To determine how the boiling point of hydrogen fluoride (HF) compares with the boiling points of hydrogen ([tex]$H_2$[/tex]) and fluorine ([tex]$F_2$[/tex]), we need to review the nature of their intermolecular forces.
1. Hydrogen ([tex]$H_2$[/tex]): Hydrogen is a non-polar diatomic molecule. The primary intermolecular forces here are weak Van der Waals forces (also known as London dispersion forces). These forces do not require a lot of energy to overcome, leading to a relatively low boiling point.
2. Fluorine ([tex]$F_2$[/tex]): Similar to hydrogen, fluorine is also a non-polar diatomic molecule and has weak Van der Waals forces. Therefore, it also has a relatively low boiling point.
3. Hydrogen Fluoride (HF): Hydrogen fluoride has a significantly different type of intermolecular force. HF molecules are polar and can form hydrogen bonds, which are much stronger than Van der Waals forces. The presence of hydrogen bonding in HF requires more energy to break these bonds during the phase transition from liquid to gas, resulting in a higher boiling point.
Based on this understanding:
- The boiling point of HF will be higher than the boiling point of [tex]$H_2$[/tex] due to the strong hydrogen bonding in HF compared to the weak Van der Waals forces in [tex]$H_2$[/tex].
- Similarly, the boiling point of HF will be higher than the boiling point of [tex]$F_2$[/tex] for the same reason, as [tex]$F_2$[/tex] only has weak Van der Waals forces.
Therefore, the correct answers are:
- The boiling point of HF is higher than the boiling point of [tex]$H_2$[/tex].
- The boiling point of HF is higher than the boiling point of [tex]$F_2$[/tex].
So, the final selection should be:
- The boiling point of HF is higher than the boiling point of [tex]$H_2$[/tex], and it is higher than the boiling point of [tex]$F_2$[/tex].
