Question

At room temperature, n-pentane is a liquid and neopentane (2, 2-dimethylpropane) is a gas. Both compounds are nonpolar hydrocarbons. Which of the following forces is primarily responsible for this difference?

A. London dispersion forces
B. Hydrogen bonding
C. Dipole-dipole interactions
D. Ionic bonding

Answer 1

The difference in states between n-pentane (liquid) and neopentane (gas) at room temperature is attributed to London dispersion forces. These forces are stronger in n-pentane due to its larger surface area compared to the spherical neopentane, resulting in a higher boiling point for n-pentane.

Step-by-step explanation:

The primary force responsible for the difference in states of matter (liquid or gas) between n-pentane and neopentane at room temperature is London dispersion forces. Both compounds are nonpolar hydrocarbons that exhibit only van der Waals forces of which London dispersion is a type. The difference in physical state is due to the differences in molecular shape and surface area; n-pentane is a linear molecule with a larger surface area that can interact more with adjacent molecules, resulting in stronger dispersion forces. On the other hand, neopentane is almost spherical with a smaller surface area for intermolecular interactions, thus experiencing weaker London dispersion forces, and is a gas at room temperature with a lower boiling point of 9.5°C as opposed to n-pentane's higher boiling point of 36.1°C.