OCR GCSE Chemistry A: Bonding (J248) – 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 covers sub-topic C2.2 Bonding (C2.2a to C2.2i, with CM2.2i to CM2.2iii) of the OCR GCSE (9–1) Chemistry A (Gateway Science) J248 specification, Version 4.0 (August 2026), for first assessment in 2018. No C2.2 statement is Higher Tier only, so every question suits both Foundation and Higher candidates. C2 content is assessed on Paper 1 and Paper 3.

Learn the content first with the Bonding study guide and the Bonding revision notes. Course links: OCR GCSE Chemistry hub and J248 checklist. For a timed check across topics, use the free diagnostics.

Questions

1. Aluminium has atomic number 13.

(a) Give the electron arrangement of an aluminium atom. [1] (b) State the group and the period of aluminium. [2] (c) Use the electron arrangement to explain whether aluminium is a metal or a non-metal. [2]

2. A student tests two unknown elements, A and B.

• A is shiny, conducts electricity, and bends without breaking. Its oxide reacts with dilute hydrochloric acid.
• B is a dull powder that crumbles and does not conduct electricity. Its oxide dissolves in water.

(a) Identify which element is the metal. Give two pieces of evidence from the list. [2] (b) Element A is magnesium. Write a balanced equation for magnesium burning in oxygen. [1] (c) Predict whether the solution made from the oxide of B is acidic or alkaline. [1]

3. Aluminium nitride is a binary ionic compound.

(a) State the number of protons and electrons in an Al³⁺ ion and in an N³⁻ ion. [2] (b) Deduce the formula of aluminium nitride and explain your answer using the ion charges. [2]

4. Describe the dot and cross diagram for aluminium fluoride, AlF₃. Show outer shells only. [3]

5. Carbon disulfide has the structure S=C=S. Describe its dot and cross diagram (outer shells only), stating the number of shared pairs and the number of lone pairs on each atom. [3]

6. A student draws lithium fluoride as two atoms sharing a pair of electrons, with no brackets or charges. State two things wrong with this diagram. [2]

7. Describe and compare the bonding in solid magnesium and in solid magnesium oxide. [4]

8. A ball and stick model of one layer of hexagonal boron nitride shows every boron atom joined to 3 nitrogen atoms and every nitrogen atom joined to 3 boron atoms. The layer continues in every direction.

(a) Use the model to deduce the formula of boron nitride. Show your reasoning. [2] (b) Name the type of structure shown. [1] (c) Give one limitation of using a ball and stick model to represent this layer. [1]

9. A gold nanoparticle is 5.8 nm across. Take the diameter of a gold atom as 0.29 nm.

(a) Estimate the number of gold atoms across the nanoparticle. [2] (b) A fibre is 1 × 10⁻⁴ m wide. Calculate how many times wider the fibre is than the nanoparticle. Give your answer in standard form to 2 significant figures. [2]

10. The table gives data for two elements.

Element	Relative atomic mass	Atomic number
Cobalt	58.9	27
Nickel	58.7	28

(a) Describe how Mendeleev arranged the elements in his table, and two other things he did when he built it. [3] (b) Explain why ordering strictly by relative atomic mass would cause a problem for this pair. [2] (c) Explain how the modern Periodic Table removes this problem. [2]

11. Fluorine (Z = 9) and chlorine (Z = 17) are both in group 7. Argon (Z = 18) is in group 0.

(a) Give the electron arrangements of fluorine and chlorine. [2] (b) Explain why both elements form ions with a 1− charge. [1] (c) Explain why fluorine is the more reactive of the two. [3] (d) Explain why argon does not form ions in this way. [1]

12. Magnesium (Z = 12), phosphorus (Z = 15) and chlorine (Z = 17) are in period 3.

(a) Chlorine reacts with each of the other two elements. Identify which product is ionic, and give its formula. [2] (b) Describe the dot and cross diagram for phosphorus trichloride, PCl₃ (outer shells only). [3] (c) Explain, in terms of electrostatic forces, what holds the particles together in each of the two products. [2] (d) Give one limitation of a dot and cross diagram for PCl₃. [1]

Answers

1. (a) 2,8,3 [1] (b) Group 3 (three outer electrons) [1]; period 3 (three occupied shells) [1] (c) It has only 3 outer electrons, which it loses in reactions [1], forming a positive ion, Al³⁺, so it is a metal [1]. Examiner insight: For “explain” you need the link from electron arrangement to behaviour; “it is on the left” alone does not use the arrangement as asked.

2. (a) A is the metal [1]: it is shiny and conducts electricity, or its oxide reacts with an acid (any two pieces of evidence) [1]. (b) 2Mg + O₂ → 2MgO [1] (c) Acidic [1]: B behaves as a non-metal, and oxides of non-metals give acidic solutions. Examiner insight: Evidence must be quoted from the information given; adding properties not in the list (such as high density) gains nothing.

3. (a) Al³⁺: 13 protons, 10 electrons [1]. N³⁻: 7 protons, 10 electrons [1]. (b) The charges +3 and −3 cancel with one of each ion [1], so the formula is AlN [1]. Examiner insight: A common slip is to swap charges into Al₃N₃; always simplify to the lowest whole-number ratio.

4. One Al³⁺ ion in square brackets with charge 3+ at top right, with no electrons left in its old outer shell [1]. Three separate F⁻ ions, each in square brackets with charge − [1]. Each fluoride ion’s outer shell holds 8 electrons: 7 dots (its own) and 1 cross (from aluminium) [1]. Examiner insight: An ionic diagram with no brackets or charges reads as a covalent one, so include both even when the electron counts are right.

