# How to predict which substance in each of the following pairs would have the greater intermolecular forces ? a) CO_2 or OCS; b) SeO_2 or SO_2; c) CH_3CH_2CH_2NH_2 or H_2NCH_2CH_2NH_2; d) CH_3CH_3 or H_2CO; e) CH_3OH or H2CO.

May 15, 2015

The strength of the intermolecular forces exhibited by a certain molecule goes hand in hand with its polarity and with its ability to form hydrogen bonds.

Right from the get-go, nonpolar molecules will have weaker intermolecular forces compared with polar molecules of comparable size.

So, here's a brief analysis of each pair (the molecule with the greater IMFs will be written in green)

• $C {O}_{2}$ and $\textcolor{g r e e n}{O C S}$

You're dealing with two linear molecules, the only difference between the two being that $C {O}_{2}$ is nonpolar, while $O C S$ is polar.

In $C {O}_{2}$'s case, the bond dipole moments are equal in magnitude and point in opposite directions, so the net dipole moment will be zero. In $O C S$'s case, the bond dipoles are not equal in magnitude because sulfur and oxygen has different electronegativity values.

• $\textcolor{g r e e n}{S e {O}_{2}}$ and $S {O}_{2}$

This one is a little more subtle. From an electronegativity stanpoint, selenium and sulfur are very similar; moreover, both molecules have a bent molecular geometry, which implies that both are polar.

However, selenium has a bigger radius than sulfur, which implies that it also has a bigger electron cloud. That translates into greater polarizability.

The positive charge that will arise on the selenium atom will be slightly bigger than that on the sulfur atom, which implies a slightly greater net dipole moment.

• $C {H}_{3} C {H}_{2} C {H}_{2} N {H}_{2}$ or $\textcolor{g r e e n}{{H}_{2} N C {H}_{2} C {H}_{2} N {H}_{2}}$

This is where the ability to form hydrogen bonds comes into play. The difference between these two amines will be made by the additional ${\text{-NH}}_{2}$ functional group present on ethylenediamine.

This second ${\text{-NH}}_{2}$ group will provide ethylenediamine with the capability to form more hydrogen bonds with neighbouring molecules when compared with propylamine, the compound that only has one ${\text{-NH}}_{2}$ group attached.

• $C {H}_{3} C {H}_{3}$ or $\textcolor{g r e e n}{{H}_{2} C O}$

Methane, or $C {H}_{3} C {H}_{3}$, is a nonpolar molecule because the $\text{C-H}$ bonds are considered to be nonpolar. As a result, methane will only exhibit weak London dispersion forces.

By comparison, the electronegative oxygen will create a permanent dipole moment on the formaldehyde molecule. This will allow the molecule to exhibit dipole-dipole interactions, in addition to the London dispersion forces that every molecule exhibits.

• $\textcolor{g r e e n}{C {H}_{3} O H}$ or ${H}_{2} C O$

Once again, this comes down to the ability to form hydrogen bonds. Both molecules exhibit London dispersion forces and dipole-dipole interactions, but the fact that ethanol, $C {H}_{3} O H$, has a hydrogen atom directly attached to an oxygen atom will allow it to engage in hydrogen bonding.