEPR to explain the geometry.绘制 $\text{CO}_2$ 的路易斯结构并写出其分子几何形状。用 VSEPR 解释该几何形状。[3]
(b)Explain why $\text{CO}_2$ is nonpolar despite having polar bonds. Use the concept of dipole moment cancellation in your answer.解释为什么 $\text{CO}_2$ 尽管含有极性键，但整体上是非极性的。在答案中使用偶极矩相消的概念。[3]
(c)Explain why $\text{H}_2\text{O}$ is polar. Predict which of the two molecules ($\text{CO}_2$ or $\text{H}_2\text{O}$) has a higher boiling point and justify your answer.解释为什么 $\text{H}_2\text{O}$ 是极性分子。预测 $\text{CO}_2$ 和 $\text{H}_2\text{O}$ 中哪个沸点更高，并说明理由。[2]
Copper ($\text{Cu}$) is a metal used in electrical wiring. Sodium chloride ($\text{NaCl}$) and silicon dioxide ($\text{SiO}_2$, a network covalent solid) are also used in various applications.铜（$\text{Cu}$）是一种用于电线的金属。氯化钠（$\text{NaCl}$）和二氧化硅（$\text{SiO}_2$，共价网状固体）也用于各种场合。
(a)Describe the metallic bonding model in copper. Explain why copper is a good conductor of electricity and also malleable (can be shaped without breaking).描述铜中的金属键模型。解释为什么铜既是良导体，又具有延展性（可成形而不断裂）。[4]
(b)$\text{SiO}_2$ has a very high melting point ($1713\ ^\circ\text{C}$) even though the bonds in $\text{SiO}_2$ are covalent. Explain why. Compare this to the melting point of a simple molecular covalent substance such as $\text{CO}_2$ (sublimes at $-78.5\ ^\circ\text{C}$).$\text{SiO}_2$ 的熔点非常高（$1713\ ^\circ\text{C}$），尽管其键是共价键。解释原因，并与简单分子共价物质（如 $\text{CO}_2$，升华温度为 $-78.5\ ^\circ\text{C}$）进行比较。[3]
(c)A student claims: "All ionic compounds conduct electricity." Evaluate this claim and provide a specific counterexample or clarification.一名学生声称："所有离子化合物都能导电。"评价这一说法，并提供一个具体的反例或澄清。[3]
🇺🇸 US NGSS美国 NGSSHS-PS1-1 · HS-PS1-2
🇨🇦 Ontario安大略SCH3U B3.4 · B3.5 · SCH4U C2.3
🇨🇦 British Columbia不列颠哥伦比亚Chemistry 11: bonding, Lewis structures, polarity化学 11：成键、路易斯结构、极性
🇨🇦 Alberta阿尔伯塔Chem 20 GO1 · GO2 · Chem 30
Full Syllabus Map lives in ../Study Guides/Unit_3_Chemical_Bonding.html. VSEPR and named IMFs are Honors-flagged (above NGSS assessed floor; core for ON SCH3U / BChem 11 / AB Chem 20).完整大纲对照表见 ../Study Guides/Unit_3_Chemical_Bonding.html。VSEPR 与具名分子间作用力为荣誉级（超出 NGSS 考查范围；为 ON SCH3U / 卑诗化学 11 / 阿省化学 20 核心内容）。