High School Chemistry

The Periodic Table and Periodic Trends周期表与周期性

The periodic table is chemistry's master map: every element's position encodes its electron count, and that position predicts radius, ionization energy, electron affinity, electronegativity, and reactivity. This guide walks through the table's organization and history, the block structure (s, p, d, f), and the seven periodic trends — each explained through effective nuclear charge $Z_\text{eff}$ and shielding. Reactivity patterns in Groups 1, 17, and 18 close the unit. Worked examples and KaTeX formulas are used throughout.周期表（periodic table，元素周期表）是化学的总体地图：每种元素的位置编码了其电子数，而该位置能预测原子半径（atomic radius，原子半径）、电离能（ionization energy，电离能）、电子亲和能（electron affinity，电子亲和能）、电负性（electronegativity，电负性）和反应活性。本指南系统介绍周期表的组织结构与历史、s/p/d/f 区块划分，以及七大周期规律——每一项均通过有效核电荷 $Z_\text{eff}$ 和屏蔽效应加以解释。第 1、17、18 族的反应活性趋势作为本单元收尾。全程使用例题与公式。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB Connects to: Chemical Bonding (Unit 3) and IB Chemistry HL Structure 2衔接：化学键（第 3 单元）与 IB Chemistry HL Structure 2
Read me first先读这里

How to use this guide如何使用本指南

The periodic table and periodic trends appear in every curriculum we map to, and the four curricula agree on a core scope: organization of the table, group and period trends in atomic radius, ionization energy, and electronegativity. They diverge on depth. US NGSS (HS-PS1-1) keeps ionization energy qualitative — "relative trends" only — and limits scope to main-group elements. Ontario SCH3U (B2.1, B2.2, B3.3) adds analytical data comparison and names electron affinity explicitly. BC Chemistry 11 integrates electronegativity into its bonding content. Alberta Chemistry 20 uses the periodic table to predict bonding in Unit A. The table below maps each curriculum to the seven sections; each row cites the extract used.元素周期表与周期趋势出现在我们所对照的每套大纲中，四套大纲在核心范围上一致：周期表的组织结构、原子半径、电离能和电负性的族与周期趋势。它们的分歧在于深度。US NGSS（HS-PS1-1）保持电离能的定性——仅要求"相对趋势"——并将范围限于主族元素。安大略 SCH3U（B2.1、B2.2、B3.3）加入分析性数据比较，并明确命名电子亲和能。BC Chemistry 11 将电负性整合到其成键内容中。阿尔伯塔 Chemistry 20 在 A 单元中使用元素周期表预测成键。下表将各大纲与七个章节对应；每行注明所用的提取文件。

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 (organization), §3 (radius), §4 (IE qualitative), §5 (electronegativity), §6 (metallic/nonmetallic), §7 (Group 1 and 17 reactivity) — all under HS-PS1-1 and HS-PS1-2. Main-group elements only; IE is qualitative (relative trends only).§1（组织结构）、§3（半径）、§4（IE 定性）、§5（电负性）、§6（金属性/非金属性）、§7（第 1、17 族反应活性）——均在 HS-PS1-1 和 HS-PS1-2 下。仅限主族元素；IE 是定性的（仅相对趋势）。 §2 (d-block detail) and quantitative $Z_\text{eff}$ calculations: valuable context but above the NGSS qualitative floor.§2（d 区细节）和 $Z_\text{eff}$ 定量计算：很有价值，但高于 NGSS 定性要求的下限。 NGSS HS-PS1 (Chemistry) — HS-PS1-1 and HS-PS1-2 PE + Clarification + Assessment Boundary— HS-PS1-1 与 HS-PS1-2 表现期望 + 澄清 + 评估边界
🇨🇦 ON SCH3U安大略 SCH3U All seven sections. B3.3 names periodic law + trends in atomic radius, IE, electron affinity, and electronegativity as specific expectations. B2.1 lists "atomic radius, effective nuclear charge, electronegativity, ionization energy, and electron affinity" as assessed terminology. B2.2 requires analysing trend data within a period.全部七节。B3.3 将周期律及原子半径、IE、电子亲和能和电负性趋势列为具体期望。B2.1 将"原子半径、有效核电荷、电负性、电离能和电子亲和能"列为被评估术语。B2.2 要求分析同一周期内的趋势数据。 SCH4U C3.3 s/p/d-block analysis (§2 deeper): honors-flag for SCH3U students.SCH4U C3.3 s/p/d 区分析（§2 深度）：SCH3U 学生标荣誉级。 Ontario SCH3U/4U Chemistry — SCH3U B2.1, B2.2, B3.3; SCH4U C3.3— SCH3U B2.1、B2.2、B3.3；SCH4U C3.3
🇨🇦 BC Chemistry 11BC Chemistry 11 All seven sections. Periodic trends are embedded in the Big Idea "Atoms and molecules are building blocks of matter" and in the elaborations "chemical bonding based on electronegativity" and "chemical bonding: Lewis structures of compounds, polarity." Electronegativity and atomic radius are core, not honors.全部七节。周期趋势嵌入大概念"原子和分子是物质的构建单元"及细化"基于电负性的化学键"和"化学键：化合物的路易斯结构、极性"中。电负性和原子半径是核心内容，不是荣誉级。 Nothing — BC integrates periodic trends throughout the Chemistry 11 bonding content.无 — BC 将周期趋势贯穿整个 Chemistry 11 成键内容。 BC Chemistry 11/12 — Chemistry 11 Big Idea; Content elaborations on electronegativity and bonding— Chemistry 11 大概念；电负性和成键细化内容
🇨🇦 AB Chemistry 20阿尔伯塔 Chemistry 20 All seven sections. Unit A Key Concepts list "electronegativity" and "valence electron" as central; knowledge outcomes include: "define valence electron, electronegativity, ionic bond"; "use the periodic table and electron dot diagrams to support and explain ionic bonding theory"; "describe bonding as a continuum ranging from complete electron transfer to equal sharing of electrons." Periodic trends directly feed bonding predictions.全部七节。A 单元关键概念将"电负性"和"价电子"列为核心；知识结果包括："定义价电子、电负性、离子键"；"用元素周期表和电子点图支持和解释离子成键理论"；"将成键描述为从完全电子转移到等量共享电子的连续体。"周期趋势直接支撑成键预测。 Nothing — AB Diploma-exam questions build directly on periodic-trend reasoning for bonding prediction.无 — AB 文凭考试直接考查用周期趋势推理预测成键。 Alberta Chemistry 20/30 — Chemistry 20 Unit A GO1/GO2, Key Concepts, knowledge outcome text— Chemistry 20 A 单元 GO1/GO2，关键概念，知识结果文本
🇺🇸 IB / AP feeder trackIB / AP 衔接轨道 All seven sections including the going-deeper derivations. IB Chemistry HL Structure 3 and AP Chemistry Unit 1 assume fluent periodic-trend reasoning and expect quantitative $Z_\text{eff}$, successive ionization energies, and ionic radius comparisons from the first week.全部七节，包括深入推导。IB Chemistry HL Structure 3 与 AP Chemistry Unit 1 第一周就默认熟练掌握周期趋势推理，并要求 $Z_\text{eff}$ 定量、连续电离能和离子半径比较。 Nothing — periodic trends are the conceptual foundation for all bonding, geometry, and reactivity units.无 — 周期趋势是所有成键、几何构型和反应活性单元的概念基础。 NGSS HS-PS1 (Chemistry) — see also the IB Chemistry HL Structure 1 feeder link in "What This Feeds Into"— 另见"本单元的去向"中的 IB Chemistry HL Structure 1 衔接链接

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。

!
If you are cramming the night before如果你在临阵磨枪

Memorise the four trend directions: atomic radius decreases across a period and increases down a group; ionization energy increases across a period and decreases down a group; electronegativity increases toward the upper right (F is highest); metallic character increases toward the lower left. Read every cram-cheat box. The single explanatory engine for all four: effective nuclear charge $Z_\text{eff}$.背熟四个趋势方向：原子半径在同一周期减小，在同一族向下增大；电离能在同一周期增大，在同一族向下减小；电负性向右上角增大（F 最高）；金属性向左下角增大。读每个速记框。所有四个趋势的唯一解释引擎：有效核电荷 $Z_\text{eff}$。

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

Be precise about $Z_\text{eff}$ vs $Z$; explain ionic radius anomalies (isoelectronic series); know the dips in the IE trend (B lower than Be; O lower than N) and give the quantum-mechanical reason; rank reactivity within and across groups using the periodic-trend framework; connect every trend to its consequence in bonding. SCH3U B2.2 and AB Chemistry 20 expect you to analyse data and give a reason, not just quote the trend direction.精确区分 $Z_\text{eff}$ 与 $Z$；解释离子半径异常（等电子系列）；了解 IE 趋势中的低谷（B 低于 Be；O 低于 N）并给出量子力学原因；用周期趋势框架排列同族内和跨族的反应活性；将每种趋势与其在成键中的后果联系起来。SCH3U B2.2 和 AB Chemistry 20 要求你分析数据并给出原因，而不仅仅是引用趋势方向。

Section 1 · Organization and history of the periodic table第 1 节 · 周期表的组织与历史

Organization and History of the Periodic Table周期表的组织与历史

The periodic table arranges elements by increasing atomic number so that elements with similar properties fall in the same column.元素周期表按原子序数递增的顺序排列元素，使性质相似的元素位于同一列。
  • Mendeleev (1869)门捷列夫（1869年） — arranged elements by atomic mass, left gaps for undiscovered elements, and predicted their properties. When those elements were found (Ga, Sc, Ge), the predictions matched. This validated the table as a predictive model — exactly what NGSS HS-PS1-1 calls it: "Use the periodic table as a model to predict the relative properties of elements."— 按原子质量排列元素，为未发现的元素留空，并预测其性质。当这些元素（Ga、Sc、Ge）被发现时，预测结果一致。这证明了周期表作为预测模型的有效性——正是 NGSS HS-PS1-1 所述："以元素周期表为模型预测元素的相对性质。"
  • Moseley (1913)莫塞莱（1913年） — used X-ray spectra to show atomic number $Z$ (proton count), not atomic mass, is the fundamental ordering principle. This resolved the anomalies in Mendeleev's table (e.g., Ar before K).— 用 X 射线光谱证明原子序数 $Z$（质子数）而非原子质量是基本排列原则。这解决了门捷列夫周期表中的异常（如 Ar 排在 K 之前）。
  • Periodic law周期律 — stated by SCH3U B3.3: "the physical and chemical properties of the elements are periodic functions of their atomic numbers." Repeat every 8 (periods 1–3) or 18 (periods 4–5) elements.— SCH3U B3.3 表述："元素的物理和化学性质是其原子序数的周期函数。"每 8 个（第 1–3 周期）或 18 个（第 4–5 周期）元素重复一次。
Worked Example 1 · Predicting a missing element (Mendeleev style)例题 1 · 预测缺失元素（门捷列夫方式）

In 1869 Mendeleev left a gap below silicon (Si) in his table. He predicted the missing element ("eka-silicon") would have: atomic mass $\approx 72\ \mathrm{u}$; density $\approx 5.5\ \mathrm{g/cm^3}$; gray metallic appearance; four valence electrons; formula of oxide EX$_2$O$_4$. Germanium (Ge, $Z = 32$) was discovered in 1886. Its actual properties: atomic mass $72.6\ \mathrm{u}$; density $5.32\ \mathrm{g/cm^3}$; gray metalloid; four valence electrons; formula GeO$_2$. Explain how the periodic table functioned as a predictive model here.1869 年门捷列夫在他的表中留下了硅（Si）下方的空格。他预测缺失元素（"类硅"）将具有：原子质量 $\approx 72\ \mathrm{u}$；密度 $\approx 5.5\ \mathrm{g/cm^3}$；灰色金属外观；四个价电子；氧化物分子式 EX$_2$O$_4$。锗（Ge，$Z = 32$）于 1886 年被发现。其实际性质：原子质量 $72.6\ \mathrm{u}$；密度 $5.32\ \mathrm{g/cm^3}$；灰色类金属；四个价电子；分子式 GeO$_2$。解释元素周期表在此如何作为预测模型。

