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Marking Guidance
- Alternative responses are indicated by a slash (/).
- Underlined terms are key marking points that must be included.
- Brackets () indicate words that are not required for marks.
- Error Carried Forward (ECF) allows credit for correct calculation steps using an incorrect previous answer.
Compare the plum pudding model and the nuclear model.
M1 Similarities: Both models contain negatively charged electrons / both have an overall neutral charge (in their standard states) [1 mark].
M2 Differences (Plum Pudding): The plum pudding model describes the atom as a single ball of positive charge with electrons embedded inside it [1 mark].
M3 Differences (Nuclear Model): The nuclear model states that the positive charge is concentrated in a tiny central nucleus [1 mark].
M4 Differences (Nuclear Model Structure): The nuclear model states that the atom is mostly empty space, with electrons orbiting the nucleus [1 mark].
Reject
Do not award M3 if they state neutrons are in the nucleus in the plum pudding model, or if they claim the plum pudding model has a nucleus.
(a) Electronic configurations of Mg and Ca, and Group 2 placement.
M1 Magnesium electronic configuration: 2,8,2 [1 mark].
M2 Calcium electronic configuration: 2,8,8,2 [1 mark].
M3 Group explanation: Both elements have 2 electrons in their outer shell [1 mark].
M4 Periodic table rule: The group number in the periodic table corresponds to the number of electrons in the outer shell [1 mark].
M5 Subatomic particles: Magnesium has 12 protons/electrons, calcium has 20 protons/electrons [1 mark].
M6 Shell structure: Magnesium has outer electrons in the 3rd shell (3 energy levels), calcium has outer electrons in the 4th shell (4 energy levels) [1 mark].
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(b) Group 7 reactivity down the group.
M1 Reactivity of Group 7 elements decreases because they need to gain one electron to form a full outer shell / form a 1- ion [1 mark].
M2 Down the group, the atoms get larger / have more electron shells [1 mark].
M3 The outer shell is further from the nucleus [1 mark].
M4 There is more shielding by inner electron shells [1 mark].
M5 There is a weaker electrostatic attraction between the nucleus and the incoming electron [1 mark].
M6 Therefore, it is harder to attract/gain an electron down the group [1 mark].
Compare structure/bonding and explain melting points and conductivity.
M1 Both have a giant covalent lattice / macromolecular structure [1 mark].
M2 In diamond, each carbon atom is bonded to 4 other carbon atoms; in silicon dioxide, each silicon is bonded to 4 oxygen atoms and each oxygen to 2 silicon atoms [1 mark].
M3 Both contain strong covalent bonds [1 mark].
M4 High melting point because a large amount of energy is required to break these strong covalent bonds [1 mark].
M5 Do not conduct electricity because all outer electrons are shared / localise in covalent bonds [1 mark].
M6 Thus, there are no delocalised electrons / no free ions to carry charge [1 mark].
(a) Dot-and-cross diagram of H2O and electrostatic attractions.
M1 Diagram: Two hydrogen circles overlapping with one oxygen circle, showing 1 pair of shared electrons (one dot, one cross) in each overlap area [1 mark].
M2 Diagram: Oxygen shown with 2 lone pairs (4 non-bonding electrons) [1 mark].
M3 Diagram: Hydrogen atoms have no other electrons besides the shared pair [1 mark].
M4 Electrostatic attraction: Covalent bonds are formed by the electrostatic attraction between the shared pair of electrons and the positive nuclei of the hydrogen and oxygen atoms [1 mark].
(b) Low boiling point of water vs MgO.
M1 Water is simple molecular, so it only has weak intermolecular forces between molecules which require little energy to overcome [1 mark].
M2 MgO has a giant ionic lattice with strong electrostatic forces of attraction between oppositely charged ions, requiring much more energy to break [1 mark].
Reject
Do not award M1 if they state covalent bonds are broken when water boils.
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Evaluate student's claims about A, B, and C.
M1 Substance A: Claim is incorrect. It is metallic because it conducts electricity in both solid and liquid states, and has a high melting point [1 mark].
M2 Reason: Giant covalent structures do not conduct electricity in the solid state (except graphite) [1 mark].
M3 Substance B: Claim is incorrect. It is giant covalent because it has a very high melting/boiling point, is insoluble in water, and does not conduct electricity in either solid or liquid states [1 mark].
M4 Reason: Simple molecular structures have low melting and boiling points / weak intermolecular forces [1 mark].
M5 Substance C: Claim is correct. It is giant ionic because it conducts electricity when liquid/molten but not as a solid, and has a high melting point [1 mark].
M6 Reason: Ionic compounds are soluble in water, and ions are free to move and carry charge only in liquid/solution form [1 mark].
