High School Chemistry

Atomic Structure and the Quantum Model原子结构与量子模型

Every property of matter — why gold is shiny, why water is a liquid at room temperature, why sodium explodes in water — traces back to how electrons are arranged inside atoms. This guide builds that picture from the ground up: from the three subatomic particles and nuclear notation, through isotopes and average atomic mass, to the Bohr planetary model and its quantised energy levels, then to the full quantum-mechanical model (s, p, d, f orbitals), electron configuration using the aufbau principle, Hund's rule and the Pauli exclusion principle, and finally to atomic emission spectra and the periodic trends that electron structure explains. Worked examples and KaTeX formulas are used throughout.物质的一切性质——为什么黄金有光泽、为什么水在室温下是液体、为什么钠在水中爆炸——都追溯到电子在原子内部的排布方式。本指南从最基础开始逐步搭建这幅图景：从三种亚原子粒子（subatomic particles，亚原子粒子）与核符号出发，经同位素（isotopes，同位素）与平均原子质量（atomic mass，原子质量），到玻尔（Bohr model，玻尔模型）行星模型及其量子化能级（energy levels，能级），再到完整的量子力学模型（s、p、d、f 轨道，orbitals，轨道），用构建原理、洪特规则与泡利不相容原理写出电子排布（electron configuration，电子排布），最后落脚于原子发射光谱（emission spectrum，发射光谱）与电子结构所解释的各种周期规律。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: orbital filling rules (ON SCH4U / AB Chem 20)荣誉级：轨道填充规则（ON SCH4U / AB Chem 20）
Read me first先读这里

How to use this guide如何使用本指南

Atomic structure is the first chemistry topic in every curriculum we map to, and the four curricula agree on a core scope: subatomic particles, atomic notation, isotopes, average atomic mass, and the Bohr energy-level model. They diverge on depth. US NGSS (HS-PS1-1) keeps electron arrangement qualitative — "patterns of electrons in the outermost energy level" — and does not assess the orbital-filling rules. Ontario SCH3U (B3.1, B3.2) adds isotopic abundance and atomic mass calculation; the full quantum-mechanical model (aufbau, Hund, Pauli) enters at SCH4U Grade 12. BC Chemistry 11 makes "quantum mechanical model and electron configuration" core content. Alberta Chemistry 20 builds from valence electrons and electron dot diagrams through the periodic table. The table below tells you which sections are core for you; each row cites the curriculum document it was checked against.原子结构是我们所对照的所有大纲中的第一个化学主题，四套大纲在核心范围上一致：亚原子粒子、原子符号、同位素、平均原子质量与玻尔能级模型。它们的分歧在于深度。US NGSS（HS-PS1-1）保持电子排布的定性描述——"最外能级的电子规律"——不考查轨道填充规则。安大略 SCH3U（B3.1、B3.2）加入同位素丰度与原子质量计算；完整的量子力学模型（构建原理、洪特规则、泡利原理）在 SCH4U 12 年级才引入。BC Chemistry 11 把"量子力学模型与电子排布"列为核心内容。阿尔伯塔 Chemistry 20 从价电子与电子点图起步，贯通元素周期表。下表告诉你哪些节是你的核心；每行均注明所依据的课纲文件。

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–§3 (subatomic particles, isotopes, Bohr model) — the core under HS-PS1-1 and HS-PS1-8. NGSS keeps electron arrangement qualitative: "outermost energy level" patterns, no orbital-filling rules.§1–§3（亚原子粒子、同位素、玻尔模型）—— HS-PS1-1 与 HS-PS1-8 下的核心。NGSS 保持电子排布的定性：最外能级的规律，无轨道填充规则。 §4–§5 (orbital model, aufbau/Hund/Pauli): valuable context but above the NGSS assessed floor, which does not require sub-shell notation.§4–§5（轨道模型、构建/洪特/泡利）：很有价值，但高于 NGSS 不要求亚壳层符号的评估下限。 NGSS HS-PS1 (Chemistry) — HS-PS1-1 PE + Clarification + Assessment Boundary; HS-PS1-8 PE + nuclear scope— HS-PS1-1 表现期望 + 澄清 + 评估边界；HS-PS1-8 表现期望 + 核范围
🇨🇦 ON Grade 10/11 — SNC2D / SCH3U安大略 10/11 年级 — SNC2D / SCH3U §1–§3 in full (SNC2D foundations + SCH3U B3.1, B3.2 for atomic/mass number, isotopes, and average atomic mass calculation); §6 (emission spectra qualitatively).§1–§3 完整学习（SNC2D 基础 + SCH3U B3.1、B3.2：原子序数/质量数、同位素与平均原子质量计算）；§6（发射光谱，定性）。 §4–§5 (orbital notation, aufbau/Hund/Pauli): Grade 12 SCH4U C2.2 / C3.2 honors depth — flag with Honors for SCH3U students.§4–§5（轨道符号、构建/洪特/泡利）：12 年级 SCH4U C2.2 / C3.2 荣誉深度 — SCH3U 学生标 Honors。 Ontario SCH3U/4U Chemistry — SCH3U Strand B Overall Expectations B1–B3, Specific Expectations B3.1, B3.2; SCH4U Strand C C2.1, C2.2, C3.1, C3.2— SCH3U B 单元总体期望 B1–B3，具体期望 B3.1、B3.2；SCH4U C 单元 C2.1、C2.2、C3.1、C3.2
🇨🇦 BC Science 10 / Chemistry 11BC Science 10 / Chemistry 11 §1–§7 in full. BC Chemistry 11 names "quantum mechanical model and electron configuration" as core Content, so the orbital model (§4) and electron configuration (§5) are assessed, not honors. Science 10 covers nuclear/radiation foundations (§1–§2).§1–§7 完整学习。BC Chemistry 11 把"量子力学模型与电子排布"列为核心内容，故轨道模型（§4）与电子排布（§5）在 BC 是被评估的，而非荣誉。Science 10 覆盖核/辐射基础（§1–§2）。 Nothing — BC assesses the full scope including orbital model and electron configuration in Grade 11.无 — BC 在 11 年级就评估完整范围，包括轨道模型与电子排布。 BC Chemistry 11/12 — Chemistry 11 Big Idea "Atoms and molecules are building blocks of matter"; Content "quantum mechanical model and electron configuration"; Science 10 Content "nuclear energy / radiation"— Chemistry 11 大概念"原子和分子是物质的构建单元"；内容"量子力学模型与电子排布"；Science 10 内容"核能/辐射"
🇨🇦 AB Science 10 / Chemistry 20阿尔伯塔 Science 10 / Chemistry 20 §1–§5 in full. Chemistry 20 Unit A (GO1, GO2) requires: valence electrons, electron dot diagrams, the periodic table as a predictor, and electron configuration as the foundation for bonding. Science 10 Unit A lays the prerequisite nuclear/atomic layer (§1–§2).§1–§5 完整学习。Chemistry 20 A 单元（GO1、GO2）要求：价电子、电子点图、元素周期表作为预测工具，以及以电子排布为成键基础。Science 10 A 单元奠定先修核/原子层（§1–§2）。 Nothing — AB expects electron dot diagrams and valence-electron reasoning in Grade 11; Diploma-exam problems build directly on §5 electron configuration.无 — AB 在 11 年级就要求电子点图与价电子推理；文凭考题直接建立在 §5 电子排布之上。 Alberta Chemistry 20/30 — Chemistry 20 Unit A GO1/GO2, Key Concepts, knowledge outcome text; Science 10 Unit A prerequisite note— Chemistry 20 A 单元 GO1/GO2，关键概念，知识结果文本；Science 10 A 单元先修说明
🇺🇸 IB / AP feeder trackIB / AP 衔接轨道 All seven sections plus every going-deeper derivation. IB Chemistry HL Structure 1 and AP Chemistry Unit 1 both assume fluent electron configuration, orbital notation, and emission-spectra interpretation from day one.全部 7 节，并完成每个"深入"推导。IB Chemistry HL Structure 1 与 AP Chemistry Unit 1 第一天就默认你熟练电子排布、轨道符号与发射光谱解读。 Nothing — this is the conceptual foundation for all bonding, periodicity, and reactivity units that follow.无 — 这是后续所有成键、周期性与反应活性单元的概念基础。 NGSS HS-PS1 (Chemistry) — the IB/AP feeder reads the full orbital-model depth; see the IB Chemistry HL link in "What This Feeds Into"— IB/AP 衔接读到完整轨道模型深度；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 speed 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 five things: protons determine the element (atomic number $Z$); mass number $A = Z + N$; isotopes have the same $Z$ but different $N$; average atomic mass is a weighted average by natural abundance; and the Bohr model gives $E_n = -13.6/n^2$ eV for hydrogen. Read every cram-cheat box. Skip the going-deeper quantum derivations if time is short.背熟五件事：质子数决定元素（原子序数 $Z$）；质量数 $A = Z + N$；同位素的 $Z$ 相同但 $N$ 不同；平均原子质量是按天然丰度加权的平均值；玻尔模型给出氢的 $E_n = -13.6/n^2$ eV。读每个速记框。若时间紧，可跳过量子力学的深入推导。

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

Be precise about what each quantum number means; know why the energy-level formula $E_n \propto -1/n^2$ makes emission lines converge at high $n$; write electron configurations for elements 1–36 without a table, including exceptions like Cr and Cu; connect each spectral line to a specific pair of energy levels using $E = h\nu$; and explain every periodic trend (§7) in terms of effective nuclear charge and shielding. ON SCH4U C2.2 and AB Chemistry 20 expect you to apply the three filling rules, not just quote them.精确理解每个量子数的含义；了解为何能级公式 $E_n \propto -1/n^2$ 使高 $n$ 时发射谱线收敛；不看表默写 1–36 号元素的电子排布，包括 Cr 和 Cu 等例外情况；用 $E = h\nu$ 把每条谱线与一对具体的能级相联系；并从有效核电荷与屏蔽效应解释每种周期规律（§7）。ON SCH4U C2.2 与 AB Chemistry 20 要求你应用三条填充规则，而非仅仅背诵。

