In the structure of $\ce{HSO3-}$, we see that the H is bonded directly to the S, not to the O's, as would be the case in other oxyacids. Why is this the case?
1 Answer
There are two distinct forms of $\ce{HSO3-}$ (see picture below; all horizontal bars are formal $-$ signs).
Only the leftmost of these two forms can be called hydrogen sulphite (or ‘bisulphite’, if you really insist); the correct name of the rightmost form is sulphonate. The two differ in their chemical characteristics; hydrogen sulphite is, for example still acidic and has a lone pair on sulphur, while sulphonate is not.
Johansson, Lindqvist and Vannerberg from Göteborg solved the crystal structure of $\ce{CsHSO3}$ in 1980 and concluded that:
- the $\ce{S-O}$ bond lengths are much closer to those in sulphate than those in sulphite ($\ce{Na2SO3}$)
- the $\ce{O-S-O}$ angle is $113°$ and thus larger than the tetrahedral angle $\approx 109°$ (expected in sulphate) and much larger than the angle in $\ce{SO3^2-}$
- the ion’s point group is $\mathrm{C_{3v}}$, so all three oxygens are equal
- there is IR evidence of a $\ce{S-H}$ bond.
From this they deduced that $\ce{CsHSO3}$ be, in fact, cesium sulphonate, not cesium hydrogen sulphite.[1]
The first reference I found for the structure of a true hydrogen sulphite was the crystal structure of cis-$\ce{[Ru(bpy)2(HSO3)2]}$ by Allen, Jeter, Cordes and Durham from Arkansas in 1988. In their structure, $\ce{HSO3-}$ coordinates to the central $\ce{Ru(II)}$ ion via the sulphur atom: $\ce{[Ru(bpy)2(HSO3}\unicode{x2D}\unicode[Times]{x3BA}S\ce{)2]}$. This coordination is, of course, only possible if there is a lone pair on sulphur, and thus the structure must be that of a hydrogen sulphite.[2]
(In my introductory chemistry class, Prof. Klüfers stated that structures for both forms be known and published.[3] Alas, I didn’t find the crystal structure of a simple metal hydrogen sulphite anywhere quickly.)
It gets easier when dealing with the corresponding acids’ esters. The hydrogen on sulphonic acid can be replaced with organic residues creating the vast class of organosulphonic acids. Sulphonic esters are also possible; they would have an organic residue attached to both the sulphur directly (replacing the hydrogen atom) and one of the oxygens. Every mesylate or tosylate (in formulae: $\ce{OMs}$ or $\ce{OTs}$) is, in fact, a sulphonic ester. Sulphurous esters are less prevalent in organic chemistry but do exist (e.g. dimethyl sulphite, $\ce{OS(OCH3)2}$). Of course, the chemical properties of methyl mesylate ($\ce{H3C-SO2-OCH3}$) — a sulphonate — and dimethyl sulphite are substantially different.
After discussing $\ce{H2SO3}$ and its salts, it is important to mention the wrong assumption you proposed in your question:
we see that the $\ce{H}$ is bonded directly to the $\ce{S}$, not to the $\ce{O}$’s, as would be the case in other oxyacids.
This is not generally true. Phosphorous at least displays a similar chemistry, even to the extent that ‘hydrogen phosphite salts’ cannot be isolated and phosphonate salts are isolated instead. (One can synthesise both trimethyl phosphite and dimethyl methylphosphonate, the corresponding phosphorous and phosphonic esters, though.) I am not firm in the oxyacid chemistry of heavier homologues of phosphorous and sulphur, but it would deeply astound me if this chemistry is not observed there at least to some degree.
Note that the halogens do indeed follow your proposed rule: all their oxyacids are monoprotic and contain only a single, oxygen-bound hydrogen in neutral state.
References:
[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.
[3] Refer to the internet scriptum of general and inorganic chemistry 1 (in German) as taught 2006 to 2008 and likely will be taught again in 2015.
(Note that the articles are most likely behind a paywall)
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$\begingroup$ Arsenous acid is truly arsenic hydroxide. I guess, that for heavier analogues inert pair is in full effect. $\endgroup$ Commented Apr 29, 2016 at 18:10
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$\begingroup$ @permeakra That’s actually a consideration I didn’t make. $\endgroup$– JanCommented Apr 29, 2016 at 18:28
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2$\begingroup$ Just to add on, I suppose in the specific case of As the so-called "inert pair" arises due to the d-block contraction. Two more rows down, the bismuth analogue also has the structure Bi(OH)3 and it is actually basic - this is probably lanthanide contraction + relativistic effects at work. Unfortunately the Sb analogue has "not been well-characterised". $\endgroup$ Commented Apr 29, 2016 at 23:55
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1$\begingroup$ The "inert pair" here could be weak bonding to hydrogen, which would come about when atoms just have large orbitals and overlap poorly with the hydrogen. For a different form of this effect, see this answer, in which electropositive metals still bond to hydrogen but the expected covalency is not there. $\endgroup$ Commented Sep 22, 2021 at 9:45
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1$\begingroup$ Another very good point, @OscarLanzi. $\endgroup$– JanCommented Sep 22, 2021 at 9:54