14
$\begingroup$

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?

enter image description here

$\endgroup$
  • $\begingroup$ Think about - lone pair electrons require greater stabilization (more s-character in the orbital) than bonding pair electrons. $\endgroup$ – ron Sep 28 '17 at 23:00
  • $\begingroup$ 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 $\endgroup$ – Uday Feb 9 '18 at 15:07
  • $\begingroup$ 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." $\endgroup$ – jonsca Feb 9 '18 at 23:38
  • 1
    $\begingroup$ pubs.acs.org/doi/abs/10.1021/ed077p1062 $\endgroup$ – arya_stark Feb 27 '18 at 5:27
1
$\begingroup$

If you look at an orbital diagram of nitrogen you will see that each p-orbital has one electron in it, if you want to add a second electron to any one of them, it makes the orbital higher energy. Atoms would generally like to be in the lowest possible energy state, and half or completely filled shells are far more low energy, and thus, more stable, this is most clearly seen with the electron configuration of copper and chromium.

Cu: [Ar] 4s1 3d10

Cr: [Ar] 4s1 3d5

Nitrogen has a half filled 2p sub level, so it has to move to a substantially higher energy state to add a fourth electron. If it gets one in a bond with Hydrogen, Nitrogen finds it energetically cheaper to dissolve the bond and return to its base configuration. The reverse applies to Carbon. Because it now is able to get another electron from Hydrogen, it now has a half filled 2p sub-level and is in a fairly stable configuration that it would like to keep.

Most of this comes from Chemistry by Maciel, Traficante, and Lavallee. The book was published in 1978, so I am sorry if anything is dated.

$\endgroup$
  • $\begingroup$ But this comparison isn't fair because you are looking at a single bond, possibly in some molecule. Clearly, in the usual molecule, like methane or ammonia, the 2p subshell would technically be fully populated. But even if you were to take this into consideration, your explanation is still flawed because bonding involves hybrid orbitals and/or molecular orbitals. So it is incorrect to still talk about the atomic orbitals. $\endgroup$ – Tan Yong Boon Sep 27 '18 at 1:24
-1
$\begingroup$

This is because there is high electron charge density on nitigen atom. It repels the electrons in sigma bond causing the bond length to increase. So the NH bond weakens.

$\endgroup$
  • $\begingroup$ Do you have references that support your claim? The bond lengths in $\ce{CH4}$ ($\pu{108.7 pm}$), $\ce{NH3}$ ($\pu{101.7 pm}$) and $\ce{H2O}$ ($\pu{95.8 pm}$) seem to contradict you. Additionally, $\ce{H2O}$ might have an even higher electron density on the oxygen atom. $\endgroup$ – aventurin Aug 13 '18 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.