Step 1: Understanding the Question:
The problem asks us to first determine the oxidation number (or oxidation state) of the element Manganese (Mn) in two different chemical compounds: potassium permanganate (\( \text{KMnO}_4 \)) and manganese dioxide (\( \text{MnO}_2 \)). After finding both oxidation numbers, we need to calculate the difference between them.
Step 2: Key Formula or Approach:
To find the oxidation number of an element in a compound, we use a set of standard rules. The key rules for this problem are:
The sum of all oxidation numbers in a neutral compound is zero.
The oxidation number of an alkali metal (like Potassium, K) in a compound is +1.
The oxidation number of Oxygen (O) in most compounds is -2.
Step 3: Detailed Explanation:
Part A: Calculate the oxidation number of Mn in \( \text{KMnO}_4 \).
Let the oxidation number of Mn be \(x\). We set up an equation where the sum of oxidation numbers equals zero:
\[
(\text{Oxidation number of K}) + (\text{Oxidation number of Mn}) + 4 \times (\text{Oxidation number of O}) = 0
\]
Substituting the known values:
\[
(+1) + (x) + 4(-2) = 0
\]
\[
1 + x - 8 = 0
\]
\[
x - 7 = 0 \implies x = +7
\]
So, the oxidation number of Mn in \( \text{KMnO}_4 \) is +7.
Part B: Calculate the oxidation number of Mn in \( \text{MnO}_2 \).
Let the oxidation number of Mn be \(y\). We again set up an equation for this neutral compound:
\[
(\text{Oxidation number of Mn}) + 2 \times (\text{Oxidation number of O}) = 0
\]
Substituting the known values:
\[
(y) + 2(-2) = 0
\]
\[
y - 4 = 0 \implies y = +4
\]
So, the oxidation number of Mn in \( \text{MnO}_2 \) is +4.
Part C: Find the difference.
The difference is the oxidation number in \( \text{KMnO}_4 \) minus the oxidation number in \( \text{MnO}_2 \):
\[
\text{Difference} = (+7) - (+4) = 3
\]
Step 4: Final Answer:
Our calculation shows that the difference in the oxidation numbers is 3. This corresponds to option (C).