Idea of isoelectronic species

Isoelectronic species have the same number of electrons.

Find electrons in each ion

Use:

• For a neutral atom, number of electrons = atomic number
• For an anion, add electrons equal to the negative charge
• For a cation, subtract electrons equal to the positive charge

Atomic numbers:

• $\mathrm{Li} = 3$
• $\mathrm{N} = 7$
• $\mathrm{O} = 8$
• $\mathrm{F} = 9$
• $\mathrm{Na} = 11$
• $\mathrm{Mg} = 12$

Check each option

Option A: $F^-$, $Li^+$, $Na^+$ and $Mg^{2+}$

• $F^-: 9 + 1 = 10$ electrons
• $Li^+: 3 - 1 = 2$ electrons
• $Na^+: 11 - 1 = 10$ electrons
• $Mg^{2+}: 12 - 2 = 10$ electrons

These are $10, 2, 10, 10$ → not all same.

Option B: $Li^+$, $Na^+$, $O^{2-}$ and $F^-$

• $Li^+: 3 - 1 = 2$ electrons
• $Na^+: 11 - 1 = 10$ electrons
• $O^{2-}: 8 + 2 = 10$ electrons
• $F^-: 9 + 1 = 10$ electrons

These are $2, 10, 10, 10$ → not all same.

Option C: $N^{3-}$, $O^{2-}$, $F^-$ and $Na^+$

• $N^{3-}: 7 + 3 = 10$ electrons
• $O^{2-}: 8 + 2 = 10$ electrons
• $F^-: 9 + 1 = 10$ electrons
• $Na^+: 11 - 1 = 10$ electrons

These are $10, 10, 10, 10$ → all are isoelectronic.

Option D: $N^{3-}$, $Li^+$, $Mg^{2+}$ and $O^{2-}$

• $N^{3-}: 7 + 3 = 10$ electrons
• $Li^+: 3 - 1 = 2$ electrons
• $Mg^{2+}: 12 - 2 = 10$ electrons
• $O^{2-}: 8 + 2 = 10$ electrons

These are $10, 2, 10, 10$ → not all same.

Conclusion

The correct isoelectronic set is: $\boxed{N^{3-},\ O^{2-},\ F^-,\ Na^+}$

So, the correct option is C.