Column position encodes valence electrons.列的位置编码价电子数。 Si (Group 14) has 4 valence electrons; the element below it must also have 4. Four valence electrons predict oxide formula EO$_2$ and tetravalent bonding — exactly GeO$_2$.Si（第 14 族）有 4 个价电子；其下方的元素也必须有 4 个。四个价电子预测氧化物分子式 EO$_2$ 和四价成键——与 GeO$_2$ 完全吻合。

Row position encodes shell count and approximate size/density.行的位置编码壳层数和近似大小/密度。 Below Si (Period 3) and Sn (Period 5), the unknown was Period 4 — intermediate size and density. Mendeleev interpolated from neighbors to predict density $\approx 5.5$ vs actual $5.32$ — a $3.4\%$ error.在 Si（第 3 周期）和 Sn（第 5 周期）之间，未知元素在第 4 周期——中等大小和密度。门捷列夫从相邻元素插值，预测密度 $\approx 5.5$，实际值 $5.32$——误差 $3.4\%$。

Conclusion.结论。 Periodic patterns in the table — not curve-fitting but structure from electron arrangement — allow quantitative predictions about undiscovered elements. This is the "model" invoked in NGSS HS-PS1-1 and stated as the periodic law in SCH3U B3.3.元素周期表中的周期规律——不是曲线拟合，而是来自电子排布的结构——允许对未发现元素进行定量预测。这就是 NGSS HS-PS1-1 所援引的"模型"，以及 SCH3U B3.3 所述的周期律。

What is the fundamental principle that determines an element's position in the modern periodic table?决定元素在现代元素周期表中位置的基本原则是什么？
§1 · Q1
Increasing atomic mass递增的原子质量
Increasing number of neutrons递增的中子数
Increasing atomic number (proton count)递增的原子序数（质子数）
Alphabetical order of element symbols元素符号的字母顺序
Moseley showed in 1913 that atomic number $Z$ (not atomic mass) is the true ordering principle. This fixed Mendeleev's anomalies (e.g., Ar, $Z=18$, properly precedes K, $Z=19$, even though Ar is slightly heavier).莫塞莱在 1913 年证明原子序数 $Z$（而非原子质量）是真正的排列原则。这修正了门捷列夫的异常（如 Ar，$Z=18$，正确地排在 K，$Z=19$ 之前，尽管 Ar 略重）。
Mendeleev used atomic mass, which worked mostly but had anomalies (Ar/K, Co/Ni). Moseley's X-ray work revealed atomic number $Z$ as the correct ordering. Neutron count is not the basis; it varies among isotopes of the same element.门捷列夫使用原子质量，大致有效但有异常（Ar/K、Co/Ni）。莫塞莱的 X 射线研究揭示了原子序数 $Z$ 才是正确的排列依据。中子数不是基础；同一元素的不同同位素中子数不同。
The periodic law states that properties of elements are periodic functions of their:周期律指出，元素的性质是以下哪项的周期函数：
§1 · Q2
Atomic numbers原子序数
Atomic masses原子质量
Neutron counts中子数
Electron masses电子质量
SCH3U B3.3 states: "physical and chemical properties of the elements are periodic functions of their atomic numbers." Every 8 or 18 elements (depending on the period), properties repeat because the valence-electron count repeats.SCH3U B3.3 规定："元素的物理和化学性质是其原子序数的周期函数。"每 8 或 18 个元素（取决于周期），性质重复，因为价电子数重复。
Mendeleev used atomic mass, but Moseley replaced it with atomic number $Z$. The modern periodic law is about atomic numbers, not masses. Mass periodicity would fail for isotope mixtures.门捷列夫使用原子质量，但莫塞莱用原子序数 $Z$ 替代了它。现代周期律是关于原子序数的，而非质量。质量周期性会在同位素混合物中失效。
Going deeper — why the table has the shape it does: period lengths and electron shells深入 — 为何周期表是现在的形状：周期长度与电子壳层

Period 1 has 2 elements (filling the 1s subshell: 2 electrons). Periods 2 and 3 have 8 elements each (filling s + p subshells: 2 + 6 = 8). Periods 4 and 5 have 18 elements each (s + d + p: 2 + 10 + 6 = 18). Periods 6 and 7 have 32 elements (s + f + d + p: 2 + 14 + 10 + 6 = 32). The formula is $2n^2$ electrons per shell — the same capacity formula from the quantum-mechanical model. SCH4U C3.3 explicitly links element position to electron configuration and block assignment, making the shape of the table a direct consequence of quantum mechanics.第 1 周期有 2 种元素（填充 1s 亚壳层：2 个电子）。第 2、3 周期各有 8 种元素（填充 s + p 亚壳层：2 + 6 = 8）。第 4、5 周期各有 18 种元素（s + d + p：2 + 10 + 6 = 18）。第 6、7 周期有 32 种元素（s + f + d + p：2 + 14 + 10 + 6 = 32）。公式为每个壳层 $2n^2$ 个电子——与量子力学模型中相同的容量公式。SCH4U C3.3 明确将元素位置与电子排布和区块分配联系起来，使周期表的形状成为量子力学的直接结果。

Section 2 · Periods, groups, and s/p/d/f blocks第 2 节 · 周期、族与 s/p/d/f 区块

Periods, Groups, and s/p/d/f Blocks周期（族）与 s/p/d/f 区块

Period = row (shell number); Group = column (valence-electron count for main-group elements).周期（period）= 行（壳层编号）；族（group）= 列（主族元素的价电子数）。
  • Period number周期数 = number of electron shells (e.g., Period 3 elements have electrons in shells 1, 2, and 3). Moving right across a period adds one proton and one electron to the same shell.= 电子壳层数（如第 3 周期元素在壳层 1、2、3 中有电子）。在同一周期向右移动，每步增加一个质子和一个电子到同一壳层。
  • Group number (IUPAC 1–18)族编号（IUPAC 1–18） For main-group (s and p block) elements: Group 1 has 1 valence electron; Group 18 has 8 (or 2 for He). Group number = valence electrons is the shortcut for Groups 1–2 and 13–18.对于主族（s 和 p 区）元素：第 1 族有 1 个价电子；第 18 族有 8 个（He 为 2 个）。对于第 1–2 族和第 13–18 族，族编号 = 价电子数是捷径。
  • s-block (Groups 1–2)s 区（第 1–2 族） — valence electrons in an s subshell. Alkali metals (1) and alkaline earth metals (2).— 价电子在 s 亚壳层中。碱金属（1）和碱土金属（2）。
  • p-block (Groups 13–18)p 区（第 13–18 族） — valence electrons filling p subshells. Includes metalloids, nonmetals, and noble gases.— 价电子填充 p 亚壳层。包括类金属、非金属和惰性气体。
  • d-block (Groups 3–12)d 区（第 3–12 族） — transition metals; d orbitals filling. SCH4U C3.3 names s, p, d block identification as assessed content.— 过渡金属；d 轨道正在填充。SCH4U C3.3 将 s、p、d 区识别列为被评估内容。
  • f-block (lanthanides and actinides)f 区（镧系和锕系） — f orbitals filling; placed separately below the main table. Not assessed in HS curricula beyond recognition.— f 轨道正在填充；单独放在主表下方。在高中大纲中仅要求认识，不作深度评估。
Block区块Groups族Filling subshell填充亚壳层Examples示例
s1–21–2nsNa, Ca, MgNa、Ca、Mg
p13–1813–18npC, O, Cl, NeC、O、Cl、Ne
d3–123–12(n−1)dFe, Cu, Zn, TiFe、Cu、Zn、Ti
fLa–Lu, Ac–LrLa–Lu、Ac–Lr(n−2)fCe, UCe、U
Phosphorus (P, $Z = 15$) is in Period 3, Group 15. How many valence electrons does it have?磷（P，$Z = 15$）在第 3 周期、第 15 族。它有多少个价电子？
§2 · Q1
33
55
1515
88
For main-group elements, the Group number (1–18 IUPAC) gives the valence electron count for Groups 13–18 by subtracting 10: Group 15 $- 10 = 5$ valence electrons. Equivalently, P has configuration $[\text{Ne}]\, 3s^2\, 3p^3$; the $3s^2 3p^3$ outer shell has 5 electrons.对于主族元素，第 13–18 族的族编号（IUPAC 1–18）减去 10 给出价电子数：第 15 族 $- 10 = 5$ 个价电子。等价地，P 的排布为 $[\text{Ne}]\, 3s^2\, 3p^3$；外层 $3s^2 3p^3$ 有 5 个电子。
For Groups 13–18: valence electrons = Group number $- 10$. Group 15: $15 - 10 = 5$. Do not confuse $Z = 15$ (total electrons) with valence electrons (just the outermost shell).对于第 13–18 族：价电子数 = 族编号 $- 10$。第 15 族：$15 - 10 = 5$。不要把 $Z = 15$（总电子数）与价电子数（仅最外层壳层）混淆。
Iron (Fe, $Z = 26$) belongs to which block of the periodic table? 🇨🇦 SCH4U C3.3铁（Fe，$Z = 26$）属于元素周期表的哪个区块？🇨🇦 SCH4U C3.3
§2 · Q2
s-blocks 区
p-blockp 区
d-blockd 区
f-blockf 区
Fe ($Z = 26$, Group 8) is a transition metal in the d-block (Groups 3–12). Its electron configuration is $[\text{Ar}]\, 3d^6\, 4s^2$ — the $3d$ subshell is being filled. SCH4U C3.3 requires identifying characteristic properties of s, p, and d block elements.Fe（$Z = 26$，第 8 族）是 d 区（第 3–12 族）的过渡金属。其电子排布为 $[\text{Ar}]\, 3d^6\, 4s^2$——$3d$ 亚壳层正在填充。SCH4U C3.3 要求识别 s、p 和 d 区元素的特征性质。
The d-block is Groups 3–12 (transition metals). Fe is in Group 8, Period 4 — firmly in the d-block. The s-block is Groups 1–2; p-block is Groups 13–18; f-block is the lanthanides/actinides.d 区是第 3–12 族（过渡金属）。Fe 在第 8 族、第 4 周期——牢固地在 d 区中。s 区是第 1–2 族；p 区是第 13–18 族；f 区是镧系/锕系。
Section 3 · Atomic and ionic radius trends第 3 节 · 原子半径与离子半径趋势

