High School Chemistry — The Periodic Table and Periodic Trends — Solutions

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Periodic table history周期表历史 · HS-PS1-1 [3 marks][3 分]

Mendeleev arranged elements in his original periodic table primarily by which property?门捷列夫在其最初的元素周期表中，主要依据哪种性质排列元素？

Answer:答案： (B) Atomic mass相对原子质量

(a) Identify Mendeleev's organising principle明确门捷列夫的排列原则 M1·A1·A1

Mendeleev published his periodic table in 1869, arranging elements in order of increasing atomic mass while grouping those with similar chemical properties. He had no knowledge of atomic number (proton count), which was not established until Moseley's work in 1913. Therefore option (B) is correct.门捷列夫于 1869 年发表元素周期表，按相对原子质量递增顺序排列元素，同时将性质相似的元素归入同一族。他当时对原子序数（质子数）一无所知，原子序数直到 1913 年莫斯利的工作才被确立。因此选 (B)。

Why the distractors fail.干扰项分析。

(A) Atomic number: Moseley proved this is the true organising principle in 1913, after Mendeleev.原子序数：莫斯利于 1913 年证明这是真正的排列原则，晚于门捷列夫。

(D) Electronegativity: a derived, calculated property not used as an organising axis.电负性：这是一个推导计算量，从未用作排列轴。

Mendeleev vs. modern table: mass order vs. atomic number order.门捷列夫周期表与现代周期表：质量排序 vs. 原子序数排序。 Mendeleev left gaps for undiscovered elements and predicted their properties. In a few cases he reversed the mass order (e.g., Te before I) to preserve chemical similarity, foreshadowing the primacy of atomic number. The modern periodic table, based on atomic number, resolves these inversions. Knowing this history helps answer AP and ON questions about why the periodic table was revised after Mendeleev.门捷列夫为未发现元素留下空位并预测其性质。他在少数情况下颠倒了质量顺序（例如将 Te 排在 I 前），以保持化学性质的相似性，预示了原子序数的核心地位。基于原子序数的现代周期表解决了这些倒置问题。了解这段历史有助于回答 AP 和安大略省考中关于"门捷列夫之后为何修订周期表"的问题。

Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §2 Periods and groups周期与族 · HS-PS1-1 [3 marks][3 分]

Which of the following elements belongs to the p-block and Period 3?下列哪种元素属于 p 区且位于第三周期？

Answer:答案： (C) Cl (chlorine) / Cl（氯）

(a) Locate each element by period and block按周期和区块定位各元素 M1·A1·A1

The p-block spans Groups 13-18. Period 3 elements have their highest-energy electrons in the $n = 3$ shell. Checking each option:p 区涵盖第 13 至 18 族。第三周期元素的最高能量电子位于 $n = 3$ 壳层。逐项检查：

(A) Na (Z = 11): Period 3, Group 1, s-block. Eliminated.(A) Na（Z = 11）：第三周期，第 1 族，s 区。排除。
(B) Fe (Z = 26): Period 4, Group 8, d-block. Eliminated (wrong period and block).(B) Fe（Z = 26）：第四周期，第 8 族，d 区。排除（周期和区块均不符）。
(C) Cl (Z = 17): Period 3, Group 17, p-block. Correct. Configuration [Ne] 3s$^2$ 3p$^5$.(C) Cl（Z = 17）：第三周期，第 17 族，p 区。正确。电子构型 [Ne] 3s$^2$ 3p$^5$。
(D) Ca (Z = 20): Period 4, Group 2, s-block. Eliminated (wrong period).(D) Ca（Z = 20）：第四周期，第 2 族，s 区。排除（周期不符）。

Block identity from electron configuration: last subshell filled determines the block.从电子构型判断区块：最后填充的亚层决定所属区块。 s-block: last electron enters an s orbital (Groups 1-2). p-block: last electron enters a p orbital (Groups 13-18). d-block: last electron enters a d orbital (Groups 3-12, transition metals). f-block: last electron enters an f orbital (lanthanides/actinides). For Cl, the configuration ends in 3p$^5$, confirming p-block Period 3. This block-reading skill is tested on every AP Chemistry exam.s 区：最后一个电子填入 s 轨道（第 1-2 族）。p 区：最后一个电子填入 p 轨道（第 13-18 族）。d 区：最后一个电子填入 d 轨道（第 3-12 族，过渡金属）。f 区：最后一个电子填入 f 轨道（镧系/锕系）。Cl 的构型末尾为 3p$^5$，确认属于 p 区第三周期。这一区块判断技能在每次 AP 化学考试中均有考查。

Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3 Atomic radius trend原子半径趋势 · SCH3U [4 marks][4 分]

Consider the elements Li, Na, K in Group 1 and the elements Na, Mg, Al in Period 3.考虑第 1 族中的 Li、Na、K，以及第三周期中的 Na、Mg、Al。

Answer:答案： (a) increases down Group 1: Li < Na < K在第 1 族中从上到下增大：Li < Na < K · (b) decreases across Period 3: Na > Mg > Al在第三周期中从左到右减小：Na > Mg > Al

(a) Trend down Group 1第 1 族向下的趋势 A1·A1

Atomic radius increases from Li to Na to K. Each successive element adds a new electron shell: Li has 2 shells ($n = 1, 2$), Na has 3 shells ($n = 1, 2, 3$), K has 4 shells ($n = 1, 2, 3, 4$). The valence electron is further from the nucleus and is more effectively shielded by inner shells, so the effective nuclear charge experienced by the outermost electron barely increases while the principal quantum number grows.原子半径从 Li 到 Na 到 K 依次增大。每个元素依次增加一个新的电子层：Li 有 2 层（$n = 1, 2$），Na 有 3 层，K 有 4 层。价电子距核越来越远，并被内层电子更有效地屏蔽，因此最外层电子所感受到的有效核电荷几乎不变，而主量子数却在增大。

(b) Trend across Period 3第三周期中的趋势 A1·A1

Atomic radius decreases from Na to Mg to Al. All three are in the same period ($n = 3$ valence shell), so the number of electron shells is the same. However, nuclear charge increases: Na has $Z = 11$, Mg has $Z = 12$, Al has $Z = 13$. The added protons increase the pull on the same shell, contracting the radius.原子半径从 Na 到 Mg 到 Al 依次减小。三者均在第三周期（价层 $n = 3$），电子层数相同。但核电荷数递增：Na 的 $Z = 11$，Mg 的 $Z = 12$，Al 的 $Z = 13$。增加的质子对同一壳层的吸引力增强，使半径收缩。

Two independent causes: shell number governs the group trend; nuclear charge governs the period trend.两个独立原因：电子层数决定族内趋势；核电荷数决定周期内趋势。 When writing exam answers, always name both the cause and its mechanism. For group trends: state that adding a new principal energy level increases the average distance of the valence electron from the nucleus, despite increasing $Z$. For period trends: state that all valence electrons are in the same shell but face a higher nuclear charge, so electrostatic attraction is stronger and the electron cloud is pulled inward. Provincial examiners award the second mark specifically for the mechanism (shielding / nuclear charge), not just the direction.写考试答案时，务必同时说明原因和机制。对于族内趋势：指出增加新的主能层使价电子距核的平均距离增大，即便 $Z$ 也在增大。对于周期内趋势：指出所有价电子都在同一壳层，但面临更高的核电荷数，静电吸引力更强，电子云被向内压缩。省考阅卷人专门为机制（屏蔽效应/核电荷数）给第二个标记分，而非仅凭趋势方向。

