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, calculating the steric number and identifying the hybridization.

### Step-by-Step Solution:

1. **Compound (i) SO₄²⁻ (Sulfate ion)**:
- Sulfur (S) has 6 valence electrons.
- In SO₄²⁻, sulfur forms 4 bonds with oxygen (2 double bonds and 2 single bonds).
- Steric Number = Number of bonded atoms + Number of lone pairs = 4 + 0 = 4.
- Hybridization = `sp³` (since steric number is 4).

2. **Compound (ii) SO₅²⁻**:
- Sulfur (S) again has 6 valence electrons.
- In SO₅²⁻, sulfur forms 5 bonds with oxygen (1 double bond and 4 single bonds).
- Steric Number = 5 + 0 = 5.
- Hybridization = `sp³d` (since steric number is 5).

3. **Compound (iii) PO₄³⁻ (Phosphate ion)**:
- Phosphorus (P) has 5 valence electrons.
- In PO₄³⁻, phosphorus forms 4 bonds with oxygen (1 double bond and 3 single bonds).
- Steric Number = 4 + 0 = 4.
- Hybridization = `sp³` (since steric number is 4).

4. **Compound (iv) PO₅³⁻**:
- Phosphorus (P) has 5 valence electrons.
- In PO₅³⁻, phosphorus forms 5 bonds with oxygen (1 double bond and 4 single bonds).
- Steric Number = 5 + 0 = 5.
- Hybridization = `sp³d` (since steric number is 5).

5. **Compound (v) I₃⁻ (Triiodide ion)**:
- Iodine (I) has 7 valence electrons.
- In I₃⁻, there are 3 iodine atoms with 2 bonded pairs and 3 lone pairs on the central iodine.
- Steric Number = 2 + 3 = 5.
- Hybridization = `sp³d` (since steric number is 5).

6. **Compound (vi) CO₃²⁻ (Carbonate ion)**:
- Carbon (C) has 4 valence electrons.
- In CO₃²⁻, carbon forms 3 bonds with oxygen (1 double bond and 2 single bonds).
- Steric Number = 3 + 0 = 3.
- Hybridization = `sp²` (since steric number is 3).

7. **Compound (vii) CO₄²⁻**:
- Carbon (C) has 4 valence electrons.
- In CO₄²⁻, carbon forms 4 bonds with oxygen (1 double bond and 3 single bonds).
- Steric Number = 4 + 0 = 4.
- Hybridization = `sp³` (since steric number is 4).

### Summary of Hybridization:
- SO₄²⁻: `sp³`
- SO₅²⁻: `sp³d`
- PO₄³⁻: `sp³`
- PO₅³⁻: `sp³d`
- I₃⁻: `sp³d`
- CO₃²⁻: `sp²`
- CO₄²⁻: `sp³`

### Conclusion:
The compounds with `sp³` hybridization are:
1. SO₄²⁻
2. PO₄³⁻
3. CO₄²⁻

Thus, **3 compounds** have `sp³` hybridization.