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Practice Questions

AQA GCSE Chemistry: Atomic Structure and the Periodic Table — Practice Questions

Original exam-style practice questions with full worked answers on atomic structure, isotopes, electronic structure and group trends for AQA GCSE Chemistry.

Subject
Chemistry
Level
GCSE
Topic
Topic 4.1 – Atomic Structure and the Periodic Table
Updated

Aligned to AQA GCSE Chemistry (8462), For teaching from September 2016. Official specification .

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These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Atomic Structure and the Periodic Table revision notes, covering electron configuration, group trends and separation techniques in full.


Section A

1. State the relative mass and relative charge of a proton, a neutron and an electron. [3]

2. An atom has 17 protons and 20 neutrons. Give its atomic number, mass number and electronic structure. [3]

Section B

3. Describe how the model of the atom changed as a result of the alpha scattering experiment. [4]

4. Chlorine has two isotopes, ³⁵Cl (75%) and ³⁷Cl (25%).

(a) Define isotope. [2] (b) Calculate the relative atomic mass of chlorine. [2] (c) Explain why isotopes of the same element have identical chemical properties. [2]

5. Explain the trend in reactivity down Group 1, referring to electronic structure. [4]

6. Explain the trend in reactivity down Group 7, and predict the observation when chlorine is added to potassium iodide solution. [5]

7. State two properties of the noble gases and explain them in terms of electronic structure. [3]

8. Explain how Mendeleev arranged the elements known in his time into an early periodic table, and how the later discovery of gallium and germanium supported his arrangement. [3]

9. Describe the trend in melting and boiling points of the halogens (Group 7) down the group, and explain it in terms of intermolecular forces. [3]

10. A student has a mixture of sand and salt solution. Describe how they could obtain (a) dry sand and (b) pure, dry salt crystals from this mixture, naming the technique used in each case. [4]


Answers

1. Proton — mass 1, charge +1 [1]. Neutron — mass 1, charge 0 [1]. Electron — mass very small (1/1836), charge −1 [1].

2. Atomic number 17 [1]; mass number 37 [1]; electronic structure 2,8,7 [1].

3. Before, the plum pudding model described a ball of positive charge with electrons embedded in it [1]. Most alpha particles passed straight through, showing the atom is mostly empty space [1]. A few were deflected or bounced back, showing there is a small, dense, positively charged nucleus [1]. This led to the nuclear model, later refined by Bohr with electrons in fixed shells [1].

4. (a) Atoms of the same element with the same number of protons but different numbers of neutrons [1] [1]. (b) (35 × 75 + 37 × 25) ÷ 100 [1] = 35.5 [1]. (c) Chemical properties depend on the number and arrangement of electrons [1], which is identical in isotopes — only the neutron number differs [1].

5. Reactivity increases down the group [1]. Each element down the group has more electron shells, so the outer electron is further from the nucleus [1] and is shielded by more inner shells [1]. The attraction between the nucleus and the outer electron is therefore weaker, so it is lost more easily and the reaction is more vigorous [1]. All three parts of the explanation — distance, shielding, and the resulting weaker attraction — are needed for full marks; naming only “more shells” earns partial credit at best.

6. Reactivity decreases down the group [1], because the outer shell is further from the nucleus and more shielded [1], so the atom attracts an incoming electron less strongly and gains one less readily [1]. Chlorine is more reactive than iodine, so it displaces iodine from potassium iodide solution, and the solution turns brown as iodine forms [1] [1].

7. They are unreactive/inert [1] and exist as single atoms (monatomic) [1], because they have a full outer shell of electrons, so they have no tendency to lose, gain or share electrons [1].

8. Mendeleev arranged the known elements in order of atomic weight, but grouped them so that elements with similar properties fell in the same column [1]. Where this meant leaving gaps, he did so anyway, and used the gaps to predict the properties of undiscovered elements [1]. When gallium and germanium were later discovered, their actual properties matched his predictions closely, confirming his arrangement was based on a real underlying pattern rather than coincidence [1].

9. Melting and boiling points increase down the group [1], because the molecules get larger as you go down [1], so the intermolecular forces between molecules become stronger and more energy is needed to overcome them [1].

10. (a) Filtration [1] separates the insoluble sand from the salt solution, exploiting the difference in particle size [1]. (b) Crystallisation [1] of the filtered salt solution — heating to evaporate water until the solution becomes saturated, then leaving it to evaporate further (or cool) so the salt it can no longer hold crystallises out [1].


Where marks are usually lost

  • Confusing atomic number with mass number.
  • Forgetting to divide by 100 in the relative atomic mass calculation.
  • Explaining group trends by “more shells” alone without mentioning shielding or distance.
  • Saying noble gases have no electrons in their outer shell.
  • Explaining Group 7’s melting/boiling point trend using the same “shielding and distance” argument as reactivity — that argument is about electron loss/gain, not intermolecular forces, and answers the wrong question if used here.
  • Naming a separation technique without stating the property it exploits (particle size, solubility, or boiling point) — the technique name alone is a partial answer.

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