OxfordAQA International GCSE Chemistry: Structure and bonding of carbon, and nanoparticles (9202) – 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.

These questions cover sections 3.2.3 (Structure and bonding of carbon) and 3.2.4 (Nanoparticles) of the OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards. There are no tiers, so all of these questions are relevant to you. Short recall questions come first; the last few combine ideas and include calculations.

Revise first with the study guide or the revision notes. The course hub and printable checklist show where this unit fits. Use Aᵣ of C = 12 and Avogadro’s constant = 6.02 × 10²³ per mole where needed.

Questions

1. State the number of covalent bonds that a carbon atom can form. [1]

2. Explain why diamond is very hard. [2]

3. This question is about graphite.

(a) State the number of other carbon atoms that each carbon atom is bonded to in graphite. [1] (b) Explain why graphite is soft and slippery. [2]

4. Explain why graphite conducts electricity, and state one way in which graphite is similar to a metal. [3]

5. Fullerenes are another form of carbon.

(a) State what the structure of fullerenes is based on. [1] (b) Give two uses of fullerenes. [2]

6. Three particles have these diameters: particle P, 0.07 µm; particle R, 2.5 × 10⁻⁷ m; particle S, 8.0 × 10⁻⁹ m.

(a) State the size range of structures studied in nanoscience. [1] (b) Convert each diameter to nanometres and identify which particles are nanoparticles. [3]

7. A catalyst can be made as cube-shaped nanoparticles of side 15 nm or as cube-shaped grains of side 1.5 µm.

(a) Calculate the surface area to volume ratio of one nanoparticle. [2] (b) Calculate the ratio for one grain, and how many times larger the nanoparticle ratio is. [2] (c) Suggest why the same mass of catalyst works faster as nanoparticles. [1]

8. Read this information.

Compound W is a white solid that absorbs ultraviolet light. As a powder of 30 µm particles it leaves a visible white layer on skin. As 30 nm particles it absorbs ultraviolet light just as well but cannot be seen on the skin.

(a) Treating the particles as cubes, show that the 30 nm particles have a surface area to volume ratio 1000 times that of the 30 µm particles. [2] (b) Use the information to suggest why a sun cream maker chooses the 30 nm form. [2] (c) Apart from suntan creams, name one other type of cosmetic that the specification gives as a possible use of nanoparticles. [1]

9. Substance J is an element made only of carbon atoms. It has a very high melting point, conducts electricity as a solid, and leaves a dark mark when rubbed on paper because it rubs off easily.

(a) Identify J. Explain each of its three properties in terms of its structure and bonding. [5] (b) Explain why diamond, rather than J, is used to coat the tips of drill bits. [2]

10. A graphite electrode has a mass of 0.36 g.

(a) Calculate the amount, in moles, of carbon atoms in the electrode. [1] (b) Calculate the number of delocalised electrons in the electrode. [2] (c) Explain why 0.36 g of diamond does not conduct electricity. [2] (d) Graphite also conducts heat. State what is responsible for this. [1]

11. A chemist compares 1.0 cm³ of a catalyst as one cube with the same volume divided into cubes of side 10 nm.

(a) Convert 1.0 cm to nanometres. [1] (b) Calculate the number of 10 nm cubes that 1.0 cm³ makes. [2] (c) Calculate the total surface area of these cubes, first in nm² and then in cm². [3] (d) The single 1.0 cm cube has a surface area of 6.0 cm². Compare this with your answer to (c) and explain why nanoparticle catalysts can save metal. [2]

12. Read this information.

Fullerene nanotubes are tubes made of hexagonal rings of carbon atoms. They are very strong but have a low density. Some other fullerenes form hollow cages large enough to hold a small molecule.

(a) Suggest why nanotubes are added to the frames of tennis rackets. [2] (b) Suggest how cage-shaped fullerenes could be used to deliver a drug into the body. [2]

Answers

1. Four [1] Examiner insight: “State” needs only the number; an added reason that is wrong can contradict an otherwise correct answer.

2. Each carbon atom forms four covalent bonds in a giant covalent structure [1]. Many strong bonds must be broken, needing a lot of energy [1]. Examiner insight: “Strong bonds” alone is too thin for two marks; name the four bonds per atom or the giant structure as well.

3. (a) Three [1] (b) There are no covalent bonds between the layers, only weak forces [1], so the layers slide over each other [1]. Examiner insight: Writing that the covalent bonds in graphite are weak is chemically wrong and loses the first mark.

4. One electron from each carbon atom is delocalised [1]. These electrons move through the structure and carry the current [1]. Like a metal, graphite has delocalised electrons [1]. Examiner insight: Writing that ions or atoms move contradicts the explanation; only electrons move in graphite.