M1 Metallic bonding is the electrostatic attraction between a lattice of positive metal ions and a sea of delocalised electrons [1 mark].
M2 Conducts electricity because the delocalised electrons are free to move throughout the structure and carry charge [1 mark].
M3 Malleable because the metal ions are arranged in regular layers [1 mark].
M4 These layers of ions can slide over each other when a force is applied, without breaking the metallic bond [1 mark].
(a) Separation in terms of mobile and stationary phases.
M1 The mobile phase (solvent) moves up the paper, carrying the dissolved dyes with it [1 mark].
M2 The stationary phase (paper) does not move [1 mark].
M3 Dyes separate because different components have different solubilities in the mobile phase / different attractions to the stationary phase [1 mark].
(b) Pencil start line.
M1 Pencil is insoluble in the solvent / will not dissolve and run up the paper (unlike ink, which would separate and interfere with the results) [1 mark].
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(c) Rf calculations.
M1 Rf formula: distance moved by substance / distance moved by solvent [1 mark].
M2 Dye A working: 2.4 cm / 8.0 cm = 0.30 (allow 0.3) [1 mark].
M3 Dye B working: 6.4 cm / 8.0 cm = 0.80 (allow 0.8) [1 mark].
M4 Both answers given with no units (Rf is a ratio) [1 mark].
M1 Calculate moles of lithium: 3.50 g / 7.0 = 0.50 mol [1 mark].
M2 Calculate moles of water: 18.0 g / 18.0 = 1.00 mol [1 mark].
M3 Limiting reactant proof: The molar ratio from the equation is 2 Li : 2 H2O (1:1). 0.50 mol of Li requires 0.50 mol of H2O. Since 1.00 mol of H2O is present, H2O is in excess and lithium is the limiting reactant [1 mark].
M4 Moles of LiOH produced = Moles of Li reacted = 0.50 mol [1 mark].
M5 Mass of LiOH: 0.50 mol * 24.0 g/mol (Mr of LiOH) = 12.0 g [1 mark].
Error Carried Forward (ECF)
Allow ECF from M1/M2/M3 to calculate final mass in M5. For example, if they identify H2O as limiting (1.00 mol), theoretical mass is 1.00 * 24.0 = 24.0 g (score max 3 marks: M2, M4-equivalent, M5-equivalent).
M1 Percentage yield formula: (actual yield / theoretical yield) * 100 [1 mark].
M2 Calculation: (9.00 g / 12.0 g) * 100 = 75.0% (allow 75%) [1 mark].
Error Carried Forward (ECF)
Allow ECF from Question 8 theoretical yield. If theoretical yield was calculated as 24.0 g: (9.00 / 24.0) * 100 = 37.5%.
M1 From equation: 2 moles of Li produce 1 mole of H2 (2:1 ratio) [1 mark].
M2 Moles of H2 produced: 0.50 mol Li / 2 = 0.25 mol [1 mark].
M3 Volume calculation: 0.25 mol * 24.0 dm3/mol [1 mark].
M4 Final volume = 6.0 dm3 (allow 6.00 / 6) [1 mark].
Error Carried Forward (ECF)
Allow ECF from Question 8 moles of lithium.
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M1 Calculate moles of HNO3: (25.0 / 1000) * 0.050 = 0.00125 mol [1 mark].
M2 From equation: 2 moles of HNO3 react with 1 mole of Ba(OH)2. Moles of Ba(OH)2 = 0.00125 / 2 = 0.000625 mol [1 mark].
M3 Calculate concentration of Ba(OH)2 in mol/dm3: 0.000625 mol / (20.0 / 1000) = 0.03125 mol/dm3 [1 mark].
M4 Calculate Mr of Ba(OH)2: 137.0 + 2 * (16.0 + 1.0) = 171.0 [1 mark].
M5 Calculate concentration of Ba(OH)2 in g/dm3: 0.03125 * 171.0 = 5.34375 g/dm3 [1 mark].
M6 Round both values to 3 significant figures: concentration = 0.0313 mol/dm3 and 5.34 g/dm3 [1 mark].
Error Carried Forward (ECF)
Allow full ECF throughout. If M2 ratio is missed (1:1 used), concentration is 0.0625 mol/dm3 (M3) and 10.7 g/dm3 (M5/M6), scoring max 5 marks.
(a) Four chemical ions in solution.
M1 Potassium ions (K+) and Bromide ions (Br-) [1 mark].
M2 Hydrogen ions (H+) and Hydroxide ions (OH-) [1 mark].
(b) Positive electrode (anode) product and half-equation.
