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Atoms, Elements and Compounds - Giant structures (Graphite, Diamond, Silicon Dioxide)

Grade 11IGCSEChemistry

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

🔑Concepts

Giant covalent structures, also known as macromolecules, consist of a vast number of atoms held together by strong covalent bonds in a regular lattice arrangement.

Diamond: Each carbon (CC) atom is covalently bonded to four other carbon atoms in a rigid tetrahedral structure. This results in extreme hardness and a high melting point (approx. 3550C3550^\circ C). It does not conduct electricity as there are no free electrons.

Graphite: Each carbon (CC) atom is covalently bonded to three other carbon atoms, forming hexagonal layers. One electron per carbon atom is delocalized, allowing graphite to conduct electricity. The layers are held by weak intermolecular forces, allowing them to slide over one another.

Silicon Dioxide (SiO2SiO_2): Also known as silica or quartz. It has a structure similar to diamond where each silicon (SiSi) atom is bonded to four oxygen (OO) atoms, and each oxygen atom is bonded to two silicon atoms.

Properties shared by giant structures: Very high melting and boiling points due to the large amount of energy required to break numerous strong covalent bonds, and general insolubility in water and organic solvents.

📐Formulae

C(s, diamond)C_{(s, \text{ diamond})}

C(s, graphite)C_{(s, \text{ graphite})}

SiO2SiO_2

Melting Point of Diamond3823K\text{Melting Point of Diamond} \approx 3823\,K

💡Examples

Problem 1:

Explain why graphite can conduct electricity while diamond cannot, despite both being made of carbon (CC) atoms.

Solution:

In graphite, each carbon atom forms only three covalent bonds. The fourth valence electron is delocalized and free to move throughout the structure, carrying an electrical charge. In diamond, all four valence electrons of each carbon atom are involved in covalent bonding, leaving no free electrons to conduct electricity.

Explanation:

Electrical conductivity in giant covalent structures requires the presence of mobile charge carriers, such as delocalized electrons in the π\pi-bonding systems of graphite.

Problem 2:

Describe the similarities between the structures of diamond and silicon dioxide (SiO2SiO_2).

Solution:

Both diamond and SiO2SiO_2 form giant tetrahedral lattices. In diamond, each CC atom is bonded to four other CC atoms. In SiO2SiO_2, each SiSi atom is bonded to four OO atoms in a tetrahedral arrangement, while each OO atom links two SiSi atoms.

Explanation:

Because of these extensive networks of strong covalent bonds, both substances exhibit high melting points and significant hardness.

Problem 3:

Why is graphite used as a lubricant in machinery?

Solution:

Graphite consists of layers of carbon atoms held together by weak van der Waals forces. These weak forces allow the layers to slide over each other easily when pressure is applied.

Explanation:

While the CCC-C bonds within a layer are very strong, the lack of strong bonding between the layers allows for the slippery physical property required for lubrication.

Giant structures (Graphite, Diamond, Silicon Dioxide) Revision - Grade 11 Chemistry IGCSE