To compare the melting points of butane [tex]\(\left( C_4H_{10} \right)\)[/tex] and octane [tex]\(\left( C_8H_{18} \right)\)[/tex], we need to understand some basic principles about hydrocarbons and their physical properties.
1. Molecular Size and Structure:
- Butane: A smaller hydrocarbon with the molecular formula [tex]\(C_4H_{10}\)[/tex].
- Octane: A larger hydrocarbon with the molecular formula [tex]\(C_8H_{18}\)[/tex].
2. Intermolecular Forces:
- In hydrocarbons, the primary intermolecular force is the van der Waals force (also known as London dispersion forces), which are generally stronger in larger molecules because there are more electrons and a larger surface area for these forces to act.
3. General Trend for Hydrocarbons:
- As the molecular size and mass increase, the strength of the van der Waals forces also increases. This means that larger hydrocarbons typically have higher melting points compared to smaller hydrocarbons because more energy is required to overcome these intermolecular forces.
Given these principles:
- Butane, being a smaller molecule with fewer carbon atoms, typically has weaker van der Waals forces and thus a lower melting point.
- Octane, being a larger molecule with more carbon atoms, has stronger van der Waals forces and thus a higher melting point.
Based on this understanding, we can infer that:
The correct statement is: The melting point is higher for octane.
Thus, the correct choice in the provided options is:
- The melting point is higher for octane.
