The ionic radii of F- and O2- respectively are 1.33 A and 1.4 A, while the covalent radius of N is 0.74 A. The correct…

🧠 $\ce{N^{3-}}$ has only 7 protons holding 10 electrons — that is the weakest pull in the set The given values: $\ce{F^-} = 1.33\,\text{Å}$, $\ce{O^{2-}} = 1.40\,\text{Å}$, $\ce{N}$ atom $= 0.74\,\text{Å}$. Now you need to place $\ce{N^{3-}}$ in this picture.

$\ce{N^{3-}}$, $\ce{O^{2-}}$, $\ce{F^-}$ are all isoelectronic (10 electrons each). Z values are 7, 8, 9 respectively. Lowest Z = biggest ion. So $\ce{N^{3-}} > \ce{O^{2-}} > \ce{F^-}$.

🗺️ Use the data to confirm $\ce{F^-}$ (Z=9): 1.33 Å. $\ce{O^{2-}}$ (Z=8): 1.40 Å — already bigger than $\ce{F^-}$ because it has one less proton pulling the same 10 electrons. $\ce{N^{3-}}$ (Z=7): even fewer protons. Must be bigger than 1.40 Å.

So $\ce{N^{3-}}$ is bigger than both $\ce{O^{2-}}$ and $\ce{F^-}$.

⚠️ The trap Option (a) and (d) both compare $\ce{N^{3-}}$ to a neutral $\ce{N}$ atom. Students see N atom = 0.74 Å (small) and think $\ce{N^{3-}}$ should also be small. That is backwards. When N picks up 3 extra electrons to become $\ce{N^{3-}}$, the cloud expands massively because the same 7 protons must now hold 10 electrons. The anion is far bigger than the neutral atom — about three times the radius. Never compare a neutral atom radius directly with an ionic radius.

$\boxed{\text{Answer: (b)}}$