Answer
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Hint: Oxidation number is also known as oxidation state. It is a number which tells us how many electrons have been lost or gained by an atom. Single elements have the oxidation state of zero always in every reaction.
Complete step by step solution:
-To determine the oxidation state of an atom, we will consider it as 'X'.
- So to calculate the oxidation state of manganese in \[\text{KMn}{{\text{O}}_{4}}\], we will consider the oxidation state of manganese as 'X'.
-We know that the oxidation state of potassium and oxygen are +1 & -2.
-So, the equation will become:
$\begin{align}
& \text{+1 + x - 8 = 0} \\
& \text{x = +7} \\
\end{align}$
-Hence, the oxidation of Mn is +7.
-Similarly, the oxidation state of Mn in ${{\text{K}}_{2}}\text{Mn}{{\text{O}}_{4}}$ will be
$\begin{align}
& +2\text{ + x - 8 = 0} \\
& \text{x = +6} \\
\end{align}$
-So, the oxidation state of Mn is +6
-Now, the oxidation state of Mn in $\text{MnS}{{\text{O}}_{4}}$ will be
$\begin{align}
& \text{x + 6 - 8 = 0} \\
& \text{x - 2 = 0} \\
& \text{x = +2} \\
\end{align}$
-We know that the oxidation state of sulphur in sulphate ion is +6 so we have put +6 as an oxidation state of sulphur.
-So, the oxidation state of Mn is +2.
- Now, we will calculate the oxidation state of Mn in $\text{Mn}{{\text{O}}_{2}}$
$\begin{align}
& \text{x - 4 = 0} \\
& \text{x = +4} \\
\end{align}$
-So, the oxidation state of Mn is +2.
-Now, we will calculate the oxidation state of Mn in \[\text{M}{{\text{n}}_{3}}{{\text{O}}_{4}}\]
\[\begin{align}
& 3\text{x - 8 = 0} \\
& \text{x = +}\dfrac{8}{3} \\
\end{align}\]
-So, the oxidation state of Mn is + 8/3.
Therefore, option A. is the correct answer.
Note: Potassium permanganate comes under the category of strong oxidising agents. Sulphur has many oxidation states such as -2, 0, +2, +4, +6 and it depends on the bonding of the sulphur with other atoms. Like in Manganese sulfate sulphur has an oxidation state of +6.
Complete step by step solution:
-To determine the oxidation state of an atom, we will consider it as 'X'.
- So to calculate the oxidation state of manganese in \[\text{KMn}{{\text{O}}_{4}}\], we will consider the oxidation state of manganese as 'X'.
-We know that the oxidation state of potassium and oxygen are +1 & -2.
-So, the equation will become:
$\begin{align}
& \text{+1 + x - 8 = 0} \\
& \text{x = +7} \\
\end{align}$
-Hence, the oxidation of Mn is +7.
-Similarly, the oxidation state of Mn in ${{\text{K}}_{2}}\text{Mn}{{\text{O}}_{4}}$ will be
$\begin{align}
& +2\text{ + x - 8 = 0} \\
& \text{x = +6} \\
\end{align}$
-So, the oxidation state of Mn is +6
-Now, the oxidation state of Mn in $\text{MnS}{{\text{O}}_{4}}$ will be
$\begin{align}
& \text{x + 6 - 8 = 0} \\
& \text{x - 2 = 0} \\
& \text{x = +2} \\
\end{align}$
-We know that the oxidation state of sulphur in sulphate ion is +6 so we have put +6 as an oxidation state of sulphur.
-So, the oxidation state of Mn is +2.
- Now, we will calculate the oxidation state of Mn in $\text{Mn}{{\text{O}}_{2}}$
$\begin{align}
& \text{x - 4 = 0} \\
& \text{x = +4} \\
\end{align}$
-So, the oxidation state of Mn is +2.
-Now, we will calculate the oxidation state of Mn in \[\text{M}{{\text{n}}_{3}}{{\text{O}}_{4}}\]
\[\begin{align}
& 3\text{x - 8 = 0} \\
& \text{x = +}\dfrac{8}{3} \\
\end{align}\]
-So, the oxidation state of Mn is + 8/3.
Therefore, option A. is the correct answer.
Note: Potassium permanganate comes under the category of strong oxidising agents. Sulphur has many oxidation states such as -2, 0, +2, +4, +6 and it depends on the bonding of the sulphur with other atoms. Like in Manganese sulfate sulphur has an oxidation state of +6.
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