OxfordAQA A-Level Chemistry: Periodicity (9620) – 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.

This set tests section 3.2.1 Periodicity of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: 3.2.1.1 (Classification) together with 3.2.1.2 (the Physical properties of Period 3 elements). The whole section is International AS content; none of it is International A-level only. Unit 1 (Inorganic 1 and Physical 1) is where the specification examines it.

Before you start, work through the study guide and the revision notes. The questions use rounded values from standard data tables (phosphorus as white P₄, sulfur as S₈). If you would rather see where you stand before you begin, a free 10-minute diagnostic will tell you.

Questions

1. State the block of the Periodic Table that contains each element.

(a) the element with proton number 9 [1] (b) the element with proton number 23 [1] (c) the element with proton number 19 [1]

2. Uranium is placed in the actinide row of the Periodic Table.

(a) State the block that contains uranium. [1] (b) Explain why you can classify uranium without writing out its electron configuration. [1]

3. Magnesium and aluminium are next to each other in Period 3.

(a) Explain why aluminium is a p-block element but magnesium is an s-block element. [2] (b) Use your answer to (a) to explain why magnesium’s first ionisation energy is higher than aluminium’s. [2]

4. The first ionisation energies of the Period 3 elements are shown below.

Element	Na	Mg	Al	Si	P	S	Cl	Ar
First ionisation energy / kJ mol⁻¹	496	738	578	787	1012	1000	1251	1521

(a) Give the equation, including state symbols, that represents the first ionisation of phosphorus. [1] (b) Name the two elements that break the rising pattern in the table. [1] (c) Explain the general increase in first ionisation energy from sodium to argon. [3] (d) Explain why phosphorus has a slightly higher first ionisation energy than sulfur. [2]

5. Arrange the atoms of sodium, aluminium, phosphorus and chlorine in order of increasing atomic radius. Explain the order. [4]

6. Five Period 3 elements are labelled J, K, L, M and N. They are magnesium, silicon, phosphorus, chlorine and argon, in no particular order. Their melting points are:

Element	J	K	L	M	N
Melting point / K	1687	172	923	84	317

(a) Identify J, L and N. [3] (b) K and M are chlorine and argon. State which is which, and explain the difference in their melting points. [3] (c) Explain why J has the highest melting point of the five. [2]

7. Sodium atoms are the largest in Period 3, and sodium has the lowest first ionisation energy in the period. Explain both facts. [3]

8. Explain why magnesium has a much higher melting point than sulfur. [4]

9. The melting points of sodium, magnesium and aluminium are 371 K, 923 K and 933 K.

(a) Calculate the increase in melting point from sodium to magnesium. [1] (b) Explain why the melting point increases from sodium to magnesium to aluminium. [3]

10. This question is about phosphorus, sulfur and chlorine. Use the data in questions 4 and 6, and a melting point of 388 K for sulfur.

(a) Describe and explain how the first ionisation energy changes from phosphorus to sulfur, and from sulfur to chlorine. [4] (b) Describe and explain how the melting point changes from phosphorus to sulfur to chlorine. [3]

Answers

1. (a) p block: 1s² 2s² 2p⁵, so the last electron enters 2p (fluorine) [1]. (b) d block: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s², so the last electron enters 3d (vanadium) [1]. (c) s block: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ (potassium) [1]. [3] Examiner insight: Write the configuration in filling order before you decide; for proton number 23 the 3d electrons are the last added, even though 4s is the outer shell.

2. (a) f block [1]. (b) The block follows from position in the Periodic Table, which is set by proton number; every element in the actinide row is in the f block [1]. [2] Examiner insight: An answer that only says “it is in the actinides” restates the question; link position to proton number to make the reason count.

3. (a) Magnesium (3s²) has its highest-energy electron in the 3s sub-shell, so it is s block [1]. Aluminium (3s² 3p¹) has its highest-energy electron in the 3p sub-shell, so it is p block [1]. (b) The electron removed from aluminium is in 3p, which is higher in energy than the 3s electron removed from magnesium [1]. The filled 3s sub-shell also partly screens the 3p electron from the nucleus; both effects reduce the energy needed to remove it [1]. [4] Examiner insight: In (b), say which electron is removed from each atom; an answer about “aluminium’s electrons” in general does not show the sub-shell idea.

4. (a) P(g) → P⁺(g) + e⁻ [1]. (b) Aluminium and sulfur (both needed) [1]. (c) Nuclear charge increases across the period, as each element has one more proton [1]. The electron always leaves the third shell, behind the same inner electrons, so shielding barely changes [1]. With a stronger pull on it (and a smaller atom), the outer electron takes more energy to remove [1]. (d) In sulfur, two of the 3p electrons share one orbital [1]. These paired electrons repel each other, so one of them is removed more easily than an unpaired 3p electron in phosphorus [1]. [7] Examiner insight: In (c), “more protons” and “similar shielding” are separate points, so state both explicitly rather than folding them into one phrase about a “stronger nucleus”.

