AQA化学 ALevel原子结构 电子排布 电离能
原子结构是 AQA A-Level Chemistry 7405 规范 3.1.1 的第一块内容，也是 Paper 1 每年都会碰到的题目来源。知识点看着基础，分数却集中在几个固定动作上：用同位素丰度算相对原子质量，按四步描述飞行时间质谱，写电子排布式，解释第一电离能的趋势和例外，再从逐级电离能数据判断电子层数。规范原文还要求考生能说明电离能数据如何支持电子层与亚层模型，所以只会背结论往往答不满分。
Atomic structure opens section 3.1.1 of the AQA A-Level Chemistry 7405 specification and supplies Paper 1 questions every year. The content looks basic, yet the marks sit in a small set of fixed skills: calculating relative atomic mass from isotope abundances, describing the four stages of time of flight mass spectrometry, writing electron configurations, explaining the trend and the exceptions in first ionisation energy, and using successive ionisation energies to count shells. The specification also asks you to explain how ionisation energy data support the shell and sub-shell model, so memorised conclusions alone rarely fill the mark scheme.
质子数 中子数与相对原子质量计算
原子核里的质子数决定元素种类，中子数不同就构成同位素。AQA 常用镁的三个同位素考相对原子质量：24Mg 占 79.0%，25Mg 占 10.0%，26Mg 占 11.0%，按丰度加权得到 (24 x 79.0 + 25 x 10.0 + 26 x 11.0) / 100 = 24.32。很多学生会把三个质量数直接相加除以三，忽略丰度权重，这是本节最常见的失分点。同位素的化学性质几乎相同，因为化学反应只涉及最外层电子，质量差异只影响扩散速率一类的物理过程。
The number of protons fixes the element, and a different number of neutrons gives an isotope. AQA questions usually use magnesium: 24Mg at 79.0%, 25Mg at 10.0% and 26Mg at 11.0% abundance, which gives (24 x 79.0 + 25 x 10.0 + 26 x 11.0) / 100 = 24.32. Adding the three mass numbers and dividing by three ignores the weighting and is the standard error in this section. Isotopes behave almost identically in chemical reactions, since only the outer electrons are involved; the mass difference shows up in physical processes such as rates of diffusion.
飞行时间质谱 电离 加速 漂移 检测
飞行时间质谱在规范里要求按四步描述，缺一步就丢一个分点。电离分两条路径：电子轰击适用于 Mr 较小的分子，电子枪射出高能电子，把样品分子中的一个电子打掉，形成 1+ 离子；电喷雾适用于 Mr 较大或结构易碎的分子，样品溶于挥发性溶剂，经高压针头喷成细雾并获得质子，形成 [M+H]+。第二步用电场加速，所有离子获得相同的动能，质量小的离子速度更大。第三步在无电场的飞行管内漂移，管内抽真空以避免与空气分子碰撞。第四步离子打在检测器上得到一个电子变成中性粒子，产生的电流大小对应它的丰度。
The specification expects four stages, and each missing stage costs a mark. Ionisation takes one of two routes. Electron impact suits molecules with a small Mr: an electron gun fires high energy electrons that remove one electron from the molecule and leave a 1+ ion. Electrospray suits larger or fragile molecules: the sample is dissolved in a volatile solvent and pushed through a fine needle at high voltage, picking up a proton to give [M+H]+. An electric field then accelerates the ions so that all of them carry the same kinetic energy, which makes the lighter ions travel faster. They drift through a flight tube with no field applied, held under vacuum so that collisions with air molecules are avoided. At the detector each ion gains an electron, and the current recorded is proportional to its abundance.
电喷雾路径多了一个质子，所以读谱时 m/z 等于 Mr + 1。题目若给出 m/z 值要求反推相对分子质量，要减掉这个质子，这一步在含氮化合物和蛋白质题里扣分很多。电子轰击的谱图还会出现 M+ 峰与碎片峰，碎片峰对应分子断裂后的片段，可以用来推断结构。
Electrospray adds a proton, so the peak appears at m/z equal to Mr + 1. If a question gives m/z and asks for the relative molecular mass, subtract that proton, a step that loses marks in protein and nitrogen containing compounds. Electron impact spectra also show an M+ peak alongside fragment peaks, and the fragments correspond to pieces of the molecule, which is what makes the method useful for structure determination.
