To answer the question about the boiling points of hydrogen ([tex]\( \text{H}_2 \)[/tex]), fluorine ([tex]\( \text{F}_2 \)[/tex]), and hydrogen fluoride ([tex]\( \text{HF} \)[/tex]), we need to consider the intermolecular forces present in each substance.
1. Hydrogen ([tex]\( \text{H}_2 \)[/tex]) and Fluorine ([tex]\( \text{F}_2 \)[/tex]):
- Both [tex]\( \text{H}_2 \)[/tex] and [tex]\( \text{F}_2 \)[/tex] are non-polar molecules.
- The primary intermolecular force in non-polar molecules is the London dispersion force (or dispersion force), which is relatively weak.
- Therefore, both [tex]\( \text{H}_2 \)[/tex] and [tex]\( \text{F}_2 \)[/tex] have relatively low boiling points.
2. Hydrogen Fluoride ([tex]\( \text{HF} \)[/tex]):
- [tex]\( \text{HF} \)[/tex] is a polar molecule and exhibits hydrogen bonding.
- Hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly electronegative atom (such as fluorine) and is attracted to another electronegative atom.
- Hydrogen bonding is one of the strongest intermolecular forces, significantly increasing the boiling point of the substance.
Given these points:
- Comparison of HF and H[tex]\(_2\)[/tex]:
- [tex]\( \text{HF} \)[/tex] has hydrogen bonding, a much stronger intermolecular force compared to the London dispersion forces in [tex]\( \text{H}_2 \)[/tex].
- Thus, the boiling point of [tex]\( \text{HF} \)[/tex] is higher than that of [tex]\( \text{H}_2 \)[/tex].
- Comparison of HF and F[tex]\(_2\)[/tex]:
- Similarly, [tex]\( \text{HF} \)[/tex] has hydrogen bonding, whereas [tex]\( \text{F}_2 \)[/tex] only has London dispersion forces.
- Therefore, the boiling point of [tex]\( \text{HF} \)[/tex] is higher than that of [tex]\( \text{F}_2 \)[/tex].
In conclusion:
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].
