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A-LevelChemistryChemical periodicity9701/22 · May/June 2023 · Q213 Marks

Chlorine is a reactive element. It forms many compounds.

  1. (a)(i)Complete Table 2.1 to show the maximum oxidation number of the elements Na to P in their chlorides. Table 2.1 element | Na | Mg | Al | Si | P ---|---|---|---|---|--- maximum oxidation number | | | | |[1]
  2. (a)(ii)State what determines the maximum oxidation number of elements in Period 3.[1]
  3. (b)(i)An excess of cold water is added to the chloride of silicon. Write an equation for the reaction between an excess of cold water and the chloride of silicon.[1]
  4. (b)(ii)Suggest the pH of the solution produced in (b)(i).[1]
  5. (c)(i)An excess of cold water is added to the chloride of phosphorus. Write an equation for the reaction between an excess of cold water and the chloride of phosphorus.[1]
  6. (c)(ii)Suggest the pH of the solution produced in (c)(i).[1]
  7. (d)(i)Write an equation for the reaction of chlorine with water.[1]
  8. (d)(ii)Write an equation for the reaction of chlorine with hot NaOH(aq).[1]
  9. (e)(i)Bleach is used as a cleaning product to kill bacteria. It is made by adding compounds like sodium chlorate(I), NaClO, to water. Identify the formula of the ion present in bleach that kills bacteria.[1]
  10. (e)(ii)Sodium chlorate(I), NaClO, reacts with hydrogen peroxide to produce sodium chloride, water and oxygen gas. Construct an equation for this reaction.[1]
  11. (e)(iii)A sample of bleach W contains an unknown concentration of sodium chlorate(I). 10.0 cm³ of W is diluted with distilled water to make a total volume of 100 cm³ of bleach solution. 25.0cm³ of this diluted bleach solution is added to an excess of hydrogen peroxide and the volume of gas produced measured under room conditions. The experiment is repeated and on average 25.0 cm³ of diluted bleach solution produces 42.0 cm³ of gas. Calculate the concentration, in gdm⁻³, of sodium chlorate(I) in W. concentration of NaClO in W =[3]

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

  1. [1] element | Na | Mg | Al | Si | P ---|---|---|---|---|--- maximum oxidation number | (+)1 | (+)2 | (+)3 | (+)4 | (+)5 All correct for one mark
  2. [1] number of outer/valence electrons
  3. [1] SiCl4 + 2H2O → SiO2 + 4HCl
  4. [1] 1–4
  5. [1] PCl5 + 4H2O → 5HCl + H3PO4
  6. [1] 1–4
  7. [1] Cl2 + H2O ⇌ HOCl + HCl
  8. [1] 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O
  9. [1] ClO⁻
  10. [1] NaClO + H2O2 → O2 + NaCl + H2O
  11. [3] M1 (show how to calculate) mol O2 gas 42 cm³ gas = 42 / 24000 mol oxygen gas (= 1.75 × 10⁻³ mol) M2 (process to find) amount NaClO in 10 cm³ sample = 100 cm³ (diluted) bleach M1 x 4 (= 4 × 1.75 × 10⁻³ mol in 10 cm⁻³ = 0.007) M3 correct calculation using M2 to find concentration = (M2 /10) × 1000 × 74.5 = 52.2 (g / dm³)

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About this question

Topic
Chemical periodicity, A-Level Chemistry (syllabus 9701). Structured question, 13 marks.
Source
A-Level Chemistry, May/June 2023 Paper 2 Variant 2, question 2.

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