The number of species which consists of `sp^(3)d` hybridised central atom for the underlined atoms in the following species/molecules is/are :
`ul(X)eF_(4),ul(I)CI_(2)^(Theta),ul(X)eO_(3),F_(2),ul(P)Cl_(4)^(o+),ul(P)Cl_(6)^(Theta)ul(S)F_(4),ul(S)OF_(4),ul(X)eOF_(4)`

Explanation

To determine the number of species with an `sp^3d` hybridized central atom from the given list, we will analyze each molecule one by one.

### Step-by-Step Solution:

1. **XeF₄ (Xenon Tetrafluoride)**:
- Valence electrons in Xenon (Xe): 8
- Valence electrons in Fluorine (F): 1 (4 Fluorine atoms contribute 4 electrons)
- Total valence electrons = 8 + 4 = 12
- Bond pairs: 4 (one for each F)
- Lone pairs: 2 (12 - 8 = 4 electrons left, which means 2 lone pairs)
- Hybridization: `sp^3d2` (due to 4 bond pairs and 2 lone pairs)
- **Conclusion**: Not `sp^3d`.

2. **ICl₂⁻ (Iodine Dichloride Ion)**:
- Valence electrons in Iodine (I): 7 + 1 (for the negative charge) = 8
- Valence electrons in Chlorine (Cl): 1 (2 Chlorine atoms contribute 2 electrons)
- Total valence electrons = 8 + 2 = 10
- Bond pairs: 2 (one for each Cl)
- Lone pairs: 3 (10 - 4 = 6 electrons left, which means 3 lone pairs)
- Hybridization: `sp^3d` (due to 2 bond pairs and 3 lone pairs)
- **Conclusion**: It is `sp^3d`.

3. **XeO₃ (Xenon Trioxide)**:
- Valence electrons in Xenon (Xe): 8
- Valence electrons in Oxygen (O): 2 (3 Oxygen atoms contribute 6 electrons)
- Total valence electrons = 8 + 6 = 14
- Bond pairs: 3 (one for each O)
- Lone pairs: 1 (14 - 8 = 6 electrons left, which means 1 lone pair)
- Hybridization: `sp^3d` (due to 3 bond pairs and 1 lone pair)
- **Conclusion**: It is `sp^3d`.

4. **F₂ (Fluorine Molecule)**:
- This is a diatomic molecule with a single bond.
- Hybridization: `sp` (only 1 bond pair)
- **Conclusion**: Not `sp^3d`.

5. **PCl₄⁺ (Phosphorus Tetrachloride Cation)**:
- Valence electrons in Phosphorus (P): 5 - 1 (for the positive charge) = 4
- Valence electrons in Chlorine (Cl): 1 (4 Chlorine atoms contribute 4 electrons)
- Total valence electrons = 4 + 4 = 8
- Bond pairs: 4 (one for each Cl)
- Lone pairs: 0 (8 - 8 = 0)
- Hybridization: `sp^3` (due to 4 bond pairs and 0 lone pairs)
- **Conclusion**: Not `sp^3d`.

6. **PCl₆ (Phosphorus Hexachloride)**:
- Valence electrons in Phosphorus (P): 5
- Valence electrons in Chlorine (Cl): 1 (6 Chlorine atoms contribute 6 electrons)
- Total valence electrons = 5 + 6 = 11
- Bond pairs: 6 (one for each Cl)
- Lone pairs: 0 (11 - 12 = -1, which is not possible)
- Hybridization: `sp^3d2` (due to 6 bond pairs)
- **Conclusion**: Not `sp^3d`.

7. **SF₄ (Sulfur Tetrafluoride)**:
- Valence electrons in Sulfur (S): 6
- Valence electrons in Fluorine (F): 1 (4 Fluorine atoms contribute 4 electrons)
- Total valence electrons = 6 + 4 = 10
- Bond pairs: 4 (one for each F)
- Lone pairs: 1 (10 - 8 = 2 electrons left, which means 1 lone pair)
- Hybridization: `sp^3d` (due to 4 bond pairs and 1 lone pair)
- **Conclusion**: It is `sp^3d`.

8. **SOF₄ (Sulfur Oxyfluoride)**:
- Valence electrons in Sulfur (S): 6
- Valence electrons in Oxygen (O): 2 (1 Oxygen contributes 2 electrons)
- Valence electrons in Fluorine (F): 1 (4 Fluorine atoms contribute 4 electrons)
- Total valence electrons = 6 + 2 + 4 = 12
- Bond pairs: 5 (4 F and 1 O)
- Lone pairs: 0 (12 - 10 = 2 electrons left, which means 0 lone pairs)
- Hybridization: `sp^3d` (due to 5 bond pairs)
- **Conclusion**: It is `sp^3d`.

9. **XeOF₄ (Xenon Oxyfluoride)**:
- Valence electrons in Xenon (Xe): 8
- Valence electrons in Oxygen (O): 2 (1 Oxygen contributes 2 electrons)
- Valence electrons in Fluorine (F): 1 (4 Fluorine atoms contribute 4 electrons)
- Total valence electrons = 8 + 2 + 4 = 14
- Bond pairs: 5 (4 F and 1 O)
- Lone pairs: 0 (14 - 10 = 4 electrons left, which means 0 lone pairs)
- Hybridization: `sp^3d` (due to 5 bond pairs)
- **Conclusion**: It is `sp^3d`.

### Summary of Findings:
The species with `sp^3d` hybridized central atoms are:
- ICl₂⁻
- XeO₃
- SF₄
- SOF₄
- XeOF₄

### Final Answer:
**Total number of species with `sp^3d` hybridized central atom: 5**

---