In case of `XeO_(2) F_(2)` and `XeF_(6)` , Xe is with

Options

A. Same hybridization but with different geometry
B. Different hybridization with same geometry
C. Different hybridization and different geometry
D. Same geometry and same hybridization

Answer

To determine the hybridization and geometry of the molecules \(XeO_2F_2\) and \(XeF_6\), we will follow these steps:

### Step 1: Determine the Valence Electrons of Xenon
Xenon (Xe) is a noble gas and has 8 valence electrons.

### Step 2: Calculate the Hybridization for \(XeO_2F_2\)
1. Identify the number of monovalent atoms attached to the central atom (Xe):
- There are 2 fluorine (F) atoms and 2 oxygen (O) atoms attached to xenon, making a total of 4 monovalent atoms.

2. Use the formula for hybrid number:
\[ \text{Hybrid Number} = \frac{1}{2} (V + M - C + A) \]
where:
- \(V = 8\) (valence electrons of Xe)
- \(M = 4\) (2 F + 2 O)
- \(C = 0\) (no cationic charge)
- \(A = 0\) (no anionic charge)

Plugging in the values:
\[ \text{Hybrid Number} = \frac{1}{2} (8 + 4 - 0 + 0) = \frac{1}{2} (12) = 6 \]

3. Determine the hybridization:
- A hybrid number of 6 corresponds to \(sp^3d^2\) hybridization.

### Step 3: Determine the Geometry for \(XeO_2F_2\)
- The geometry corresponding to \(sp^3d^2\) hybridization is octahedral.

### Step 4: Calculate the Hybridization for \(XeF_6\)
1. Identify the number of monovalent atoms attached to the central atom (Xe):
- There are 6 fluorine (F) atoms attached to xenon.

2. Use the formula for hybrid number:
\[ \text{Hybrid Number} = \frac{1}{2} (V + M - C + A) \]
where:
- \(V = 8\) (valence electrons of Xe)
- \(M = 6\) (6 F)
- \(C = 0\) (no cationic charge)
- \(A = 0\) (no anionic charge)

Plugging in the values:
\[ \text{Hybrid Number} = \frac{1}{2} (8 + 6 - 0 + 0) = \frac{1}{2} (14) = 7 \]

3. Determine the hybridization:
- A hybrid number of 7 corresponds to \(sp^3d^3\) hybridization.

### Step 5: Determine the Geometry for \(XeF_6\)
- The geometry corresponding to \(sp^3d^3\) hybridization is pentagonal bipyramidal.

### Conclusion
In summary:
- For \(XeO_2F_2\):
- Hybridization: \(sp^3d^2\)
- Geometry: Octahedral
- For \(XeF_6\):
- Hybridization: \(sp^3d^3\)
- Geometry: Pentagonal bipyramidal

Thus, the correct answer is that Xe is with different hybridization and different geometry in \(XeO_2F_2\) and \(XeF_6\).