Arrange these compounds: CO2, CH3OH, RbF, CH3Br in order of increasing boiling points

Question

Arrange these compounds: $\ce{CO2}$, $\ce{CH3OH}$, $\ce{RbF}$, $\ce{CH3Br}$ in order of increasing boiling points.

I think I should consider the forces between them, that is:

• $\ce{CO2}$: dispersion forces
• $\ce{RbF}$: dispersion and ionic forces
• $\ce{CH3OH}$: Dipole-dipole interactions, Hydrogen bonding and dispersion forces
• $\ce{CH3Br}$: Dipole-dipole interactions and dispersion forces

It is obvious that $\ce{CO2}$ is the smallest one and because $\ce{CH3OH}$ stronger than $\ce{CH3Br}$ it will have higher boiling point

But how to arrange the rest? Or how to compare ionic forces with Dipole?

Answer 1

You know $\ce{CO_2}$ is gaseous at room temperature, so let's put that at the bottom. Methanol forms hydrogen bonds, so that will be above bromomethane which does not. At last we have rubidium fluoride which is a salt. Salts generally have a very high boiling point (> 1000 °C, much higher than molecular structures) because of the ionic (electrostatic) interaction between the ions, so that one will be at the top. Ionic forces can be seen as extreme dipoles in a certain way, there is a grey area when electronegativity becomes large enough, that it can be seen either as a molecular structure or ionic structure. Consulting online information about the boiling points of these compounds (i.e. just check Wikipedia or some MSDS site) confirms the theory.

Answer 2

The following is the order from lowest boiling point to highest based on the types of forces these compounds have:

$\ce{CO2}$ - dispersion forces

$\ce{CH3Br}$ - dispersion and dipole-dipole

$\ce{CH3OH}$ - dispersion, dipole-dipole, and hydrogen bonding

$\ce{RbF}$ - ionic