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One of the canonical structures for sulfur dioxide - $\ce{SO2}$ - has sulfur (with a lone electron pair) double bonded to each oxygen atom to form a total of 4 bonds for sulfur - which can be achieved via valence expansion into empty d-orbitals.

What then is the hybridization of the valence-expanded sulfur? It is described as sp². But how can that be? This seems unlikely because d-orbitals are involved since the sulfur underwent valence expansion.

On might imagine a pair of electrons from the 3s/3p oribital(s) being promoted to an empty d-orbital and then having the 3s and 3p orbitals hybridize in to sp². If this is true it would mean that the lone electron pair of the valence-expanded sulfur consists of 2 electrons occupying and unhybridized d-orbital. But is this correct?

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  • $\begingroup$ This photo should help you. $\endgroup$ – Kenny Lau May 14 '16 at 5:19
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    $\begingroup$ The role of d orbital in sulfur bonding is a controversial issue. It seems to me that the problem is not about if d orbital helps, rather than whether the role of d orbital should be regarded as a polarization function. $\endgroup$ – Rodriguez May 14 '16 at 5:56
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The structure of sulfur dioxide ($\ce{SO_2}$) is quite complicated.

This photo from this website explains it quite well: SO2 hybridization diagram

As seen, all the atoms have $\ce{sp^2}$ hybridization.

I'll only focus on the central sulfur atom.

  • Two $\ce{sp^2}$ orbitals form $\ce{\sigma}$-bonds with the two oxygens.
  • The other $\ce{sp^2}$ orbital is where the lone pair lives in.

Now, we have dealt with 4 electrons, and only have 2 electrons to deal with.

The 2 remaining electrons actually live in the unhybridized $\ce{p}$ orbital.

At this point, the two oxygen atoms will also have 2 electrons left to pair with the sulfur atom.

  • The two oxygen atoms get to keep their own electron.
  • Sulfur shares two electrons among itself and the other two oxygen atoms.

Thus, no electron lives in the $\ce{d}$ orbital.

SO2 diagram

  • $\color{Red}{\mbox{red}}$ represents number of electrons (I am too lazy to draw the fish-hooks).
  • $\color{Green}{\mbox{green}}$ represents $\ce{sp^2}$ orbital.
  • $\color{blue}{\mbox{Blue}}$ represents $\ce{p}$ orbital.

Note that the two $\ce{sp^2}$ orbitals between sulfur and the two oxygen atoms are in $\ce{\sigma}$-bond.

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  • $\begingroup$ We have been taught that $SO_2$ had one $ d\pi - p\pi$ bond and one $p\pi - p\pi$ bond. So is this wrong? $\endgroup$ – samjoe Jul 16 '17 at 13:50
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    $\begingroup$ @samjoe $d$ orbital is hardly used, even in $\ce{SO3}$. $\endgroup$ – Kenny Lau Jul 16 '17 at 16:09
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The canonical structure for sulphur dioxide nowadays has charge separation, one oxygen bonded to sulphur in a single bond, the other in a double bond. But not too long ago the canonical structure was indeed what you proposed with one pp-π bond and one dp-π bond.

This was generally explained with one of sulphur’s 3d-orbitals taking part in hybridisation giving rise to a ‘sp²d’ state. Taking Kenny’s scheme, though, sulphur’s 3d-orbitals are a good step above the 3p orbitals in energy. Technically, the 4s orbital should almost come before it. So this hybridisation is highly unlikely.

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Simple method for determining the hybridisation of any molecule is- Hybridisation=no.of lone pair on atom+no. of bond Ex.in the molecule NH3, there is one line pair on nitrogen atom+ there are three bonds =4 .therefore it is sp3 hybridised. If addition comes - 1)2 -sp hybridisation 2)3- sp2 hybridisation 3)4-sp3 hybridisation 4)5-sp3d hybridisation 5)6-sp3d2 hybridisation It is important to note that measure 1 point for double as well tripple bond

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