1. Check whether the ions are isoelectronic

We count the number of electrons in each species:

• $\mathrm{O}^{2-}$: Oxygen has $8$ electrons, so $8 + 2 = 10$
• $\mathrm{F}^{-}$: Fluorine has $9$ electrons, so $9 + 1 = 10$
• $\mathrm{Na}^{+}$: Sodium has $11$ electrons, so $11 - 1 = 10$
• $\mathrm{Mg}^{2+}$: Magnesium has $12$ electrons, so $12 - 2 = 10$

Thus, all four species have $10$ electrons.

So, (A) is correct.

2. Check whether all have the same nuclear charge

Nuclear charge depends on the number of protons, i.e. atomic number $Z$:

• Oxygen: $Z=8$
• Fluorine: $Z=9$
• Sodium: $Z=11$
• Magnesium: $Z=12$

These are clearly different.

So, (B) is incorrect.

3. Compare ionic radii in an isoelectronic series

For an isoelectronic series, radius decreases as nuclear charge increases, because the same number of electrons is attracted more strongly by a greater number of protons.

Here the nuclear charges increase as: $\mathrm{O}^{2-} < \mathrm{F}^{-} < \mathrm{Na}^{+} < \mathrm{Mg}^{2+}$ with $8 < 9 < 11 < 12$

Therefore, the ionic radii decrease in the order: $\mathrm{O}^{2-} > \mathrm{F}^{-} > \mathrm{Na}^{+} > \mathrm{Mg}^{2+}$

So:

• $\mathrm{O}^{2-}$ has the largest ionic radius → (C) is correct
• $\mathrm{Mg}^{2+}$ has the smallest ionic radius → (D) is correct

4. Select the correct option

Correct statements are: $\boxed{(A), (C) \text{ and } (D)}$

This matches Option B.

5. Comparison with stored correct answer

Stored correct answer = B

My derived answer = B

So, they agree.