Consider the following sets of H-bonds:

$$ \begin{align} &\text{P:} &\ce{-O-H\bond{....}N} \\ &\text{Q:} &\ce{-O-H\bond{....}O} \\ &\text{R:} &\ce{-N-H\bond{....}N} \\ &\text{S:} &\ce{-N-H\bond{....}O} \end{align} $$

Since oxygen is more electronegative $(\mathrm{EN})$ and polarises the hydrogen more than nitrogen, I think the order of decreasing strength is Q > P > S > R.

But the correct answer is P > Q > R > S. The reason given is that for H-bonding in


to be strong, $\mathrm{EN}(\ce{A}) > \mathrm{EN}(\ce{B})$ and $\ce{B}$ should be electron-rich.

I am not able to understand where I went wrong.

  • $\begingroup$ A hint: Which one is a stronger base, $\ce{NH3}$ or $\ce{H2O}$? $\endgroup$
    – Poutnik
    Feb 7, 2022 at 7:35
  • $\begingroup$ @Poutnik NH3? I still don't get it, $\endgroup$ Feb 7, 2022 at 8:04
  • $\begingroup$ Which ( Broensted-Lawry ) base is supposed to be more willing to bond to hydrogen by its free electron pair? A stronger one or a weaker one? $\endgroup$
    – Poutnik
    Feb 7, 2022 at 8:06
  • $\begingroup$ @Poutnik Thank you! I understand it now. $\endgroup$ Feb 7, 2022 at 8:08
  • 1
    $\begingroup$ @Intermechanic If you got your answer, feel free to self-answer it. $\endgroup$ Feb 7, 2022 at 9:12

1 Answer 1


One way to understand this is through partial charges. A-H in −A−H⋅⋅⋅⋅B is acting as a hydrogen donor and B is acting as a hydrogen acceptor. H is partially positively charged in −A−H⋅⋅⋅⋅B. The more electron is withdrawn from the H, the more positively charged it is, and the more attraction there will be between H and B. Again, the H in −A−H⋅⋅⋅⋅B is partially positively charged, lacks electron density or is electron-poor, therefore it is attracted towards an electron-rich B.

Oxygen is more electronegative than nitrogen, it can make H to have a bigger partial positive charge than nitrogen can. Nitrogen is less electronegative than oxygen, can put its partial negative charge into H⋅⋅⋅⋅B more readily than oxygen can.


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