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The boiling point of chlorine is −34 °C. Which of the following best predicts the boiling point of iodine?

a) Higher than −34 °C because dispersion forces are stronger in iodine due to a greater number of electrons.

b) Lower than −34 °C because chlorine is more polar than iodine on account of its higher electronegativity.

c) Higher than −34 °C because dipole-dipole interactions in iodine are stronger than dispersion forces in chlorine.

d) Lower than −34 °C because permanent dipoles created in chlorine are stronger than temporary dipoles in iodine.

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Answer:

a) Higher than −34 °C because dispersion forces are stronger in iodine due to a greater number of electrons.

Explanation:

Both chlorine and Iodine are halogens. They belong to the 7th group on the periodic table.

To examine the trend of boiling point in this group we must consider the nature of the intermolecular bonds that would be formed by the molecules of these elements. We know that London dispersion forces, a type of van der Waals attraction would be more prevalent. This is so because the molecules here would be non-polar and the uneven distribution of the constantly moving electrons would initiate the intermolecular bonding.

The uneven charge distribution leads to the formation of a temporary dipole.

This makes boiling point increase down the group because more electrons becomes involved and the bond becomes stronger.

Answer:

Higher than −34 °C because dispersion forces are stronger in iodine due to a greater number of electrons.

Explanation:

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