Atomic and Ionic Radius Trends原子半径与离子半径趋势

Two competing forces determine size: nuclear charge pulling in, additional shells pushing out.两种竞争力决定大小：核电荷向内拉，额外壳层向外推。
  • Effective nuclear charge $Z_\text{eff}$有效核电荷 $Z_\text{eff}$ = net nuclear charge felt by a valence electron after core electrons shield part of the attraction: $Z_\text{eff} \approx Z - S$ (S = shielding constant). Valence electrons in the same shell shield each other poorly; inner shells shield effectively.= 价电子在芯电子屏蔽掉部分吸引力后感受到的净核电荷：$Z_\text{eff} \approx Z - S$（S = 屏蔽常数）。同一壳层的价电子相互屏蔽效果差；内层壳层屏蔽效果好。
  • Across a period (left to right)在同一周期内（从左到右） : same shell, proton count rises, so $Z_\text{eff}$ increases. The valence electrons are pulled inward. Atomic radius decreases.：相同壳层，质子数增加，因此 $Z_\text{eff}$ 升高。价电子被向内拉。原子半径减小。
  • Down a group在同一族向下 : a new shell is added, dramatically increasing atomic radius despite rising $Z$. Radius increases.：增加一个新壳层，尽管 $Z$ 升高，原子半径也急剧增大。半径增大。
  • Ionic radius离子半径 : cations (lost electrons) are smaller than the parent atom — fewer electrons, same $Z$, so $Z_\text{eff}$ per electron rises. Anions (gained electrons) are larger — more electrons, same $Z$, so repulsion expands the cloud. Isoelectronic series (same electron count): radius decreases as $Z$ increases (higher $Z_\text{eff}$ on the same electron cloud).：阳离子（失去电子）比母体原子小——电子更少，$Z$ 相同，故每个电子的 $Z_\text{eff}$ 升高。阴离子（获得电子）更大——电子更多，$Z$ 相同，故排斥膨胀了电子云。等电子体系列（相同电子数）：随 $Z$ 增大，半径减小（相同电子云上 $Z_\text{eff}$ 更高）。
NGSS HS-PS1-1 grounds radius in "patterns of electrons in the outermost energy level." SCH3U B2.1 lists "atomic radius" and "effective nuclear charge" as assessed terminology. AB Chemistry 20 GO2 knowledge outcome: "use the periodic table and electron dot diagrams to support and explain ionic bonding theory" — ionic radius comparisons underpin this.NGSS HS-PS1-1 将原子半径建立在"最外能级的电子规律"上。SCH3U B2.1 将"原子半径"和"有效核电荷"列为被评估术语。AB Chemistry 20 GO2 知识结果："用元素周期表和电子点图支持和解释离子成键理论"——离子半径比较是其基础。
Worked Example 3 · Ranking atomic and ionic radii例题 3 · 排列原子半径与离子半径

Rank the following species in order of increasing radius: $\text{Na}^+$, $\text{Mg}^{2+}$, $\text{F}^-$, $\text{O}^{2-}$, $\text{Ne}$. All are isoelectronic with 10 electrons.按增大的半径顺序排列以下粒子：$\text{Na}^+$、$\text{Mg}^{2+}$、$\text{F}^-$、$\text{O}^{2-}$、$\text{Ne}$。均与 10 个电子等电子。

All have 10 electrons.所有粒子均有 10 个电子。 The size difference comes entirely from $Z_\text{eff} = Z - S$, with $S$ approximately the same for all (same 10-electron core). Higher $Z$ → higher $Z_\text{eff}$ → smaller radius.大小差异完全来自 $Z_\text{eff} = Z - S$，所有粒子的 $S$ 近似相同（相同的 10 电子核心）。$Z$ 越高 → $Z_\text{eff}$ 越高 → 半径越小。

Atomic numbers:原子序数：

$\text{O}^{2-}\ (Z=8) \quad \text{F}^-\ (Z=9) \quad \text{Ne}\ (Z=10) \quad \text{Na}^+\ (Z=11) \quad \text{Mg}^{2+}\ (Z=12)$

Ranking (smallest to largest radius):排名（从小到大）：

$$ r(\text{Mg}^{2+}) < r(\text{Na}^+) < r(\text{Ne}) < r(\text{F}^-) < r(\text{O}^{2-}) $$

$\text{Mg}^{2+}$ ($Z=12$) has the highest $Z_\text{eff}$ on those 10 electrons, so the tightest cloud. $\text{O}^{2-}$ ($Z=8$) has the lowest $Z_\text{eff}$, so the most diffuse cloud.$\text{Mg}^{2+}$（$Z=12$）对那 10 个电子的 $Z_\text{eff}$ 最高，因此电子云最紧。$\text{O}^{2-}$（$Z=8$）的 $Z_\text{eff}$ 最低，因此电子云最弥散。

Which of the following has the largest atomic radius? 🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B2.1以下哪个粒子的原子半径最大？🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B2.1
§3 · Q1
Cl (Period 3, Group 17)Cl（第 3 周期，第 17 族）
Na (Period 3, Group 1)Na（第 3 周期，第 1 族）
F (Period 2, Group 17)F（第 2 周期，第 17 族）
K (Period 4, Group 1)K（第 4 周期，第 1 族）
K has the most electron shells (4) of these four elements and the lowest $Z_\text{eff}$ relative to shell count — Group 1 means only 1 valence electron, minimal effective pull. Na has 3 shells but is in the same group; K adds a 4th shell, dramatically increasing radius. K is the largest.K 在这四种元素中电子壳层最多（4 个），且相对于壳层数的 $Z_\text{eff}$ 最低——第 1 族意味着只有 1 个价电子，有效拉力最小。Na 有 3 个壳层但在同一族；K 增加了第 4 个壳层，使半径急剧增大。K 最大。
Radius increases down a group (more shells) and decreases across a period (rising $Z_\text{eff}$). K (Period 4, Group 1) has the largest radius: maximum shells, minimum $Z_\text{eff}$ per electron. Na (Period 3) is smaller; Cl pulls valence electrons tighter; F is smallest (Period 2, high $Z_\text{eff}$, few shells).半径在同一族向下增大（壳层更多），在同一周期减小（$Z_\text{eff}$ 升高）。K（第 4 周期，第 1 族）半径最大：壳层最多，每个电子的 $Z_\text{eff}$ 最小。Na（第 3 周期）更小；Cl 将价电子拉得更紧；F 最小（第 2 周期，高 $Z_\text{eff}$，壳层少）。
Which species is larger: Na or Na$^+$? Why?哪个粒子更大：Na 还是 Na$^+$？为什么？
§3 · Q2
Na, because losing an electron reduces electron-electron repulsion and the remaining electrons are pulled closer by the same nuclear chargeNa，因为失去一个电子减少了电子间排斥，剩余电子被相同核电荷拉得更近
Na$^+$, because the positive charge attracts surrounding electrons from other atomsNa$^+$，因为正电荷从其他原子吸引周围电子
They are the same size because they have the same nuclear charge它们大小相同，因为核电荷相同
Na$^+$, because the positive ion repels electron clouds of neighboring atoms and appears largerNa$^+$，因为正离子排斥相邻原子的电子云而显得更大
Na has 11 protons and 11 electrons; Na$^+$ has 11 protons and 10 electrons. Same $Z$, fewer electrons → $Z_\text{eff}$ per electron is higher in Na$^+$ → electrons are held closer → Na$^+$ is smaller. So Na (11e$^-$) > Na$^+$ (10e$^-$).Na 有 11 个质子和 11 个电子；Na$^+$ 有 11 个质子和 10 个电子。相同 $Z$，电子更少 → Na$^+$ 中每个电子的 $Z_\text{eff}$ 更高 → 电子被固定得更近 → Na$^+$ 更小。故 Na（11 e$^-$）$>$ Na$^+$（10 e$^-$）。
Cations are always smaller than their neutral parent. Same nuclear charge ($Z=11$) but fewer electrons in Na$^+$ means each remaining electron experiences a higher $Z_\text{eff}$, drawing the cloud tighter. This is why ionic radii matter for predicting bond lengths.阳离子总是比其中性母体更小。Na$^+$ 中核电荷相同（$Z=11$）但电子更少，意味着每个剩余电子经历更高的 $Z_\text{eff}$，将电子云拉得更紧。这就是为什么离子半径对预测键长很重要。
Section 4 · Ionization energy trend第 4 节 · 电离能趋势

Ionization Energy Trend电离能趋势

Ionization energy (IE) = energy needed to remove the outermost electron from a gas-phase atom.电离能（IE）= 从气相原子中移除最外层电子所需的能量。
  • Across a period (left to right)在同一周期内（从左到右） : $Z_\text{eff}$ rises (same shell, more protons), so valence electrons are held more tightly. IE increases. NGSS HS-PS1-1 Assessment Boundary: "does not include quantitative understanding of ionization energy beyond relative trends" — know the direction, not the number.：$Z_\text{eff}$ 升高（相同壳层，质子更多），故价电子被固定得更紧。IE 增大。NGSS HS-PS1-1 评估边界："不包括超出相对趋势的电离能定量理解"——了解方向，而非具体数值。
  • Down a group在同一族向下 : a new shell is added, valence electron is farther from the nucleus and better shielded. IE decreases.：增加一个新壳层，价电子离核更远且屏蔽更好。IE 减小。
  • Two notable dips within a period一个周期内的两个显著低谷 : (1) B (Group 13) has lower IE than Be (Group 2) — B's valence electron is in a 2p orbital (higher energy, easier to remove) vs Be's 2s. (2) O (Group 16) has lower IE than N (Group 15) — O has a paired 2p electron that experiences extra repulsion and is easier to remove. SCH3U B2.2 requires analysing trend data within a period; these dips are the key data-analysis talking points.：(1) B（第 13 族）的 IE 低于 Be（第 2 族）——B 的价电子在 2p 轨道（能量更高，更易移除）而非 Be 的 2s。(2) O（第 16 族）的 IE 低于 N（第 15 族）——O 有一个配对的 2p 电子，额外排斥使其更易移除。SCH3U B2.2 要求分析一个周期内的趋势数据；这些低谷是关键的数据分析要点。
  • Successive ionization energies连续电离能 : removing each successive electron costs more energy (fewer electrons, same $Z$). A large jump in IE signals removal of a core electron — the quantum number of that large jump identifies the number of valence electrons. (IB/AP-depth; flagged here for feeder students.)：移除每个连续的电子需要更多能量（电子更少，$Z$ 相同）。IE 的大幅跳升标志着移除了芯电子——那个大跳升的量子数可识别价电子数。（IB/AP 深度；为衔接学生在此标注。）
Worked Example 4 · Explaining the dip at Group 13 (B vs Be)例题 4 · 解释第 13 族的低谷（B 与 Be）

First ionization energies: Be $= 900\ \mathrm{kJ/mol}$; B $= 801\ \mathrm{kJ/mol}$. Both are in Period 2. Moving right should increase IE. Why is B lower than Be?第一电离能：Be $= 900\ \mathrm{kJ/mol}$；B $= 801\ \mathrm{kJ/mol}$。两者均在第 2 周期。向右移动应增大 IE。为何 B 低于 Be？

Electron configurations:电子排布： Be is $1s^2\, 2s^2$; B is $1s^2\, 2s^2\, 2p^1$.Be 为 $1s^2\, 2s^2$；B 为 $1s^2\, 2s^2\, 2p^1$。

Key difference:关键差异： The electron removed from Be is a $2s$ electron; the electron removed from B is a $2p$ electron. The $2p$ subshell is slightly higher in energy than $2s$ (more shielded by the filled $2s$) and farther from the nucleus on average. Therefore $2p$ electrons are easier to remove than $2s$ electrons at the same period.从 Be 移除的是 $2s$ 电子；从 B 移除的是 $2p$ 电子。$2p$ 亚壳层能量略高于 $2s$（被填满的 $2s$ 更多屏蔽），平均距核更远。因此，在同一周期内，$2p$ 电子比 $2s$ 电子更容易移除。

Conclusion:结论： Despite B having a higher $Z$, the lower subshell energy of $2s$ in Be wins: IE(Be) $>$ IE(B). The general left-to-right trend resumes after B. SCH3U B2.2 requires students to explain period-trend anomalies from data — this is the prototypical example.尽管 B 的 $Z$ 更高，Be 中 $2s$ 更低的亚壳层能量占优：IE(Be) $>$ IE(B)。一般从左到右的趋势在 B 之后恢复。SCH3U B2.2 要求学生从数据解释周期趋势异常——这是典型示例。

