As `PF_(5)` molecule is `sp^(3)` d hybridised and is trigonal bipyramidal (TbP) shape Which d-orbital is involved in `sp^(3)` d hyridisation .

Options

A. `dx^(2)-y^(2)`
B. `dz^(2)`
C. `dxy`
D. `dzx`

Answer

To determine which d-orbital is involved in the `sp^3d` hybridization of the `PF5` molecule, we can follow these steps:

### Step 1: Understand the Electronic Configuration of Phosphorus
Phosphorus (atomic number 15) has the electronic configuration:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^3 \]
This indicates that phosphorus has 5 valence electrons available for bonding.

**Hint:** Remember that the valence electrons are the ones involved in bonding and hybridization.

### Step 2: Identify the Ground State and Excited State Configuration
In the ground state, the 3d orbitals are empty. To form bonds in `PF5`, one of the 3s electrons is promoted to the empty 3d orbital, resulting in the excited state configuration:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^1 \, 3p^3 \, 3d^1 \]
This gives us a total of 5 unpaired electrons.

**Hint:** Excitation of electrons is necessary for hybridization to occur.

### Step 3: Determine the Hybridization
With 5 unpaired electrons, the hybridization of phosphorus in `PF5` is `sp^3d`. This involves one s orbital, three p orbitals, and one d orbital.

**Hint:** The number of hybrid orbitals corresponds to the number of atomic orbitals mixed together.

### Step 4: Identify the Geometry of `PF5`
The geometry of `PF5` is trigonal bipyramidal, which is characteristic of `sp^3d` hybridization. In this arrangement, three fluorine atoms occupy the equatorial positions, while two fluorine atoms occupy the axial positions.

**Hint:** The geometry can help confirm the type of hybridization.

### Step 5: Identify the Involved d-Orbital
Among the five d orbitals (dxy, dyz, dzx, dx²-y², dz²), the dz² orbital is the most suitable for `PF5`. This is because the dz² orbital has no components in the xy-plane, which minimizes repulsion with the p orbitals in the equatorial plane.

**Hint:** The choice of the d orbital is crucial for minimizing electron repulsion and stabilizing the molecular structure.

### Conclusion
The d-orbital involved in the `sp^3d` hybridization of the `PF5` molecule is the **dz² orbital**.