3
$\begingroup$

Order of hydrogen bond strength:

$\ce{O-H...N}$ > $\ce{O-H...O}$ > $\ce{N-H...N}$ > $\ce{N-H...O}$. Why is that so?

According to me $\ce{O-H...O}$ should have the maximum strength while $\ce{N-H...N}$ should have the weakest hydrogen bonds based on the fact that oxygen is more electronegative than nitrogen. It would develop a higher $\delta^-$ charge and the hydrogen attached to it will have a higher $\delta^+$ charge which would make the hydrogen bonding stronger.

I can't reason out the order given in my textbook. After observing it, all I can guess is that oxygen being highly electronegative leads to higher $\delta^+$ charge on hydrogen while nitrogen being less electronegative easily shares its electron cloud with hydrogen, making the hydrogen bond stronger.

I am not sure about this reason as according to this it is more stable for hydrogen to make a hydrogen bond with a less electronegative element and possibly even an electropositive element, (unless and until the hydrogen is attached to a highly electronegative element), which does not occur in normal circumstances.

It would be great if anybody could help me figure out the reason for the above trend.

||improve this question||||
$\endgroup$
  • $\begingroup$ Whoever asked this question in the first place did not understand a single thing about hydrogen bonding. There are obviously too many parameters to consider when comparing "strengths". Not even if we were to keep all distances the same the comparison would work. Any reasoning trying to reproduce this "trend" is similarly bound to fail. $\endgroup$ – Martin - マーチン♦ Dec 12 '17 at 15:24
  • 1
    $\begingroup$ I don't understand why you say that any reasoning to understand this trend is bound to fail, given that there are several ways to measure hydrogen bond strenght. For example, one can measure dissociation enthalpies in gas phase or dissociation constants in a reference solvent. On the computational side, one could use electronic interaction energies of fully optimized dimers. What is the problem with trying to rationalize hydrogen bond strength? $\endgroup$ – diogom Feb 7 '18 at 4:30
  • 1
    $\begingroup$ @Martin-マーチン This question now has ~3k views. Whichever textbook is popularising this concept, does seem to be very popular... $\endgroup$ – Gaurang Tandon Apr 5 '18 at 2:30
  • 1
    $\begingroup$ @GaurangTondon this concept is used to solve a few questions in jee, and India is full of jee aspirants which is why this question is becoming popular. $\endgroup$ – oshhh Apr 6 '18 at 4:07
0
$\begingroup$

As a short answer, Hydrogen bonds (for elementary chemistry) are formed with donation of lone pair of a lewis base into LUMO of H attached to a highly electronegative atom. A better donor makes a better bond and more electronegative atom makes a better bond. From my experience, the second factor overrides the first one in most cases.

N is a better donor than O and H in $\ce{O-H}$ is a better acceptor than H in $\ce{N-H}$.

||improve this answer||||
$\endgroup$
  • 7
    $\begingroup$ "From my experience" simply is not a good enough reason. Also this kind of generalisation is dangerous, and if not completely wrong then at least not right for the right reasons. This backwards fitting to some supposedly found experimental data is wrong and leads to a completely distorted understanding of chemistry. $\endgroup$ – Martin - マーチン♦ Dec 12 '17 at 15:27

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.