I recently came across this sentence in my textbook:

the bonds between sulphur and oxygen in oxides of sulphur ($\ce{SO2}$ and $\ce{SO3}$) are much shorter than might be expected for a single bond.



I feel it could be due to partial double bond character due to resonance.

And the same textbook gives another explanation:

In these molecules, in addition to normal π bond, a π bond is also formed by the sidewise overlap of a filled 2p orbital of oxygen with a vacant 3d orbital on the sulphur). This is called pπ - dπ bond and results in bringing the two atoms closer and thus accounts for shorter bond length of $\ce{S-O}$ bond.

The reason provided in the textbook (as mentioned above) is definitely untrue and incorrect

Because of reasons mentioned in these posts

Why is the bond order in the SO₃ molecule 1.33 and not 2? . Hybridization of sulfur in sulfur dioxide

Please provide a explaination for the same (the title question).

  • 4
    $\begingroup$ Well, maybe this is because they are not single bonds? $\endgroup$ Commented Jan 14, 2019 at 8:17
  • $\begingroup$ Please cite the sources for the quotes and images you use, even (or especially) it they are from of own network. $\endgroup$ Commented Apr 14, 2019 at 19:32
  • $\begingroup$ "Which is definitely untrue and incorrect" Why do you think so? $\endgroup$
    – Karl
    Commented Apr 14, 2019 at 19:56
  • 1
    $\begingroup$ @Karl “That is thoroughly incorrect. chemistry.stackexchange.com/questions/51168/… chemistry.stackexchange.com/questions/29101/… – Mithoron” $\endgroup$
    – Chemist
    Commented Apr 14, 2019 at 21:07
  • $\begingroup$ Please explain this (in one or two scentences) in your question. It's not obvious. $\endgroup$
    – Karl
    Commented Apr 14, 2019 at 21:12

1 Answer 1


If you would consider it more carefully, you would see that the bond order in SO2 and SO3 are, respectively, 1.5 and 4/3. This means that they are shorter than a single bond of sulphur and oxygen.


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