What type(s) of intermolecular force is (are) common to each of the following. (Select all that apply.)

1) Xe and methanol (CH3OH)
a)london-dispersion forces
b)dipole-dipole
c)hydrogen bonding

i put a and b but it was wrong

2)CH3OH and acetonitrile (CH3CN)
a)london-dispersion forces
b)dipole-dipole
c)hydrogen bonding

i got a b and c but it was wrong.

please explain why the answer that is correct is correct

thanks:):)

Both have London forces.

Xe can't be dipole-dipole although it can be dipole-induced dipole and your instructor might be counting that as b (but I don't think that is very probable). CH3OH has hydrogen bonds; obviously Xe has no H and it can't have hydrogen bonding. So I would go with a only.

For b.
Everything has London forces.
Both are polar (therefore dipoles) but not strong ones.
CH3OH forms hydrogen bonds with other molecules of CH3OH. I don't think CH3CN is likely to form H bonds with itself so I would go with a and b.

To determine the types of intermolecular forces present in each of the given scenarios, we need to analyze the chemical structures and polarities of the molecules involved.

1) Xe and methanol (CH3OH):
Xe (Xenon) is a noble gas that exists as individual atoms, which are nonpolar. Methanol (CH3OH) is a polar molecule since it has a slightly positive charge on the hydrogen atoms and a slightly negative charge on the oxygen atom.

The only intermolecular force that can occur between Xe and methanol is London-dispersion forces. London-dispersion forces are temporary shifts in electron distribution that lead to an instantaneous dipole, even in nonpolar molecules. These forces occur between all molecules, but they are generally weaker than other intermolecular forces.

So, the correct answer for this scenario is (a) London-dispersion forces.

2) CH3OH and acetonitrile (CH3CN):
Both CH3OH and CH3CN are polar molecules due to differences in electronegativity between the atoms involved.

In this case, there are three types of intermolecular forces to consider:
- London-dispersion forces: Present in all molecules, including polar ones.
- Dipole-dipole interactions: These occur between polar molecules, where the positive end of one molecule attracts the negative end of another molecule.
- Hydrogen bonding: A special type of dipole-dipole interaction that occurs when hydrogen is bonded directly to a highly electronegative atom, such as oxygen, nitrogen, or fluorine.

In CH3OH, the hydrogen atom is bonded to an oxygen atom, making it capable of forming hydrogen bonds. In CH3CN, there is no hydrogen atom directly bonded to an electronegative atom, so hydrogen bonding is not possible.

Therefore, the correct answer for this scenario is (a) London-dispersion forces and (b) dipole-dipole interactions, omitting the option for hydrogen bonding.

1) Xe and methanol (CH3OH)

The correct answer is (a) london-dispersion forces.

Xe is an element and exists as individual atoms. It is a noble gas with a full valence electron shell, so it does not have a permanent dipole moment. Therefore, there are no dipole-dipole forces or hydrogen bonding between Xe atoms.

Methanol (CH3OH) is a small polar molecule with a slightly positive carbon and slightly negative oxygen. It can form hydrogen bonding due to the presence of an electronegative (oxygen) atom bonded to a hydrogen atom.

However, since Xe does not have a permanent dipole moment, there are no dipole-dipole forces or hydrogen bonding between Xe and methanol. The only intermolecular force present between Xe and methanol is london-dispersion forces, which result from temporary fluctuations in the electron distribution that induce a temporary dipole in nearby molecules.

2) CH3OH and acetonitrile (CH3CN)

The correct answer is (a) london-dispersion forces.

Both CH3OH and CH3CN are polar molecules since they have atoms with different electronegativities that create permanent dipole moments.

CH3OH can form hydrogen bonding due to the presence of an electronegative oxygen atom bonded to a hydrogen atom.

CH3CN cannot form hydrogen bonding since it does not have an electronegative atom bonded to a hydrogen atom.

However, both CH3OH and CH3CN experience london-dispersion forces, which are present in all molecules. London-dispersion forces result from temporary fluctuations in the electron distribution that induce temporary dipoles in nearby molecules.