Honors flag.荣誉级标记。 Sections 4–5 (quantum-mechanical orbital model, aufbau/Hund/Pauli electron configuration) carry the Honors chip for US NGSS and Ontario SCH3U, where the assessed depth is the Bohr energy-level model and qualitative electron arrangement. They are core, not honors, in BC Chemistry 11 (Content: "quantum mechanical model and electron configuration") and Ontario SCH4U (C2.2). AB Chemistry 20 expects valence-electron dot diagrams as core; the full sub-shell notation is Alberta honors depth. If your row sends you to §4–§5, treat them as required content.§4–§5（量子力学轨道模型、构建/洪特/泡利电子排布）在 US NGSS 与安大略 SCH3U 轨道上标 Honors，这两个大纲的评估深度是玻尔能级模型与定性电子排布。在 BC Chemistry 11（内容："量子力学模型与电子排布"）与安大略 SCH4U（C2.2）中，它们是核心而非荣誉内容。AB Chemistry 20 把价电子点图列为核心；完整的亚壳层符号是阿尔伯塔的荣誉深度。如果你的行指向 §4–§5，就把它们视为必学内容。
Section 1 · Subatomic particles and atomic notation第 1 节 · 亚原子粒子与原子符号

The Atom: Subatomic Particles and Atomic Notation原子：亚原子粒子与原子符号

Three particles, two numbers — this whole guide rests on them.三种粒子，两个数字 — 本指南的全部基础。
  • Proton (p)质子（p） — in the nucleus, charge $+1$, mass $\approx 1\ \mathrm{u}$. The number of protons $=$ atomic number $Z$; it defines which element you have.— 在原子核中，电荷 $+1$，质量 $\approx 1\ \mathrm{u}$。质子数 $=$ 原子序数 $Z$；它决定你所拥有的是哪种元素。
  • Neutron (n)中子（n） — in the nucleus, charge $0$, mass $\approx 1\ \mathrm{u}$. Number of neutrons $N = A - Z$.— 在原子核中，电荷 $0$，质量 $\approx 1\ \mathrm{u}$。中子数 $N = A - Z$。
  • Electron (e)电子（e） — outside the nucleus (in shells/orbitals), charge $-1$, mass $\approx 1/1836\ \mathrm{u}$ (negligible). In a neutral atom, electrons $= Z$.— 在原子核外（壳层/轨道中），电荷 $-1$，质量 $\approx 1/1836\ \mathrm{u}$（可忽略）。中性原子中，电子数 $= Z$。

Standard nuclear notation:标准核符号：

$$ {}^{A}_{Z}\text{X} \qquad A = \text{mass number} = Z + N $$ Example: ${}^{23}_{11}\text{Na}$ has $Z = 11$ protons, $A = 23$, so $N = 23 - 11 = 12$ neutrons, and (neutral) $11$ electrons. NGSS HS-PS1-8 and SCH3U B3.1 both name the mass-number / atomic-number relationship as assessed content.示例：${}^{23}_{11}\text{Na}$ 有 $Z = 11$ 个质子，$A = 23$，故 $N = 23 - 11 = 12$ 个中子，（中性）有 $11$ 个电子。NGSS HS-PS1-8 与 SCH3U B3.1 均把质量数/原子序数关系列为被评估内容。
Worked Example 1 · Reading nuclear notation例题 1 · 解读核符号

An atom is written ${}^{56}_{26}\text{Fe}$. State (a) the atomic number, (b) the mass number, (c) the number of neutrons, and (d) the number of electrons in the neutral atom.某原子写作 ${}^{56}_{26}\text{Fe}$。请说明 (a) 原子序数，(b) 质量数，(c) 中子数，(d) 中性原子的电子数。

(a) Atomic number $Z = 26$(a) 原子序数 $Z = 26$ — the subscript. This identifies the element as iron (Fe).— 下标。这标识元素为铁（Fe）。

(b) Mass number $A = 56$(b) 质量数 $A = 56$ — the superscript. Total protons + neutrons = 56.— 上标。质子数 + 中子数 = 56。

(c) Neutrons $N = A - Z = 56 - 26 = 30$.(c) 中子数 $N = A - Z = 56 - 26 = 30$。

(d) Electrons $= Z = 26$(d) 电子数 $= Z = 26$ (neutral atom: protons = electrons).（中性原子：质子数 = 电子数）。

An atom of chlorine-35 is written ${}^{35}_{17}\text{Cl}$. How many neutrons does it contain?氯-35 的原子写作 ${}^{35}_{17}\text{Cl}$。它含有多少个中子？
§1 · Q1
1717
1818
3535
5252
$N = A - Z = 35 - 17 = 18$ neutrons. The atomic number (17) gives the proton count; the mass number (35) gives protons + neutrons.$N = A - Z = 35 - 17 = 18$ 个中子。原子序数（17）给出质子数；质量数（35）给出质子数 + 中子数。
Neutrons $= $ mass number $-$ atomic number. The subscript is $Z$ (protons); the superscript is $A$ (protons + neutrons); $N = A - Z$.中子数 $= $ 质量数 $-$ 原子序数。下标是 $Z$（质子数）；上标是 $A$（质子数 + 中子数）；$N = A - Z$。
Which subatomic particle determines which element an atom is?哪种亚原子粒子决定原子属于哪种元素？
§1 · Q2
Neutron中子
Electron电子
Proton质子
Photon光子
The atomic number $Z$ is the proton count. Every atom with $Z = 6$ is carbon, every atom with $Z = 79$ is gold — regardless of neutron count or charge state.原子序数 $Z$ 就是质子数。$Z = 6$ 的原子都是碳，$Z = 79$ 的原子都是金 — 无论中子数或电荷状态如何。
Neutrons determine mass (isotopes); electrons determine charge (ions). Only the proton count — the atomic number — identifies the element.中子决定质量（同位素）；电子决定电荷（离子）。只有质子数——原子序数——才标识元素。
Going deeper — the nuclear model: Rutherford's gold-foil experiment深入 — 核模型：卢瑟福金箔实验

Before 1909, the dominant model was Thomson's "plum-pudding": negative electrons embedded in a diffuse positive charge. Rutherford fired alpha particles (${}^{4}_{2}\text{He}^{2+}$) at gold foil. Most passed straight through — consistent with a mostly-empty atom. But a tiny fraction bounced back at large angles. Rutherford's conclusion: nearly all the atom's mass and all its positive charge are concentrated in a tiny dense nucleus at the centre, with electrons orbiting outside. The ratio of nuclear radius to atomic radius is roughly $10^{-5}$ — if the atom were the size of a stadium, the nucleus would be a marble on the centre line. SCH4U C3.1 cites Rutherford's and Bohr's contributions explicitly; this is the historical foundation for §3.1909 年之前，主流模型是汤姆孙的"葡萄干布丁"：带负电的电子嵌入弥散的正电荷中。卢瑟福将 α 粒子（${}^{4}_{2}\text{He}^{2+}$）射向金箔。大多数 α 粒子直接穿过 — 与大部分空旷的原子一致。但极少数粒子以大角度反弹。卢瑟福的结论：原子几乎全部的质量和全部正电荷都集中在中心的一个微小致密原子核中，电子在外侧绕核运动。原子核半径与原子半径之比约为 $10^{-5}$ — 如果原子有体育场那么大，原子核只是中线上的一颗弹珠。SCH4U C3.1 明确引用卢瑟福与玻尔的贡献；这是 §3 的历史基础。

Section 2 · Isotopes and atomic mass第 2 节 · 同位素与原子质量

Isotopes and Atomic Mass同位素与原子质量

Isotopes: same element, different mass.同位素：相同元素，不同质量。
  • Isotopes同位素 are atoms of the same element (same $Z$) with different numbers of neutrons (different $A$). Example: ${}^{35}_{17}\text{Cl}$ and ${}^{37}_{17}\text{Cl}$ are both chlorine, but one has 18 neutrons and the other 20.是相同元素（相同 $Z$）但中子数不同（不同 $A$）的原子。示例：${}^{35}_{17}\text{Cl}$ 与 ${}^{37}_{17}\text{Cl}$ 都是氯，但前者有 18 个中子，后者有 20 个中子。
  • Average atomic mass平均原子质量 on the periodic table is a weighted average of all naturally occurring isotopes' masses, weighted by their natural abundances (as decimals, summing to 1):元素周期表上的平均原子质量是所有天然同位素质量的加权平均，权重为天然丰度（用小数表示，总和为 1）：
$$ \bar{m} = \sum_i (\text{isotopic mass}_i \times \text{abundance}_i) $$ SCH3U B3.2 names the "relationship between isotopic abundance and relative atomic mass" as an assessed specific expectation. NGSS HS-PS1-8 covers isotopes qualitatively through the nuclear model (alpha, beta, gamma decays; fission, fusion).SCH3U B3.2 把"同位素丰度与相对原子质量的关系"列为具体的被评估期望。NGSS HS-PS1-8 通过核模型定性涵盖同位素（α、β、γ 衰变；裂变、聚变）。
Worked Example 2 · Average atomic mass of chlorine例题 2 · 氯的平均原子质量

Chlorine has two stable isotopes: ${}^{35}\text{Cl}$ (mass $34.969\ \mathrm{u}$, natural abundance $75.77\%$) and ${}^{37}\text{Cl}$ (mass $36.966\ \mathrm{u}$, natural abundance $24.23\%$). Calculate the average atomic mass of chlorine.氯有两种稳定同位素：${}^{35}\text{Cl}$（质量 $34.969\ \mathrm{u}$，天然丰度 $75.77\%$）和 ${}^{37}\text{Cl}$（质量 $36.966\ \mathrm{u}$，天然丰度 $24.23\%$）。计算氯的平均原子质量。

Convert percentages to decimals.将百分数转换为小数。 $75.77\% = 0.7577$; $24.23\% = 0.2423$. (Check: $0.7577 + 0.2423 = 1.000$ ✓)$75.77\% = 0.7577$；$24.23\% = 0.2423$。（核验：$0.7577 + 0.2423 = 1.000$ ✓）

Apply the weighted-average formula.应用加权平均公式。

$$ \bar{m} = (34.969)(0.7577) + (36.966)(0.2423) $$ $$ \bar{m} = 26.496 + 8.957 = 35.453\ \mathrm{u}. $$

Compare to the periodic table.与元素周期表对照。 The listed atomic mass of Cl is $35.45\ \mathrm{u}$ — matching to four significant figures. Notice the value is closer to $35$ than $37$ because ${}^{35}\text{Cl}$ is about three times more abundant.元素周期表中 Cl 的原子质量为 $35.45\ \mathrm{u}$ — 四位有效数字吻合。注意该值更接近 $35$ 而非 $37$，因为 ${}^{35}\text{Cl}$ 的丰度约为 ${}^{37}\text{Cl}$ 的三倍。

