Which of the following would most likely be a gas at room temperature? Justify your choice. (a) \(\mathrm{NaCl}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{2}\) (c) Na metal (d) \(\mathrm{CH}_{3} \mathrm{~F}\)

Short Answer

Expert verified
The most likely substance to be a gas at room temperature is \(\mathrm{C}_{2} \mathrm{H}_{2}\) due to its weak Vander Waals dispersion forces. Other substances such as \(\mathrm{NaCl}\), Na metal, and \(\mathrm{CH}_{3} \mathrm{~F}\) have stronger types of bonding and intermolecular forces, resulting in higher melting and boiling points which make them less likely to exist as gases at room temperature.

Step by step solution

01

Identify bonding type and intermolecular forces in each substance

(a) \(\mathrm{NaCl}\): Ionic bonding - very strong electrostatic attraction between positively charged metal cations (Na+) and negatively charged nonmetal anions (Cl-). (b) \(\mathrm{C}_{2} \mathrm{H}_{2}\): Covalent bonding - consists of carbon-hydrogen bonds, primarily experiencing Vander Waals dispersion forces which are very weak. (c) Na metal: Metallic bonding - consists of a sea of delocalized electrons surrounding closely packed metal cations. (d) \(\mathrm{CH}_{3} \mathrm{~F}\): Covalent bonding - consists of carbon-hydrogen and carbon-fluorine bonds, experiencing Vander Waals dispersion forces as well as dipole-dipole interactions due to the polar nature of the carbon-fluorine bond.
02

Compare the forces present in each substance

(a) \(\mathrm{NaCl}\): Ionic bonding is very strong, resulting in a high melting and boiling point. This substance is unlikely to be a gas at room temperature. (b) \(\mathrm{C}_{2} \mathrm{H}_{2}\): Vander Waals dispersion forces are very weak, which often result in low melting and boiling points. This substance has a high possibility of being a gas at room temperature. (c) Na metal: Metallic bonding is strong and results in high melting and boiling points for metals. This substance is unlikely to be a gas at room temperature. (d) \(\mathrm{CH}_{3} \mathrm{~F}\): The presence of dipole-dipole interactions due to the polar nature of the carbon-fluorine bond makes the intermolecular forces stronger than in \(\mathrm{C}_{2} \mathrm{H}_{2}\). However, these forces are still weaker compared to ionic or metallic bonding. This substance could be a gas at room temperature, but less likely than \(\mathrm{C}_{2} \mathrm{H}_{2}\).
03

Choose the most likely substance to be a gas at room temperature

Considering the type of bonding and the intermolecular forces present in each of the given substances, \(\mathrm{C}_{2} \mathrm{H}_{2}\), with its weak Vander Waals dispersion forces, is the most likely substance to be a gas at room temperature.

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