Answer

Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state.

Answer

Chromium (Cr) : The atomic number of chromium is 24 . The electron configuration for neutral chromium is $ (Ar)3d^54s^1 $ . For $ Cr^{2+} $ , we remove two electrons , resulting in $ (Ar)3d^4 $ .

Iron (Fe) : The atomic number of iron is 26 . The electron configuration for neutral iron is $ (Ar)3d^64s^2 $ . For $ Fe^{2+} $ , we remove two electrons , resulting in $ (Ar)3d^6 $ .

When $ Cr^{2+} $ oxidizes to $ Cr^{3+} $ , it changes from $ (Ar)3d^4 $ to $ (Ar)3d^3 $ . The $ 3d^3 $ configuration is stable because it has a half-filled $ t_2g $ subshell . When $ Fe^{2+} $ oxidizes to $ Fe^{3+} $ , it changes from $ (Ar)3d^6 $ to $ (Ar)3d^5 $ . The $ 3d^5 $ configuration is also stable due to its has a half-filled nature .

Although both $ Cr^{3+} $ and $ Fe^{3+} $ are stable , the half-filled $ t_2g $ configuration of $ Cr^{3+} $ provides additional stability compared to $ Fe^{3+} $ .

A stronger reducing agent is one that more readily loses electrons (oxidizes) . $ Cr^{2+} $ has a greater tendency to oxidize to $ Cr^{3+} $ due to the stability of the resulting half-filled $ t_2g $ configuration . In contrast , $ Fe^{2+} $ is relatively more stable and does not oxidize as readily .

Thus , $ Cr^{2+} $ is a stronger reducing agent than $ Fe^{2+} $ because it more readily oxidizes to $ Cr^{3+} $ , which has a stable half-filled electronic configuration .