How Does SO3 Form 3 Pi Bonds Without an Expanded Octet Structure?

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SUMMARY

SO3 contains three double bonds with oxygen, resulting in a total of three pi bonds formed by sulfur. The central sulfur atom in SO3 is sp2 hybridized, which allows for the formation of these bonds despite the common misconception of an "expanded octet structure." The discussion clarifies that the concept of an expanded octet in main group element chemistry has been disproven, indicating that sulfur can only form one double bond that resonates among the oxygen atoms.

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  • Understanding of molecular hybridization, specifically sp2 hybridization.
  • Knowledge of pi bond formation and resonance structures.
  • Familiarity with the concept of octet rule in main group elements.
  • Basic principles of chemical bonding and electron configuration.
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  • Research the principles of sp2 hybridization in detail.
  • Study resonance structures and their significance in molecular stability.
  • Explore the limitations of the octet rule in main group chemistry.
  • Investigate the bonding characteristics of sulfur in various compounds.
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Since SO3 has an expanded octet structure, the central S should form three double bonds with O, meaning S forms 3 pi bonds in total. However, after hybridisation of S, there are less than three unpaired electrons in unhybridised p-orbitals of S. How can S form 3 pi bonds when there aren’t enough electrons from unhybridised p-orbitals? Thanks.
 
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SO3 does contain three double bonds with O. Sulfur is SO3 is sp2 hybridized. I count six electrons? And of course two for each oxygen atom for a total of 12 electrons in the system.

Did I misunderstand your question?
 
There is no such thing as "expanded octet structure" in main group element chemistry. This has long been disproven. So at most one double bond, which may resonate between the different O-atoms.
 

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