How many of the following compounds have `sp^(3)` hybridisation
(i) `SO_(4)^(2-)` (ii) `SO_(5)^(2-)` (iii) `PO_(4)^(3-)` (iv) `PO_(5)^(3-)`
(v) `I_(3)^(Theta)` (iv) `CO_(3)^(2-)` (vii) `CO_(4)^(2-) `.

Explanation

To determine how many of the given compounds have `sp^3` hybridization, we will analyze each compound one by one based on their steric number, which is the sum of the number of bond pairs and lone pairs around the central atom.

### Step-by-Step Solution:

1. **Compound (i) SO₄²⁻**
- Sulfur (S) is the central atom.
- It forms 4 bonds with oxygen atoms (4 bond pairs) and has no lone pairs.
- Steric number = 4 (bond pairs) + 0 (lone pairs) = 4.
- Hybridization = `sp^3`.
- **Conclusion**: SO₄²⁻ has `sp^3` hybridization.

2. **Compound (ii) SO₅²⁻**
- Sulfur (S) is the central atom.
- It forms 5 bonds with oxygen atoms (5 bond pairs) and has no lone pairs.
- Steric number = 5 (bond pairs) + 0 (lone pairs) = 5.
- Hybridization = `sp^3d`.
- **Conclusion**: SO₅²⁻ does not have `sp^3` hybridization.

3. **Compound (iii) PO₄³⁻**
- Phosphorus (P) is the central atom.
- It forms 4 bonds with oxygen atoms (4 bond pairs) and has no lone pairs.
- Steric number = 4 (bond pairs) + 0 (lone pairs) = 4.
- Hybridization = `sp^3`.
- **Conclusion**: PO₄³⁻ has `sp^3` hybridization.

4. **Compound (iv) PO₅³⁻**
- Phosphorus (P) is the central atom.
- It forms 5 bonds with oxygen atoms (5 bond pairs) and has no lone pairs.
- Steric number = 5 (bond pairs) + 0 (lone pairs) = 5.
- Hybridization = `sp^3d`.
- **Conclusion**: PO₅³⁻ does not have `sp^3` hybridization.

5. **Compound (v) I₃⁻**
- Iodine (I) is the central atom.
- It forms 2 bonds with iodine atoms (2 bond pairs) and has 3 lone pairs.
- Steric number = 2 (bond pairs) + 3 (lone pairs) = 5.
- Hybridization = `sp^3d`.
- **Conclusion**: I₃⁻ does not have `sp^3` hybridization.

6. **Compound (vi) CO₃²⁻**
- Carbon (C) is the central atom.
- It forms 3 bonds with oxygen atoms (3 bond pairs) and has no lone pairs.
- Steric number = 3 (bond pairs) + 0 (lone pairs) = 3.
- Hybridization = `sp²`.
- **Conclusion**: CO₃²⁻ does not have `sp^3` hybridization.

7. **Compound (vii) CO₄²⁻**
- Carbon (C) is the central atom.
- It forms 4 bonds with oxygen atoms (4 bond pairs) and has no lone pairs.
- Steric number = 4 (bond pairs) + 0 (lone pairs) = 4.
- Hybridization = `sp^3`.
- **Conclusion**: CO₄²⁻ has `sp^3` hybridization.

### Final Count:
- Compounds with `sp^3` hybridization: SO₄²⁻, PO₄³⁻, CO₄²⁻.
- Total = 3 compounds.

### Summary:
The total number of compounds with `sp^3` hybridization is **3**.