亚层能级顺序 4s 为何先于 3d 填充
亚层按 1s 2s 2p 3s 3p 4s 3d 4p 的顺序填充，原因是 4s 亚层的能量低于 3d。s 亚层有 1 个轨道，p 有 3 个，d 有 5 个，每个轨道最多放 2 个自旋相反的电子。铁的排布写作 1s2 2s2 2p6 3s2 3p6 3d6 4s2，简写为 [Ar] 3d6 4s2。常见的错误是写排布式时把 3d 提到 4s 前面，或者把铁的 d 电子数记成 8，选择题里这类错误很快会被判掉，解释题里则直接丢全部分点。
Sub-shells fill in the order 1s 2s 2p 3s 3p 4s 3d 4p because 4s lies lower in energy than 3d. An s sub-shell contains one orbital, p contains three and d contains five, each holding two electrons of opposite spin. Iron is 1s2 2s2 2p6 3s2 3p6 3d6 4s2, or [Ar] 3d6 4s2 in shorthand. Putting 3d ahead of 4s, or recalling eight d electrons for iron, is an error that disappears quickly in multiple choice but costs the whole explanation mark in a written answer.
周期表分区 s区 p区 d区与最外层电子
周期表的分区取决于最后一个电子进入哪个亚层。第一、二族属于 s 区，最外层电子在 s 亚层；13 到 18 族属于 p 区，最后一个电子进入 p 亚层；中间三到十二族的过渡元素属于 d 区，电子填在 3d 或 4d。主族元素的族号个位数等于最外层电子数，这条关系在推断元素身份的题目里经常用到：先看排布式中最外层电子所在的亚层定区，再数外层电子定族。常见的错误是把氦归进 p 区，氦的排布是 1s2，属于 s 区，尽管它摆在第十八族。
写分区时还要分清 d 区与过渡元素两个说法。锌的排布是 [Ar] 3d10 4s2，d 亚层已经全满，AQA 把它算作 d 区元素，但涉及催化、配位和可变氧化态的内容通常不把锌当作过渡元素处理，题目问 why zinc is not a transition element 时答案就落在 d 亚层全满这一点上。
The block of the periodic table depends on the sub-shell that receives the last electron. Groups 1 and 2 form the s block, where the outer electrons sit in an s sub-shell; groups 13 to 18 form the p block, with the final electron entering a p sub-shell; the transition elements in groups 3 to 12 form the d block. For main group elements the final digit of the group number equals the number of outer electrons, a link used constantly in identify-the-element questions: read the block from the last sub-shell in the configuration, then count the outer electrons to fix the group. Putting helium in the p block is a common slip, since helium is 1s2 and belongs to the s block even though it sits in group 18.
Zinc is a reminder that d block and transition element are not synonyms. Its configuration, [Ar] 3d10 4s2, leaves the d sub-shell full, so AQA treats zinc as a d block element but not as a transition element in questions on catalysis, complex formation or variable oxidation states. Asked why zinc is not a transition element, the answer is that full d sub-shell.
第一电离能 定义 周期趋势与两个例外
第一电离能指从一摩尔气态原子中各自移走一个电子，形成一摩尔气态 1+ 离子所需的能量，方程式写作 Na(g) -> Na+(g) + e-。定义里气态这个词不能省，写成固态或水溶液都不得分。沿周期从左到右电离能总体上升，因为核电荷增加而屏蔽基本不变，半径缩小，核对外层电子的吸引增强。沿族向下则下降，因为电子层数增加，屏蔽增强，外层电子离核更远。两个例外要能解释：铝低于镁，因为铝失去的是 3p 电子，3p 能量高于 3s，且被 3s 电子部分屏蔽；硫低于磷，因为硫的 3p 轨道里有一对成对电子，电子间排斥让其中一个更容易被移走。
First ionisation energy is the energy needed to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions: Na(g) -> Na+(g) + e-. Dropping the word gaseous loses the mark, and writing solid or aqueous loses it as well. Across a period the value rises overall because nuclear charge increases while shielding stays roughly constant, so the radius falls and the outer electron is held more strongly. Down a group the value falls, since extra shells bring more shielding and greater distance from the nucleus. Both exceptions need an explanation. Aluminium is lower than magnesium because the electron removed comes from 3p, higher in energy than 3s and partly shielded by the 3s pair. Sulfur is lower than phosphorus because one 3p orbital in sulfur holds a pair, and repulsion between those two electrons makes one of them easier to remove.
