Structures & properties
Structures & properties
Diamond and graphite
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Diamond and graphite both contain giant covalent networks, but each carbon atom bonds differently. That difference explains their different properties.
In diamond, each carbon forms four bonds in a rigid three-dimensional network. There are no mobile electrons, so it does not conduct.
In graphite, each carbon forms three bonds within a layer. One electron per carbon is delocalised and can carry charge. Weak van der Waals attractions between layers allow the layers to slide.
Both structures have high melting points because many strong covalent bonds must be broken. Graphite is not made of separate molecules, even if an older diagram calls the between-layer attractions 'intermolecular'. Silicon and graphene are also giant covalent networks; their fuller comparisons belong to Periodicity.
- Giant covalent structure
- A continuous network of atoms joined by many strong covalent bonds.
| Feature | Diamond | Graphite |
|---|---|---|
| Bonding | Four covalent bonds per carbon | Three covalent bonds per carbon |
| Structure | Rigid three-dimensional network | Flat layers with weak attractions between them |
| Melting | Very high: many strong bonds must be broken | Very high: many strong bonds within layers must be broken |
| Conductivity | No mobile charge carriers | Delocalised electrons move along layers |
| Mechanical property | Hard, rigid network | Layers can slide; relatively soft |
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Diamond and graphite are giant covalent forms of carbon.
(a) Explain why diamond has a very high melting point. [2 marks]
(b) Explain why diamond does not conduct electricity. [2 marks]
(c) Explain why graphite conducts electricity. [2 marks]
(d) Explain why the layers in graphite can slide over one another. [2 marks]
Total: 8 marks
Show answer & marking points
(a) Diamond is a giant three-dimensional covalent network [1]; many strong covalent bonds must be broken, requiring much energy [1]. [2]
(b) All four outer electrons of each carbon are localised in covalent bonds [1], so there are no mobile charged particles [1]. [2]
(c) Each carbon in graphite forms three covalent bonds, leaving delocalised electrons [1] that move along the layers and carry charge [1]. [2]
(d) Only weak attractions act between graphite layers [1], so little energy is needed for layers to slide [1]. [2]
Graphite is used as an electrode. Diamond is used on cutting tools.
(a) Explain why graphite is suitable for use as an electrode. [3 marks]
(b) Explain why diamond is suitable for use on a cutting tool. [3 marks]
(c) Explain why graphite is softer than diamond. [2 marks]
Total: 8 marks
Show answer & marking points
(a) Graphite contains delocalised electrons [1] that are mobile along the layers [1] and carry electrical charge through the electrode [1]. [3]
(b) Diamond is a rigid three-dimensional network [1] in which each carbon forms four strong covalent bonds [1], making it very hard and resistant to deformation [1]. [3]
(c) Graphite layers are held together by weak attractions [1], allowing the layers to slide over one another [1]. [2]
Simple molecular lattices and state changes
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A simple molecular substance contains separate molecules. Each molecule has strong covalent bonds inside it, while neighbouring molecules attract through intermolecular forces.
Melting and boiling overcome attractions between molecules; the molecules usually stay intact. This is why a molecular substance can have a low boiling point despite having strong covalent bonds.
Pure molecular substances normally have no mobile ions or delocalised electrons, so they do not conduct. To explain solubility, compare the attractions disrupted with the new attractions formed with the solvent.
- Simple molecular solid
- Discrete covalently bonded molecules held together by intermolecular attractions.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Iodine is a simple molecular substance made from I2 molecules.
(a) Describe the particles and attractions in solid iodine. [2 marks]
(b) Explain why iodine has a much lower melting point than diamond. [2 marks]
(c) State what is overcome when iodine melts and what remains intact. [1 mark]
(d) Explain why iodine does not conduct electricity in the solid or liquid state. [1 mark]
Total: 6 marks
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(a) Solid iodine contains neutral I2 molecules [1] attracted to one another by intermolecular forces [1]. [2]
(b) Melting iodine overcomes only weak intermolecular (van der Waals) forces between I2 molecules [1]. Much more energy is required to break the strong C–C covalent bonds in diamond [1]. [2]
(c) Intermolecular forces are overcome; the I-I covalent bonds remain intact [1]. [1]
(d) Neutral molecules provide no mobile ions or delocalised electrons [1]. [1]
Identify structures from physical properties
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Start a structure question by identifying the particles: ions, atoms or molecules. Then name the attraction between them. This stops you using the right explanation for the wrong structure.
