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IGCSE Chemistry Unit Test: Atomic Structure & The Periodic Table | IGCSE 化学单元测试卷:原子结构与元素周期表

2026-07-14 02:31 · 本文部分内容由AI辅助创作

📚 IGCSE Chemistry Unit Test: Atomic Structure & The Periodic Table | IGCSE 化学单元测试卷:原子结构与元素周期表

This Unit Test is designed to help IGCSE Chemistry students assess their understanding of atomic structure and the periodic table. It includes multiple-choice questions, structured questions, a calculation and a completion task. Work through each question carefully, then check the detailed explanations and answers provided. Use this resource to identify areas for revision before your real exam.

这份单元测试卷旨在帮助 IGCSE 化学学生评估他们对原子结构和元素周期表的理解。测试包括选择题、结构化问题、计算题和填表题。仔细完成每道题,然后查看提供的详细解释和答案。利用这份资源找出需要复习的薄弱环节,为真正的考试做好准备。

1. Multiple Choice – Subatomic Particles | 选择题 – 亚原子粒子

An atom of element X has an atomic number of 19 and a mass number of 39. How many neutrons are there in one atom of X?

A 19
B 20
C 39
D 58

一个 X 元素的原子序数为 19,质量数为 39。一个 X 原子中有多少个中子?

A 19
B 20
C 39
D 58

Answer: B c number gives the number of protons, not neutrons. Option A is the proton number, C is the mass number, and D is the sum of protons and mass number, which is incorrect.

答案:B。中子数 = 质量数 − 原子序数 = 39 − 19 = 20。原子序数表示质子数,并非中子数。选项 A 是质子数,C 是质量数,D 是质子数与质量数之和,均是错误的理解。


2. Multiple Choice – Isotopes | 选择题 – 同位素

Which statement correctly describes isotopes?

A Atoms of the same element with different numbers of protons.
B Atoms of different elements with the same number of neutrons.
C Atoms of the same element with different numbers of neutrons.
D Atoms of different elements with the same mass number.

哪一项正确地描述了同位素?

A 同一种元素中质子数不同的原子。
B 不同元素中中子数相同的原子。
C 同一种元素中中子数不同的原子。
D 不同元素中质量数相同的原子。

Answer: C. Isotopes are atoms of the same element (same number of protons) but with different numbers of neutrons, leading to different mass numbers. Option A is wrong because the number of protons defines the element; B and D do not define isotopes.

答案:C。同位素是质子数相同(属于同种元素)但中子数不同的原子,因此质量数不同。选项 A 错误,因为质子数决定了元素种类;B 和 D 都不能准确地定义同位素。


3. Multiple Choice – Electronic Configuration & Groups | 选择题 – 电子排布与族

An element has the electronic configuration 2,8,2. In which group of the periodic table is it found?

A Group I
B Group II
C Group VIII
D Group 0

某元素的电子排布为 2,8,2。它位于元素周期表的哪一个族?

A I 族
B II 族
C VIII 族
D 0 族

Answer: B. The number of electrons in the outer shell is 2. Elements with 2 outer-shell electrons belong to Group II. Group I has 1 outer electron, Group VIII is a transition metal group with variable outer shells, and Group 0 (noble gases) has a full outer shell.

答案:B。最外层电子数为 2。最外层有 2 个电子的元素属于 II 族。I 族元素最外层有 1 个电子,VIII 族是过渡金属,最外层电子数不固定,而 0 族(稀有气体)的最外层为全满状态。


4. Multiple Choice – Periodic Trends | 选择题 – 周期律趋势

As you go down Group I (the alkali metals), what happens to the reactivity?

A It decreases because the outer electron is closer to the nucleus.
B It stays the same.
C It increases because the outer electron is more easily lost.
D It increases because the atomic radius decreases.

沿着 I 族(碱金属)从上往下,反应性会怎样变化?

A 反应性减弱,因为最外层电子离核更近。
B 反应性保持不变。
C 反应性增强,因为最外层电子更容易失去。
D 反应性增强,因为原子半径减小。

Answer: C. Reactivity increases down Group I because the outer electron is further from the nucleus, the shielding effect increases, and the electron is more easily lost. Atomic radius actually increases down the group, so D is incorrect.

答案:C。I 族从上往下反应性增强,因为最外层电子离核更远,屏蔽效应增强,电子更容易失去。实际上原子半径是增大的,因此选项 D 错误。


5. Structured Question – Sodium Atom & Ion | 结构化问题 – 钠原子与离子

Sodium (Na) has an atomic number of 11.
(a) Draw the electronic structure of a sodium atom, showing all shells and electrons.
(b) Explain what happens to the electronic structure when a sodium atom forms a sodium ion, Na⁺.