Q4MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §4 Ionization energy电离能 · Chem 11 [5 marks][5 分]

The successive ionization energies (in kJ/mol) for an unknown element X are: $IE_1 = 578$, $IE_2 = 1817$, $IE_3 = 2745$, $IE_4 = 11578$, $IE_5 = 14831$. (a) Identify the group. (b) Explain the large jump between $IE_3$ and $IE_4$. (c) Predict the group if the largest jump is between $IE_2$ and $IE_3$.某未知元素 X 的逐级电离能（kJ/mol）为：$IE_1 = 578$，$IE_2 = 1817$，$IE_3 = 2745$，$IE_4 = 11578$，$IE_5 = 14831$。(a) 确定所在族。(b) 解释 $IE_3$ 与 $IE_4$ 间大幅跃升的原因。(c) 若最大跃升在 $IE_2$ 与 $IE_3$ 间，预测该元素所在族。

Answer:答案： (a) Group 13 (3 valence electrons)第 13 族（3 个价电子） · (b) $IE_4$ removes the first core electron$IE_4$ 移除第一个核心电子 · (c) Group 12 (2 valence electrons)第 12 族（2 个价电子）

(a) Identify the group from the large jump从大幅跃升判断所在族 M1·A1

The large jump occurs between $IE_3$ and $IE_4$: $\Delta IE = 11578 - 2745 = 8833$ kJ/mol, roughly four times any earlier gap. This means the first three electrons are removed from the valence shell, and the fourth comes from the core (inner) shell, which is far closer to the nucleus and requires enormously more energy. Having 3 valence electrons places element X in Group 13. A likely candidate is Al (Z = 13), whose actual $IE_1 = 577$ kJ/mol matches closely.大幅跃升出现在 $IE_3$ 与 $IE_4$ 之间：$\Delta IE = 11578 - 2745 = 8833$ kJ/mol，约为前几次跨度的四倍。这说明前三个电子从价层移除，第四个电子来自核层（内层），距核远近得多，需要大得多的能量。拥有 3 个价电子说明元素 X 属于第 13 族。最可能的候选是 Al（Z = 13），其实际 $IE_1 = 577$ kJ/mol 与题目数据高度吻合。

(b) Explain the jump between $IE_3$ and $IE_4$解释 $IE_3$ 与 $IE_4$ 间的跳跃 A1·A1

The first three ionization energies remove valence electrons (from the $n = 3$ shell for Al), which experience significant shielding from inner shells and are relatively far from the nucleus. The fourth electron must be removed from the $n = 2$ core shell, which is (1) much closer to the nucleus, (2) experiences less shielding, and therefore (3) has a much higher effective nuclear charge pulling on it. This produces the dramatic energy jump.前三次电离能移除的是价层电子（对 Al 而言为 $n = 3$ 层），这些电子受到内层较大屏蔽且距核较远。第四个电子须从 $n = 2$ 核层移除，该层 (1) 距核近得多，(2) 受到的屏蔽更少，因此 (3) 感受到的有效核电荷更大。这就造成了能量的骤然跃升。

(c) Predict group from jump between $IE_2$ and $IE_3$从 $IE_2$ 与 $IE_3$ 间的跃升预测族别 A1

If the largest jump is between $IE_2$ and $IE_3$, the element has 2 valence electrons and belongs to Group 12 (or Group 2 for s-block). The jump signals that $IE_3$ is the first core-electron removal.若最大跃升出现在 $IE_2$ 与 $IE_3$ 之间，则该元素有2 个价电子，属于第 12 族（或 s 区第 2 族）。该跃升表明 $IE_3$ 是第一次移除核层电子。

The jump in successive ionization energies reveals the number of valence electrons and thus the group.逐级电离能的跃升揭示了价电子数，进而确定所属族别。 Rule: the large jump occurs after removing all valence electrons; the number of removals before the jump equals the group number for main-group elements. This is one of the most powerful periodic-trends tools on AP Chemistry FRQs and BC/ON provincial exams. Note that $IE$ values roughly double from one ionization to the next within the same shell, but jump by a factor of 4-10 when crossing from valence to core. Always report the group, not just the jump location, for full marks.规律：大幅跃升发生在所有价电子被移除之后；跃升前的移除次数即为主族元素的族号。这是 AP 化学简答题和卑诗/安大略省考中最有力的周期趋势工具之一。注意：在同一壳层内，$IE$ 值从一次电离到下一次大约翻倍；而从价层跨越到核层时，则会跃升 4 至 10 倍。作答时务必报出族别，而非仅说明跃升位置，才能得满分。

Q5MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §5 Electronegativity电负性 · HS-PS1-2 [4 marks][4 分]

Arrange Na, Cl, F, Mg in order of increasing electronegativity. Which sequence is correct?将 Na、Cl、F、Mg 按电负性从小到大排列，哪个顺序正确？

Answer:答案： (A) Na < Mg < Cl < F

(a) Apply electronegativity trends to rank the four elements运用电负性趋势对四种元素排序 M1·A1·A1·A1

Electronegativity increases across a period (left to right) and up a group (bottom to top). Using Pauling values: Na = 0.93, Mg = 1.31, Cl = 3.16, F = 3.98. Ranked in increasing order: Na < Mg < Cl < F, which is option (A).电负性在同周期中从左到右增大，在同族中从下到上增大。鲍林电负性值：Na = 0.93，Mg = 1.31，Cl = 3.16，F = 3.98。按从小到大排列：Na < Mg < Cl < F，即选项 (A)。

Why the distractors fail.干扰项分析。

(B) Reverses the order entirely; F is the most electronegative element.完全颠倒了顺序；F 是电负性最高的元素。

(D) Places Mg below Na, which is incorrect; Mg has higher nuclear charge in the same period.将 Mg 排在 Na 之下，不正确；Mg 在同周期中核电荷数更高。

Electronegativity correlates with effective nuclear charge and inversely with atomic radius.电负性与有效核电荷正相关，与原子半径负相关。 F is the most electronegative element (no element exceeds it) because it combines a very high effective nuclear charge with a very small radius, pulling shared electrons strongly toward itself. Na and Mg are both metals with low electronegativity; Mg's slightly higher nuclear charge gives it a modest edge. Cl vs. F: both are halogens (Group 17), but F is one period higher and has a smaller radius, making it more electronegative despite lower Z. This Group 17 anomaly (F more electronegative than Cl even though Cl has higher Z) is a standard AP and Ontario exam question.F 是电负性最高的元素（没有元素超过它），因为它同时具有很高的有效核电荷和极小的原子半径，能将共用电子强烈地拉向自身。Na 和 Mg 均为金属，电负性较低；Mg 略高的核电荷数使其稍占优势。Cl 与 F 的比较：两者都是卤素（第 17 族），但 F 在更高的周期，半径更小，因此即便 Z 更低仍比 Cl 电负性更高。这一第 17 族特例（F 的电负性高于 Cl，尽管 Cl 的 Z 更大）是 AP 和安大略省考的标准考题。

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §3+4 Radius and ionization energy半径与电离能 · HS-PS1-1 [7 marks][7 分]