5. Carbon shares two pairs with each sulfur atom, forming two double bonds: 4 shared pairs, each with one dot and one cross [1]. Carbon has no lone pairs [1]. Each sulfur has 2 lone pairs, so every atom has 8 outer electrons [1]. Examiner insight: Check your total: 4 + 6 + 6 = 16 outer electrons, which must equal 4 shared pairs plus 4 lone pairs; if your diagram does not add up, a pair is missing or misplaced, so fix it before moving on.

6. Lithium fluoride is ionic, so electrons should be transferred (one electron from Li to F), not shared [1]. The ions should be drawn in brackets with charges, Li⁺ and F⁻ [1]. Examiner insight: “Two things wrong” needs two separate points; restating the same error in two ways counts as only one point.

7. In magnesium, there is a lattice of positive magnesium ions [1] attracted to a sea of delocalised electrons (metallic bonding) [1]. In magnesium oxide, electrons have transferred to form Mg²⁺ and O²⁻ ions [1]. These oxide and magnesium ions attract each other in all directions in a giant ionic lattice; both structures are giant and both are held by electrostatic attraction [1]. Examiner insight: “Compare” needs a similarity or a difference stated directly; two separate descriptions with no link usually fall short of full marks.

8. (a) Each boron makes 3 bonds to nitrogen and each nitrogen makes 3 bonds to boron, so the bond totals only match with equal numbers of each atom [1]. Formula BN [1]. (b) Giant covalent structure [1]. (c) One from: real bonds are not rigid sticks; atom sizes are not to scale; gaps between atoms are exaggerated; a model shows only a small part of a layer that goes on in every direction [1]. Examiner insight: “Show your reasoning” means the bond-count argument is needed; a bare formula without it is unlikely to gain both marks.

9. (a) Number of atoms = 5.8 ÷ 0.29 [1] = 20 atoms [1]. (b) 5.8 nm = 5.8 × 10⁻⁹ m; 1 × 10⁻⁴ ÷ 5.8 × 10⁻⁹ [1] = 1.7 × 10⁴ times [1]. Examiner insight: Converting nanometres to metres is the step most likely to go wrong; write the conversion as a separate line so your method can still earn credit after a slip.

10. (a) He ordered the elements by atomic weight (relative atomic mass) [1]. He left gaps for undiscovered elements and predicted their properties [1]. He swapped some pairs so elements sat with others of similar properties [1]. (b) Cobalt has the larger relative atomic mass, so mass order would put nickel first [1]. That would place each element in a group whose properties do not match its own [1]. (c) The modern table is ordered by atomic number (number of protons) [1]. Cobalt (27) comes before nickel (28), which fits their properties without any swap [1]. Examiner insight: Mixing up “atomic weight” and “atomic number” is the usual way to lose marks here; say which quantity each table uses.

11. (a) Fluorine 2,7 [1]; chlorine 2,8,7 [1]. (b) Each has seven outer electrons and gains one more to complete its outer shell [1]. (c) Fluorine’s outer shell is the second shell, closer to the nucleus than chlorine’s third shell [1]. Fewer inner electrons screen it from the nucleus [1]. So an incoming electron feels a stronger pull and is gained more easily [1]. (d) Argon (2,8,8) already has a full outer shell, so it has no tendency to gain or lose electrons [1]. Examiner insight: For group 7 the argument is about gaining an electron; reusing the group 1 “electron is lost more easily” wording reverses the logic and makes the explanation wrong.

12. (a) The product with magnesium, magnesium chloride, is ionic (metal with non-metal) [1]. Formula MgCl₂ [1]. (b) Phosphorus shares one pair with each of three chlorine atoms: 3 shared pairs, each a dot and a cross [1]. Phosphorus keeps 1 lone pair [1]. Each chlorine has 3 lone pairs, giving every atom 8 outer electrons [1]. (c) In MgCl₂, the electrostatic attraction between oppositely charged Mg²⁺ and Cl⁻ ions holds the lattice together [1]. In PCl₃, both nuclei in each bond attract the shared pair of electrons [1]. (d) One from: it does not show the 3D shape; it suggests the dot and cross electrons are different, but all electrons are identical; it does not show relative atom sizes [1]. Examiner insight: In part (c) name the attracting particles for each substance; “strong bonds” says nothing about what attracts what, so it misses the point of the question.

Where marks are usually lost

• Ionic dot and cross diagrams missing square brackets or charges.
• Electrons left on the metal ion’s old outer shell.
• Lone pairs forgotten on the central atom (PCl₃) or on the outer atoms (CS₂).
• Formulas left unsimplified, or charges not balanced.
• Saying an ion gains or loses protons.
• Mixing up nanometres and metres in scale estimates.
• Answering “compare” questions with two separate descriptions.
• Explaining group 7 reactivity with electrons being lost, or with “more electrons” instead of distance and attraction.
• Stating that Mendeleev used atomic number.

Next steps

• Recap with the Bonding revision notes.
• Return to the Bonding study guide for any question you found hard.
• See the OCR GCSE Chemistry hub and the J248 checklist.
• Take one of the all free 10-minute diagnostics.
• Try the C2.1 purity and separating mixtures practice.
• Book a free trial class with a Marlbridge teacher.

Official syllabus

OCR GCSE (9–1) Chemistry A (Gateway Science) J248 specification, Version 4.0 (August 2026), for first assessment in 2018, published by OCR. Sub-topic C2.2 Bonding.