Q.Carbon monoxide ($\text{CO}$) and nitrogen ($\text{N}_2$) have almost identical molar masses ($28\ \text{g mol}^{-1}$), yet $\text{CO}$ boils at $-192^{\circ}\text{C}$ while $\text{N}_2$ boils at $-196^{\circ}\text{C}$. Explain this difference in terms of the intermolecular forces present in each.

[!TLDR] $\text{CO}$ is polar (dipole-dipole + dispersion forces); $\text{N}_2$ is non-polar (dispersion forces only), despite near-identical molar mass. [!ANSWER] $\text{CO}$ boils higher than $\text{N}_2$ because CO's extra dipole-dipole attraction (on top of similar dispersi …

Molar mass is a reasonable rough predictor of the strength of London dispersion forces (heavier, more electron-rich molecules are generally more polarisable), and since $\text{CO}$ and $\text{N}_2$ have almost identical molar masses ($28\ \text{g mol}^{-1}$ each), their dispersion forces are expected to be very similar in strength. The key difference is that carbon and oxygen have different electronegativities, so $\text{CO}$ is a polar molecule with a genuine (if modest) permanent dipole moment, giving CO molecules an additional dipole-dipole attraction to one another; nitrogen, $\text{N}_2$, is a homonuclear diatomic molecule and therefore has zero dipole moment by symmetry, relying on dispersion forces alone. This extra dipole-dipole contribution in CO, stacked on top of dispersion forces comparable to those in $\text{N}_2$, makes CO's overall inter …

Note the similar molar mass implies similar dispersion forces, then identify the polarity difference (CO polar, N2 non-polar by symmetry) as the source of th …

Attributing the boiling-point difference to molar mass, which is nearly identical for the two molecule …