Which of the following correctly orders Period 2 elements by increasing first ionization energy? 🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B2.2 / AB Chem 20以下哪项正确地按增大的第一电离能顺序排列了第 2 周期元素？🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B2.2 / AB Chem 20
§4 · Q1
Ne < F < O < N < C < B < Be < LiNe < F < O < N < C < B < Be < Li
Li < Be < B < C < N < O < F < NeLi < Be < B < C < N < O < F < Ne
Li < B < Be < C < O < N < F < NeLi < B < Be < C < O < N < F < Ne
Li < Be < B < C < O < N < F < NeLi < Be < B < C < O < N < F < Ne
General trend: IE increases left to right. Two dips: B < Be (B removes a 2p electron, Be a 2s) and O < N (O has a paired 2p electron with extra repulsion). So the correct order is Li < B < Be < C < O < N < F < Ne. The dips swap Be/B and N/O from the naive left-to-right order.一般趋势：IE 从左到右增大。两个低谷：B < Be（B 移除 2p 电子，Be 移除 2s 电子）和 O < N（O 有配对的 2p 电子，额外排斥）。故正确顺序为 Li < B < Be < C < O < N < F < Ne。这两个低谷将 Be/B 和 N/O 的位置从单纯的从左到右顺序中互换。
There are two anomalies in Period 2: B is lower than Be (2p vs 2s removal), and O is lower than N (paired 2p electron extra repulsion). Option (b) misses both anomalies; option (d) swaps Be/B correctly but reverses N/O incorrectly.第 2 周期有两个异常：B 低于 Be（移除 2p 对 2s），O 低于 N（配对 2p 电子额外排斥）。选项 (b) 遗漏了两个异常；选项 (d) 正确交换了 Be/B 但错误地反转了 N/O。
Why does ionization energy decrease going down Group 1 (Li → Na → K → Rb)?为什么电离能沿第 1 族向下减小（Li → Na → K → Rb）？
§4 · Q2
The nuclear charge decreases down the group核电荷沿族向下减小
Each successive element adds a new electron shell, placing the valence electron farther from the nucleus and increasing shielding每种后续元素增加一个新电子壳层，将价电子放置得距核更远且屏蔽增强
More neutrons in larger atoms repel the valence electron较大原子中更多的中子排斥价电子
The number of valence electrons increases down the group价电子数沿族向下增大
Down Group 1: Li has 2 shells, Na has 3, K has 4, Rb has 5. Each new shell adds a complete inner-electron shielding layer. The valence electron is farther from the nucleus and more shielded, so less energy is needed to remove it. Nuclear charge ($Z$) does increase, but the shell distance effect dominates.沿第 1 族向下：Li 有 2 个壳层，Na 有 3 个，K 有 4 个，Rb 有 5 个。每个新壳层增加了完整的内层电子屏蔽层。价电子离核更远且屏蔽更强，因此移除它需要的能量更少。核电荷（$Z$）确实增大，但壳层距离效应占主导。
Nuclear charge increases down a group (more protons) — so option (a) is wrong. Neutrons don't repel electrons (they're neutral) — option (c) is wrong. All Group 1 elements have 1 valence electron — option (d) is wrong. The correct answer is the shell addition argument.核电荷沿族向下增大（质子更多）——所以选项 (a) 错误。中子不排斥电子（它们是中性的）——选项 (c) 错误。所有第 1 族元素都有 1 个价电子——选项 (d) 错误。正确答案是壳层增加的论点。
Going deeper — successive ionization energies and valence electron count深入 — 连续电离能与价电子数

When you strip electrons one at a time, each successive IE$_n$ is larger than the previous (fewer electrons, same $Z$, higher $Z_\text{eff}$ per electron). There is always a dramatic jump in IE when you remove the first core electron (the one that crosses from the valence shell to the next lower shell). For sodium: IE$_1 = 496\ \mathrm{kJ/mol}$ (removing 3s$^1$), IE$_2 = 4562\ \mathrm{kJ/mol}$ (removing 2p — a core electron). The jump factor of ~9 reveals that Na has exactly 1 valence electron — matching Group 1. This is the experimental basis for identifying group membership from ionization-energy data, as tested in IB Chemistry HL and AP Chemistry. SCH3U B2.2 uses period-trend data comparison; the successive IE technique extends that to group identification.逐个移除电子时，每个连续的 IE$_n$ 都大于前一个（电子更少，$Z$ 相同，每个电子的 $Z_\text{eff}$ 更高）。当你移除第一个芯电子（从价层跨越到下一个更低层的那个）时，IE 总会出现急剧跳升。对于钠：IE$_1 = 496\ \mathrm{kJ/mol}$（移除 3s$^1$），IE$_2 = 4562\ \mathrm{kJ/mol}$（移除 2p——芯电子）。约 9 倍的跳升因子揭示 Na 恰好有 1 个价电子——与第 1 族一致。这是从电离能数据识别族成员的实验依据，在 IB Chemistry HL 和 AP Chemistry 中均有考查。SCH3U B2.2 使用周期趋势数据比较；连续 IE 技术将其扩展到族识别。

Section 5 · Electron affinity and electronegativity第 5 节 · 电子亲和能与电负性

Electron Affinity and Electronegativity电子亲和能与电负性

Both measure attraction to electrons — EA for a free electron added; EN for a shared electron in a bond.两者都衡量对电子的吸引力——EA 针对加入的自由电子；EN 针对键中的共享电子。
  • Electron affinity (EA)电子亲和能（EA） = energy change when a neutral gas-phase atom gains one electron: $\text{X}(g) + e^- \to \text{X}^-(g) + \text{EA}$. A large (more negative) EA means the atom strongly attracts an added electron. SCH3U B2.1 names "electron affinity" as assessed terminology.= 中性气相原子获得一个电子时的能量变化：$\text{X}(g) + e^- \to \text{X}^-(g) + \text{EA}$。较大（更负）的 EA 意味着原子强烈吸引加入的电子。SCH3U B2.1 将"电子亲和能"列为被评估术语。
  • EA trendEA 趋势 : generally more negative (larger magnitude) across a period (higher $Z_\text{eff}$, stronger pull) and less negative down a group (electron added to a larger, more shielded shell). Noble gases have positive (endothermic) EA — they resist gaining electrons.：一般在同一周期越来越负（幅度更大）（$Z_\text{eff}$ 更高，吸引力更强），在同一族向下越来越不负（电子加入更大、屏蔽更好的壳层）。惰性气体的 EA 为正（吸热）——它们抵制获得电子。
  • Electronegativity (EN)电负性（EN） = an atom's relative ability to attract shared electrons within a covalent bond (Pauling scale: 0.7 for Cs to 3.98 for F). It is not directly measurable but derived from bond energies. NGSS HS-PS1-2 uses it to predict bonding outcomes; SCH3U B2.1, B2.5; AB Chemistry 20 GO2 (knowledge outcome: "define valence electron, electronegativity, ionic bond").= 原子在共价键中吸引共享电子的相对能力（泡利标度：Cs 为 0.7，F 为 3.98）。它不能直接测量，而是从键能推导而来。NGSS HS-PS1-2 用它预测成键结果；SCH3U B2.1、B2.5；AB Chemistry 20 GO2（知识结果："定义价电子、电负性、离子键"）。
  • EN trendEN 趋势 : increases across a period and decreases down a group — same drivers as IE and EA. Highest: F (3.98); lowest: Fr (~0.7). Upper-right corner = high EN; lower-left corner = low EN.：在同一周期增大，在同一族向下减小——驱动因素与 IE 和 EA 相同。最高：F（3.98）；最低：Fr（约 0.7）。右上角 = 高 EN；左下角 = 低 EN。
  • EN difference predicts bond typeEN 差预测键的类型 : $\Delta\text{EN} < 0.4$ — nonpolar covalent; $0.4 \le \Delta\text{EN} < 1.7$ — polar covalent; $\Delta\text{EN} \ge 1.7$ — ionic. (Thresholds are approximate; exact values vary by source.) AB Chemistry 20 GO2: "describe bonding as a continuum ranging from complete electron transfer to equal sharing of electrons."：$\Delta\text{EN} < 0.4$ — 非极性共价；$0.4 \le \Delta\text{EN} < 1.7$ — 极性共价；$\Delta\text{EN} \ge 1.7$ — 离子键。（阈值是近似值；确切数值因资料而异。）AB Chemistry 20 GO2："将成键描述为从完全电子转移到等量共享电子的连续体。"
Worked Example 5 · Using electronegativity to classify bonds 🇨🇦 SCH3U B2.5 / AB Chem 20 GO2例题 5 · 用电负性分类化学键🇨🇦 SCH3U B2.5 / AB Chem 20 GO2

Classify the bonds in (a) HCl, (b) NaCl, (c) Cl$_2$ as ionic, polar covalent, or nonpolar covalent. Use EN values: H = 2.20, Cl = 3.16, Na = 0.93.将 (a) HCl、(b) NaCl、(c) Cl$_2$ 中的化学键分类为离子键、极性共价键或非极性共价键。使用 EN 值：H = 2.20，Cl = 3.16，Na = 0.93。

(a) HCl:(a) HCl： $\Delta\text{EN} = 3.16 - 2.20 = 0.96$. Falls in the range $0.4$–$1.7$ → polar covalent. Cl is the partial negative end ($\delta-$); H is partial positive ($\delta+$).$\Delta\text{EN} = 3.16 - 2.20 = 0.96$。在 $0.4$–$1.7$ 范围内 → 极性共价键。Cl 为部分负端（$\delta-$）；H 为部分正端（$\delta+$）。

(b) NaCl:(b) NaCl： $\Delta\text{EN} = 3.16 - 0.93 = 2.23 \ge 1.7$ → ionic. Na transfers its valence electron to Cl.$\Delta\text{EN} = 3.16 - 0.93 = 2.23 \ge 1.7$ → 离子键。Na 将其价电子转移给 Cl。

(c) Cl$_2$:(c) Cl$_2$： $\Delta\text{EN} = 3.16 - 3.16 = 0 < 0.4$ → nonpolar covalent. Both atoms are identical; the bond is perfectly symmetric.$\Delta\text{EN} = 3.16 - 3.16 = 0 < 0.4$ → 非极性共价键。两个原子相同；键完全对称。

Fluorine (F) has the highest electronegativity of all elements. Which best explains this? 🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U B2.1氟（F）是所有元素中电负性最高的。以下哪项最能解释这一点？🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U B2.1
§5 · Q1
F has the most neutrons in Period 2, giving it a dense nucleusF 在第 2 周期中子数最多，赋予其致密的原子核
F has more total electrons than most elements in Period 2F 的总电子数比第 2 周期大多数元素多
F's electrons are far from the nucleus because it is in Period 2由于 F 在第 2 周期，其电子距核较远
F has a high $Z_\text{eff}$ and a very small atomic radius, making the bonding electrons feel a very strong attraction to the F nucleusF 的 $Z_\text{eff}$ 高且原子半径极小，使成键电子感受到对 F 原子核极强的吸引力
F is at the top right of the main table: Period 2 gives it a tiny radius (no additional shells shielding) and Group 17 gives it $Z=9$ with high $Z_\text{eff}$. Both factors maximise the attractive pull on shared electrons. Electronegativity correlates with IE and small radius — all three peak at the upper-right corner.F 位于主表的右上角：第 2 周期赋予它极小的半径（没有额外的壳层屏蔽），第 17 族赋予它 $Z=9$ 和高 $Z_\text{eff}$。这两个因素使对共享电子的吸引力最大化。电负性与 IE 和小半径相关——三者都在右上角达到峰值。
Neutrons do not affect electronegativity. Period 2 means F is close to the nucleus, not far. F has only 9 electrons. The driver is high $Z_\text{eff}$ (strong nuclear pull) combined with small radius (shared electrons close to the F nucleus).中子不影响电负性。第 2 周期意味着 F 离核近，而非远。F 只有 9 个电子。驱动因素是高 $Z_\text{eff}$（强核吸引力）与小半径（共享电子靠近 F 原子核）的结合。
A bond between atoms with $\Delta\text{EN} = 1.2$ is best classified as: 🇨🇦 SCH3U B2.5 / AB Chem 20 GO2$\Delta\text{EN} = 1.2$ 的原子间的化学键最好分类为：🇨🇦 SCH3U B2.5 / AB Chem 20 GO2
§5 · Q2
Ionic离子键
Polar covalent极性共价键
Nonpolar covalent非极性共价键
Metallic金属键
$\Delta\text{EN} = 1.2$ falls in the range $0.4$–$1.7$: polar covalent. The electron pair is shared but unequally — it spends more time near the more electronegative atom, creating partial charges ($\delta+$ and $\delta-$). AB Chemistry 20 GO2 describes bonding as a continuum: ionic ($\Delta$EN large) → polar covalent → nonpolar covalent ($\Delta$EN $\approx 0$).$\Delta\text{EN} = 1.2$ 在 $0.4$–$1.7$ 范围内：极性共价键。电子对被共享，但不均等——它在电负性更高的原子附近停留更多时间，产生部分电荷（$\delta+$ 和 $\delta-$）。AB Chemistry 20 GO2 将成键描述为连续体：离子键（$\Delta$EN 大）→ 极性共价 → 非极性共价（$\Delta$EN $\approx 0$）。
Ionic requires $\Delta$EN $\ge 1.7$ (approximate). Nonpolar covalent requires $\Delta$EN $< 0.4$. At $\Delta$EN $= 1.2$, the electrons are shared but unequally — that is polar covalent. Metallic bonding is a different type involving a sea of delocalized electrons in metals.离子键要求 $\Delta$EN $\ge 1.7$（近似）。非极性共价键要求 $\Delta$EN $< 0.4$。在 $\Delta$EN $= 1.2$ 时，电子被共享但不均等——这是极性共价键。金属键是涉及金属中离域电子海的不同类型。
Section 6 · Metallic vs nonmetallic character第 6 节 · 金属性与非金属性

