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Chemical Bonding - Macromolecules

Grade 9IB

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

🔑Concepts

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Macromolecules, or giant covalent structures, consist of a vast number of atoms held together by strong covalent bonds in a regular three-dimensional lattice.

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In Diamond, each carbon atom is covalently bonded to 44 other carbon atoms in a tetrahedral arrangement. This rigid structure results in extreme hardness and a very high melting point (>3500∘C> 3500^{\circ}C).

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In Graphite, each carbon atom is bonded to 33 other carbon atoms, forming hexagonal layers. The fourth valence electron is delocalized between layers, allowing graphite to conduct electricity.

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The layers in graphite are held together by weak intermolecular forces (van der Waals forces), allowing them to slide over each other, which makes graphite soft and useful as a lubricant.

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Silicon dioxide (SiO2SiO_2), also known as silica, has a giant covalent structure similar to diamond. Each Silicon (SiSi) atom is bonded to 44 Oxygen (OO) atoms, and each Oxygen atom is bonded to 22 Silicon atoms.

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Giant covalent structures do not dissolve in water or organic solvents because the attraction between the solvent molecules and the atoms is not strong enough to break the covalent bonds.

📐Formulae

C(s,graphite)→3000∘CC(l)C_{(s, graphite)} \xrightarrow{3000^{\circ}C} C_{(l)}

SiO2SiO_2

Number of valence electrons=Group Number (for non-metals)\text{Number of valence electrons} = \text{Group Number (for non-metals)}

💡Examples

Problem 1:

Explain why Diamond does not conduct electricity, whereas Graphite does, despite both being made of Carbon atoms.

Solution:

In Diamond, all 44 valence electrons of each Carbon atom are involved in covalent bonding (sp3sp^3 hybridization), leaving no free electrons. In Graphite, only 33 out of 44 valence electrons are used for bonding (sp2sp^2 hybridization). The 4th4^{th} electron is delocalized and free to move throughout the structure.

Explanation:

Electrical conductivity requires the presence of mobile charge carriers, such as delocalized electrons or ions. Diamond lacks these, while Graphite has delocalized electrons (e−e^-).

Problem 2:

Compare the melting points of Silicon Dioxide (SiO2SiO_2) and Carbon Dioxide (CO2CO_2).

Solution:

Silicon Dioxide has a very high melting point (approx. 1710∘C1710^{\circ}C) because it is a giant macromolecular structure. Carbon Dioxide has a very low melting point (−78∘C-78^{\circ}C via sublimation) because it exists as simple discrete molecules.

Explanation:

To melt SiO2SiO_2, strong covalent bonds must be broken. To melt CO2CO_2, only weak intermolecular forces between O=C=OO=C=O molecules must be overcome, requiring much less energy (EE).