OxfordAQA A-Level Chemistry: Atomic structure (9620) – 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 from your board or school.

This set tests section 3.1.1 Atomic structure from the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, across 3.1.1.1 (fundamental particles), 3.1.1.2 (mass number and isotopes) and 3.1.1.3 (electron configuration). The whole section is International AS content, so no question here is International A-level only.

Learn the content first from the atomic structure study guide and the atomic structure revision notes. See also the course hub, the printable checklist and a free 10-minute diagnostic. Where you meet the mole next, see amount of substance.

Questions

1. A particle contains 17 protons, 18 neutrons and 18 electrons. Give its symbol including the mass number, atomic number and charge. [3]

2. Chlorine-35 and chlorine-37 are isotopes. Explain what the term means and why the two behave identically in reactions. [3]

3. Describe how the results of firing alpha particles at thin gold foil led scientists to change the plum pudding model of the atom. [3]

4. (a) For ⁵⁵Mn²⁺ (atomic number 25), give the numbers of protons, neutrons and electrons. [3] (b) Give the relative charge and relative mass of an electron. [2]

5. Describe what happens to the ions in the acceleration and ion drift stages of a time of flight (TOF) mass spectrometer, and explain why ions of different mass arrive at the detector at different times. [4]

6. An element contains three isotopes. Its mass spectrum has these peaks.

m/z	64	66	68
Relative abundance	48.0	28.0	24.0

Calculate the relative atomic mass to 4 significant figures. [3]

7. An element R has two isotopes, ⁷⁹R and ⁸¹R. Its relative atomic mass is 80.2. Calculate the percentage abundance of each isotope, and state which peak is taller in the mass spectrum. [4]

8. The mass spectrum of a compound has its peak of highest m/z at 72. State what this peak represents and what it shows about the compound. Write an equation, with state symbols, for the formation of this ion from a molecule M in the ionisation stage. [3]

9. Write the full electron configuration of each of the following, using s, p and d notation.

(a) Vanadium atom (Z = 23) [1] (b) Chromium atom (Z = 24) [1] (c) Co²⁺ (Z = 27 for the atom) [1] (d) Se²⁻ (Z = 34 for the atom) [1]

10. (a) Define the first ionisation energy. [2] (b) Give the equation, including state symbols, for the removal of the third electron from aluminium. [1] (c) Explain why the first ionisation energy of barium is lower than that of beryllium, and state what this is evidence for. [4]

11. Account for the drop in first ionisation energy from magnesium to aluminium, and again from phosphorus to sulfur. State what each trend suggests about the structure of the atom. [6]

12. The successive ionisation energies of a Period 3 element (invented data) are given in kJ mol⁻¹.

Number removed	1	2	3	4	5	6	7	8
Ionisation energy	760	1480	3200	4300	16000	19500	23300	29400

(a) Deduce the group of the element and give its electron configuration. [3] (b) Write the equation for the fifth ionisation energy. [1] (c) Explain the large increase between the fourth and fifth values. [2] (d) State which successive ionisation energy shows the next very large increase, and explain why. [2]

Answers

1. Protons = 17 so the element is chlorine, atomic number 17 [1]; mass number = 17 + 18 = 35 [1]; electrons exceed protons by one so the charge is 1−: ³⁵Cl⁻ [1] [3] Examiner insight: Give the mass number as a superscript on the left and the charge on the right; a symbol with the numbers the wrong way round is easy to misread.

2. Isotopes have nuclei with an equal number of protons (so they are one element) [1] but different numbers of neutrons [1]. Their electron arrangements are identical, and reactions are governed by the electrons [1] [3] Examiner insight: “Same number of protons” and “different number of neutrons” both need stating; saying “different mass” alone describes the effect and not the cause.

3. The great majority of alpha particles went undeviated, so most of the atom is empty [1]. A small fraction rebounded through big angles, so positive charge is concentrated in one place [1]. This showed a small, dense, positive nucleus, not positive charge spread through the whole atom as the plum pudding model said [1] [3] Examiner insight: Link each observation to its conclusion; listing observations with no inference usually leaves marks unclaimed.

4. (a) Protons 25 [1]; neutrons 55 − 25 = 30 [1]; electrons 25 − 2 = 23 [1] (b) Relative charge −1 [1]; relative mass about 1/1840 (very small compared with 1 for a proton) [1] [5] Examiner insight: For ions, subtract a positive charge from the proton number to get electrons. Either “1/1836” or “1/1840” shows the right idea, but “0” is too loose.

5. In acceleration an electric field acts on the positive ions [1], giving every ion the same kinetic energy [1]. In the drift region there is no field and ions move at constant speed [1]. With equal kinetic energy, lighter ions are faster so reach the detector sooner and heavier ions take longer [1] [4] Examiner insight: The key phrase is “the same kinetic energy”. Saying “the same speed” contradicts the explanation of why times differ.

