I thought of two approaches to solve this question . Since the electronegativity of Oxygen is lower than Fluorine it would be a better donor than Fluorine to the electron deficient Hydrogen . Another approach was since the ∆- charge of fluorine would be more than oxygen it would attract the ∆+ charge on hydrogen leading to more stability . I could not decide which approach was right . I saw this question in a book in which the correct ans was given as F-H- -F

  • $\begingroup$ F---H---F is stronger. This is an example of hydrogen bonding. More the partial charges generated on each atom better the interaction. The hydrogen bond is an effect. The cause is the high seperation of partial charge, which is maximum in H---F bond. Similarly comparing corresponding say O-H and and F-H the latter will be more polar leading to stronger interaction. $\endgroup$ Aug 8, 2020 at 13:36
  • 1
    $\begingroup$ @AdityaRoychowdhury Hydrogen bonding is not just dipolar interaction. I think the answer might be a bit of a matter of opinion if you have to answer it with textbook knowledge. $\endgroup$
    – Karl
    Aug 8, 2020 at 16:03
  • 2
    $\begingroup$ I do not understand this question. Are we talking about $\ce{[FHF]-}$ and $\ce{[FHO]-}$? I wouldn't expect the latter to be stable at all, and I would not want to guess (if it was stable) which one the hydrogen bond was, where the radical density is located, etc., and the overall bonding situation. If this is an exercise from a book, please quote it verbatim (in the body) and cite the source. In general: $\ce{[FHF]-}$ is often used as a benchmark case for the strongest hydrogen bond. See Do symmetric hydrogen bonds in neutral molecules exist? $\endgroup$ Aug 8, 2020 at 17:01
  • 1
    $\begingroup$ Naturally related: Is KHF2 an ionic compound or a covalent compound? $\endgroup$ Aug 8, 2020 at 17:02
  • 1
    $\begingroup$ Experimentally determined , the $\ce{F^-\cdots HF}$ bond has energy 163 kJ/mol and $\ce{F^-\cdots HOH}$, 96 kJ/mol , according to Jeffery 'Introduction to Hydrogen Bonding', (OUP 1997). $\endgroup$
    – porphyrin
    Aug 9, 2020 at 5:50

1 Answer 1


A hydrogen bond is defined as the attractive interaction between an X–H unit (where the electronegativity of X is greater than of H) and a donor atom, or a Lewis base (B).

Hydrogen bond strength is a combination of factors:

  • An electrostatic contribution
  • The formation of a covalent interaction (charge transfer from B to H–X)
  • Dispersion forces

The electrostatic component (the attraction of a bonded hydrogen atom to a region of relatively high electron density of the base (B)) is considered the dominant contribution to most hydrogen bonds.

Regarding the X–H unit, the greater the electronegativity of X, the stronger is the hydrogen bonding donor (higher hydrogen bonding acidity). However, regarding the donor atoms, electronegativity influences two factors that go in opposite directions: greater electronegativity = stronger dipole (stronger electrostatic interaction), but smaller charge transfer (less electron pair donation and less covalent bonding).

Gilli developed a pKa equalization approach for predicting the strength of hydrogen bonds in aqueous solution. It tries to rationalize, for instance, the fact that a water-water hydrogen bond is six times weaker than a water-hydronium hydrogen bond, despite both being O–H···O hydrogen bonds. According to this approach, a $\Delta$pKa value should be calculated to estimate the strength of the hydrogen bond

$\Delta$pKa = (XH···B) = pKa(XH) - pKa(BH+)

The closer $\Delta$pKa is to 0, the stronger the predicted hydrogen bond in aqueous solution. Therefore, considering:

$$\begin{array}{c|c|c|} \text{Species} & {pK_a} \\ \hline \text{HF} & \text{3.15} \\ \hline {H_3O^+} & \text{-1.7} \\ \hline {H_2O} & \text{14} \\ \hline \end{array}$$

$\Delta$pKa(H-F···F-) = 3.2 - (3.2) = 0 (strong)

$\Delta$pKa(H-F···HOH) = 3.2 - (-1.7) = 4.9 (medium strong)

$\Delta$pKa(H-F···HO-) = 3.2 - (14) = -10.8 (medium strong)

Therefore, regarding your question, it seems that F- would form a stronger hydrogen bonding interaction with HF due to a strong electrostatic contribution, even though hydroxide is a better donor than fluoride.

Source: Inorganic Chemistry 5th ed - Miessler, Fischer, Tarr

Please note that in neutral organic molecules, basicity outweighs the dipolar contribution; therefore, amines are the strongest hydrogen bond acceptors (or electron pair donors), followed by alcohols and fluoro-compounds

More on this: http://fbdd-lit.blogspot.com/2017/11/hydrogen-bonding-and-electroengativity.html


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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