- Choose the structure type from the evidence: giant ionic, metallic, giant covalent or simple molecular.
- For melting or boiling, name the attractions that must be overcome and connect their strength to the energy needed.
- For conductivity, identify the charged particles and say whether they can move.
- Use all the evidence together. A solid conductor could be a metal or graphite, so one property alone may not identify it.
| Structure | Particles | Main attraction | Conductivity | Melting / boiling |
|---|---|---|---|---|
| Giant ionic | Positive and negative ions | Electrostatic attraction between ions | Molten / aqueous only | High |
| Metallic | Cations and delocalised electrons | Ions attracted to delocalised electrons | Solid and liquid | Often high |
| Giant covalent | Atoms | Strong covalent bonds throughout a network | Usually none; graphite is an exception | Very high |
| Simple molecular | Molecules | Intermolecular forces | None in pure substance | Usually lower |
An unknown solid
High melting point; does not conduct as a solid; conducts when molten.
Deduce the likely structure type of the solid and explain how its properties support your answer.
Worked answer
Charged particles exist, but they become mobile only after the solid lattice breaks down.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
The properties of four substances are shown.
| Substance | Melting point / °C | Conducts as solid? | Conducts as liquid? |
|---|---|---|---|
| A | 801 | No | Yes |
| B | 650 | Yes | Yes |
| C | 3550 | No | No |
| D | 114 | No | No |
- Identify the most likely structure type for A–D. [4]
- For each substance, use the conductivity evidence together with its melting point to justify the identification. [4]
Total: 8 marks
Show answer & marking points
(a) A ionic [1]; B metallic [1]; C giant covalent [1]; D simple molecular [1]. [4]
(b) A conducts only when its ions become mobile in the liquid [1]. B conducts in both states because delocalised electrons remain mobile [1]. C has no mobile charged particles [1]. D contains neutral molecules and no mobile charge carriers [1]. [4]
Conductivity alone cannot distinguish C from D: neither conducts. Their very different melting points provide the additional evidence.
Four unfamiliar substances have the following properties.
| Substance | Melting point / °C | Solid conductivity | Liquid conductivity |
|---|---|---|---|
| P | 80 | None | None |
| Q | 1414 | None | None |
| R | 770 | None | Good |
| S | 1085 | Good | Good |
- Identify the most likely structure type for P–S. [4]
- Explain the change in conductivity when R melts. [3]
- Explain why Q has a much higher melting point than P. [2]
Total: 9 marks
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(a) P simple molecular [1]; Q giant covalent [1]; R ionic [1]; S metallic [1]. [4]
(b) In solid R, ions are fixed and cannot carry charge [1]. On melting, ions become mobile [1] and carry charge through the liquid [1]. [3]
(c) Q requires many strong covalent bonds in a giant network to be broken [1], whereas P requires only intermolecular forces between molecules to be overcome [1]. [2]
Build complete property explanations
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Build an explanation as a linked chain. Each sentence should explain why the next part follows.
- Name the structure and the particles present.
- Identify the attraction between those particles.
- For a melting-point question, link attraction strength to the energy needed to separate the particles.
- For a conductivity question, name the charge carrier and say whether it can move.
Use specific terms such as ions, delocalised electrons or intermolecular forces. A statement that a substance 'has strong bonds' does not identify the attraction.
Sodium chloride versus iodine
Explain why sodium chloride has a much higher melting point than iodine. Refer to their structures and the attractions overcome on melting.
Worked answer
- NaCl is giant ionic; iodine is simple molecular.
- Strong electrostatic attractions act between ions throughout the NaCl lattice.
- I2 molecules are attracted by weaker intermolecular forces.
- Much more energy is needed to overcome the ionic attractions.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
A student compares barium fluoride, BaF2, potassium metal and iodine.
- State the structure type of each substance. [3]
- Explain why solid potassium conducts electricity but solid barium fluoride and solid iodine do not. [3]
- Explain why barium fluoride has a much higher melting point than iodine. [2]
Total: 8 marks
Show answer & marking points
- Barium fluoride: giant ionic [1]; potassium: giant metallic [1]; iodine: simple molecular [1].
- Solid BaF2 has fixed ions and no mobile charge carriers [1]. Potassium has mobile delocalised electrons that carry charge [1]. Neutral I2 molecules provide no mobile ions or delocalised electrons [1].
- BaF2 has strong electrostatic attractions between Ba2+ and F− ions throughout its lattice [1]. These require much more energy to overcome than the weak intermolecular forces between I2 molecules [1].