Answer: Option (d) is the correct answer.
Explanation:
In a \(KMnO_{4}\) molecule, there is one potassium atom, one manganese atom and four oxygen atoms.
Let the oxidation state of manganese is x. And it is known that oxidation number of potassium is +1 and oxidation number of oxygen is -2.
Therefore, calculate the oxidation number of manganese (Mn) in potassium permanganate (\(KMnO_{4}\)) as follows.
\(1 \times oxidation no. of K + oxidation no. of Mn + 4 \times oxidation no. of oxygen\) = charge on molecule
\((1 \times 1) + x + (4 \times -2)\) = 0
1 + x - 8 = 0
x = +7
Thus, we can conclude that oxidation state of number of manganese (Mn) in potassium permanganate (\(KMnO_{4}\)) is +7.
