The hybridisation of Xe and Sb in product when `XeF_(4)` react with `SbF_(5)` :-

Options

A. `sp^(3)d, sp^(3)d^(2)`
B. `sp^(3)d^(2), sp^(3)d`
C. `sp^(3), sp^(3)d^(2)`
D. `sp^(3)d^(3), sp^(3)d`

Answer

To determine the hybridization of xenon (Xe) and antimony (Sb) in the product formed when \(XeF_4\) reacts with \(SbF_5\), we can follow these steps:

### Step 1: Analyze the structure of \(XeF_4\)
- **Xenon (Xe)** in \(XeF_4\) has 4 fluorine atoms bonded to it and 2 lone pairs of electrons.
- The total number of electron pairs around xenon is 6 (4 bond pairs + 2 lone pairs).
- The hybridization can be determined using the formula:
\[ \text{Hybridization} = \text{Number of electron pairs} \]
Since there are 6 electron pairs, the hybridization of xenon in \(XeF_4\) is \(sp^3d^2\).

### Step 2: Analyze the structure of \(SbF_5\)
- **Antimony (Sb)** in \(SbF_5\) has 5 fluorine atoms bonded to it and no lone pairs of electrons.
- The total number of electron pairs around antimony is 5 (5 bond pairs + 0 lone pairs).
- Using the same formula for hybridization:
\[ \text{Hybridization} = \text{Number of electron pairs} \]
Since there are 5 electron pairs, the hybridization of antimony in \(SbF_5\) is \(sp^3d\).

### Step 3: Conclusion
- The hybridization of xenon in \(XeF_4\) is \(sp^3d^2\).
- The hybridization of antimony in \(SbF_5\) is \(sp^3d\).

### Final Answer
- The hybridization of Xe in \(XeF_4\) is \(sp^3d^2\).
- The hybridization of Sb in \(SbF_5\) is \(sp^3d\).