逐级电离能数据如何证明电子层结构
逐级电离能是判断电子层数最直接的工具。钠的数值依次是 496、4562、6912、9543、13354、16613、20114、25496、28932、141362、159076 kJ/mol，第一次与第二次之间出现十倍以上的跃升，说明第二个电子来自内层，钠只有 1 个外层电子。镁的跃升出现在第二次之后，铝出现在第三次之后，位置每右移一格正好对应外层电子多一个。错得最多的是只描述数值大小而不指出跃升位置，或者把跃升归因于核电荷变化，实际核电荷没有变，变的是被移走电子所在的电子层以及它受到的屏蔽。
Successive ionisation energies give the number of shells directly. For sodium the values run 496, 4562, 6912, 9543, 13354, 16613, 20114, 25496, 28932, 141362 and 159076 kJ/mol. The jump of more than ten times between the first and second shows that the second electron comes from an inner shell, so sodium has a single outer electron. Magnesium jumps after the second value and aluminium after the third, each step to the right adding one outer electron. The usual error is describing the size of the numbers without saying where the jump falls, or blaming the jump on a change in nuclear charge. Nuclear charge is unchanged; what changes is the shell the electron leaves and the shielding around it.
离子的电子排布 铬与铜两个特例
形成阳离子时先失去 4s 电子，再失去 3d 电子。铁失去两个电子得到 [Ar] 3d6，失去三个得到 [Ar] 3d5。铬和铜是两个特例，铬写作 [Ar] 3d5 4s1，铜写作 [Ar] 3d10 4s1，原因是半充满和全充满的 d 亚层更稳定。写阴离子时电子加在最高能量的亚层上，氯离子与氩等电子，排布为 1s2 2s2 2p6 3s2 3p6。阅卷时经常见到把 Fe2+ 写成 [Ar] 3d4 4s2 的答案，这等于把 4s 当成先填后留，而 4s 先填只是因为能量低，失电子时它反而先离开。
Cations lose 4s electrons before 3d electrons. Iron gives [Ar] 3d6 as Fe2+ and [Ar] 3d5 as Fe3+. Chromium and copper are the two exceptions, [Ar] 3d5 4s1 and [Ar] 3d10 4s1, explained by the extra stability of half filled and fully filled d sub-shells. Anions gain electrons in the highest energy sub-shell available, so the chloride ion is 1s2 2s2 2p6 3s2 3p6 and is isoelectronic with argon. Scripts regularly show Fe2+ written as [Ar] 3d4 4s2, which treats 4s as filled and kept. The 4s sub-shell fills first only because it is lower in energy, and it empties first when electrons are lost.
解释题答题结构 核电荷 屏蔽与距离
电离能解释题的评分点通常分成三块：核电荷、屏蔽、距离或半径。只写 attraction is stronger 而不说明屏蔽不变、半径变小，一般只能拿到一个分点。定义题要逐字对照规范用语，气态、一摩尔、各自移走一个电子这三处都是判分点。质谱题问 m/z，电喷雾路径要加 1；问相对分子质量，则要把这个质子减掉。这一步最容易在两种电离方式之间混用，题干出现 macromolecule 或 protein 就选电喷雾，出现 small Mr 或 organic 就选电子轰击。做历年真题时把评分点抄在题目旁边，比反复看笔记更能发现自己漏在哪一句。
Explanation questions on ionisation energy are marked on three ideas: nuclear charge, shielding, and distance or radius. Writing that the attraction is stronger without stating that shielding is unchanged and the radius smaller normally earns one mark. Definition questions are marked word by word, so gaseous, one mole and one electron from each atom each carry credit. If a mass spectrum question asks for m/z after electrospray, add one; if it asks for relative molecular mass, subtract that proton. Mixing up the two ionisation methods is the most frequent slip: a macromolecule or protein in the stem points to electrospray, a small Mr or organic sample points to electron impact. Copying the mark scheme points beside each past paper question shows which sentence of your answer is missing far more clearly than rereading notes.
把 7405 规范的 3.1.1 小节对着这三类题走一遍，会发现题目换的只是元素和数据，答题结构几乎不变。同位素计算看丰度权重，质谱看四步与加一减一，排布看 4s 与 3d 的先后，电离能看三块评分点。每周挑两道真题写完整答案，再和评分方案逐条对照，比一次性背完整个章节更稳。
Working through section 3.1.1 of the 7405 specification against these question types shows that only the element and the numbers change. Isotope calculations turn on weighting by abundance, mass spectra on the four stages and the plus or minus one, configurations on the order of 4s and 3d, and ionisation energy on the three marking points. Two past paper questions written out in full each week, checked line by line against the mark scheme, hold up better than reading the whole chapter once.
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