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    Module 3 – Periodic table and energy — OCR A-Level Chemistry

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    Module 3 – Periodic table and energy explained

    Module 1 focuses on the development of practical skills in chemistry, which are fundamental to understanding the subject.

    Read the full explanation

    It covers planning, implementing, analysing, and evaluating experimental work, with skills assessed both through written examinations and a mandatory Practical Endorsement.

    What to demonstrate

    1. Experimental design including selection of suitable apparatus and techniques
    2. Identification of variables to be controlled
    3. Correct use of practical apparatus and techniques
    Show all 9 objectives
    1. Accurate recording of measurements with appropriate units
    2. Processing and analysis of qualitative and quantitative data
    3. Use of appropriate mathematical skills and significant figures
    4. Plotting and interpreting graphs including gradients and intercepts
    5. Evaluation of results, identification of anomalies, and limitations of procedures
    6. Calculation of percentage errors and uncertainties

    Module 3 – Periodic table and energy exam tips

    Quick Revision Summary (Key Takeaway)

    Module 3 – Periodic table and energy covers the physical and chemical properties of Period 3 elements, the reactions of their oxides and chlorides with water, and the underlying trends in ionisation energy, electronegativity, and structure. It also introduces key energy concepts including enthalpy changes, Hess's law, and calorimetry, which are essential for understanding reaction feasibility and bond energetics.

    Topic Overview

    Module 3 – Periodic table and energy is a core component of OCR A-Level Chemistry, bridging the gap between atomic structure and chemical reactivity. It begins by exploring the trends in physical properties across Period 3, such as atomic radius, ionisation energy, and electronegativity, and explains how these trends arise from increasing nuclear charge and shielding. This section also examines the structures and bonding of the elements, from metallic sodium to giant covalent silicon and simple molecular chlorine, and how these determine melting points and electrical conductivity.

    The module then shifts focus to the reactions of Period 3 oxides and chlorides with water, highlighting the acidic, basic, or amphoteric nature of these compounds. This links to the broader concepts of periodicity and acid-base behaviour, which are essential for understanding inorganic chemistry. The energy part of the module introduces thermodynamics, covering enthalpy changes, calorimetry, and Hess's law. These concepts are fundamental for calculating energy transfers in chemical reactions and for predicting whether reactions are exothermic or endothermic.

    Mastery of this module is crucial because it underpins many other topics in A-Level Chemistry, such as rates of reaction, equilibrium, and redox chemistry. The skills developed here—interpreting data, performing calculations, and constructing Hess cycles—are directly assessed in exams and are valuable for any future study in chemistry or related sciences.

