(03.06 HC)

The boiling point of bromine is 59 °C. Which of the following best predicts the boiling point of iodine monochloride, a polar compound?

Group of answer choices

Lower than 59 °C because ionic bonding in bromine is stronger than covalent bonding in iodine monochloride.

Lower than 59 °C because dipole-dipole interactions in iodine monochloride are weaker than in bromine.

Higher than 59 °C because dipole-dipole interactions in iodine monochloride are stronger than dispersion forces in bromine.

Higher than 59 °C because ionic bonding in iodine monochloride is stronger than H-bonding in bromine.

Respuesta :

Answer:

higher than 59 c because dipole- dipole interactions in iodine monocholoride are stronger than dispersion forces in bromine

The boiling point is the temperature at which the pressure of the liquid is equal to the vapor pressure. The b.p of iodine monochloride will be higher than 59 °C because dipole-dipole interactions are stronger. Thus, option C is correct.

What is a dipole-dipole interaction?

A dipole-dipole interaction is the type of electrostatic attractive force that exists between the positive and the negative side of the polar molecule. They are strong interactions between polar molecules.

The Iodine monochloride (ICl) has a higher boiling point as compared to bromine as the former is a polar molecule with a dipole-dipole bond. The bromine shows the dispersion forces that are the weakest forces.

The polar molecules have differences in their electronegativities that make them have strong dipole interactions. As a result, will need higher temperatures to break bonds.

Therefore, option C. ICl will need a higher boiling temperature than bromine due to the presence of strong dipole forces.

Learn more about dipole interactions here:

https://brainly.com/question/15887295

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