In the structure of $\ce{H2SO3}$

H2SO3 pic

there is a lone pair on sulfur.

Why can't sulfur make a bond with one of the hydrogen atoms and a double bond with oxygen to avoid having a lone pair?

Such as given below:

enter image description here

Because here there will be sp3 hybridisation..... no lone pair, 4 sigma and 2 pi bonds.

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    $\begingroup$ What's wrong with having a lone pair though? $\endgroup$ Apr 29, 2016 at 13:24
  • 1
    $\begingroup$ related chemistry.stackexchange.com/questions/38129/… $\endgroup$
    – Mithoron
    Apr 29, 2016 at 15:19
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    $\begingroup$ there is nothing wrong but also whats wrong in sulfur having a direct bond with hydrogen and a double bond with oxygen. $\endgroup$
    – jasnoor
    Apr 30, 2016 at 5:28
  • 1
    $\begingroup$ I would very much doubt the existence of pi bonds in this molecule. $\endgroup$ Apr 30, 2016 at 7:55

1 Answer 1


Most of what I would be saying now is stuff I already highlighted in another answer of mine to a related question.

$\ce{H2SO3}$ exists in two forms, one of which is equivalent to the structure you drew, the other is the structure you proposed. Taking the monoanions (because that was the focus of the other question and I am too lazy to redraw the picture) will give you these two structures:

hydrogen sulphite and sulphonate

(All horizontal bars are formal $-$ signs, never lone pairs.)

The crystal structure of $\ce{CsHSO3}$ was solved in 1980 by Johansson, Lindqvist and Vannerberg; they conclude that $\ce{CsHSO3}$ is not cesium hydrogen sulphite but rather cesium sulphonate.[1] See the other answer for details.

In cis-bisbipyridylbishydrogensulphito-κS-ruthenium(II) $\ce{[Ru(bpy)2(HSO3}\unicode{x2D}\unicode[Times]{x3BA}S\ce{)2]}$, a crystal structure solved by Allen, Jeter, Cordes and Durham, the sulphur atom directly coordinates the ruthenium centre which is only possible if it has a lone pair; thus we must be dealing with hydrogensulphite.[2]

Therefore, depending on the conditions both forms are possible and observed.

In case of the free acid, it is likely more stable in one of the two forms. I do not have analytical or thermodynamic data at hand to tell you which tautomer is preferred.


[1]: L.-G. Johansson, O. Lindqvist, N.-G. Vannerberg, Acta Cryst. 1980, B36, 2523. DOI: 10.1107/S0567740880009351.
[2]: L. R. Allen, D. Y. Jeter, A. W. Cordes, B. Durham, Inorg. Chem. 1988, 27, 3880. DOI: 10.1021/ic00295a003.

  • $\begingroup$ Jan you are right but maybe you got my proposed structure wrong . i have just edited my question ,have a look at my proposed structure. $\endgroup$
    – jasnoor
    Apr 30, 2016 at 5:42
  • $\begingroup$ While not entirely rigorous, I guess a simple approach would be to just look at the bond strengths to gauge which tautomer is more stable. In the tautomerisation $\ce{SO(OH)2 -> HSO2(OH)}$, we lose one S-O and one O-H bond, but gain one S-H and one "S=O" bond. Now, we just need the data... which I am lazy to find. (The S-O bond is rather elusive.) $\endgroup$ Apr 30, 2016 at 13:53
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    $\begingroup$ @orthocresol I’m looking forward to your competing answer ;D $\endgroup$
    – Jan
    Apr 30, 2016 at 14:59
  • $\begingroup$ @jasnoor No, except for the fact that I chose the conjugate base our structures are identical. Martin showed that the sulphite ion is better represented with three $\ce{S-O}$ single bonds though. I’m sure there are other questions/answers on the topic of $\ce{S=O}$ versus $\ce{{S+}-O^-}$ on here that I can’t find atm. $\endgroup$
    – Jan
    Apr 30, 2016 at 15:03
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    $\begingroup$ @ortho but me wants sportsmanship ó.ò $\endgroup$
    – Jan
    Apr 30, 2016 at 16:57

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