Consider the elements Na, Al, and Cl, all in Period 3. (a) Rank in order of increasing atomic radius, explain. (b) Rank in order of increasing first ionization energy, explain. (c) Compare radii of Na and Na$^+$.考虑均处于第三周期的元素 Na、Al 和 Cl。(a) 按原子半径从小到大排序并解释。(b) 按第一电离能从小到大排序并解释。(c) 比较 Na 与 Na$^+$ 的半径。

Answer:答案： (a) Cl < Al < Na · (b) Na < Al < Cl · (c) Na is larger than Na$^+$Na 比 Na$^+$ 大

(a) Rank by increasing atomic radius按原子半径从小到大排序 M1·A1·A1

All three elements are in Period 3, so they have valence electrons in the $n = 3$ shell. Moving from Na (Z = 11) to Al (Z = 13) to Cl (Z = 17), nuclear charge increases with no new shells added. The increasing nuclear charge pulls the electron cloud inward, decreasing radius. Increasing radius order: Cl < Al < Na. Approximate atomic radii: Cl = 99 pm, Al = 143 pm, Na = 186 pm.三者均在第三周期，价电子均位于 $n = 3$ 层。从 Na（Z = 11）到 Al（Z = 13）再到 Cl（Z = 17），核电荷数递增而未增加新壳层。核电荷增大将电子云向内拉，使半径减小。按半径从小到大：Cl < Al < Na。参考原子半径：Cl = 99 pm，Al = 143 pm，Na = 186 pm。

(b) Rank by increasing first ionization energy按第一电离能从小到大排序 M1·A1

The general trend across Period 3 is increasing $IE_1$: Na (496 kJ/mol) < Al (577 kJ/mol) < Cl (1251 kJ/mol). The trend mirrors decreasing atomic radius: smaller radius means the valence electron is held more tightly. Note: Mg (738 kJ/mol) is anomalously higher than Al because Al's outermost electron is in a 3p subshell, which is higher in energy and more easily removed than Mg's 3s$^2$ pair. This Al/Mg anomaly is not in play here since Mg is not listed. Increasing $IE_1$: Na < Al < Cl.第三周期的总体趋势是 $IE_1$ 递增：Na（496 kJ/mol）< Al（577 kJ/mol）< Cl（1251 kJ/mol）。该趋势与原子半径递减相对应：半径越小，价电子被束缚得越紧。注意：Mg（738 kJ/mol）异常高于 Al，因为 Al 的最外层电子在 3p 亚层，能量更高且比 Mg 的 3s$^2$ 电子对更易移除。此处未列出 Mg，该反常不涉及。按 $IE_1$ 从小到大：Na < Al < Cl。

(c) Na vs. Na$^+$ radiusNa 与 Na$^+$ 的半径比较 A1·A1

Na (Z = 11) has the electron configuration [Ne] 3s$^1$ (3 shells, 11 electrons). Na$^+$ loses the 3s$^1$ electron: configuration is [Ne] = 1s$^2$ 2s$^2$ 2p$^6$ (2 shells, 10 electrons). The same 11 protons now attract only 10 electrons distributed in 2 shells rather than 3. The result is a dramatically smaller ionic radius: Na$^+$ radius = 102 pm vs. Na radius = 186 pm. Na is larger than Na$^+$.Na（Z = 11）的电子构型为 [Ne] 3s$^1$（3 个壳层，11 个电子）。Na$^+$ 失去 3s$^1$ 电子后，构型为 [Ne] = 1s$^2$ 2s$^2$ 2p$^6$（2 个壳层，10 个电子）。同样的 11 个质子现在只吸引 10 个电子，且电子分布在 2 个壳层而非 3 个。结果是离子半径大幅缩小：Na$^+$ = 102 pm，Na = 186 pm。Na 比 Na$^+$ 大。

Cations are always smaller than their parent atoms; anions are always larger.阳离子始终比其原子小；阴离子始终比其原子大。 When Na forms Na$^+$, it loses an entire electron shell, collapsing from 3 shells to 2. This is a dramatic size reduction. The same proton count now governs fewer electrons spread over fewer shells, increasing the effective nuclear charge per electron and contracting the cloud. The reverse happens for anions: Cl gains an electron to form Cl$^-$, adding electron-electron repulsion in the same shell, which expands the cloud against the same nuclear charge. This cation/anion size rule is worth memorising for AP FRQs on periodic trends.Na 形成 Na$^+$ 时失去一整个电子层，从 3 层坍缩至 2 层，半径急剧减小。同样的质子数现在主导更少的电子，分布在更少的壳层，每个电子所感受到的有效核电荷增大，电子云收缩。阴离子则相反：Cl 得到一个电子形成 Cl$^-$，在同一壳层产生额外的电子间排斥，在同等核电荷作用下使电子云膨胀。这一阳离子/阴离子大小规律值得在 AP 周期趋势简答题中熟记。

Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §5+6 Electron affinity, electronegativity, metallic character电子亲和能、电负性、金属性 · SCH3U [8 marks][8 分]

Pauling electronegativity values: Li = 1.0, Be = 1.6, B = 2.0, N = 3.0, O = 3.4, F = 4.0, Na = 0.9, Cl = 3.2, Br = 2.9. (a) Describe and explain the electronegativity trend across Period 2 and down Group 17. (b) Explain metallic character trends. (c) Classify Si, Sr, Br as metal, metalloid, or nonmetal.鲍林电负性值：Li = 1.0，Be = 1.6，B = 2.0，N = 3.0，O = 3.4，F = 4.0，Na = 0.9，Cl = 3.2，Br = 2.9。(a) 描述并解释第二周期和第 17 族的电负性趋势。(b) 解释金属性趋势。(c) 将 Si、Sr、Br 分类。

Answer:答案： (a) increases across Period 2; decreases down Group 17第二周期中增大；第 17 族向下减小 · (b) increases down group; decreases across period族内向下增大；周期内向右减小 · (c) Si = metalloid; Sr = metal; Br = nonmetal

(a) Electronegativity trend across Period 2 (Li to F)第二周期（Li 至 F）电负性趋势 M1·A1·M1·A1

Data: Li = 1.0, Be = 1.6, B = 2.0, N = 3.0, O = 3.4, F = 4.0. Electronegativity increases from Li to F. Explanation: all six elements have valence electrons in $n = 2$ shells. Moving left to right, nuclear charge increases from $Z = 3$ to $Z = 9$ without adding new shells. The higher nuclear charge pulls the shared electron pair more strongly toward the nucleus, increasing electronegativity. Simultaneously, atomic radius decreases, placing bonding electrons even closer to the pulling nucleus.数据：Li = 1.0，Be = 1.6，B = 2.0，N = 3.0，O = 3.4，F = 4.0。电负性从 Li 到 F 依次增大。解释：六种元素的价电子均在 $n = 2$ 层。从左到右，核电荷数从 $Z = 3$ 增至 $Z = 9$，未增加新壳层。更高的核电荷数更强烈地将共用电子对拉向核，使电负性增大。同时，原子半径减小，使成键电子距拉力更强的核更近。 Electronegativity trend down Group 17 (F to Br): F = 4.0, Cl = 3.2, Br = 2.9. Electronegativity decreases going down the group. Explanation: adding new electron shells increases atomic radius and the shielding effect. The bonding electrons are further from the nucleus and are more screened from it, so the pull on shared electrons weakens.第 17 族（F 至 Br）电负性趋势：F = 4.0，Cl = 3.2，Br = 2.9。电负性在族内向下依次减小。解释：增加新的电子层使原子半径增大，屏蔽效应增强。成键电子距核更远，对核的屏蔽更强，因此对共用电子的吸引力减弱。

