Question

Manganese forms oxide and fluoride with highest oxidation state. The difference in the highest oxidation state of Mn in the oxide and fluoride is:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This question requires knowledge of the typical chemical compounds formed by manganese (Mn) with oxygen and fluorine to determine their maximum achievable oxidation states.
Step 2: Detailed Explanation:
Manganese (Mn) exhibits its highest oxidation state with oxygen in the compound \(Mn_{2}O_{7}\). In this oxide, the oxidation state of Mn is \(+7\). Oxygen is capable of stabilizing such high oxidation states because it can form multiple \(d\pi - p\pi\) bonds with the metal.

With fluorine, the highest oxidation state manganese can reach in a binary compound is \(+4\), as seen in \(MnF_{4}\). Higher fluorides like \(MnF_{7}\) are not known. Although fluorine is more electronegative than oxygen, its inability to form multiple bonds prevents it from stabilizing oxidation states higher than \(+4\) for manganese.

The difference between these maximum oxidation states is:
\[ \text{Difference} = 7 - 4 = 3 \]
Step 3: Final Answer:
The difference in the highest oxidation state of Mn in the oxide and fluoride is 3, corresponding to option (C).