To determine the number of lone pairs on the central atom, we first need to draw the Lewis structure for each species and apply VSEPR theory.

Step 1: Determine the number of valence electrons and lone pairs for each central atom.

• A) ${\mathrm{ClO}}_3^{-}$ (Chlorate ion):
  Central atom: Cl. Valence electrons of Cl = 7. Oxygen atoms are terminal. One negative charge.
  Total valence electrons = 7 (Cl) + 3 $\times$ 6 (O) + 1 (charge) = 7 + 18 + 1 = 26.
  Number of bond pairs = 3 (Cl-O bonds). Number of electrons used in bonds = 3 $\times$ 2 = 6.
  Remaining electrons = 26 - 6 = 20. These form lone pairs on oxygen atoms and the central chlorine atom.
  Each oxygen needs 6 electrons (3 lone pairs). So, 3 $\times$ 6 = 18 electrons for oxygen lone pairs.
  Remaining electrons for Cl = 20 - 18 = 2. These 2 electrons form 1 lone pair on Cl.
• B) ${\mathrm{XeF}}_4$ (Xenon tetrafluoride):
  Central atom: Xe. Valence electrons of Xe = 8. Fluorine atoms are terminal.
  Total valence electrons = 8 (Xe) + 4 $\times$ 7 (F) = 8 + 28 = 36.
  Number of bond pairs = 4 (Xe-F bonds). Number of electrons used in bonds = 4 $\times$ 2 = 8.
  Remaining electrons = 36 - 8 = 28. These form lone pairs on fluorine atoms and the central xenon atom.
  Each fluorine needs 6 electrons (3 lone pairs). So, 4 $\times$ 6 = 24 electrons for fluorine lone pairs.
  Remaining electrons for Xe = 28 - 24 = 4. These 4 electrons form 2 lone pairs on Xe.
• C) ${\mathrm{SF}}_4$ (Sulfur tetrafluoride):
  Central atom: S. Valence electrons of S = 6. Fluorine atoms are terminal.
  Total valence electrons = 6 (S) + 4 $\times$ 7 (F) = 6 + 28 = 34.
  Number of bond pairs = 4 (S-F bonds). Number of electrons used in bonds = 4 $\times$ 2 = 8.
  Remaining electrons = 34 - 8 = 26. These form lone pairs on fluorine atoms and the central sulfur atom.
  Each fluorine needs 6 electrons (3 lone pairs). So, 4 $\times$ 6 = 24 electrons for fluorine lone pairs.
  Remaining electrons for S = 26 - 24 = 2. These 2 electrons form 1 lone pair on S.
• D) ${\mathrm{I}}_3^{-}$ (Triiodide ion):
  Central atom: I. Valence electrons of I = 7. Two terminal iodine atoms. One negative charge.
  Total valence electrons = 3 $\times$ 7 (I) + 1 (charge) = 21 + 1 = 22.
  The central iodine forms two single bonds with the other two iodine atoms.
  Number of bond pairs = 2 (I-I bonds). Number of electrons used in bonds = 2 $\times$ 2 = 4.
  Remaining electrons = 22 - 4 = 18. These form lone pairs on terminal iodine atoms and the central iodine atom.
  Each terminal iodine needs 6 electrons (3 lone pairs). So, 2 $\times$ 6 = 12 electrons for terminal iodine lone pairs.
  Remaining electrons for central I = 18 - 12 = 6. These 6 electrons form 3 lone pairs on the central I atom.

Comparing the number of lone pairs on the central atom for each species:

• ${\mathrm{ClO}}_3^{-}$: 1 lone pair
• ${\mathrm{XeF}}_4$: 2 lone pairs
• ${\mathrm{SF}}_4$: 1 lone pair
• ${\mathrm{I}}_3^{-}$: 3 lone pairs

Thus, ${\mathrm{I}}_3^{-}$ has the maximum number of lone pairs on its central atom.

Correct Answer: (D)