For Na$$^+$$, Mg$$^{2+}$$, F$$^-$$ and O$$^{2-}$$; the correct order of increasing ionic radii is:

We begin by noting that the four given ions $$\text{Na}^+,\; \text{Mg}^{2+},\; \text{F}^-,\; \text{O}^{2-}$$ are isoelectronic. This term means that every one of them possesses the same total number of electrons. Let us verify this explicitly.

• Sodium atom has atomic number $$11$$, so neutral Na has $$11$$ electrons. After losing one electron to form $$\text{Na}^+$$, the ion has $$11-1 = 10$$ electrons.
• Magnesium atom has atomic number $$12$$, thus neutral Mg has $$12$$ electrons. Losing two electrons to become $$\text{Mg}^{2+}$$ leaves $$12-2 = 10$$ electrons.
• Fluorine atom has atomic number $$9$$, hence neutral F has $$9$$ electrons. Gaining one electron to form $$\text{F}^-$$ gives $$9+1 = 10$$ electrons.
• Oxygen atom has atomic number $$8$$, so neutral O has $$8$$ electrons. Gaining two electrons to create $$\text{O}^{2-}$$ yields $$8+2 = 10$$ electrons.

Thus, all the species contain $$10$$ electrons and belong to the same electronic configuration as the noble-gas neon: $$1s^2\,2s^2\,2p^6$$.

For a set of isoelectronic species, the ionic radius depends primarily on the effective nuclear charge, denoted $$Z_{\text{eff}}$$. The rule is:

Greater $$Z_{\text{eff}}$$ $$\Rightarrow$$ stronger attraction of the nucleus for the same electron cloud $$\Rightarrow$$ smaller ionic radius.

The formal nuclear charge $$Z$$ for each ion equals the atomic number of the element because the number of protons in the nucleus does not change during ion formation. We therefore list the actual nuclear charges:

$$Z(\text{Mg}^{2+}) = 12,\quad Z(\text{Na}^+) = 11,\quad Z(\text{F}^-) = 9,\quad Z(\text{O}^{2-}) = 8.$$

As $$Z$$ increases, the electrons feel a stronger pull. Since the shielding is identical (same electron count and arrangement), the effective nuclear charge order mirrors the actual nuclear charge order:

$$Z_{\text{eff}}(\text{Mg}^{2+}) > Z_{\text{eff}}(\text{Na}^+) > Z_{\text{eff}}(\text{F}^-) > Z_{\text{eff}}(\text{O}^{2-}).$$

Applying our radius rule—higher $$Z_{\text{eff}}$$ means smaller radius—we invert this sequence to obtain the order of increasing ionic radii:

$$\text{Mg}^{2+} \;<\; \text{Na}^+ \;<\; \text{F}^- \;<\; \text{O}^{2-}.$$

We now compare this derived order with the options provided. Option C reads:

$$\text{Mg}^{2+} < \text{Na}^+ < \text{F}^- < \text{O}^{2-}.$$

This exactly matches our calculated sequence.

Hence, the correct answer is Option C.