High School Chemistry — Atomic Structure and the Quantum Model — 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 🇨🇦 ON安 AP-style MCQAP 风格选择题 §1 Subatomic particles亚原子粒子 · HS-PS1-1 [3 marks][3 分]

An atom is represented as ${}^{40}_{20}\text{Ca}$. How many neutrons does this atom contain?某原子表示为 ${}^{40}_{20}\text{Ca}$。该原子含有多少个中子？

Answer:答案： (A) $20$ neutrons / 个中子

Decode the nuclear symbol解读核符号 M1·A1·A1

In the symbol ${}^{A}_{Z}\text{X}$, the mass number $A = 40$ is the total count of protons and neutrons, and the atomic number $Z = 20$ is the number of protons:在符号 ${}^{A}_{Z}\text{X}$ 中，质量数 $A = 40$ 是质子数与中子数的总和，原子序数 $Z = 20$ 是质子数： $$ \text{neutrons} = A - Z = 40 - 20 = 20. $$ Option (A).选 (A)。

Why the distractors fail.干扰项分析。

(B) $40$: this is the mass number $A$, not the neutron count.这是质量数 $A$，而非中子数。

(D) $18$: the number of electrons in the isoelectronic ion $\text{Ca}^{2+}$, not the neutral atom's neutron count.这是等电子离子 $\text{Ca}^{2+}$ 的电子数，并非中性原子的中子数。

Mass number minus atomic number gives neutron count.质量数减去原子序数等于中子数。 The subscript (bottom number) in a nuclear symbol is always the atomic number $Z$, equal to the number of protons and, for a neutral atom, also the number of electrons. The superscript (top number) is the mass number $A$, the total of protons and neutrons. Neutrons are therefore $A - Z$. Calcium has 20 protons regardless of the isotope; ${}^{40}\text{Ca}$ and ${}^{44}\text{Ca}$ differ only in neutron count (20 vs 24). The periodic table lists calcium at $Z = 20$, confirming the subscript.核符号中的下标（底部数字）始终是原子序数 $Z$，等于质子数，对于中性原子也等于电子数。上标（顶部数字）是质量数 $A$，即质子数与中子数之和。因此中子数 $= A - Z$。钙的质子数恒为 20，与同位素无关；${}^{40}\text{Ca}$ 与 ${}^{44}\text{Ca}$ 只在中子数上不同（分别为 20 和 24）。元素周期表中钙的 $Z = 20$，与下标一致。

Q2EASY易 🇺🇸 US美 🇨🇦 BC卑 AP-style MCQAP 风格选择题 §2 Isotopes同位素 · SCH3U B3.2 [3 marks][3 分]

Which pair of species represents isotopes of the same element?下列哪对物种是同一元素的同位素？

Answer:答案： (B) ${}^{14}_{6}\text{C}$ and ${}^{12}_{6}\text{C}$

Apply the definition of isotopes应用同位素定义 M1·A1·A1

Isotopes are atoms of the same element (same $Z$, same number of protons) with different mass numbers (different numbers of neutrons). Check each pair:同位素是同一元素的原子（相同的 $Z$，即相同的质子数），但质量数不同（中子数不同）。逐项检查：

(A) ${}^{12}_{6}\text{C}$ vs ${}^{14}_{7}\text{N}$: $Z = 6$ vs $Z = 7$, so these are different elements.(A) ${}^{12}_{6}\text{C}$ 与 ${}^{14}_{7}\text{N}$：$Z = 6$ 与 $Z = 7$，不同元素。
(B) ${}^{14}_{6}\text{C}$ vs ${}^{12}_{6}\text{C}$: both have $Z = 6$, but $A = 14$ vs $A = 12$, so same element with different neutron counts. These are isotopes. Correct.(B) ${}^{14}_{6}\text{C}$ 与 ${}^{12}_{6}\text{C}$：均有 $Z = 6$，但 $A = 14$ 与 $A = 12$，相同元素、中子数不同。这是同位素。正确。
(C) ${}^{16}_{8}\text{O}$ vs ${}^{18}_{9}\text{F}$: $Z = 8$ vs $Z = 9$, so these are different elements.(C) ${}^{16}_{8}\text{O}$ 与 ${}^{18}_{9}\text{F}$：$Z = 8$ 与 $Z = 9$，不同元素。
(D) ${}^{23}_{11}\text{Na}$ vs ${}^{24}_{12}\text{Mg}$: $Z = 11$ vs $Z = 12$, so these are different elements.(D) ${}^{23}_{11}\text{Na}$ 与 ${}^{24}_{12}\text{Mg}$：$Z = 11$ 与 $Z = 12$，不同元素。

Isotopes: same $Z$, different $A$ (and different neutron count).同位素：相同 $Z$，不同 $A$（中子数不同）。 The atomic number $Z$ is the identity of an element. Two atoms with the same $Z$ are the same element by definition; if they also differ in $A$, they are isotopes. Carbon-12 and carbon-14 are the classic pair: both have 6 protons, but C-12 has 6 neutrons while C-14 has 8. C-14 is radioactive and forms the basis of radiocarbon dating. A quick check for any multiple-choice pair: look at the subscripts first. If they differ, the pair cannot be isotopes regardless of what the superscripts say.原子序数 $Z$ 是元素的身份标识。两个原子若 $Z$ 相同，根据定义就是同一种元素；若同时 $A$ 不同，则互为同位素。碳-12 与碳-14 是经典的同位素对：两者均有 6 个质子，但 C-12 有 6 个中子而 C-14 有 8 个。C-14 具有放射性，是放射性碳定年法的基础。对于任何选择题，先检查下标：若下标不同，无论上标如何，该对物种都不可能是同位素。

Q3MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3 Bohr model玻尔模型 · SCH4U C3.1 [5 marks][5 分]

A hydrogen electron transitions from energy level $n = 4$ to $n = 2$. Energy levels: $E_n = -13.6/n^2\ \text{eV}$.氢原子中电子从能级 $n = 4$ 跃迁到 $n = 2$。能级：$E_n = -13.6/n^2\ \text{eV}$。

Answer:答案： (a) $E_4 = -0.850\ \text{eV}$, $E_2 = -3.40\ \text{eV}$ · (b) $\Delta E = 2.55\ \text{eV}$ · (c) Yes, visible是，可见光

(a) Calculate $E_4$ and $E_2$计算 $E_4$ 和 $E_2$ A1·A1

$$ E_4 = \frac{-13.6}{4^2} = \frac{-13.6}{16} = -0.850\ \text{eV} $$ $$ E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.40\ \text{eV} $$

(b) Energy of emitted photon发射光子的能量 M1·A1

The electron falls from $n = 4$ to $n = 2$, releasing energy. The photon energy equals the magnitude of the energy difference:电子从 $n = 4$ 跃迁到 $n = 2$，释放能量。光子能量等于能量差的绝对值： $$ E_{\text{photon}} = E_4 - E_2 = (-0.850) - (-3.40) = 2.55\ \text{eV} $$

