3 Nitrogen, \( \mathrm{N}_{2} \), and carbon monoxide, CO , both have \( \mathrm{M}_{\mathrm{r}}=28 \). The boiling point of \( \mathrm{N}_{2} \) is 77 K . The boiling point of CO is 82 K . What could be responsible for this difference in boiling points? A CO molecules have a permanent dipole, the \( \mathrm{N}_{2} \) molecules are not B \( \mathrm{N}_{2} \) has \( \sigma \) and \( \pi \) bonding, CO has \( \sigma \) bonding only. C \( \mathrm{N}_{2} \) has a strong \( \mathrm{N}=\mathrm{N} \) bond, CO has a \( \mathrm{C}=\mathrm{O} \) bond. D The CO molecule has more electrons than the \( \mathrm{N}_{2} \) molecule.

1. Both \(\mathrm{N}_2\) and CO have the same molecular weight (\(M_r = 28\)), so the dominant intermolecular forces in the absence of any polarity would be dispersion forces. 2. \(\mathrm{N}_2\) is a homonuclear molecule and, therefore, nonpolar. This means it only exhibits London dispersion forces. 3. CO, in contrast, is a heteronuclear molecule. It has a slight separation of charge (a permanent dipole moment) because the electronegativity of oxygen is greater than that of carbon. In addition to dispersion forces, CO molecules experience dipole-dipole interactions. 4. The presence of dipole-dipole interactions in CO leads to stronger intermolecular attractions compared to the nonpolar \(\mathrm{N}_2\). Stronger attraction between molecules corresponds to a higher boiling point. 5. Therefore, the difference in boiling points (77 K for \(\mathrm{N}_2\) and 82 K for CO) is primarily due to the fact that CO molecules have a permanent dipole, whereas \(\mathrm{N}_2\) molecules do not. The correct answer is: A CO molecules have a permanent dipole, the \(\mathrm{N}_2\) molecules are not.