PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 25 marksAP 风格选择题 + 安/卑省考短答 · 共 25 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter and show enough work to justify your choice. For short-answer items, name units where applicable. No calculator on Q1-Q3; calculator permitted on Q4-Q5.本节包含选择题与短答题。选择题请圈出字母答案并写出足以支持选项的推理。短答题需写明适用单位。Q1-Q3 不可使用计算器;Q4-Q5 可用计算器。
Arrange the following elements in order of increasing electronegativity: Na, Cl, F, Mg. Which sequence is correct?将以下元素按电负性从小到大排列:Na、Cl、F、Mg。哪个顺序正确?
(A) Na < Mg < Cl < F
(B) F < Cl < Mg < Na
(C) Na < Cl < Mg < F
(D) Mg < Na < F < Cl
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 30 marksAP 衔接简答题 + 荣誉级 · 共 30 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. Justify all trend predictions with reference to nuclear charge and electron shielding. State units where applicable. Calculator permitted on Q6-Q9.每一步推理都要写出。所有趋势预测均需结合核电荷数与电子屏蔽效应加以论证。需写明适用单位。Q6-Q9 可用计算器。
Consider the elements Na, Al, and Cl, all in Period 3.考虑均处于第三周期的元素 Na、Al 和 Cl。
(a)Rank Na, Al, Cl in order of increasing atomic radius. Explain in terms of nuclear charge and shielding.将 Na、Al、Cl 按原子半径从小到大排序,并结合核电荷数和屏蔽效应加以解释。[3]
(b)Rank Na, Al, Cl in order of increasing first ionization energy. Explain the general trend and the anomaly between Al and Si (where applicable).将 Na、Al、Cl 按第一电离能从小到大排序,并解释总体趋势及 Al 与 Si 之间的反常(如适用)。[2]
(c)Predict whether Na or Na$^+$ has a larger radius. Explain.预测 Na 与 Na$^+$ 哪个半径更大,并解释原因。[2]
The Pauling electronegativity values for selected elements are: 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.部分元素的鲍林电负性值如下: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)Using the data above, describe the trend in electronegativity across Period 2 (Li to F) and down Group 17 (F to Br). Explain each in terms of nuclear charge and atomic radius.利用以上数据,描述电负性在第二周期(Li 到 F)和第 17 族(F 到 Br)的变化趋势,并分别结合核电荷数与原子半径加以解释。[4]
(b)Explain why metallic character increases down a group and decreases across a period.解释为什么金属性在同族中从上到下增大,在同周期中从左到右减小。[2]
(c)Classify the following as metals, metalloids, or nonmetals and justify: Si, Sr, Br.将以下元素分类为金属、类金属或非金属,并说明理由:Si、Sr、Br。[2]
Group 1 alkali metals (Li, Na, K) all react with water. Group 17 halogens (F$_2$, Cl$_2$, Br$_2$, I$_2$) react with hydrogen gas. Group 18 noble gases are unreactive under standard conditions.第 1 族碱金属(Li、Na、K)均与水反应;第 17 族卤素(F$_2$、Cl$_2$、Br$_2$、I$_2$)与氢气反应;第 18 族惰性气体在标准条件下不发生反应。
(a)Write a balanced equation for the reaction of potassium (K) with water. Identify the products and their states at room temperature.写出钾(K)与水反应的配平方程式。注明产物及其在室温下的状态。[3]
(b)Explain why K reacts more vigorously with water than Li does. Use the concept of ionization energy and atomic radius.解释为什么 K 与水的反应比 Li 更剧烈,需结合电离能和原子半径的概念。[2]
(c)Rank F$_2$, Cl$_2$, Br$_2$ in order of decreasing reactivity with H$_2$. Justify in terms of bond energy and electron affinity.将 F$_2$、Cl$_2$、Br$_2$ 按与 H$_2$ 的反应活性从大到小排列,并从键能和电子亲和能角度加以论证。[2]
(d)Explain why noble gases are unreactive. What orbital configuration causes this?解释惰性气体为何不活泼。是什么轨道构型导致这一性质?[1]
The isoelectronic series N$^{3-}$, O$^{2-}$, F$^-$, Ne, Na$^+$, Mg$^{2+}$ all have 10 electrons.等电子体系 N$^{3-}$、O$^{2-}$、F$^-$、Ne、Na$^+$、Mg$^{2+}$ 均含 10 个电子。
(a)Rank the species in order of increasing ionic (or atomic) radius. Explain using the proton-to-electron ratio.将以上物种按离子(或原子)半径从小到大排序,并用质子-电子比加以解释。[3]