(c) Visible range check可见光范围判断 A1

The visible range is $1.77$ to $3.10\ \text{eV}$. Since $1.77 < 2.55 < 3.10$, the photon is within the visible range. This line corresponds to the Balmer series (blue-green light, approximately $486\ \text{nm}$).可见光范围为 $1.77$ 至 $3.10\ \text{eV}$。由于 $1.77 < 2.55 < 3.10$，该光子在可见光范围内。这条谱线属于巴尔末系（蓝绿色光，约 $486\ \text{nm}$）。

Downward transitions release photons; energy = gap between levels.向下跃迁释放光子；光子能量等于能级间隔。 In the Bohr model, energy levels are negative because electrons are bound (zero energy corresponds to a free electron at rest infinitely far from the nucleus). When an electron drops from a higher level (less negative) to a lower level (more negative), the difference in energy is released as a photon. Always compute $E_{\text{photon}} = E_{\text{upper}} - E_{\text{lower}}$, which is positive for emission. The Balmer series ($n \to 2$) produces lines in the visible spectrum. The $n = 4 \to 2$ line at $2.55\ \text{eV}$ is the blue-green H-beta line at $486\ \text{nm}$.在玻尔模型中，能级为负值，因为电子处于束缚态（零能量对应无穷远处静止的自由电子）。电子从较高能级（较小负值）跃迁到较低能级（较大负值）时，能量差以光子形式释放。始终用 $E_{\text{光子}} = E_{\text{上}} - E_{\text{下}}$ 计算，对于发射过程该值为正。巴尔末系（$n \to 2$）的谱线落在可见光范围内。$n = 4 \to 2$ 跃迁产生能量 $2.55\ \text{eV}$ 的蓝绿色 H-beta 谱线，波长约 $486\ \text{nm}$。

Q4MEDIUM中 🇨🇦 AB阿 🇨🇦 ON安 AB Diploma-style阿尔伯塔毕业考风格 §2 Average atomic mass平均原子质量 · SCH3U B3.2 [6 marks][6 分]

Copper: ${}^{63}\text{Cu}$ (mass $62.930\ \text{u}$, abundance $69.17\%$) and ${}^{65}\text{Cu}$ (mass $64.928\ \text{u}$, abundance $30.83\%$).铜：${}^{63}\text{Cu}$（质量 $62.930\ \text{u}$，丰度 $69.17\%$）和 ${}^{65}\text{Cu}$（质量 $64.928\ \text{u}$，丰度 $30.83\%$）。

Answer:答案： (a) Cu-63: 29p, 34n, 29e; Cu-65: 29p, 36n, 29eCu-63：29 质子、34 中子、29 电子；Cu-65：29 质子、36 中子、29 电子 · (b) $63.55\ \text{u}$ · (c) Cu-63 is more abundantCu-63 丰度更高

(a) Subatomic particle counts亚原子粒子数 A1·A1

Copper has $Z = 29$ (29 protons). For a neutral atom, electrons $=$ protons $= 29$. Neutrons $= A - Z$:铜的 $Z = 29$（29 个质子）。对于中性原子，电子数 $=$ 质子数 $= 29$。中子数 $= A - Z$：

${}^{63}\text{Cu}$: 29 protons, $63 - 29 = 34$ neutrons, 29 electrons.29 个质子，$63 - 29 = 34$ 个中子，29 个电子。
${}^{65}\text{Cu}$: 29 protons, $65 - 29 = 36$ neutrons, 29 electrons.29 个质子，$65 - 29 = 36$ 个中子，29 个电子。

(b) Average atomic mass calculation平均原子质量计算 M1·A1·A1

Average atomic mass is the abundance-weighted sum of isotope masses:平均原子质量是同位素质量按丰度加权的总和： $$ \bar{m} = (0.6917 \times 62.930) + (0.3083 \times 64.928) $$ $$ = 43.528 + 20.026 = 63.554\ \text{u} \approx 63.55\ \text{u} $$ (The periodic table value for Cu is $63.546\ \text{u}$, confirming this calculation.)（元素周期表中铜的原子质量为 $63.546\ \text{u}$，与计算结果吻合。）

(c) Why the average is closer to 63为何平均值更接近 63 A1

Because ${}^{63}\text{Cu}$ has a much higher natural abundance ($69.17\%$) than ${}^{65}\text{Cu}$ ($30.83\%$), the weighted average is pulled toward the mass of ${}^{63}\text{Cu}$.由于 ${}^{63}\text{Cu}$ 的天然丰度（$69.17\%$）远高于 ${}^{65}\text{Cu}$（$30.83\%$），加权平均值向 ${}^{63}\text{Cu}$ 的质量靠拢。

Average atomic mass is a weighted average, not an arithmetic average.平均原子质量是加权平均，而非算术平均。 The simple arithmetic mean of 62.930 and 64.928 would be $63.929\ \text{u}$, which is wrong because it treats both isotopes as equally abundant. The weighted average accounts for the fact that a random copper atom is roughly twice as likely to be ${}^{63}\text{Cu}$ as ${}^{65}\text{Cu}$. This is why the reported atomic mass on the periodic table ($63.55\ \text{u}$) is not a whole number: it reflects the isotopic mixture found in nature. Note that no individual copper atom has a mass of $63.55\ \text{u}$: every copper atom is either ${}^{63}\text{Cu}$ or ${}^{65}\text{Cu}$.62.930 与 64.928 的简单算术平均为 $63.929\ \text{u}$，这是错误的，因为它把两种同位素的丰度视为相等。加权平均考虑了随机抽取一个铜原子时，${}^{63}\text{Cu}$ 的概率约为 ${}^{65}\text{Cu}$ 的两倍。这正是元素周期表上铜的原子质量（$63.55\ \text{u}$）不是整数的原因：它反映了自然界中同位素的混合比例。注意，没有任何一个铜原子的质量恰好等于 $63.55\ \text{u}$，每个铜原子要么是 ${}^{63}\text{Cu}$，要么是 ${}^{65}\text{Cu}$。

Q5MEDIUM中 🇨🇦 BC卑 🇨🇦 AB阿 BC Provincial-style卑诗省考风格 §5 Electron configuration电子排布 · HS-PS1-1 [9 marks][9 分]

Full ground-state electron configurations and valence electron counts for P, Cl$^-$, and explanation of isoelectronic Cl$^-$/Ar.写出 P、Cl$^-$ 的基态全电子排布式、价电子数，并解释 Cl$^-$ 与 Ar 等电子的原因。

Answer:答案： (a) $1s^2 2s^2 2p^6 3s^2 3p^3$, 5 valence e5 个价电子 · (b) $1s^2 2s^2 2p^6 3s^2 3p^6$, 8 valence e8 个价电子 · (c) Cl gains 1 electron, reaching 18 electrons = ArCl 得到 1 个电子，共 18 个电子，与 Ar 相同