Metallic vs Nonmetallic Character金属性与非金属性

Metallic character = tendency to lose electrons and form cations. Nonmetallic character = tendency to gain electrons and form anions (or share electrons in covalent bonds).金属性（metallic character）= 失去电子形成阳离子的倾向。非金属性（nonmetallic character）= 获得电子形成阴离子（或在共价键中共享电子）的倾向。
  • Metallic character金属性 decreases across a period (IE increases, harder to remove electrons) and increases down a group (IE decreases, easier to remove electrons). Highest metallic character: lower-left corner (Fr, Cs, Rb).在同一周期减小（IE 升高，电子更难失去），在同一族向下增大（IE 降低，电子更易失去）。金属性最高：左下角（Fr、Cs、Rb）。
  • Nonmetallic character非金属性 increases across a period and decreases down a group — same direction as electronegativity. Highest: upper-right corner (F, O, N, Cl).在同一周期增大，在同一族向下减小——方向与电负性相同。最高：右上角（F、O、N、Cl）。
  • Metalloids (semiconductors)类金属（半导体） — elements along the staircase line (B, Si, Ge, As, Sb, Te) show intermediate properties: semiconducting behaviour, amphoteric oxides, and both ionic and covalent bonding tendencies. BC Chemistry 11 content elaboration includes "chemical bonding based on electronegativity," which positions metalloids at the crossover.— 位于阶梯线的元素（B、Si、Ge、As、Sb、Te）显示出中间性质：半导体行为、两性氧化物，以及既有离子键又有共价键的倾向。BC Chemistry 11 内容细化包括"基于电负性的化学键"，将类金属置于交叉点。
  • Oxide chemistry氧化物化学 : metallic oxides are basic (react with water to give bases, e.g. Na$_2$O + H$_2$O → 2NaOH); nonmetallic oxides are acidic (e.g. SO$_3$ + H$_2$O → H$_2$SO$_4$). NGSS HS-PS1-2: "examples of chemical reactions could include the reaction of sodium and chlorine, of carbon and oxygen, or of carbon and hydrogen."：金属氧化物是碱性的（与水反应生成碱，如 Na$_2$O + H$_2$O → 2NaOH）；非金属氧化物是酸性的（如 SO$_3$ + H$_2$O → H$_2$SO$_4$）。NGSS HS-PS1-2："化学反应示例包括钠和氯的反应、碳和氧的反应或碳和氢的反应。"
Worked Example 6 · Placing elements on the metal–nonmetal spectrum 🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U B3.3 / AB Chem 20 GO2例题 6 · 将元素定位于金属–非金属谱上🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U B3.3 / AB Chem 20 GO2

Rank the following by decreasing metallic character and justify: Na (Period 3, Group 1), Mg (Period 3, Group 2), Al (Period 3, Group 13), Si (Period 3, Group 14), K (Period 4, Group 1).按减小的金属性顺序排列以下元素并说明理由：Na（第 3 周期，第 1 族）、Mg（第 3 周期，第 2 族）、Al（第 3 周期，第 13 族）、Si（第 3 周期，第 14 族）、K（第 4 周期，第 1 族）。

Within Period 3:在第 3 周期内： Metallic character decreases left to right (increasing $Z_\text{eff}$, higher IE): Na > Mg > Al > Si. Si is a metalloid.金属性从左到右减小（$Z_\text{eff}$ 升高，IE 更高）：Na > Mg > Al > Si。Si 是类金属。

Down Group 1:沿第 1 族向下： K (Period 4) has a lower IE than Na (Period 3) — the 4th shell is farther from the nucleus. So K > Na in metallic character.K（第 4 周期）的 IE 低于 Na（第 3 周期）——第 4 壳层离核更远。故 K > Na（金属性）。

Ranking (most to least metallic):排名（金属性从大到小）：

$$ K > Na > Mg > Al > Si $$
Which element has the greatest metallic character? 🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B3.3哪种元素的金属性最强？🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B3.3
§6 · Q1
Cs (Period 6, Group 1)Cs（第 6 周期，第 1 族）
Li (Period 2, Group 1)Li（第 2 周期，第 1 族）
Mg (Period 3, Group 2)Mg（第 3 周期，第 2 族）
Cl (Period 3, Group 17)Cl（第 3 周期，第 17 族）
Metallic character is highest in the lower-left of the table — maximum shells (lowest IE) and minimum $Z_\text{eff}$. Cs (Period 6, Group 1) has 6 shells and only 1 valence electron, giving the lowest first IE of a stable element. It is the most metallic element (excluding radioactive Fr).金属性在表的左下角最高——最多壳层（IE 最低）和最低 $Z_\text{eff}$。Cs（第 6 周期，第 1 族）有 6 个壳层且只有 1 个价电子，给出稳定元素中最低的第一 IE。它是金属性最强的元素（不含放射性 Fr）。
Metallic character: more shells = lower IE = more metallic; Group 1 = one valence electron = most easily lost. Cs (Period 6, Group 1) wins over Li (Period 2). Mg has 2 valence electrons (higher IE than Cs). Cl is a nonmetal (Group 17, high EN, gains electrons).金属性：壳层更多 = IE 更低 = 金属性更强；第 1 族 = 一个价电子 = 最易失去。Cs（第 6 周期，第 1 族）胜过 Li（第 2 周期）。Mg 有 2 个价电子（IE 高于 Cs）。Cl 是非金属（第 17 族，EN 高，得电子）。
Sodium oxide (Na$_2$O) dissolves in water. What type of solution results? 🇺🇸 NGSS HS-PS1-2氧化钠（Na$_2$O）溶于水。会产生什么类型的溶液？🇺🇸 NGSS HS-PS1-2
§6 · Q2
Acidic酸性
Neutral中性
Basic (alkaline)碱性
It does not dissolve in water它不溶于水
Na is a highly metallic element (Group 1, Period 3; low IE, readily loses electrons). Metal oxides react with water to form bases: Na$_2$O + H$_2$O → 2NaOH. NaOH is a strong base; the solution is alkaline. This contrasts with nonmetal oxides (e.g. SO$_3$ → H$_2$SO$_4$, acidic). NGSS HS-PS1-2 uses Na + Cl reactions as an example of trend-based prediction.Na 是高度金属性的元素（第 1 族，第 3 周期；IE 低，易失电子）。金属氧化物与水反应生成碱：Na$_2$O + H$_2$O → 2NaOH。NaOH 是强碱；溶液是碱性的。这与非金属氧化物（如 SO$_3$ → H$_2$SO$_4$，酸性）形成对比。NGSS HS-PS1-2 将 Na + Cl 反应用作基于趋势预测的示例。
Metal oxides + water → basic solutions (metal hydroxides). Na$_2$O + H$_2$O gives NaOH, a strong base. It's not acidic (that's nonmetal oxide behavior) and it dissolves readily (Na compounds are very soluble).金属氧化物 + 水 → 碱性溶液（金属氢氧化物）。Na$_2$O + H$_2$O 生成 NaOH，一种强碱。它不是酸性的（那是非金属氧化物的行为），且很容易溶解（Na 化合物溶解性很强）。
Section 7 · Reactivity trends: Groups 1, 17, and 18第 7 节 · 反应活性趋势：第 1、17、18 族

Reactivity Trends: Groups 1, 17, and 18反应活性趋势：第 1、17、18 族

Reactivity is the chemical consequence of all the periodic trends we have seen.反应活性是我们所见的所有周期趋势的化学结果。
  • Group 1 — Alkali metals (Li, Na, K, Rb, Cs, Fr)第 1 族——碱金属（Li、Na、K、Rb、Cs、Fr）
    • 1 valence electron; very low IE; lose that electron readily to form $\text{M}^+$ ions.1 个价电子；IE 极低；极易失去该电子形成 $\text{M}^+$ 离子。
    • Reactivity increases down the group: K is more reactive than Na; Cs is more reactive than K. Each added shell lowers IE, making electron loss easier.反应活性沿族向下增大：K 的反应性强于 Na；Cs 的反应性强于 K。每个增加的壳层降低 IE，使电子损失更容易。
    • With water: $2\text{M}(s) + 2\text{H}_2\text{O}(l) \to 2\text{MOH}(aq) + \text{H}_2(g)$. Li fizzes; Na fizzes vigorously; K ignites the H$_2$; Cs explodes. NGSS HS-PS1-1: "reactivity of metals" as a pattern predicted from electron structure.与水反应：$2\text{M}(s) + 2\text{H}_2\text{O}(l) \to 2\text{MOH}(aq) + \text{H}_2(g)$。Li 冒泡；Na 剧烈冒泡；K 点燃 H$_2$；Cs 爆炸。NGSS HS-PS1-1："金属反应活性"是从电子结构预测的规律。
  • Group 17 — Halogens (F, Cl, Br, I, At)第 17 族——卤素（F、Cl、Br、I、At）
    • 7 valence electrons; high EA; gain 1 electron to form $\text{X}^-$ ions (or share in covalent bonds).7 个价电子；EA 高；获得 1 个电子形成 $\text{X}^-$ 离子（或在共价键中共享）。
    • Reactivity decreases down the group: F$_2$ is the most reactive nonmetal known; Cl$_2$ is very reactive; Br$_2$ is less so; I$_2$ is only mildly reactive. Each added shell lowers EA and EN, reducing the drive to gain electrons.反应活性沿族向下减小：F$_2$ 是已知反应性最强的非金属；Cl$_2$ 反应性很强；Br$_2$ 较弱；I$_2$ 只有轻微反应性。每个增加的壳层降低 EA 和 EN，减弱获得电子的驱动力。
    • Displacement reactions confirm the trend: Cl$_2$ displaces Br$^-$ and I$^-$; Br$_2$ displaces I$^-$ but not Cl$^-$; I$_2$ displaces neither. NGSS HS-PS1-2: "reaction of sodium and chlorine" as a predictable outcome from the periodic table.置换反应确认了趋势：Cl$_2$ 置换 Br$^-$ 和 I$^-$；Br$_2$ 置换 I$^-$ 但不置换 Cl$^-$；I$_2$ 都不置换。NGSS HS-PS1-2："钠和氯的反应"是元素周期表可预测的结果。
  • Group 18 — Noble gases (He, Ne, Ar, Kr, Xe, Rn)第 18 族——惰性气体（He、Ne、Ar、Kr、Xe、Rn）
    • Full outer shell (8 electrons, or 2 for He); very high IE and positive EA — they neither lose nor gain electrons easily. Under normal conditions, noble gases are chemically inert.外层充满（8 个电子，He 为 2 个）；IE 很高且 EA 为正——它们既不容易失去也不容易获得电子。在正常条件下，惰性气体化学上是惰性的。
    • Xe and Kr can form compounds under extreme conditions (e.g. XeF$_2$, XeF$_4$) — at IB/AP depth, not assessed in NGSS or SCH3U.Xe 和 Kr 在极端条件下可以形成化合物（如 XeF$_2$、XeF$_4$）——在 IB/AP 深度，NGSS 或 SCH3U 不考查。
Worked Example 7 · Predicting a halogen displacement reaction 🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U B3.3例题 7 · 预测卤素置换反应🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U B3.3

