To answer the question of how the boiling points of CH and CH₂CH compare based on their intermolecular forces, we need to consider the nature and strength of these forces in each molecule.
### Step-by-Step Analysis
1. Molecular Size and Structure:
- CH: This hypothetical molecule is relatively small.
- CH₂CH: This hypothetical molecule is larger than CH due to the additional functional groups attached to the base CH structure.
2. Intermolecular Forces:
- CH: Primarily experiences dispersion forces (London dispersion forces), which are the weakest type of van der Waals forces. Given that CH is a small molecule, these dispersion forces will be relatively minor.
- CH₂CH: Being a larger molecule, it still primarily experiences dispersion forces, but these forces are stronger due to the larger electron cloud and the increase in the number of atoms contributing to the temporary dipoles.
3. Boiling Point Trends:
- Molecules with stronger intermolecular forces will generally have higher boiling points, as it requires more energy to break these forces.
- Larger molecules with more complex structures tend to have higher boiling points because their larger surface areas and electron clouds create more significant dispersion forces.
### Conclusion
By comparing the sizes and types of intermolecular forces, we expect CH₂CH to have stronger intermolecular forces due to being a larger molecule with greater dispersion forces. As a result, CH would have a lower boiling point compared to CH₂CH because its intermolecular forces are weaker and easier to overcome.
Therefore, the correct choice is:
A. a lower
The detailed analysis shows that CH has a lower boiling point compared to CH₂CH based on the strength of their intermolecular forces.