(a) Phosphorus ($Z = 15$)磷（$Z = 15$） A1·A1·A1

Fill subshells in order of increasing energy (1s, 2s, 2p, 3s, 3p, ...):按能量由低到高填充各分层（1s, 2s, 2p, 3s, 3p, ...）： $$ \text{P}: 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^3 \quad (2 + 2 + 6 + 2 + 3 = 15\ \checkmark) $$ Valence electrons are in the outermost shell ($n = 3$): $3s^2 3p^3 = 5$ valence electrons.价电子位于最外层（$n = 3$）：$3s^2 3p^3$，共 $5$ 个价电子。

(b) Chloride ion ($\text{Cl}^-$, $Z = 17$, gains 1 electron)氯离子（$\text{Cl}^-$，$Z = 17$，得到 1 个电子） A1·A1·A1

Neutral Cl has 17 electrons; $\text{Cl}^-$ has $17 + 1 = 18$ electrons:中性 Cl 有 17 个电子；$\text{Cl}^-$ 有 $17 + 1 = 18$ 个电子： $$ \text{Cl}^-: 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6 \quad (2 + 2 + 6 + 2 + 6 = 18\ \checkmark) $$ Valence electrons in $n = 3$ shell: $3s^2 3p^6 = 8$ valence electrons (a full octet).最外层 $n = 3$ 中的价电子：$3s^2 3p^6$，共 $8$ 个价电子（满八隅体）。

(c) Why $\text{Cl}^-$ and Ar are isoelectronic为何 $\text{Cl}^-$ 与 Ar 等电子 A1·A1·A1

Argon ($Z = 18$) has 18 electrons with configuration $1s^2 2s^2 2p^6 3s^2 3p^6$. Chlorine ($Z = 17$) has only 17 electrons, but when it gains one electron to form $\text{Cl}^-$, it too has 18 electrons arranged identically. The electron configurations are the same because electron count determines the arrangement, not the number of protons. However, they are chemically different: $\text{Cl}^-$ has 17 protons (a greater proton-to-electron ratio of 17:18) so it is slightly smaller than Ar, which has an 18:18 ratio and a larger effective nuclear charge per electron than $\text{Cl}^-$.氩（$Z = 18$）有 18 个电子，构型为 $1s^2 2s^2 2p^6 3s^2 3p^6$。氯（$Z = 17$）只有 17 个电子，但当它获得一个电子形成 $\text{Cl}^-$ 后，同样拥有 18 个电子，排布方式完全相同。电子排布相同是因为排布由电子数决定，而非质子数。然而两者化学性质不同：$\text{Cl}^-$ 有 17 个质子（质子与电子之比为 17:18），因此比 Ar 略小；而 Ar 的比值为 18:18，每个电子所受的有效核电荷比 $\text{Cl}^-$ 大。

Isoelectronic species share electron count but differ in nuclear charge and chemical behavior.等电子体共享电子数，但核电荷数和化学行为不同。 The series $\text{S}^{2-}$, $\text{Cl}^-$, $\text{Ar}$, $\text{K}^+$, $\text{Ca}^{2+}$ all have 18 electrons and the same electron configuration $[\text{Ne}]\ 3s^2 3p^6$. They differ only in the number of protons, which sets the effective nuclear charge. More protons pull the same 18 electrons closer, reducing ionic radius. This is why ionic radius decreases across an isoelectronic series as nuclear charge increases: $\text{S}^{2-}$ is largest, $\text{Ca}^{2+}$ is smallest.$\text{S}^{2-}$、$\text{Cl}^-$、$\text{Ar}$、$\text{K}^+$、$\text{Ca}^{2+}$ 均有 18 个电子，电子排布相同（$[\text{Ne}]\ 3s^2 3p^6$）。它们的差异仅在质子数，进而决定有效核电荷。质子越多，对相同的 18 个电子吸引力越大，离子半径越小。因此，在等电子体系列中，随核电荷增大，离子半径依次减小：$\text{S}^{2-}$ 最大，$\text{Ca}^{2+}$ 最小。

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 🇨🇦 ON安 AP-feeder FRQAP 衔接简答题 §6 Atomic emission spectra原子发射光谱 · SCH4U C3.1 [7 marks][7 分]

Hydrogen gas excited by electrical discharge produces a line spectrum. Balmer series: transitions to $n = 2$.氢气受电放电激发产生线状光谱。巴尔末系：跃迁到 $n = 2$。

Answer:答案： (a) Quantized energy levels量子化能级 · (b) $E_{\text{photon}} = 1.89\ \text{eV}$ · (c) Emission; atom was previously excited发射；原子此前已被激发

(a) Why hydrogen produces a line spectrum氢为何产生线状光谱 A1·A1·A1

Electrons in hydrogen can only occupy specific, quantized energy levels (not a continuous range). When electrons transition between these discrete levels, they emit photons of only those specific energies that match the gaps between levels. Since only certain energy gaps exist, only certain photon wavelengths are produced, yielding a series of discrete lines rather than a continuous spectrum. A continuous spectrum would require all possible energy values to be allowed, which they are not in the quantum model.氢原子中的电子只能处于特定的量子化能级（而非连续的能量范围）。当电子在这些离散能级之间跃迁时，只释放与能级间隔精确对应的特定能量光子。由于只存在特定的能量间隔，只产生特定波长的光子，从而形成一系列离散谱线而非连续光谱。连续光谱要求所有可能的能量值都被允许，而在量子模型中并非如此。

(b) Energy of the $n = 3 \to n = 2$ Balmer line$n = 3 \to n = 2$ 巴尔末系谱线的光子能量 M1·A1·A1

$$ E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9} = -1.511\ \text{eV} $$ $$ E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.400\ \text{eV} $$ $$ E_{\text{photon}} = E_3 - E_2 = (-1.511) - (-3.400) = 1.889\ \text{eV} \approx 1.89\ \text{eV} $$ This is the H-alpha line (red, approximately $656\ \text{nm}$), the most prominent line in the Balmer series.这是 H-alpha 谱线（红色，约 $656\ \text{nm}$），是巴尔末系中最显著的一条。

(c) Emission vs absorption; prior process发射还是吸收；发生此过程前原子经历了什么 A1

This is emission: the electron is falling from a higher energy level ($n = 3$) to a lower one ($n = 2$), releasing a photon. Before emitting the photon, the atom must have been excited (absorbed energy, either from the electrical discharge or a photon), promoting an electron from $n = 2$ to $n = 3$ or higher.这是发射过程：电子从较高能级（$n = 3$）跃迁到较低能级（$n = 2$），释放光子。在发射光子之前，原子必须已被激发（通过电放电或吸收光子获得能量），使电子从 $n = 2$ 跃迁到 $n = 3$ 或更高能级。

