Identify the set of species that are isoelectronic with one another.

Isoelectronic species are those which have the same number of electrons.

For the given options, the number of electrons in each species is calculated as follows:

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

$P^{3-}$ has $15 + 3 = 18$ electrons, $S^{2-}$ has $16 + 2 = 18$ electrons, $Cl^{-}$ has $17 + 1 = 18$ electrons, $K$ has $19$ electrons.

$Cl^{-}$ has $17 + 1 = 18$ electrons, $K^{+}$ has $19 - 1 = 18$ electrons, $Ca^{2+}$ has $20 - 2 = 18$ electrons, $Sc^{3+}$ has $21 - 3 = 18$ electrons.

$N^{3-}$ has $7 + 3 = 10$ electrons, $C^{4-}$ has $6 + 4 = 10$ electrons, $O^{-}$ has $8 + 1 = 9$ electrons, $F^{-}$ has $9 + 1 = 10$ electrons.

The species $Cl^{-}$, $K^{+}$, $Ca^{2+}$, and $Sc^{3+}$ all contain exactly 18 electrons and are therefore isoelectronic with one another.

Answer: $Cl^{-}, K^{+}, Ca^{2+}, Sc^{3+}$