    Key Concepts
    • →Trends across Period 3: atomic radius decreases, first ionisation energy generally increases, and electronegativity increases across the period due to increasing nuclear charge and constant shielding.
    • →Structure and bonding of Period 3 elements: Na, Mg, Al are metallic; Si is giant covalent; P₄, S₈, Cl₂ are simple molecular. These explain variations in melting points and conductivity.
    • →Reactions of Period 3 oxides with water: Na₂O and MgO are basic, Al₂O₃ is amphoteric, and SiO₂, P₄O₁₀, SO₂, SO₃, and Cl₂O are acidic. The pH of the resulting solutions reflects this.
    • →Reactions of Period 3 chlorides with water: NaCl and MgCl₂ dissolve to form neutral or slightly acidic solutions; AlCl₃ and SiCl₄ hydrolyse vigorously to form acidic solutions with fumes of HCl.
    • →Enthalpy changes: definitions of standard enthalpy of formation, combustion, neutralisation, and atomisation. Calorimetry experiments measure heat changes, and Hess's law allows calculation of enthalpy changes that are difficult to measure directly.
    Marking Points
    • Experimental design including selection of suitable apparatus and techniques
    • Identification of variables to be controlled
    • Correct use of practical apparatus and techniques
    • Accurate recording of measurements with appropriate units
    • Processing and analysis of qualitative and quantitative data
    • Use of appropriate mathematical skills and significant figures
    • Plotting and interpreting graphs including gradients and intercepts
    • Evaluation of results, identification of anomalies, and limitations of procedures
    • Calculation of percentage errors and uncertainties
    Examiner Tips
    • 💡Ensure all measurements are recorded with the correct SI units
    • 💡Always show working in calculations and state the final answer to the correct number of significant figures
    • 💡When evaluating experiments, focus on specific limitations of the procedure rather than generic errors
    • 💡Be prepared to suggest improvements to experimental designs to increase accuracy or precision
    • 💡Practice interpreting data from unfamiliar practical contexts
    • 💡When explaining trends, always mention the three factors: nuclear charge, shielding, and atomic radius. For ionisation energy, also consider subshell stability.
    • 💡For reactions of oxides and chlorides, write balanced equations and state the pH of the resulting solution. Use observations like 'fizzes' or 'steamy fumes' to secure marks.
    • 💡In Hess's law calculations, draw a clear cycle and label arrows. Check the sign of ΔH: if you reverse a reaction, change the sign. Always include units in your final answer.
    Common Mistakes
    • Failure to use appropriate significant figures in calculations
    • Incorrect selection of apparatus for specific experimental techniques
    • Inability to identify and control all relevant variables
    • Poor evaluation of experimental limitations or sources of error
    • Incorrect labelling of graph axes or failure to use appropriate scales
    • Misconception: First ionisation energy increases smoothly across Period 3. Correction: There are drops at Al and S due to the change in subshell (3p vs 3s) and electron-electron repulsion in paired p orbitals.
    • Misconception: All Period 3 oxides are acidic. Correction: Na₂O and MgO are basic, Al₂O₃ is amphoteric, and only the non-metal oxides are acidic.
    • Misconception: In calorimetry, the temperature change in °C is not the same as in K. Correction: A change of 1°C equals a change of 1 K, so ΔT is numerically identical, but absolute temperatures must be in K for gas calculations.
    Revision Plan
    1. 1Week 1: Focus on Periodicity. Review electron configurations of Period 3 elements. Learn the trends in atomic radius, ionisation energy, and electronegativity. Practice explaining the dips in ionisation energy.
    2. 2Week 1: Study the structures and melting points of Period 3 elements. Create a table summarising bonding, structure, and properties. Use diagrams to visualise giant covalent and simple molecular structures.
    3. 3Week 2: Learn the reactions of oxides and chlorides with water. Write balanced equations and note the pH of solutions. Use flashcards for the acidic/basic/amphoteric classification.
    4. 4Week 2: Move to energy. Define all enthalpy changes and practise calorimetry calculations. Then master Hess's law with past paper questions. Finally, attempt mixed questions that combine periodicity and energy.
    Exam Question Types
    • 📋Multiple choice questions testing trends in ionisation energy or melting points across Period 3. Advice: Eliminate options that contradict the general trend, but watch for exceptions.
    • 📋Short-answer questions asking to explain a trend, e.g., 'Explain why the first ionisation energy of aluminium is lower than that of magnesium.' Advice: Use the mark scheme structure: state the factor (subshell), then explain the effect.
    • 📋Calculation questions on enthalpy changes using calorimetry data or Hess's law. Advice: Show all working, include units, and check the sign of ΔH.
    • 📋6-mark extended response questions on the reactions of Period 3 oxides with water, requiring balanced equations and explanations of acidity. Advice: Plan your answer, use correct terminology, and link structure to properties.
    Command Word Expectations (OCR)
    Explain

    Give a reason or cause for a trend or observation. In OCR A-Level, you must link the observation to underlying principles (e.g., nuclear charge, shielding, structure). For example, 'Explain the trend in first ionisation energy across Period 3' requires you to state the increase in nuclear charge and constant shielding, leading to a stronger attraction for outer electrons.

    Calculate

    Perform a numerical calculation, showing all steps and units. For enthalpy changes, you must use the correct formula (q=mcΔT) and convert to kJ mol⁻¹. Include the sign of ΔH. Marks are awarded for method, correct substitution, and final answer with units.

    State

    Give a brief, precise answer without explanation. For example, 'State the pH of a solution of sodium oxide' – answer: 'pH 13-14' or 'strongly alkaline'. No reasoning is required.