Two isotopes of boron are ${}^{10}\text{B}$ (abundance $19.9\%$, mass $10.013\ \mathrm{u}$) and ${}^{11}\text{B}$ (abundance $80.1\%$, mass $11.009\ \mathrm{u}$). Which value is closest to boron's average atomic mass?硼的两种同位素为 ${}^{10}\text{B}$（丰度 $19.9\%$，质量 $10.013\ \mathrm{u}$）和 ${}^{11}\text{B}$（丰度 $80.1\%$，质量 $11.009\ \mathrm{u}$）。下列哪个值最接近硼的平均原子质量？
§2 · Q1
$10.0\ \mathrm{u}$
$10.5\ \mathrm{u}$
$11.0\ \mathrm{u}$
$10.8\ \mathrm{u}$
$\bar m = (10.013)(0.199) + (11.009)(0.801) = 1.993 + 8.818 = 10.81\ \mathrm{u}$. Because ${}^{11}\text{B}$ dominates ($80\%$), the average is pulled close to $11$ but not all the way there.$\bar m = (10.013)(0.199) + (11.009)(0.801) = 1.993 + 8.818 = 10.81\ \mathrm{u}$。由于 ${}^{11}\text{B}$ 占主导（$80\%$），平均值被拉向 $11$，但还没到那里。
The weighted average pulls toward the more abundant isotope. Since ${}^{11}\text{B}$ has $80\%$ abundance, the answer must be closer to $11$ than to $10$, but not equal to $11$.加权平均向丰度较大的同位素靠拢。由于 ${}^{11}\text{B}$ 丰度为 $80\%$，答案必须更接近 $11$ 而非 $10$，但不等于 $11$。
Which statement correctly defines isotopes?下列哪句话正确定义了同位素？
§2 · Q2
Same atomic number, different mass number原子序数相同，质量数不同
Same mass number, different atomic number质量数相同，原子序数不同
Same number of neutrons, different number of protons中子数相同，质子数不同
Same number of electrons, different atomic number电子数相同，原子序数不同
Isotopes are atoms of the same element ($Z$ is the same, so the same number of protons) with different numbers of neutrons — hence different mass numbers $A = Z + N$.同位素是同一元素（$Z$ 相同，即质子数相同）的原子，但中子数不同 — 因此质量数 $A = Z + N$ 不同。
Same element means same $Z$ (same atomic number). Isotopes differ only in neutron count $N$, so $A$ differs while $Z$ is fixed.同一元素意味着相同的 $Z$（相同的原子序数）。同位素只有中子数 $N$ 不同，因此在 $Z$ 固定的情况下 $A$ 不同。
Section 3 · The Bohr model and energy levels第 3 节 · 玻尔模型与能级

The Bohr Model and Energy Levels玻尔模型与能级

Electrons live only in specific energy levels; transitions emit or absorb light.电子只能存在于特定能级；跃迁时发射或吸收光。
  • Bohr's model (1913)玻尔模型（1913年） — electrons orbit the nucleus in fixed circular orbits labelled by the principal quantum number $n = 1, 2, 3, \ldots$. Energy is quantised: only certain values are allowed.— 电子在固定的圆形轨道上绕核运动，轨道用主量子数 $n = 1, 2, 3, \ldots$ 标记。能量是量子化的：只有某些值被允许。
  • Energy levels for hydrogen:氢的能级：
$$ E_n = -\frac{13.6\ \mathrm{eV}}{n^2} \qquad (n = 1, 2, 3, \ldots) $$
  • Ground state基态 $n = 1$: $E_1 = -13.6\ \mathrm{eV}$ (lowest energy, most stable).$n = 1$：$E_1 = -13.6\ \mathrm{eV}$（能量最低，最稳定）。
  • Excited state激发态 $n \ge 2$: electron has absorbed energy. As $n \to \infty$, $E_n \to 0$ (ionisation).$n \ge 2$：电子已吸收能量。当 $n \to \infty$ 时，$E_n \to 0$（电离）。
  • Transition energy:跃迁能量： when an electron falls from level $n_\text{high}$ to $n_\text{low}$, it releases a photon of energy $\Delta E = E_{n_\text{high}} - E_{n_\text{low}}$, obeying $E = h\nu$.当电子从 $n_\text{high}$ 跃迁到 $n_\text{low}$ 时，释放能量为 $\Delta E = E_{n_\text{high}} - E_{n_\text{low}}$ 的光子，满足 $E = h\nu$。
SCH4U C3.1 cites Rutherford's and Bohr's models as assessed historical context. NGSS HS-PS1-1 references "outermost energy level" — Bohr levels are the simplest framework for that idea.SCH4U C3.1 把卢瑟福与玻尔模型列为被评估的历史背景。NGSS HS-PS1-1 提到"最外能级"——玻尔能级是理解这一概念最简单的框架。
Worked Example 3 · Photon energy from a Bohr transition例题 3 · 玻尔跃迁释放的光子能量

A hydrogen electron falls from the $n = 4$ level to the $n = 2$ level. Using $E_n = -13.6/n^2\ \mathrm{eV}$ and $h = 6.626 \times 10^{-34}\ \mathrm{J{\cdot}s}$, find (a) the energy of the emitted photon in eV, and (b) state whether this photon is in the visible range. (Visible light: $1.77$–$3.10\ \mathrm{eV}$.)氢原子中电子从 $n = 4$ 能级跃迁到 $n = 2$ 能级。用 $E_n = -13.6/n^2\ \mathrm{eV}$ 和 $h = 6.626 \times 10^{-34}\ \mathrm{J{\cdot}s}$，求 (a) 发射光子的能量（eV），(b) 判断该光子是否在可见光范围内。（可见光：$1.77$–$3.10\ \mathrm{eV}$。）

(a) Energy levels.(a) 能级。

$$ E_4 = -\frac{13.6}{16} = -0.850\ \mathrm{eV}, \qquad E_2 = -\frac{13.6}{4} = -3.40\ \mathrm{eV}. $$

Photon energy = energy difference.光子能量 = 能量差。

$$ \Delta E = E_4 - E_2 = -0.850 - (-3.40) = +2.55\ \mathrm{eV}. $$

(b) Visible range.(b) 可见光范围。 $2.55\ \mathrm{eV}$ falls between $1.77$ and $3.10\ \mathrm{eV}$, so this is a visible-light photon (blue-green, part of the Balmer series). ✓$2.55\ \mathrm{eV}$ 落在 $1.77$ 到 $3.10\ \mathrm{eV}$ 之间，因此这是一个可见光光子（蓝绿色，属于巴尔末系）。✓

In the Bohr model, what happens when an electron absorbs a photon whose energy exactly matches the gap between two energy levels?在玻尔模型中，当电子吸收的光子能量恰好等于两个能级之间的间隔时，会发生什么？
§3 · Q1
The electron is ejected from the atom电子从原子中被逐出
The electron emits a photon and drops to a lower level电子发射光子并跃迁到更低能级
The electron jumps to a higher energy level电子跃迁到更高能级
Nothing happens; the photon passes through什么都没发生；光子直接穿过
Bohr quantisation: an electron can only absorb a photon whose energy exactly matches an allowed $\Delta E$. Absorbing that energy promotes the electron to the higher level (excited state).玻尔量子化：电子只能吸收能量恰好等于允许的 $\Delta E$ 的光子。吸收该能量使电子跃迁到更高能级（激发态）。
Absorption promotes the electron upward; emission (the reverse) sends it downward. Ionisation requires enough energy to reach $n \to \infty$ ($E = 0$).吸收使电子向上跃迁；发射（反过程）使其向下跃迁。电离需要足够的能量以到达 $n \to \infty$（$E = 0$）。
For hydrogen, $E_1 = -13.6\ \mathrm{eV}$ and $E_3 = -1.51\ \mathrm{eV}$. What is the energy of the photon emitted when an electron transitions from $n = 3$ to $n = 1$?对于氢，$E_1 = -13.6\ \mathrm{eV}$，$E_3 = -1.51\ \mathrm{eV}$。电子从 $n = 3$ 跃迁到 $n = 1$ 时发射光子的能量是多少？
§3 · Q2
$1.51\ \mathrm{eV}$
$12.09\ \mathrm{eV}$
$13.6\ \mathrm{eV}$
$15.11\ \mathrm{eV}$
$\Delta E = E_3 - E_1 = -1.51 - (-13.6) = +12.09\ \mathrm{eV}$. This is an ultraviolet photon in the Lyman series.$\Delta E = E_3 - E_1 = -1.51 - (-13.6) = +12.09\ \mathrm{eV}$。这是莱曼系中的紫外光子。
Photon energy $= E_\text{upper} - E_\text{lower}$. Subtract the upper level's energy from the lower level: $-1.51 - (-13.6)$.光子能量 $= E_\text{upper} - E_\text{lower}$。用上层能量减去下层能量：$-1.51 - (-13.6)$。
Section 4 · The quantum-mechanical model Honors — US NGSS / ON SCH3U第 4 节 · 量子力学模型 荣誉 — US NGSS / ON SCH3U

The Quantum-Mechanical Model: Orbitals and Quantum Numbers量子力学模型：轨道与量子数

Curriculum note.课纲提示。 This section is core in BC Chemistry 11 (Content: "quantum mechanical model and electron configuration" with elaboration "electron configuration: molecular geometry, VSEPR theory") and Ontario SCH4U (C2.1, C2.2, C3.2). It carries the Honors chip for US NGSS (HS-PS1-1 assesses only qualitative "outermost energy level" patterns) and Ontario SCH3U (the Bohr model and periodic trends are assessed; the orbital model enters at SCH4U Grade 12). Alberta Chemistry 20 expects electron dot diagrams and valence electrons as core; the full sub-shell orbital notation (1s², 2s², 2p⁶...) is the honors-depth layer used to support bonding predictions.本节在 BC Chemistry 11（内容："量子力学模型与电子排布"，细化为"电子排布：分子几何、VSEPR 理论"）和安大略 SCH4U（C2.1、C2.2、C3.2）中是核心内容。在 US NGSS（HS-PS1-1 仅评估定性的"最外能级"规律）和安大略 SCH3U（玻尔模型与周期趋势被评估；轨道模型在 SCH4U 12 年级引入）中标 Honors。阿尔伯塔 Chemistry 20 把电子点图和价电子列为核心；完整的亚壳层轨道符号（1s²、2s²、2p⁶……）是用于支持成键预测的荣誉深度层。
Orbitals replace Bohr orbits: probability clouds, not fixed paths.轨道取代玻尔轨道：概率云，而非固定路径。
  • Principal quantum number $n$主量子数 $n$ $(1, 2, 3, \ldots)$ — shell number (energy level). Higher $n$ = higher energy, larger orbital.$(1, 2, 3, \ldots)$ — 壳层编号（能级）。$n$ 越大 = 能量越高，轨道越大。
  • Subshell (angular momentum quantum number $\ell$)亚壳层（角动量量子数 $\ell$） $\ell = 0, 1, 2, 3$ named s, p, d, f. Each subshell has a characteristic shape: s is spherical; p is dumbbell; d has four-lobed and other shapes.$\ell = 0, 1, 2, 3$ 分别命名为 s、p、d、f。每个亚壳层有特征形状：s 是球形；p 是哑铃形；d 有四叶形等形状。
  • Orbitals per subshell:每个亚壳层的轨道数： s has 1, p has 3, d has 5, f has 7. Each orbital holds at most 2 electrons (with opposite spins).s 有 1 个，p 有 3 个，d 有 5 个，f 有 7 个。每个轨道最多容纳 2 个电子（自旋相反）。

