Q91 E - Chemistry (OpenStax)

Question

Give the hybridization and oxidation state for sulfur in \({\rm{S}}{{\rm{O}}_2}\), in \({\rm{S}}{{\rm{O}}_3}\), and  in \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\).

Step-by-Step Solution

Answer

Therefore,

 

1. The  oxidation number of is \( + 4\).
2. The  oxidation number of \({\rm{S}}{{\rm{O}}_3}\)is \( + 6\).
3. The  oxidation number of \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\)is \( + 6\).

Step 1: What is the oxidation state of \({\rm{S}}{{\rm{O}}_2}\) .

a) In\({\rm{S}}{{\rm{O}}_2}\)the oxidation number of oxygen is \( - 2\). 

 

So, the  oxidation number of sulfur is \( + 4\).

             \(s{p^2}\)hybridization

Step 2: What is the oxidation state of .

b) In\({\rm{S}}{{\rm{O}}_3}\)the oxidation number of oxygen is \( - 2\). 

 

So, the oxidation number of sulfur is \( + 6\).

                 \(s{p^2}\)hybridization

Step 3: What is the oxidation state of \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\) .

c) In\({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\)the oxidation number of hydrogen is \( + 1\), while oxygen is \( - 2\). 

 

So, the  oxidation number of sulfur is \( + 6\).

            \(s{p^3}\)hybridization