To determine which species has the maximum number of lone pairs of electrons on the central atom, we'll examine the electron configuration of each molecule's central atom:

Option A: $$[ClO_3]^-$$

The central atom is chlorine (Cl).

Chlorine has 7 valence electrons. In the $$[ClO_3]^-$$ ion, chlorine forms three bonds with oxygen atoms and there is an additional negative charge (one extra electron). Thus, chlorine has:

$$7 + 1 - 3 = 5$$ valence electrons left, which form 2.5 lone pairs. However, due to the requirement to minimize formal charges, typically chlorine in $$[ClO_3]^-$$ is involved in resonance structures reducing the average lone pairs per resonance structure to 1.

So, $$[ClO_3]^-$$ has 1 lone pair on the central atom.

Option B: $$XeF_4$$

The central atom is xenon (Xe).

Xenon has 8 valence electrons. In $$XeF_4$$, xenon forms four bonds with four fluorine atoms. Thus, xenon has:

$$8 - 4 = 4$$ valence electrons left, which form 2 lone pairs.

So, $$XeF_4$$ has 2 lone pairs on the central atom.

Option C: $$SF_4$$

The central atom is sulfur (S).

Sulfur has 6 valence electrons. In $$SF_4$$, sulfur forms four bonds with four fluorine atoms. Thus, sulfur has:

$$6 - 4 = 2$$ valence electrons left, which form 1 lone pair.

So, $$SF_4$$ has 1 lone pair on the central atom.

Option D: $$[I_3]^-$$

The central atom is iodine (I).

Iodine has 7 valence electrons. In the $$[I_3]^-$$ ion, iodines form a linear structure where the central iodine forms two bonds with two other iodine atoms and there is an additional negative charge (one extra electron). Thus, the central iodine has:

$$7 + 1 - 2 = 6$$ valence electrons left, which form 3 lone pairs.

So, $$[I_3]^-$$ has 3 lone pairs on the central atom.

Hence, the species with the maximum number of lone pairs of electrons on the central atom is:

Option D: $$[I_3]^-$$