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Practice Questions

AQA A Level Chemistry: Atomic Structure — Practice Questions

Original exam-style practice questions with full worked answers on time-of-flight mass spectrometry, electron configuration and ionisation energy for AQA A Level Chemistry 7405.

Subject
Chemistry
Level
AS LEVEL
Topic
Section 3.1.1 – Atomic Structure
Updated

Aligned to AQA A Level Chemistry (7405), For teaching from September 2015. 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 revision notes


Questions

1. Name the five stages of time-of-flight mass spectrometry in order. [5]

2. Compare electron impact and electrospray ionisation, stating when each is used. [4]

3. A sample analysed by electrospray ionisation gives a peak at m/z 181.

(a) State the M_r of the molecule, explaining your reasoning. [2] (b) Explain why this correction is not needed with electron impact ionisation. [1]

4. In the acceleration stage all ions are given the same kinetic energy.

(a) Explain why lighter ions reach the detector first. [2] (b) An ion of mass 1.4 × 10⁻²⁵ kg has kinetic energy 4.8 × 10⁻¹⁶ J. Calculate its speed. [3]

5. Magnesium has isotopes of mass 24, 25 and 26 with abundances 79.0%, 10.0% and 11.0%. Calculate the relative atomic mass to one decimal place. [3]

6. Explain why the first ionisation energy of aluminium is lower than that of magnesium. [2]

7. Explain why the first ionisation energy of sulfur is lower than that of phosphorus. [2]

8. An element’s successive ionisation energies show a very large jump between the third and fourth values.

(a) State the group of the element, explaining your reasoning. [2] (b) Explain, in terms of electron shells, why a large jump occurs at this point rather than a gradual increase throughout. [2]

9. Write the electron configuration, in noble-gas shorthand, of chromium (Z = 24) and of copper (Z = 29), and explain why each does not follow the expected 4s-before-3d filling pattern predicted by aufbau order alone. [4]


Answers

1. Ionisation [1], acceleration [1], ion drift [1], detection [1], data analysis [1].

2. Electron impact: a high-energy electron knocks an electron out, forming X⁺ [1]; used for elements and low-M_r compounds [1]. Electrospray: the sample gains a proton, forming XH⁺ [1]; used for large biological molecules, which would fragment under electron impact [1].

3. (a) 180 [1] — the ion detected is XH⁺, so its mass is one greater than the molecule’s [1]. (b) Electron impact forms X⁺ by removing an electron, whose mass is negligible, so the m/z equals the M_r [1].

4. (a) All ions have the same kinetic energy [1], and since KE = ½mv², a smaller mass must have a greater speed [1]. (b) v = √(2KE ÷ m) = √((2 × 4.8 × 10⁻¹⁶) ÷ 1.4 × 10⁻²⁵) [1] = √(6.857 × 10⁹) [1] = 8.28 × 10⁴ m s⁻¹ [1].

5. A_r = [(24 × 79.0) + (25 × 10.0) + (26 × 11.0)] ÷ 100 [1] = (1896 + 250 + 286) ÷ 100 [1] = 2432 ÷ 100 = 24.3 [1].

6. Aluminium’s outer electron is in a 3p sub-shell, which is higher in energy than 3s [1] and is slightly shielded by the 3s electrons, so less energy is needed to remove it [1].

7. In sulfur, two electrons occupy the same 3p orbital [1] and the repulsion between them makes one easier to remove than an unpaired 3p electron in phosphorus [1].

8. (a) Group 3 [1]. The first three electrons removed come from the outer shell, so this element has 3 outer-shell electrons [1]. (b) The fourth electron removed comes from a shell closer to the nucleus, with less shielding and greater nuclear attraction [1], so considerably more energy is needed to remove it than the outer electrons [1].

9. Chromium: [Ar]3d⁵4s¹ [1]. Copper: [Ar]3d¹⁰4s¹ [1]. Both are exceptions caused by the extra stability of a half-filled (3d⁵) or fully-filled (3d¹⁰) sub-shell [1]: one electron moves from 4s into 3d so that the d sub-shell reaches this specially stable arrangement, rather than following the 3d⁴4s² or 3d⁹4s² pattern the simple aufbau filling order would otherwise predict [1].


Where marks are usually lost

  • Forgetting to subtract 1 from the m/z after electrospray ionisation.
  • Giving the mass spectrometry stages out of order or omitting ion drift.
  • Explaining the Al dip by nuclear charge rather than sub-shell energy.
  • Not dividing by 100 in a relative atomic mass calculation.
  • Miscounting which ionisation energy the jump occurs after when deducing a group from successive ionisation energies — the jump comes immediately after all the outer-shell electrons have been removed, not before.
  • Writing chromium as [Ar]3d⁴4s² or copper as [Ar]3d⁹4s² — both are the two standard exceptions to the aufbau filling order, caused by the stability of a half-filled or filled d sub-shell.
  • Forgetting that isotopes have identical chemical properties (same electron arrangement) despite differing physical properties such as mass.

Successive ionisation energies — reading the group off a graph

A graph or table of successive ionisation energies is really a shell-by-shell map of an atom’s electron structure: within one shell, energies rise gradually as each successive electron is removed from an increasingly positive ion, but a sharp jump appears the moment removal starts on the next shell in, since that shell sits closer to the nucleus with less shielding. Counting how many ionisation energies come before the first big jump tells you directly how many electrons are in the outer shell, and therefore the element’s group — a technique worth practising on several different jump positions (after the 1st, 2nd, 3rd electron, and so on) rather than only the specific example seen in class. For the full electron-configuration filling order, including both anomalies in period 4, see the Atomic Structure revision notes.

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