Idea: Isoelectronic species have the same number of electrons.

Find electrons in each species using:

• Neutral atom: electrons = atomic number
• Cation: electrons = atomic number $-$ positive charge
• Anion: electrons = atomic number $+$ magnitude of negative charge

Atomic numbers needed:

• $\mathrm{O} = 8$
• $\mathrm{F} = 9$
• $\mathrm{Na} = 11$
• $\mathrm{Mg} = 12$
• $\mathrm{Al} = 13$

Count electrons for each species:

1. $\mathrm{O}^{2-}$ : $8 + 2 = 10$
2. $\mathrm{F}^{-}$ : $9 + 1 = 10$
3. $\mathrm{Al}$ : $13$
4. $\mathrm{Mg}^{2+}$ : $12 - 2 = 10$
5. $\mathrm{Na}^{+}$ : $11 - 1 = 10$
6. $\mathrm{O}^{+}$ : $8 - 1 = 7$
7. $\mathrm{Mg}$ : $12$
8. $\mathrm{Al}^{3+}$ : $13 - 3 = 10$
9. $\mathrm{F}$ : $9$

Group the isoelectronic species:

Species having $10$ electrons are: $\mathrm{O}^{2-},\ \mathrm{F}^{-},\ \mathrm{Mg}^{2+},\ \mathrm{Na}^{+},\ \mathrm{Al}^{3+}$

These are all isoelectronic with Ne.

Therefore, the total number of isoelectronic species in the set is $5$

Comparison with stored answer: Stored correct answer = $5$, which matches our result.