Capacity per shell:每个壳层的容量：

$$ \text{Max electrons in shell } n = 2n^2 $$ BC Chemistry 11 explicitly lists "quantum mechanical model and electron configuration" as core Content, with elaboration "VSEPR theory" built on top. SCH4U C2.1 names "orbital" and "energy level" as assessed terminology.BC Chemistry 11 明确把"量子力学模型与电子排布"列为核心内容，细化为建立在其上的"VSEPR 理论"。SCH4U C2.1 把"轨道"和"能级"列为被评估的术语。
Worked Example 4 · Identifying subshells例题 4 · 识别亚壳层

For the third shell ($n = 3$): (a) list all the subshells present, (b) state how many orbitals are in each, (c) state the maximum number of electrons the whole shell can hold.对于第三壳层（$n = 3$）：(a) 列出所有亚壳层，(b) 说明每个亚壳层中的轨道数，(c) 说明整个壳层能容纳的最大电子数。

(a) Subshells with $n = 3$:(a) $n = 3$ 的亚壳层： $\ell$ runs from $0$ to $n-1 = 2$, giving $3s$ ($\ell=0$), $3p$ ($\ell=1$), $3d$ ($\ell=2$).$\ell$ 从 $0$ 到 $n-1 = 2$，给出 $3s$（$\ell=0$）、$3p$（$\ell=1$）、$3d$（$\ell=2$）。

(b) Orbitals:(b) 轨道数： $3s$: 1 orbital; $3p$: 3 orbitals; $3d$: 5 orbitals. Total = 9 orbitals.$3s$：1 个轨道；$3p$：3 个轨道；$3d$：5 个轨道。共 9 个轨道。

(c) Capacity:(c) 容量： $2n^2 = 2(3)^2 = 18$ electrons. (Or: 9 orbitals $\times$ 2 electrons each $= 18$. ✓)$2n^2 = 2(3)^2 = 18$ 个电子。（或：9 个轨道 $\times$ 每个 2 个电子 $= 18$。✓）

How many electrons can the 3p subshell hold at most?3p 亚壳层最多能容纳多少个电子？
§4 · Q1
66
22
1010
33
A p subshell has 3 orbitals, each holding 2 electrons: $3 \times 2 = 6$ electrons. (s holds 2, d holds 10, f holds 14.)p 亚壳层有 3 个轨道，每个容纳 2 个电子：$3 \times 2 = 6$ 个电子。（s 容纳 2 个，d 容纳 10 个，f 容纳 14 个。）
Number of orbitals: s=1, p=3, d=5, f=7. Each orbital holds 2 electrons. So p subshell $= 3 \times 2 = 6$.轨道数：s=1，p=3，d=5，f=7。每个轨道 2 个电子。故 p 亚壳层 $= 3 \times 2 = 6$。
Which subshell shape is described as "dumbbell-shaped"?哪种亚壳层形状被描述为"哑铃形"？
§4 · Q2
ss
dd
ff
pp
The p orbital ($\ell = 1$) is dumbbell-shaped: two lobes oriented along an axis (x, y, or z). The s orbital is spherical; d orbitals have four-lobed and other shapes.p 轨道（$\ell = 1$）呈哑铃形：两个叶沿轴（x、y 或 z）方向分布。s 轨道是球形；d 轨道有四叶形等形状。
s is spherical; p is dumbbell; d has four-lobed and other complex shapes; f is even more complex.s 是球形；p 是哑铃形；d 有四叶形等复杂形状；f 更加复杂。
Going deeper — why orbitals replaced Bohr orbits: the de Broglie wavelength and Heisenberg uncertainty深入 — 为何轨道取代了玻尔轨道：德布罗意波长与海森堡不确定性

The Bohr model treats electrons as classical particles in fixed circular orbits. Two results from the 1920s demolished this picture. First, de Broglie (1924) proposed that particles with momentum $p$ have an associated wavelength $\lambda = h/p$. For electrons in an atom, this wavelength is comparable to the orbital size — electrons are waves, not billiard balls. Second, Heisenberg's uncertainty principle states:玻尔模型把电子视为经典固定圆形轨道上的粒子。1920 年代的两个结果打破了这幅图景。首先，德布罗意（1924 年）提出，动量为 $p$ 的粒子具有关联波长 $\lambda = h/p$。对于原子中的电子，这个波长与轨道尺寸相当——电子是波，而不是台球。其次，海森堡不确定性原理指出：

$$ \Delta x \cdot \Delta p \ge \frac{h}{4\pi} $$

You cannot simultaneously know an electron's exact position and exact momentum. Precise circular orbits require knowing both — they are forbidden. The quantum-mechanical solution replaces orbits with orbitals: mathematical functions $\psi$ whose squared magnitude $|\psi|^2$ gives the probability density of finding the electron at each point in space. The shapes in §4 (s spheres, p dumbbells, d four-lobes) are the surfaces containing $90\%$ of the probability. SCH4U C2.1 names "orbital" as a term students must know; BC Chemistry 11 requires the orbital picture for molecular geometry and VSEPR.你无法同时精确知道电子的位置和动量。精确的圆形轨道需要同时知道两者——这是被禁止的。量子力学的解决方案用轨道取代轨道：数学函数 $\psi$，其平方模 $|\psi|^2$ 给出在空间每个点发现电子的概率密度。§4 中的形状（s 球形、p 哑铃形、d 四叶形）是包含 $90\%$ 概率的曲面。SCH4U C2.1 把"轨道"列为学生必须掌握的术语；BC Chemistry 11 要求用轨道图来理解分子几何与 VSEPR。

Section 5 · Electron configuration第 5 节 · 电子排布

Electron Configuration: Aufbau, Hund, and Pauli电子排布：构建原理、洪特规则与泡利原理

Three rules fill the orbitals in a predictable order.三条规则以可预测的顺序填充轨道。
  • Aufbau principle构建原理 (Aufbau = "building up" in German): fill orbitals from lowest energy upward. Energy order: $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, \ldots$ (use the diagonal mnemonic).（Aufbau = 德语"建立"）：从最低能量起向上填充轨道。能量顺序：$1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, \ldots$（使用对角线记忆法）。
  • Pauli exclusion principle泡利不相容原理 : each orbital holds at most 2 electrons, and they must have opposite spins (↑↓). No two electrons in an atom can have the same set of all four quantum numbers.：每个轨道最多容纳 2 个电子，且它们必须自旋相反（↑↓）。原子中没有两个电子能拥有完全相同的四个量子数。
  • Hund's rule洪特规则 : when filling orbitals of equal energy (degenerate orbitals, e.g. the three 2p orbitals), place one electron in each before pairing any. All unpaired electrons have the same spin.：填充等能量（简并）轨道（如三个 2p 轨道）时，先在每个轨道各放一个电子，再开始配对。所有未配对电子的自旋相同。
ON SCH4U C2.2 specifically requires: "use the Pauli exclusion principle, Hund's rule, and the aufbau principle to write electron configurations." AB Chemistry 20 GO2 requires electron dot diagrams and valence-electron reasoning, which are the direct output of electron configuration.ON SCH4U C2.2 明确要求："用泡利不相容原理、洪特规则和构建原理书写电子排布"。AB Chemistry 20 GO2 要求电子点图和价电子推理，这是电子排布的直接产物。
Worked Example 5 · Writing the electron configuration of phosphorus (Z = 15)例题 5 · 书写磷（Z = 15）的电子排布

Write the full electron configuration of phosphorus and identify the number of valence electrons. State its electron-dot (Lewis dot) representation.写出磷的完整电子排布，并确定价电子数。说明其电子点（路易斯点）表示。

Fill orbitals in order.按顺序填充轨道。 15 electrons total. Following aufbau:共 15 个电子。按构建原理：

$$ 1s^2\; 2s^2\; 2p^6\; 3s^2\; 3p^3 $$

Check total:总数核验： $2 + 2 + 6 + 2 + 3 = 15$ ✓.$2 + 2 + 6 + 2 + 3 = 15$ ✓。

Valence electrons价电子 = electrons in the outermost shell (shell $n = 3$): $3s^2 3p^3 = 5$ valence electrons. Phosphorus sits in Group 15 of the periodic table — the group number equals the valence electron count for main-group elements.= 最外层（$n = 3$ 壳层）的电子：$3s^2 3p^3 = 5$ 个价电子。磷位于元素周期表第 15 族 — 对于主族元素，族数等于价电子数。

Hund's rule for 3p:3p 的洪特规则： The three 3p orbitals each receive one electron before pairing: $3p_x^{\uparrow}\ 3p_y^{\uparrow}\ 3p_z^{\uparrow}$. Phosphorus therefore has 3 unpaired electrons.三个 3p 轨道各放一个电子后再配对：$3p_x^{\uparrow}\ 3p_y^{\uparrow}\ 3p_z^{\uparrow}$。因此磷有 3 个未配对电子。

