A Level H2 Chemistry Periodic Table Quiz

A-Level Chemistry H2 Quiz - Periodic Table: Answer Key

Total Marks: 50

Section A: Periodicity and Trends (Questions 1–5)

Question 1

(a) [2 marks]

• Across Period 3, nuclear charge (number of protons) increases while electrons are added to the same principal quantum shell (n=3) [1 mark].
• The increased nuclear charge exerts a greater attractive force on the outer electrons, pulling them closer to the nucleus, so atomic radius decreases [1 mark].

(b) [2 marks]

• Magnesium has electronic configuration 1s² 2s² 2p⁶ 3s²; aluminium has 1s² 2s² 2p⁶ 3s² 3p¹ [1 mark].
• The 3p electron in aluminium is in a higher energy subshell than the 3s electrons in magnesium. The 3p electron is better shielded from the nucleus by the 3s electrons, so the effective nuclear charge experienced is lower, resulting in a larger atomic radius / the 3p orbital is more diffuse than the 3s orbital [1 mark].

Question 2

(a) [2 marks]

• Magnesium has a greater nuclear charge (+12 vs +11 for sodium) [1 mark].
• Both have outer electrons in the 3s orbital, but the increased nuclear charge in magnesium exerts a stronger attractive force on the outer electrons, requiring more energy to remove an electron [1 mark].

(b) [2 marks]

• The outer electron in aluminium is in a 3p orbital, whereas in magnesium it is in a 3s orbital [1 mark].
• The 3p electron is at a higher energy level and is better shielded from the nucleus by the 3s electrons, so less energy is required to remove it [1 mark].

Question 3

(a) [2 marks]

• Silicon has a giant covalent / macromolecular structure [1 mark].
• A large amount of energy is required to break the many strong covalent bonds between silicon atoms throughout the lattice, resulting in a very high melting point [1 mark].

(b) [2 marks]

• Phosphorus exists as discrete P₄ molecules with a simple molecular structure [1 mark].
• The intermolecular forces between P₄ molecules are weak van der Waals' forces, which require little energy to overcome, resulting in a low melting point [1 mark].

Question 4

(a) [1 mark]

• Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l) (Accept: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O)

(b) [1 mark]

• Chlorine (oxidation state 0) is simultaneously oxidised to ClO⁻ (Cl: +1) and reduced to Cl⁻ (Cl: −1), so it undergoes both oxidation and reduction in the same reaction [1 mark].

Question 5

(a) [1 mark]

• Sodium has a giant metallic lattice with delocalised electrons that are free to move and carry charge throughout the structure [1 mark].

(b) [1 mark]

• Sulfur has a simple molecular structure with no free-moving charged particles (electrons or ions); all electrons are localised in covalent bonds or lone pairs [1 mark].

Section B: Group 2 Chemistry (Questions 6–10)

Question 6

(a) [2 marks]

• Ba(s) + 2H₂O(l) → Ba(OH)₂(aq) + H₂(g) [2 marks] (1 mark for correct reactants and products, 1 mark for correct state symbols and balancing)

(b) [2 marks]

• The vigour of reaction increases down the group [1 mark].
• Down the group, atomic radius increases and ionisation energy decreases. It becomes easier to remove the two outer electrons to form M²⁺ ions, so the metal reacts more readily with water [1 mark].

Question 7

(a) [2 marks]

• CaCO₃(s) → CaO(s) + CO₂(g) [2 marks] (1 mark for correct products, 1 mark for state symbols and balancing)

(b) [3 marks]

• The carbonate ion (CO₃²⁻) is polarised by the metal cation [1 mark].
• Down Group 2, the charge density of the cation decreases (larger ionic radius, same charge), so the polarising power decreases [1 mark].
• Magnesium ions polarise the carbonate ion more strongly, weakening the C–O bonds within the carbonate ion, making it easier to decompose. Barium ions have lower charge density, so polarise less, requiring higher temperature for decomposition [1 mark].

Question 8

(a) [1 mark]

• Solubility decreases down the group from magnesium sulfate (soluble) to barium sulfate (insoluble) [1 mark].

(b) [2 marks]

• Add aqueous barium chloride (or barium nitrate) to each solution [1 mark].
• Magnesium sulfate: a white precipitate of barium sulfate forms (Ba²⁺ + SO₄²⁻ → BaSO₄). Barium sulfate: no visible change as it is already insoluble / no further precipitation [1 mark]. (Accept: Add dilute HCl followed by BaCl₂; white precipitate confirms sulfate in MgSO₄ solution.)

Question 9

(a) [2 marks]

• Barium gives an apple-green / pale green flame colour [1 mark].
• When heated, electrons in the metal ion absorb energy and are promoted to higher energy levels. As they fall back to lower energy levels, they emit energy in the form of visible light of characteristic wavelengths [1 mark].

Question 10

(a) [1 mark]

• MgO(s) + H₂O(l) → Mg(OH)₂(s) [1 mark] (Accept: MgO(s) + H₂O(l) → Mg(OH)₂(aq) for partial solubility)

(b) [2 marks]

• Down Group 2, the ionic radius of the cation increases, so the charge density decreases [1 mark].
• The lattice energy of the hydroxide decreases more rapidly than the hydration energy, so the enthalpy of solution becomes more exothermic / less endothermic, resulting in increased solubility [1 mark].

Section C: Group 17 Chemistry (Questions 11–15)

Question 11

(a) [3 marks]

• Boiling points increase down the group from fluorine to iodine [1 mark].
• The halogen molecules are non-polar and held together by van der Waals' forces (instantaneous dipole–induced dipole interactions) [1 mark].
• Down the group, the number of electrons per molecule increases, so the electron cloud is more polarisable. This results in stronger van der Waals' forces, requiring more energy to overcome [1 mark].

