The size of ions is generally influenced by two main factors: the number of protons in the nucleus (nuclear charge) and the number of electrons around the nucleus (electronic configuration). Additionally, ions of the same period on the periodic table can be compared based on their charge and number of electrons.

In this set of options, all the ions are isoelectronic, meaning they all possess the same number of electrons (10 electrons each). They correspond to the electronic configuration of neon (Ne), a noble gas:

• N^3- is a nitride ion with a -3 charge.
• O^2- is an oxide ion with a -2 charge.
• F^- is a fluoride ion with a -1 charge.
• Na⁺ is a sodium ion with a +1 charge.

To determine which ion is smallest, we look at the effective nuclear charge, which increases across a period from left to right due to an increase in protons while maintaining the same electronic shield (same number of electrons). Therefore, as the number of protons increases, the attraction between the nucleus and the electrons increases, pulling the electrons closer to the nucleus and decreasing the radius of the ion.

Given the ions listed:

• Nitrogen (N) in N^3- has 7 protons.
• Oxygen (O) in O^2- has 8 protons.
• Fluorine (F) in F^- has 9 protons.
• Sodium (Na) in Na⁺ has 11 protons.

Based on this count, Na⁺ with the highest number of protons will have the highest effective nuclear charge. This leads to a greater attraction of electrons towards the nucleus, making Na⁺ the smallest ion among those listed.

Thus, the answer is Option D: Na⁺.