What is the state of hybridisation of Xe in cationic part of solid `XeF_6`

Options

A. `sp^3d^3`
B. `sp^3d^2`
C. `sp^3d`
D. `sp^3`

Answer

To determine the state of hybridization of xenon (Xe) in the cationic part of solid XeF6, we can follow these steps:

### Step 1: Determine the Valence Electrons
- Xenon (Xe) has 8 valence electrons.
- Each fluorine (F) atom has 7 valence electrons, but in the context of bonding, we consider only the 1 valence electron that participates in the bond formation.

### Step 2: Count the Bonding Electrons
- In XeF6, xenon forms bonds with 6 fluorine atoms. Therefore, it uses 6 of its valence electrons to form 6 bond pairs with fluorine atoms.

### Step 3: Calculate Remaining Electrons
- After forming 6 bonds, xenon has 8 - 6 = 2 valence electrons left.
- These 2 remaining electrons will form a lone pair.

### Step 4: Determine the Steric Number
- The steric number is calculated by adding the number of bond pairs and lone pairs:
- Number of bond pairs = 6 (from the 6 Xe-F bonds)
- Number of lone pairs = 1 (the lone pair we calculated)
- Therefore, the steric number = 6 + 1 = 7.

### Step 5: Identify the Hybridization
- The hybridization can be determined from the steric number:
- A steric number of 6 corresponds to sp³d² hybridization.
- However, since we have a lone pair, we consider the geometry and the hybridization state that accommodates the lone pair.

### Conclusion
- The hybridization of xenon in the cationic part of solid XeF6 is **sp³d²**.