A molecule that has hydrogen bonding usually follows these two premises.

1.) There is a hydrogen atom involved

2.) Hydrogen must be bonded to a highly electronegative element which are nitrogen ($\ce{N}$), fluorine ($\ce{F}$) and oxygen ($\ce{O}$).

Seeing that both oxygen and chlorine have a small difference in their electronegativity (oxygen being roughly 3.5 and chlorine being roughly 3.0), why does chlorine in a hydrogen chloride molecule ($\ce{HCl}$) have a dipole-dipole interaction, while the oxygen in a water molecule ($\ce{H2O}$) causes the water molecule to have a stronger form of dipole-dipole interaction called hydrogen bonding? I do not understand this since chlorine is just as electronegative as oxygen and nitrogen?

Edit: I should also added that nitrogen and chlorine have the same EN value (3.0)

  • 2
    $\begingroup$ Related. Short answer: $\ce{Cl}$ is too large. $\endgroup$
    – Asker123
    May 17, 2015 at 17:13
  • $\begingroup$ @Asker123 Could you go a little more in depth? $\endgroup$ May 17, 2015 at 17:14
  • $\begingroup$ Also, check out the example from this answer. $\endgroup$
    – andselisk
    Jan 6, 2018 at 14:31

2 Answers 2


Very, very related. Check it for a more in depth answer.

Well, first-of-all you need to understand that the $\ce{H}$ bonding isn't actually bonding. It is just a covalent attraction. Also since $\ce{Cl}$ is larger than $\ce{N}$, $\ce{F}$ and $\ce{O}$ it does not make a strong $\ce{H}$ bond. The size of the $\ce{Cl}$ makes the dipole-dipole attraction weaker. However $\ce{N}$, $\ce{F}$ and $\ce{O}$ are smaller and thus have an $\ce{H}$ bond.

Although in reality, compared to other covalently bonded structures, $\ce{HCl}$ has a very strong covalent bond.


Despite its electronegativity, size of chlorine atom is large and hence, electron density of chlorine is not sufficient to form Hydrogen Bonding. Hence, HCl does not have hydrogen Bonding.


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