To determine the decreasing order of ionic size for the ions Mg²⁺, O^2-, Na⁺, and F^-, we need to consider several factors such as nuclear charge, electrons’ configurations, and their positions in the periodic table.

Firstly, when ions have the same electronic configuration (which we can think of as isoelectronic), the nuclear charge and its effective pull on the electrons will determine their sizes. A higher positive charge (more protons in the nucleus) results in a smaller ionic radius because the protons pull the electron cloud more tightly towards the nucleus.

The ions Mg²⁺, O^2-, Na⁺, and F^- are all approximately isoelectronic, having the same number of electrons as the noble gas Ne (10 electrons). Thus, the order of their ionic sizes will mainly depend on their nuclear charge:

O^2- (8 protons)
F^- (9 protons)
Na⁺ (11 protons)
Mg²⁺ (12 protons)

Higher proton count correlates with greater effective nuclear charge, leading to a more strongly pulled-in electron cloud, and thus a smaller ionic radius.

Therefore, the decreasing order of ionic size from largest to smallest is:

$$ \text{O}^{2-} > \text{F}^{-} > \text{Na}^{+} > \text{Mg}^{2+} $$

O^2-, having the fewest protons among these ions, exhibits the largest ionic size due to a lesser effective nuclear pull on its electron cloud, compared to the other ions with more protons. Conversely, Mg²⁺ is the smallest as it has the highest number of protons exerting the strongest pull on the same number of electrons.