5. (a) Hexagonal rings of carbon atoms [1] (b) Any two of: drug delivery into the body; lubricants; catalysts; nanotubes for reinforcing materials [1] [1]. Examiner insight: Vague answers such as “medicine” or “sport” are weaker than the specification wording; write “drug delivery” and “reinforcing tennis rackets”.

6. (a) 1–100 nm [1] (b) P: 0.07 × 1000 = 70 nm; R: 2.5 × 10⁻⁷ ÷ 1 × 10⁻⁹ = 250 nm; S: 8.0 × 10⁻⁹ ÷ 1 × 10⁻⁹ = 8 nm [1]. P is a nanoparticle [1]. S is a nanoparticle; R is not [1]. Examiner insight: Show each converted value; a bare list of letters gives the examiner nothing to credit if one conversion slips.

7. (a) Surface area = 6 × 15² = 1350 nm²; volume = 15³ = 3375 nm³ [1]. Ratio = 1350 ÷ 3375 = 0.4 nm⁻¹ [1] (b) 1.5 µm = 1500 nm, so ratio = 6 ÷ 1500 = 0.004 nm⁻¹ [1]. 0.4 ÷ 0.004 = 100 times larger [1] (c) More of the catalyst is exposed at the surface, so more reacting particles can make contact with it [1]. Examiner insight: Convert both lengths to the same unit before dividing; using 1.5 instead of 1500 gives a factor that is badly wrong.

8. (a) 30 nm: 6 ÷ 30 = 0.2 nm⁻¹; 30 µm = 30 000 nm: 6 ÷ 30 000 = 0.0002 nm⁻¹ [1]. 0.2 ÷ 0.0002 = 1000 [1] (b) The 30 nm form absorbs ultraviolet light as well as the powder [1] but cannot be seen, so it does not leave a white layer on the skin [1]. (c) Deodorants [1] Examiner insight: In a “show that” the final value is given, so every step of the working must appear to earn the marks.

9. (a) J is graphite [1]. Very high melting point: many strong covalent bonds must break [1]. Conducts: one electron per carbon atom is delocalised and moves through the structure [1]. Rubs off: each atom bonds to three others in layers [1] with only weak forces between the layers, so layers slide off [1]. (b) Diamond has four covalent bonds per carbon atom, giving a rigid giant structure [1], so diamond is very hard and is not worn away like graphite [1]. Examiner insight: Each property needs its own structural reason; one general sentence about “strong bonds” cannot explain all three.

10. (a) n = 0.36 ÷ 12 = 0.030 mol [1] (b) Atoms = 0.030 × 6.02 × 10²³ [1] = 1.81 × 10²², one delocalised electron per atom [1] (c) In diamond, all four outer electrons of every carbon atom are used in covalent bonds [1], so there are no delocalised electrons to move [1]. (d) The delocalised electrons [1] Examiner insight: State the one-electron-per-atom link in (b); a number without it does not show why atoms equal electrons.

11. (a) 1 cm = 1 × 10⁻² m ÷ 1 × 10⁻⁹ m = 1.0 × 10⁷ nm [1] (b) Volume = (1.0 × 10⁷)³ = 1.0 × 10²¹ nm³ [1]. Number = 1.0 × 10²¹ ÷ 10³ = 1.0 × 10¹⁸ [1] (c) Area of one cube = 6 × 10² = 600 nm² [1]. Total = 1.0 × 10¹⁸ × 600 = 6.0 × 10²⁰ nm² [1]. 1 nm² = 1 × 10⁻¹⁴ cm², so total = 6.0 × 10⁶ cm² [1] (d) The nanoparticles have 1 000 000 (10⁶) times the surface area [1]. A much smaller mass of catalyst gives the same surface for reactions, so less metal is needed [1]. Examiner insight: Area units square the conversion factor; using 10⁻⁷ instead of 10⁻¹⁴ for nm² to cm² is an easy slip to make.

12. (a) Nanotubes are very strong [1], and their low density means they reinforce the frame without adding much mass [1]. (b) The drug molecule is held inside the hollow cage [1], which carries it through the body to where it is needed [1]. Examiner insight: “Suggest” questions reward answers built from the information given; quote the property, then link it to the use.

Where marks are usually lost

• Saying graphite is soft because covalent bonds are weak, instead of weak forces between layers.
• Saying ions, atoms or “particles” carry the current in graphite.
• Explaining diamond’s hardness with “strong bonds” but not four bonds per atom or a giant structure.
• Missing the lower limit when quoting the nanoscale range.
• Dividing lengths in different units in a surface area to volume calculation.
• Converting area with the length factor (10⁻⁷) instead of its square (10⁻¹⁴).
• Ignoring the information given in “suggest” questions and writing a memorised answer.
• Giving one structural reason for several properties.

Next steps

• Revision notes
• Study guide
• Quantitative chemistry practice for more mole calculations
• OxfordAQA International GCSE Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards, published by OxfordAQA. Sections 3.2.3 and 3.2.4.