Sulfuric acid is manufactured by a series of reactions represented by the following equations:

\(\begin{array}{l}{{\rm{S}}_{\rm{8}}}{\rm{(s) + 8}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{8S}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}}\\{\rm{2S}}{{\rm{O}}_{\rm{2}}}{\rm{(g) + }}{{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2S}}{{\rm{O}}_{\rm{3}}}{\rm{(g)}}\\{\rm{S}}{{\rm{O}}_{\rm{3}}}{\rm{(g) + }}{{\rm{H}}_{\rm{2}}}{\rm{O(l)}} \to {{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{(l)}}\end{array}\)

Draw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following:

(a) circular \({{\rm{S}}_{\rm{8}}}\)molecule

(b) \({\rm{S}}{{\rm{O}}_{\rm{2}}}\)molecule

(c) \({\rm{S}}{{\rm{O}}_{\rm{3}}}\)molecule

(d) \({{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\)molecule (the hydrogen atoms are bonded to oxygen atoms)

Lewis structure: Tthe Lewis dot structure depicts bonding in molecules and ions.

Hybridization: When two or more different pure orbitals with comparable energy are mixed together to produce an equivalent amount of impure orbitals with equal energy and definite geometry.

We know that a sulphur atom has six valence electrons, so two electrons will participate in bond formation and the remaining four electrons will be lone pairs in the given question because we are to find the Lewis dot structure for circular\({{\rm{S}}_{\rm{8}}}\). If we join the eight sulphur atoms together, we will form a ring with two bond pair electrons and two lone pair electrons, giving each sulphur atom a total of four valence electrons. As a result, the hybridization is\({\rm{s}}{{\rm{p}}^{\rm{3}}}\), and each sulphur atom will achieve a bent molecular geometry.

The following is the Lewis dot structure:

Therefore, the required hybridization is \({\rm{s}}{{\rm{p}}^{\rm{3}}}\)and the molecular geometry is bent shape.

Since we are looking for the Lewis dot structure for \({\rm{S}}{{\rm{O}}_{\rm{2}}}\), we know that a sulphur atom has six valence electrons, and an oxygen atom also has six valence electrons, so two pair electrons from sulphur and one oxygen will participate in bond formation and form a double bond, and one pair electron from the remaining four sulphur atoms will form a coordinate bond with another oxygen atom, thus the hybridization is \({\rm{s}}{{\rm{p}}^{\rm{2}}}\).

The structure of the Lewis dot is as follows:

Therefore, the required hybridization is \({\rm{s}}{{\rm{p}}^{\rm{2}}}\)and the molecular geometry is bent shape.

Since we are looking for the Lewis dot structure for\({\rm{S}}{{\rm{O}}_{\rm{3}}}\), we know that a sulphur atom has six valence electrons and an oxygen atom has six valence electrons, so two pairs of sulphur atom electrons share two pairs of oxygen atom electrons and form a double bond, and one pair of sulphur atom electrons bonds with the other two oxygen atoms, resulting in three bonds.

The structure of the Lewis dot is as follows:

Therefore, the required hybridization is sp\(^{\rm{2}}\) and the molecular geometry is trigonal planar.

We know that a sulphur atom has six valence electrons, a hydrogen atom has one valence electron, and an oxygen atom has six valence electrons, so the sulphur atom shares two pairs of electrons with the two oxygen atoms and forms covalent bond, and it mutually shares two pairs of electrons with the other two oxygen atoms and forms coordinate bond in the given question. Because oxygen and hydrogen share a pair of electrons, the hybridization is\({\rm{s}}{{\rm{p}}^{\rm{3}}}\), and each sulphur atom achieves a tetrahedral geometry with no lone electrons.

As follows is the Lewis dot structure:

Therefore, the required hybridization is \({\rm{s}}{{\rm{p}}^{\rm{3}}}\) and the molecular geometry is tetrahedral.