Predict whether the following reaction will occur, and explain why using periodic trends: $\text{Br}_2(aq) + 2\text{KCl}(aq) \to ?$预测以下反应是否会发生，并用周期趋势解释：$\text{Br}_2(aq) + 2\text{KCl}(aq) \to ?$

Identify the halogens:识别卤素： Br$_2$ is the molecular halogen (oxidizing agent); Cl$^-$ is the halide ion already formed. For displacement to occur, Br$_2$ must be a stronger oxidizing agent than Cl$_2$ — it must attract electrons more strongly than Cl.Br$_2$ 是分子卤素（氧化剂）；Cl$^-$ 是已形成的卤素离子。要发生置换，Br$_2$ 必须是比 Cl$_2$ 更强的氧化剂——它必须比 Cl 更强地吸引电子。

Apply the reactivity trend:应用反应活性趋势： Reactivity of halogens decreases down Group 17: F > Cl > Br > I. So Cl is more reactive (better oxidizer) than Br. Br$_2$ cannot displace Cl$^-$.卤素的反应活性沿第 17 族向下减小：F > Cl > Br > I。故 Cl 比 Br 更具反应性（更好的氧化剂）。Br$_2$ 不能置换 Cl$^-$。

Conclusion:结论： No reaction occurs: Br$_2$ + 2KCl → no reaction. Compare with Cl$_2$ + 2KBr → 2KCl + Br$_2$ (Cl$_2$ does displace Br$^-$ because Cl is more reactive).不发生反应：Br$_2$ + 2KCl → 无反应。与 Cl$_2$ + 2KBr → 2KCl + Br$_2$ 对比（Cl$_2$ 确实置换 Br$^-$，因为 Cl 更具反应性）。

Why does potassium (K) react more vigorously with water than lithium (Li)? 🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B3.3 / AB Chem 20 GO1为什么钾（K）与水的反应比锂（Li）更剧烈？🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B3.3 / AB Chem 20 GO1
§7 · Q1
K has more neutrons, giving it a larger nuclear massK 有更多中子，赋予其更大的核质量
K has more valence electrons than LiK 比 Li 有更多价电子
K has more electron shells, so its valence electron is farther from the nucleus, held less tightly, and more easily lostK 有更多电子壳层，因此其价电子距核更远，被固定得更弱，更易失去
K has a higher electronegativity than LiK 的电负性高于 Li
K is in Period 4 (4 shells); Li is in Period 2 (2 shells). More shells = valence electron farther from nucleus + more shielding = lower IE = easier to lose the electron = more vigorous reaction with water. This is the group-trend driver: reactivity of alkali metals increases down Group 1. NGSS HS-PS1-1: "reactivity of metals" as a pattern predicted from electron structure.K 在第 4 周期（4 个壳层）；Li 在第 2 周期（2 个壳层）。壳层更多 = 价电子离核更远 + 屏蔽更强 = IE 更低 = 电子更易失去 = 与水的反应更剧烈。这是族趋势的驱动因素：碱金属的反应活性沿第 1 族向下增大。NGSS HS-PS1-1："金属反应活性"是从电子结构预测的规律。
Neutrons don't affect reactivity. Both K and Li have exactly 1 valence electron (Group 1). K's electronegativity is lower than Li's (lower EN = less attraction for electrons = more metallic). The correct reason is the extra electron shells in K lowering its first ionization energy.中子不影响反应活性。K 和 Li 都恰好有 1 个价电子（第 1 族）。K 的电负性低于 Li（较低的 EN = 对电子的吸引力较小 = 金属性更强）。正确原因是 K 中额外的电子壳层降低了其第一电离能。
Which halogen is the most reactive oxidizing agent? 🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C2.5哪种卤素是反应活性最强的氧化剂？🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C2.5
§7 · Q2
F$_2$F$_2$
Cl$_2$Cl$_2$
Br$_2$Br$_2$
I$_2$I$_2$
Halogen reactivity (as an oxidizing agent) decreases down Group 17: F > Cl > Br > I. F has the highest EN and EA, the smallest radius and the most tightly held nuclear charge — it most strongly attracts electrons. F$_2$ is so reactive it will oxidize water and even some noble gas compounds under extreme conditions.卤素的反应活性（作为氧化剂）沿第 17 族向下减小：F > Cl > Br > I。F 的 EN 和 EA 最高，半径最小，核电荷被固定得最紧——它最强烈地吸引电子。F$_2$ 的反应性如此之强，以至于在极端条件下它会氧化水甚至某些惰性气体化合物。
Halogen reactivity (oxidizing ability) decreases down the group: F$_2$ > Cl$_2$ > Br$_2$ > I$_2$. The trend mirrors electronegativity and electron affinity — both decrease down Group 17 due to additional electron shells and shielding. I$_2$ is the weakest oxidizer of the stable halogens.卤素的反应活性（氧化能力）沿族向下减小：F$_2$ > Cl$_2$ > Br$_2$ > I$_2$。该趋势与电负性和电子亲和能一致——两者都因额外的电子壳层和屏蔽而在第 17 族向下减小。I$_2$ 是稳定卤素中最弱的氧化剂。
Exam Tactics考试策略

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

Setup discipline for every periodic-trend question每道周期趋势题的解题纪律
  • Locate each element on the table first.首先在周期表上定位每种元素。 State the period (shell count) and group (valence electron count). The trend direction follows from these two numbers — do not try to memorise individual values.说明周期（壳层数）和族（价电子数）。趋势方向从这两个数字推出——不要试图记忆单个数值。
  • Give the mechanism, not just the direction.给出机制，而不仅仅是方向。 SCH3U B2.2 and AB require reasoning: "same shell, rising $Z_\text{eff}$ → radius decreases" or "new shell added → valence electron farther → IE decreases." A bare trend statement earns partial credit at best.SCH3U B2.2 和 AB 要求推理："相同壳层，$Z_\text{eff}$ 升高 → 半径减小"或"增加新壳层 → 价电子更远 → IE 减小"。单纯陈述趋势最多得部分分。
  • Isoelectronic species: nuclear charge wins.等电子体：核电荷决定胜负。 When comparing species with the same electron count, rank by $Z$: higher $Z$ → smaller radius (higher $Z_\text{eff}$ on the same electron cloud). Example: O$^{2-}$ > F$^-$ > Ne > Na$^+$ > Mg$^{2+}$ in radius.比较相同电子数的粒子时，按 $Z$ 排列：$Z$ 越高 → 半径越小（相同电子云上 $Z_\text{eff}$ 更高）。示例：O$^{2-}$ > F$^-$ > Ne > Na$^+$ > Mg$^{2+}$（半径）。
The two period-trend anomalies you must know你必须了解的两个周期趋势异常
  • IE: B < Be (Group 13 lower than Group 2).IE：B < Be（第 13 族低于第 2 族）。 B removes a 2p electron (higher subshell energy) vs Be's 2s electron (lower subshell). The 2p is easier to remove despite B having a higher $Z$.B 移除 2p 电子（亚壳层能量更高），而 Be 移除 2s 电子（亚壳层能量更低）。尽管 B 的 $Z$ 更高，2p 更容易移除。
  • IE: O < N (Group 16 lower than Group 15).IE：O < N（第 16 族低于第 15 族）。 O has a paired 2p electron — the extra repulsion from spin pairing lowers the IE. N's three 2p electrons are all unpaired (Hund's rule) and experience no extra repulsion.O 有一个配对的 2p 电子——自旋配对的额外排斥降低了 IE。N 的三个 2p 电子全部未配对（洪特规则），没有额外排斥。
Reactivity trap: Group 1 increases down, Group 17 decreases down反应活性陷阱：第 1 族沿族向下增大，第 17 族沿族向下减小
  • Alkali metals (Group 1): more reactive as you go down (lower IE → easier to lose electron). K > Na > Li in water reaction vigor.碱金属（第 1 族）：向下反应活性增大（IE 更低 → 更易失电子）。K > Na > Li（与水反应的剧烈程度）。
  • Halogens (Group 17): less reactive as you go down (lower EA, lower EN → weaker drive to gain electron). F$_2$ > Cl$_2$ > Br$_2$ > I$_2$ in oxidizing strength.卤素（第 17 族）：向下反应活性减小（EA 更低，EN 更低 → 获得电子的驱动力更弱）。F$_2$ > Cl$_2$ > Br$_2$ > I$_2$（氧化能力）。
  • These two groups are on opposite sides of the trend: memorise the reason (lose vs gain), not just the direction.这两个族的趋势方向相反：记住原因（失去 vs 获得），而不仅仅是方向。
Electronegativity and bond classification (§5) — answer hygiene电负性与键分类（§5）——作答规范
  • Quote $\Delta$EN and the threshold.引用 $\Delta$EN 和阈值。 Always show the subtraction: $\Delta\text{EN} = |\text{EN}_A - \text{EN}_B|$. Then state the bond type using the thresholds ($< 0.4$: nonpolar covalent; $0.4$–$1.7$: polar covalent; $\ge 1.7$: ionic). AB Chemistry 20 GO2 expects the bonding continuum framing.始终显示减法：$\Delta\text{EN} = |\text{EN}_A - \text{EN}_B|$。然后使用阈值说明键的类型（$< 0.4$：非极性共价；$0.4$–$1.7$：极性共价；$\ge 1.7$：离子键）。AB Chemistry 20 GO2 期望成键连续体框架。
  • Mark the partial charges.标记部分电荷。 In a polar covalent bond, the more electronegative atom carries $\delta-$ and the less electronegative carries $\delta+$. For HCl: $\text{H}^{\delta+}$–$\text{Cl}^{\delta-}$. SCH3U B2.5 and BC Chemistry 11 ("chemical bonding: polarity") both require identifying polarity.在极性共价键中，电负性更高的原子带 $\delta-$，电负性较低的带 $\delta+$。对于 HCl：$\text{H}^{\delta+}$–$\text{Cl}^{\delta-}$。SCH3U B2.5 和 BC Chemistry 11（"化学键：极性"）都要求识别极性。
Active Recall主动复习

