Question

What are the oxidation states of manganese in the following compounds?
\[MnO_{4}^{-}\],\[M{{n}_{2}}{{O}_{3}}\] and \[Mn{{O}_{2}}\]

Answer 1

Hint: To calculate the oxidation number of Mn in all of the above compounds, we should know the oxidation number of the other atoms in them i.e. the oxidation number of O= -2. Now, with the help of these oxidation numbers we can easily find the oxidation number of Mn. Now solve the given statement accordingly.

Complete answer:
First of all, let’s discuss what an oxidation number is. By the term oxidation number, we simply mean the total number electrons which are present in the valence shell of an atom when it loses or gains the electrons while undergoing the chemical reaction.
For metals, the number of valence electrons is always equal to the number of the electrons in the outermost shell of the metal.
For non-metals, the number of valence electrons= eight minus the number of electrons in the outermost shell of the non-metal.
Now, let’s find the oxidation states of the given compounds.
1. \[MnO_{4}^{-}\]
Let, the Oxidation state of \[Mn=x\]
Oxidation state of \[O=\text{ 4}\left( -2 \right)=-8\]
Oxidation state of Mn in \[MnO_{4}^{-}\] is
\[\begin{align}
  & \Rightarrow \text{x-8}=-1 \\
 & \Rightarrow x=+7 \\
\end{align}\]
2. \[M{{n}_{2}}{{O}_{3}}\]
Let, the Oxidation state of \[Mn=x\]
Oxidation state of \[O=\text{ 3}\left( -2 \right)=-6\]
Oxidation state of Mn in \[M{{n}_{2}}{{O}_{3}}\] is
\[\begin{align}
  & \Rightarrow 2\text{x-6}=0 \\
 & \Rightarrow 2x=+6 \\
 & \Rightarrow x=\dfrac{6}{2} \\
 & \Rightarrow x=3 \\
\end{align}\]
3. \[Mn{{O}_{2}}\]
Let, the Oxidation state of \[Mn=x\]
Oxidation state of \[O=\text{ 2}\left( -2 \right)=-4\]
Oxidation state of Mn in \[Mn{{O}_{2}}\] is
\[\begin{align}
  & \Rightarrow \text{x-4=0} \\
 & \Rightarrow x=+4 \\
\end{align}\]
So, the oxidation states of Mn in \[MnO_{4}^{-}\],\[M{{n}_{2}}{{O}_{3}}\] and \[Mn{{O}_{2}}\]are \[+7,+3\text{ }and\text{ }+5\] respectively.

Note:
The oxidation state of any free element is always zero, for monatomic ions it is the same as the charge on them. In peroxides , it has an oxidation number as 1and for polyatomic ions it is equal to the net charge of the ion.