Step 1: Understanding the Concept:
The ability of an element to exhibit a high oxidation state depends heavily on the bonding characteristics of the electronegative atoms it coordinates with.
Step 2: Key Formula or Approach:
Analyze the valency and bonding capabilities (single vs. multiple bonds) and the resulting steric crowding effects of fluorine compared to oxygen.
Step 3: Detailed Explanation:
1. Fluorine is a monovalent halogen and can only form a single covalent bond (sigma bond) with the central manganese atom. To reach an oxidation state of $+7$ solely with fluorine, Mn would theoretically need to be surrounded by 7 bulky fluorine atoms ($MnF_7$), which creates impossible steric hindrance and repulsion, making such a molecule highly unstable.
2. Oxygen, being divalent, has the ability to form multiple bonds (double bonds utilizing $p\pi-d\pi$ interactions). This allows oxygen to stabilize the highest oxidation state of Mn ($+7$ in $Mn_{2}O_{7}$ or $MnO_{4}^{-}$) using significantly fewer coordinating atoms. Only 4 oxygen atoms are needed to reach the valence requirements, vastly reducing steric crowding.
Step 4: Final Answer:
The ability of oxygen to form double bonds is the primary reason it can stabilize higher oxidation states than fluorine.