    How Students Lose Marks (Examiner Pitfalls)
    Pitfall: Students often confuse the trend in melting points across Period 3 by ignoring the different bonding types (metallic, giant covalent, simple molecular) and instead try to explain it solely by increasing nuclear charge.
    ❌ Weak Answer (Loses Marks):Melting point increases across Period 3 because the nuclear charge increases and more energy is needed to overcome the stronger attraction.
    Example improved answer:Across Period 3, melting points vary due to changes in structure and bonding. Na, Mg, Al have metallic bonding; the strength increases with the number of delocalised electrons per atom and the charge on the ion, so melting point increases from Na to Al. Silicon has a giant covalent structure with strong covalent bonds, giving a very high melting point. P4, S8, Cl2 are simple molecular with weak van der Waals forces, so melting points are low; sulfur has a higher melting point than phosphorus because S8 molecules are larger, leading to stronger van der Waals forces.
    Examiner Tip: Always link melting point to the type of structure and bonding, not just nuclear charge. Mention the specific particles (atoms, ions, molecules) and the forces between them.
    Pitfall: In calorimetry questions, students often forget to convert temperature change from °C to K (though the change is the same) or they use the mass of the solution incorrectly, ignoring the mass of any added solid or the density of the solution.
    ❌ Weak Answer (Loses Marks):q = mcΔT = 50 × 4.18 × 20 = 4180 J, then divide by moles of water.
    Example improved answer:First, calculate the heat change: q = mcΔT, where m is the mass of the solution (assume 1.00 g cm⁻³, so 50.0 g for 50.0 cm³), c = 4.18 J g⁻¹ K⁻¹, and ΔT = 20.0 K (since a change of 20°C equals 20 K). Thus q = 50.0 × 4.18 × 20.0 = 4180 J = 4.18 kJ. Then determine the moles of the limiting reactant (e.g., if using 0.0500 mol of acid, that is the limiting reactant). Finally, enthalpy change = -q/moles = -4.18 kJ / 0.0500 mol = -83.6 kJ mol⁻¹. The sign is negative because the temperature increased (exothermic).
    Examiner Tip: Always state the sign of the enthalpy change. Use the mass of the solution, not the mass of the solid reactant added. Convert kJ/mol correctly and quote units.
    Step-by-Step Worked Solutions

    Question: A student added 4.00 g of anhydrous copper(II) sulfate to 50.0 cm³ of water in a polystyrene cup. The temperature rose from 20.0°C to 26.5°C. Calculate the enthalpy change of solution of anhydrous copper(II) sulfate, in kJ mol⁻¹. Assume the specific heat capacity of the solution is 4.18 J g⁻¹ K⁻¹ and the density of the solution is 1.00 g cm⁻³.

    1. 1.Step 1: Calculate the heat absorbed by the solution: q = mcΔT. Mass of solution = 50.0 g (since density = 1.00 g cm⁻³ and volume = 50.0 cm³). ΔT = 26.5 - 20.0 = 6.5 K. So q = 50.0 × 4.18 × 6.5 = 1358.5 J = 1.3585 kJ.
    2. 2.Step 2: Calculate moles of CuSO₄: Molar mass of CuSO₄ = 63.5 + 32.1 + 4×16.0 = 159.6 g mol⁻¹. Moles = 4.00 / 159.6 = 0.02506 mol.
    3. 3.Step 3: Calculate enthalpy change per mole: ΔH = -q / moles = -1.3585 kJ / 0.02506 mol = -54.2 kJ mol⁻¹. The sign is negative because the temperature increased (exothermic process).
    Final Answer: ΔH = -54.2 kJ mol⁻¹ (exothermic)

    Question: Using Hess's law, calculate the standard enthalpy change of formation of propane (C₃H₈) given the following standard enthalpy changes of combustion: ΔHc°(C(s)) = -393.5 kJ mol⁻¹, ΔHc°(H₂(g)) = -285.8 kJ mol⁻¹, ΔHc°(C₃H₈(g)) = -2219.9 kJ mol⁻¹.