Emission spectra and absorption spectra are complementary: same lines, opposite processes.发射光谱与吸收光谱互补：谱线相同，过程相反。 When atoms are excited (heated, discharged), electrons jump up; they then fall back down emitting photons of specific wavelengths (emission spectrum: bright lines on dark background). When white light passes through a cool gas, the same wavelengths are absorbed (absorption spectrum: dark lines on continuous background). The Balmer series for hydrogen has four visible lines: H-alpha (red, 656 nm, 1.89 eV), H-beta (blue-green, 486 nm, 2.55 eV), H-gamma (violet, 434 nm), and H-delta (violet, 410 nm). The electrical discharge in a discharge tube provides the energy to excite electrons to higher levels; the observed light comes from the subsequent downward transitions.当原子被激发（加热、放电）时，电子向上跃迁；随后落回低能级，发射特定波长的光子（发射光谱：暗背景上的亮线）。当白光通过冷气体时，相同波长被吸收（吸收光谱：连续背景上的暗线）。氢的巴尔末系有四条可见光谱线：H-alpha（红，656 nm，1.89 eV）、H-beta（蓝绿，486 nm，2.55 eV）、H-gamma（紫，434 nm）和 H-delta（紫，410 nm）。放电管中的电放电提供能量将电子激发到更高能级；观察到的光来自随后的向下跃迁。

Q7MEDIUM中 🇨🇦 ON安 🇨🇦 AB阿 ON Provincial-style安大略省考风格 §7 Periodic patterns周期规律 · HS-PS1-1 [8 marks][8 分]

Electron configuration and periodic trends: Na, Cl, K atomic radius; Na vs Mg ionization energy; electronegativity vs atomic radius trends.电子排布与周期规律：Na、Cl、K 的原子半径；Na 与 Mg 的电离能；电负性与原子半径趋势。

Answer:答案： (a) Cl < Na < KCl < Na < K · (b) Mg has higher first IEMg 的第一电离能更高 · (c) Both driven by nuclear charge increase across period均由同周期核电荷增大驱动

(a) Atomic radius order: Na, Cl, K原子半径排序：Na、Cl、K A1·A1·A1

Electron configurations: Na ($Z = 11$): $[\text{Ne}]\ 3s^1$; Cl ($Z = 17$): $[\text{Ne}]\ 3s^2 3p^5$; K ($Z = 19$): $[\text{Ar}]\ 4s^1$.电子排布：Na（$Z = 11$）：$[\text{Ne}]\ 3s^1$；Cl（$Z = 17$）：$[\text{Ne}]\ 3s^2 3p^5$；K（$Z = 19$）：$[\text{Ar}]\ 4s^1$。

Na vs Cl (same period, Period 3): Both have electrons in $n = 3$ as the outermost shell. But Cl has $Z = 17$ vs Na's $Z = 11$. More protons pull the same-shell electrons closer, so Cl is smaller than Na.Na 与 Cl（同周期，第三周期）：两者最外层均在 $n = 3$，但 Cl 的 $Z = 17$，Na 的 $Z = 11$。质子越多，对同层电子的吸引力越大，故 Cl 比 Na 小。
K vs Na: K has its outermost electron in $n = 4$, which is farther from the nucleus than Na's $n = 3$ shell, making K much larger.K 与 Na：K 的最外层电子在 $n = 4$，比 Na 的 $n = 3$ 层离核更远，故 K 远大于 Na。

Order of increasing radius: $\text{Cl} < \text{Na} < \text{K}$.原子半径从小到大：$\text{Cl} < \text{Na} < \text{K}$。

(b) First ionization energy: Na vs Mg第一电离能：Na 与 Mg A1·A1·A1

Na ($Z = 11$): $[\text{Ne}]\ 3s^1$ (one valence electron). Mg ($Z = 12$): $[\text{Ne}]\ 3s^2$ (two valence electrons). Moving from Na to Mg, nuclear charge increases by 1 while both elements still have electrons only in $n = 3$ as the outermost shell and similar shielding from core electrons. The increased nuclear charge of Mg pulls the outermost electron more strongly. Therefore, Mg has a higher first ionization energy than Na ($738\ \text{kJ/mol}$ vs $496\ \text{kJ/mol}$).Na（$Z = 11$）：$[\text{Ne}]\ 3s^1$（一个价电子）。Mg（$Z = 12$）：$[\text{Ne}]\ 3s^2$（两个价电子）。从 Na 到 Mg，核电荷增加 1，而两者的最外层均在 $n = 3$，内层屏蔽相近。Mg 更大的核电荷对最外层电子吸引力更强，因此Mg 的第一电离能更高（$738\ \text{kJ/mol}$ 对比 $496\ \text{kJ/mol}$）。

(c) Electronegativity is the reverse of atomic radius across a period同周期内电负性与原子半径趋势相反 A1·A1

Across a period (left to right), nuclear charge increases while the number of electron shells remains the same. The greater nuclear charge pulls all electrons (including those in bonds) closer and more strongly. This simultaneously decreases atomic radius (electrons pulled in) and increases electronegativity (greater attraction for bonding electrons). The two trends are thus directly linked to the same underlying cause: increasing nuclear charge across a period.同周期从左到右，核电荷增大而电子层数不变。更大的核电荷对所有电子（包括成键电子）的吸引力更强。这同时导致原子半径减小（电子被拉近）和电负性增大（对成键电子的吸引力增强）。两种趋势都源于同一根本原因：同周期核电荷的增大。

Periodic trends all trace back to nuclear charge and shielding.所有周期规律都源于核电荷与屏蔽效应。 The two main factors controlling atomic size are (1) the principal quantum number $n$ of the outermost electrons (more shells = larger atom) and (2) the effective nuclear charge $Z_{\text{eff}} = Z - \sigma$, where $\sigma$ is the shielding constant from inner electrons. Going across a period, $n$ stays the same but $Z$ increases, so $Z_{\text{eff}}$ increases, shrinking the atom and raising the ionization energy and electronegativity. Going down a group, $n$ increases by 1 each row, which dominates over the increasing $Z$, making the atom larger and lowering IE and electronegativity.控制原子大小的两个主要因素是：(1) 最外层电子的主量子数 $n$（层数越多原子越大）；(2) 有效核电荷 $Z_{\text{eff}} = Z - \sigma$，其中 $\sigma$ 是内层电子的屏蔽常数。同周期从左到右，$n$ 不变而 $Z$ 增大，故 $Z_{\text{eff}}$ 增大，原子缩小，电离能和电负性升高。同族从上到下，$n$ 每行增加 1，这一效应主导，原子增大，电离能和电负性降低。

Q8HARD难 🇺🇸 US美 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §1 + §5 Ions and electron configuration离子与电子排布 · HS-PS1-1 [7 marks][7 分]

Iron ion with electron configuration $[\text{Ar}]\ 3d^6$. Iron: $Z = 26$; $[\text{Ar}] = 18$ electrons.铁离子电子排布为 $[\text{Ar}]\ 3d^6$。铁：$Z = 26$；$[\text{Ar}] = 18$ 个电子。

