Hybridization of P in the phosphate ion (PO₄³⁻) is the same as in:

(A) I in ICl₄⁻
(B) S in SO₃
(C) N in NO₃⁻
(D) S in SO₃²⁻

To determine the hybridization of phosphorus (P) in the phosphate ion (PO4^3-), you can use the formula:

Hybridization = [V + N - C + A]/2

Where:
V = Number of valence electrons of the central atom (P)
N = Number of monovalent atoms (Oxygen atoms) bonded to the central atom
C = Charge on the ion
A = Number of additional atoms (hydrogen atoms, if any)

For the phosphate ion (PO4^3-), we have:

Number of valence electrons of P = 5 (group 15 element in the periodic table)
Number of monovalent atoms (Oxygen atoms) bonded to P = 4
Charge on the ion (C) = -3 (3- charge on the ion)
No additional atoms (hydrogen atoms)
Now, let's calculate the hybridization for P:
Hybridization = [5 + 4 - (-3) + 0]/2
Hybridization = (5 + 4 + 3)/2
Hybridization = 12/2
Hybridization = 6

So, the hybridization of phosphorus (P) in the phosphate ion (PO4^3-) is sp3.

Now, let's determine the hybridization for the options given:
(A) I in ICl4^-: Iodine (I) is in group 17 of the periodic table, so it has 7 valence electrons. In ICl4^-, I is bonded to 4 Cl atoms and has a -1 charge, so the hybridization would also be sp3.

(B) S in SO3: Sulfur (S) is in group 16 of the periodic table, so it has 6 valence electrons. In SO3, S is bonded to 3 O atoms, so the hybridization would be sp2.

(C) N in NO3^-: Nitrogen (N) is in group 15 of the periodic table, so it has 5 valence electrons. In NO3^-, N is bonded to 3 O atoms and has a -1 charge, so the hybridization would be sp2.

(D) S in SO3^2-: We already determined that the hybridization of sulfur (S) in SO3 is sp2. Adding two additional electrons (the 2- charge) doesn't change the hybridization.

So, the correct answer is (A) I in ICl4^-. The hybridization of iodine in ICl4^- is the same as the hybridization of phosphorus in the phosphate ion (PO4^3-), which is sp3.