1. Identify the ions and their electron counts

We compare the ionic radii of:

• $O^{2-}$
• $B^{3+}$
• $Li^+$
• $F^-$

Let us find the number of electrons in each ion:

• $O$ has atomic number $8$, so $O^{2-}$ has $8+2=10$ electrons.
• $B$ has atomic number $5$, so $B^{3+}$ has $5-3=2$ electrons.
• $Li$ has atomic number $3$, so $Li^+$ has $3-1=2$ electrons.
• $F$ has atomic number $9$, so $F^-$ has $9+1=10$ electrons.

So:

• $O^{2-}$ and $F^-$ are isoelectronic with $10$ electrons.
• $Li^+$ has only $2$ electrons.
• $B^{3+}$ also has only $2$ electrons.

2. Use periodic trends for ionic size

(a) Among isoelectronic species

For isoelectronic ions, the ion with smaller nuclear charge has larger radius, because the same number of electrons are held less strongly.

Compare $O^{2-}$ and $F^-$:

• $O^{2-}$: nuclear charge $=8$
• $F^-$: nuclear charge $=9$

Since $O^{2-}$ has the smaller nuclear charge, it has the larger ionic radius.

Therefore, $O^{2-} > F^-$

(b) Compare cations with anions

• Cations like $Li^+$ and $B^{3+}$ are much smaller because losing electrons reduces electron-electron repulsion and often removes the outer shell.
• Anions like $O^{2-}$ and $F^-$ are larger because gaining electrons increases electron-electron repulsion.

So both $Li^+$ and $B^{3+}$ are much smaller than $O^{2-}$ and $F^-$.

(c) Between $Li^+$ and $B^{3+}$

Both are isoelectronic with $2$ electrons.

• $Li^+$ has nuclear charge $3$
• $B^{3+}$ has nuclear charge $5$

Hence, $Li^+ > B^{3+}$

3. Overall order

Combining the above results: $O^{2-} > F^- > Li^+ > B^{3+}$

Thus, the ion with the highest ionic radius is: $\boxed{O^{2-}}$

4. Compare with stored correct answer

Stored correct answer: A

Our derived answer: A ($O^{2-}$)

They match.