AQA化学 原子结构 质谱 电子构型 电离能
AQA A-Level 化学 7405 的第一章就是原子结构，官方编号 3.1.1，它不属于选修内容，三张卷子都可以考它。很多学生以为这一章只是初中知识的重复，翻两页就过去，结果在质谱计算与逐级电离能上反复丢分。这一节真正要求掌握的只有四块内容：三种基本粒子的相对质量与电荷、同位素与相对原子质量、飞行时间质谱的读数与计算、电子构型与电离能趋势。AQA 官方 spec 把 3.1.1 放在物理化学的开头，后面的物质的量、键能与化学平衡都要用到这里的结论。Paper 1 与 Paper 2 各 2 小时、105 分、占整科 35%，Paper 3 是 2 小时、90 分、占 30%，三张卷都可能出现这一节的题。下面的内容按这四块拆开，每一块都给出卷面考法、必背定义与可以核对的实例数字。
Section 3.1.1 of the AQA A-level Chemistry specification (7405) opens the physical chemistry content, and all three written papers can examine it, since Papers 1 and 2 are each 2 hours for 105 marks and 35%, while Paper 3 runs 2 hours for 90 marks and 30%. Candidates often treat atomic structure as GCSE revision and skim it, which is exactly where mass spectrometry calculations begin to go wrong. Four blocks matter in practice, namely the relative mass and charge of the subatomic particles, isotopes and relative atomic mass, time of flight mass spectrometry, and electron configuration with its ionisation energy trends; the numbers you meet here feed every later topic from amounts of substance to equilibrium.
3.1.1 在试卷中的位置与分值
7405 是线性课程，2015 年 9 月起施教、2017 年首次考试，全部考试在课程结束的同一考季完成。Paper 1 考 3.1.1 到 3.1.4、3.1.6 到 3.1.8、3.1.10 到 3.1.12 这些物理化学主题，加上整个无机化学，2 小时 105 分占 35%。Paper 2 考 3.1.2 到 3.1.6 与 3.1.9，配上有机化学，同样是 2 小时 105 分 35%。Paper 3 是综合卷，任何内容都可以出现，2 小时 90 分占 30%，其中 30 分是选择题。原子结构在 Paper 1 里通常落在 3 到 6 分的题上，很少单独成为一道大题，但它常常作为计算题的背景数据出现。硫元素与过渡金属的题一旦涉及电子构型，就又回到了 3.1.1。阅卷时最常见的丢分不是算错数字，而是定义写漏了气态或者一摩尔这两个限定。
The course is linear, first taught from September 2015 with first assessment in 2017, so all papers are taken in the same series at the end of two years. Paper 1 carries atomic structure among physical topics 3.1.1 to 3.1.4, 3.1.6 to 3.1.8 and 3.1.10 to 3.1.12 together with the whole of inorganic chemistry, while Paper 2 pairs 3.1.2 to 3.1.6 and 3.1.9 with organic chemistry; both are 2 hours, 105 marks and 35% of the qualification. Paper 3 is the synoptic paper of 90 marks in 2 hours, including 30 marks of multiple choice, and questions from this topic tend to be worth 3 to 6 marks there; the marks lost are usually definitional, because students leave out the words gaseous or one mole that the mark scheme insists on.
三种基本粒子的相对质量与电荷
原子由中心的原子核与核外电子组成，原子核内有质子与中子。质子的相对质量是 1、相对电荷是 +1；中子的相对质量是 1、相对电荷是 0；电子的相对质量约为 1/1836、相对电荷是 -1。原子序数等于质子数，在中性原子中也等于电子数；质量数等于质子数与中子数之和。同位素是同一种元素的原子，质子数相同而中子数不同，化学性质基本一致，相对质量与部分物理性质有差别。离子由中性原子得失电子得到，例如镁失去两个电子成为 Mg2+，电子构型随之变成 1s2 2s2 2p6。很多学生以为质量数也要写成带小数的值，其实质量数一定是整数，带小数的是相对原子质量。写核符号时左上角写质量数、左下角写原子序数，卷面上这两个位置不能互换。
An atom has a small central nucleus of protons and neutrons surrounded by electrons, and the relative masses and charges are worth memorising exactly: a proton is 1 and +1, a neutron is 1 and 0, and an electron is about 1/1836 and -1. The atomic number equals the number of protons and, in a neutral atom, the number of electrons, while the mass number is protons plus neutrons; isotopes therefore share an atomic number but differ in mass number, which is why Mg2+ is 1s2 2s2 2p6 and not the configuration of neutral magnesium. A common slip is treating relative atomic mass as a mass number, and the two never share a value.