Answer:答案： (a) $\text{Fe}^{2+}$ · (b) $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2$ · (c) Period 4, Group 8第四周期，第 8 族 · (d) 6 valence electrons (3d electrons)6 个价电子（3d 电子）

(a) Identify the ion确定该离子 M1·A1

The ion has $[\text{Ar}]\ 3d^6$, meaning $18 + 6 = 24$ electrons. Neutral iron ($Z = 26$) has 26 electrons. Lost electrons $= 26 - 24 = 2$. The ion is $\text{Fe}^{2+}$.该离子有 $[\text{Ar}]\ 3d^6$，即 $18 + 6 = 24$ 个电子。中性铁（$Z = 26$）有 26 个电子。失去电子数 $= 26 - 24 = 2$，故该离子为 $\text{Fe}^{2+}$。

(b) Full electron configuration of neutral Fe中性铁原子的全电子排布式 M1·A1

Fill 26 electrons in order: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6$ (total 18, $=$ [Ar]), then $3d^6, 4s^2$ (note: $4s$ fills before $3d$ but $4s$ electrons are lost first when forming cations):依序填入 26 个电子：$1s^2, 2s^2, 2p^6, 3s^2, 3p^6$（共 18 个，即 [Ar]），然后 $3d^6, 4s^2$（注意：$4s$ 在 $3d$ 之前填充，但形成阳离子时 $4s$ 电子先失去）： $$ \text{Fe}: 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^6\ 4s^2 \quad (2+2+6+2+6+6+2=26\ \checkmark) $$ Abbreviated: $[\text{Ar}]\ 3d^6\ 4s^2$.简写为：$[\text{Ar}]\ 3d^6\ 4s^2$。

(c) Period and group of iron铁在周期表中的周期和族 A1·A1

Period 4: the highest principal quantum number in Fe's configuration is $n = 4$ (the $4s^2$ electrons), placing it in Period 4. Group 8: Iron is a d-block element. For d-block elements, the group number $=$ (number of $d$ electrons) $+$ (number of $s$ electrons in the outermost $s$ subshell) $= 6 + 2 = 8$. Iron is in Group 8 (transition metals).第四周期：铁的电子排布中最高主量子数为 $n = 4$（$4s^2$ 电子），因此位于第四周期。第 8 族：铁是 d 区元素。对于 d 区元素，族数 $=$ d 电子数 $+$ 最外层 s 亚层电子数 $= 6 + 2 = 8$。铁位于第 8 族（过渡金属）。

(d) Valence electrons of $\text{Fe}^{2+}$$\text{Fe}^{2+}$ 的价电子数 A1

For $\text{Fe}^{2+}$ with configuration $[\text{Ar}]\ 3d^6$: the $4s$ electrons have been removed (they are lost first when forming cations). The remaining valence electrons are the 6 electrons in the $3d$ subshell. For transition metal ions, the d electrons participate in bonding and chemical reactions, so $\text{Fe}^{2+}$ has 6 valence electrons.$\text{Fe}^{2+}$ 的电子排布为 $[\text{Ar}]\ 3d^6$：$4s$ 电子已被移除（形成阳离子时 $4s$ 电子先失去）。剩余价电子即 $3d$ 分层中的 6 个电子。对于过渡金属离子，d 电子参与成键和化学反应，故 $\text{Fe}^{2+}$ 有 6 个价电子。

Transition metals lose $s$ electrons before $d$ electrons when forming cations.过渡金属形成阳离子时，先失去 s 电子，再失去 d 电子。 Although $4s$ fills before $3d$ in neutral atoms (Aufbau principle), the $4s$ electrons are higher in energy in the presence of the ion's reduced electron count. Therefore, when Fe loses 2 electrons, it loses both $4s$ electrons to give $[\text{Ar}]\ 3d^6$ rather than losing $3d$ electrons to give $[\text{Ar}]\ 3d^4\ 4s^2$. This is why $\text{Fe}^{2+}$ retains all 6 d electrons. An important consequence: $\text{Fe}^{2+}$ ($3d^6$) has 4 unpaired d electrons, making it paramagnetic and contributing to the magnetic properties of iron compounds.尽管在中性原子中 $4s$ 先于 $3d$ 填充（Aufbau 原理），但在离子的较少电子数情况下，$4s$ 电子能量更高。因此，铁失去 2 个电子时，失去的是 $4s$ 电子，得到 $[\text{Ar}]\ 3d^6$，而非失去 $3d$ 电子得到 $[\text{Ar}]\ 3d^4\ 4s^2$。这就是 $\text{Fe}^{2+}$ 保留全部 6 个 d 电子的原因。一个重要推论：$\text{Fe}^{2+}$（$3d^6$）有 4 个未配对 d 电子，使其具有顺磁性，这也是铁化合物磁性的来源。

Q9HARD难Honors荣誉级 🇺🇸 US美 🇨🇦 ON安 AP-feeder FRQAP 衔接简答题 §4 Quantum-mechanical model量子力学模型 · HS-PS1-1 (above Chem 30 floor)（超出 Chem 30 基准） [8 marks][8 分]

Quantum numbers $n$, $\ell$, $m_\ell$, $m_s$. Pauli exclusion principle.量子数 $n$, $\ell$, $m_\ell$, $m_s$。泡利不相容原理。

Answer:答案： (a) $\ell = 0, 1, 2$ (s, p, d) · (b) 5 orbitals, $m_\ell = -2, -1, 0, +1, +2$5 个轨道，$m_\ell = -2, -1, 0, +1, +2$ · (c) Only 2 values of $m_s$ per orbital每个轨道只有 2 个 $m_s$ 值 · (d) Bohr fails for multi-electron atoms玻尔模型对多电子原子失效

(a) Allowed values of $\ell$ when $n = 3$$n = 3$ 时 $\ell$ 的允许值 A1·A1

The rule: $\ell$ ranges from $0$ to $n - 1$. For $n = 3$: $\ell = 0, 1, 2$, corresponding to the subshells:规则：$\ell$ 的范围为 $0$ 到 $n - 1$。当 $n = 3$ 时：$\ell = 0, 1, 2$，对应的分层为：

$\ell = 0$: $s$ subshell (spherical)$s$ 分层（球形）
$\ell = 1$: $p$ subshell (dumbbell-shaped)$p$ 分层（哑铃形）
$\ell = 2$: $d$ subshell (more complex shapes)$d$ 分层（更复杂形状）

