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

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.

As mentioned earlier, every compound exhibits London dispersion 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