To determine the hybridization and shape, we first calculate the steric number (SN) for the central atom, which is the sum of the number of sigma bonds and lone pairs.

Option A: SO₃, CO₃^2-

• SO₃: Central atom S. SN = 3 sigma bonds + 0 lone pairs = 3. Hybridization = sp². Shape = Trigonal planar.
• CO₃^2-: Central atom C. SN = 3 sigma bonds + 0 lone pairs = 3. Hybridization = sp². Shape = Trigonal planar.

Hybridization is same (sp²), and shape is same (Trigonal planar). So, A is incorrect.

Option B: SO₃^2-, NH₃

• SO₃^2-: Central atom S. SN = 3 sigma bonds + 1 lone pair = 4. Hybridization = sp³. Shape = Trigonal pyramidal.
• NH₃: Central atom N. SN = 3 sigma bonds + 1 lone pair = 4. Hybridization = sp³. Shape = Trigonal pyramidal.

Hybridization is same (sp³), and shape is same (Trigonal pyramidal). So, B is incorrect.

Option C: PCl₅, POCl₃

• PCl₅: Central atom P. SN = 5 sigma bonds + 0 lone pairs = 5. Hybridization = sp³d. Shape = Trigonal bipyramidal.
• POCl₃: Central atom P. SN = 4 sigma bonds + 0 lone pairs = 4. Hybridization = sp³. Shape = Tetrahedral.

Hybridization is not the same. So, C is incorrect.

Option D: XeF₂, ICl₃

• XeF₂: Central atom Xe. Xe is in Group 18, so 8 valence electrons. Two F atoms form 2 sigma bonds. Remaining electrons = 8 - 2 = 6, which form 3 lone pairs. SN = 2 sigma bonds + 3 lone pairs = 5. Hybridization = sp³d. Shape = Linear (due to lone pair repulsion).
• ICl₃: Central atom I. I is in Group 17, so 7 valence electrons. Three Cl atoms form 3 sigma bonds. Remaining electrons = 7 - 3 = 4, which form 2 lone pairs. SN = 3 sigma bonds + 2 lone pairs = 5. Hybridization = sp³d. Shape = T-shaped (due to lone pair repulsion).

Hybridization is same (sp³d), but shape is not the same (Linear vs T-shaped). So, D is correct.

Correct Answer: (D)