Q93 E - Chemistry (OpenStax)

Question

Determine the oxidation state of sulfur in \({\rm{S}}{{\rm{F}}_6},{\rm{S}}{{\rm{O}}_2}\;{{\rm{F}}_2}\), and KHS.

Step-by-Step Solution

Answer

Therefore,

1. The oxidation number of\({\rm{S}}{{\rm{F}}_6}\)is 6.
2. The oxidation number of\({\rm{S}}{{\rm{O}}_2}\)is 6.
3. The oxidation number of \({{\rm{F}}_2}\) is \( - 2\).

Step 1: To determine the oxidation state of \({\rm{S}}{{\rm{F}}_6}\) .

a) In one molecule of \(S{F_6}\) there are \(6\;{\rm{F}}\)atoms with oxidation number \( - 1\). 

Since the molecule is neutral, the total charge of all of the atom should be zero. Oxidation number of\(S\)in unknown \(x\).

\(\begin{array}{l}0 = x + 6 \cdot ( - 1)\\0 = x - 6\\x = 6\end{array}\)

The oxidation number of\({\rm{S}}{{\rm{F}}_6}\)is 6.

Step 2: To determine the oxidation state of \({\rm{S}}{{\rm{O}}_2}\) .

b) Oxidation number of oxygen is \(( - 2)\) and there are two atoms of \({\rm{O}}\) in \({\rm{S}}{{\rm{O}}_2}\;{{\rm{F}}_2}\), as there are two atoms of \({\rm{F}}\)in oxidation state \(( - 1)\) and with neutral total charge:

\(\begin{array}{l}0 = x + 2 \cdot ( - 2) + 2 \cdot ( - 1)\\0 = x - 6\\x = 6.\end{array}\)

The oxidation number of \({\rm{S}}{{\rm{O}}_2}\) is 6.

Step 3: To determine the oxidation state of \({{\rm{F}}_2}\) .

c) Oxidation number of \({\rm{K}}\)and of \({\rm{H}}\) is \(( + 1)\), and there is only one atom of each in \({\rm{KHS}}\) with total charge of the molecule 0 .

\(\begin{array}{l}0 = x + 1 \cdot 1 + 1 \cdot 1\\0 = x + 2\\x =  - 2.\end{array}\)

The oxidation number of \({{\rm{F}}_2}\) is \( - 2\).