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Consider the substances hydrogen [tex]\(\left( H_2 \right)\)[/tex], fluorine [tex]\(\left( F_2 \right)\)[/tex], and hydrogen fluoride (HF). 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 :

GinnyAnsweridn
Sure! Let's compare the boiling points of the substances hydrogen [tex]\((H_2)\)[/tex], fluorine [tex]\((F_2)\)[/tex], and hydrogen fluoride [tex]\((HF)\)[/tex] based on their molecular structures and intermolecular forces.

1. Hydrogen (H₂):
- Hydrogen molecules are nonpolar.
- They exhibit London dispersion forces, which are relatively weak.

2. Fluorine (F₂):
- Fluorine molecules are also nonpolar.
- Like hydrogen, they exhibit London dispersion forces.
- However, fluorine has a greater molar mass compared to hydrogen.

3. Hydrogen Fluoride (HF):
- Hydrogen fluoride is a polar molecule.
- HF molecules exhibit hydrogen bonding, a much stronger intermolecular force than London dispersion forces.

Boiling Point Comparison:

- Hydrogen fluoride (HF) has hydrogen bonding, which is a strong intermolecular force. As a result, it has a higher boiling point than both hydrogen ([tex]\(H_2\)[/tex]) and fluorine ([tex]\(F_2\)[/tex]), which only exhibit the weaker London dispersion forces.
- Among nonpolar molecules, fluorine ([tex]\(F_2\)[/tex]), due to its greater molar mass, has a higher boiling point than hydrogen ([tex]\(H_2\)[/tex]).

Therefore, the boiling point of HF is greater than the boiling point of [tex]\(H_2\)[/tex], and it is also greater than the boiling point of [tex]\(F_2\)[/tex].

So the completed statement will be:
The boiling point of HF is greater than the boiling point of [tex]\(H_2\)[/tex], and it is greater than the boiling point of [tex]\(F_2\)[/tex].
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