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 working in the margin that a marker can verify. For short-answer items, state units where applicable. No calculator on Q1-Q3; calculator permitted on Q4-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。短答题适用时须写出单位。Q1-Q3 不可使用计算器；Q4-Q5 可用计算器。
A hydrogen electron transitions from energy level $n = 4$ to $n = 2$. The energy levels for hydrogen are given by $E_n = -13.6/n^2\ \text{eV}$.氢原子中电子从能级 $n = 4$ 跃迁到 $n = 2$。氢的能级由 $E_n = -13.6/n^2\ \text{eV}$ 给出。
(a)Calculate $E_4$ and $E_2$ in eV.计算 $E_4$ 和 $E_2$（单位：eV）。[2]
(b)Calculate the energy of the emitted photon in eV.计算发射光子的能量（单位：eV）。[2]
(c)State whether the photon is in the visible range. (Visible light: $1.77$ to $3.10\ \text{eV}$.)判断该光子是否在可见光范围内。（可见光：$1.77$ 至 $3.10\ \text{eV}$。）[1]
Copper has two naturally occurring isotopes: ${}^{63}\text{Cu}$ (mass $62.930\ \text{u}$, natural abundance $69.17\%$) and ${}^{65}\text{Cu}$ (mass $64.928\ \text{u}$, natural abundance $30.83\%$).铜有两种天然同位素：${}^{63}\text{Cu}$（质量 $62.930\ \text{u}$，天然丰度 $69.17\%$）和 ${}^{65}\text{Cu}$（质量 $64.928\ \text{u}$，天然丰度 $30.83\%$）。
(a)State the number of protons, neutrons, and electrons in each isotope (neutral atom).写出每种同位素（中性原子）的质子数、中子数和电子数。[2]
(b)Calculate the average atomic mass of copper in u. Show your working.计算铜的平均原子质量（单位：u）。需写出计算过程。[3]
(c)Explain why the average atomic mass is closer to $63$ than to $65$.解释为何平均原子质量更接近 $63$ 而非 $65$。[1]
Q5MEDIUM中🇨🇦 BC卑🇨🇦 AB阿BC Provincial-style卑诗省考风格§5 Electron configuration电子排布 · HS-PS1-1[9 marks][9 分]
Write the full ground-state electron configurations for each of the following atoms and ions, and identify the number of valence electrons in each.写出下列各原子及离子的基态全电子排布式，并指出各自的价电子数。
(a)Phosphorus (P, $Z = 15$)磷（P，$Z = 15$）[3]
(b)Chloride ion ($\text{Cl}^-$, $Z = 17$)氯离子（$\text{Cl}^-$，$Z = 17$）[3]
(c)Explain why $\text{Cl}^-$ and $\text{Ar}$ (Z = 18) have the same electron configuration even though they are different species.解释为何 $\text{Cl}^-$ 与 $\text{Ar}$（Z = 18）具有相同的电子排布，尽管它们是不同的物种。[3]
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 units in every final answer. Define any symbol you introduce. Calculator permitted on Q6-Q9.每一步推理都要写出。每个最终答案都要写单位。需定义所引入的符号。Q6-Q9 可用计算器。
When hydrogen gas is excited by an electrical discharge and the emitted light is passed through a prism, a line spectrum (not a continuous spectrum) is observed. The Balmer series involves transitions to $n = 2$ from higher levels.当氢气受到电放电激发时，将发出的光通过棱镜，可观察到线状光谱（而非连续光谱）。巴尔末系涉及从更高能级跃迁到 $n = 2$ 的过程。
(a)Explain, in terms of energy levels, why hydrogen produces a line spectrum rather than a continuous spectrum.用能级的概念解释为何氢会产生线状光谱而非连续光谱。[3]
(b)A line in the Balmer series corresponds to the transition from $n = 3$ to $n = 2$. Using $E_n = -13.6/n^2\ \text{eV}$, calculate the energy of this photon in eV.巴尔末系中某条谱线对应从 $n = 3$ 到 $n = 2$ 的跃迁。用 $E_n = -13.6/n^2\ \text{eV}$ 计算该光子的能量（eV）。[3]
