Atomic orbitals involved in hybridisation of `SF_(6)` molecule

Options

A. `3 s , 3p_(x) , 3 p_(y) , 3p_(z) , 3d_(z^(2)) , d_(x^(2) - y^(2))`
B. `3 s 3 p_(x) , 3p_(y) , 3 p_(z) , 3 d_(z^(2))`
C. ` 3 s , 3p_(x) , 3p_(y) , 3 p_(z) , 3 d_(x^(2) - y^(2)) , d_(xy)`
D. `3s , 3p_(x) , 3p_(y) , 3d_(xy) , 3d_(yz) , 3d_(xz)`

Answer

To determine the atomic orbitals involved in the hybridization of the `SF_(6)` molecule, we can follow these steps:

### Step 1: Identify the central atom and its electronic configuration
- The central atom in `SF_(6)` is sulfur (S).
- The electronic configuration of sulfur is `1s² 2s² 2p⁶ 3s² 3p⁴`.

### Step 2: Determine the oxidation state of sulfur in `SF_(6)`
- In `SF_(6)`, sulfur is bonded to six fluorine atoms. Each fluorine atom has a -1 oxidation state.
- Therefore, the oxidation state of sulfur in `SF_(6)` is +6.

### Step 3: Analyze the hybridization process
- In its ground state, sulfur has the following orbitals filled: `3s² 3p⁴`.
- To accommodate six fluorine atoms, sulfur must undergo hybridization.
- In the excited state, two electrons from the `3s` and `3p` orbitals are promoted to the `3d` orbital.

### Step 4: Write the excited state configuration
- The excited state configuration of sulfur becomes:
- `3s¹ 3p³ 3d²`
- This indicates that sulfur now has one electron in the `3s`, three electrons in the `3p`, and two electrons in the `3d` orbitals.

### Step 5: Determine the type of hybridization
- The hybridization that occurs is `sp³d²`, which involves:
- 1 `3s` orbital
- 3 `3p` orbitals
- 2 `3d` orbitals
- This results in a total of six hybrid orbitals.

### Step 6: Identify the molecular geometry
- The geometry of `SF_(6)` is octahedral due to the arrangement of the six hybrid orbitals around the sulfur atom.

### Conclusion
- The atomic orbitals involved in the hybridization of `SF_(6)` are:
- 1 `3s`, 3 `3p`, and 2 `3d` orbitals.
- Therefore, the hybridization of `SF_(6)` is `sp³d²`.