OCR A Level Chemistry A: Periodicity (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

Periodicity, section 3.1.1 of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, is the focus here. The questions test all of outcomes 3.1.1(a) to (g) in Module 3 of this full A level. The last two mix several outcomes.

Revise first with the Periodicity study guide and the Periodicity revision notes. Find other Module 3 topics on the OCR chemistry hub, tick outcomes you are confident with on the printable checklist, and use the free diagnostics to test the rest of the course.

All data needed is given; use it rather than values you remember.

Questions

1. This question is about how the periodic table is organised.

(a) State what is meant by the term periodicity. [1] (b) Lithium and sodium react with water in a similar way. Explain this in terms of their electrons. [1] (c) Mendeleev published his table in 1869, before many elements were known. Explain how his approach led to the Periodic Law being accepted by other scientists. [2]

2. The proton numbers of three elements are 55, 52 and 25.

(a) For each element, state the block of the periodic table it is in. [3] (b) State the group and period of the element with proton number 52. [1]

3.

(a) Define the term first ionisation energy. [2] (b) Write an equation, with state symbols, for the third ionisation energy of phosphorus. [1]

4. The first ionisation energies of three Group 16 elements are: oxygen 1314, sulfur 1000 and selenium 941 kJ mol⁻¹. Describe and explain the trend shown. [3]

5. Magnesium has a higher first ionisation energy than aluminium, and phosphorus has a higher first ionisation energy than sulfur.

(a) A student says: “Aluminium has one more proton than magnesium, so its first ionisation energy must be higher.” Evaluate this statement. [3] (b) Account for the fall in first ionisation energy from phosphorus to sulfur. [2]

6. Element Q is in Period 3. Its first six ionisation energies, in kJ mol⁻¹, are 1012, 1907, 2914, 4964, 6274 and 21267.

(a) Explain why the 2nd ionisation energy of Q is larger than the 1st. [1] (b) Use the data to find the group of Q, showing your reasoning. [2] (c) Identify Q. [1]

7. A graph of log₁₀(ionisation energy) against the number of electrons removed is plotted for all twelve electrons of magnesium. There is a large rise between the 2nd and 3rd electrons removed, and another between the 10th and 11th.

(a) State the number of electrons in each shell of a magnesium atom. [1] (b) Explain the large rise between the 2nd and 3rd electrons removed. [2] (c) Suggest how graphs like this supported the Bohr model of the atom. [1]

8. Rubidium melts at 39 °C and strontium at 777 °C.

(a) Describe the bonding and structure in strontium. [2] (b) Explain why strontium has the higher melting point. [3] (c) Explain why solid strontium conducts electricity. [1]

9. Diamond, graphite and graphene are all forms of carbon.

(a) Graphite is used to make electrodes; diamond could not be. Explain this difference in terms of bonding. [3] (b) Explain why graphene is very strong, and suggest one possible application of graphene. [2] (c) Explain why diamond does not dissolve in water. [1]

10. The melting points of the Period 3 elements, in °C, are: Na 98, Mg 650, Al 660, Si 1414, P 44, S 115, Cl −102, Ar −189.

Describe and explain the variation in melting point across Period 3 in terms of structure and bonding. [8]

11. D, E and F are elements in Period 3. Their successive ionisation energies, in kJ mol⁻¹, are given below.

Element	Successive ionisation energies
D	496, 4562, 6910
E	787, 1577, 3232, 4356, 16091
F	1251, 2298, 3822, 5159, 6542, 9362, 11018, 33604

(a) Deduce the group of each element. Show how you used the data. [3] (b) Identify D, E and F. [1] (c) Predict the order of melting points of D, E and F, from highest to lowest. Explain your order. [3] (d) State which of D, E and F conducts electricity as a solid, and give the reason. [1]

Answers

1. (a) A repeating pattern (trend) in physical and chemical properties across each period [1] (b) Both have one electron in an outer s sub-shell (2s¹ and 3s¹), the same outer arrangement [1] (c) He left spaces in his table for elements not yet found and forecast what they would be like [1]; when elements such as germanium were found with the predicted properties, this supported the law [1] Examiner insight: In (c), a description of how the table looks earns little; the credit is for the predictions and their later confirmation by experiment.

2. (a) 55: [Xe] 6s¹, s-block [1]; 52: [Kr] 4d¹⁰ 5s² 5p⁴, p-block [1]; 25: [Ar] 3d⁵ 4s², d-block [1] (b) Group 16, Period 5 [1] Examiner insight: Write the configuration first; the block follows from the sub-shell the last electron enters, so a correct configuration makes the answer easy to justify.

3. (a) The energy required for 1 mol of atoms in the gas phase to lose 1 mol of electrons [1], producing 1 mol of 1+ ions, also gaseous [1] (b) P²⁺(g) → P³⁺(g) + e⁻ [1] Examiner insight: Missing “gaseous” or the mole quantities loses credit in (a); in (b) the left-hand species must already be the 2+ ion, with (g) on both ions.

4. First ionisation energy decreases down the group [1]. Each element has an extra shell, so its outer electron sits further out [1] and more inner electrons shield it; these effects outweigh the extra protons, so less energy is required [1] Examiner insight: Mention nuclear charge and explain why it is outweighed; an answer that only lists radius and shielding leaves the extra protons unaccounted for.