(c)State whether emission or absorption is occurring in this process, and identify what the atom must have done before emitting this photon.判断此过程是发射还是吸收，并说明原子在发射该光子之前必须经历了什么过程。[1]
Use your knowledge of electron configuration and periodic trends to answer the following.运用电子排布和周期规律的知识回答以下问题。
(a)Arrange the following atoms in order of increasing atomic radius: Na, Cl, K. Explain your reasoning in terms of electron shells and nuclear charge.按原子半径从小到大排列以下原子：Na、Cl、K。根据电子层数和核电荷解释你的推理。[3]
(b)Predict which has the higher first ionization energy: Na or Mg. Explain in terms of electron configuration and nuclear charge.预测 Na 与 Mg 中哪个具有更高的第一电离能。用电子排布和核电荷解释。[3]
(c)Explain why the trend in electronegativity is the reverse of the trend in atomic radius across a period.解释为何同周期内电负性的变化趋势与原子半径的变化趋势相反。[2]
Q8HARD难🇺🇸 US美🇨🇦 BC卑AP-feeder FRQAP 衔接简答题§1 + §5 Ions and electron configuration离子与电子排布 · HS-PS1-1[7 marks][7 分]
An ion of iron has the electron configuration $[\text{Ar}]\ 3d^6$. Use this to answer the following.某铁离子具有电子排布 $[\text{Ar}]\ 3d^6$。用此回答以下问题。
(a)Identify the ion (including charge). Iron has atomic number $Z = 26$; $[\text{Ar}]$ corresponds to $18$ electrons.确定该离子（包括电荷数）。铁的原子序数 $Z = 26$；$[\text{Ar}]$ 对应 $18$ 个电子。[2]
(b)Write the full ground-state electron configuration of the neutral iron atom ($Z = 26$) using subshell notation.用分层符号写出中性铁原子（$Z = 26$）的基态全电子排布式。[2]
(c)State the period and group of iron in the periodic table, and justify your answer using the electron configuration.写出铁在元素周期表中的周期和族，并用电子排布说明理由。[2]
(d)State how many valence electrons the $\text{Fe}^{2+}$ ion has, explaining your reasoning.说明 $\text{Fe}^{2+}$ 离子有多少个价电子，并解释理由。[1]
The quantum-mechanical model describes electrons using four quantum numbers: $n$ (principal), $\ell$ (angular momentum), $m_\ell$ (magnetic), and $m_s$ (spin). The Pauli exclusion principle states that no two electrons in the same atom can have identical sets of all four quantum numbers.量子力学模型用四个量子数描述电子：$n$（主量子数）、$\ell$（角动量量子数）、$m_\ell$（磁量子数）和 $m_s$（自旋量子数）。泡利不相容原理规定，同一原子中不存在四个量子数完全相同的两个电子。
(a)State the allowed values of $\ell$ when $n = 3$, and name the corresponding subshell types.写出 $n = 3$ 时 $\ell$ 的允许值，并命名对应的分层类型。[2]
(b)How many orbitals are in the $3d$ subshell? State the allowed values of $m_\ell$ for a $d$ subshell.$3d$ 分层有多少个轨道？写出 $d$ 分层 $m_\ell$ 的允许值。[2]
(c)Using the Pauli exclusion principle, explain why each orbital can hold at most two electrons.利用泡利不相容原理，解释为何每个轨道最多容纳两个电子。[2]
(d)Compare the Bohr model and the quantum-mechanical model: state one key limitation of the Bohr model that the quantum-mechanical model corrects.比较玻尔模型与量子力学模型：写出玻尔模型的一个关键局限性，以及量子力学模型如何纠正它。[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 symbols (with units) at the start of each question. Show all steps. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.每题开始时定义符号（含单位）。写出所有步骤。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)Calculate the average atomic mass of neon in u. Show full working.计算氖的平均原子质量（单位：u）。需写出完整计算过程。[4]
