3
$\begingroup$

This question already has an answer here:

I'm asked to find the Lewis Structure of sulfuric acid, $SO_4H^-$. After struggling for a while, I sumply looked it up.

Source: Wikipedia

How can this be? Sulfur has 6 electrons in its valence shell, so it should form bonds so as to collect 2 more electrons and satisfy the octet rule. So, surely it should form 2 bonds?

But it's involved in no less than six bonds here. Huh?

$\endgroup$

marked as duplicate by Ben Norris, ManishEarth Apr 1 '13 at 14:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Welcome to Chemistry.SE! This question has already been answered before (not explicitly for sulfuric acid, but expanded valence on sulfur is covered). Check out the answer to this question: chemistry.stackexchange.com/questions/444/…. I had trouble search for it using the title of your question, so its understandable how you might have missed it. $\endgroup$ – Ben Norris Apr 1 '13 at 10:53
2
$\begingroup$

The octet rule only applies firmly to elements below main energy level 3 (in fact, there are rare instances in which even that isn't entirely accurate, but molecular orbtal theory needs to be invoked to explain those situations). At main energy level 3, which Sulfur occupies, unfilled d orbitals become available, and those d orbitals serve to accommodate the additional electrons in Sulfur's bonds. In reality, Sulfur is a much larger atom than Oxygen, and the π orbital overlap between the two elements is fairly poor, giving the bonds significant single-bond character. You might therefore see resonance structures in which single bonds are drawn instead, with formal positive charges assigned to Sulfur and formal negatives assigned to Oxygen.

$\endgroup$
  • $\begingroup$ I notice that in this molecule, the Oxygens and Hydrogens are still following the octet rule. Will this be true in general? That is, an element below energy level 3 will always follow the octet rule, even when in a compound with elements that don't? $\endgroup$ – Jack M Mar 31 '13 at 16:15
  • $\begingroup$ @JackM, yes, that's generally the case. There are, however, specific instances in which valence bond theory breaks down and cannot explain the observed bond order and geometry of certain atoms even below the third main energy level (e.g., diborane). Molecular orbital theory is necessary to adequately rationalize those situations. $\endgroup$ – Greg E. Mar 31 '13 at 19:28
  • $\begingroup$ But usually, a good strategy for drawing a Lewis Structure would be to make sure all the octet-rule-following atoms satisfy the octet rule, and if that means that the other atoms end up with "too many" electrons, then just so be it? $\endgroup$ – Jack M Mar 31 '13 at 19:35
  • $\begingroup$ @JackM, the normal strategy for drawing a LEDS given no info other than the formula is to select a central atom and create single bonds to the other atoms, aiming for a maximally symmetrical skeletal molecule. Then, fill in lone pairs to the outer atoms (using octet rule), and then to the central atoms. Then, convert lone pairs to double and/or triple bonds as necessary with the aim of satisfying all octets and minimizing charge distribution. Exceeding the octet of centrals atoms is acceptable, assuming the atom is at the third main energy level or higher and doing so minimizes charges. $\endgroup$ – Greg E. Mar 31 '13 at 20:35
  • $\begingroup$ @JackM, for H2SO4 (32 valence electrons total), the procedure would be: 1) for maximum symmetry, select S as the central atom. 2) draw single bonds from S to each of the four O atoms (which satisfies S octet for the moment) and to two of the O add single bonds to H. 3) add remaining electrons as lone pairs to each O, stopping when octet is full. This leaves you with a structure in which all octets are satisfied, but two of the O atoms have -1 formal charge, while the S has +2 formal charge, so eliminate the formal charges by using one lone pair from each O to create a double bond to S. $\endgroup$ – Greg E. Mar 31 '13 at 20:43

Not the answer you're looking for? Browse other questions tagged or ask your own question.