1. Definition of isoelectronic species

   Isoelectronic species are atoms/ions/molecules having the same number of total electrons.

2. Check each option by counting electrons

Option A: $N_3^-$, $F^-$, $Na^+$

• For $N_3^-$: Each nitrogen has atomic number $7$. $3 \times 7 = 21$ Negative charge $-1$ means one extra electron: $21 + 1 = 22$

• For $F^-$: Fluorine has atomic number $9$. $9 + 1 = 10$

• For $Na^+$: Sodium has atomic number $11$. $11 - 1 = 10$

These are not all equal. So option A is not isoelectronic.

Note: If the intended species were $N^{3-}$ instead of $N_3^-$, then it would have $7+3=10$ electrons and option A would become correct. But as written, $N_3^-$ has $22$ electrons.

Option B: $Be$, $Al^{3+}$, $Cl^-$

• $Be$: atomic number $4$ $4 \text{ electrons}$
• $Al^{3+}$: atomic number $13$ $13 - 3 = 10$
• $Cl^-$: atomic number $17$ $17 + 1 = 18$

Not equal, so not isoelectronic.

Option C: $Ca^{2+}$, $Cs^+$, $Br$

• $Ca^{2+}$: atomic number $20$ $20 - 2 = 18$
• $Cs^+$: atomic number $55$ $55 - 1 = 54$
• $Br$: atomic number $35$ $35$

Not equal, so not isoelectronic.

Option D: $Na^+$, $Ca^{2+}$, $Mg^{2+}$

• $Na^+$: atomic number $11$ $11 - 1 = 10$
• $Ca^{2+}$: atomic number $20$ $20 - 2 = 18$
• $Mg^{2+}$: atomic number $12$ $12 - 2 = 10$

Not all equal, so not isoelectronic.

3. Conclusion

As written, none of the options contains a collection in which all species are isoelectronic.

The stored answer says A, which would be correct only if the first species were $N^{3-}$ instead of $N_3^-$.

Thus, there appears to be a misprint in the question/options.