5. (a) The statement is wrong: magnesium’s first ionisation energy is higher than aluminium’s [1]. Aluminium’s electron is removed from the 3p sub-shell, which is higher in energy than magnesium’s 3s [1], so less energy is needed even though aluminium has an extra proton [1] (b) Sulfur’s electron is removed from a 3p orbital that holds a pair of electrons [1]; repulsion between the paired electrons makes it easier to remove [1] Examiner insight: With “evaluate”, say plainly whether the claim is right before giving reasons; also keep the two reasons separate, as sub-shell energy explains (a) and pair repulsion explains (b).

6. (a) The 2nd electron is removed from a Q⁺ ion, which attracts its electrons more strongly than the neutral atom does [1] (b) The largest jump is between the 5th and 6th values (21267 ÷ 6274 = 3.39) [1]; so Q has five outer-shell electrons and is in Group 15 [1] (c) Phosphorus [1] Examiner insight: Quote the two values either side of the jump; stating “Group 15” with no reference to the data does not show how you used it.

7. (a) 2, 8, 2 [1] (b) The 3rd electron is taken from the second shell, nearer the nucleus [1]; fewer inner electrons shield it, so the attraction is far stronger [1] (c) The jumps show electrons are arranged in separate shells (energy levels), as the Bohr model proposed [1] Examiner insight: In (b), say which shell the electron comes from; “it is harder to remove” only restates the data.

8. (a) Giant metallic lattice of Sr²⁺ cations [1] held by strong electrostatic attraction to delocalised electrons [1] (b) Strontium forms 2+ ions and rubidium 1+ ions; Sr²⁺ is also smaller [1]. Strontium gives two delocalised electrons per atom, rubidium one [1]. So the attraction between cations and delocalised electrons is stronger in strontium and needs more energy to overcome [1] (c) Delocalised electrons can move through the lattice and carry charge [1] Examiner insight: Use “cations” and “delocalised electrons” in every metallic bonding answer; “atoms” or “nuclei” in place of cations does not describe the bonding.

9. (a) In graphite each carbon is bonded to three others [1], so the fourth outer electron is delocalised and can move along the layers [1]. In diamond all four outer electrons are in covalent bonds, so there are no mobile charged particles [1] (b) Every atom is held by strong covalent bonds across the whole sheet [1]; one use, such as flexible electronics, sensors or strong lightweight composites [1] (c) The covalent bonds are too strong for water to break, so the lattice does not break up [1] Examiner insight: For conductivity, name the mobile charged particle (delocalised electrons); “free ions” in graphite is wrong chemistry.

10. Sodium, magnesium and aluminium have giant metallic lattices [1]. Melting overcomes the attraction between cations and delocalised electrons [1]. From Na⁺ to Mg²⁺ to Al³⁺ the charge increases, the ions are smaller and more electrons are delocalised, so the bonding is stronger and the melting point rises [1]. Silicon has a giant covalent lattice [1]; melting requires many strong covalent bonds throughout the lattice to break, which gives the highest value in the period [1]. Phosphorus (P₄), sulfur (S₈) and chlorine (Cl₂) are simple molecular, and argon exists as single atoms [1]. Melting overcomes only weak London forces between molecules, not the covalent bonds, so melting points are low [1]. S₈ has 128 electrons, P₄ 60, Cl₂ 34 and Ar 18; more electrons give stronger London forces, so the order is S > P > Cl > Ar [1] Examiner insight: In an extended response, link each structure to what is overcome and how strong it is; a list of structures with no explanation of the order of values gains only part of the credit.

11. (a) D: largest jump after the 1st value (4562 ÷ 496 = 9.20), so Group 1 [1]. E: after the 4th (16091 ÷ 4356 = 3.69), so Group 14 [1]. F: after the 7th (33604 ÷ 11018 = 3.05), so Group 17 [1] (b) D is sodium, E is silicon and F is chlorine [1] (c) E > D > F. E (silicon) is giant covalent, so many strong covalent bonds must be broken [1]. D (sodium) is giant metallic, with attraction between Na⁺ and delocalised electrons that is strong but weaker than silicon’s network of bonds [1]. F (chlorine) is simple molecular, Cl₂, with only weak London forces between molecules [1] (d) D, because it has delocalised electrons that can move [1] Examiner insight: In (c), give the order and also the structure of each element; a correct order with no reasons, or reasons with no order, earns limited credit.

Where marks are usually lost

• Leaving out “gaseous” or the mole amounts when defining first ionisation energy.
• Writing successive ionisation equations that start from the wrong ion.
• Swapping the two explanations for the small falls in first ionisation energy across a period.
• Ignoring nuclear charge when explaining the decrease down a group.
• Counting electrons after the largest jump in successive ionisation energies.
• Describing metallic bonding with “atoms” instead of cations and delocalised electrons.
• Saying covalent bonds break when a simple molecular element melts.
• Saying graphite conducts because of ions, or that diamond conducts.
• Ranking simple molecular elements without referring to the number of electrons and London forces.

Next steps

• Recap the tables and methods: Periodicity revision notes
• Full explanations and worked examples: Periodicity study guide
• Check the whole course: all free 10-minute diagnostics
• The background to orbitals and configurations: Electron structure
• Simple molecular lattices and London forces: Bonding and structure
• More OCR chemistry topics: course hub
• Printable checklist: OCR A Level Chemistry checklist
• Book a free trial class: /trial/

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 3.1.1 Periodicity, learning outcomes (a) to (g). Published by OCR.