Question

What is the oxidation number of $\text{Mn}$ in $\text{MnS}{{\text{O}}_{\text{4}}},\text{ KMn}{{\text{O}}_{4}}$ and ${{\text{K}}_{\text{2}}}\text{Mn}{{\text{O}}_{4}}$?

Answer 1

Hint: The oxidation number, in simple terms, is basically the number of electrons that atoms in a molecule can lose, gain, or share while forming chemical bonds with other atoms of a different element. The sum of oxidation numbers of all atoms in a neutral molecule is always equal to 0.

Complete answer:
The atoms form bonds with other atoms to gain stability and they do so by transferring or sharing electrons. The number of electrons that an atom can either lose, gain or share during bond formation is known as its oxidation number.
When an atom loses electrons, it has a positive oxidation number while on the other hand when an atom gains an electron, it has a negative oxidation state.
There are a few points that need to be considered before finding out the oxidation number of an atom in a molecule.
- The net charge on neutral atoms or molecules is zero. So, the overall oxidation state is also zero. For example, the oxidation state of elemental atoms such as sodium and the oxidation state of neutral molecules such as oxygen molecules is zero.
- The oxidation state of charged ions is equal to the net charge of the ion.
Now, let’s calculate the oxidation number of manganese in $\text{MnS}{{\text{O}}_{\text{4}}}$, $\text{KMn}{{\text{O}}_{4}}$ and ${{\text{K}}_{\text{2}}}\text{Mn}{{\text{O}}_{4}}$:
1 – All of these are neutral molecules. So, the net charge and overall oxidation number must be 0 in all of them.
2 – For $\text{MnS}{{\text{O}}_{\text{4}}}$we can write:
\[\begin{align}
  & \text{O}\text{.N}\text{. of Mn}+\text{O}\text{.N}\text{. of S}{{\text{O}}_{\text{4}}}=0 \\
 & \Rightarrow \text{O}\text{.N}\text{. of Mn}=-\text{O}\text{.N}\text{. of S}{{\text{O}}_{\text{4}}} \\
 & \because \text{O}\text{.N}\text{. of S}{{\text{O}}_{\text{4}}}=-2 \\
 & \Rightarrow \text{O}\text{.N}\text{. of Mn}=-\left( -2 \right)=+2 \\
\end{align}\]
2 – For $\text{KMn}{{\text{O}}_{4}}$we can write:
\[\begin{align}
  & \text{O}\text{.N}\text{. of K}+\text{O}\text{.N}\text{. of Mn}+4\left( \text{O}\text{.N}\text{. of O} \right)=0 \\
 & \Rightarrow \text{O}\text{.N}\text{. of Mn}=-\left\{ \text{O}\text{.N}\text{. of K}+4\left( \text{O}\text{.N}\text{. of O} \right) \right\} \\
 & \because \text{O}\text{.N}\text{. of K}=+1\text{ and O}\text{.N}\text{. of O}=-2 \\
 & \Rightarrow \text{O}\text{.N}\text{. of Mn}=-\left\{ +1+4\left( -2 \right) \right\}=+7 \\
\end{align}\]
2 – For ${{\text{K}}_{\text{2}}}\text{Mn}{{\text{O}}_{4}}$ we can write:
\[\begin{align}
  & \text{2}\left( \text{O}\text{.N}\text{. of K} \right)+\text{O}\text{.N}\text{. of Mn}+4\left( \text{O}\text{.N}\text{. of O} \right)=0 \\
 & \Rightarrow \text{O}\text{.N}\text{. of Mn}=-\left\{ \text{2}\left( \text{O}\text{.N}\text{. of K} \right)+4\left( \text{O}\text{.N}\text{. of O} \right) \right\} \\
 & \because \text{O}\text{.N}\text{. of K}=+1\text{ and O}\text{.N}\text{. of O}=-2 \\
 & \Rightarrow \text{O}\text{.N}\text{. of Mn}=-\left\{ 2\left( +1 \right)+4\left( -2 \right) \right\}=+6 \\
\end{align}\]
Hence, the oxidation number of $\text{Mn}$ in $\text{MnS}{{\text{O}}_{\text{4}}},\text{ KMn}{{\text{O}}_{4}}$ and ${{\text{K}}_{\text{2}}}\text{Mn}{{\text{O}}_{4}}$ is +2, +7 and +6 respectively.

Note:
Since the numbers of electrons are whole numbers, the oxidation number of individual atoms also has to be a whole integer. But, some molecules contain atoms that possess a fractional oxidation number. The fractional oxidation state is always an average oxidation number of the same atoms in a molecule and does not reflect the true oxidation number.