同位素与相对原子质量的计算
相对原子质量是按天然丰度加权的平均值，可以写成各同位素相对质量与丰度乘积之和除以丰度之和。卷面给出的丰度有时是百分数，有时是质谱峰的相对高度，两种都要会算。举例来说，若铜的两个峰落在 m/z 63 与 m/z 65，丰度分别是 69.2% 与 30.8%，则相对原子质量等于 (63 x 69.2 + 65 x 30.8) / 100 = 63.6。计算时先乘再相加，最后除以丰度总和，中间不要提前四舍五入，收尾再按题目要求保留有效数字。很多学生把百分数当成小数直接乘，结果差出十倍，题目写 30.8% 就先用 30.8。如果给的是没有刻度的质谱图，就先量峰高再按比例换算，这一步最容易把相对高度当成绝对数量。相对原子质量通常印在考试局提供的周期表上，可以用它反过来核对计算结果。
Relative atomic mass is a weighted mean, calculated as the sum of each isotope's relative mass multiplied by its abundance, divided by the total of the abundances. Copper makes a useful check, since peaks at m/z 63 and m/z 65 with abundances of 69.2% and 30.8% give (63 x 69.2 + 65 x 30.8) / 100 = 63.6. Do not round until the final line, read the significant figure instruction before writing the answer, and remember that scaling peak heights from an unlabelled spectrum is the step students most often get wrong, because they treat relative height as an absolute count.
飞行时间质谱的六个步骤
飞行时间质谱仪的流程可以拆成六步，写作时要按顺序写出来并说明每一步的作用。第一步是汽化，把样品从液态或固态变成气态，只有气态粒子才能被电离并被加速。第二步是电离，用高能电子轰击或者电喷雾让粒子带上正电荷，仪器只检测正离子。第三步是加速，所有离子在电场中获得相同的动能，随后进入没有电场的飞行管。第四步是漂移，离子在飞行管里按质量分离，重的慢、轻的快。第五步是检测，离子打到检测器上产生电流，电流大小与到达的离子数量成正比。第六步是分析，横坐标是质荷比 m/z，纵坐标是丰度，仪器给出的是相对丰度而不是绝对数量。常见的错误是把加速与漂移并成一步，或者忘记说明真空环境，空气分子与离子碰撞会改变飞行路径。
The instrument works through six stages, and the order matters when you write them out. First the sample is vaporised, because only gaseous particles can be ionised and accelerated, and then it is ionised to give a positive charge, since the detector records positive ions only. Acceleration gives every ion the same kinetic energy before the ions drift through a field-free tube where lighter ones arrive first; detection converts arriving ions into a current proportional to their number, and the spectrum plots abundance against mass-to-charge ratio. Two errors appear again and again, namely merging acceleration and drift into one stage, and forgetting the vacuum that stops ions colliding with air molecules and losing their path.
