# Relative strength of intermolecular forces [duplicate]

My attempt at a solution:

First I look for hydrogen bonds:

The only compounds with hydrogen bonds are the second and fourth ones, so they have the strongest intermolecular forces. The second must have stronger intermolecular forces since it harbors three hydrogen bonds, while the fourth only has room for two.

Then I look for dipole-dipole interactions:

The first compound has the potential for a dipole-dipole moment. It would be the third strongest.

Then I look for potential for London-dispersion forces:

The third compound and The fourth experience London-dispersion forces. But I'm stuck on discerning which is stronger.

So, my working solution is: 3, 5, x, 4, x

• HINT: Induced dipole forces depend upon surface area of the compound. Aug 27, 2020 at 2:04

I would say that your analysis of first, second and fourth one is correct, as H-bonds are stronger than dipole-dipole interactions. However, you made a mistake while filtering compounds having dipole-dipole interaction. The $$1$$ and $$3$$, both have dipoles!
The $$\ce{C=O}$$ bonds in $$\ce{CO2}$$ are polar and hence they possess dipoles, but unlike a water molecule, $$\ce{CO2}$$ shape is linear, and therefore the equal dipoles cancel each other, resulting in a non-polar molecule as seen in the following figure,
That's the reason $$\ce{CO2}$$ possess dipole-dipole interactions whereas $$\ce{F2}$$ doesn't as it has only one bond which is non-polar!