Graphite: strong layers, weak interlayer attractions and mobile electrons
مُقدَّم| English | المعنى |
|---|---|
| graphite/ˈɡræfaɪt/ | غرافيت |
| hexagonal ring/ˈheksəɡənl rɪŋ/ | حلقة سداسية |
What would explain this observation?
- Graphite 石墨 can be soft enough to leave a pencil mark while still having very strong covalent bonds. Different properties depend on connections within and between its layers.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Each graphite carbon makes three covalent bonds to other carbons in layers of hexagonal rings 六元环. Strong bonds inside the layers require much energy to overcome, giving a very high melting point in the GCSE account. There are no covalent bonds between the layers; weaker attractions allow layers to slide past one another, explaining softness and lubricating use. One electron from each carbon is delocalised and can carry charge through the structure.
- graphite: A carbon form with covalently bonded hexagonal layers and one delocalised electron per carbon; hexagonal ring: A ring of six connected atoms in the stated graphite layer model.
What explains graphite layers sliding?
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Within each layer the covalent bonds are strong, but the layers are held by weak intermolecular attractions with no covalent bonds between them, so they slide easily. The ring bonds themselves are strong.
Match each technical term to its precise meaning.
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Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- Do not use weak interlayer forces to explain the high melting point: that property involves strong covalent bonding. Conversely, naming strong covalent bonds alone does not explain easy layer sliding. Graphite resembles metals in having delocalised electrons, but its bonding network is layered covalent rather than metallic. Conductivity directions in real graphite are not represented fully by one simple flat drawing.
- Inspect a hexagonal-layer model and identify three neighbours for an interior carbon, with bonds continuing at cut edges. Compare layers stacked without covalent connecting lines. A teacher-approved low-voltage test of a prepared graphite rod can provide conductivity evidence. Record contact quality and geometry, and distinguish the rod material from a pencil’s mixed graphite/clay core.
Which two habits make the investigation or model in this case more defensible?
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Stating the conditions and assumptions lets others repeat the work exactly, and recording the supporting evidence lets them check the conclusion. Hiding disagreeing results removes the evidence that could challenge a wrong conclusion, and one measurement can never establish a wider claim.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: a stated interior graphite-layer model has 90 carbon atoms contributing one delocalised electron each, so it represents 90 such electrons. Its 90×3=270 bond ends correspond to 135 bonds after dividing by two under a boundary-free convention. The electron count and covalent bond count describe different aspects of the same structure.
In the stated graphite model, 55 carbon atoms each contribute one delocalised electron. How many are represented? Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
In the stated graphite model, 55 carbon atoms each contribute one delocalised electron. How many are represented?
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The result is 55 electrons. Known: 55 carbon atoms each contribute one delocalised electron. Delocalised electrons = 55 x 1 = 55 electrons. This count is separate from the covalent bond count of the same structure.
Check the conclusion and its limits
- Graphite does not have four covalent neighbours per carbon like diamond. Weak layer attractions are not absent attraction altogether. A cut edge in a model has fewer drawn neighbours and should not be treated as evidence against the interior three-bond pattern.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Graphite’s high melting point is explained solely by its weak interlayer attractions. This claim is false: The melting point is set by the strong covalent bonds WITHIN the layers, which must be broken to melt graphite - hence it is very high. The weak attractions act only between layers and explain slipperiness, not melting.
Graphite: strong layers, weak interlayer attractions and mobile electrons: Do not use weak interlayer forces to explain the high melting point: that property involves strong covalent bonding. Conversely, naming strong covalent bonds alone does not explain easy layer sliding. Graphite resembles metals in having delocalised electrons, but its bonding network is layered covalent rather than metallic. Conductivity directions in real graphite are not represented fully by one simple flat drawing.
Graphite’s high melting point is explained solely by its weak interlayer attractions.
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The melting point is set by the strong covalent bonds WITHIN the layers, which must be broken to melt graphite - hence it is very high. The weak attractions act only between layers and explain slipperiness, not melting.
A carbon form with covalently bonded hexagonal layers and one delocalised electron per carbon: write the technical term.
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graphite means A carbon form with covalently bonded hexagonal layers and one delocalised electron per carbon.