电子轰击与电喷雾两种电离方式
电子轰击电离用电子枪发出的高能电子撞击气态样品，把一个电子从粒子中打出，形成 1+ 离子，并常常伴随碎片峰，适合相对分子质量低于一千的样品。电喷雾电离把样品溶在挥发性溶剂里，经细针头在高电压下喷成带电液滴，溶剂蒸发时粒子从溶剂中得到一个质子 H+，所以离子带一个正电荷，质量比原来的分子多 1。电喷雾几乎不产生碎片，适合蛋白质这类高分子，谱图上会出现一组相邻峰，相邻两峰的间距对应不同的电荷数。判断方式很直接，碎片峰多就是电子轰击，出现整分子峰加一就是电喷雾。常见的错误是把电喷雾说成失去电子，它得到的其实是质子。如果电喷雾给出的是 M+1 峰，从 m/z 反推相对分子质量时要把那 1 减回去。
Electron impact fires high energy electrons from an electron gun at a vaporised sample, knocking out an electron to leave a 1+ ion and often a set of fragment peaks, which suits samples with a relative molecular mass below about 1000. Electrospray dissolves the sample in a volatile solvent, sprays it from a fine needle under high voltage, and the evaporating droplets donate a proton to the particle, so the ion carries one positive charge and one extra mass unit. Because electrospray is gentler it produces little fragmentation and is used for large molecules such as proteins, and the giveaway on a spectrum is a group of adjacent peaks whose spacings reveal the charge carried.
质荷比与飞行时间的计算
飞行管里所有离子的动能相同，所以速度只由质量决定，公式写成 KE = 1/2 mv^2。把等式变形得到 v = sqrt(2KE/m)，质量越大速度越小，而飞行时间 t = d/v，因此重离子的飞行时间更长。举例来说，动能给定 1.5 x 10^-15 J 时，m/z 20 的离子速度是 3.0 x 10^5 m/s，m/z 22 的离子速度是 2.9 x 10^5 m/s，两者相差约 4.7%。若飞行管长 1.50 m，两个离子的到达时间分别是 5.0 微秒与 5.2 微秒，相差 0.24 微秒，仪器必须能分辨这个差值。代入公式前要把质量单位换成千克，一个原子质量单位约等于 1.661 x 10^-27 kg，不换单位是这一节最典型的错误。m/z 相同的离子飞行时间相同，所以质谱无法区分同分异构体，结构信息要靠碎片图样推断。题目通常给出动能与飞行距离，让你比较两个峰的先后，比较时只需比较质量的平方根，不必算出具体时间。
In the drift tube all ions carry the same kinetic energy, so their speed depends only on mass, since KE = 1/2 mv^2 rearranges to v = sqrt(2KE/m) and the flight time t = d/v is longer for heavier ions. Take KE = 1.5 x 10^-15 J, where an ion of m/z 20 travels at 3.0 x 10^5 m/s against 2.9 x 10^5 m/s for m/z 22, a difference of roughly 4.7% that becomes 5.0 microseconds against 5.2 microseconds over a 1.50 m tube. Convert mass to kilograms before substituting, because one atomic mass unit is about 1.661 x 10^-27 kg, and note that ions with the same m/z arrive together, so a mass spectrum separates fragments rather than isomers.
电子构型子层与轨道容量
电子按能量从低到高的顺序填入子层，四个子层的容量是 s 层 2 个、p 层 6 个、d 层 10 个、f 层 14 个。每个子层由若干轨道组成，s 有一个轨道、p 有三个、d 有五个，每个轨道最多容纳两个自旋方向相反的电子。s 轨道是球形对称的，p 轨道呈哑铃形，三个 p 轨道互相垂直。AQA 使用的填充顺序是 1s 2s 2p 3s 3p 4s 3d 4p，4s 的能量低于 3d，所以先填 4s。举例来说，钾的电子构型是 1s2 2s2 2p6 3s2 3p6 4s1，而不是把最后一个电子放进 3d。写成简式时用稀有气体做核心，钾写作 [Ar] 4s1。学生最常把 4s 与 3d 的顺序写反，画轨道箭头图时也容易漏掉配对面。形成离子时先失去最外层电子，过渡金属离子往往先失去 4s 电子，这是后面无机化学题的前提。
Electrons fill sub-levels in order of increasing energy, with capacities of 2 for s, 6 for p, 10 for d and 14 for f. Each sub-level holds a fixed number of orbitals, one spherical s orbital, three dumbbell-shaped p orbitals at right angles and five d orbitals, each taking two electrons of opposite spin. The filling order used by AQA is 1s 2s 2p 3s 3p 4s 3d 4p, because 4s lies lower in energy than 3d in a neutral atom, so potassium is 1s2 2s2 2p6 3s2 3p6 4s1 and writing it as [Ar] 4s1 is both accepted and safer.
