Question

In the gaseous phase, the sigma O – S – O bond angle in the $S{O}_3$ molecule is degrees.

Answer 1

First let us draw the lewis structure for the $S{O}_3$ molecule:

Do note that in the above structure, we have not highlighted the lone pairs on the peripheral oxygen atoms. Each atom has two lone pairs of electrons.
Let us calculate the formal charge on the central S atom:
Formal charge = Valence electrons – non-bonding valence electrons -$\frac{1}{2}$ (Bonding electrons)
=6 - 0 - 6 = 0
Similarly, we can also calculate the formal charge on the oxygen atoms. Please try that as an exercise. Oxygen cannot violate the octet rule. Its octet must be complete while Sulphur can hold more than 8 electrons due to the availability of vacant d orbitals.
Thus we have minimized the formal charge difference between the central atom and the peripheral atoms. We can see that there are three $\sigma$ S-O bonds three $\pi$ – O bonds
From VSEPR theory, with no lone pair of electrons on the central S atom and three $\sigma$ S – O bond pairs, we have $s{p}^2$ as the hybridization. What would be the dipole moment for the $S{O}_3$ gaseous molecule? Due to the symmetry it is 0.
The shape is trigonal planar with each O – S – O angle = 120 degrees.