The boiling point of ammonia is −33 °C while that of $\ce{HCN}$ is 25 °C. In a recent AP (Advanced Placement) Chemistry test, a free response question asked why this is the case. Can someone shine light on this?
Based on Jan's answer:
- Although the $\ce{C-H}$ bond does not usually exhibit good hydrogen bond, $\ce{H-CN}$ is a special case in which the bond is polar enough to provide better hydrogen bonding than $\ce{NH3}$. I agree with Jan on this point.
- However, Jan's explanation is for the polarity of the $\ce{H-CN}$ bond is based on $\mathrm pK_\mathrm a$ (lower $\mathrm pK_\mathrm a$ suggests higher acidity and good hydrogen bonding). As one user pointed out, $\mathrm pK_\mathrm a$ is only defined in aqueous solution, so stating pK$_a$ as an evidence is not suitable for the system I have which is a pure $\ce{HCN}$ liquid. In such system, $\ce{HCN}$ does not significantly dissociate. A suggestion was made on using proton affinity as the evidence for the acidity of $\ce{HCN}$.
I perceive that the answers (reasons and evidence) are mostly targeted at the chemical process of dissociation of $\ce{HCN}$. Should this question be addressed more in terms of a physical process? I mean there is little dissociation of pure-liquid $\ce{HCN}$.
What I have been thinking about is a MO picture in which the cyano group withdraws electron density from the hydrogen atom, in a similar way to how acetate group withdraws electron from the hydrogen in acetic acid. What is your thought on this?