To determine the correct order of ionic radii, we analyze the species given:

1. Isoelectronic Species: The ions $N^{3-}$ ($Z=7$), $Mg^{2+}$ ($Z=12$), and $Al^{3+}$ ($Z=13$) are isoelectronic, meaning they all possess the same number of electrons (10 electrons).
2. Trend: For isoelectronic species, the ionic radius decreases as the nuclear charge (atomic number, Z) increases. This is because a higher nuclear charge exerts a stronger attraction on the same number of electrons, pulling the electron cloud closer to the nucleus.
3. Comparison: $N^{3-}$: $Z=7$ (Smallest nuclear charge $\rightarrow$ Largest radius) $Mg^{2+}$: $Z=12$

• $Al^{3+}$: $Z=13$ (Largest nuclear charge $\rightarrow$ Smallest radius)

4. Conclusion: The correct order is $N^{3-} > Mg^{2+} > Al^{3+}$.

(Note: Option 1 is incorrect because the radius of a hydride ion $H^-$ is larger than the hydrogen atom $H$, which is larger than a proton $H^+$. Options 2 and 3 are incorrect because for the isoelectronic series $O^{2-}, F^-, Na^+$, the order should be $O^{2-} > F^- > Na^+$).