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When I was drawing the lewis structure for $\ce{H2SO4}$ I got this configuration, now I know this isn't the best configuration but I don't see whats wrong with it. The number of valence electrons is right, the free electrons are on the most electronegative element and the formal charges are all 0. I'm not sure if there's a rule I'm missing.

  • $\begingroup$ related chemistry.stackexchange.com/questions/42675/… $\endgroup$
    – Mithoron
    Commented Dec 22, 2017 at 1:14
  • $\begingroup$ Obviously, what is wrong with it is that sulphur is exceeding its octet. $\endgroup$
    – Jan
    Commented Dec 22, 2017 at 3:39
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    $\begingroup$ BTW, I think this is a good question because I see many students asking similar things - how do I know what is and what is not a valid Lewis structure? $\endgroup$ Commented Dec 30, 2017 at 0:14

2 Answers 2


As Jan mentions in the other answer, in general, a Lewis structure is preferred if it retains the octet rule. There can be more complicated bonding, and sulfur and phosphorous are two common examples (e.g., $\ce{SF6}$).

I find that many students can come up with "non-traditional" Lewis structures that, like yours, satisfy the number of valence electrons and minimize the formal charge. I usually consider these correct responses on an exam.

The "missing piece" is ring strain, which is not typically discussed until organic chemistry courses.

Note that your diagram as two O-S-O three-membered rings. These are extremely high in energy, because the O-O-S angle won't be anywhere near the expected 109.5°.

I tried a few quick calculations using the PM7 method and MOPAC. The best ring structure I could find looks like this (a distorted octahedral shape) and is estimated to have a $\Delta H_f^0$ = +33.05 kcal/mol. The O-O-S ring angles are ~65°.


The lowest energy geometry is the traditional $\ce{H2SO4}$ Lewis structure, estimated at $\Delta H_f^0$ = -177.88 kcal/mol. While PM7 isn't a highly accurate method, the difference in stability isn't close.


  • $\begingroup$ Sir so here the problem is just the ring strain, we can have hypervalecy shown in lewis structures right ? $\endgroup$
    – Orion_Pax
    Commented Apr 1, 2022 at 12:19
  • $\begingroup$ Personally, I don't like the term "hypervalency" because the real picture is less clear but typically we draw Lewis structures as I depict, and the problem with the structure in the question is of ring strain. When I see logical (albeit incorrect) Lewis structures, I typically give full marks. $\endgroup$ Commented Apr 2, 2022 at 14:38

You are missing the octet rule. In your proposed structure, sulphur would have a total of twelve valence electrons which just doesn’t happen.

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    $\begingroup$ Except, there are plenty of "hypervalent" sulfur compounds - it's one of the more common examples, e.g. octahedral $\ce{SF6}$ $\endgroup$ Commented Dec 29, 2017 at 23:46
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    $\begingroup$ Agree with Geoff. The traditional Lewis structure of sulfuric acid also has 12 electrons. You can argue you draw that with formal charge on S and O and whatnot, but you could easily do the same with OP's structure - just break two of the S–O single bonds and tack on + and – charges appropriately. Heck, you can even have resonance forms. $\endgroup$ Commented Dec 30, 2017 at 0:11
  • $\begingroup$ @GeoffHutchison Hypervalent =/= violating the octet rule. We know about 4e3c bonds and the likes and we should teach students to follow these principles rather than outdated and wrong ways such as d orbital participation. $\endgroup$
    – Jan
    Commented Dec 30, 2017 at 16:15
  • $\begingroup$ @orthocresol True, the short answer I gave does not go into all those details which I consider not relevant for this question. But I could add a note on all those other possibilities. $\endgroup$
    – Jan
    Commented Dec 30, 2017 at 16:16
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    $\begingroup$ @Jan - I used quotes around "hypervalent" very intentionally. If students are aware that $\ce{SF6}$ is a known, stable, sulfur compound, they can and will come up with this possibility for $\ce{H2SO4}$. Yes, my preferred structure for $\ce{H2SO4}$ satisfies the octet. But IMHO, the main problem with this structure is ring strain. $\endgroup$ Commented Dec 30, 2017 at 16:33

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