Problem 121 - Chemistry An Atoms First Approach

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

The compounds that exhibit only London Dispersion forces are CO2, ICl5, and XeF4. The compounds that exhibit Hydrogen bonding forces are CH3CH2OH and HF. Among the compounds without Hydrogen bonding interactions, only SF4 exhibits Dipole-Dipole intermolecular forces.

Step 1: Identify molecular polarities

First, let's identify whether each of the molecules is polar or nonpolar. To do this, we need to consider each compound's molecular geometry and electronegativity difference. a. SF4: This molecule is polar since its molecular geometry (see-saw) results in a net dipole moment. b. CO2: This molecule is nonpolar, despite the C=O bond being polar. The molecular geometry (linear) results in the cancellation of the dipoles. c. CH3CH2OH: The molecule is polar, due to the presence of the polar O-H bond. The molecular geometry results in a net dipole moment. d. HF: The molecule is polar due to the presence of the highly electronegative fluorine atom. The linear geometry results in a net dipole moment. e. ICl5: This molecule is polar, although no dipole moment emerges since it has a nonpolar geometry (square pyramidal). f. XeF4: This molecule is nonpolar. It has a planar square geometry, and the dipoles cancel out.

Step 2: Determine the intermolecular forces

Now, let's determine the types of IMFs present in each molecule, based on their polarities and the presence of hydrogen bonding. a. SF4: Polar molecule with no H-bonding atoms. So, it has London Dispersion + Dipole-Dipole forces. b. CO2: Nonpolar molecule with no H-bonding atoms. So, it has only London Dispersion forces. c. CH3CH2OH: Polar molecule with an O-H bond (H-bond donor). So, it has London Dispersion + Dipole-Dipole + Hydrogen Bonding forces. d. HF: Polar molecule with an H-F bond (H-bond donor). So, it has London Dispersion + Dipole-Dipole + Hydrogen Bonding forces. e. ICl5: Nonpolar molecule with no H-bonding atoms. So, it has only London Dispersion forces. f. XeF4: Nonpolar molecule with no H-bonding atoms. So, it has only London Dispersion forces. To summarize and answer the initial questions: 1. Compounds that exhibit only London Dispersion forces: CO2, ICl5, and XeF4. 2. Compounds that exhibit Hydrogen bonding forces: CH3CH2OH, and HF. 3. Compounds without Hydrogen bonding interactions but with Dipole-Dipole forces: SF4.

Key Concepts

London Dispersion Forces

London Dispersion Forces, sometimes referred to as Van der Waals forces, are one of the weakest intermolecular forces. Yet, they play a crucial role in the interaction between molecules, especially for nonpolar molecules. These forces arise from the temporary dipoles which occur when electron clouds around an atom or molecule are distorted. This distortion leads to a temporary dipole in one molecule, inducing another dipole in a neighboring molecule, resulting in a weak attraction.

These forces are universal and occur in all molecular interactions, but are the only forces present in nonpolar molecules. For example, molecules like \(\mathrm{CO}_2\), \(\mathrm{ICl}_5\), and \(\mathrm{XeF}_4\) primarily exhibit London Dispersion Forces. Despite their weak nature, these forces are significant when large numbers of interactions occur, allowing substances like gases to condense into liquids and solids at low temperatures. Remember, the bigger the electron cloud, the stronger the London Dispersion Forces, which is why heavier atoms or molecules typically exhibit stronger attractions.

Hydrogen Bonding

Hydrogen Bonding is a special type of dipole-dipole interaction. It occurs when hydrogen is covalently bonded to a highly electronegative atom, such as oxygen, nitrogen, or fluorine. This makes the hydrogen atom highly electropositive, allowing it to be attracted towards lone pairs of electrons on neighboring electronegative atoms. This type of bonding is much stronger than regular dipole-dipole interactions but still weaker than covalent or ionic bonds.

In this scenario, compounds like \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}\) and \(\mathrm{HF}\) exhibit hydrogen bonding due to the presence of the O-H and H-F bonds respectively. This type of interaction significantly affects the properties of a substance, such as increasing its boiling and melting points. Hydrogen bonds are what give water its high boiling point relative to other molecules of similar size. Therefore, the presence of hydrogen bonding in a molecule implies stronger interactions within molecular assemblies.

Dipole-Dipole Interactions

Dipole-Dipole Interactions are attractive forces that occur between the positive end of one polar molecule and the negative end of another. These interactions are generally stronger than London Dispersion Forces, though not as strong as hydrogen bonds. They come into play when molecules have a permanent dipole moment due to differences in electronegativity between the atoms forming the molecules.

For molecules without hydrogen bonding interactions, those that still have a net dipole can exhibit dipole-dipole interactions. For instance, \(\mathrm{SF}_4\) has these interactions due to its polar nature and see-saw geometry. These forces cause deviations from ideal behavior in gases and significantly influence the boiling and melting points of compounds. Moreover, even though \(\mathrm{SF}_4\) does not have hydrogen bonds, its dipole-dipole interactions along with London Dispersion Forces account for its intermolecular characteristics. Hence, in polar molecules, dipole-dipole interactions act alongside any other forces at play to dictate physical properties.