A Level H2 Chemistry Periodic Table Quiz

A-Level Chemistry H2 Quiz - Periodic Table: ANSWER KEY

Total Marks: 50

Section A: Multiple Choice (10 marks)

Section B: Multiple Choice (10 marks)

Section C: Structured Questions (15 marks)

Question 11

(a) [2 marks]

• Al has electron configuration [Ne] 3s² 3p¹; Mg has [Ne] 3s². [1]
• The 3p electron in Al is at a higher energy level than the 3s electron in Mg, and is further from the nucleus / more shielded / experiences less effective nuclear attraction. Therefore, less energy is required to remove the 3p electron from Al. [1]

(b) [2 marks]

• P has electron configuration [Ne] 3s² 3p³ (half-filled p subshell); S has [Ne] 3s² 3p⁴. [1]
• In S, the 3p electron being removed is paired in a 3p orbital, leading to electron-electron repulsion which makes it easier to remove. In P, the 3p electrons are unpaired (half-filled subshell stability), so more energy is required. [1]

(c) [2 marks]

• General trend: first ionisation energy increases across Period 3. [1]
• Explanation: nuclear charge increases across the period while electrons are added to the same principal quantum shell (n=3), so shielding remains approximately constant. The increased nuclear attraction makes it harder to remove an electron. [1]

Question 12

(a) [3 marks]

• Boiling points increase from fluorine to iodine. [1]
• All halogens exist as diatomic molecules (F₂, Cl₂, Br₂, I₂) with simple molecular structures and van der Waals' forces between molecules. [1]
• Down the group, the number of electrons increases, leading to stronger instantaneous dipole-induced dipole (London) forces. More energy is required to overcome these intermolecular forces, hence boiling points increase. [1]

(b) [3 marks]

• Oxidising ability decreases down the group (F₂ > Cl₂ > Br₂ > I₂). [1]
• Oxidising ability depends on the tendency to gain an electron: X₂ + 2e⁻ → 2X⁻. [1]
• Down the group, atomic radius increases and shielding increases, so the attraction for an added electron decreases. The electron affinity becomes less exothermic, and the halogens become weaker oxidising agents. [1]

(c) [2 marks]

• Ionic equation: Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq) [1]
• Colour change: colourless/pale yellow-green solution turns orange/brown (due to formation of bromine). [1]

Question 13

(a) [3 marks]

• Transition metals have partially filled d-orbitals. [1]
• They can use these d-orbitals to form weak bonds with reactant molecules on their surface (chemisorption). [1]
• This brings reactant molecules together in the correct orientation, weakens existing bonds, and lowers the activation energy of the reaction. After reaction, products desorb, freeing the surface for further catalysis. [1]

(b) [2 marks]

• Any valid example, e.g.:
  • Iron in the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) [1 mark for catalyst, 1 mark for equation]
  • Vanadium(V) oxide in the Contact process: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
  • Nickel in hydrogenation of alkenes: C₂H₄(g) + H₂(g) → C₂H₆(g)

Question 14

(a)(i) [1 mark]

• Na₂O: ionic (giant ionic lattice) [1]

(a)(ii) [1 mark]

• SiO₂: giant covalent / macromolecular (network covalent) [1]

(b) [1 mark]

• Na₂O(s) + H₂O(l) → 2NaOH(aq) [1]

(c) [3 marks]

• Aluminium oxide is amphoteric because it reacts with both acids and bases. [1]
• With acid: Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l) OR Al₂O₃(s) + 6H⁺(aq) → 2Al³⁺(aq) + 3H₂O(l) [1]
• With base: Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2NaAl(OH)₄(aq) OR Al₂O₃(s) + 2OH⁻(aq) + 3H₂O(l) → 2[Al(OH)₄]⁻(aq) [1]

Question 15

(a) [3 marks]

• In transition metal complexes, the five d-orbitals are split into two sets of different energies (e.g., t₂g and eg in octahedral complexes) due to the approach of ligands. [1]
• The energy difference (ΔE) between these sets of d-orbitals corresponds to the energy of visible light. [1]
• An electron in a lower-energy d-orbital can absorb a photon of visible light and be promoted to a higher-energy d-orbital (d-d transition). The colour observed is the complementary colour of the light absorbed. [1]

(b) [2 marks]