5. Increasing radius: Cl < P < Al < Na [1]. Proton number increases from 11 (Na) to 17 (Cl), so nuclear charge increases [1]. All four atoms keep their outer electrons in the third shell, behind the same ten inner electrons, so shielding hardly differs [1]. The outer electrons are pulled closer as nuclear charge rises, so the radius falls from Na to Cl [1]. [4] Examiner insight: Check the direction asked for: “increasing” starts with the smallest atom, and a reversed list can lose the order mark even when the explanation is right.

6. (a) J is silicon [1]; L is magnesium [1]; N is phosphorus [1]. (b) K is chlorine and M is argon [1]. Chlorine is made of Cl₂ molecules with 34 electrons each; argon is single atoms with 18 electrons each [1]. More electrons give stronger van der Waals forces, so more energy is needed to overcome them in chlorine, which melts at the higher temperature [1]. (c) Silicon is macromolecular, so melting breaks many strong covalent bonds [1]. Breaking them takes more energy than disrupting magnesium’s metallic bonding or the van der Waals forces in the other three [1]. [8] Examiner insight: In (b), name the particles (Cl₂ molecules, Ar atoms) before comparing forces; a comparison of “size” with no reference to electrons is hard to credit.

7. Sodium has the smallest nuclear charge in the period (11 protons) [1]. Its outer electron is in the third shell with the same inner-shell shielding as the other elements, so it is the least strongly attracted and lies furthest from the nucleus [1]. This weakly held 3s electron needs the least energy to remove, so sodium has the lowest first ionisation energy [1]. [3] Examiner insight: One chain explains both facts; tie the radius point and the energy point to the same weak attraction rather than giving two unconnected statements.

8. Magnesium has a giant metallic structure [1]. Melting must overcome the strong attraction between Mg²⁺ ions and the delocalised electrons [1]. Sulfur is molecular (S₈); melting only overcomes van der Waals forces between molecules, and the S–S covalent bonds do not break [1]. Van der Waals forces are much weaker than metallic bonding, so far less energy is needed and sulfur’s melting point is lower [1]. [4] Examiner insight: Mentioning S–S bonds breaking on melting contradicts the molecular point, so leave them out or say clearly that they stay intact.

9. (a) 923 − 371 = 552 K [1]. (b) All three are giant metallic lattices [1]. The ion charge rises (Na⁺, Mg²⁺, Al³⁺) and the number of delocalised electrons per atom rises (1, 2, 3), while the ions get smaller [1]. Each of these makes the metallic bonding stronger, so melting the metal takes more energy [1]. [4] Examiner insight: “Stronger metallic bonding” alone is a claim, not an explanation; credit comes from giving the ion charge or delocalised electron count that makes it stronger.

10. (a) From phosphorus to sulfur the first ionisation energy falls slightly, from 1012 to 1000 kJ mol⁻¹ [1]. The electron taken from sulfur comes from a doubly filled 3p orbital, and repulsion inside that pair lowers the energy needed [1]. From sulfur to chlorine it rises, to 1251 kJ mol⁻¹ [1]. Chlorine has one more proton, with the electron removed from the same shell and similar shielding, so it is held more strongly [1]. (b) The melting point rises from phosphorus (317 K) to sulfur (388 K), then falls to chlorine (172 K) [1]. All three are molecular, so melting only overcomes van der Waals forces between molecules [1]. S₈ has the most electrons per molecule (128), then P₄ (60), then Cl₂ (34), so the forces are strongest between S₈ molecules and weakest between Cl₂ molecules [1]. [7] Examiner insight: “Describe and explain” asks for two things at each step: state the direction of change, with the data, and then give the reason; a reason on its own leaves the description unanswered.

Where marks are usually lost

• Picking the block from the sub-shell written last rather than the one filled last.
• Giving “Group 0” as a reason for argon or helium being in a particular block.
• Writing X(s) or X(aq) in a first ionisation energy equation.
• Naming only one of the two ionisation energy dips when two are asked for.
• Explaining the Mg → Al drop with electron pairing, or the P → S drop with a new sub-shell: the two reasons are often swapped.
• Saying atomic radius falls because “electrons are added to new shells”.
• Comparing molecular elements by “size of molecule” with no mention of the number of electrons.
• Treating aluminium’s p-block position as a reason for covalent bonding.
• Giving a melting point comparison that names the structure of only one element.

Next steps

• Study guide: periodicity
• Revision notes: periodicity
• Atomic structure
• Bonding and structure
• OxfordAQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Source document: OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. Section 3.2.1 Periodicity (International AS): point 3.2.1.1, Classification; point 3.2.1.2, on the Physical properties of Period 3 elements.