Question

If a molecule can hydrogen bond, does it guarantee that it will have a higher boiling point than a molecule that cannot? Explain why notb) If a molecule has a stronger dipole moment than another molecule, does it guarantee it will have a higher boiling point? Explain why not

Answer 1

Answer:

a): not necessarily due to London Dispersion Forces and dipole-dipole interactions.  

b): not necessarily due to London Dispersion Forces.

Explanation:

There are three major types of intermolecular interaction:

• Hydrogen bonding between molecules with H-O, H-N, or H-F bonds and molecules with lone pairs.
• Dipole-dipole interactions between all molecules.
• London dispersion forces between all molecules.

The melting point of a substance is a result of all three forces, combined.

Note that the more electrons in each molecule, the stronger the London Dispersion Force. Generally, that means the more atoms in each molecule, the stronger the London dispersion force. The strength of London dispersion force between large molecules can be surprisingly strong.

For example, $\rm H_2O$ (water) molecules are capable of hydrogen bonding. The melting point of $\rm H_2O$ at $\rm 1\; atm$ is around $0 \; ^{\circ}\rm C$. That's considerably high when compared to other three-atom molecules.

In comparison, the higher alkane hexadecane ($\rm C_{16}H_{34}$, straight-chain) isn't capable of hydrogen bonding. However, under a similar pressure, hexadecane melts at around $18\; ^{\circ}\rm C$ above the melting point of water. The reason is that with such a large number of atoms (and hence electrons) per molecule, the London dispersion force between hexadecane molecules could well be stronger than that the hydrogen bonding between water molecules.

Similarly, the dipole moments in HCl (due to the highly-polar H-Cl bonds) are much stronger than those in hexadecane (due to the C-H bonds.) However, the boiling point of hexadecane under standard conditions is much higher (at around $287\; \rm ^\circ C$ than that of HCl.