(b) Orbitals in $3d$ and allowed $m_\ell$ values$3d$ 分层的轨道数和 $m_\ell$ 允许值 A1·A1

For $\ell = 2$ (d subshell), $m_\ell$ ranges from $-\ell$ to $+\ell$:对于 $\ell = 2$（d 分层），$m_\ell$ 的范围为 $-\ell$ 到 $+\ell$： $$ m_\ell = -2,\ -1,\ 0,\ +1,\ +2 $$ That gives $2\ell + 1 = 2(2) + 1 = 5$ distinct values, so the $3d$ subshell contains 5 orbitals and can hold up to $5 \times 2 = 10$ electrons.共有 $2\ell + 1 = 2(2) + 1 = 5$ 个不同值，故 $3d$ 分层含 5 个轨道，最多容纳 $5 \times 2 = 10$ 个电子。

(c) Pauli exclusion principle: max 2 electrons per orbital泡利不相容原理：每个轨道最多 2 个电子 A1·A1

Each orbital is specified by a set of three quantum numbers $(n, \ell, m_\ell)$. By the Pauli exclusion principle, no two electrons in the same atom can have the same four quantum numbers. Since $n$, $\ell$, and $m_\ell$ are fixed for a given orbital, the only quantum number that can differ between two electrons in that orbital is $m_s$. The spin quantum number has exactly two allowed values: $m_s = +\tfrac{1}{2}$ (spin-up) and $m_s = -\tfrac{1}{2}$ (spin-down). Therefore, at most 2 electrons (with opposite spins) can occupy any single orbital.每个轨道由一组三个量子数 $(n, \ell, m_\ell)$ 确定。根据泡利不相容原理，同一原子中不存在四个量子数完全相同的两个电子。由于 $n$、$\ell$、$m_\ell$ 对同一轨道是固定的，该轨道中两个电子之间唯一可能不同的量子数是 $m_s$。自旋量子数恰好只有两个允许值：$m_s = +\tfrac{1}{2}$（自旋向上）和 $m_s = -\tfrac{1}{2}$（自旋向下）。因此，每个轨道最多容纳 2 个电子（自旋方向相反）。

(d) Bohr model limitation corrected by the quantum-mechanical model量子力学模型纠正的玻尔模型关键局限 A1·A1

Limitation of the Bohr model: It only works for hydrogen and hydrogen-like ions (one electron). It assumes electrons travel in fixed, circular orbits and fails to predict the spectra of multi-electron atoms because it ignores electron-electron repulsions and the three-dimensional nature of electron motion. Quantum-mechanical model correction: Electrons are described by probability distributions (orbitals) in three dimensions, with four quantum numbers that encode energy, shape, orientation, and spin. This correctly accounts for electron-electron interactions and predicts spectra for all elements.玻尔模型的局限：它只适用于氢原子和类氢离子（单电子）。它假设电子在固定的圆形轨道上运动，由于忽略了电子间排斥力和电子运动的三维性质，无法预测多电子原子的光谱。量子力学模型的纠正：用三维概率分布（轨道）描述电子，用四个量子数编码能量、形状、取向和自旋。这正确考虑了电子间相互作用，并能预测所有元素的光谱。

The four quantum numbers completely specify an electron's state; no two electrons in the same atom can share all four.四个量子数完整描述电子的状态；同一原子中不存在四个量子数完全相同的两个电子。 Quantum number summary: $n$ (shell, $1, 2, 3, \ldots$) sets the energy and size; $\ell$ (subshell, $0$ to $n-1$) sets the shape; $m_\ell$ (orbital, $-\ell$ to $+\ell$) sets the spatial orientation; $m_s$ ($\pm \tfrac{1}{2}$) sets the spin. The total number of states in a shell $n$ is $2n^2$ (e.g., $n = 3$: $2 \times 9 = 18$ states). The Pauli principle is what prevents all electrons from collapsing into the $1s$ orbital and is the quantum-mechanical foundation for the shell structure of the periodic table.量子数汇总：$n$（壳层，$1, 2, 3, \ldots$）决定能量和大小；$\ell$（分层，$0$ 到 $n-1$）决定形状；$m_\ell$（轨道，$-\ell$ 到 $+\ell$）决定空间取向；$m_s$（$\pm \tfrac{1}{2}$）决定自旋。一个壳层 $n$ 中的状态总数为 $2n^2$（如 $n = 3$：$2 \times 9 = 18$ 个状态）。泡利原理阻止了所有电子塌缩到 $1s$ 轨道，是元素周期表壳层结构的量子力学基础。

Section C · Worked SolutionsC 部分 · 详细解答

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §2 + §3 Isotopes + Bohr energy同位素 + 玻尔能量 · Chem 30 Unit A [8 marks][8 分]

Neon isotopes: ${}^{20}\text{Ne}$ ($19.992\ \text{u}$, $90.48\%$), ${}^{21}\text{Ne}$ ($20.994\ \text{u}$, $0.27\%$), ${}^{22}\text{Ne}$ ($21.991\ \text{u}$, $9.25\%$). Also: Ne$^{9+}$ hydrogen-like ion energy levels.氖同位素：${}^{20}\text{Ne}$（$19.992\ \text{u}$，$90.48\%$），${}^{21}\text{Ne}$（$20.994\ \text{u}$，$0.27\%$），${}^{22}\text{Ne}$（$21.991\ \text{u}$，$9.25\%$）。以及：Ne$^{9+}$ 类氢离子能级。

Answer:答案： (a) $\bar{m} \approx 20.18\ \text{u}$ · (b) $\Delta E = 1020\ \text{eV}$

(a) Average atomic mass of neon氖的平均原子质量 M1·A1·A1·A1

Convert percentages to fractions and compute the weighted sum:将百分比转换为小数，计算加权和： $$ \bar{m} = (0.9048 \times 19.992) + (0.0027 \times 20.994) + (0.0925 \times 21.991) $$ $$ = 18.093 + 0.05668 + 2.0342 $$ $$ = 20.184\ \text{u} \approx 20.18\ \text{u} $$ (Published value: $20.180\ \text{u}$, confirming the calculation. The average is close to 20 because ${}^{20}\text{Ne}$ dominates at $90.48\%$.)（公布值：$20.180\ \text{u}$，计算正确。平均值接近 20，因为 ${}^{20}\text{Ne}$ 占主导，丰度高达 $90.48\%$。）

(b) Energy released by Ne$^{9+}$ transition $n = 2 \to n = 1$Ne$^{9+}$ 从 $n = 2$ 跃迁到 $n = 1$ 释放的能量 M1·A1·A1·A1