• The two complexes have different colours because the energy gap (ΔE) between the split d-orbitals is different. [1]
• The magnitude of ΔE depends on the oxidation state of the metal ion; Fe³⁺ has a higher charge density than Fe²⁺, leading to stronger ligand-metal interactions and a larger ΔE. This results in absorption of different wavelengths of visible light, hence different observed colours. [1]

Section D: Data-Based and Extended Questions (15 marks)

Question 16

(a) [3 marks]

• Silicon has a giant covalent (macromolecular) structure. [1]
• Each silicon atom is covalently bonded to four other silicon atoms in a tetrahedral arrangement, forming an extended three-dimensional network. [1]
• Melting silicon requires breaking many strong covalent bonds throughout the structure, which requires a large amount of energy, hence the very high melting point. [1]

(b) [2 marks]

• Both aluminium and magnesium have giant metallic structures. [1]
• Aluminium has three delocalised electrons per atom (Al³⁺) while magnesium has two (Mg²⁺). The greater number of delocalised electrons in aluminium results in stronger metallic bonding and a higher melting point. Additionally, Al³⁺ ions are smaller and have a higher charge density than Mg²⁺, further strengthening the metallic bonds. [1]

(c) [2 marks]

• Phosphorus (P₄), sulfur (S₈), chlorine (Cl₂), and argon (Ar) all have simple molecular structures. [1]
• The intermolecular forces between their molecules are weak van der Waals' forces. Only these weak forces need to be overcome during melting, so the melting points are low. [1]

Question 17

(a) [3 marks]

• NaCl has a giant ionic lattice structure with strong electrostatic forces of attraction between Na⁺ and Cl⁻ ions throughout the lattice. A large amount of energy is required to overcome these forces, resulting in a high melting point. [1]
• AlCl₃ has a simple molecular structure in the solid state (it exists as Al₂Cl₆ dimers) with covalent bonding within the molecules. [1]
• The intermolecular forces between Al₂Cl₆ molecules are weak van der Waals' forces, which require relatively little energy to overcome, resulting in a low melting/sublimation point. [1]

(b) [1 mark]

• SiCl₄(l) + 2H₂O(l) → SiO₂(s) + 4HCl(aq) OR SiCl₄(l) + 4H₂O(l) → Si(OH)₄(s) + 4HCl(aq) [1]

(c) [2 marks]

• The solution is strongly acidic (low pH, e.g., pH 0-1). [1]
• PCl₅(s) + 4H₂O(l) → H₃PO₄(aq) + 5HCl(aq) Both phosphoric acid and hydrochloric acid are produced, giving a strongly acidic solution. [1]

Question 18

(a) [2 marks]

• Transition metals have partially filled d-orbitals. [1]
• The 4s and 3d electrons are relatively close in energy, so varying numbers of electrons can be lost from both the 4s and 3d subshells, resulting in variable oxidation states. [1]

(b) [2 marks]

• Highest oxidation state: +7. [1]
• Example: KMnO₄ (potassium manganate(VII)) or Mn₂O₇. [1]

Question 19

(a) [1 mark]

• 2Mg(NO₃)₂(s) → 2MgO(s) + 4NO₂(g) + O₂(g) [1]

(b) [2 marks]

• Thermal stability of Group 2 nitrates increases down the group. [1]
• Down the group, the cation size increases and charge density decreases, so the polarising power of the cation decreases. This causes less distortion of the nitrate ion, making it more difficult to decompose the nitrate. Hence, a higher temperature is required for decomposition. [1]

Question 20

(a) [3 marks]

• NaCl is an ionic compound that dissolves in water to give Na⁺ and Cl⁻ ions. Neither ion hydrolyses significantly, so the solution remains neutral (pH ≈ 7). [1]
• AlCl₃ is largely covalent in nature and hydrolyses in water. [1]
• The Al³⁺ ion (or AlCl₃) reacts with water: AlCl₃ + 6H₂O → [Al(H₂O)₆]³⁺ + 3Cl⁻, and the hydrated Al³⁺ ion undergoes hydrolysis: [Al(H₂O)₆]³⁺ + H₂O ⇌ [Al(H₂O)₅(OH)]²⁺ + H₃O⁺, producing H₃O⁺ ions and making the solution acidic. [1]

(b) [1 mark]

• PCl₅(s) + 4H₂O(l) → H₃PO₄(aq) + 5HCl(aq) [1]