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A-LevelChemistryAtomic structure9701/21 · May/June 2019 · Q311 Marks

Construct an equation for the second ionisation energy of argon.

  1. (a)Construct an equation for the second ionisation energy of argon.[1]
  2. (b)The graph shows successive ionisation energies for the element argon. Complete the graph with predictions for the eighth and ninth ionisation energies of argon. Use a cross (x) for each data point. [Figure showing ionisation energy graph][2]
  3. (c)The energy value required to remove the first electron from an atom of argon is circled on the graph. Sketch the shape of the orbital that contains this electron.[1]
  4. (d)(i)Calculate the mass, in grams, of Cl2(g) formed. For this calculation, assume that chlorine behaves as an ideal gas under these conditions. mass of Cl2(g) = ...................................... g[3]
  5. (d)(ii)Calculate the number of chlorine atoms in this sample of Cl2(g). You may find it helpful to use your answer to (d)(i). If you are unable to calculate an answer to (d)(i), use 0.36 g of Cl2. This is not the correct answer. number of chlorine atoms = ......................................[2]
  6. (d)(iii)Cl2(g) does not behave as an ideal gas under these conditions. Explain why Cl2(g) behaves even less ideally at: • very high pressures • very low temperatures.[2]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

  1. [1] Ar+(g) → Ar²+(g) + e(-) OR Ar+(g) – e(-) → Ar²+(g)
  2. [1] at x = 8, within range 13000–20000
  3. [1] at x = 9, within range 35000–45000
  4. [1] M1 correct conversions of data to SI/consistent units (p = 404 000 Pa; V = 20 × 10–6 m3; T = 298 K)
  5. [1] M2 calculation of n (= pV/RT) from M1 values (n = (404000 × 20 × 10-6) / (8.31 × 298) = 3.263 × 10-3 mol of Cl2)
  6. [1] M3 finding the mass of Cl2 (= 3.263 × 10-3 × 71.0 = 0.23 (g))
  7. [1] M1 = 3.263 × 10-3 × 2 (moles of Cl atoms)
  8. [1] M2 = 6.02 × 1023 x M1 = 3.93 x 1021 atoms of Cl
  9. [1] M1 size / volume of molecule / particle becomes significant / non-negligible OR IMFs become significant / non-negligible
  10. [1] M2 IMFs becomes significant /non-negligible / collisions are not elastic

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About this question

Topic
Atomic structure, A-Level Chemistry (syllabus 9701). Structured question, 11 marks.
Source
A-Level Chemistry, May/June 2019 Paper 2 Variant 1, question 3.

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