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Questions
Describe the structure of graphite with the help of a labelled diagram.
Discuss the structures of graphite. How would you explain the properties of the graphite on the basis of its structure?
Explain the conducting properties of graphite on the basis of its structure.
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Solution

Structure of graphite
- Graphite is an allotropic form of carbon, distinct from diamond in structure and properties.
- Each carbon atom in graphite undergoes sp2 hybridization.
- Each carbon atom is covalently bonded to three other carbon atoms in the same plane, forming flat hexagonal rings.
- These hexagonal rings join to form extended layers or sheets of carbon atoms, as seen in a honeycomb-like structure.
- The C–C bond length within a layer is 142 pm (1.42 Å).
- The distance between adjacent layers is 340 pm (3.4 Å).
- The layers are held together by weak van der Waals forces, which allows them to slide over each other easily.
- This sliding of layers makes graphite soft and slippery.
- Each carbon atom has one delocalized electron (not involved in bonding) that moves freely within the layers.
- These free electrons enable graphite to be a good conductor of electricity.
- Graphite has a low density (2.26 g/cm3) due to the large spacing between layers.
Properties of graphite
1. Physical Appearance:
- Dark grey in colour with a metallic lustre.
- Graphite feels soft and greasy to the touch.
- Graphite, when used in pencils, leaves a black mark on paper.
2. Electrical Conductivity:
- Conducts electricity due to delocalized π-electrons in its layers.
- Each carbon is sp² hybridized, bonding with 3 other carbon atoms.
- The fourth electron in each carbon remains free in a 2p orbital, forming a π-bonding system across the layer.
- These free-moving electrons enable excellent electrical conductivity.
3. Chemical Resistance:
- Chlorine, alkalis, and diluted acids do not harm it.
- It gradually oxidises to carbon dioxide in a solution containing sulphuric acid and potassium dichromate.
Notes
Students should refer to the answer according to the question.
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- \[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\phantom{....}|\phantom{....}|\\
\ce{H - C - C - C - C - H}\\
|\phantom{....}|\phantom{....}|\phantom{....}|\\
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
\end{array}\] - \[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\phantom{....}|\\
\ce{H - C - C - C - H}\\
|\phantom{.....}|\phantom{.....}|\\
\ce{H}\ce{H-C-H}\ce{H}\\
|\\
\ce{H}\\
\end{array}\] - \[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\phantom{....}|\\
\ce{H - C - C - C - H}\\
|\phantom{.....}\backslash\phantom{..}|\\
\phantom{....}\ce{H}\phantom{......}\ce{C - H}\phantom{}\\
\phantom{.......}|\\
\phantom{.......}\ce{H}\\
\end{array}\] - \[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\\
\ce{H - C - C - H}\\
|\phantom{....}|\\
\ce{H - C - C - H}\\
|\phantom{....}|\\
\ce{H}\phantom{...}\ce{H}\\
\end{array}\]
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