My attempt to the question: We know that more polar the $\ce{O-H}$ bond, the more acidic is the compound. According to electronegativity, the $\ce{O-H}$ bond of $\ce{HOF}$ should be more polar than $\ce{O-H}$ bond of $\ce{HOCl}$ since electronegativity order is $\ce{F} > \ce{Cl}$. But we should not miss another factor! That is the backbonding of 2p electrons (lone pairs) of oxygen with the 3d vacant orbital of chlorine. This is not possible in $\ce{HOF}$ due to the absence of vacant orbitals. Backbonding reduces (maybe, I do not know) electron density in oxygen and therefore polarity of $\ce{O-H}$ bond increases.

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    $\begingroup$ For future reference, if deploying Latin script, do not skip spaces after the period. This renders recognition where a phrase ends and a new one starts easier. $\endgroup$
    – Buttonwood
    Commented Aug 29, 2019 at 20:19
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    $\begingroup$ The 3d orbitals in 3rd row p block elements are not available for bonding, see any number of questions here, e.g. the first answer to chemistry.stackexchange.com/questions/13949/…. Thus there is no back bonding. $\endgroup$
    – Ian Bush
    Commented Aug 30, 2019 at 5:35

2 Answers 2


If we compare the HOF and HOCl 1) In HOF , the oxygen atom cannot do backbonding since fluorine has no vacant orbital while in HOCl it can do hydrogen bonding it can do backbonding which reduces electron density on oxygen by a slight extent( Remember 2pπ-3pπ backbonding is not very much effective due to size difference of 2p and 3p orbitals) . But yeah this factor makes HOCl bond more polar . But there are two more important factors-:

2[)INDUCTIVE EFFECT]- among oxygen and fluorine , fluorine is more electronegative , so it partially pulls the sigma bonded electrons towards itself giving a delta+ charge to oxygen . While in HOCl , O is more electronegative than Chlorine so the opposite mechanism happens and O gets a slight negative charge. So according to this HOF has increased bond polarity than HOCl. 3) This factor is the most important and sufficient to answer almost any acid comparison . Compare the stability of the ion formed after releasing H+ ( formally called as the conjugate base of the acid). So the question left is simply the comparison between OF- and OCl- , The first two factors again work here but another important factor is the 2p-2p electron repulsions( due to smalls ize of sigma bond the electrons in the orbitals of O and F repel each other. This factor in itself is quiet dominating) present in OF- makes OF- less stable than OCl-

So according to me HOCl must be more acidic than HOF.

Another important thing i wanted to draw attention to is that HOF is not the common acid that can be kept in laboratory. It is quite reactive and reacts vigorously even at 0°C , measuring its acidity is quite diffcult . I also searched online but couldn't find any data concerning Ka values for ionisation of HOF. I hope this helps.

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    $\begingroup$ Fluorine has exactly the available orbitals of chlorine (one s, three p) so either both or none of the two can back-bond based on an orbital discussion. $\endgroup$
    – Jan
    Commented Oct 1, 2019 at 14:22

According to me, $\ce{HOF}$ is more acidic than $\ce{HOCl}$.

As the $\ce{O-H}$ bond length decreases, the acidic nature of the compound increases.

In $\ce{HOF}$ , the $\ce{O-H}$ bond length is $\pu{96 pm}$ while in $\ce{HOCl}$, the $\ce{O-H}$ bond length is $\pu{97 pm}$ and so $\ce{HOF}$ is more acidic than $\ce{HOCl}$.


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