# Why is the bond dissociation energy of C-H bond higher than that of a N-H bond?

Looking at a chart of BDEs (bond dissociation energies) regarding elements bonded with hydrogen, the general trend seems to be that the BDE increases as we go to the top and to the right. This can be explained by the fact that the elements to the right have stronger attraction for the covalent bonds, making homolytic cleavage harder. Likewise, those elements near the top have shorter radii, which means that their attraction is larger too.

However, there is a single anomaly in the $\ce{N-H}$ and $\ce{C-H}$ bonds, where the BDE of $\ce{C-H}$ is higher than $\ce{N-H}$. Why is this?

• Think about - lone pair electrons require greater stabilization (more s-character in the orbital) than bonding pair electrons. – ron Sep 28 '17 at 23:00
• It is experimental data and one could only say according to data they have. You have to be careful in comparing bond strengths even in diatomic molecules, but especially when the bond is in a polyatomic molecule. In the case of polyatomic molecules, the bond strengths are even more scattered depending upon the structure of the rest of the molecule. A somewhat more sensitive parameter of bond strength is actually the bond length (short bonds -> stronger bonds & long bonds -> weaker bonds) but data on the bond lengths in polyatomic molecules is much more scarce than bond energies because the – Uday Feb 9 '18 at 15:07
• con't from above: "detailed molecular structures are more difficult to obtain.The bond energies in CH and NH are essentially the same, but even so, the electronic states of the diatomics may be different so that comparisons may be mismatched. So I think we cannot clearly tell why it is so." – jonsca Feb 9 '18 at 23:38
• pubs.acs.org/doi/abs/10.1021/ed077p1062 – arya_stark Feb 27 '18 at 5:27