Q109 E - Chemistry (OpenStax)

Question

What is the oxidation state of the halogen in each of the following?

\(\begin{array}{l}(a){{\rm{H}}_5}{\rm{I}}{{\rm{O}}_6}\\(b){\rm{IO}}_4^ - \\(c){\rm{Cl}}{{\rm{O}}_2}\\(d){\rm{IC}}{{\rm{l}}_3}\\(e){F_2}\end{array}\)

Step-by-Step Solution

Answer

Various oxidation states are given below:

\(\begin{array}{l}a) + 7\\b) + 7\\c) + 4.\\d) + 3.\\e){\rm{ }}0.\end{array}\)

 

Step 1: Define Oxidation state

The overall number of electrons that an atom either gains or loses such as to form a chemical bond with another atom.

Step 2: To find the oxidation state of \({{\rm{H}}_5}{\rm{I}}{{\rm{O}}_6}\) .

Oxidation state of \(O\) is \(( - 2)\) and of \({\rm{H}}\) is\(( + 1)\). 

The total charge on the molecule is 0. 

Oxida on state of \(I\) is\(x\); find \(x\).

\(0 = x + 6 \cdot ( - 2) + 5 \cdot ( + 1)\)

\(0 = x - 7\)

\(x =   + 7\).

The oxidation number of \({\rm{I}}\) in \({{\rm{H}}_5}{\rm{I}}{{\rm{O}}_6}\) is \( + 7\).

Step 2: To find the oxidation state of \({\rm{IO}}_4^ - \) .

 

b) Oxidation state of \(O\) is \(( - 2)\) and the total charge on the molecule is \(( - 1)\).

\( - 1 = x + 4 \cdot ( - 2)\)

 

\( - 1 = x - 8\)

 

\(x =  + 7\).

 

The oxidation number of \({\rm{I}}\) in \({\rm{IO}}_4^ - \) is \( + 7\).

 

Step 3: To find the oxidation state of \({\rm{Cl}}{{\rm{O}}_2}\) .

Oxidation state of \(O\) is \(( - 2)\) and the total charge on the molecule is 0.

\(0 = x - 4\)

\(x =   + 4\).

The oxidation number of \({\rm{Cl}}\) in \({\rm{Cl}}{{\rm{O}}_2}\) is \( + 4\).

Step 4: To find the oxidation state of \({\rm{IC}}{{\rm{l}}_3}\) .

Oxidation state of \({\rm{Cl}}\) is\(( - 1)\)and the total charge on the molecule is \(0\).

\(0 = x + 3 \cdot ( - 1)\)

\(0 = x - 3\)

\(x = 3\).

The oxidation number of \({\rm{I}}\) in \({\rm{IC}}{{\rm{l}}_3}\) is \( + 3\).

Step 5: To find the oxidation state of \({F_2}\) .

e) Oxidation state of atoms in elementary state is zero, therefore, oxidation number of \({\rm{F}}\) in \({F_2}\) is \(0\).