These are real past-paper questions. Think of your answer first, then open the explanation.
May/June 2021 · Paper 1 · Atoms/molecules/stoichiometry
Which contains the largest number of hydrogen atoms?
- 0.10 mol of pentane
- 0.20 mol of but-2-ene
- 1.00 mol of hydrogen molecules
- 6.02 × 10²³ hydrogen atoms
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Answer: C. 1.00 mol of hydrogen molecules
1.00 mol of H2 contains 2.00 mol of H atoms, more than the 1.2, 1.6 or 1.0 mol of H atoms given by the other options.
May/June 2021 · Paper 1 · Atomic structure
In which pair of species do both species have only one unpaired p electron?
- Ar⁺ and C⁻
- B and Ti⁺
- F and Ga
- Se⁻ and Si⁻
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Answer: C. F and Ga
F (2p⁵) has electron arrangement px²py²pz¹ (1 unpaired); Ga (4p¹) also has just 1 unpaired p electron.
May/June 2021 · Paper 1 · Chemical bonding
Phosphorus forms two chlorides. Phosphorus(III) chloride, PCl3, is a covalent liquid. Phosphorus(V) chloride is an ionic solid. One of the ions present is [PCl4]⁺. What is the shape of the PCl3 molecule and the [PCl4]⁺ ion?
- PCl3 pyramidal, [PCl4]⁺ square planar
- PCl3 pyramidal, [PCl4]⁺ tetrahedral
- PCl3 tetrahedral, [PCl4]⁺ square planar
- PCl3 trigonal planar, [PCl4]⁺ tetrahedral
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Answer: B. PCl3 pyramidal, [PCl4]⁺ tetrahedral
PCl3 has 3 bonding pairs + 1 lone pair (pyramidal); [PCl4]⁺ has 4 bonding pairs and no lone pair (tetrahedral, isoelectronic with SiCl4).
May/June 2021 · Paper 2 · Group 17
Sodium halide salts react with concentrated sulfuric acid at room temperature.
(a)(i) Write an equation to represent the reaction of NaCl(s) with concentrated sulfuric acid. [1]
(ii) Name this type of reaction. [1]
(b) NaI(s) reacts with concentrated sulfuric acid, at room temperature, to form steamy fumes.
(i) Identify the chemical responsible for the steamy fumes. [1]
(ii) The reaction of NaI(s) with concentrated sulfuric acid continues, forming several other products, including a dark grey solid. Identify the chemical responsible for the dark grey solid and one other product of this further reaction. [2]
(c) Explain the differences in observations, at room temperature, when NaI(s) reacts with concentrated sulfuric acid compared to those for NaCl(s). [2]
(d) Complete the equation for the reaction of Br⁻ with excess concentrated H2SO4 at room temperature: ...Br⁻ + ...H⁺ + ...H2SO4 → ... [1]
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(a)(i) NaCl + H2SO4 → NaHSO4 + HCl (or 2NaCl + H2SO4 → Na2SO4 + 2HCl)
(ii) Displacement (acid–base) reaction.
(b)(i) Hydrogen iodide, HI.
(ii) Dark grey solid: I2 (iodine). Other product: S (sulfur) or H2S (hydrogen sulfide) or H2O/steam.
(c) Iodide ions are stronger reducing agents than chloride ions, so HI/iodide is oxidised by the concentrated H2SO4 (further redox reactions occur, giving I2, S/H2S) whereas HCl/chloride is not oxidised (only an acid–base reaction occurs with NaCl).
(d) 2Br⁻ + 2H⁺ + H2SO4 → Br2 + 2H2O + SO2
Reducing power of halide ions increases down Group 17 (I⁻ > Br⁻ > Cl⁻), so only I⁻ and Br⁻ are strong enough to reduce concentrated H2SO4.