(b) Metallic character trends金属性趋势 A1·A1

Metallic character increases down a group because each successive element has a new shell, making valence electrons further from the nucleus, easier to lose, and therefore more metal-like in behaviour (low ionization energy, positive oxidation states, lustre). Metallic character decreases across a period because increasing nuclear charge holds valence electrons more tightly (higher ionization energy), making elements increasingly nonmetallic.金属性在族内向下增大，因为每个元素依次增加新壳层，价电子距核越来越远，越容易失去，金属特性越强（电离能低、正氧化态、金属光泽）。金属性在周期内向右减小，因为核电荷数增大使价电子被束缚得更紧（电离能更高），元素的非金属性越来越强。

(c) Classify Si, Sr, Br分类 Si、Sr、Br A1·A1

Si (Z = 14): Period 3, Group 14. Located on the metalloid staircase. Has semiconductor properties, brittle, moderate conductivity. Metalloid. Sr (Z = 38): Period 5, Group 2. Alkaline earth metal, shiny, conducts electricity, forms Sr$^{2+}$. Metal. Br (Z = 35): Period 4, Group 17. Halogen, liquid at room temperature, poor conductor, high electronegativity. Nonmetal.Si（Z = 14）：第三周期，第 14 族。位于类金属阶梯线上。具有半导体性质，质脆，导电性中等。类金属。Sr（Z = 38）：第五周期，第 2 族。碱土金属，有金属光泽，能导电，形成 Sr$^{2+}$。金属。Br（Z = 35）：第四周期，第 17 族。卤素，室温下为液体，导电性差，电负性高。非金属。

Electronegativity, ionization energy, and metallic character are all governed by the same two variables: nuclear charge and number of shells.电负性、电离能和金属性均由同两个变量决定：核电荷数和电子层数。 This is the unifying insight for all periodic trend questions. High nuclear charge and few shells = high electronegativity, high ionization energy, nonmetal. Low nuclear charge and many shells = low electronegativity, low ionization energy, metal. The metalloids (B, Si, Ge, As, Sb, Te) occupy the diagonal transition zone. For exam efficiency, memorise the staircase boundary and the rule that elements to the upper right are nonmetals, lower left are metals, and the boundary elements are metalloids.这是回答所有周期性趋势问题的统一洞见。高核电荷数且壳层少 = 高电负性、高电离能、非金属。低核电荷数且壳层多 = 低电负性、低电离能、金属。类金属（B、Si、Ge、As、Sb、Te）占据对角线过渡区域。为提高考试效率，记住阶梯分界线和这一规律：右上方元素为非金属，左下方为金属，边界元素为类金属。

Q8HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §7 Reactivity trends反应活性趋势 · Chem 11 [8 marks][8 分]

Group 1 alkali metals and Group 17 halogens react; Group 18 noble gases are unreactive. (a) Balanced equation for K + water. (b) Why K reacts more vigorously than Li. (c) Rank F$_2$, Cl$_2$, Br$_2$ reactivity with H$_2$. (d) Why noble gases are unreactive.第 1 族碱金属与第 17 族卤素均可发生反应；第 18 族惰性气体不活泼。(a) K 与水反应的配平方程式。(b) K 比 Li 反应更剧烈的原因。(c) F$_2$、Cl$_2$、Br$_2$ 与 H$_2$ 的反应活性排列。(d) 惰性气体不活泼的原因。

Answer:答案： (a) $2\text{K}(s) + 2\text{H}_2\text{O}(l) \to 2\text{KOH}(aq) + \text{H}_2(g)$ · (b) larger radius, lower IE in KK 半径更大、电离能更低 · (c) $\text{F}_2 > \text{Cl}_2 > \text{Br}_2$ · (d) full valence octet, zero tendency to gain or lose electrons满价层八隅体，得失电子倾向为零

(a) Balanced equation: K with water配平方程式：K 与水 M1·A1·A1

Potassium reacts vigorously with water to produce potassium hydroxide (a soluble strong base) and hydrogen gas:钾与水剧烈反应，生成氢氧化钾（可溶强碱）和氢气： $$ 2\text{K}(s) \;+\; 2\text{H}_2\text{O}(l) \;\longrightarrow\; 2\text{KOH}(aq) \;+\; \text{H}_2(g) $$ The reaction is exothermic and may ignite the H$_2$ gas produced. State symbols: K is solid, water is liquid, KOH dissolves in the aqueous phase, H$_2$ is a gas.该反应放热，可能点燃生成的 H$_2$ 气体。状态符号：K 为固体，水为液体，KOH 溶于水相，H$_2$ 为气体。

(b) Why K reacts more vigorously than LiK 比 Li 反应更剧烈的原因 A1·A1

K (Period 4) has a larger atomic radius and more electron shells than Li (Period 2). Its single valence electron (4s$^1$) is further from the nucleus, experiences greater shielding from inner shells, and therefore has a lower first ionization energy ($IE_1$: Li = 520 kJ/mol vs. K = 419 kJ/mol). The valence electron is more easily lost, making K a stronger reducing agent and increasing its reactivity with water. The reaction also releases more heat per mole for K.K（第四周期）的原子半径比 Li（第二周期）更大，拥有更多电子层。其唯一的价电子（4s$^1$）距核更远，受内层电子更大的屏蔽，因此具有更低的第一电离能（$IE_1$：Li = 520 kJ/mol，K = 419 kJ/mol）。价电子更易失去，使 K 成为更强的还原剂，与水的反应活性更强。K 的反应每摩尔释放的热量也更多。

(c) Rank F$_2$, Cl$_2$, Br$_2$ reactivity with H$_2$F$_2$、Cl$_2$、Br$_2$ 与 H$_2$ 的反应活性排列 A1·A1

Decreasing reactivity: F$_2$ > Cl$_2$ > Br$_2$. Justification: (1) Electron affinity decreases down Group 17 (F has the highest tendency to gain an electron). (2) The H-X bond formed becomes weaker down the group (H-F = 565 kJ/mol > H-Cl = 432 kJ/mol > H-Br = 366 kJ/mol), though F$_2$ itself has an anomalously weak F-F bond (159 kJ/mol) which also facilitates its reaction. F$_2$ reacts explosively with H$_2$ even in the dark; Cl$_2$ reacts on ignition or UV light; Br$_2$ requires heating and reacts slowly.活性递减顺序：F$_2$ > Cl$_2$ > Br$_2$。论证：(1) 第 17 族向下电子亲和能递减（F 获得电子的倾向最强）。(2) 生成的 H-X 键在族内向下变弱（H-F = 565 kJ/mol > H-Cl = 432 kJ/mol > H-Br = 366 kJ/mol）；此外 F$_2$ 本身的 F-F 键异常弱（159 kJ/mol），也有助于其发生反应。F$_2$ 即使在黑暗中也会与 H$_2$ 爆炸性反应；Cl$_2$ 需点火或紫外线；Br$_2$ 需加热且反应缓慢。