What is the correct electron configuration for sodium ($Z = 11$)?钠（$Z = 11$）的正确电子排布是什么？
§5 · Q1
$1s^2\ 2s^2\ 2p^6\ 3s^2$
$1s^2\ 2s^2\ 2p^6\ 3s^1$
$1s^2\ 2s^2\ 2p^5\ 3s^2$
$1s^2\ 2s^2\ 2p^4\ 3s^3$
$11$ electrons: fill $1s$ (2), $2s$ (2), $2p$ (6) — that's 10 — then put the last one in $3s^1$. Total: $2 + 2 + 6 + 1 = 11$ ✓. Sodium has 1 valence electron, matching Group 1.11 个电子：填充 $1s$（2）、$2s$（2）、$2p$（6）——共 10 个——再将最后一个放入 $3s^1$。总计：$2 + 2 + 6 + 1 = 11$ ✓。钠有 1 个价电子，与第 1 族一致。
Count to 11 electrons: $1s^2$ (2), $2s^2$ (4), $2p^6$ (10), then one more goes into $3s$. Option (a) has 12 electrons; options (c) and (d) violate aufbau order.数到 11 个电子：$1s^2$（2）、$2s^2$（4）、$2p^6$（10），再一个放入 $3s$。选项 (a) 有 12 个电子；选项 (c) 和 (d) 违反构建原理顺序。
Hund's rule states that when filling orbitals of equal energy, electrons should:洪特规则规定，填充等能量轨道时，电子应：
§5 · Q2
Pair up in the first available orbital before moving to the next先在第一个可用轨道中配对，再移到下一个
Fill the highest-energy orbital first先填充能量最高的轨道
Occupy each orbital singly before any pairing occurs在发生任何配对之前，每个轨道各占据一个电子
Always pair with opposite spin regardless of available orbitals无论有无可用轨道，始终与反向自旋电子配对
Hund's rule: maximise the number of unpaired electrons in degenerate orbitals. Fill one electron per orbital first (all with the same spin), then pair. This minimises electron-electron repulsion.洪特规则：在简并轨道中使未配对电子数最大化。先每个轨道各放一个电子（自旋相同），再配对。这样可以最小化电子间排斥。
Electrons prefer to spread out (lower repulsion). Hund says: one per orbital first, then pair. The aufbau principle handles the low-to-high energy order; Pauli handles the two-electrons-per-orbital limit.电子倾向于分散（降低排斥）。洪特规定：先每个轨道一个，再配对。构建原理处理由低到高的能量顺序；泡利处理每个轨道两个电子的上限。
Going deeper — exceptions to the aufbau order: chromium and copper深入 — 构建顺序的例外：铬与铜

The aufbau order predicts Cr ($Z = 24$) as $[\text{Ar}]\, 3d^4\, 4s^2$, but the actual configuration is $[\text{Ar}]\, 3d^5\, 4s^1$. Similarly, Cu ($Z = 29$) is predicted as $[\text{Ar}]\, 3d^9\, 4s^2$ but is actually $[\text{Ar}]\, 3d^{10}\, 4s^1$. In both cases, one electron migrates from the $4s$ to the $3d$ to achieve either a half-filled ($3d^5$, all five d orbitals singly occupied) or fully-filled ($3d^{10}$) subshell. These configurations are slightly lower in energy because: (1) the exchange energy stabilises a maximally-spin-aligned half-filled set (Cr), and (2) a completely filled subshell eliminates all unpaired-electron repulsion (Cu). Exam boards (ON SCH4U, AP Chemistry) expect you to know these two exceptions and explain them in terms of extra stability from half-filled or full d subshells.构建顺序预测 Cr（$Z = 24$）的排布为 $[\text{Ar}]\, 3d^4\, 4s^2$，但实际排布是 $[\text{Ar}]\, 3d^5\, 4s^1$。类似地，Cu（$Z = 29$）预测为 $[\text{Ar}]\, 3d^9\, 4s^2$，但实际上是 $[\text{Ar}]\, 3d^{10}\, 4s^1$。在这两种情况下，一个电子从 $4s$ 迁移到 $3d$，以实现半充满（$3d^5$，五个 d 轨道各占一个电子）或全充满（$3d^{10}$）的亚壳层。这些排布能量稍低，因为：(1) 交换能稳定了最大自旋对齐的半充满组（Cr），(2) 全充满亚壳层消除了所有未配对电子的排斥（Cu）。考试机构（ON SCH4U、AP Chemistry）期望你知道这两个例外，并用半充满或全充满 d 亚壳层的额外稳定性来解释。

Section 6 · Atomic emission spectra第 6 节 · 原子发射光谱

Atomic Emission Spectra原子发射光谱

Each element's spectrum is a unique fingerprint of its energy levels.每种元素的光谱是其能级的唯一指纹。
  • Continuous spectrum连续光谱 — all wavelengths present (e.g. sunlight through a prism). Produced by dense, hot solids or liquids.— 所有波长均存在（如阳光透过棱镜）。由致密的高温固体或液体产生。
  • Emission (bright-line) spectrum发射（明线）光谱 — only specific wavelengths (bright lines on a dark background). Produced when a gas is excited electrically or by heat: electrons jump to higher levels, then fall back, emitting photons whose energy equals the level gap.— 只有特定波长（暗背景上的亮线）。当气体被电气或热激发时产生：电子跃迁到更高能级，然后落回，发射能量等于能级间隔的光子。
  • Absorption spectrum吸收光谱 — dark lines on a continuous background at the same wavelengths as the emission lines. Produced when a cool gas sits in front of a hot source.— 在连续背景上出现暗线，波长与发射谱线相同。当冷气体位于热源前时产生。
$$ E_\text{photon} = h\nu = \frac{hc}{\lambda} = \Delta E_\text{level} $$ SCH4U C2.1 names "emission spectrum" and "photon" as assessed terminology. Emission spectra are direct experimental evidence for the quantised energy levels of the Bohr model (§3) and the quantum-mechanical model (§4).SCH4U C2.1 把"发射光谱"和"光子"列为被评估的术语。发射光谱是玻尔模型（§3）和量子力学模型（§4）中量子化能级的直接实验证据。
Worked Example 6 · Wavelength of a hydrogen emission line例题 6 · 氢发射谱线的波长

The $n = 3 \to n = 2$ transition in hydrogen emits a photon of energy $\Delta E = 1.89\ \mathrm{eV}$. Using $h = 6.626 \times 10^{-34}\ \mathrm{J{\cdot}s}$, $c = 3.00 \times 10^8\ \mathrm{m/s}$, and $1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$, find the wavelength of this spectral line and identify whether it is visible.氢原子 $n = 3 \to n = 2$ 的跃迁发射能量为 $\Delta E = 1.89\ \mathrm{eV}$ 的光子。利用 $h = 6.626 \times 10^{-34}\ \mathrm{J{\cdot}s}$、$c = 3.00 \times 10^8\ \mathrm{m/s}$、$1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$，求该谱线的波长并判断是否可见。

Convert energy to Joules.将能量转换为焦耳。

$$ \Delta E = 1.89 \times 1.602 \times 10^{-19} = 3.03 \times 10^{-19}\ \mathrm{J}. $$

Solve for wavelength.求波长。

$$ \lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{3.03 \times 10^{-19}} = 6.56 \times 10^{-7}\ \mathrm{m} = 656\ \mathrm{nm}. $$

Identify.识别。 $656\ \mathrm{nm}$ is red visible light. This is the H-alpha line, the brightest line in hydrogen's visible (Balmer) series. ✓$656\ \mathrm{nm}$ 是红色可见光。这是 H-alpha 线，氢可见光（巴尔末）系中最亮的谱线。✓

Why does each element produce a unique emission spectrum?为什么每种元素产生独特的发射光谱？
§6 · Q1
Each element has a unique set of electron energy levels, so the photon energies emitted are unique每种元素有独特的电子能级组，因此发射的光子能量是独特的
Each element has a different number of neutrons, which changes the photon colour每种元素有不同数量的中子，这改变了光子颜色
Larger atoms emit higher-energy photons because they have more electrons较大的原子因为有更多电子而发射能量更高的光子
The spectrum depends on the temperature at which the gas is heated光谱取决于气体被加热的温度
Spectral lines arise from electron transitions between specific energy levels. Each element has a unique set of allowed energy levels (determined by its nuclear charge and electron arrangement), so the set of possible $\Delta E$ values — and hence wavelengths — is unique to each element.谱线来自电子在特定能级间的跃迁。每种元素有由其核电荷和电子排布决定的独特的允许能级组，因此可能的 $\Delta E$ 值集合——以及因此的波长——对每种元素都是独特的。
Neutrons do not determine spectral lines; electrons and their energy levels do. The temperature affects intensity and which levels are populated, but not the positions (wavelengths) of the lines.中子不决定谱线；电子及其能级才决定谱线。温度影响强度和哪些能级被占据，但不影响谱线的位置（波长）。
An electron in hydrogen falls from $n = 4$ to $n = 1$. Compared with the $n = 4 \to n = 2$ transition, the photon emitted has:氢原子中电子从 $n = 4$ 跃迁到 $n = 1$。与 $n = 4 \to n = 2$ 跃迁相比，发射的光子：
§6 · Q2
Lower energy and longer wavelength能量更低，波长更长
The same energy能量相同
Higher energy and longer wavelength能量更高，波长更长
Higher energy and shorter wavelength能量更高，波长更短
The $n = 4 \to n = 1$ drop spans a much larger energy gap (the electron falls all the way to the ground state) than $n = 4 \to n = 2$. Higher $\Delta E \Rightarrow$ higher frequency $\Rightarrow$ shorter wavelength ($E = h\nu = hc/\lambda$). The $n \to 1$ series (Lyman) is ultraviolet; the $n \to 2$ series (Balmer) is visible/UV.$n = 4 \to n = 1$ 跨越的能量间隔（电子一路落到基态）远大于 $n = 4 \to n = 2$。更高的 $\Delta E \Rightarrow$ 更高的频率 $\Rightarrow$ 更短的波长（$E = h\nu = hc/\lambda$）。$n \to 1$ 系（莱曼系）是紫外线；$n \to 2$ 系（巴尔末系）是可见光/紫外线。
A larger energy drop produces a higher-energy photon. Since $E = hc/\lambda$, higher energy means shorter wavelength. Falling to $n=1$ (ground state) is always a larger gap than falling to $n=2$.能量跌落更大产生能量更高的光子。由于 $E = hc/\lambda$，能量更高意味着波长更短。跌落到 $n=1$（基态）始终是比跌落到 $n=2$ 更大的间隔。
Section 7 · Periodic patterns from electron structure第 7 节 · 电子结构带来的周期规律

