To determine the hybridization of the central atom in each species, we first need to calculate the steric number (SN), which is the sum of the number of sigma bonds and lone pairs around the central atom. Hybridization is then determined as follows: SN = 2 (sp), SN = 3 (sp²), SN = 4 (sp³), SN = 5 (sp³d), SN = 6 (sp³d²).

Step 1: Determine the hybridization of P in PO₄^3-.
The central atom is P. The total number of valence electrons is 5 (P) + 4 * 6 (O) + 3 (charge) = 5 + 24 + 3 = 32. In PO₄^3-, phosphorus forms four single bonds with four oxygen atoms. There are no lone pairs on phosphorus. Thus, the number of sigma bonds = 4, and the number of lone pairs = 0. Steric number = 4 + 0 = 4. Therefore, the hybridization of P in PO₄^3- is sp³.

Step 2: Analyze option A) I in ICl₄^-.
The central atom is I. The total number of valence electrons is 7 (I) + 4 * 7 (Cl) + 1 (charge) = 7 + 28 + 1 = 36. Iodine forms four single bonds with four chlorine atoms. The remaining electrons form lone pairs: (36 - 4*2) / 2 = 28 / 2 = 14 electrons, which means 2 lone pairs. Thus, the number of sigma bonds = 4, and the number of lone pairs = 2. Steric number = 4 + 2 = 6. Therefore, the hybridization of I in ICl₄^- is sp³d².

Step 3: Analyze option B) S in SO₃.
The central atom is S. The total number of valence electrons is 6 (S) + 3 * 6 (O) = 6 + 18 = 24. In SO₃, sulfur forms three double bonds with three oxygen atoms (or resonance structures with one double and two single bonds, but for hybridization, we count sigma bonds). There are no lone pairs on sulfur. Thus, the number of sigma bonds = 3, and the number of lone pairs = 0. Steric number = 3 + 0 = 3. Therefore, the hybridization of S in SO₃ is sp².

Step 4: Analyze option C) N in NO₃^-.
The central atom is N. The total number of valence electrons is 5 (N) + 3 * 6 (O) + 1 (charge) = 5 + 18 + 1 = 24. In NO₃^-, nitrogen forms one double bond and two single bonds with oxygen atoms (resonance structures). There are no lone pairs on nitrogen. Thus, the number of sigma bonds = 3, and the number of lone pairs = 0. Steric number = 3 + 0 = 3. Therefore, the hybridization of N in NO₃^- is sp².

Step 5: Analyze option D) S in SO₃^2-.
The central atom is S. The total number of valence electrons is 6 (S) + 3 * 6 (O) + 2 (charge) = 6 + 18 + 2 = 26. Sulfur forms three single bonds with three oxygen atoms. The remaining electrons form lone pairs: (26 - 3*2) / 2 = 20 / 2 = 10 electrons, which means 1 lone pair on sulfur. Thus, the number of sigma bonds = 3, and the number of lone pairs = 1. Steric number = 3 + 1 = 4. Therefore, the hybridization of S in SO₃^2- is sp³.

Comparing the hybridization of P in PO₄^3- (sp³) with the options, we find that S in SO₃^2- also has sp³ hybridization.

Correct Answer: (D)