Question

Rank the following compounds in order of decreasing boiling point: potassium fluoride (KF), carbon tetrafluoride (CF4), and iodomethane (CH3I) Rank from highest to lowest boiling point. To rank items as equivalent, overlap them.

Answer 1

In decreasing order of boiling points, dimethyl sulfoxide has the highest boiling point due to its polarity, followed by ethyl methyl sulfide and 2-methylbutane based on their molecular sizes and London dispersion forces, with carbon tetrafluoride having the lowest.

Step-by-step explanation:

To determine the boiling point of these compounds, we need to consider intermolecular forces. For carbon tetrafluoride (CF4), only London dispersion forces are present as it is a nonpolar molecule. Ethyl methyl sulfide (CH3SC2H5), also presents London dispersion forces, yet it will have a higher boiling point than CF4 due to its greater molecular weight and size, which contribute to stronger London forces. Dimethyl sulfoxide ((CH3)2S=O) is a polar molecule due to the presence of a sulfoxide group, which induces dipole-dipole interactions in addition to London dispersion forces, therefore, it will have an even higher boiling point. 2-methylbutane is a nonpolar molecule like CF4 but with a larger size, contributing to stronger London forces.

The boiling point ranking from highest to lowest is:

1. Dimethyl sulfoxide ((CH3)2S=O)
2. Ethyl methyl sulfide (CH3SC2H5)
3. 2-methylbutane (isopentane, (CH3)2CHCH2CH3)
4. Carbon tetrafluoride (CF4)

Answer 2

Step-by-step explanation:

Ionic compounds have high boiling point due to the presence of strong intermolecular forces present within the molecules due to the partial opposite charges on its atoms.

For example, KF is an ionic compound so it has high boiling point.

Whereas covalent compounds have weak forces because of sharing of electrons taking place between the molecules of a substance.

Whereas
`CF_(4)` is a covalent compound as both the atoms are non-metals. Hence, it has low boiling point.

In
`CH_(3)I`, there is dipole moment. Hence, it will have a boiling point greater than
`CF_(4)`.

Therefore, we can conclude that order of decreasing boiling point of given compounds is as follows.

`KF > CH_(3) > CF_(4)`