For a hydrogen-like ion with nuclear charge $Z$, the energy levels are $E_n = -13.6\,Z^2/n^2\ \text{eV}$. Ne$^{9+}$ has $Z = 10$ (neon) and only one electron remaining:对于核电荷为 $Z$ 的类氢离子，能级为 $E_n = -13.6\,Z^2/n^2\ \text{eV}$。Ne$^{9+}$ 的 $Z = 10$（氖），只剩一个电子： $$ E_1 = \frac{-13.6 \times 10^2}{1^2} = \frac{-13.6 \times 100}{1} = -1360\ \text{eV} $$ $$ E_2 = \frac{-13.6 \times 100}{2^2} = \frac{-1360}{4} = -340\ \text{eV} $$ $$ \Delta E = E_2 - E_1 = (-340) - (-1360) = 1020\ \text{eV} $$ The ion releases $1020\ \text{eV}$ as a photon when the electron drops from $n = 2$ to $n = 1$. This is in the extreme ultraviolet / X-ray range, far above visible light, consistent with the very high nuclear charge of neon compressing the energy levels enormously compared to hydrogen.电子从 $n = 2$ 跃迁到 $n = 1$ 时，该离子释放 $1020\ \text{eV}$ 的光子。此能量处于极紫外/X 射线范围，远高于可见光，与氖极高的核电荷将能级大幅压缩（相比氢原子）一致。

Hydrogen-like ion energy levels scale as $Z^2$: high $Z$ produces enormous energies.类氢离子能级与 $Z^2$ 成正比：高 $Z$ 产生巨大能量。 For hydrogen ($Z = 1$), the $n = 2 \to 1$ transition releases $10.2\ \text{eV}$ (Lyman alpha). For Ne$^{9+}$ ($Z = 10$), the same transition releases $10.2 \times Z^2 = 10.2 \times 100 = 1020\ \text{eV}$. The $Z^2$ scaling reflects the stronger Coulomb attraction between the electron and the nucleus: doubling $Z$ quadruples the binding energy. This is why inner-shell X-ray emission (involving high-$Z$ nuclei) is used in medical and materials imaging: the emitted X-ray photons have the right energy to penetrate tissue or resolve crystal structures.对于氢（$Z = 1$），$n = 2 \to 1$ 跃迁释放 $10.2\ \text{eV}$（赖曼 alpha）。对于 Ne$^{9+}$（$Z = 10$），同一跃迁释放 $10.2 \times Z^2 = 10.2 \times 100 = 1020\ \text{eV}$。$Z^2$ 的比例关系反映了电子与核之间更强的库仑引力：$Z$ 加倍则束缚能增大四倍。正因如此，内壳层 X 射线发射（涉及高 $Z$ 核）被用于医学和材料成像：发射的 X 射线光子能量恰好足以穿透组织或分辨晶体结构。

Q11MEDIUM中 🇺🇸 US美 🇨🇦 ON安 AP-feeder FRQAP 衔接简答题 §5 + §7 Electron config + periodic trends电子排布 + 周期规律 · HS-PS1-1 [9 marks][9 分]

Elements: S ($Z = 16$), Ar ($Z = 18$), K ($Z = 19$). Electron configurations, IE ranking, and K/Ar anomaly.元素：S（$Z = 16$）、Ar（$Z = 18$）、K（$Z = 19$）。电子排布、电离能排序以及 K/Ar 异常。

Answer:答案： (a) S: $[\text{Ne}]3s^2 3p^4$; Ar: $[\text{Ne}]3s^2 3p^6$; K: $[\text{Ar}]4s^1$S：$[\text{Ne}]3s^2 3p^4$；Ar：$[\text{Ne}]3s^2 3p^6$；K：$[\text{Ar}]4s^1$ · (b) K < S < Ar · (c) K's 4s electron is shielded by a full $n=3$ shellK 的 4s 电子被完整的 $n=3$ 层屏蔽

(a) Noble-gas shorthand configurations稀有气体简写电子排布式 A1·A1·A1

S ($Z = 16$): core $=$ Ne (10 electrons), then $3s^2 3p^4$: $\quad [\text{Ne}]\ 3s^2\ 3p^4$S（$Z = 16$）：核心为 Ne（10 个电子），再填 $3s^2 3p^4$：$\quad [\text{Ne}]\ 3s^2\ 3p^4$
Ar ($Z = 18$): core $=$ Ne, then $3s^2 3p^6$ (completes Period 3): $\quad [\text{Ne}]\ 3s^2\ 3p^6$Ar（$Z = 18$）：核心为 Ne，再填 $3s^2 3p^6$（完成第三周期）：$\quad [\text{Ne}]\ 3s^2\ 3p^6$
K ($Z = 19$): core $=$ Ar (18 electrons), then $4s^1$: $\quad [\text{Ar}]\ 4s^1$K（$Z = 19$）：核心为 Ar（18 个电子），再填 $4s^1$：$\quad [\text{Ar}]\ 4s^1$

(b) Rank by first ionization energy (lowest to highest)按第一电离能从低到高排序 A1·A1·A1

Order: K < S < Ar.排序：K < S < Ar。

K is lowest ($419\ \text{kJ/mol}$): its single $4s^1$ electron is in a new, higher shell ($n = 4$) far from the nucleus and shielded by 18 core electrons. Very easy to remove.K 最低（$419\ \text{kJ/mol}$）：其唯一的 $4s^1$ 电子处于更高的新壳层（$n = 4$），离核较远，且被 18 个内层电子屏蔽，极易移除。
S is middle ($1000\ \text{kJ/mol}$): outermost electrons are in $n = 3$, with $Z = 16$ pulling them in. However, the $3p^4$ configuration has one paired $p$ orbital, so one electron experiences slight repulsion from its partner, making it marginally easier to remove than if all $3p$ electrons were unpaired.S 居中（$1000\ \text{kJ/mol}$）：最外层电子在 $n = 3$，$Z = 16$ 将其拉近。但 $3p^4$ 构型中有一个配对的 $p$ 轨道，其中一个电子受到伴电子的轻微排斥，移除略微容易。
Ar is highest ($1521\ \text{kJ/mol}$): noble gas with $Z = 18$ and a full $3p^6$ subshell. Maximum effective nuclear charge in Period 3 for electrons in the same shell, plus a complete and stable electron configuration strongly resisting removal.Ar 最高（$1521\ \text{kJ/mol}$）：稀有气体，$Z = 18$，$3p^6$ 满壳层。在第三周期同层电子中有效核电荷最大，且完整稳定的电子构型强烈抵抗电子的移除。

(c) Why K has lower IE than Ar despite more protons为何 K 的电离能低于 Ar，尽管质子数更多 A1·A1·A1

Potassium's $19^{\text{th}}$ electron enters a new shell, $n = 4$. The 18 electrons of the argon core (all in $n = 1, 2, 3$) very effectively shield this $4s^1$ electron from the nuclear charge. The effective nuclear charge felt by the $4s^1$ electron is approximately $Z_{\text{eff}} \approx 19 - 18 = 1$ (crude estimate), compared to argon's outermost $3p$ electrons experiencing $Z_{\text{eff}} \approx 18 - 10 = 8$. The $4s^1$ electron of K is both farther from the nucleus and much more shielded, making it far easier to remove than any electron in Ar.钾的第 19 个电子进入新壳层 $n = 4$。氩核心的 18 个电子（全部在 $n = 1, 2, 3$）对这个 $4s^1$ 电子形成非常有效的屏蔽。$4s^1$ 电子感受到的有效核电荷约为 $Z_{\text{eff}} \approx 19 - 18 = 1$（粗略估计），而氩的最外层 $3p$ 电子感受的 $Z_{\text{eff}} \approx 18 - 10 = 8$。K 的 $4s^1$ 电子既离核更远，又被更多地屏蔽，因此比 Ar 中任何电子都容易移除。

