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What is the strongest type of intermolecular force present in CHF3?

A) ion-dipole
B) dispersion
C) hydrogen bonding
D) dipole-dipole
E) none of the above

User Arbaz Alam
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2 Answers

4 votes

Final answer:

The strongest intermolecular force in CHF3 is the dispersion force, despite the presence of minimal polarity in the molecule, because the molecule lacks the necessary features for stronger dipole-dipole interactions or hydrogen bonding.

Step-by-step explanation:

The strongest type of intermolecular force present in CHF3 (trifluoromethane) is dipole-dipole interactions. This is because CHF3 is a polar molecule with a permanent dipole due to the electronegative fluorine atoms creating a separation of positive and negative charges within the molecule. These dipoles interact electrostatically with each other, which are generally stronger than London dispersion forces that are present due to temporary fluctuations in electron charge distribution.

However, in the case of CHF3, dispersion forces are the most significant intermolecular force rather than dipole-dipole interactions or hydrogen bonding. This is because the C-H bond while polar, is only minimally polar and not strong enough to produce a significant dipole-dipole interaction. Also, CHF3 does not have the requisite F-H, O-H, or N-H bonds necessary for hydrogen bonding, which is why C-HF3 mainly exhibits dispersion forces.

User Nikita Mazur
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5.0k points
4 votes

Answer:

D) dipole-dipole

Step-by-step explanation:

  • The strongest type of intermolecular force present in CHF3 is a dipole-dipole interaction.
  • CHF3 is a polar molecule and has a dipole-dipole interaction along with the weak dispersion forces.
  • Fluorine is highly electronegative, therefore, it will attract the electrons to itself, leaving the hydrogen end of the molecule positive.
  • This allows for other CHF3 molecules to be attracted to the positive side of the molecule, by allowing the negatively-charged fluorine side get attracted to the hydrogen side of the molecule.
User William Roberts
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