周期表区块与两个反常构型
电子构型决定元素在周期表中的位置，最后一个电子进入 s 层的是 s 区，进入 p 层的是 p 区，进入 d 层的是 d 区，过渡金属全部落在 d 区。铬与铜是两个必须记住的例外，铬写作 [Ar] 3d5 4s1，而不是按顺序推出的 3d4 4s2；铜写作 [Ar] 3d10 4s1，而不是 3d9 4s2。原因是半满与全满的 d 子层能量更低、排布更稳定，这条理由要写在卷面上才能拿到解释分。很多学生以为例外只能靠死记，其实 3d 与 4s 的能量差很小，电子重新排布带来的稳定化足以补偿跃迁的代价。处理离子构型时先失去 4s 电子，例如 Fe2+ 是 [Ar] 3d6，Fe3+ 是 [Ar] 3d5。这一条在无机化学的颜色与磁性题里直接决定成败，也会影响电极电位的判断。写构型时数字要用上标，卷面上 3d5 与 3d 5 表达的是完全不同的意思。
The block of an element follows from where the final electron goes, so the s, p and d blocks correspond to the sub-level being filled and the transition metals sit in the d block. Chromium and copper are the two configurations worth memorising, [Ar] 3d5 4s1 and [Ar] 3d10 4s1, rather than the 3d4 4s2 and 3d9 4s2 that the filling order alone would predict; half-filled and fully filled d sub-levels are more stable, and that reason has to appear in your answer to earn the explanation mark. For ions, remove the 4s electrons first, so Fe2+ is [Ar] 3d6 while Fe3+ is [Ar] 3d5.
第一电离能的定义与方程式
AQA 要求的第一电离能定义是：从一摩尔气态原子中各移走一个电子，形成一摩尔气态 1+ 离子所需的能量。定义里有三个信息点不能省，一摩尔、气态、各移走一个电子。方程式要带状态符号，钠写成 Na(g) -> Na+(g) + e-，漏掉状态符号会被扣分。第二电离能是从一摩尔气态 1+ 离子中各移走一个电子，形成一摩尔气态 2+ 离子所需的能量，方程式是 Na+(g) -> Na2+(g) + e-。电离一定是吸热过程，因为电子与原子核之间存在静电吸引，把它移走要输入能量。常见的错误是只写钠失去电子变成离子，既没写气态也没写一摩尔，化学上不错但拿不到分。题目问原子时状态符号要写成 g，写成 s 或者 l 都不给分。定义题的分值通常是 2 到 3 分，三个信息点刚好对应三个给分点。
The definition AQA requires is the energy needed to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions, and three elements cannot be left out, namely one mole, gaseous, and one electron from each atom. Equations must carry state symbols, so sodium is Na(g) -> Na+(g) + e-, while the second ionisation energy starts from one mole of gaseous 1+ ions and gives 2+ ions. Ionisation is always endothermic because the electron is held by attraction to the nucleus, and an answer that says sodium loses an electron without mentioning the gaseous state or the mole will lose marks even when the chemistry is correct.