(d) Why noble gases are unreactive惰性气体为何不活泼 A1

Noble gases have a complete valence shell (octet for He is a duet: 1s$^2$; for Ne, Ar, etc.: ns$^2$np$^6$). This configuration is energetically extremely stable. Noble gases have no tendency to gain electrons (very negative electron affinity would be required to force an extra electron into an already-full shell) and essentially no tendency to lose electrons (very high ionization energies). Therefore they do not form bonds under standard conditions.惰性气体具有完整的价层（He 为两电子构型：1s$^2$；Ne、Ar 等为 ns$^2$np$^6$ 八隅体）。这种构型在能量上极为稳定。惰性气体既无得电子的倾向（将额外电子填入已满壳层需要极高的负值电子亲和能），也几乎无失电子的倾向（电离能极高）。因此在标准条件下不形成化学键。

Reactivity of alkali metals and halogens are mirror images: metals want to lose electrons; halogens want to gain them.碱金属与卤素的活性规律互为镜像：金属倾向失电子；卤素倾向得电子。 For alkali metals, reactivity increases down the group because valence electrons are progressively easier to lose (lower $IE_1$). For halogens, reactivity decreases down the group because the electron-attracting pull weakens (lower electron affinity, larger radius). Noble gases represent the chemical "finish line" of electron configuration stability. Understanding why F$_2$ is anomalously reactive (weak F-F bond despite high electronegativity) is a classic AP Chemistry curveball: the question tests whether you can balance multiple factors, not just apply a single trend.对碱金属而言，活性在族内向下增大，因为价电子越来越容易失去（$IE_1$ 更低）。对卤素而言，活性在族内向下减小，因为吸引电子的能力减弱（电子亲和能更低，半径更大）。惰性气体代表电子构型稳定性的化学"终点线"。理解 F$_2$ 为何异常活泼（尽管电负性高，但 F-F 键弱）是 AP 化学的经典难题：该问题考查学生能否综合多重因素，而非仅套用单一趋势。

Q9HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §3+4 Isoelectronic series + successive IE等电子体系 + 逐级电离能 · HS-PS1-2 (Honors extension)（荣誉级延伸） [7 marks][7 分]

Isoelectronic series N$^{3-}$, O$^{2-}$, F$^-$, Ne, Na$^+$, Mg$^{2+}$ all have 10 electrons. (a) Rank in order of increasing ionic/atomic radius. (b) Analyze successive IEs of phosphorus (Z = 15).等电子体系 N$^{3-}$、O$^{2-}$、F$^-$、Ne、Na$^+$、Mg$^{2+}$ 均含 10 个电子。(a) 按离子/原子半径从小到大排序。(b) 分析磷（Z = 15）的逐级电离能。

Answer:答案： (a) Mg$^{2+}$ < Na$^+$ < Ne < F$^-$ < O$^{2-}$ < N$^{3-}$ · (b) IE$_1$ through IE$_3$ remove 3p valence electrons; IE$_4$ and IE$_5$ remove 3s valence; IE$_6$ onward removes core 2p electronsIE$_1$ 至 IE$_3$ 移除 3p 价电子；IE$_4$ 和 IE$_5$ 移除 3s 价电子；IE$_6$ 起移除核层 2p 电子

(a) Rank isoelectronic species by increasing radius按半径从小到大对等电子体系排序 M1·A1·A1

All six species have exactly 10 electrons. The only variable is the nuclear charge (proton count Z). Higher Z means more protons pulling the same 10 electrons closer to the nucleus, so smaller radius. Lower Z means fewer protons, less pull, larger radius.六种物种均恰好含 10 个电子。唯一变量是核电荷数（质子数 Z）。Z 越高，质子越多，将同样的 10 个电子拉得越近，半径越小。Z 越低，质子越少，拉力越弱，半径越大。

Mg$^{2+}$: Z = 12, 10 e$^-$, proton/electron ratio = 1.20 (highest) → smallest
Na$^+$: Z = 11, 10 e$^-$, ratio = 1.10
Ne: Z = 10, 10 e$^-$, ratio = 1.00
F$^-$: Z = 9, 10 e$^-$, ratio = 0.90
O$^{2-}$: Z = 8, 10 e$^-$, ratio = 0.80
N$^{3-}$: Z = 7, 10 e$^-$, ratio = 0.70 (lowest) → largest

Increasing radius: Mg$^{2+}$ < Na$^+$ < Ne < F$^-$ < O$^{2-}$ < N$^{3-}$. Approximate values: Mg$^{2+}$ = 72 pm, Na$^+$ = 102 pm, Ne = 154 pm (van der Waals), F$^-$ = 133 pm, O$^{2-}$ = 140 pm, N$^{3-}$ = 146 pm.半径从小到大：Mg$^{2+}$ < Na$^+$ < Ne < F$^-$ < O$^{2-}$ < N$^{3-}$。参考值：Mg$^{2+}$ = 72 pm，Na$^+$ = 102 pm，Ne = 154 pm（范德华半径），F$^-$ = 133 pm，O$^{2-}$ = 140 pm，N$^{3-}$ = 146 pm。

(b) Successive ionization energies of phosphorus (Z = 15)磷（Z = 15）的逐级电离能 M1·A1·A1·A1

P configuration: [Ne] 3s$^2$ 3p$^3$. Valence electrons: five (3s$^2$ 3p$^3$). Core electrons: ten (in $n = 1$ and $n = 2$ shells). Given: $IE_1 = 1012$, $IE_2 = 1907$, $IE_3 = 2914$, $IE_4 = 4964$ kJ/mol.P 的构型：[Ne] 3s$^2$ 3p$^3$。价电子：5 个（3s$^2$ 3p$^3$）。核心电子：10 个（位于 $n = 1$ 和 $n = 2$ 层）。给定：$IE_1 = 1012$，$IE_2 = 1907$，$IE_3 = 2914$，$IE_4 = 4964$ kJ/mol。

$IE_1$ (1012) and $IE_2$ (1907) and $IE_3$ (2914): remove the three 3p electrons one by one. The jumps are moderate ($\Delta \approx 895$ and $1007$ kJ/mol) because all three come from the same $n = 3$ p-subshell. The increase reflects increasing effective nuclear charge felt by each successive 3p electron as electron-electron repulsion decreases.$IE_1$（1012）、$IE_2$（1907）、$IE_3$（2914）：依次移除三个 3p 电子。跃升幅度适中（$\Delta \approx 895$ 和 1007 kJ/mol），因为三者均来自同一 $n = 3$ p 亚层。增大反映了随着电子间排斥减小，每次后续 3p 电子所感受到的有效核电荷增大。
$IE_4$ (4964): removes the first 3s electron. The jump from $IE_3$ to $IE_4$ ($\Delta = 2050$ kJ/mol, roughly double) is notable: 3s electrons are lower in energy and closer to the nucleus than 3p, so require more energy. This is a subshell jump, not a core jump.$IE_4$（4964）：移除第一个 3s 电子。从 $IE_3$ 到 $IE_4$ 的跃升（$\Delta = 2050$ kJ/mol，约为前者两倍）明显：3s 电子能量低于 3p，距核更近，因此需要更多能量。这是一次亚层跃升，而非核层跃升。

The major core jump (factor of 4-10) would appear between $IE_5$ and $IE_6$ (after removing both 3s electrons and then hitting the $n = 2$ core), not visible in the data given.真正的核层大跃升（4 至 10 倍）将出现在 $IE_5$ 与 $IE_6$ 之间（移除两个 3s 电子后触及 $n = 2$ 核层），在题目给定的数据范围内尚未出现。