Periodic Patterns from Electron Structure电子结构带来的周期规律

Every periodic trend traces back to two competing factors: nuclear charge and shielding.每种周期趋势都追溯到两个相互竞争的因素：核电荷与屏蔽效应。
  • Effective nuclear charge $Z_\text{eff}$有效核电荷 $Z_\text{eff}$ = the net positive charge felt by a valence electron after inner (core) electrons shield part of the nuclear attraction. $Z_\text{eff} \approx Z - S$ (where $S$ is the shielding constant). Across a period, $Z$ increases faster than shielding, so $Z_\text{eff}$ rises left to right.= 价电子在内层（芯）电子屏蔽掉部分核吸引力后感受到的净正电荷。$Z_\text{eff} \approx Z - S$（$S$ 为屏蔽常数）。在同一周期中，$Z$ 增加的速度快于屏蔽的增加，因此 $Z_\text{eff}$ 从左到右升高。
  • Atomic radius原子半径 : decreases across a period (higher $Z_\text{eff}$ pulls electrons closer); increases down a group (new shells added).：在同一周期减小（更高的 $Z_\text{eff}$ 把电子拉得更近）；在同一族向下增大（增加新壳层）。
  • Ionisation energy (IE)电离能（IE） : energy to remove the outermost electron. Increases across a period (higher $Z_\text{eff}$, electron harder to remove); decreases down a group (valence electron farther from nucleus).：移除最外层电子所需的能量。在同一周期增大（更高的 $Z_\text{eff}$，电子更难移除）；在同一族向下减小（价电子离核更远）。
  • Electronegativity电负性 : an atom's ability to attract electrons in a bond. Same trends as IE: highest at top-right (F is highest of all), lowest at bottom-left.：原子在键中吸引电子的能力。趋势与 IE 相同：右上角最高（F 是所有元素中最高的），左下角最低。
NGSS HS-PS1-1 grounds all these trends in "patterns of electrons in the outermost energy level." SCH3U B3.3 names "periodic law" and patterns in atomic radius, ionization energy, electron affinity, and electronegativity as assessed content. AB Chemistry 20 GO2 requires using the periodic table and electron dot diagrams to explain bonding trends.NGSS HS-PS1-1 把所有这些趋势建立在"最外能级电子的规律"上。SCH3U B3.3 把"周期律"以及原子半径、电离能、电子亲和能和电负性的规律列为被评估内容。AB Chemistry 20 GO2 要求用元素周期表和电子点图解释成键趋势。
Worked Example 7 · Comparing periodic trends例题 7 · 比较周期趋势

Rank the following in order of increasing atomic radius: Na ($Z=11$), Mg ($Z=12$), K ($Z=19$), Ca ($Z=20$). Explain your reasoning using effective nuclear charge and shell number.按增大的原子半径顺序排列以下元素：Na（$Z=11$）、Mg（$Z=12$）、K（$Z=19$）、Ca（$Z=20$）。用有效核电荷和壳层数解释你的推理。

Identify periods and groups.确定周期和族。 Na and Mg are in Period 3 (3 shells); K and Ca are in Period 4 (4 shells). K is directly below Na (Group 1); Ca is directly below Mg (Group 2).Na 和 Mg 在第 3 周期（3 个壳层）；K 和 Ca 在第 4 周期（4 个壳层）。K 在 Na 正下方（第 1 族）；Ca 在 Mg 正下方（第 2 族）。

Within a period.在同一周期内。 Na ($Z=11$, $Z_\text{eff}$ lower) $>$ Mg ($Z=12$, $Z_\text{eff}$ higher): same 3 shells, but Mg pulls valence electrons tighter. So $r(\text{Na}) > r(\text{Mg})$, and $r(\text{K}) > r(\text{Ca})$.Na（$Z=11$，$Z_\text{eff}$ 较低）$>$ Mg（$Z=12$，$Z_\text{eff}$ 较高）：相同的 3 个壳层，但 Mg 把价电子拉得更紧。故 $r(\text{Na}) > r(\text{Mg})$，$r(\text{K}) > r(\text{Ca})$。

Down a group.在同一族向下。 K has 4 shells vs Na's 3: $r(\text{K}) > r(\text{Na})$; same for Ca vs Mg.K 有 4 个壳层而 Na 只有 3 个：$r(\text{K}) > r(\text{Na})$；Ca 与 Mg 同理。

Ranking (smallest to largest).排名（从小到大）。

$$ r(\text{Mg}) < r(\text{Na}) < r(\text{Ca}) < r(\text{K}). $$
Which element has the highest first ionisation energy among Na, Al, Cl, and Ar (all in Period 3)?在 Na、Al、Cl 和 Ar（均在第 3 周期）中，哪种元素的第一电离能最高？
§7 · Q1
NaNa
ArAr
AlAl
ClCl
Ionisation energy increases across a period. Moving left to right in Period 3: Na, Al, Cl, Ar — Ar has the highest $Z_\text{eff}$ and a full outer shell, making it the hardest to remove an electron from. (Noble gases have exceptionally high IE.)电离能在同一周期内从左到右增大。在第 3 周期从左到右：Na、Al、Cl、Ar——Ar 的 $Z_\text{eff}$ 最高且外层充满，使其最难被移走一个电子。（惰性气体有极高的电离能。）
IE increases left to right in a period (increasing $Z_\text{eff}$). Na is leftmost (lowest IE); Ar is rightmost (highest IE). Na has the lowest IE of the four.电离能在同一周期从左到右增大（$Z_\text{eff}$ 增大）。Na 最靠左（IE 最低）；Ar 最靠右（IE 最高）。Na 的 IE 在这四个元素中最低。
Why does atomic radius decrease across a period (left to right)?为什么原子半径在同一周期从左到右减小？
§7 · Q2
More neutrons are added, compressing the nucleus增加了更多中子，压缩了原子核
Electrons are added to smaller inner shells电子被加入到更小的内层壳层中
Increasing effective nuclear charge pulls the valence electrons closer to the nucleus增大的有效核电荷将价电子拉向更靠近原子核的位置
Each new element has fewer electrons than the previous one每种新元素的电子数比前一种少
Across a period, electrons are added to the same shell but the nuclear charge increases each step. The shielding from core electrons stays roughly constant, so $Z_\text{eff}$ rises. The higher $Z_\text{eff}$ attracts valence electrons more strongly, pulling the electron cloud inward and shrinking the radius.在同一周期中，电子被加入同一壳层，但核电荷每步递增。芯电子的屏蔽保持大致不变，因此 $Z_\text{eff}$ 升高。更高的 $Z_\text{eff}$ 更强地吸引价电子，将电子云向内拉，从而缩小半径。
Neutrons don't affect electron distribution. Across a period, electrons go into the same shell (not inner ones), and the count increases. The key is rising $Z_\text{eff}$: more protons attract the same-shell electrons more strongly.中子不影响电子分布。在同一周期中，电子进入同一壳层（而非内层），且数量增加。关键在于 $Z_\text{eff}$ 升高：更多质子更强地吸引同壳层电子。
Exam Tactics考试策略

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

Setup discipline for every atomic-structure question每道原子结构题的解题纪律
Bohr model and spectra (§3, §6)玻尔模型与光谱（§3、§6）
  • Photon energy equals the level gap.光子能量等于能级间隔。 $\Delta E = E_\text{high} - E_\text{low}$ (always positive since $E_\text{high} > E_\text{low}$, i.e. less negative). Then $E = h\nu = hc/\lambda$. Higher energy gap $\Rightarrow$ shorter wavelength.$\Delta E = E_\text{high} - E_\text{low}$（始终为正，因为 $E_\text{high} > E_\text{low}$，即负值较小）。然后 $E = h\nu = hc/\lambda$。更大的能量间隔 $\Rightarrow$ 更短的波长。
  • Emission vs absorption.发射与吸收。 Electron falls down $\Rightarrow$ photon emitted (bright line). Electron absorbs $\Rightarrow$ jumps up (dark line in absorption spectrum). Same wavelengths appear in both.电子向下跃迁 $\Rightarrow$ 发射光子（亮线）。电子吸收 $\Rightarrow$ 向上跃迁（吸收光谱中的暗线）。两种光谱的波长相同。
Electron configuration (§4–§5) Honors — US NGSS / ON SCH3U电子排布（§4–§5）荣誉 — US NGSS / ON SCH3U
  • Know the fill order cold.熟记填充顺序。 $1s,\ 2s,\ 2p,\ 3s,\ 3p,\ 4s,\ 3d,\ 4p,\ 5s,\ 4d,\ 5p, \ldots$ Note that $4s$ fills before $3d$.$1s,\ 2s,\ 2p,\ 3s,\ 3p,\ 4s,\ 3d,\ 4p,\ 5s,\ 4d,\ 5p, \ldots$ 注意 $4s$ 在 $3d$ 之前填充。
  • Cr and Cu are the two classic exceptions.Cr 和 Cu 是两个经典例外。 Both "steal" one electron from $4s$ to achieve a half-filled ($3d^5$) or full ($3d^{10}$) d subshell. Memorise: Cr is $[\text{Ar}]\, 3d^5\, 4s^1$; Cu is $[\text{Ar}]\, 3d^{10}\, 4s^1$.两者都从 $4s$ "借走"一个电子以实现半充满（$3d^5$）或全充满（$3d^{10}$）的 d 亚壳层。记住：Cr 是 $[\text{Ar}]\, 3d^5\, 4s^1$；Cu 是 $[\text{Ar}]\, 3d^{10}\, 4s^1$。
  • Apply all three rules in order.按顺序应用三条规则。 Aufbau (fill low to high), Pauli (max 2 per orbital, opposite spins), Hund (one per degenerate orbital before pairing). Missing a rule in an orbital-box diagram loses marks.构建原理（从低到高填充）、泡利（每个轨道最多 2 个，自旋相反）、洪特（简并轨道先各放一个再配对）。在轨道方格图中漏掉一条规则会失分。
Periodic trends (§7) and answer hygiene周期趋势（§7）与作答规范
  • Always explain with $Z_\text{eff}$ and shielding.始终用 $Z_\text{eff}$ 和屏蔽效应解释。 Stating a trend without a reason earns partial credit at best. Across a period: same shell, rising $Z_\text{eff}$ → radius shrinks, IE rises. Down a group: new shell → radius grows, IE falls.仅陈述趋势而不给出原因最多得部分分。在同一周期：同一壳层，$Z_\text{eff}$ 升高 → 半径缩小，IE 升高。在同一族向下：新壳层 → 半径增大，IE 降低。
  • Significant figures for atomic mass calculations.原子质量计算的有效数字。 Match the least number of decimal places in the isotopic masses given. Keep at least 3 significant figures in your answer.与给定同位素质量中最少的小数位数对齐。答案中保留至少 3 位有效数字。
  • Know what each quantum number controls.了解每个量子数控制什么。 $n$ = shell (energy); $\ell$ = subshell (shape); $m_\ell$ = orbital orientation; $m_s$ = spin. The Pauli exclusion principle says all four must differ between any two electrons in the same atom.$n$ = 壳层（能量）；$\ell$ = 亚壳层（形状）；$m_\ell$ = 轨道取向；$m_s$ = 自旋。泡利不相容原理指出，同一原子中任意两个电子的所有四个量子数必须不同。
Active Recall主动复习

