Final answer:
The n-hexane and n-octane pair as well as the I2 and CCl4 pair primarily experience London dispersion forces due to their nonpolar nature. Methanol (CH3OH) and ethanol (C2H5OH), being polar substances, primarily exhibit hydrogen bonding as the key intermolecular attractive interaction.
Step-by-step explanation:
When considering the most important types of intermolecular attractive interactions between the given pairs of substances, it's fundamental to take into account the molecular structures and properties of the substances involved. In the case of n-hexane and n-octane, both are nonpolar alkanes which mean they will predominantly exhibit London dispersion forces as their intermolecular attraction. This is due to the instantaneous and induced dipoles that occur as the electrons move around the nuclei of these molecules.
Regarding the pair involving I2 and CCl4, both of these molecules are also nonpolar; thus, they will similarly experience London dispersion forces as the primary type of intermolecular interaction. London dispersion forces are generally stronger in molecules with more significant molecular masses and surface areas, which implies that these forces are the most robust amongst larger and more elongated nonpolar molecules.
Last but not least, considering substances like methanol CH3OH and ethanol C2H5OH, which both have polar molecules with hydrogen bonding capabilities, the most significant intermolecular force would be hydrogen bonding. This type of interaction is particularly strong compared to other dipole-dipole interactions and London dispersion forces due to the highly electronegative atoms like oxygen creating a large dipole and the hydrogen being attracted to the lone pair electrons on the oxygen.