IB Chemistry HL Structure 2 Bonding and Structure: IB-Style Practice

IB-Style Practice Questions: Structure 2.1 to 2.4IB 风格练习题：覆盖 Structure 2.1 至 2.4

Multiple Choice选择题

Each item carries 1 mark. No calculator or data booklet required unless electronegativity values are given directly in the question. Items tagged HL test content beyond the standard-level syllabus (resonance, expanded octets, formal charge, sigma/pi bonds, hybridization, transition-metal bonding).每题 1 分。除非题目中直接给出电负性数值，否则不需要计算器或数据手册（data booklet）。标记 HL 的题目考查超出标准级别（SL）大纲的内容：共振（resonance）、扩展八隅体、形式电荷、σ/π 键、杂化、过渡金属成键。

Q2MEDIUM Paper 1 2.1 Lattice Enthalpy [1]

Which combination of ionic charge and ionic radius would give a compound the highest lattice enthalpy?哪一种离子电荷与离子半径的组合，会使化合物的晶格焓（lattice enthalpy）最高？

(A) Small charge, small radius电荷小、半径小
(B) Large charge, large radius电荷大、半径大
(C) Large charge, small radius电荷大、半径小
(D) Small charge, large radius电荷小、半径大

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Q3MEDIUM Paper 1 2.2 VSEPR [1]

Predict the electron-domain geometry and the molecular shape of the sulfite ion, $\mathrm{SO_3^{2-}}$.预测亚硫酸根离子 $\mathrm{SO_3^{2-}}$ 的电子域几何与分子几何。

(A) Trigonal planar electron domains; trigonal planar molecule电子域为平面三角形；分子几何也是平面三角形
(B) Tetrahedral electron domains; trigonal pyramidal molecule电子域为正四面体；分子几何为三角锥形
(C) Tetrahedral electron domains; tetrahedral molecule电子域为正四面体；分子几何也是正四面体
(D) Trigonal bipyramidal electron domains; seesaw molecule电子域为三角双锥；分子几何为跷跷板形

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Q4MEDIUM Paper 1 2.2 Molecular Polarity [1]

$\mathrm{BF_3}$ contains three polar B-F bonds. Which statement correctly explains its overall polarity?$\mathrm{BF_3}$ 含有三条极性 B-F 键。下列哪一项正确解释了它的整体极性？

(A) Polar overall, because all bonds in the molecule are polar.整体是极性的，因为分子中所有键都是极性键。
(B) Non-polar overall, because the trigonal planar symmetry makes the three bond dipoles cancel.整体是非极性的，因为平面三角形的对称性使三条键偶极相互抵消。
(C) Non-polar overall, because boron has an incomplete octet.整体是非极性的，因为硼未达到八隅体。
(D) Polar overall, because fluorine is the most electronegative element.整体是极性的，因为氟是电负性最大的元素。

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Q5MEDIUM Paper 1 2.2 Covalent Network [1]

Graphite conducts electricity along its layers, but diamond does not conduct at all, even though both are covalent network forms of carbon. What best explains this difference?石墨能沿层面导电，而金刚石完全不导电，尽管两者都是碳的共价网状结构。以下哪一项最能解释这一差异？

(A) Graphite has ionic bonding between layers; diamond does not.石墨层间是离子键；金刚石没有。
(B) Each C in graphite is sp² hybridized, leaving one delocalized electron per atom that is free to move within the layer; each C in diamond is sp³ hybridized with all four electrons localized in sigma bonds.石墨中每个 C 为 sp² 杂化，每个原子留下一个可在层内自由移动的离域电子；金刚石中每个 C 为 sp³ 杂化，四个电子全部定域在 σ 键中。
(C) Diamond has a lower density than graphite.金刚石的密度比石墨低。
(D) Graphite atoms carry a partial positive charge.石墨中的原子带部分正电荷。

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Q6HARD Paper 1 2.2 Intermolecular Forces [1]

Which lists $\mathrm{CH_4}$, $\mathrm{PH_3}$, and $\mathrm{NH_3}$ in order of increasing boiling point?下列哪一项按沸点由低到高正确排列了 $\mathrm{CH_4}$、$\mathrm{PH_3}$ 和 $\mathrm{NH_3}$？

