Oxidation Number And Oxidation States

We will calculate the oxidation state of Mn in each compound:
(1) MnO $_2$ :
Let oxidation state of Mn be $x$ : $x%20%2B%202(-2)%20%3D%200%20%5CRightarrow%20x%20%3D%20%2B4$ (2) MnO $_4$ :
This is likely referring to the MnO $_4%5E-$ ion. $x%20%2B%204(-2)%20%3D%20-1%20%5CRightarrow%20x%20%3D%20%2B7$ (3) KMnO $_4$ :
Potassium is +1, and each oxygen is -2: $%2B1%20%2B%20x%20%2B%204(-2)%20%3D%200%20%5CRightarrow%20x%20%3D%20%2B7$ (4) Mn $_2$ O $_3$ : $2x%20%2B%203(-2)%20%3D%200%20%5CRightarrow%202x%20%3D%20%2B6%20%5CRightarrow%20x%20%3D%20%2B3$ So, the highest oxidation state is $%2B7$ , which occurs in both MnO $_4%5E-$ and KMnO $_4$ . Among the given named compounds, KMnO $_4$ is the correct compound.

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Oxidation Number And Oxidation States

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs