1. We need to find the number of lone pairs on the central atom in each species, then take the maximum.

The species are:

• $\mathrm{ClO_3^-}$
• $\mathrm{XeF_4}$
• $\mathrm{SF_4}$
• $\mathrm{I_3^-}$

2. For $\mathrm{ClO_3^-}$

Let chlorine be the central atom.

Total valence electrons: $7 + 3(6) + 1 = 26$

In chlorate ion, chlorine forms three bonds with oxygen and has one lone pair in the usual Lewis structure.

Using VSEPR, steric number around Cl is 4:

• 3 bond pairs
• 1 lone pair

So, lone pairs on central atom in $\mathrm{ClO_3^-}$: $1$

3. For $\mathrm{XeF_4}$

Xenon is the central atom.

Total valence electrons: $8 + 4(7) = 36$

Xe forms 4 bonds with F atoms. After completing octets, xenon has 2 lone pairs.

VSEPR: steric number $= 6$:

• 4 bond pairs
• 2 lone pairs

So, lone pairs on central atom in $\mathrm{XeF_4}$: $2$

4. For $\mathrm{SF_4}$

Sulfur is the central atom.

Total valence electrons: $6 + 4(7) = 34$

Sulfur forms 4 bonds with F atoms and retains 1 lone pair.

VSEPR: steric number $= 5$:

• 4 bond pairs
• 1 lone pair

So, lone pairs on central atom in $\mathrm{SF_4}$: $1$

5. For $\mathrm{I_3^-}$

The central iodine is the middle atom.

Total valence electrons: $3(7) + 1 = 22$

The central iodine forms 2 bonds with terminal iodine atoms.

After completing octets, the central iodine has 3 lone pairs.

VSEPR: steric number $= 5$:

• 2 bond pairs
• 3 lone pairs

So, lone pairs on central atom in $\mathrm{I_3^-}$: $3$

6. Now compare all values:

• $\mathrm{ClO_3^-} : 1$
• $\mathrm{XeF_4} : 2$
• $\mathrm{SF_4} : 1$
• $\mathrm{I_3^-} : 3$

Therefore, the maximum number of lone pairs on the central atom is: $\boxed{3}$

7. Comparison with stored answer

Stored correct answer = $3$

Our derived answer = $3$

So they agree.