Step 1: Understanding the Concept:
The oxidation number (or oxidation state) is the formal charge an atom would have if all bonds were completely ionic.
We need to calculate the oxidation state of Manganese (\(\text{Mn}\)) in both given compounds and then find the absolute difference between them.
Step 2: Key Formula or Approach:
Use the formula \(\sum \text{Oxidation states} = \text{Overall charge}\).
Assign standard oxidation states (K = +1, O = -2) to find the oxidation number of Mn in each compound.
Step 3: Detailed Explanation:
Let's calculate the oxidation number of \(\text{Mn}\) in potassium permanganate, \(\text{KMnO}_4\).
Let the oxidation state of \(\text{Mn}\) be \(x\).
\[ (+1) + x + 4(-2) = 0 \]
\[ 1 + x - 8 = 0 \]
\[ x - 7 = 0 \]
\[ x = +7 \]
So, the oxidation number of \(\text{Mn}\) in \(\text{KMnO}_4\) is \(+7\).
Now, let's calculate the oxidation number of \(\text{Mn}\) in manganese dioxide, \(\text{MnO}_2\).
Let the oxidation state of \(\text{Mn}\) be \(y\).
\[ y + 2(-2) = 0 \]
\[ y - 4 = 0 \]
\[ y = +4 \]
So, the oxidation number of \(\text{Mn}\) in \(\text{MnO}_2\) is \(+4\).
Finally, find the difference between the two oxidation numbers.
\[ \text{Difference} = |(+7) - (+4)| \]
\[ \text{Difference} = 7 - 4 = 3 \]
Step 4: Final Answer:
The difference in the oxidation numbers is 3.