(A) $\mathrm{CH_4} < \mathrm{PH_3} < \mathrm{NH_3}$
(B) $\mathrm{CH_4} < \mathrm{NH_3} < \mathrm{PH_3}$
(C) $\mathrm{NH_3} < \mathrm{PH_3} < \mathrm{CH_4}$
(D) $\mathrm{PH_3} < \mathrm{CH_4} < \mathrm{NH_3}$

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Q7HARD Paper 1 HL 2.3 Transition Metal Bonding [1]

Iron has a considerably higher melting point (1538°C) than potassium (63.5°C), even though both are metals. What is the best explanation?铁的熔点（1538°C）远高于钾（63.5°C），尽管两者都是金属。最佳解释是？

(A) Fe has a smaller ionic radius and contributes delocalized d-electrons in addition to s-electrons, giving stronger metallic bonding than K's single delocalized electron per atom.Fe 的离子半径更小，且除 s 电子外还贡献可离域的 d 电子，因此金属键比 K（每原子仅贡献一个离域电子）更强。
(B) Fe atoms are held together by covalent bonds, while K atoms are held together by metallic bonds.Fe 原子间以共价键结合，而 K 原子间以金属键结合。
(C) K has a higher nuclear charge than Fe.K 的核电荷数比 Fe 大。
(D) Fe conducts heat by ion movement, while K conducts heat by electron movement.Fe 靠离子移动导热，而 K 靠电子移动导热。

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Q8HARD Paper 1 HL 2.2 Hybridization & σ/π Bonds [1]

In the ethene molecule, $\mathrm{CH_2{=}CH_2}$, what is the hybridization of each carbon atom, and how many sigma ($\sigma$) and pi ($\pi$) bonds does the whole molecule contain?在乙烯分子 $\mathrm{CH_2{=}CH_2}$ 中，每个碳原子的杂化方式是什么？整个分子共含多少个 σ 键和 π 键？

(A) sp²; 5$\sigma$, 1$\pi$sp²；5σ、1π
(B) sp³; 6$\sigma$, 0$\pi$sp³；6σ、0π
(C) sp; 4$\sigma$, 2$\pi$sp；4σ、2π
(D) sp²; 4$\sigma$, 2$\pi$sp²；4σ、2π

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Q9MEDIUM Paper 1 2.2 Coordinate Bond [1]

When $\mathrm{NH_3}$ reacts with $\mathrm{H^+}$ to form $\mathrm{NH_4^+}$, the fourth N-H bond is best described as当 $\mathrm{NH_3}$ 与 $\mathrm{H^+}$ 反应生成 $\mathrm{NH_4^+}$ 时，第四条 N-H 键最恰当的描述是

(A) An ionic bond between $\mathrm{N}$ and $\mathrm{H^+}$.N 与 H⁺ 之间的离子键。
(B) A coordinate (dative) covalent bond, since both bonding electrons come from the nitrogen lone pair; once formed, it is identical to the other three N-H bonds.配位（coordinate / dative）共价键，因为成键的两个电子都来自 N 的孤对电子；一旦形成，它与另外三条 N-H 键完全相同。
(C) A metallic bond.金属键。
(D) A coordinate bond that remains permanently weaker than the other three N-H bonds.一条永久比另外三条 N-H 键弱的配位键。

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Structured Response结构化解答题

Show all working and reasoning in the space provided. Marks for correct method are awarded even if a final statement is incomplete. State units where relevant.在指定区域写出全部推理过程。即便最终表述不完整，方法正确仍可得分。在相关处注明单位。

SR 1MEDIUM Paper 2 2.1 The Ionic Model [18]

This question concerns the ionic bonding model.本题考查离子键模型。

(a) Deduce the formula and name of the ionic compound formed between calcium ions and phosphate ions.推导由钙离子与磷酸根离子形成的离子化合物的化学式与名称。 [2]

(b) Explain, in terms of structure and bonding, why ionic compounds conduct electricity when molten but not when solid.从结构与成键的角度解释：为何离子化合物熔融时导电，而固态时不导电。 [3]

(c) The lattice enthalpies of magnesium oxide and sodium chloride are $+3795~\mathrm{kJ\,mol^{-1}}$ and $+787~\mathrm{kJ\,mol^{-1}}$ respectively. Explain, using ionic charge and ionic radius, why the lattice enthalpy of MgO is so much greater than that of NaCl.氧化镁与氯化钠的晶格焓分别为 $+3795~\mathrm{kJ\,mol^{-1}}$ 和 $+787~\mathrm{kJ\,mol^{-1}}$。用离子电荷与离子半径解释为何 MgO 的晶格焓远大于 NaCl。 [3]

