Is the S–O bond strength in sulfur trioxide or sulfite anion larger?

Question

Can you please explain the difference between the bond enthalpies in sulfurtrioxide $\ce{SO3}$ and the sulfite anion $\ce{SO3^{2-}}$?

Answer 1

Due to symmetry constraints ($D_\mathrm{3h}$) in $\ce{SO3}$ there 6 electrons in $\pi$ type orbitals. In a wider sense of the term this molecule is Y-aromatic, but the HOMO actually represents two electrons in in-plane lone pair orbitals of oxygen. The LUMO is an antibonding $\pi$ orbital with respect to all $\ce{S-O}$ bonds.

The following valence molecular orbitals were derived at BP86/cc-pVTZ, Energy level decreases from top to bottom (enlarge).

The bonding picture is usually trivialised as each $\ce{S-O}$ bond being a double bond. But this is actually far away from the truth, as it does not respect the charge of $q=+2$ at the sulfur atom and the charges at the oxygens with $q=-\frac23$. This is due to the fact, that the sulfur atom actually only contributes to one of the three $\pi$ bonding orbitals, resulting in electron density deficiency.

However, the $\ce{S-O}$ bond in sulfurtrioxide is still stronger and shorter, than in the sulfite anion. (BO means bond order)

\begin{array}{lrr} & \mathbf{d}(\ce{S-O}) & \text{BO}(\ce{S-O})\\\hline \ce{SO3} & 1.458 & 1.4 \\ \ce{SO3^{2-}} & 1.578 & 1.1 \\\hline \end{array}

The main reason for this is, that upon reduction of sulfurtrioxide to sulfite, $$\ce{SO3 + 2e- <=> SO3^{2-}},$$ you will fill the LUMO. As already stated, this orbital is antibonding with respect to the $\ce{S-O}$ bonds. This is energetically unfavourable for the molecule, therefore the molecule distorts to get rid of the symmetry constraints, i.e. it adapts the lower $C_\mathrm{3v}$ point group. This results in pyramidalization and a different bonding pattern.

You will now no longer find any $\pi$ type bonding orbitals. More accurately this molecule is best described consisting of three equal $\ce{S-O}$ covalent single bonds, and three lone pairs at each oxygen. This results in a charge at each oxygen of $q=-1$ and at the sulfur of $q=1$ (which does not fit the traditional Lewis type bonding description.)

This is also reflected in the molecular orbitals (BP86/cc-pVTZ, energy decreases from top to bottom, enlarge).

It is also very important to understand that the latter molecule would not be stable in gas phase, since it is too highly charged. Without compensation of solvent and/ or counter ions, this molecule will most certainly fragmentate.

Answer 2

My hypothesis is that the bond enthalpy of the S-O bond in the sulfate anion is smaller than that of the S-O bond enthalpy in sulfur trioxide. I would say that the S-O bond strength lies with atomic size.

In sulfate anion, we have the electronegative oxygen atoms stabilizing the bulk of the negative 2 formal charge.

Bearing in mind that anions are generally bigger than their respective neutral counterparts, and that cations are smaller than their respective neutral counterparts. Why? This lies with electron-electron shielding and $\ce{Z_{eff}}$, or effective nuclear charge.

Having more electrons in a valence, such as in the case of the oxygens in the sulfate anion as opposed to the oxygens in sulfur dioxide, decreases $\ce{Z_{eff}}$. This in turn allows the electrons to be further from the nucleus, and this obviously increases size.

The increased size of the oxygens in the sulfate anion in turn lengthen the S-O bonds, and this in turn decreases the bond enthalpy of the S-O bonds in the sulfate anion.

Answer 3

Sulfur Trioxide is typically represented with Sulfur double bonded to each of three oxygens. In this context, the oxygens have full octets, while the Sulfur has an 'expansion'.

Because the oxygens are full (and sulfur is pretty close to 'full'), to add electrons to this molecule requires the placement of electrons into 'antibonding orbitals'. These orbitals would be pi-antibonding orbitals, weakening the S-O bonds somewhat.

Please be advised, it is difficult to speak about the enthalpy of either bond(sigma or pi) within a double bond. Typically the enthalpy of a double bond is compared to a single bond and the difference is assigned to the 'pi-bond'.