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Consider the substances hydrogen [tex]$\left( H_2 \right)$[/tex], fluorine [tex]$\left( F_2 \right)$[/tex], and hydrogen fluoride [tex]$( HF )$[/tex]. Based on their molecular structures, how does the boiling point of HF compare with the boiling points of [tex]$H_2$[/tex] and [tex]$F_2$[/tex]?

The boiling point of HF is [tex]$\square$[/tex] the boiling point of [tex]$H_2$[/tex], and it is [tex]$\square$[/tex] the boiling point of [tex]$F_2$[/tex].

Sagot :

GinnyAnswer
The boiling point of a substance is greatly influenced by the type and strength of intermolecular forces present in the substance.

1. Hydrogen ([tex]$H_2$[/tex]): Hydrogen molecules exhibit weak London dispersion forces as they are non-polar and there are no permanent dipoles.

2. Fluorine ([tex]$F_2$[/tex]): Similar to hydrogen, fluorine molecules are also non-polar and experience London dispersion forces, which are relatively weak.

3. Hydrogen fluoride (HF): Hydrogen fluoride has very strong hydrogen bonding due to the high electronegativity of fluorine, which makes the H-F bond highly polar. Hydrogen bonds are much stronger compared to London dispersion forces.

Given these points, we can compare the boiling points:

- The boiling point of HF is much higher than the boiling point of [tex]$H_2$[/tex], because hydrogen bonding is significantly stronger than the weak dispersion forces in [tex]$H_2$[/tex].
- The boiling point of HF is also higher than the boiling point of [tex]$F_2$[/tex], for the same reason: hydrogen bonding in HF is stronger than the dispersion forces in [tex]$F_2$[/tex].

Therefore, the correct answers from each drop-down menu are:
- 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].
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