A new outer shell resets the effective nuclear charge to near 1, regardless of how many protons the nucleus has.进入新外层壳使有效核电荷重置到接近 1，无论原子核有多少质子。 This is why Group 1 elements always have much lower first ionization energies than the noble gas just before them in atomic number. The shielding provided by a complete electron shell is highly efficient. The same logic explains why K ($Z = 19$, IE $= 419\ \text{kJ/mol}$) has a lower IE than Na ($Z = 11$, IE $= 496\ \text{kJ/mol}$): K's $4s^1$ electron is one shell farther from the nucleus and even more shielded. The dominant factor going down Group 1 is increasing principal quantum number, not increasing nuclear charge.这就是第 1 族元素的第一电离能始终远低于原子序数紧排其前的稀有气体的原因。完整电子层提供的屏蔽效应非常高效。同理可解释为何 K（$Z = 19$，IE $= 419\ \text{kJ/mol}$）的电离能低于 Na（$Z = 11$，IE $= 496\ \text{kJ/mol}$）：K 的 $4s^1$ 电子又多了一层离核距离和更多屏蔽。在第 1 族从上到下，主导因素是主量子数的增大，而非核电荷的增大。

Q12HARD难 🇨🇦 AB阿 🇺🇸 US美 AB Diploma-style阿尔伯塔毕业考风格 §1, 2, 3, 6 Integration: structure to spectra综合：从结构到光谱 · Chem 30 Unit A [9 marks][9 分]

Two gas-discharge tubes: hydrogen and helium (${}^{4}_{2}\text{He}$). Each produces a unique line spectrum.两个气体放电管：一个含氢气，一个含氦气（${}^{4}_{2}\text{He}$）。每种气体产生独特的线状光谱。

Answer:答案： (a) 2 protons, 2 neutrons, 2 electrons2 个质子、2 个中子、2 个电子 · (b) Different electron counts give different allowed transitions and photon energies电子数不同导致允许跃迁和光子能量不同 · (c) Ground state absorbs energy, rises to excited state, falls back emitting a specific-color photon基态吸收能量，升至激发态，跃回基态并发射特定颜色光子

(a) Subatomic particles in ${}^{4}_{2}\text{He}$${}^{4}_{2}\text{He}$ 中的亚原子粒子 A1·A1

From the nuclear symbol ${}^{4}_{2}\text{He}$:由核符号 ${}^{4}_{2}\text{He}$：

Protons: $Z = 2$质子数：$Z = 2$
Neutrons: $A - Z = 4 - 2 = 2$中子数：$A - Z = 4 - 2 = 2$
Electrons (neutral atom): $2$电子数（中性原子）：$2$

(b) Why H and He produce different line spectra为何 H 和 He 产生不同的线状光谱 A1·A1·A1

Hydrogen has 1 electron; helium has 2 electrons. The allowed energy levels differ between these atoms for two reasons:氢有 1 个电子；氦有 2 个电子。两种原子的允许能级不同，原因有两点：

Different nuclear charge: He ($Z = 2$) has a stronger nuclear attraction than H ($Z = 1$), pulling electrons to lower (more negative) energy levels. All energy levels in He are lower than the corresponding levels in H.核电荷不同：He（$Z = 2$）的核引力比 H（$Z = 1$）更强，将电子拉至更低（更负）的能级。He 的所有能级都比 H 中对应能级更低。
Electron-electron repulsion: He's two electrons repel each other, further modifying the allowed energy levels in ways not present in hydrogen. This creates a different set of allowed transitions and therefore different photon energies and wavelengths.电子间排斥：He 的两个电子相互排斥，以氢中不存在的方式进一步修改了允许能级。这产生了一组不同的允许跃迁，因而产生不同的光子能量和波长。

Since every element has a unique electron count and nuclear charge, every element produces a unique set of allowed energy-level transitions and thus a unique line spectrum. This is the basis of spectroscopic identification of elements.由于每种元素有独特的电子数和核电荷数，每种元素产生独特的允许能级跃迁集合，从而产生独特的线状光谱。这是光谱法鉴别元素的基础。

(c) Flame test explanation: ground state, excited state, photon emission焰色反应解释：基态、激发态、光子发射 A1·A1·A1·A1

In a flame test:在焰色反应中：

Energy absorption (up in energy levels): The thermal energy of the flame excites the electrons from their lowest-energy state (the ground state) to a higher-energy state (an excited state). The electron jumps upward to a specific higher energy level.吸收能量（能级升高）：火焰的热能将电子从最低能量状态（基态）激发到更高能量状态（激发态）。电子跃迁到特定的更高能级。
Photon emission (down in energy levels): The excited state is unstable. The electron rapidly falls back down to a lower energy level (often back to the ground state), releasing the energy difference as a photon of specific wavelength (color).发射光子（能级降低）：激发态不稳定。电子迅速跌落回较低能级（通常回到基态），以特定波长（颜色）光子的形式释放能量差。
Characteristic color: Because each element has unique, quantized energy levels, the photons emitted have specific wavelengths unique to that element, producing a characteristic color. For example, sodium emits at $589\ \text{nm}$ (yellow); copper emits blue-green; lithium emits red.特征颜色：由于每种元素具有独特的量子化能级，发射的光子具有该元素特有的特定波长，产生特征颜色。例如：钠发射 $589\ \text{nm}$（黄色）；铜发射蓝绿色；锂发射红色。

The flame test is a qualitative emission spectroscopy technique: the color directly reveals the element.焰色反应是一种定性发射光谱技术：颜色直接揭示元素种类。 The sequence is always: ground state absorbs energy (electron goes up) then excited state decays by emitting a photon (electron goes down). The photon energy exactly equals the energy gap between the two levels. In a flame, many different transitions can occur simultaneously (different electrons, different initial and final levels), producing multiple spectral lines. The dominant visible color observed depends on which transitions are most probable and emit photons in the visible range. Sodium's flame test (yellow) is used in labs to detect even trace amounts of Na$^+$ ions, since it is one of the most intense and distinctive flame colors.过程顺序始终为：基态吸收能量（电子向上跃迁），激发态通过发射光子衰减（电子向下跃迁）。光子能量恰好等于两能级之间的能量差。在火焰中，许多不同的跃迁可同时发生（不同电子、不同初态和末态），产生多条光谱线。观察到的主要可见颜色取决于哪些跃迁概率最高且发射可见光光子。钠的焰色反应（黄色）用于实验室检测痕量 Na$^+$ 离子，因为它是最强烈、最具特征的焰色之一。