    1. 1.Step 1: Write the formation equation: 3C(s) + 4H₂(g) → C₃H₈(g).
    2. 2.Step 2: Construct a Hess cycle: The formation reaction can be achieved by combusting the elements to CO₂ and H₂O, then reversing the combustion of propane. The enthalpy change is: ΔHf° = [3 × ΔHc°(C) + 4 × ΔHc°(H₂)] - ΔHc°(C₃H₈).
    3. 3.Step 3: Substitute values: ΔHf° = [3 × (-393.5) + 4 × (-285.8)] - (-2219.9) = [-1180.5 - 1143.2] + 2219.9 = -2323.7 + 2219.9 = -103.8 kJ mol⁻¹.
    Final Answer: ΔHf°(C₃H₈) = -103.8 kJ mol⁻¹
    Active Recall Memory Test
    What is the trend in atomic radius across Period 3 and why?
    Key Fact: Atomic radius decreases across Period 3 because the nuclear charge increases, pulling the outer electrons closer to the nucleus, while shielding remains constant.
    Why is the first ionisation energy of sulphur lower than that of phosphorus?
    Key Fact: In sulphur, the electron removed is from a doubly occupied 3p orbital, and the electron-electron repulsion makes it easier to remove, so the ionisation energy is lower.
    What is the pH of a solution formed when aluminium oxide reacts with water?
    Key Fact: Aluminium oxide is amphoteric, so it does not react with water to form a strongly acidic or basic solution; it is insoluble, so the pH remains around 7.
    State Hess's law in your own words.
    Key Fact: The enthalpy change for a reaction is independent of the route taken, as long as the initial and final conditions are the same.
    Frequently Asked Questions
    Why does the melting point of silicon is so high compared to other Period 3 elements?
    Silicon has a giant covalent structure, where each silicon atom is bonded to four others by strong covalent bonds. These bonds require a large amount of energy to break, resulting in a very high melting point (about 1414°C). In contrast, metals like sodium have weaker metallic bonds, and simple molecular substances like chlorine have weak van der Waals forces between molecules, so they melt at much lower temperatures.
    How do I remember which oxides are acidic, basic, or amphoteric in Period 3?
    A useful rule is that metal oxides tend to be basic, non-metal oxides are acidic, and metalloid oxides (like aluminium) are amphoteric. Specifically, Na₂O and MgO are basic, Al₂O₃ is amphoteric, and SiO₂, P₄O₁₀, SO₂, SO₃, and Cl₂O are acidic. You can also remember that the pH of the solution formed with water reflects this: basic oxides give pH > 7, acidic oxides give pH < 7, and amphoteric oxides are insoluble so pH stays neutral.
    What is the difference between enthalpy change of formation and enthalpy change of combustion?
    The standard enthalpy change of formation (ΔHf°) is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions. The standard enthalpy change of combustion (ΔHc°) is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions. For example, the formation of water is H₂(g) + ½O₂(g) → H₂O(l), while the combustion of hydrogen is the same equation, but for a compound like methane, combustion is CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).
    Why do we use Hess's law in chemistry?
    Hess's law is useful because many enthalpy changes, such as the enthalpy of formation of a compound directly from its elements, are difficult or impossible to measure experimentally. By using known enthalpy changes of combustion or other reactions, we can calculate the desired enthalpy change indirectly. This is based on the principle that enthalpy is a state function, meaning it only depends on the initial and final states, not the route taken.
    How do I calculate the enthalpy change of neutralisation from a calorimetry experiment?
    First, calculate the heat released or absorbed using q = mcΔT, where m is the mass of the solution (usually the total volume of acid and alkali, assuming density 1 g/cm³), c is the specific heat capacity (4.18 J g⁻¹ K⁻¹), and ΔT is the temperature change. Then, determine the number of moles of water formed (which is limited by the moles of acid or alkali). Finally, divide the heat change by the moles of water and convert to kJ mol⁻¹. The sign is negative for exothermic reactions.
    What are the key things to include in a 6-mark answer on the reactions of Period 3 oxides with water?
    You should include balanced chemical equations for each oxide you discuss, state whether the oxide is acidic, basic, or amphoteric, and give the pH of the resulting solution. For example, for sodium oxide: Na₂O + H₂O → 2NaOH, solution is strongly alkaline with pH 13-14. For phosphorus(V) oxide: P₄O₁₀ + 6H₂O → 4H₃PO₄, solution is strongly acidic with pH 1-2. Also, mention any observations such as fizzing or dissolving. Make sure to use correct formulas and state symbols.