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. \(\mathrm{XeF}_{4}\)

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.

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.