Flashcards闪卡

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Atomic number $Z$?原子序数 $Z$？
Number of protons in the nucleus. Defines the element. In a neutral atom, electrons $= Z$.原子核中的质子数。决定元素种类。在中性原子中，电子数 $= Z$。
Mass number $A$?质量数 $A$？
$$A = Z + N$$ Protons + neutrons. Neutrons $= A - Z$.质子数 + 中子数。中子数 $= A - Z$。
Isotopes?同位素？
Same element (same $Z$), different neutron count $N$, so different mass number $A$. Chemical properties are nearly identical.相同元素（相同 $Z$），不同中子数 $N$，故质量数 $A$ 不同。化学性质几乎相同。
Average atomic mass formula?平均原子质量公式？
$$\bar{m} = \sum_i m_i \times f_i$$ $m_i$ = isotopic mass, $f_i$ = fractional abundance (sum to 1)$m_i$ = 同位素质量，$f_i$ = 分数丰度（总和为 1）
Bohr energy levels (hydrogen)?玻尔能级（氢）？
$$E_n = -\frac{13.6\,\text{eV}}{n^2}$$ Ground state $n=1$; higher $n$ = higher energy; $E \to 0$ at ionisation.基态 $n=1$；$n$ 越大能量越高；电离时 $E \to 0$。
Photon energy from electron transition?电子跃迁释放的光子能量？
$$E = h\nu = \frac{hc}{\lambda} = \Delta E_\text{levels}$$ Larger gap $\Rightarrow$ higher $E$ $\Rightarrow$ shorter $\lambda$.间隔越大 $\Rightarrow$ $E$ 越高 $\Rightarrow$ $\lambda$ 越短。
s, p, d, f: how many orbitals each?s、p、d、f 各有多少轨道？
s: 1 (2 e⁻), p: 3 (6 e⁻), d: 5 (10 e⁻), f: 7 (14 e⁻).s: 1（2 e⁻），p: 3（6 e⁻），d: 5（10 e⁻），f: 7（14 e⁻）。
Max electrons per shell?每个壳层最多容纳多少电子？
$$2n^2$$ Shell 1: 2; Shell 2: 8; Shell 3: 18; Shell 4: 32.第 1 壳层：2；第 2 壳层：8；第 3 壳层：18；第 4 壳层：32。
Aufbau principle?构建原理？
Fill orbitals from lowest energy up: $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, \ldots$ Note $4s$ fills before $3d$.从最低能量起向上填充：$1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, \ldots$ 注意 $4s$ 在 $3d$ 前填充。
Pauli exclusion principle?泡利不相容原理？
No two electrons in an atom can have the same four quantum numbers. Max 2 electrons per orbital, with opposite spins (↑↓).原子中没有两个电子能拥有完全相同的四个量子数。每个轨道最多 2 个电子，自旋相反（↑↓）。
Hund's rule?洪特规则？
Fill degenerate orbitals with one electron each (same spin) before any pairing. Maximises unpaired electrons.在配对之前，每个简并轨道各放一个电子（相同自旋）。使未配对电子数最大化。
Atomic radius trend across a period?同一周期原子半径趋势？
Decreases left to right: same shell, rising $Z_\text{eff}$ pulls electrons closer. Increases down a group: new shell added.从左到右减小：相同壳层，$Z_\text{eff}$ 升高将电子拉近。在同一族向下增大：增加新壳层。
Ionisation energy trend?电离能趋势？
Increases left to right (rising $Z_\text{eff}$, electron harder to remove). Decreases down a group (valence electron farther from nucleus).从左到右增大（$Z_\text{eff}$ 升高，电子更难移除）。在同一族向下减小（价电子离核更远）。
Emission spectrum: what and why?发射光谱：是什么，为什么？
Bright lines on dark background. Electrons in excited gas fall to lower levels, releasing photons of specific $\Delta E$. Each element has a unique set.暗背景上的亮线。激发气体中的电子落回低能级，释放具有特定 $\Delta E$ 的光子。每种元素的发射光谱是唯一的。
Mixed Practice综合练习