Isoelectronic series: same electron count, different nuclear pull. Successive IEs: each removal probes a different subshell depth.等电子体系：电子数相同，核拉力不同。逐级电离能：每次移除探测不同亚层的深度。 The isoelectronic analysis teaches a fundamental principle: radius and energy levels depend on the balance between nuclear charge and electron count, not on electron count alone. For the successive IE analysis, the key is mapping each IE to its source electron (which subshell, which shell). The moderate jump between $IE_3$ and $IE_4$ for P (3p to 3s, same $n$) is distinct from the dramatic jump that would come between $IE_5$ and $IE_6$ (3s to 2p, different $n$). AP Chemistry FRQs frequently ask students to identify core vs. valence removals by inspection of the IE data, without explicitly giving the configuration.等电子体系分析揭示了一个基本原理：半径和能级取决于核电荷数与电子数之间的平衡，而非仅取决于电子数。对于逐级电离能分析，关键在于将每个 $IE$ 对应到其来源电子（哪个亚层、哪个壳层）。P 从 $IE_3$ 到 $IE_4$（3p 至 3s，同一 $n$）的适中跃升，有别于 $IE_5$ 至 $IE_6$（3s 至 2p，不同 $n$）将会出现的剧烈跃升。AP 化学简答题经常要求学生通过观察 $IE$ 数据来判断核层与价层的移除，而不显式给出电子构型。

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §1+2 Electron configuration + block classification电子构型 + 区块分类 · Chem 20-B [8 marks][8 分]

Element X: [Ar] 3d$^{10}$ 4s$^2$ 4p$^3$. Element Y: [Kr] 5s$^1$. (a) Identify period, group, block for each. (b) Which has larger radius and higher IE? (c) Classify X and give its common ion formula.元素 X：[Ar] 3d$^{10}$ 4s$^2$ 4p$^3$。元素 Y：[Kr] 5s$^1$。(a) 分别确定周期、族和区块。(b) 哪个半径更大、电离能更高？(c) 分类 X 并给出其常见离子化学式。

Answer:答案： (a) X = Period 4, Group 15, p-block (As); Y = Period 5, Group 1, s-block (Rb) · (b) Y (Rb) has larger radius; X (As) has higher IEY（Rb）半径更大；X（As）电离能更高 · (c) X is a metalloid; As$^{3-}$ or As$^{3+}$ (As$^{3-}$ in ionic compounds)X 为类金属；As$^{3-}$ 或 As$^{3+}$（离子化合物中常见 As$^{3-}$）

(a) Identify period, group, block for X and Y确定 X 和 Y 的周期、族和区块 M1·A1·M1·A1

Element X [Ar] 3d$^{10}$ 4s$^2$ 4p$^3$: the highest principal quantum number is $n = 4$ (Period 4). The last electrons fill 4p, so X is in the p-block. There are 3 electrons beyond the filled 4s and 3d subshells (4s$^2$ 4p$^3$ = 5 valence electrons counting from the s subshell in the same period), placing X in Group 15. X is arsenic (As), Z = 33.元素 X [Ar] 3d$^{10}$ 4s$^2$ 4p$^3$：最高主量子数为 $n = 4$（第四周期）。最后填充的是 4p，故 X 属于 p 区。计入同周期 s 和 p 亚层共有 5 个价电子，X 属于第 15 族。X 为砷（As），Z = 33。 Element Y [Kr] 5s$^1$: the highest principal quantum number is $n = 5$ (Period 5). The last electron fills 5s, so Y is in the s-block. One electron beyond the noble gas core places Y in Group 1. Y is rubidium (Rb), Z = 37.元素 Y [Kr] 5s$^1$：最高主量子数为 $n = 5$（第五周期）。最后一个电子填入 5s，故 Y 属于 s 区。在惰性气体核之外仅有 1 个电子，Y 属于第 1 族。Y 为铷（Rb），Z = 37。

(b) Compare atomic radius and IE比较原子半径和电离能 A1·A1

Larger radius: Y (Rb). Rb is Period 5, Group 1 (s-block), with 5 electron shells and low nuclear charge pulling on its outer electrons. As is Period 4, Group 15, with a much higher nuclear charge (Z = 33 vs. Z = 37 in the same period context but Rb has the extra 5th shell). Rb radius = 248 pm; As radius = 119 pm. Higher IE: X (As). As has a higher nuclear charge in a smaller atom, so its valence electrons are held more tightly ($IE_1$ As = 947 kJ/mol vs. Rb = 403 kJ/mol).半径更大：Y（Rb）。Rb 属于第五周期第 1 族（s 区），有 5 个电子层，外层电子受到的核拉力较小。As 属于第四周期第 15 族，核电荷数更高，且 Rb 多了第 5 个壳层。Rb 半径 = 248 pm；As 半径 = 119 pm。电离能更高：X（As）。As 原子更小且核电荷数更高，价电子被束缚得更紧（$IE_1$：As = 947 kJ/mol，Rb = 403 kJ/mol）。

(c) Classify X; give common ion formula分类 X；给出常见离子化学式 A1·A1

As (arsenic) lies on the metalloid staircase in Group 15. It shows intermediate properties: brittle solid with semiconductor behaviour, forms both covalent and ionic compounds. Classification: metalloid. Common ion: in ionic compounds with metals, As often forms As$^{3-}$ (arsenide, e.g., Na$_3$As). In oxyanion compounds it shows +3 or +5 oxidation states. For the ion most commonly formed in simple ionic bonding: As$^{3-}$.As（砷）位于第 15 族的类金属阶梯线上。它表现出中间性质：质脆固体，具有半导体行为，既形成共价化合物也形成离子化合物。分类：类金属。常见离子：在与金属形成离子化合物时，As 常形成 As$^{3-}$（砷化物，例如 Na$_3$As）。在含氧酸根化合物中显 +3 或 +5 氧化态。最常见的简单离子化合物离子为 As$^{3-}$。

Reading the periodic table from electron configuration: period = highest n; block = last subshell; group = number of valence electrons (s and p only for main group).从电子构型读取周期表信息：周期 = 最高 n；区块 = 最后填充的亚层；族 = 价电子数（主族仅计 s 和 p）。 This three-step decode works for all main-group elements. For d-block (transition metals), group = (d + s valence electrons). For X, count 4s$^2$ + 4p$^3$ = 5 valence electrons, Group 15. For Y, count 5s$^1$ = 1 valence electron, Group 1. This AB diploma-exam style question tests whether students can reverse-engineer the configuration into the periodic table location without memorising the element, a skill more powerful than rote recall.这三步解码法适用于所有主族元素。对于 d 区（过渡金属），族号 = d + s 价电子数。对于 X：计算 4s$^2$ + 4p$^3$ = 5 个价电子，属第 15 族。对于 Y：计算 5s$^1$ = 1 个价电子，属第 1 族。这道阿省毕业考风格题考查学生能否在不死记元素的情况下，从电子构型反推在周期表中的位置，这是比死记更强大的技能。

Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §6+7 Halogen displacement reactions卤素置换反应 · Chem 30 [8 marks][8 分]