M1 Product is bromine (gas/liquid) [1 mark].
M2 Half-equation: 2Br- → Br2 + 2e- / 2Br- - 2e- → Br2 [1 mark].
M3 Explanation: Both Br- and OH- ions are attracted to the anode [1 mark].
M4 Halide ions (Br-) are discharged preferentially to hydroxide ions (OH-) / halide ions are easier to oxidise than OH- [1 mark].
(c) Negative electrode (cathode) product and half-equation.
M1 Product is hydrogen (gas) [1 mark].
M2 Half-equation: 2H+ + 2e- → H2 [1 mark].
M3 Explanation: Both K+ and H+ ions are attracted to the cathode [1 mark].
M4 Hydrogen is less reactive than potassium, so hydrogen ions are discharged preferentially / reduced preferentially [1 mark].
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M1 State that pH is related to hydrogen ion concentration by the base-10 relationship: [H+] = 10^-pH [1 mark].
M2 Calculate [H+] at pH 1.0 = 10^-1 = 0.1 mol/dm3 [1 mark].
M3 Calculate [H+] at pH 3.0 = 10^-3 = 0.001 mol/dm3 [1 mark].
M4 Ratio of concentrations: 0.1 / 0.001 = 100. Therefore, as pH increases from 1.0 to 3.0, the [H+] decreases by a factor of 100 / 10^2 [1 mark].
(a) Polystyrene cup vs glass beaker.
M1 Polystyrene is a good thermal insulator [1 mark].
M2 It reduces heat loss to the surroundings, making the recorded maximum temperature rise more accurate [1 mark].
(b) Ionic equation.
M1 Correct formula and species: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) [1 mark].
M2 All state symbols correct: (s), (aq), (aq), (s) [1 mark].
(c) Temperature rise and leveling off.
M1 Temperature increases because the reaction is exothermic, releasing thermal energy to the solution [1 mark].
M2 As more zinc is added, more copper ions react, releasing more energy [1 mark].
M3 The temperature stops rising because copper(II) sulfate is the limiting reactant and has completely reacted. Any excess zinc does not react, and heat is lost to the surroundings [1 mark].
(d) Zinc oxidation half-equation.
M1 Zinc atoms lose electrons: Zn → Zn2+ + 2e- [1 mark].
M2 Identify this process as oxidation (loss of electrons) [1 mark].
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(a) Endothermic reaction definition.
M1 A reaction that takes in heat/energy from the surroundings, leading to a decrease in temperature of the surroundings [1 mark].
(b) Identify labels W, X, Y, Z.
M1 W: Reactants / Calcium carbonate / CaCO3(s) [1 mark].
M2 X: Products / Calcium oxide and carbon dioxide / CaO(s) + CO2(g) [1 mark].
M3 Y: Activation energy / Ea [1 mark].
M4 Z: Overall energy change / Enthalpy change / ΔH [1 mark].
(c) Endothermic reaction in terms of bond breaking and bond making.
M1 Energy is taken in to break chemical bonds in the reactants (CaCO3) [1 mark].
M2 Energy is released when new bonds are made in the products (CaO + CO2) [1 mark].
M3 The reaction is endothermic because more energy is absorbed to break bonds than is released when making them [1 mark].
(a) Calculate overall enthalpy change.
M1 Energy absorbed to break reactant bonds: H-H + Br-Br = 436 + 193 = 629 kJ/mol [1 mark].
M2 Energy released to make product bonds: 2 * H-Br = 2 * 366 = 732 kJ/mol [1 mark].
M3 Enthalpy change (ΔH) = 629 - 732 = -103 kJ/mol [1 mark].
Error Carried Forward (ECF)
Allow ECF from M1 and M2. If they calculate 629 - 366 = +263 kJ/mol, award 1 mark. If they calculate 732 - 629 = +103 kJ/mol, award 2 marks (missing negative sign).
(b) Exothermic explanation.
M1 Bond breaking absorbs energy (endothermic) and bond making releases energy (exothermic) [1 mark].
M2 The energy released when making 2 H-Br bonds (732 kJ/mol) is greater than the energy absorbed to break H-H and Br-Br bonds (629 kJ/mol) [1 mark].
M3 Therefore, there is a net release of energy to the surroundings / negative enthalpy change, making it exothermic [1 mark].
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This resource is an independent educational tool created to support student revision. It is completely independent and is not endorsed by, affiliated with, or sponsored by any official examination board. All trademarked terms are used under Nominative Fair Use purely for descriptive compatibility indexing. Licensed for individual personal use only.
Chemistry Made Easy is an independent resource. Not affiliated with or endorsed by AQA, Pearson Edexcel, or the IBO.