Problem 131 - Chemistry

Question

Which of the following compound(s) exhibit only London dispersion intermolecular forces? Which compound(s) exhibit hydrogen-bonding forces? Considering only the compounds without hydrogen-bonding interactions, which compounds have dipole–dipole intermolecular forces? a. \(\mathrm{SF}_{4}\) b. \(\mathrm{CO}_{2}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) d. \(\mathrm{HF}\) e. \(\mathrm{ICl}_{5}\) f. \(\mathrm{XeF}_{4}\)

Step-by-Step Solution

Answer

In summary, compounds c (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\)) and d (\(\mathrm{HF}\)) exhibit hydrogen-bonding forces, compound a (\(\mathrm{SF}_{4}\)) exhibits dipole-dipole forces, and compounds b (\(\mathrm{CO}_{2}\)), e (\(\mathrm{ICl}_{5}\)), and f (\(\mathrm{XeF}_{4}\)) exhibit only London dispersion forces.

Step 1: Identifying Hydrogen-Bonding Forces

To determine if a compound exhibits hydrogen-bonding forces, we need to check for polar covalent bonds involving hydrogen and highly electronegative atoms like fluorine, nitrogen, or oxygen. a. \(\mathrm{SF}_{4}\) - No hydrogen atoms present. b. \(\mathrm{CO}_{2}\) - No hydrogen atoms present. c. \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\) - The O-H bond has hydrogen bonding force. d. \(\mathrm{HF}\) - The H-F bond has hydrogen bonding force. e. \(\mathrm{ICl}_{5}\) - No hydrogen atoms present. f. \(\mathrm{XeF}_{4}\) - No hydrogen atoms present. Thus, compounds c and d exhibit hydrogen-bonding forces.

Step 2: Identifying London Dispersion Forces

As mentioned earlier, every compound exhibits London dispersion forces.

Step 3: Identifying Dipole-Dipole Forces

To determine if a compound has dipole-dipole intermolecular forces, we need to ensure the compound is somewhat polar. We are only considering compounds without hydrogen-bonding forces (a, b, e, f). a. \(\mathrm{SF}_{4}\) - The molecule is polar, exhibiting dipole-dipole forces. b. \(\mathrm{CO}_{2}\) - The molecule is non-polar, so it only exhibits London dispersion forces. e. \(\mathrm{ICl}_{5}\) - The molecule is non-polar, so it only exhibits London dispersion forces. f. \(\mathrm{XeF}_{4}\) - The molecule is non-polar, so it only exhibits London dispersion forces. Only compound a, \(\mathrm{SF}_{4}\), exhibits dipole-dipole intermolecular forces. In summary: Hydrogen-bonding forces: c and d Dipole-dipole forces: a London dispersion forces only: b, e, and f

Key Concepts

London Dispersion Forces

London dispersion forces, often called Van der Waals forces, are the weakest form of intermolecular attraction. They arise because of temporary shifts in electron density within molecules. This creates a momentary polarity, causing molecules to be attracted to one another.

These forces are universal, existing in all atoms and molecules, whether they are polar or non-polar. However, they are particularly significant in non-polar compounds like

• \( \mathrm{CO}_{2} \),
• \( \mathrm{ICl}_{5} \), and
• \( \mathrm{XeF}_{4} \).

These molecules lack a permanent dipole moment, so London dispersion forces become crucial in holding them together.

Factors affecting these forces include:

• The size of the molecule: Larger molecules have more electrons, which can lead to stronger dispersion forces.
• The shape of the molecule: More surface area allows molecules to better "stick" to each other.

Thus, even though London dispersion forces are weak, they play a vital role, particularly in large or heavy molecules.

Hydrogen Bonding

Hydrogen bonding is a type of strong dipole-dipole interaction. It occurs specifically when hydrogen is directly bonded to highly electronegative elements such as nitrogen, oxygen, or fluorine. The presence of these elements creates a significant electronegativity difference, leading to a highly polar bond.

This interaction can be seen in compounds such as:

• \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \) (ethanol), where the O-H bond provides sites for hydrogen bonding.
• \( \mathrm{HF} \), where the H-F bond results in strong hydrogen bonding.

The strength of hydrogen bonds gives these molecules unique properties, like higher boiling and melting points compared to molecules primarily joined by other forces. For example, water, with extensive hydrogen bonding, has a high boiling point compared to other molecules of similar size.

Understanding hydrogen bonding helps explain why certain substances dissolve in water and why some liquids are more viscous or have higher surface tension.

Dipole-Dipole Interactions

When molecules contain polar bonds, they can exhibit dipole-dipole interactions due to their permanent dipole moments. This type of intermolecular force occurs between the positive end of one polar molecule and the negative end of another, leading to an attraction between the two.

In our example, \( \mathrm{SF}_{4} \) shows this kind of interaction because the molecule is polar. The lone pairs and the arrangement of fluorine atoms create an uneven charge distribution, allowing dipole-dipole interactions to occur.

Dipole-dipole forces are generally stronger than London dispersion forces but weaker than hydrogen bonds. They play an important role in determining the physical properties of substances. For instance, they can influence the boiling and melting points of polar compounds, usually resulting in higher values compared to non-polar substances held only by London dispersion forces. Understanding these interactions gives insight into molecular behavior in different chemical environments.