Flashcards闪卡

0 / 14 flipped0 / 14 已翻
Periodic law?周期律？
Physical and chemical properties of the elements are periodic functions of their atomic numbers. (SCH3U B3.3)元素的物理和化学性质是其原子序数的周期函数。（SCH3U B3.3）
Effective nuclear charge $Z_\text{eff}$?有效核电荷 $Z_\text{eff}$？
$$Z_\text{eff} \approx Z - S$$ Net nuclear pull felt by a valence electron after core-electron shielding $S$. Rises left to right across a period.价电子在芯电子屏蔽 $S$ 后感受到的净核吸引力。在同一周期从左到右升高。
Atomic radius trend?原子半径趋势？
Decreases across a period ($Z_\text{eff}$ rises, same shell, electrons pulled tighter). Increases down a group (new shell added).在同一周期减小（$Z_\text{eff}$ 升高，相同壳层，电子被拉得更紧）。在同一族向下增大（增加新壳层）。
Cation vs anion radius vs parent?阳离子和阴离子半径与母体比较？
Cation (lost e⁻) < parent atom. Anion (gained e⁻) > parent atom. Same $Z$, fewer/more electrons changes $Z_\text{eff}$ per electron.阳离子（失去 e⁻）< 母体原子。阴离子（获得 e⁻）> 母体原子。相同 $Z$，电子更少/更多改变了每个电子的 $Z_\text{eff}$。
Ionization energy (IE) trend?电离能（IE）趋势？
Increases across a period (rising $Z_\text{eff}$). Decreases down a group (new shell, electron farther away). Two dips: B<Be; O<N.在同一周期增大（$Z_\text{eff}$ 升高）。在同一族向下减小（新壳层，电子更远）。两个低谷：B<Be；O<N。
Why B IE < Be IE?为何 B 的 IE < Be 的 IE？
Be loses a $2s$ electron; B loses a $2p$ electron. The $2p$ subshell is higher in energy and easier to remove, despite B having the larger $Z$.Be 失去 $2s$ 电子；B 失去 $2p$ 电子。$2p$ 亚壳层能量更高，更易移除，尽管 B 的 $Z$ 更大。
Why O IE < N IE?为何 O 的 IE < N 的 IE？
O has a paired $2p$ electron — extra repulsion from spin pairing makes it easier to remove. N's three $2p$ electrons are all unpaired (Hund's rule).O 有一个配对的 $2p$ 电子——自旋配对的额外排斥使其更易移除。N 的三个 $2p$ 电子全部未配对（洪特规则）。
Electron affinity (EA) vs electronegativity (EN)?电子亲和能（EA）vs 电负性（EN）？
EA: energy change when a free electron is added to a gas-phase atom ($\text{X} + e^- \to \text{X}^-$). EN: relative ability to attract shared electrons in a bond (Pauling scale).EA：气相原子获得一个自由电子时的能量变化（$\text{X} + e^- \to \text{X}^-$）。EN：在键中吸引共享电子的相对能力（泡利标度）。
Electronegativity trend + highest/lowest?电负性趋势与最高/最低？
Increases across a period; decreases down a group. Highest: F (3.98). Lowest: Fr (~0.7). Upper-right = most EN; lower-left = least EN.在同一周期增大；在同一族向下减小。最高：F（3.98）。最低：Fr（约 0.7）。右上角 = EN 最高；左下角 = EN 最低。
Bond type from $\Delta$EN?从 $\Delta$EN 判断键型？
$\Delta$EN < 0.4 → nonpolar covalent. 0.4–1.7 → polar covalent. ≥ 1.7 → ionic. (Thresholds approximate.)$\Delta$EN < 0.4 → 非极性共价。0.4–1.7 → 极性共价。≥ 1.7 → 离子键。（阈值近似。）
Metallic vs nonmetallic character trend?金属性与非金属性趋势？
Metallic: increases down a group, decreases across a period. Highest metallic: lower-left (Cs). Highest nonmetallic: upper-right (F).金属性：在同一族向下增大，在同一周期减小。金属性最高：左下角（Cs）。非金属性最高：右上角（F）。
Group 1 reactivity trend?第 1 族反应活性趋势？
Increases down the group: K > Na > Li with water. More shells → lower IE → electron lost more easily → more vigorous reaction.沿族向下增大：K > Na > Li（与水）。壳层更多 → IE 更低 → 电子更易失去 → 反应更剧烈。
Group 17 reactivity trend?第 17 族反应活性趋势？
Decreases down the group: F$_2$ > Cl$_2$ > Br$_2$ > I$_2$. More shells → lower EA → weaker drive to gain electron → less reactive oxidizer.沿族向下减小：F$_2$ > Cl$_2$ > Br$_2$ > I$_2$。壳层更多 → EA 更低 → 获得电子的驱动力更弱 → 氧化剂反应性更弱。
s/p/d/f blocks — what fills?s/p/d/f 区——填充什么？
s-block (Gp 1–2): $ns$ fills. p-block (Gp 13–18): $np$ fills. d-block (Gp 3–12): $(n-1)d$ fills (transition metals). f-block: lanthanides and actinides.s 区（第 1–2 族）：$ns$ 填充。p 区（第 13–18 族）：$np$ 填充。d 区（第 3–12 族）：$(n-1)d$ 填充（过渡金属）。f 区：镧系和锕系。
Mixed Practice综合练习

