Which orbital is not involved in the formation of `PCl_(5)` molecule :-

Options

A. s
B. `d_(z^(2))`
C. `d_(x^(2)-y^(2))`
D. `P_(z)`

Answer

To determine which orbital is not involved in the formation of the PCl₅ molecule, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Central Atom**:
The central atom in PCl₅ is phosphorus (P).

2. **Determine the Valence Electrons**:
Phosphorus has 5 valence electrons, and each chlorine (Cl) atom has 7 valence electrons. Since there are 5 chlorine atoms, the total number of valence electrons contributed by chlorine is 5 × 7 = 35. Therefore, the total number of valence electrons in PCl₅ is 5 + 35 = 40.

3. **Determine the Hybridization**:
To find the hybridization, we can use the formula:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Number of valence electrons of central atom} + \text{Number of monovalent atoms} + \text{Charge} \right) \]
Here, the number of valence electrons of phosphorus is 5, the number of monovalent chlorine atoms is 5, and there is no charge on the molecule. Thus, we have:
\[ \text{Hybridization} = \frac{1}{2} (5 + 5 + 0) = \frac{10}{2} = 5 \]
This indicates that the hybridization of PCl₅ is sp³d.

4. **Identify the Involved Orbitals**:
In sp³d hybridization, the orbitals involved are:
- One s orbital
- Three p orbitals (px, py, pz)
- One d orbital (dz²)

5. **Identify the Orbital Not Involved**:
The d orbital involved in sp³d hybridization is typically the dz² orbital. The dx² - y² orbital is not involved in the hybridization for PCl₅.

6. **Conclusion**:
Therefore, the orbital that is not involved in the formation of the PCl₅ molecule is the dx² - y² orbital.

### Final Answer:
The orbital that is not involved in the formation of the PCl₅ molecule is **dx² - y²**.

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