钠 (Na) 的原子序数为 11。
(a) 画出钠原子的电子结构,需显示所有电子层和电子。
(b) 解释当钠原子形成钠离子 Na⁺ 时,其电子结构会发生什么变化。

Answer (a): The electronic configuration of a sodium atom is 2,8,1. The first shell contains 2 electrons, the second shell 8 electrons, and the third shell 1 electron. You would draw a nucleus with 11 protons, first circle with 2 dots, second circle with 8 dots, third circle with 1 dot.

答案 (a):钠原子的电子排布为 2,8,1。第一电子层有 2 个电子,第二电子层有 8 个电子,第三电子层有 1 个电子。画图时需画出含有 11 个质子的原子核,第一层画 2 个点,第二层画 8 个点,第三层画 1 个点。

Answer (b): The sodium atom loses its single outer electron to achieve a full outer shell (the original second shell becomes the outer shell with 8 electrons). The resulting Na⁺ ion has the electronic configuration 2,8. The ion has a +1 charge because it now has 11 protons but only 10 electrons.

答案 (b):钠原子失去唯一的那个最外层电子,从而获得全满的最外层结构(原来的第二电子层变成了最外层,拥有 8 个电子)。形成的 Na⁺ 离子的电子排布为 2,8。该离子带 +1 电荷,因为它有 11 个质子,但只有 10 个电子。


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6. Structured Question – Stability of Chloride Ion | 结构化问题 – 氯离子的稳定性

A chlorine atom gains one electron to form a chloride ion, Cl⁻. Explain, in terms of electronic structure, why the chloride ion is more stable than a chlorine atom.

氯原子获得一个电子形成氯离子 Cl⁻。请从电子结构的角度解释,为什么氯离子比氯原子更稳定。

Answer: A chlorine atom has the electronic configuration 2,8,7. Its outer shell is not full. By gaining one electron, it becomes a chloride ion with configuration 2,8,8, which is the same electronic structure as the noble gas argon. Having a full outer shell of 8 electrons makes the ion very stable, similar to the electronic arrangement of noble gases. The atom achieves a lower energy state by completing its octet.

答案:氯原子的电子排布为 2,8,7,最外层未满。获得一个电子后,氯离子具有 2,8,8 的排布,这与稀有气体氩的电子结构相同。拥有 8 个电子的全满最外层使离子非常稳定,类似于稀有气体的电子排布。通过达成八隅体,氯原子达到了能量更低、更稳定的状态。


7. Calculation – Relative Atomic Mass | 计算题 – 相对原子质量

Chlorine has two naturally occurring isotopes: ³⁵Cl with an abundance of 75% and ³⁷Cl with an abundance of 25%. Calculate the relative atomic mass (Aᵣ) of chlorine to one decimal place. Show your working.

氯有两种天然同位素:³⁵Cl 的丰度为 75%,³⁷Cl 的丰度为 25%。计算氯的相对原子质量 (Aᵣ),结果保留一位小数。请写出计算过程。

Answer: Aᵣ = [(mass of isotope 1 × abundance) + (mass of isotope 2 × abundance)] / 100
Aᵣ = [(35 × 75) + (37 × 25)] / 100
Aᵣ = (2625 + 925) / 100
Aᵣ = 3550 / 100 = 35.5
So the relative atomic mass of chlorine is 35.5. This value appears on the periodic table as it accounts for the weighted average of the isotopes.

答案:Aᵣ = [(同位素1质量 × 丰度) + (同位素2质量 × 丰度)] / 100
Aᵣ = [(35 × 75) + (37 × 25)] / 100
Aᵣ = (2625 + 925) / 100
Aᵣ = 3550 / 100 = 35.5
因此氯的相对原子质量为 35.5。这个数值你会在元素周期表中看到,因为它包含了同位素的加权平均值。


8. Completion Table – Nuclear Symbols | 填表题 – 核符号

Complete the table below by writing the correct nuclear symbol for each particle. The first one has been done for you as an example.

完成下表中的每一个粒子,写出正确的核符号。第一个已经作为例子为你填写好了。

Particle Protons Neutrons Electrons Nuclear Symbol
A 11 12 11 ²³₁₁Na
B 17 18 18
C 8 8 10
D 6 6 6

Answers:
₇Cl⁻ (The mass number is 17+18=35; it has one more electron than protons, so a 1− charge.)
C: ¹⁶₈O²⁻ (Mass number 8+8=16; it has two extra electrons, giving a 2− charge.)
D: ¹²₆C (Carbon-12 atom, neutral.)

答案:
B:³⁵₁₇Cl⁻(质量数为 17+18=35;电子比质子多 1 个,因此带 1− 电荷。)
C:¹⁶₈O²⁻(质量数为 8+8=16;多余 2 个电子,带 2− 电荷。)
D:¹²₆C(碳-12 原子,电中性。)


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