Among the following species which option contains all isoelectronic species?

First, we recall the definition: two or more species are called isoelectronic when they possess the same total number of electrons. Hence, to test each option we must compute the electron count for every listed particle.

For a neutral atom the number of electrons equals its atomic number $$Z$$. For an anion we add the charge magnitude to $$Z$$ because extra electrons are present. For a cation we subtract the charge magnitude from $$Z$$ because electrons have been removed.

We now write the atomic numbers once, so they can be used repeatedly: $$O: Z=8,\; F: Z=9,\; Na: Z=11,\; Mg: Z=12.$$

Let us analyse every option one by one.

Option A: $$O^{-},\; F^{-},\; Na,\; Mg^{+}$$

Electron counts:

We have $$O^{-}: 8+1=9,$$ $$F^{-}: 9+1=10,$$ $$Na: 11,$$ $$Mg^{+}: 12-1=11.$$

The counts are $$9,\;10,\;11,\;11$$ which are not all equal, so Option A fails.

Option B: $$O^{2-},\; F^{-},\; Na,\; Mg^{2+}$$

Electron counts:

We have $$O^{2-}: 8+2=10,$$ $$F^{-}: 9+1=10,$$ $$Na: 11,$$ $$Mg^{2+}: 12-2=10.$$

The counts are $$10,\;10,\;11,\;10,$$ not all equal; Option B fails.

Option C: $$O^{-},\; F^{-},\; Na^{+},\; Mg^{2+}$$

Electron counts:

We have $$O^{-}: 8+1=9,$$ $$F^{-}: 9+1=10,$$ $$Na^{+}: 11-1=10,$$ $$Mg^{2+}: 12-2=10.$$

The counts are $$9,\;10,\;10,\;10,$$ not identical, so Option C fails.

Option D: $$O^{2-},\; F^{-},\; Na^{+},\; Mg^{2+}$$

Electron counts:

We have $$O^{2-}: 8+2=10,$$ $$F^{-}: 9+1=10,$$ $$Na^{+}: 11-1=10,$$ $$Mg^{2+}: 12-2=10.$$

All four species contain exactly $$10$$ electrons. Because every member of this set shares the same electron count, they are indeed isoelectronic.

Hence, the correct answer is Option D.