When xenon hexafluoride is treated with `AsF_(5)` to form an ionic compound, the hybridisation of Xe and As will be respectively

Options

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

Answer

To determine the hybridization of xenon (Xe) and arsenic (As) when xenon hexafluoride (XeF6) is treated with arsenic pentafluoride (AsF5), we can follow these steps:

### Step 1: Identify the products formed
When XeF6 reacts with AsF5, it forms the ionic compound XeF5+ and AsF6-.

### Step 2: Analyze the xenon ion (XeF5+)
- Xenon (Xe) has 8 valence electrons.
- In forming the ion XeF5+, xenon loses one electron, resulting in 7 electrons remaining.
- Out of these 7 electrons, 5 are used to form bonds with 5 fluorine atoms, leaving 1 lone pair.

### Step 3: Calculate the steric number for XeF5+
- The steric number is calculated as the sum of the number of sigma bonds and the number of lone pairs.
- For XeF5+, there are 5 sigma bonds (with fluorine) and 1 lone pair.
- Therefore, the steric number = 5 (sigma bonds) + 1 (lone pair) = 6.

### Step 4: Determine the hybridization of xenon
- A steric number of 6 corresponds to sp3d2 hybridization.
- Thus, the hybridization of xenon in XeF5+ is **sp3d2**.

### Step 5: Analyze the arsenic ion (AsF6-)
- Arsenic (As) has 5 valence electrons.
- The AsF6- ion has one extra electron due to the negative charge, resulting in a total of 6 valence electrons.
- These 6 electrons form bonds with 6 fluorine atoms.

### Step 6: Calculate the steric number for AsF6-
- For AsF6-, there are 6 sigma bonds (with fluorine) and no lone pairs.
- Therefore, the steric number = 6 (sigma bonds) + 0 (lone pairs) = 6.

### Step 7: Determine the hybridization of arsenic
- A steric number of 6 also corresponds to sp3d2 hybridization.
- Thus, the hybridization of arsenic in AsF6- is **sp3d2**.

### Conclusion
The hybridization of xenon and arsenic in the respective ions is **sp3d2** for both.