(b)The first four successive ionization energies of phosphorus (Z = 15) are approximately: $IE_1 = 1012$, $IE_2 = 1907$, $IE_3 = 2914$, $IE_4 = 4964$ kJ/mol. Identify which electrons (core vs valence) are being removed at each step. Explain the pattern in the jumps.磷(Z = 15)前四级电离能约为:$IE_1 = 1012$,$IE_2 = 1907$,$IE_3 = 2914$,$IE_4 = 4964$ kJ/mol。判断每步移除的是核外还是价层电子,并解释跃升规律。[4]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 27 marks阿省毕业考 + 通用题型 · 共 27 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define all variables (with units) at the start of each question. Justify trend predictions with atomic theory. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.每题开始时定义所有变量(含单位)。用原子理论支撑所有趋势预测。每题以一句完整的结合情境的句子作答。第三部分全程可用计算器。
Element X has the electron configuration [Ar] 3d$^{10}$ 4s$^2$ 4p$^3$. Element Y has the configuration [Kr] 5s$^1$.元素 X 的电子构型为 [Ar] 3d$^{10}$ 4s$^2$ 4p$^3$。元素 Y 的构型为 [Kr] 5s$^1$。
(a)Identify the period, group, and block (s, p, d, or f) for each of X and Y.分别确定 X 和 Y 所在的周期、族与区块(s、p、d 或 f 区)。[4]
(b)Predict which element has the larger atomic radius and which has the higher first ionization energy. Justify each prediction.预测哪个元素的原子半径更大,哪个的第一电离能更高,并分别论证。[2]
(c)Predict whether X is more likely to be a metal, metalloid, or nonmetal. State the chemical formula of the ion X most commonly forms.预测 X 更可能是金属、类金属还是非金属,并写出 X 最常见离子的化学式。[2]
A student adds Cl$_2$(aq) to a solution of KBr(aq), then separately adds Br$_2$(aq) to a solution of KI(aq). Both solutions change colour, indicating a halogen displacement reaction.学生向 KBr(aq) 溶液中加入 Cl$_2$(aq),再分别向 KI(aq) 溶液中加入 Br$_2$(aq),两溶液均变色,表明发生了卤素置换反应。
(a)Write balanced ionic equations for both displacement reactions.写出两个置换反应的配平离子方程式。[4]
(b)Explain using periodic trends why Cl$_2$ can displace Br$^-$ but Br$_2$ cannot displace Cl$^-$.用周期性趋势解释为什么 Cl$_2$ 能置换 Br$^-$,但 Br$_2$ 不能置换 Cl$^-$。[2]
(c)Predict whether I$_2$(aq) would displace Br$^-$ from KBr(aq). Justify your answer.预测 I$_2$(aq) 是否能从 KBr(aq) 中置换出 Br$^-$,并给出理由。[2]
Use periodic trends to answer questions about four elements: Be (Z = 4), Mg (Z = 12), Ca (Z = 20), and Ba (Z = 56), all in Group 2.运用周期性趋势回答有关四种元素的问题:Be(Z = 4)、Mg(Z = 12)、Ca(Z = 20)和 Ba(Z = 56),均位于第 2 族。
(a)Rank Be, Mg, Ca, Ba in order of increasing first ionization energy. Explain the trend in terms of atomic radius and nuclear shielding.将 Be、Mg、Ca、Ba 按第一电离能从小到大排序,并从原子半径和核屏蔽效应角度解释该趋势。[3]
(b)Which element (Be or Ba) would you expect to have a higher electronegativity? Explain your choice using the concept of effective nuclear charge.Be 和 Ba 中哪个的电负性更高?用有效核电荷的概念解释你的选择。[2]
(c)Predict the relative reactivity of Be, Mg, Ca, and Ba with water. Write a general equation for the reaction of a Group 2 metal (M) with water.预测 Be、Mg、Ca、Ba 与水反应的相对活性。写出第 2 族金属(M)与水反应的通式。[3]
(d)The ionic radii of the 2+ cations are: Be$^{2+}$ = 45 pm, Mg$^{2+}$ = 72 pm, Ca$^{2+}$ = 100 pm, Ba$^{2+}$ = 135 pm. Explain why Ca$^{2+}$ has a larger radius than Be$^{2+}$ despite both having a 2+ charge, and why each cation is smaller than its neutral atom.2+ 离子的离子半径为:Be$^{2+}$ = 45 pm,Mg$^{2+}$ = 72 pm,Ca$^{2+}$ = 100 pm,Ba$^{2+}$ = 135 pm。解释为什么 Ca$^{2+}$ 半径比 Be$^{2+}$ 大(尽管均为 +2 价),以及为什么每个阳离子均比其中性原子小。[3]
🇺🇸 US NGSS美国 NGSSHS-PS1-1 · HS-PS1-2
🇨🇦 Ontario安大略SCH3U Unit 1 · SCH4U Unit 1
🇨🇦 British Columbia不列颠哥伦比亚Chemistry 11: periodic table, trends化学 11:周期表、周期性趋势
🇨🇦 Alberta阿尔伯塔Chemistry 20 Unit B · Chem 30
Full Syllabus Map in the companion Study Guide. Note: NGSS covers periodic trends (HS-PS1-1/2); AB Chem 30 extends to advanced bonding applications of trends.完整大纲对照表见配套学习指南。注:NGSS 涵盖周期性趋势(HS-PS1-1/2);阿省化学 30 将趋势延伸至高级成键应用。