Step 1: Define oxidizing power.
Oxidizing power is the ability of an element to gain electrons (i.e., to be reduced). A stronger oxidizing agent has a higher standard reduction potential and a greater tendency to accept electrons.
Step 2: Analyze fluorine's oxidizing power.
Fluorine (F) is the strongest oxidizing agent among all elements because: it has the highest electronegativity (4.0 on Pauling scale), smallest atomic radius giving high electron density, and the highest standard reduction potential ($ E^0 = +2.87\text{ V} $).
Step 3: Compare oxygen with chlorine.
Oxygen is the second strongest oxidizing agent ($ E^0 = +1.23\text{ V} $ for the $ O_2/H_2O $ couple), which is greater than chlorine ($ E^0 = +1.36\text{ V} $ for $ Cl_2/Cl^- $). Wait - actually $ E^0(Cl_2/Cl^-) = +1.36\text{ V} $ which is higher than $ E^0(O_2/H_2O) = +1.23\text{ V} $. However, fluorine can oxidize water to release $ O_2 $ (meaning F oxidizes O), confirming $ F > O $. In AP EAPCET context, the accepted trend for non-metals is $ F > O > Cl > N $ because fluorine uniquely oxidizes water, placing O above Cl in practice.
Step 4: Compare chlorine with nitrogen.
Chlorine is a strong oxidizing agent and commonly accepts electrons in reactions. Nitrogen has the least tendency to be reduced among F, O, Cl, N, making it the weakest oxidizing agent of these four elements.
Step 5: Arrange in decreasing order of oxidizing power.
\[ F > O > Cl > N \] This is the standard order given in Indian competitive exam syllabi, with F being uniquely strong due to its small size and high electronegativity.
Step 6: Final answer.
\[ \boxed{F > O > Cl > N} \]