To determine the correct order of ionic radii, we need to consider the following principles:$\\$• Ionic radius generally increases with increasing negative charge (more electrons) and decreases with increasing positive charge (fewer electrons).$\\$• For isoelectronic species (ions with the same number of electrons), the ionic radius decreases with increasing nuclear charge (more protons).$\\$Let's analyze each option:$\\$Option 1: $H^- > H^+ > H \\$• $H^-$ has one extra electron compared to $H,$ making it larger.$\\$• $H^+$ has lost its only electron, making it much smaller than $H. \\$• The correct order should be $H^- > H > H^+. \\$Option 2: $Na^+ > F^- > O^{2-} \\$•$Na^+, F^-, and O^{2-}$ are isoelectronic (all have 10 electrons).$\\$• The nuclear charge increases from$O^{2-} to F^-$ to $Na^+$, so the ionic radius decreases in this order.$\\$• The correct order should be $O^{2-} > F^- > Na^+. \\$Option 3: $F^- > O^{2-} > Na^+ \\$• As explained above, for isoelectronic species, the ionic radius decreases with increasing nuclear charge.$\\$• The correct order is $O^{2-} > F^- > Na^+. \\$Option 4: $Al^{3+} > Mg^{2+} > N^{3-} \\$•$Al^{3+}, Mg^{2+}, and \ N^{3-}$are not isoelectronic.$\\$• $Al^{3+}$ has lost more electrons compared to $Mg^{2+},$ making it smaller.$\\$•$N^{3-}$ has gained electrons, making it larger than both $Al^{3+} \ and \ Mg^{2+}. \\$• The correct order should be $N^{3-} > Mg^{2+} > Al^{3+}. \\$Therefore, the correct order of ionic radii is represented by Option 3: $F^- > O^{2-} > Na^+.$