AS Chemistry: Atomic Structure, Radius and Isotopes — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Work through these before looking at the answers. Write full answers — the marks are for the reasoning, not the conclusion.

Related: Atomic Structure: Particles, Radius and Isotopes revision notes

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Section A — short answer

1. State the relative charge and relative mass of a neutron. [2]

2. Define the term isotope. [2]

3. Explain why isotopes of the same element have identical chemical properties. [2]

4. A beam of protons and a beam of electrons, both travelling at the same velocity, pass through the same uniform electric field. State one similarity and one difference in how they behave. [3]

5. An atom of potassium has atomic (proton) number 19 and contains 20 neutrons. Write the full isotope notation for this atom. [2]

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Section B — structured

6. Atomic radius decreases across Period 3 (Na to Ar) but increases down Group 1 (Li to Cs).

(a) Explain why atomic radius decreases across Period 3. [3]

(b) Explain why atomic radius increases down Group 1. [3]

(c) State whether shielding changes significantly across Period 3, and justify your answer. [2]

7. Magnesium reacts to form Mg²⁺; sulfur reacts to form S²⁻.

(a) State and explain how the radius of Mg²⁺ compares with the radius of a Mg atom. [2]

(b) State and explain how the radius of S²⁻ compares with the radius of a S atom. [2]

8. Chlorine has two stable isotopes, ³⁵Cl and ³⁷Cl.

(a) State the number of protons, neutrons and electrons in one atom of ³⁷Cl. [3]

(b) A sample of chlorine gas, Cl₂, contains molecules with three different relative molecular masses. Explain why, and give all three Mr values. [3]

Section C

9. Explain why a beam of neutrons, passing through a uniform electric field alongside a beam of protons and a beam of electrons, is undeflected. [2]

10. Using electron configurations, explain why a sodium ion Na⁺ is smaller than a sodium atom Na, but a chloride ion Cl⁻ is larger than a chlorine atom Cl. [4]

11. State one physical property in which isotopes of the same element differ, and one chemical property in which they are identical, explaining why in each case. [4]

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12. (a) Define the term relative atomic mass. [1]

(b) A sample of boron contains only boron-10 and boron-11. Its relative atomic mass is 10.8. Calculate the percentage abundance of boron-10. [2]

Answers

1. Relative charge 0 [1]; relative mass 1 [1].

2. Atoms of the same element (same proton/atomic number) [1] with different numbers of neutrons (and therefore different mass/nucleon numbers) [1].

3. Chemical behaviour is determined by electron arrangement [1], and isotopes of the same element have the same proton number and therefore the same electron arrangement [1].

4. Similarity: both are deflected by the field, since both are charged particles [1]. Difference: they deflect in opposite directions (protons towards the negative plate, electrons towards the positive plate) [1], and the electrons deflect through a much larger angle, because their mass is so much smaller than a proton’s for the same magnitude of charge [1]. Any one valid similarity and one valid difference score full marks — both of the differences above are commonly expected.

5. Mass number = 19 + 20 = 39 [1]; notation ³⁹₁₉K [1].

6. (a) Nuclear charge increases by one proton per element [1], while each new electron is added to the same outer shell, so shielding stays roughly constant [1]; the stronger nuclear attraction on the outer electrons pulls them in more, contracting the atom [1].

(b) Each successive element adds a complete new outer shell [1], giving extra shielding and greater distance from the nucleus [1]; this outweighs the simultaneous increase in nuclear charge, so the atom is larger [1].

(c) Shielding stays roughly constant across Period 3 [1], because electrons are being added to the same shell each time — the inner, shielding shells are unchanged [1]. “More shielding” is the standard wrong explanation for the Period 3 trend — shielding is roughly constant across a period; it is the increasing nuclear charge that does the work.

7. (a) Mg²⁺ is smaller than a Mg atom [1]. Forming Mg²⁺ removes the entire outer shell, so the remaining electrons are held by the same nuclear charge with no extra shielding to oppose it [1].

(b) S²⁻ is larger than a S atom [1]. No new shell is added, but the two extra electrons increase electron–electron repulsion within the same outer shell, pushing the existing electrons further apart [1].

8. (a) Protons 17 [1]; neutrons 20 [1]; electrons 17 [1].

(b) Each Cl atom in a Cl₂ molecule can independently be either ³⁵Cl or ³⁷Cl [1], giving three possible combinations: 35+35 = 70, 35+37 = 72, and 37+37 = 74 [2].

9. Neutrons carry no charge [1], so an electric field exerts no force on them, and they pass straight through the field, undeflected [1].

10. Sodium (2,8,1) loses its single outer-shell electron to form Na⁺ (2,8) [1] — this removes the entire outer shell, so the remaining electrons are held by the same nuclear charge with no extra shielding to oppose it, and the ion is markedly smaller [1]. Chlorine (2,8,7) gains one electron to form Cl⁻ (2,8,8) [1] — no new shell is added, but the extra electron increases electron–electron repulsion within the same outer shell, pushing the existing electrons slightly further apart, so the ion is larger, but only modestly [1].

11. Physical property, any one: mass or density [1] — the extra neutrons add mass without changing the electron arrangement, so isotopes with different mass numbers have different masses (and, correspondingly, different densities) [1]. Chemical property: identical reactivity [1] — chemical behaviour is determined by electron arrangement, and isotopes of the same element share the same proton number and therefore an identical electron arrangement, so they react identically [1].

12. (a) The weighted mean mass of an atom of the element compared with one twelfth of the mass of an atom of carbon-12 [1]. (b) Let the percentage of boron-10 be x: 10x + 11(100 − x) = 10.8 × 100 [1], so 1100 − x = 1080 and x = 20% [1].

Examiner insight (June 2025): weaker answers defined relative isotopic mass instead of relative atomic mass, left out “one twelfth of the mass of a carbon-12 atom”, or did not attempt to find the abundance of the unknown isotope. Set up one equation with x and (100 − x). Try the real question next: Cambridge International AS & A Level Chemistry 9701, June 2025, Paper 22, Question 2.

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Where marks are usually lost

• Giving a neutron a relative charge other than 0, or forgetting its relative mass is 1.
• Explaining the Period 3 radius trend by “more shielding” instead of increasing nuclear charge.
• Explaining the Group 1 radius trend by nuclear charge alone, without mentioning the extra shell.
• Forgetting that Mg²⁺ loses a whole shell (not just electrons), which is why the radius drop is so large.
• Missing one of the three possible Cl₂ relative molecular masses.