Practice Quiz综合测验

Which of the following correctly states the modern periodic law? 🇨🇦 SCH3U B3.3下列哪项正确陈述了现代周期律？🇨🇦 SCH3U B3.3
Q1
Properties of elements are periodic functions of their atomic masses元素的性质是其原子质量的周期函数
Properties of elements are periodic functions of their atomic numbers元素的性质是其原子序数的周期函数
Properties of elements are periodic functions of their neutron counts元素的性质是其中子数的周期函数
Elements with the same mass number always have the same chemical properties质量数相同的元素总是具有相同的化学性质
Moseley's law (confirmed 1913): the ordering principle is atomic number $Z$ (proton count), not atomic mass. SCH3U B3.3 explicitly states: "physical and chemical properties of the elements are periodic functions of their atomic numbers."莫塞莱定律（1913 年确认）：排列原则是原子序数 $Z$（质子数），而非原子质量。SCH3U B3.3 明确指出："元素的物理和化学性质是其原子序数的周期函数。"
Mendeleev used atomic mass (mostly works but has anomalies at Ar/K, Co/Ni). Moseley's X-ray work proved atomic number $Z$ is the correct ordering principle. Neutron count varies across isotopes of the same element and is not the ordering basis.门捷列夫使用原子质量（大多有效，但 Ar/K、Co/Ni 处有异常）。莫塞莱的 X 射线工作证明原子序数 $Z$ 是正确的排列原则。中子数在同一元素的不同同位素中各不相同，不是排列依据。
Rank Na, Cl, and Ar (all Period 3) in order of increasing atomic radius. 🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B2.1按增大的原子半径顺序排列 Na、Cl 和 Ar（均在第 3 周期）。🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B2.1
Q2
Ar < Cl < NaAr < Cl < Na
Na < Cl < ArNa < Cl < Ar
Cl < Na < ArCl < Na < Ar
Ar < Na < ClAr < Na < Cl
Atomic radius decreases across a period (same shell, rising $Z_\text{eff}$). Na ($Z=11$, Group 1) has the largest radius; Cl ($Z=17$, Group 17) is smaller; Ar ($Z=18$, Group 18) is smallest. So Ar < Cl < Na.原子半径在同一周期减小（相同壳层，$Z_\text{eff}$ 升高）。Na（$Z=11$，第 1 族）半径最大；Cl（$Z=17$，第 17 族）更小；Ar（$Z=18$，第 18 族）最小。故 Ar < Cl < Na。
In any period, radius decreases from left to right: highest $Z_\text{eff}$ at the right-hand end compresses the electron cloud most. Na (leftmost, lowest $Z_\text{eff}$) is largest; Ar (rightmost, highest $Z_\text{eff}$) is smallest.在任何周期中，半径从左到右减小：右端最高的 $Z_\text{eff}$ 最大程度地压缩电子云。Na（最左，$Z_\text{eff}$ 最低）最大；Ar（最右，$Z_\text{eff}$ 最高）最小。
Which isoelectronic species has the smallest radius? All have 18 electrons. 🇨🇦 SCH3U B2.1 / AB Chem 20 GO1哪种等电子粒子的半径最小？均有 18 个电子。🇨🇦 SCH3U B2.1 / AB Chem 20 GO1
Q3
S$^{2-}$ ($Z=16$)S$^{2-}$（$Z=16$）
Cl$^-$ ($Z=17$)Cl$^-$（$Z=17$）
Ar ($Z=18$)Ar（$Z=18$）
Ca$^{2+}$ ($Z=20$)Ca$^{2+}$（$Z=20$）
All four have 18 electrons (isoelectronic). Size is determined entirely by $Z_\text{eff}$: same number of electrons, so higher $Z$ → higher $Z_\text{eff}$ → smaller radius. Ca$^{2+}$ has $Z=20$ — the highest of the four — so it has the smallest radius.四者均有 18 个电子（等电子体）。大小完全由 $Z_\text{eff}$ 决定：电子数相同，因此 $Z$ 越高 → $Z_\text{eff}$ 越高 → 半径越小。Ca$^{2+}$ 的 $Z=20$——四者中最高——所以它的半径最小。
For isoelectronic species: same electrons, different $Z$. Higher $Z$ means more protons pulling the same electron cloud inward → smaller radius. Ca$^{2+}$ ($Z=20$) < Ar ($Z=18$) < Cl$^-$ ($Z=17$) < S$^{2-}$ ($Z=16$) in size.对于等电子体：电子相同，$Z$ 不同。$Z$ 更高意味着更多质子将相同的电子云向内拉 → 半径更小。Ca$^{2+}$（$Z=20$）< Ar（$Z=18$）< Cl$^-$（$Z=17$）< S$^{2-}$（$Z=16$）（大小）。
A bond between Si ($\text{EN}=1.90$) and Cl ($\text{EN}=3.16$) has $\Delta\text{EN}=1.26$. How is it classified? 🇨🇦 SCH3U B2.5 / AB Chem 20 GO2Si（EN=1.90）与 Cl（EN=3.16）之间的键 $\Delta\text{EN}=1.26$。如何分类？🇨🇦 SCH3U B2.5 / AB Chem 20 GO2
Q4
Nonpolar covalent非极性共价键
Ionic离子键
Polar covalent极性共价键
Metallic金属键
$\Delta\text{EN} = 1.26$ falls in the 0.4–1.7 range: polar covalent. The bond is shared but unequal — Cl ($\delta-$) pulls the electrons more strongly than Si ($\delta+$). This classification matches SCH3U B2.5 and AB Chemistry 20 GO2's bonding continuum.$\Delta\text{EN} = 1.26$ 在 0.4–1.7 范围内：极性共价键。键被共享，但不均等——Cl（$\delta-$）比 Si（$\delta+$）更强地拉动电子。这一分类与 SCH3U B2.5 和 AB Chemistry 20 GO2 的成键连续体一致。
Nonpolar covalent requires $\Delta$EN < 0.4. Ionic requires $\Delta$EN ≥ 1.7. At $\Delta$EN = 1.26, the bond is between these thresholds — polar covalent. Metallic bonding occurs in metal lattices, not in Si–Cl bonds.非极性共价键要求 $\Delta$EN < 0.4。离子键要求 $\Delta$EN ≥ 1.7。在 $\Delta$EN = 1.26 时，键在这两个阈值之间——极性共价键。金属键发生在金属晶格中，而不是在 Si–Cl 键中。
Which of the following correctly predicts the outcome of mixing Cl$_2$(aq) with NaBr(aq)? 🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C2.5以下哪项正确预测了 Cl$_2$(aq) 与 NaBr(aq) 混合的结果？🇺🇸 NGSS HS-PS1-2 / 🇨🇦 SCH3U C2.5
Q5
No reaction; Cl$_2$ is less reactive than Br$_2$无反应；Cl$_2$ 的反应性弱于 Br$_2$
Cl$_2$ displaces Br$^-$: Cl$_2$ + 2NaBr → 2NaCl + Br$_2$Cl$_2$ 置换 Br$^-$：Cl$_2$ + 2NaBr → 2NaCl + Br$_2$
Br$^-$ displaces Cl$_2$: 2Br$^-$ + Cl$_2$ → Br$_2$ + 2Cl$_2$Br$^-$ 置换 Cl$_2$：2Br$^-$ + Cl$_2$ → Br$_2$ + 2Cl$_2$
An explosive reaction occurs because both halogens react with sodium发生爆炸性反应，因为两种卤素都与钠反应
Halogen reactivity decreases down Group 17: Cl is more reactive than Br. A more reactive halogen displaces a less reactive one from solution. Cl$_2$ + 2Br$^-$ → 2Cl$^-$ + Br$_2$. The solution turns orange-brown (Br$_2$ colour). NGSS HS-PS1-2: predicting outcomes from trends in the periodic table.卤素的反应活性沿第 17 族向下减小：Cl 比 Br 更具反应性。反应性更强的卤素将反应性较弱的卤素从溶液中置换出来。Cl$_2$ + 2Br$^-$ → 2Cl$^-$ + Br$_2$。溶液变为橙棕色（Br$_2$ 的颜色）。NGSS HS-PS1-2：从元素周期表中的趋势预测结果。
Cl is above Br in Group 17 → Cl$_2$ is more reactive. A more reactive halogen will displace a less reactive halide ion. Option (a) reverses the reactivity. Option (c) reverses the reaction direction.Cl 在第 17 族中位于 Br 之上 → Cl$_2$ 反应性更强。反应性更强的卤素将置换反应性较弱的卤素离子。选项 (a) 颠倒了反应活性。选项 (c) 颠倒了反应方向。
Which sequence correctly orders elements by decreasing ionization energy? 🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B2.2下列哪个序列正确地按减小的电离能顺序排列了元素？🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B2.2
Q6
Ne > F > N > O > CNe > F > N > O > C
Ne > F > O > N > CNe > F > O > N > C
C > N > O > F > NeC > N > O > F > Ne
Ne > O > N > F > CNe > O > N > F > C
General trend: IE increases left to right in Period 2. Two dips: O < N (paired 2p electron in O → extra repulsion) and B < Be (2p vs 2s; not tested here). Among these five, the correct order placing N above O is: Ne > F > N > O > C. Ne has the highest (full shell); C has the lowest (leftmost of the five).一般趋势：IE 在第 2 周期从左到右增大。两个低谷：O < N（O 中配对的 2p 电子 → 额外排斥）和 B < Be（2p 与 2s；此处不考）。在这五种元素中，正确顺序是 N 高于 O：Ne > F > N > O > C。Ne 最高（充满壳层）；C 最低（五者中最左）。
Option (b) would be correct if there were no anomalies, but O has a paired 2p electron that makes it easier to remove than N's half-filled 2p (all unpaired). The dip at O is a tested anomaly in SCH3U B2.2.如果没有异常，选项 (b) 将是正确的，但 O 有一个配对的 2p 电子，使其比 N 的半充满 2p（全部未配对）更容易移除。O 处的低谷是 SCH3U B2.2 中测试的异常。
An element in Period 5, Group 2 has which of these properties? 🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B3.3第 5 周期、第 2 族的元素具有以下哪种性质？🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B3.3
Q7
7 valence electrons and nonmetallic character7 个价电子和非金属性
1 valence electron and very high reactivity with water1 个价电子和与水的高反应活性
8 valence electrons and chemical inertness8 个价电子和化学惰性
2 valence electrons, 5 shells, and metallic character2 个价电子，5 个壳层，以及金属性
Period 5 → 5 electron shells. Group 2 → 2 valence electrons. Group 2 elements are alkaline earth metals (metallic). The element is strontium (Sr, $Z=38$). Five shells means a large atomic radius and relatively low IE — metallic character confirmed. 2 valence electrons, 5 shells, metallic — option (d).第 5 周期 → 5 个电子壳层。第 2 族 → 2 个价电子。第 2 族元素是碱土金属（金属性）。该元素是锶（Sr，$Z=38$）。五个壳层意味着较大的原子半径和相对较低的 IE——金属性确认。2 个价电子，5 个壳层，金属性——选项 (d)。
Period tells you the shell count (5 shells). Group tells you the valence electrons (2 for Group 2). Group 2 = alkaline earth metals = metallic. Option (a) is Group 17 (halogens); (b) is Group 1 (alkali metals); (c) is Group 18 (noble gases).周期告诉你壳层数（5 个壳层）。族告诉你价电子数（第 2 族为 2 个）。第 2 族 = 碱土金属 = 金属性。选项 (a) 是第 17 族（卤素）；(b) 是第 1 族（碱金属）；(c) 是第 18 族（惰性气体）。
Which best explains why cesium (Cs) reacts more vigorously with water than sodium (Na)? 🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B3.3 / AB Chem 20 GO1以下哪项最好地解释了为什么铯（Cs）与水的反应比钠（Na）更剧烈？🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B3.3 / AB Chem 20 GO1
Q8
Cs has a larger atomic mass, giving it more kinetic energyCs 的原子质量更大，赋予它更多动能
Cs has more valence electrons than NaCs 比 Na 有更多价电子
Cs has more electron shells, so its valence electron is farther from the nucleus and held less tightly (lower IE)Cs 有更多电子壳层，因此其价电子距核更远，被固定得更弱（IE 更低）
Cs has a higher electronegativity than NaCs 的电负性高于 Na
Cs (Period 6) has 6 shells; Na (Period 3) has 3. Both have 1 valence electron (Group 1). More shells → valence electron farther from nucleus → more shielding → lower IE → electron lost more easily → more vigorous reaction. This is the down-group IE decrease driving Group 1 reactivity.Cs（第 6 周期）有 6 个壳层；Na（第 3 周期）有 3 个。两者都有 1 个价电子（第 1 族）。壳层更多 → 价电子距核更远 → 屏蔽更强 → IE 更低 → 电子更易失去 → 反应更剧烈。这是沿族向下的 IE 减小驱动第 1 族反应活性的机制。
Atomic mass doesn't drive reactivity. Both Cs and Na have exactly 1 valence electron. Cs's EN is lower than Na's (lower EN = less attraction for electrons = more metallic). The key is shell count → lower IE → easier electron loss.原子质量不驱动反应活性。Cs 和 Na 都恰好有 1 个价电子。Cs 的 EN 低于 Na（较低的 EN = 对电子的吸引力较小 = 金属性更强）。关键是壳层数 → IE 更低 → 电子更易失去。
Iron (Fe, $Z=26$) is in the d-block. Which of the following is also in the d-block? 🇨🇦 SCH4U C3.3铁（Fe，$Z=26$）在 d 区。以下哪种元素也在 d 区？🇨🇦 SCH4U C3.3
Q9
Ca ($Z=20$, Group 2)Ca（$Z=20$，第 2 族）
Cu ($Z=29$, Group 11)Cu（$Z=29$，第 11 族）
Ga ($Z=31$, Group 13)Ga（$Z=31$，第 13 族）
Br ($Z=35$, Group 17)Br（$Z=35$，第 17 族）
The d-block contains Groups 3–12 (transition metals). Cu is in Group 11 — firmly in the d-block. Ca (Group 2) is s-block; Ga (Group 13) is p-block; Br (Group 17) is p-block. SCH4U C3.3 requires identifying which block each element belongs to.d 区包含第 3–12 族（过渡金属）。Cu 在第 11 族——牢固地在 d 区中。Ca（第 2 族）是 s 区；Ga（第 13 族）是 p 区；Br（第 17 族）是 p 区。SCH4U C3.3 要求识别每种元素属于哪个区块。
d-block = Groups 3–12 (transition metals). Ca is Group 2 → s-block. Ga is Group 13 → p-block. Br is Group 17 → p-block. Only Cu (Group 11) is in the d-block.d 区 = 第 3–12 族（过渡金属）。Ca 是第 2 族 → s 区。Ga 是第 13 族 → p 区。Br 是第 17 族 → p 区。只有 Cu（第 11 族）在 d 区。
Which element has the greatest nonmetallic character among the following? 🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B3.3以下元素中，哪种元素的非金属性最强？🇺🇸 NGSS HS-PS1-1 / 🇨🇦 SCH3U B3.3
Q10
F (Period 2, Group 17)F（第 2 周期，第 17 族）
Cl (Period 3, Group 17)Cl（第 3 周期，第 17 族）
Br (Period 4, Group 17)Br（第 4 周期，第 17 族）
O (Period 2, Group 16)O（第 2 周期，第 16 族）
Nonmetallic character increases up a group and across a period to the right. F (Period 2, Group 17) is at the top right of the p-block — maximum EN, minimum radius, highest tendency to gain electrons. F is the most nonmetallic element (even slightly more so than O because F has fewer shells and a higher nuclear-charge-to-shell ratio).非金属性在同一族向上增大，在同一周期向右增大。F（第 2 周期，第 17 族）位于 p 区的右上角——最高 EN，最小半径，获得电子的倾向最强。F 是非金属性最强的元素（甚至略强于 O，因为 F 的壳层更少，核电荷与壳层的比率更高）。
Nonmetallic character is highest at the upper right of the table. Among Group 17 elements, reactivity (as an oxidizer) and EN decrease going down: F > Cl > Br. O is in Group 16, Period 2 — also very nonmetallic, but F (Group 17, Period 2) has one more proton on the same period, giving it a slightly higher $Z_\text{eff}$ and EN than O.非金属性在周期表右上角最高。在第 17 族元素中，反应活性（作为氧化剂）和 EN 向下减小：F > Cl > Br。O 在第 16 族，第 2 周期——也很强的非金属性，但 F（第 17 族，第 2 周期）在同一周期有一个更多的质子，使其 $Z_\text{eff}$ 和 EN 略高于 O。
Self-check自查

Readiness Checklist准备就绪清单

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。

0 / 12 mastered已掌握 0 / 12
Next Steps后续单元

What This Feeds Into本单元的去向

The periodic table and its trends are the predictive engine of all of chemistry. Every subsequent topic — chemical bonding (electronegativity determines polarity and bond type), Lewis structures (valence electron count from group position), stoichiometry (reactivity predictions), and equilibrium (acid strength from EN of the nonmetal) — uses the framework you mastered here. The cross-references below point at the college-credit feeder and the next High School Chemistry unit.元素周期表及其趋势是化学所有预测的引擎。每一个后续主题——化学键（电负性决定极性和键型）、路易斯结构（从族的位置得出价电子数）、化学计量学（反应活性预测）和平衡（从非金属的 EN 得出酸强度）——都使用你在这里掌握的框架。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。

Chemical Bonding (Unit 3) uses electronegativity ($\Delta$EN) and valence-electron counts from this unit to draw Lewis structures, classify bond polarity (polar covalent, nonpolar covalent, ionic), and predict molecular shape (VSEPR). Nomenclature (Unit 4) uses the ion charges that follow from group position. Every unit that involves a chemical reaction — from stoichiometry (Unit 5) through equilibrium (Unit 12) — uses the reactivity-trend reasoning developed in §7 of this guide.化学键（第 3 单元）使用本单元的电负性（$\Delta$EN）和价电子数来绘制路易斯结构、分类键极性（极性共价、非极性共价、离子键）并预测分子形状（VSEPR）。命名法（第 4 单元）使用从族的位置得出的离子电荷。每个涉及化学反应的单元——从化学计量学（第 5 单元）到平衡（第 12 单元）——都使用本指南 §7 中发展的反应活性趋势推理。

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 and the periodic table at IB depth)IB Chemistry HL · Structure 1：物质的微粒本质（IB 深度下原子结构与元素周期表的大学学分衔接）

If you are aiming for IB Chemistry HL or AP Chemistry, the periodic-trend framework here is assumed from the first week of the college-credit course. IB Chemistry HL Structure 3 (Periodicity) extends this with successive ionisation energy data and graphical analysis, the explanation of period lengths from quantum numbers, and the detailed reactivity of Groups 1, 2, and 17. AP Chemistry Unit 1 (Atomic Structure and Properties) adds photoelectron spectroscopy (PES) as a direct probe of shell energies and Coulomb's law as the formal basis for $Z_\text{eff}$. The electronegativity and bond-type reasoning here is the same framework used in IB Chemistry HL Structure 2 and AP Chemistry bonding units from day one.如果你目标是 IB Chemistry HL 或 AP Chemistry，这里的周期趋势框架从大学学分课程的第一周就被默认掌握。IB Chemistry HL Structure 3（周期性）通过连续电离能数据和图形分析、从量子数解释周期长度以及第 1、2、17 族的详细反应活性来延伸这部分内容。AP Chemistry Unit 1（原子结构与性质）增加了光电子能谱（PES）作为壳层能量的直接探针，以及库仑定律作为 $Z_\text{eff}$ 的正式基础。这里的电负性和键型推理与 IB Chemistry HL Structure 2 和 AP Chemistry 成键单元中从第一天起使用的框架完全相同。

DINGRUI SCHOLARS

High School Chemistry · The Periodic Table and Periodic Trends · v1高中化学 · 周期表与周期性 · v1