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p-Block Elements - Allotropes of Carbon

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Carbon exhibits allotropy, existing in both crystalline and amorphous forms. The primary crystalline allotropes are Diamond, Graphite, and Fullerenes.

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In Diamond, each carbon atom undergoes sp3sp^3 hybridization and is linked to four other carbon atoms in a tetrahedral geometry. This 3D network results in a very high melting point and makes it the hardest natural substance.

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In Graphite, carbon atoms are sp2sp^2 hybridized, forming hexagonal planar layers. These layers are held together by weak van der Waals forces with a separation of 3.40 A˚3.40 \text{ \AA}. The presence of a free fourth valence electron makes graphite a good conductor of electricity.

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Fullerenes, such as C60C_{60} (Buckminsterfullerene), are cage-like molecules. C60C_{60} contains 2020 six-membered rings and 1212 five-membered rings. All carbon atoms in fullerenes are sp2sp^2 hybridized.

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Thermodynamic stability: Graphite is the most thermodynamically stable allotrope of carbon at room temperature and pressure. The enthalpy of formation (ΔfH∘\Delta_f H^\circ) of graphite is taken as 0 kJ mol−10 \text{ kJ mol}^{-1}.

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Amorphous forms of carbon like charcoal, coke, and carbon black are actually micro-crystalline forms of graphite.

📐Formulae

Bond length in Diamond:dC−C=1.54 A˚(154 pm)Bond \, length \, in \, Diamond: d_{C-C} = 1.54 \text{ \AA} (154 \text{ pm})

Bond length in Graphite layers:dC−C=1.415 A˚(141.5 pm)Bond \, length \, in \, Graphite \, layers: d_{C-C} = 1.415 \text{ \AA} (141.5 \text{ pm})

Interplanar distance in Graphite:d=3.40 A˚(340 pm)Interplanar \, distance \, in \, Graphite: d = 3.40 \text{ \AA} (340 \text{ pm})

ΔfH∘(graphite)=0 kJ mol−1\Delta_f H^\circ (\text{graphite}) = 0 \text{ kJ mol}^{-1}

ΔfH∘(diamond)=1.90 kJ mol−1\Delta_f H^\circ (\text{diamond}) = 1.90 \text{ kJ mol}^{-1}

💡Examples

Problem 1:

Explain why graphite acts as a lubricant while diamond is used as an abrasive.

Solution:

Graphite has a layered structure where layers are held by weak van der Waals forces, allowing them to slide over each other (d=3.40 A˚d = 3.40 \text{ \AA}). In contrast, diamond has a rigid 3D covalent network of sp3sp^3 hybridized carbons.

Explanation:

The mobility of layers in graphite provides its lubricating property. The strong C−CC-C covalent bonds (154 pm154 \text{ pm}) in the tetrahedral lattice of diamond provide the hardness required for abrasive tools.

Problem 2:

Calculate the number of single and double bonds in a C60C_{60} molecule.

Solution:

In C60C_{60}, each carbon atom is sp2sp^2 hybridized. There are 6060 single bonds (linking pentagons and hexagons) and 3030 double bonds (at the junctions of two hexagons).

Explanation:

The structure consists of 2020 hexagons and 1212 pentagons. To satisfy the valency of sp2sp^2 carbon, double bonds are localized at the 6,66,6 ring junctions while single bonds are at the 5,65,6 junctions.

Problem 3:

Which allotrope of carbon is thermodynamically more stable: Diamond or Graphite?

Solution:

Graphite is more stable.

Explanation:

The standard enthalpy of formation (ΔfH∘\Delta_f H^\circ) for graphite is 0 kJ/mol0 \text{ kJ/mol}, whereas for diamond it is +1.90 kJ/mol+1.90 \text{ kJ/mol}. Lower energy implies higher thermodynamic stability.

Allotropes of Carbon - Revision Notes & Key Formulas | ICSE Class 11 Chemistry