Identify the species and their electron count

The species are:

• $\mathrm{F^-}$
• $\mathrm{Ne}$
• $\mathrm{Na^+}$

Number of electrons in each:

• $\mathrm{F^-}: 9+1 = 10$
• $\mathrm{Ne}: 10$
• $\mathrm{Na^+}: 11-1 = 10$

So, all three are isoelectronic species.

Rule for isoelectronic species

In an isoelectronic series, all species have the same number of electrons and generally the same principal shell occupancy.

Therefore, the factor that mainly decides their size is the nuclear charge $Z$.

Effect of nuclear charge

As nuclear charge increases, the attraction between nucleus and electrons increases, so the electron cloud contracts and size decreases.

Nuclear charges are:

• For $\mathrm{F^-}$: $Z=9$
• For $\mathrm{Ne}$: $Z=10$
• For $\mathrm{Na^+}$: $Z=11$

Hence, size order is: $\mathrm{F^-} > \mathrm{Ne} > \mathrm{Na^+}$

Check the options

• A: Nuclear charge $(Z)$ → Correct
• B: None of the factors because their size is the same → Incorrect, their sizes are not the same
• C: Electron-electron interaction in the outer orbitals → Not the main differentiating factor here, since all are isoelectronic
• D: Principal quantum number $(n)$ → Same outer shell pattern here, so not the reason for size difference

Final answer

The size of $\mathrm{F^-}$, $\mathrm{Ne}$ and $\mathrm{Na^+}$ is mainly affected by: $\boxed{\text{Nuclear charge }(Z)}$