Sulfuric Acid - Bond Lengths and Charge Separation

Question

So I know that the best Lewis Structure representation of sulfuric acid is one with only single bonds - i.e. sulfur isn't hypervalent.

However, why are the $\ce{S-O}$ and $\ce{S-OH}$ bonds not equivalent then? Why do their lengths differ?

I suspect this has something to do with the addition of the hydrogen on the oxygen atom.

What else could be a cause for the discrepancy?

I ask because my teacher challenged the fact that sulfur might not be hypervalent on the basis that the bond lengths differ.

In retrospect, I think that he:

1) Is placing too much weight on Lewis Structures - which are just rough renditions of what chemicals really are.

2) The $\ce{S-O}$ bonds and $\ce{S-OH}$ bonds may differ in length not because the $\ce{S-O}$ bonds are necessarily double bonded but also because the $\ce{S-OH}$ bonds are different in that S is bound to a hydroxyl group rather than a single oxygen. Also can't there be very minor resonance contributors involving 5-6 bonds on sulfur? It's not that sulfur absolutely can't access its d-orbitals but rather it doesn't to any great extent.

Answer 1

Good questions, let's step through them one at a time.

1) "So I know that the best Lewis Structure representation of sulfuric acid is one with only single bonds - i.e. sulfur isn't hypervalent. "

Look at the resonance structures below. The one on the left with sulfur hypervalent is the only uncharged resonance structure. I suspect that that counts for something. Here is a link that says it is the most important of the structures (don't get confused when you view the link, they use equilibrium arrows (wrong) in their diagram instead of resonance arrows (correct).

2) "why are the S-O and S-OH bonds not equivalent then? Why do their lengths differ?"

This is a very fundamental question. My answer is, they are not equivalent because they are different. Just as you wouldn't expect the same $\ce{C-O}$ bond length in $\ce{CH_3-O-CH_3}$ and $\ce{CH3-OH}$, you wouldn't expect the same bond lengths in an $\ce{S-O}$ bond and an $\ce{S-OH}$ bond. In fact, since these bonds are not superimposable one on the other by a symmetry operation, we can say that they MUST be different (i.e. have different lengths). As one of my favorite professors used to say, now if were to actually measure the $\ce{S-O}$ and $\ce{S-OH}$ bond lengths and our measurements gave rise to the same numeric bond length out to 10 decimal places, we could still step back and say that both we and God know that the lengths are not the same, we just need to measure out to more decimal places! Let me say the converse, the only bonds that can be identical are those that can be interconverted one with another through a symmetry operation. For example in chloroform, $\ce{CHCl3}$, there is a $\ce{C_3}$ axis containing the $\ce{C-H}$ bond. Rotation about this axis interconverts the 3 $\ce{C-Cl}$ bonds, therefor they are equivalent bonds. Their bond lengths MUST be the same.