Practice Quiz综合测验

An atom of iron has $Z = 26$ and $A = 56$. How many neutrons does it have?铁原子的 $Z = 26$，$A = 56$。它有多少个中子？
Q1
3030
2626
5656
8282
$N = A - Z = 56 - 26 = 30$ neutrons. The atomic number (26) counts the protons; the rest of the mass number is neutrons.$N = A - Z = 56 - 26 = 30$ 个中子。原子序数（26）是质子数；质量数的其余部分是中子数。
Neutrons $= A - Z$. The subscript is protons ($Z$); the superscript is the total of protons + neutrons ($A$).中子数 $= A - Z$。下标是质子数（$Z$）；上标是质子数 + 中子数的总和（$A$）。
Silicon has two common isotopes: ${}^{28}\text{Si}$ (mass $27.977\ \mathrm{u}$, abundance $92.23\%$) and ${}^{29}\text{Si}$ (mass $28.976\ \mathrm{u}$, abundance $4.67\%$) and ${}^{30}\text{Si}$ (mass $29.974\ \mathrm{u}$, abundance $3.10\%$). Which of these is closest to the average atomic mass?硅有三种常见同位素：${}^{28}\text{Si}$（质量 $27.977\ \mathrm{u}$，丰度 $92.23\%$）、${}^{29}\text{Si}$（质量 $28.976\ \mathrm{u}$，丰度 $4.67\%$）和 ${}^{30}\text{Si}$（质量 $29.974\ \mathrm{u}$，丰度 $3.10\%$）。下列哪个值最接近平均原子质量？
Q2
$27.977\ \mathrm{u}$
$28.98\ \mathrm{u}$
$28.09\ \mathrm{u}$
$29.00\ \mathrm{u}$
$(27.977)(0.9223) + (28.976)(0.0467) + (29.974)(0.0310) = 25.802 + 1.353 + 0.929 = 28.084 \approx 28.09\ \mathrm{u}$. Because ${}^{28}\text{Si}$ dominates at $92\%$, the average is pulled strongly toward $28$.$(27.977)(0.9223) + (28.976)(0.0467) + (29.974)(0.0310) = 25.802 + 1.353 + 0.929 = 28.084 \approx 28.09\ \mathrm{u}$。由于 ${}^{28}\text{Si}$ 在 $92\%$ 中占主导，平均值被强烈拉向 $28$。
Multiply each isotopic mass by its decimal abundance, then sum. Since ${}^{28}\text{Si}$ has over $92\%$ abundance, the average must be just above $28$, not $29$ or $30$.将每种同位素质量乘以其小数丰度，然后求和。由于 ${}^{28}\text{Si}$ 的丰度超过 $92\%$，平均值必须略高于 $28$，而非 $29$ 或 $30$。
In the Bohr model of hydrogen, which transition emits the photon with the most energy?在氢的玻尔模型中，哪种跃迁发射能量最大的光子？
Q3
$n = 3 \to n = 2$$n = 3 \to n = 2$
$n = 4 \to n = 3$$n = 4 \to n = 3$
$n = 2 \to n = 1$$n = 2 \to n = 1$
$n = 5 \to n = 1$$n = 5 \to n = 1$
The biggest energy gap is the fall all the way to the ground state ($n=1$). Among these, $n=5 \to n=1$ spans a larger gap than $n=2 \to n=1$: $\Delta E = -13.6/25 - (-13.6) = 13.06\ \mathrm{eV}$ vs $10.2\ \mathrm{eV}$ for $n=2 \to n=1$.最大的能量间隔是一路跌落到基态（$n=1$）。在这些选项中，$n=5 \to n=1$ 跨越的间隔大于 $n=2 \to n=1$：$\Delta E = -13.6/25 - (-13.6) = 13.06\ \mathrm{eV}$，而 $n=2 \to n=1$ 为 $10.2\ \mathrm{eV}$。
Photon energy equals the level gap. The largest gap involves the ground state ($n=1$), and the higher the starting $n$, the bigger the gap. $n=5 \to n=1$ is the largest gap offered.光子能量等于能级间隔。最大的间隔涉及基态（$n=1$），起始 $n$ 越高，间隔越大。$n=5 \to n=1$ 是提供的最大间隔。
What is the electron configuration of chlorine ($Z = 17$)?氯（$Z = 17$）的电子排布是什么？
Q4
$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^4\ 3d^1$
$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^5$
$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6$
$1s^2\ 2s^2\ 2p^5\ 3s^2\ 3p^6$
$17$ electrons: $1s^2$ (2), $2s^2$ (4), $2p^6$ (10), $3s^2$ (12), $3p^5$ (17). Total $= 17$ ✓. Chlorine has $7$ valence electrons, consistent with Group 17 (VIIA).$17$ 个电子：$1s^2$（2）、$2s^2$（4）、$2p^6$（10）、$3s^2$（12）、$3p^5$（17）。总计 $= 17$ ✓。氯有 $7$ 个价电子，与第 17 族（VIIA 族）一致。
Count to 17: 1s² (2), 2s² (4), 2p⁶ (10), 3s² (12) — that's 12, so 5 more go into 3p. 3p⁵ not 3p⁶ (that would be 18 electrons = argon). No d electrons yet.数到 17：1s²（2），2s²（4），2p⁶（10），3s²（12）——共 12 个，再 5 个放入 3p。是 3p⁵ 而非 3p⁶（那将是 18 个电子 = 氩）。尚无 d 电子。
Nitrogen has the configuration $1s^2\, 2s^2\, 2p^3$. How many unpaired electrons does it have?氮的排布为 $1s^2\, 2s^2\, 2p^3$。它有多少个未配对电子？
Q5
00
11
33
55
By Hund's rule, the 3 electrons in $2p$ each occupy a separate orbital before pairing: $2p_x^{\uparrow}\ 2p_y^{\uparrow}\ 2p_z^{\uparrow}$. All three are unpaired, giving nitrogen $3$ unpaired electrons. This makes nitrogen paramagnetic.根据洪特规则，$2p$ 中的 3 个电子在配对之前各占一个独立轨道：$2p_x^{\uparrow}\ 2p_y^{\uparrow}\ 2p_z^{\uparrow}$。三个均未配对，使氮有 $3$ 个未配对电子。这使氮具有顺磁性。
Hund's rule: the three 2p orbitals each get one electron first, no pairing yet. So all three 2p electrons are unpaired. The 1s² and 2s² electrons are paired (0 unpaired each).洪特规则：三个 2p 轨道各先放一个电子，尚未配对。因此三个 2p 电子均未配对。1s² 和 2s² 电子已配对（各 0 个未配对）。
Which element has the smallest atomic radius: Li, Na, K, or Rb?在 Li、Na、K 和 Rb 中，哪种元素的原子半径最小？
Q6
LiLi
NaNa
KK
RbRb
All four are in Group 1 (alkali metals). Atomic radius increases going down a group (each successive element adds a new shell). Li is in Period 2 (2 shells) — the fewest shells, hence the smallest radius.这四种元素都在第 1 族（碱金属）。原子半径在同一族向下增大（每种后续元素增加一个新壳层）。Li 在第 2 周期（2 个壳层）——壳层数最少，因此半径最小。
Down Group 1: Li (2 shells) → Na (3 shells) → K (4 shells) → Rb (5 shells). More shells = larger radius. Li at the top has the smallest radius.沿第 1 族向下：Li（2 个壳层）→ Na（3 个壳层）→ K（4 个壳层）→ Rb（5 个壳层）。壳层越多 = 半径越大。顶部的 Li 半径最小。
Fluorine (F) has the highest electronegativity of all elements. Which factor best explains this? 🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B3.3氟（F）是所有元素中电负性最高的。哪个因素最能解释这一点？🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B3.3
Q7
F has the most neutrons, giving it a large nuclear massF 的中子数最多，赋予其较大的核质量
F has more total electrons than most elementsF 的总电子数比大多数元素多
F is in the second period, so its electrons are far from the nucleusF 在第 2 周期，所以其电子距核较远
F has a high $Z_\text{eff}$ and a small atomic radius, so it attracts shared electrons very stronglyF 的 $Z_\text{eff}$ 高且原子半径小，因此它对共享电子的吸引力极强
F is at the top right of the periodic table (Period 2, Group 17), giving it a very high $Z_\text{eff}$ (strong nuclear pull) and a tiny atomic radius (electrons are close to the nucleus). Both factors make it extremely effective at attracting bonding electrons — hence the highest electronegativity (Pauling scale: 3.98).F 位于元素周期表的右上角（第 2 周期，第 17 族），赋予它极高的 $Z_\text{eff}$（强核吸引力）和极小的原子半径（电子靠近原子核）。这两个因素使其在吸引成键电子方面极为有效——因此电负性最高（泡利标度：3.98）。
Electronegativity is driven by nuclear pull (high $Z$) and small radius (close contact with shared electrons), not neutron count or total electron count. F has only 9 electrons — it's relatively small.电负性由核吸引力（高 $Z$）和小半径（与共享电子接触紧密）驱动，而非中子数或总电子数。F 只有 9 个电子——相对较小。
Which orbital subshell is filled immediately after $3p$ in the aufbau sequence? 🇨🇦 SCH4U C2.2 / BC Chemistry 11在构建顺序中，$3p$ 之后立即填充哪个亚壳层？🇨🇦 SCH4U C2.2 / BC Chemistry 11
Q8
$3d$$3d$
$4s$$4s$
$4p$$4p$
$3f$$3f$
The aufbau fill order is: $1s, 2s, 2p, 3s, 3p, \mathbf{4s}, 3d, 4p, \ldots$ The $4s$ orbital is lower in energy than $3d$, so it fills first. This is why potassium ($Z=19$) has configuration $[\text{Ar}]\, 4s^1$ rather than $[\text{Ar}]\, 3d^1$.构建顺序为：$1s, 2s, 2p, 3s, 3p, \mathbf{4s}, 3d, 4p, \ldots$ $4s$ 轨道能量低于 $3d$，所以先填充。这就是为什么钾（$Z=19$）的排布是 $[\text{Ar}]\, 4s^1$ 而不是 $[\text{Ar}]\, 3d^1$。
The key surprise in aufbau: $4s$ comes before $3d$. After $3p$ completes (18 electrons), the next electron goes into $4s$ (potassium), then $4s$ fills completely (calcium), then $3d$ begins (scandium).构建顺序的关键惊喜：$4s$ 在 $3d$ 之前。$3p$ 完成后（18 个电子），下一个电子进入 $4s$（钾），然后 $4s$ 完全填满（钙），然后 $3d$ 开始（钪）。
Which series of hydrogen's emission spectrum is in the visible light range? 🇨🇦 SCH4U C2.1氢的发射光谱中哪个系列在可见光范围内？🇨🇦 SCH4U C2.1
Q9
Lyman series (transitions to $n = 1$)莱曼系（跃迁到 $n = 1$）
Paschen series (transitions to $n = 3$)帕邢系（跃迁到 $n = 3$）
Balmer series (transitions to $n = 2$)巴尔末系（跃迁到 $n = 2$）
Brackett series (transitions to $n = 4$)布拉克特系（跃迁到 $n = 4$）
The Balmer series ($n \to 2$) produces visible wavelengths (approximately $400$–$700$ nm): H-alpha (red, $656$ nm), H-beta (cyan, $486$ nm), H-gamma (violet, $434$ nm). The Lyman series ($n \to 1$) is ultraviolet; Paschen and Brackett are infrared.巴尔末系（$n \to 2$）产生可见波长（约 $400$–$700$ nm）：H-alpha（红色，$656$ nm）、H-beta（青色，$486$ nm）、H-gamma（紫色，$434$ nm）。莱曼系（$n \to 1$）是紫外线；帕邢系和布拉克特系是红外线。
Lyman ($n\to1$) has the largest energy gaps — ultraviolet. Balmer ($n\to2$) gaps match visible light energies. Paschen ($n\to3$) and beyond are infrared (smaller gaps, lower-energy photons, longer wavelengths).莱曼系（$n\to1$）的能量间隔最大——紫外线。巴尔末系（$n\to2$）的间隔与可见光能量匹配。帕邢系（$n\to3$）及以上是红外线（间隔更小，光子能量更低，波长更长）。
Using the periodic table, which element has the highest ionisation energy in Period 2 (Li through Ne)? 🇺🇸 NGSS HS-PS1-1 / 🇨🇦 AB Chem 20 GO2利用元素周期表，第 2 周期（Li 到 Ne）中哪种元素的电离能最高？🇺🇸 NGSS HS-PS1-1 / 🇨🇦 AB Chem 20 GO2
Q10
LiLi
NeNe
FF
NN
Ionisation energy increases left to right across a period. Ne is at the far right of Period 2 and has the highest $Z_\text{eff}$ of the group, plus a completely filled outer shell (very stable). Ne has the highest IE in Period 2 ($2080\ \mathrm{kJ/mol}$).电离能在同一周期从左到右增大。Ne 在第 2 周期最右端，具有该组中最高的 $Z_\text{eff}$，加上完全充满的外层（非常稳定）。Ne 在第 2 周期的电离能最高（$2080\ \mathrm{kJ/mol}$）。
IE increases left to right. Li (leftmost) has the lowest IE; Ne (rightmost, noble gas) has the highest. Note: F actually has a slightly lower IE than expected due to electron-electron repulsion in the half-filled 2p, but Ne's full shell gives the maximum.电离能从左到右增大。Li（最左边）的 IE 最低；Ne（最右边，惰性气体）的 IE 最高。注：由于半充满 2p 中电子间排斥，F 的 IE 实际上略低于预期，但 Ne 的充满壳层给出最大值。
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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

Atomic structure is the conceptual root of all of chemistry. Every subsequent topic — chemical bonding (why atoms share or transfer electrons), the periodic table (why properties repeat by period), stoichiometry (why one mole of any element contains the same number of atoms), and spectroscopy — traces back to the electron arrangement you mastered here. The cross-references below point at the college-credit feeder and the next High School Chemistry unit.原子结构是化学所有概念的根基。每一个后续主题——化学键（为什么原子共享或转移电子）、元素周期表（为什么性质按周期重复）、化学计量学（为什么任何元素的一摩尔含有相同数量的原子）和光谱学——都可以追溯到你在这里掌握的电子排布。以下链接指向大学学分衔接课程和下一个高中化学单元。

Within High School Chemistry.在 HS Chemistry 内部。

The next unit, Periodic Table and Periodic Trends, extends §7 of this guide — the same $Z_\text{eff}$, shielding, and radius / IE / electronegativity framework is deepened with the d-block and transition-metal exceptions. Chemical Bonding (Unit 3) requires you to know valence-electron counts from §5 configuration to draw Lewis structures and predict bond types. Electron configuration is the backbone of every unit that follows.下一个单元《元素周期表与周期趋势》延伸了本指南的 §7——相同的 $Z_\text{eff}$、屏蔽效应以及半径/IE/电负性框架将通过 d 区和过渡金属例外加以深化。《化学键》（第 3 单元）要求你知道来自 §5 排布的价电子数，以绘制路易斯结构并预测键的类型。电子排布是后续每个单元的骨干。

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

IB Chemistry HL · Structure 1: Particulate Nature of Matter (the college-credit feeder for atomic structure, subatomic particles, isotopes, and the full quantum-mechanical model at IB depth)IB Chemistry HL · Structure 1：物质的微粒本质（原子结构、亚原子粒子、同位素以及 IB 深度下完整量子力学模型的大学学分衔接）

If you are aiming for IB Chemistry HL or AP Chemistry, the electron configuration, orbital model, and emission spectra material here is assumed from the first week of the college-credit course. IB Chemistry HL Structure 1 extends this with quantum numbers, electron density maps, and the full treatment of ionisation energies (including successive IE data and graphical analysis). AP Chemistry Unit 1 (Atomic Structure and Properties) adds mass spectrometry, photoelectron spectroscopy, and Coulomb's law framing. The electron configuration you write here is the same one used to determine bond order, molecular geometry (VSEPR), and molecular polarity in every subsequent unit.如果你目标是 IB Chemistry HL 或 AP Chemistry，这里的电子排布、轨道模型和发射光谱材料从大学学分课程的第一周就被默认掌握。IB Chemistry HL Structure 1 通过量子数、电子密度图和电离能的完整处理（包括连续电离能数据和图形分析）来延伸这部分内容。AP Chemistry Unit 1（原子结构与性质）增加了质谱法、光电子能谱和库仑定律框架。你在这里写出的电子排布与后续每个单元中确定键级、分子几何（VSEPR）和分子极性所用的排布完全相同。

DINGRUI SCHOLARS

High School Chemistry · Atomic Structure and the Quantum Model · v1高中化学 · 原子结构与量子模型 · v1