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Which of the following structures is the most preferred and hence of lowest energy for $SO_3$?
A
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B
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C
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D
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Detailed Solution
The preferred Lewis structure is the one with the lowest formal charges on the atoms (ideally zero) and the maximum number of covalent bonds.
Formal charge = valence electrons − non-bonding electrons − $\frac{1}{2}$(bonding electrons).
In the structure with three S=O double bonds, sulphur has 6 bonds and no lone pair: formal charge on S = $6 - 0 - \frac{1}{2}(12) = 0$.
Each oxygen in that structure has two lone pairs and a double bond: formal charge = $6 - 4 - \frac{1}{2}(4) = 0$.
In the structures containing one or more S–O single bonds, the singly bonded oxygen carries a formal charge of −1 and sulphur carries a positive formal charge, so they have charge separation and are of higher energy.
Hence the structure in which sulphur is joined to all three oxygen atoms by double bonds is the most preferred.
Formal charge = valence electrons − non-bonding electrons − $\frac{1}{2}$(bonding electrons).
In the structure with three S=O double bonds, sulphur has 6 bonds and no lone pair: formal charge on S = $6 - 0 - \frac{1}{2}(12) = 0$.
Each oxygen in that structure has two lone pairs and a double bond: formal charge = $6 - 4 - \frac{1}{2}(4) = 0$.
In the structures containing one or more S–O single bonds, the singly bonded oxygen carries a formal charge of −1 and sulphur carries a positive formal charge, so they have charge separation and are of higher energy.
Hence the structure in which sulphur is joined to all three oxygen atoms by double bonds is the most preferred.