(b) [2 marks]

• Astatine is predicted to be a solid at room temperature and pressure [1 mark].
• The trend in boiling points increases down the group; iodine is already a solid. Astatine, being below iodine, would have even stronger van der Waals' forces and a higher melting/boiling point, so it would be solid [1 mark].

Question 12

(a) [2 marks]

• Cl₂(g) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq) [2 marks] (1 mark for correct species, 1 mark for balancing and state symbols)

(b) [2 marks]

• E° for Cl₂/Cl⁻ = +1.36 V; E° for Br₂/Br⁻ = +1.07 V [1 mark].
• E°cell = E°(reduction) − E°(oxidation) = +1.36 − (+1.07) = +0.29 V. Since E°cell is positive, the reaction is thermodynamically feasible / chlorine is a stronger oxidising agent than bromine and can oxidise Br⁻ to Br₂ [1 mark].

Question 13

(a) [1 mark]

• White precipitate [1 mark].

(b) [1 mark]

• Ag⁺(aq) + Cl⁻(aq) → AgCl(s) [1 mark].

(c) [3 marks]

• Silver chloride dissolves in dilute ammonia solution, forming the soluble complex ion [Ag(NH₃)₂]⁺, so the white precipitate disappears [1 mark].
• Silver iodide is insoluble in dilute ammonia because the lattice energy of AgI is too large / the solubility product is too low for dissolution to occur [1 mark].
• Therefore, if the precipitate dissolves in dilute ammonia, the halide is chloride; if it remains, it is iodide [1 mark].

Question 14

(a) [2 marks]

• H₂(g) + Cl₂(g) → 2HCl(g) [1 mark]
• Sunlight / ultraviolet light / heat [1 mark].

(b) [2 marks]

• Thermal stability decreases from HCl to HI [1 mark].
• Down the group, the H–X bond length increases and bond strength decreases, so less energy is required to break the bond, making the hydrogen halide easier to decompose on heating [1 mark].

Question 15

(a) [1 mark]

• Cl₂(g) + H₂O(l) ⇌ HCl(aq) + HOCl(aq) (Accept: Cl₂ + H₂O → HCl + HOCl)

(b) [1 mark]

• Chloric(I) acid (HOCl) / hypochlorous acid is a strong oxidising agent that kills bacteria by oxidation / Chlorine reacts with water to form HOCl which destroys bacterial cell walls [1 mark].

Section D: Transition Elements (Questions 16–20)

Question 16

(a) [1 mark]

• 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ (Accept: [Ar] 3d⁶) [1 mark]

(b) [3 marks]

• Iron has the electronic configuration [Ar] 3d⁶ 4s². It can lose its two 4s electrons to form Fe²⁺ ([Ar] 3d⁶), and can also lose one 3d electron to form Fe³⁺ ([Ar] 3d⁵) because the 3d and 4s orbitals are close in energy [1 mark].
• The Fe³⁺ ion has a stable half-filled d-subshell (3d⁵), which provides additional stability [1 mark].
• Zinc has the electronic configuration [Ar] 3d¹⁰ 4s². It can only lose its two 4s electrons to form Zn²⁺ ([Ar] 3d¹⁰). The filled 3d subshell is very stable, so further ionisation would require very high energy and is not observed under normal chemical conditions [1 mark].

Question 17

(a) [3 marks]

• Transition metal ions have partially filled d-orbitals; s-block metal ions have either empty or completely filled d-orbitals [1 mark].
• In transition metal complexes, ligands cause the d-orbitals to split into two sets of different energies [1 mark].
• Electrons can absorb visible light of a specific wavelength to undergo d–d transitions from the lower energy d-orbitals to the higher energy d-orbitals. The complementary colour of the absorbed light is observed. In s-block compounds, d–d transitions are not possible (no partially filled d-orbitals), so they appear white or colourless [1 mark].

(b) [2 marks]

• The pale blue solution turns deep blue / royal blue [1 mark].
• The complex ion is [Cu(NH₃)₄(H₂O)₂]²⁺ / tetraamminediaquacopper(II) ion [1 mark].

Question 18

(a) [2 marks]

• Iron / finely divided iron [1 mark].
• Heterogeneous catalysis [1 mark].

(b) [1 mark]

• The catalyst provides an alternative reaction pathway with a lower activation energy, so a greater proportion of reactant molecules possess energy equal to or greater than the activation energy, increasing the frequency of effective collisions [1 mark].

Question 19

(a) [2 marks]

• A ligand is a species that donates a lone pair of electrons to a central metal ion to form a coordinate / dative covalent bond [1 mark].
• Examples: H₂O, NH₃, Cl⁻, CN⁻ (any one correct example) [1 mark].

(b) [2 marks]

• Transition metal ions have vacant, energetically accessible d-orbitals that can accept lone pairs from ligands [1 mark].
• s-block metal ions have no vacant d-orbitals of suitable energy; their outer orbitals are either empty or fully occupied, making complex formation less favourable [1 mark].

Question 20

(a) [1 mark]

• The purple colour of MnO₄⁻ disappears and the solution becomes colourless / very pale pink [1 mark].

(b) [1 mark]

• Potassium manganate(VII) is intensely purple in colour, but its reduction product Mn²⁺ is almost colourless. The end point is indicated by the first permanent pink colour when excess MnO₄⁻ is present, so no external indicator is needed [1 mark].

End of Answer Key

Marking Notes:

• Award marks for correct chemical principles even if wording differs from model answer.
• For calculation questions, award error-carried-forward (ECF) marks where appropriate.
• State symbols are required where specified; deduct 1 mark per omission up to the stated mark allocation.