Cl$_2$(aq) added to KBr(aq); Br$_2$(aq) added to KI(aq). Both cause colour changes indicating halogen displacement. (a) Write balanced ionic equations for both. (b) Explain using periodic trends why Cl$_2$ displaces Br$^-$ but Br$_2$ cannot displace Cl$^-$. (c) Predict whether I$_2$(aq) displaces Br$^-$ from KBr(aq).向 KBr(aq) 中加入 Cl$_2$(aq)；向 KI(aq) 中加入 Br$_2$(aq)，两溶液均变色，发生了卤素置换反应。(a) 写出两个配平离子方程式。(b) 用周期性趋势解释 Cl$_2$ 能置换 Br$^-$，但 Br$_2$ 不能置换 Cl$^-$。(c) 预测 I$_2$(aq) 是否能从 KBr(aq) 中置换 Br$^-$。

Answer:答案： (a) see below见下方 · (b) Cl$_2$ has higher oxidising power than Br$_2$ due to higher electron affinityCl$_2$ 的氧化性强于 Br$_2$，因为电子亲和能更高 · (c) No, I$_2$ cannot displace Br$^-$否，I$_2$ 不能置换 Br$^-$

(a) Balanced net ionic equations配平净离子方程式 M1·A1·M1·A1

Reaction 1: Cl$_2$(aq) displaces Br$^-$ from KBr(aq). Cl$_2$ is the stronger oxidising agent; it is reduced to Cl$^-$ while Br$^-$ is oxidised to Br$_2$. Net ionic equation:反应 1：Cl$_2$(aq) 从 KBr(aq) 中置换 Br$^-$。Cl$_2$ 是更强的氧化剂，被还原为 Cl$^-$，Br$^-$ 被氧化为 Br$_2$。净离子方程式： $$ \text{Cl}_2(aq) \;+\; 2\text{Br}^-(aq) \;\longrightarrow\; 2\text{Cl}^-(aq) \;+\; \text{Br}_2(aq) $$ Reaction 2: Br$_2$(aq) displaces I$^-$ from KI(aq). Br$_2$ is a stronger oxidising agent than I$_2$; Br$_2$ is reduced to Br$^-$ while I$^-$ is oxidised to I$_2$. Net ionic equation:反应 2：Br$_2$(aq) 从 KI(aq) 中置换 I$^-$。Br$_2$ 的氧化性强于 I$_2$；Br$_2$ 被还原为 Br$^-$，I$^-$ 被氧化为 I$_2$。净离子方程式： $$ \text{Br}_2(aq) \;+\; 2\text{I}^-(aq) \;\longrightarrow\; 2\text{Br}^-(aq) \;+\; \text{I}_2(aq) $$

(b) Why Cl$_2$ displaces Br$^-$ but Br$_2$ cannot displace Cl$^-$为何 Cl$_2$ 能置换 Br$^-$，而 Br$_2$ 不能置换 Cl$^-$ A1·A1

Oxidising power of halogens decreases down Group 17: F$_2$ > Cl$_2$ > Br$_2$ > I$_2$. This is because electron affinity decreases down the group (smaller atoms attract electrons more strongly) and the X-X bond weakens. Cl$_2$ has a higher electron affinity (349 kJ/mol) than Br$_2$ (325 kJ/mol), making it a stronger oxidising agent. It can oxidise Br$^-$ to Br$_2$ because it is thermodynamically more favourable for Cl$_2$ to gain electrons than for Br$_2$. Br$_2$ cannot oxidise Cl$^-$ because Cl$^-$ is the reduced form of a stronger oxidising agent; there is no thermodynamic driving force.卤素的氧化性在第 17 族向下递减：F$_2$ > Cl$_2$ > Br$_2$ > I$_2$。这是因为电子亲和能在族内向下递减（较小的原子对电子的吸引力更强），且 X-X 键变弱。Cl$_2$ 的电子亲和能（349 kJ/mol）高于 Br$_2$（325 kJ/mol），使其成为更强的氧化剂。Cl$_2$ 能将 Br$^-$ 氧化为 Br$_2$，因为 Cl$_2$ 得到电子在热力学上比 Br$_2$ 更有利。Br$_2$ 不能氧化 Cl$^-$，因为 Cl$^-$ 是更强氧化剂的还原形式，没有热力学驱动力。

(c) Can I$_2$ displace Br$^-$ from KBr?I$_2$ 能从 KBr 中置换 Br$^-$ 吗？ A1·A1

No. I$_2$ is a weaker oxidising agent than Br$_2$ (and far weaker than Cl$_2$). Since Br$_2$ cannot displace Cl$^-$, I$_2$ cannot displace Br$^-$, which requires an oxidising agent stronger than Br$_2$. The reaction $\text{I}_2 + 2\text{Br}^- \to 2\text{I}^- + \text{Br}_2$ is thermodynamically unfavourable; it would spontaneously run in the reverse direction (Br$_2$ displacing I$^-$), which is precisely what Reaction 2 above shows.不能。I$_2$ 的氧化性弱于 Br$_2$（远弱于 Cl$_2$）。由于 Br$_2$ 不能置换 Cl$^-$，I$_2$ 自然也无法置换 Br$^-$，因为这需要比 Br$_2$ 更强的氧化剂。反应 $\text{I}_2 + 2\text{Br}^- \to 2\text{I}^- + \text{Br}_2$ 在热力学上不利；它会自发地向逆方向进行（即 Br$_2$ 置换 I$^-$），这正是上述反应 2 所展示的。

Halogen displacement follows a strict activity series: a halogen can only displace halide ions of halogens below it in Group 17.卤素置换遵循严格的活动性顺序：卤素只能置换位于其在第 17 族中下方的卤素的卤化物离子。 This is the halogen activity series: F$_2$ > Cl$_2$ > Br$_2$ > I$_2$. The colour changes are diagnostic: Cl$_2$ turns a KBr solution orange-brown (Br$_2$ produced); Br$_2$ turns a KI solution brown-black (I$_2$ produced). In a laboratory, these colour tests identify unknown halide solutions. The same trend underlies industrial uses: Cl$_2$ is used to oxidise Br$^-$ in seawater to harvest bromine commercially. Remember that "displacement" in this context means oxidation-reduction, not physical displacement.这就是卤素活动性顺序：F$_2$ > Cl$_2$ > Br$_2$ > I$_2$。颜色变化具有诊断意义：Cl$_2$ 使 KBr 溶液变为橙棕色（生成 Br$_2$）；Br$_2$ 使 KI 溶液变为棕黑色（生成 I$_2$）。在实验室中，这些颜色测试可用于鉴别未知卤化物溶液。同一趋势也有工业应用：工业上利用 Cl$_2$ 氧化海水中的 Br$^-$ 来提取溴。请记住，此处的"置换"是氧化还原反应，而非物理置换。

Q12HARD难 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §3+5 Periodic trend cross-comparison周期性趋势综合对比 · HS-PS1-1+2 [11 marks][11 分]

Be (Z = 4), Mg (Z = 12), Ca (Z = 20), Ba (Z = 56), all in Group 2. (a) Rank by increasing IE$_1$. (b) Which of Be or Ba has higher electronegativity? (c) Predict reactivity with water; write general equation. (d) Explain why Ca$^{2+}$ is larger than Be$^{2+}$, and why cations are smaller than neutral atoms.Be（Z = 4）、Mg（Z = 12）、Ca（Z = 20）、Ba（Z = 56），均位于第 2 族。(a) 按 $IE_1$ 从小到大排序。(b) Be 和 Ba 中哪个电负性更高？(c) 预测与水的反应活性；写出通式。(d) 解释 Ca$^{2+}$ 为何比 Be$^{2+}$ 大，以及阳离子为何比中性原子小。