(b)Neon-20 has $Z = 10$. For a hydrogen-like ion with nuclear charge $Z$, the energy levels are $E_n = -13.6\,Z^2/n^2\ \text{eV}$. Calculate the energy released when a Ne$^{9+}$ ion (one electron) transitions from $n = 2$ to $n = 1$.氖-20 的 $Z = 10$。对于核电荷为 $Z$ 的类氢离子，能级为 $E_n = -13.6\,Z^2/n^2\ \text{eV}$。计算 Ne$^{9+}$ 离子（单电子）从 $n = 2$ 跃迁到 $n = 1$ 时释放的能量。[4]
Consider the elements sulfur (S, $Z = 16$), argon (Ar, $Z = 18$), and potassium (K, $Z = 19$).考虑元素硫（S，$Z = 16$）、氩（Ar，$Z = 18$）和钾（K，$Z = 19$）。
(a)Write the ground-state electron configuration for each element using noble-gas shorthand notation.用稀有气体简写符号写出每种元素的基态电子排布式。[3]
(b)Rank these three elements in order of first ionization energy (lowest to highest) and justify your ranking.将这三种元素按第一电离能从低到高排列，并说明理由。[3]
(c)Potassium ($Z = 19$) has a much lower first ionization energy than argon ($Z = 18$) even though it has more protons. Use the concept of electron shielding and the new outermost shell to explain this anomaly.钾（$Z = 19$）的第一电离能远低于氩（$Z = 18$），尽管钾的质子数更多。用电子屏蔽效应和新最外层的概念解释这一异常。[3]
Q12HARD难🇨🇦 AB阿🇺🇸 US美AB Diploma-style阿尔伯塔毕业考风格§1, 2, 3, 6 Integration: structure to spectra综合：从结构到光谱 · Chem 30 Unit A[9 marks][9 分]
A chemistry student examines two gas-discharge tubes: one contains hydrogen gas and the other contains helium gas (${}^{4}_{2}\text{He}$). Each produces a unique line spectrum.一名化学学生观察两个气体放电管：一个含氢气，另一个含氦气（${}^{4}_{2}\text{He}$）。每种气体产生独特的线状光谱。
(a)For ${}^{4}_{2}\text{He}$: state the number of protons, neutrons, and electrons in a neutral helium atom.对于 ${}^{4}_{2}\text{He}$：写出中性氦原子的质子数、中子数和电子数。[2]
(b)Explain why hydrogen and helium produce different line spectra, referring to the number of electrons and the allowed energy-level transitions.解释为何氢和氦产生不同的线状光谱，需提及电子数和允许的能级跃迁。[3]
(c)In a flame test, each element produces a characteristic colour. Explain this observation using the concept of ground state, excited state, and photon emission. Include the direction of the electron transition (up or down in energy levels).在焰色反应中，每种元素产生特定颜色。用基态、激发态和光子发射的概念解释这一现象，并注明电子跃迁的方向（能级升高还是降低）。[4]
🇺🇸 US NGSS美国 NGSSHS-PS1-1 · HS-PS1-8
🇨🇦 Ontario安大略SCH3U B3 · SCH4U C3
🇨🇦 British Columbia不列颠哥伦比亚Chemistry 11: atoms, isotopes, electron configuration, Bohr model化学 11：原子、同位素、电子排布、玻尔模型
🇨🇦 Alberta阿尔伯塔Chem 30 Unit A · atomic theory, periodic trends原子理论，周期规律
Full Syllabus Map lives in ../Study Guides/Unit_1_Atomic_Structure_and_the_Quantum_Model.html. Q9 (quantum-mechanical model) is Honors-flagged: assessed by US NGSS and ON SCH3U/SCH4U; not a core Chem 30 diploma expectation.完整大纲对照表见 ../Study Guides/Unit_1_Atomic_Structure_and_the_Quantum_Model.html。Q9（量子力学模型）标注荣誉级：由美国 NGSS 和安省 SCH3U/SCH4U 评估；非 Chem 30 毕业考核心内容。