(d) Predict, with a reason, whether the lattice enthalpy of calcium oxide would be greater than or less than that of magnesium oxide.预测氧化钙的晶格焓比氧化镁的更大还是更小，并说明理由。 [2]

(e) Two solids, X and Y, are tested. X conducts electricity as a solid and is malleable. Y does not conduct as a solid, but conducts when molten, and shatters when struck. Identify the type of bonding in X and in Y, justifying each answer using the structural properties described.测试两种固体 X 与 Y。X 固态即可导电，且具延展性。Y 固态不导电，熔融后可导电，且受撞击会碎裂。指出 X 与 Y 中的成键类型，并用所述结构性质加以说明。 [4]

(f) Equal numbers of moles of solid NaCl and solid $\mathrm{MgCl_2}$ are each dissolved in equal volumes of water. Predict which solution conducts electricity more strongly, explaining your answer in terms of the number and charge of the ions produced.将等物质的量的固体 NaCl 与固体 $\mathrm{MgCl_2}$ 分别溶于等体积的水中。预测哪一溶液导电性更强，并从生成离子的数目与电荷角度说明理由。 [4]

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SR 2HARD Paper 2 HL 2.2 Resonance, Formal Charge, Hybridization [18]

This question concerns HL extensions of the covalent bonding model.本题考查共价键模型的 HL 拓展内容。

(a) Draw a Lewis structure for the nitrate ion, $\mathrm{NO_3^-}$, showing one N=O double bond and two N-O single bonds. Calculate the formal charge on the nitrogen atom and on each type of oxygen atom, and show that the sum of the formal charges equals the overall charge on the ion.画出硝酸根离子 $\mathrm{NO_3^-}$ 的路易斯结构，其中含一条 N=O 双键与两条 N-O 单键。计算氮原子以及两类氧原子各自的形式电荷，并证明形式电荷之和等于该离子的总电荷。 [4]

(b) Explain why the nitrate ion is best described using resonance rather than a single Lewis structure, and predict how the length of each N-O bond compares with a typical N=O double bond and a typical N-O single bond.解释为何硝酸根离子最好用共振而非单一路易斯结构来描述，并预测每条 N-O 键的键长与典型的 N=O 双键、N-O 单键相比如何。 [3]

(c) Sulfur tetrafluoride, $\mathrm{SF_4}$, has an expanded octet around the central S atom. Deduce the number of bonding pairs and lone pairs around S, and hence predict its electron-domain geometry and molecular geometry.四氟化硫 $\mathrm{SF_4}$ 的中心 S 原子具有扩展八隅体。推导 S 周围的成键对与孤对数目，进而预测其电子域几何与分子几何。 [3]

(d) For the ethyne molecule, $\mathrm{H{-}C{\equiv}C{-}H}$, determine the hybridization of each carbon atom, and state the total number of sigma ($\sigma$) and pi ($\pi$) bonds in the whole molecule.对乙炔分子 $\mathrm{H{-}C{\equiv}C{-}H}$，确定每个碳原子的杂化方式，并写出整个分子中 σ 键与 π 键的总数。 [4]

(e) Explain, in terms of orbital overlap, the difference between a sigma bond and a pi bond, and explain why rotation about a C=C double bond is restricted while rotation about a C-C single bond is not.从轨道重叠的角度解释 σ 键与 π 键的区别，并解释为何绕 C=C 双键的旋转受限，而绕 C-C 单键的旋转不受限。 [4]

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SR 3MEDIUM Paper 2 HL 2.3 / 2.4 Metals, Alloys, Polymers [16]

This question concerns the metallic bonding model and materials built from it.本题考查金属键模型及以此建构的材料。

(a) Explain, in terms of the metallic bonding model, why the melting point of aluminium (933 K) is considerably higher than that of sodium (371 K), even though both are metals in period 3.用金属键模型解释：为何铝的熔点（933 K）远高于钠（371 K），尽管两者都是第三周期金属。 [3]

(b) Transition metals such as iron typically have much higher melting points than Group 1 metals. Using the metallic bonding model, explain why d-block elements have particularly strong metallic bonding.铁等过渡金属的熔点通常远高于第 1 族金属。用金属键模型解释为何 d 区元素的金属键特别强。 [3]