影响电离能的三个因素
影响电离能的因素有三个，核电荷、原子半径与电子屏蔽。核电荷越大，核对电子的吸引越强，电离能升高；原子半径越大，外层电子离核越远，电离能降低；内层电子越多，屏蔽效应越强，外层电子感受到的有效核电荷越小，电离能降低。答题时要把这三个因素按题目要求的方向组织起来，同周期比较时说核电荷增加、屏蔽基本相同、半径减小，所以电离能升高。同主族比较时说电子层数增加、半径增大、屏蔽增强，所以电离能降低。学生最常把同周期与同主族的理由写反，也有人只写核电荷一个因素就交卷。阅卷时给分点按因素逐个给，少一个因素就少一分。题目要求比较相邻两个元素时，要写明它们外层电子所在的子层，这一步最容易看出有没有真正理解能量顺序。
Three factors control ionisation energy, and each has to be named with its direction, namely nuclear charge, atomic radius and electron shielding. A larger nuclear charge pulls the outer electrons in more strongly and raises the energy, while a larger radius and more shielding both lower it by weakening the effective nuclear charge felt by the outer electron; across a period the charge rises while shielding stays much the same and the radius falls, so the energy increases, and down a group extra shells add both radius and shielding, so it falls. Mixing these two explanations up is the most common way of dropping the marks, and a comparison question also expects you to name the sub-level the outer electron occupies.
同周期趋势与两处反常
第二周期与第三周期的第一电离能总体上向右升高，但有两处必须记住的反常。镁的第一电离能是 738 kJ mol-1，铝是 578 kJ mol-1，铝反而更低，因为铝的最外层电子进入 3p 子层，3p 的能量高于 3s，同时也被 3s 电子屏蔽。磷是 1012 kJ mol-1，硫是 1000 kJ mol-1，硫反而更低，因为硫的 3p4 排布里有一对电子共占同一个轨道，电子之间的排斥让其中一个更容易被移走。两处反常的差别在于一个来自子层能量，另一个来自自旋配对，答题时不能把理由互换。很多学生以为同周期趋势是一条直线，看到图上的两处下凹就以为数据印错了，其实这两处下凹正是考点。比较两个元素时先写出它们的电子构型，反常自然解释得通。第三周期从钠到氩也有对应的两处下凹，理由与第二周期相同。
First ionisation energy rises across Period 2 and Period 3 with two well-known dips, and both need their own explanation. Magnesium is 738 kJ mol-1 against aluminium at 578 kJ mol-1, because the outer electron of aluminium occupies a 3p sub-level that lies higher in energy than 3s and is shielded by the 3s electrons. Phosphorus is 1012 kJ mol-1 while sulfur is 1000 kJ mol-1, because the 3p4 arrangement pairs two electrons in the same orbital and the repulsion between them makes one easier to remove; one dip comes from sub-level energy and the other from electron pairing, so the two reasons must never be swapped.
逐级电离能与卷面失分点
逐级电离能指从气态原子上一个接一个移走电子所需的能量，数值跳变的位置直接说明电子分成几层。钠的第一电离能是 496 kJ mol-1，第二是 4562 kJ mol-1，跳了约 9 倍，说明第二个电子要从更内层的 2p 子层移走。继续往下看，第十与第十一之间的跳变最大，因为要开始破坏 2 电子层，这也解释了钠为什么只形成 1+ 离子。判断主族元素族数有固定方法，数一数大跳变之前移走了几个电子，个数就是族数。卷面上这部分最常见的失分有四类，定义漏掉气态或者一摩尔、方程式缺状态符号、电子构型顺序写成 [Ar] 3d1 4s2 这类颠倒形式、解释趋势时只写核电荷一个因素。三周复习可以按这个节奏走，先把定义与构型写法练到不假思索，再专攻质谱读数与计算，最后用真题里的原子结构小题限时做一遍。
Successive ionisation energies measure the energy needed to remove electrons one at a time from a gaseous atom, and the position of the jumps reveals the shell structure. Sodium goes from 496 to 4562 kJ mol-1 between the first and second removals, a jump of about nine times, which shows the second electron comes from the inner 2p sub-level; the largest jump comes later, when the 2 electron shell itself is breached. To read a group number, count the electrons removed before that big jump, then build three weeks of revision around precise definitions first, mass spectrometry calculations second, and timed past paper questions last.
海外直订AQA GCSE Chemistry Student Book AQA GCSE化学学生手册¥588.0📱 去京东购买京东精选
需要一对一辅导？Need One-on-One Tutoring?
A-Level / GCSE / IB / KS2-KS3 全科辅导
Visit aleveler.com
或截图后用「扫一扫 → 相册」选择本图