When hybridisation involving d-orbitals are considered then all the five d-orbitals are not degenerate, rather `d_(x^(2)-y^(2)),d_(s^(2)) and d_(xy),d_(zx)` form two different sets of orbitals and orbitals of appropriate set is involved in the hybridisation.
In `sp^(3)d^(3)` hybridisation, which orbitals are involved ?

Options

A. `d_(x^(2)-y^(2)),d_(z^(2)),d_(xy)`
B. `d_(xy),d_(yz),d_(zx)`
C. `d_(x^(2)-y^(2)),d_(xy),d_(xz)`
D. `d_(x^(2)),d_(yz),d_(zx)`

Answer

To determine which orbitals are involved in the `sp^3d^3` hybridization, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Hybridization**: Hybridization is the mixing of atomic orbitals to form new hybrid orbitals that are degenerate (of equal energy). In `sp^3d^3` hybridization, we are combining s, p, and d orbitals.

2. **Identify the Orbitals**: In `sp^3d^3` hybridization, we have:
- 1 s orbital
- 3 p orbitals
- 3 d orbitals

3. **Determine the d Orbitals**: The d orbitals are not all degenerate. In this case, the relevant d orbitals involved in `sp^3d^3` hybridization are:
- `d_(x^2-y^2)`
- `d_(z^2)`
- `d_(xy)`

4. **Geometry of Hybridization**: The hybridization results in a geometry that is bipyramidal. This means that there are five bonds formed in the equatorial plane and two axial bonds (one above and one below the equatorial plane).

5. **Example of Application**: A common example of `sp^3d^3` hybridization is in the molecule IF7 (iodine heptafluoride). Iodine has 7 valence electrons, and when it forms bonds with fluorine, it utilizes the `sp^3d^3` hybridization.

6. **Bond Angles**: In this hybridization, the bond angles between the equatorial bonds are 90 degrees, while the axial bonds make angles of 72 degrees with the equatorial plane.

### Final Answer:
The orbitals involved in `sp^3d^3` hybridization are:
- 1 s orbital
- 3 p orbitals
- 3 d orbitals: specifically `d_(x^2-y^2)`, `d_(z^2)`, and `d_(xy)`.