(c) Stainless steel is an alloy of iron, chromium, and nickel. Explain, in terms of structure, why stainless steel is harder than pure iron.不锈钢是铁、铬、镍的合金。从结构角度解释为何不锈钢比纯铁更硬。 [3]

(d) Distinguish between addition polymers and condensation polymers, naming one monomer type for each.区分加成聚合物与缩合聚合物，并各举出一种单体类型。 [3]

(e) Classify polyethene, nylon, and PET (polyethylene terephthalate) as addition or condensation polymers. For any you classify as condensation, state the small molecule released at each linkage.将聚乙烯、尼龙与 PET（聚对苯二甲酸乙二醇酯）分类为加成聚合物或缩合聚合物。对被判为缩合聚合物的，写出每形成一个连接所释放的小分子。 [2]

(f) Explain why nylon has a higher tensile strength and melting point than polyethene, in terms of the intermolecular forces between adjacent polymer chains.从相邻聚合物链间分子间作用力的角度，解释为何尼龙的抗拉强度和熔点都高于聚乙烯。 [2]

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Data-Based Question (HL)数据题（HL）

This question tests the bonding-triangle model of Structure 2.4, using electronegativity values and physical-property data. Higher-level students should attempt all parts.本题考查 Structure 2.4 的成键三角形（bonding triangle）模型，需用到电负性数值与物理性质数据。HL 学生应作答全部小题。

P3-1HARD Paper 3 HL HL 2.4 Bonding Triangle [18]

Table 1 gives Pauling-scale electronegativity values (one decimal place) for the elements needed in this question.表 1 给出本题所需元素的鲍林（Pauling）标度电负性数值（保留一位小数）。

Na	Mg	Al	Si	H	C	O	Cl	F
0.9	1.3	1.6	1.9	2.2	2.6	3.4	3.2	4.0

Table 2 gives selected properties of five substances.表 2 给出五种物质的部分性质。

Substance物质	NaF	MgO	HCl	Cl₂	Al
$T_m$ (°C)	993	2852	−114	−101	660
Conducts as solid?固态是否导电？	No否	No否	No否	No否	Yes是
Conducts molten/liquid?熔融/液态是否导电？	Yes是	Yes是	No否	No否	Yes是

(a) Using Table 1, calculate the electronegativity difference ($\Delta\chi$) and the average electronegativity ($\bar\chi$) for NaF and for HCl.利用表 1，计算 NaF 与 HCl 各自的电负性差（$\Delta\chi$）与平均电负性（$\bar\chi$）。 [2]

(b) Both $\mathrm{Cl_2}$ and Al have $\Delta\chi = 0$. Explain why these two substances nonetheless lie in completely different regions of the bonding triangle.$\mathrm{Cl_2}$ 与 Al 的 $\Delta\chi$ 都为 0。解释为何这两种物质在成键三角形中却位于完全不同的区域。 [3]

(c) Using your calculated values from (a), Table 1, and the properties in Table 2, classify each of the five substances in Table 2 by bonding type. Justify two of your five classifications by direct reference to the data.利用 (a) 中的计算结果、表 1 以及表 2 中的性质，对表 2 中五种物质各自的成键类型进行分类。请直接引用数据，说明其中两项分类的理由。 [5]

(d) MgO has a numerically smaller $\Delta\chi$ than NaF, yet MgO's melting point is far higher. Explain this apparent contradiction in terms of the factors that determine ionic bond strength beyond electronegativity difference alone.MgO 的 $\Delta\chi$ 在数值上比 NaF 更小，但其熔点却远高于 NaF。请从电负性差之外、决定离子键强度的其他因素出发，解释这一表面上的矛盾。 [4]

(e) $\mathrm{SiCl_4}$ has polar Si-Cl bonds ($\Delta\chi = 1.3$ using Table 1). Predict, giving a reason based on bond polarity and molecular shape, whether $\mathrm{SiCl_4}$ is polar or non-polar overall.$\mathrm{SiCl_4}$ 含有极性 Si-Cl 键（由表 1 得 $\Delta\chi = 1.3$）。请基于键的极性与分子形状，预测 $\mathrm{SiCl_4}$ 整体是极性还是非极性，并说明理由。 [4]

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Bonding and Structure成键与结构

Multiple Choice选择题

Structured Response结构化解答题

Data-Based Question (HL)数据题（HL）