6. Total abundance = 48.0 + 28.0 + 24.0 = 100.0 [1]; (64 × 48.0 + 66 × 28.0 + 68 × 24.0) / 100.0 = 6552 / 100.0 [1]; 65.52 [1] [3] Examiner insight: Show the sum divided by the total abundance. Check that the answer lies between 64 and 68 and nearer the largest peak.

7. Let x be the percentage of ⁸¹R: 79(100 − x) + 81x = 80.2 × 100 [1]; 2x = 120 so x = 60 [1]; ⁸¹R is 60% and ⁷⁹R is 40% [1]; the peak at m/z 81 is taller [1] [4] Examiner insight: A quick sense check helps: 80.2 is closer to 81 than to 79, so the heavier isotope must be the more abundant one.

8. It is the molecular ion [1], so the relative molecular mass is 72 [1]. M(g) + e⁻ → M⁺(g) + 2e⁻ [1] [3] Examiner insight: Write the ion with its positive charge and balance the electrons; “M⁺” with a missing second electron on the right is a frequent slip.

9. (a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s² [1] (b) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ [1] (c) 25 electrons, with the 4s electrons lost first: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷ [1] (d) 36 electrons: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ [1] [4] Examiner insight: Count the total electrons in your answer against the atomic number and charge before moving on.

10. (a) The energy change when each atom in a mole of gaseous atoms loses one electron [1], forming a mole of gaseous 1+ ions [1] (b) Al²⁺(g) → Al³⁺(g) + e⁻ [1] (c) Barium’s outermost electron sits in a shell much more distant from the nucleus [1] and is screened by extra inner electrons [1]; this outweighs the greater nuclear charge, so less energy is needed [1]; this is evidence for electron shells [1] [7] Examiner insight: An answer that cites only “more protons” or only “more shells” is incomplete; the argument needs distance, shielding and the comparison with nuclear charge.

11. Al: its outermost electron occupies 3p, a sub-shell higher in energy and slightly further from the nucleus than 3s [1], and is shielded by the 3s electrons [1], so it is removed more easily than a 3s electron of Mg [1]. This is evidence for sub-shells [1]. S: two of its 3p electrons share one orbital [1], and mutual repulsion within that pair makes one electron leave more readily than the unpaired outer electron of P [1] [6] Examiner insight: Name the sub-shell in each explanation and say what the trend is evidence for; many answers explain the dip but never state the conclusion the question asks for.

12. (a) The biggest jump comes after the fourth electron, so there are four outer electrons [1]: Group 4(14) (silicon) [1]; 1s² 2s² 2p⁶ 3s² 3p² [1] (b) X⁴⁺(g) → X⁵⁺(g) + e⁻ [1] (c) The fifth electron comes from the second shell, which is closer to the nucleus [1] and less shielded, so it is held much more strongly [1] (d) The thirteenth [1], because it is the first electron removed from the 1s shell closest to the nucleus [1] [8] Examiner insight: Count electrons removed before the jump, not the position of the largest number. For the next jump, add the sizes of the shells: 4 outer, then 8, then 2.

Where marks are usually lost

• Calculating electrons as A − Z for an ion, or adding rather than subtracting a positive charge.
• Describing isotopes with differences in “mass” or “atomic number”.
• Saying ions are accelerated to the same speed, rather than the same kinetic energy.
• Not dividing by the total abundance when the percentages do not add to exactly 100.
• Giving a relative atomic mass outside the range of the isotopic masses.
• Writing 4s before 3d in an ion, or losing 3d electrons before 4s.
• Omitting state symbols or the electron in ionisation equations.
• Explaining the Group 2 trend without both shielding and distance.
• Describing a dip in ionisation energy without naming the sub-shell or stating what it shows.

Next steps

• Atomic structure revision notes
• Atomic structure study guide
• Amount of substance
• OxfordAQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Frequently asked questions

What does OxfordAQA A-level Chemistry 3.1.1 Atomic structure cover? Section 3.1.1 of OxfordAQA International AS and A-level Chemistry (9620) covers fundamental particles, mass number and isotopes, the time of flight mass spectrometer, electron configurations up to Z = 36 and ionisation energies. It is all International AS content.

What are common mistakes in atomic structure questions? Frequent errors include calculating electrons as A − Z for an ion, saying ions are accelerated to the same speed rather than the same kinetic energy, not dividing by the total abundance, and leaving state symbols or the electron out of ionisation equations.

How can you check a relative atomic mass calculation? A sensible relative atomic mass lies between the smallest and largest isotopic masses and closer to the most abundant isotope. An answer outside that range shows an arithmetic slip.

How do you explain a dip in first ionisation energy to get full marks? Name the sub-shell the electron is removed from and explain why it is easier to remove, then state what the dip shows about electron configuration in sub-shells. Describing the dip without naming the sub-shell loses credit.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. Section 3.1.1 Atomic structure, points 3.1.1.1 to 3.1.1.3, headed: Fundamental particles; Mass number and isotopes; Electron configuration.