Answer:答案： (a) Ba < Ca < Mg < Be · (b) Be · (c) reactivity increases down Group 2: Be < Mg < Ca < Ba活性在第 2 族向下增大：Be < Mg < Ca < Ba · (d) Ca$^{2+}$ has more shells than Be$^{2+}$; cations lost a shell vs. parent atomCa$^{2+}$ 比 Be$^{2+}$ 有更多电子层；阳离子相比中性原子失去一个壳层

(a) Rank Be, Mg, Ca, Ba by increasing first ionization energy将 Be、Mg、Ca、Ba 按第一电离能从小到大排序 M1·A1·A1

Going down Group 2, each element adds a new electron shell. The valence electrons (2s or 5s for Ba) are progressively farther from the nucleus and more shielded by inner shells, requiring less energy to remove. The effective nuclear charge on the outermost electron barely changes despite increasing Z. Therefore $IE_1$ decreases down the group:沿第 2 族向下，每个元素增加一个新的电子层。价电子（2s 或 Ba 的 5s）距核越来越远，受内层屏蔽越来越多，移除所需的能量越来越少。尽管 Z 增大，最外层电子所感受的有效核电荷几乎不变。因此 $IE_1$ 在族内向下递减：

Be: $IE_1 = 900$ kJ/mol (2 shells)
Mg: $IE_1 = 738$ kJ/mol (3 shells)
Ca: $IE_1 = 590$ kJ/mol (4 shells)
Ba: $IE_1 = 503$ kJ/mol (6 shells)

Increasing order: Ba < Ca < Mg < Be.从小到大顺序：Ba < Ca < Mg < Be。

(b) Be vs. Ba electronegativityBe 与 Ba 的电负性比较 A1·A1

Be has higher electronegativity. Be is at the top of Group 2 (Period 2), with a very small atomic radius and relatively high effective nuclear charge for its period, pulling shared electrons more strongly toward itself. Pauling values: Be = 1.57, Ba = 0.89. The principle: higher effective nuclear charge and smaller radius both raise electronegativity. Moving down Group 2, both $Z_{\text{eff}}$ and radius increase, but the radius effect dominates, lowering electronegativity.Be 的电负性更高。Be 位于第 2 族顶部（第二周期），原子半径极小，在其所在周期中有效核电荷相对较高，能更强烈地将共用电子拉向自身。鲍林电负性：Be = 1.57，Ba = 0.89。原理：有效核电荷越高、半径越小，电负性越大。沿第 2 族向下，$Z_{\text{eff}}$ 和半径均增大，但半径效应占主导，导致电负性降低。

(c) Reactivity with water; general equation与水的反应活性；通式 M1·A1·A1

Reactivity with water increases down Group 2 because lower $IE_1$ means the metal can more readily lose its two valence electrons to water. Be barely reacts with water (it forms a protective BeO layer). Mg reacts very slowly with cold water but faster with steam. Ca reacts noticeably with cold water; Ba reacts vigorously. General equation for Group 2 metal M reacting with water:与水的反应活性在第 2 族向下增大，因为 $IE_1$ 更低意味着金属更容易将两个价电子转移给水。Be 几乎不与水反应（形成保护性 BeO 层）。Mg 与冷水反应极慢，但与水蒸气较快。Ca 与冷水有明显反应；Ba 反应剧烈。第 2 族金属 M 与水反应的通式： $$ \text{M}(s) \;+\; 2\text{H}_2\text{O}(l) \;\longrightarrow\; \text{M(OH)}_2(aq) \;+\; \text{H}_2(g) $$ Predicted reactivity order: Be < Mg < Ca < Ba (increases down the group).预测活性顺序：Be < Mg < Ca < Ba（沿族向下增大）。

(d) Why Ca$^{2+}$ > Be$^{2+}$ in radius; why cations are smaller than neutral atomsCa$^{2+}$ 为何比 Be$^{2+}$ 半径大；阳离子为何比中性原子小 A1·A1·A1

Why Ca$^{2+}$ is larger than Be$^{2+}$: Both cations have lost their 2 valence electrons. Ca$^{2+}$ has configuration [Ar] = 1s$^2$ 2s$^2$ 2p$^6$ 3s$^2$ 3p$^6$ (3 shells remaining, 18 electrons, Z = 20). Be$^{2+}$ has configuration [He]$^0$ = 1s$^2$ (1 shell remaining, 2 electrons, Z = 4). Ca$^{2+}$ has more electron shells, which places its outermost electrons farther from the nucleus. The data confirm: Be$^{2+}$ = 45 pm, Ca$^{2+}$ = 100 pm.Ca$^{2+}$ 比 Be$^{2+}$ 大的原因：两个阳离子均失去了 2 个价电子。Ca$^{2+}$ 的构型为 [Ar] = 1s$^2$ 2s$^2$ 2p$^6$ 3s$^2$ 3p$^6$（剩余 3 个壳层，18 个电子，Z = 20）。Be$^{2+}$ 的构型为 1s$^2$（剩余 1 个壳层，2 个电子，Z = 4）。Ca$^{2+}$ 有更多电子层，使其最外层电子距核更远。数据印证：Be$^{2+}$ = 45 pm，Ca$^{2+}$ = 100 pm。 Why each cation is smaller than its neutral atom: When Group 2 atoms form 2+ cations, they lose their entire outermost shell (the ns$^2$ shell). The remaining electrons occupy inner shells that are closer to the nucleus. Additionally, the same nuclear charge now attracts fewer electrons, increasing the effective nuclear charge per remaining electron. Both effects reduce the radius. For example, Ca atom = 197 pm; Ca$^{2+}$ = 100 pm (almost half).每个阳离子比中性原子小的原因：第 2 族原子形成 2+ 阳离子时，失去整个最外层（ns$^2$ 层）。剩余电子分布在距核更近的内层。此外，同样的核电荷现在吸引更少的电子，使每个剩余电子所感受的有效核电荷增大。两种效应共同减小了半径。例如：Ca 原子 = 197 pm；Ca$^{2+}$ = 100 pm（约减半）。

For isogroup cation comparisons, shell count dominates radius even when nuclear charge increases; for cation vs. neutral atom, losing a shell is a far larger effect than the proton count.对同族阳离子的比较，即便核电荷数增大，壳层数对半径的影响仍占主导；对阳离子与中性原子的比较，失去一个壳层的效应远大于质子数的差别。 This question synthesises four separate periodic-trend concepts in a single context. The AP Chemistry exam frequently uses Group 2 as the test bed because (1) all four trends apply (radius, IE, electronegativity, reactivity); (2) the $M^{2+}$ cation comparison adds a fifth layer (ionic radius vs. atomic radius); (3) Be is anomalous (does not react with cold water, forms mostly covalent bonds) while Ba behaves like a typical reactive alkaline earth. Knowing these exceptions distinguishes a 5 from a 3 on the AP Chemistry exam.这道题在单一情境中综合了四个独立的周期性趋势概念。AP 化学考试频繁以第 2 族为测试载体，原因在于：(1) 四种趋势（半径、电离能、电负性、活性）全部适用；(2) $M^{2+}$ 阳离子比较增加了第五个维度（离子半径与原子半径）；(3) Be 存在反常（不与冷水反应，多形成共价键），而 Ba 表现为典型的活泼碱土金属。了解这些例外正是 AP 化学考试 5 分与 3 分的分水岭。