The state of hybridization of xenon of `XeF_(6)` is

Options

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

Answer

To determine the state of hybridization of xenon in \(XeF_6\), we can follow these steps:

### Step 1: Determine the Valence Electrons
Xenon (Xe) is a noble gas and has 8 valence electrons. Fluorine (F), being a halogen, has 7 valence electrons. Since there are 6 fluorine atoms in \(XeF_6\), we calculate the total number of valence electrons as follows:

\[ \text{Total valence electrons} = \text{Valence electrons of Xe} + 6 \times \text{Valence electrons of F} = 8 + 6 \times 7 = 8 + 42 = 50 \]

### Step 2: Determine the Number of Bond Pairs and Lone Pairs
To find the number of bond pairs and lone pairs, we need to consider that each bond involves 2 electrons. We can use the formula:

\[ \text{Number of pairs} = \frac{\text{Total valence electrons}}{2} = \frac{50}{2} = 25 \]

### Step 3: Calculate the Bond Pairs and Lone Pairs
Now, we know that xenon can form bonds with fluorine. Since xenon is the central atom, we need to account for the bonds it forms with the fluorine atoms. Each fluorine atom forms a single bond with xenon, so:

- Bond pairs = 6 (one for each \(Xe-F\) bond)
- Total pairs = 25
- Lone pairs = Total pairs - Bond pairs = 25 - 6 = 19

However, we need to consider that the maximum number of electrons around xenon is 8 (octet rule), but xenon can expand its octet due to the presence of d-orbitals.

### Step 4: Determine the Hybridization
The total number of electron pairs around xenon is 7 (6 bond pairs and 1 lone pair). To find the hybridization, we can use the following:

- 1 lone pair + 6 bond pairs = 7 pairs

The hybridization corresponding to 7 pairs of electrons is \(sp^3d^3\).

### Conclusion
Thus, the state of hybridization of xenon in \(XeF_6\) is \(sp^3d^3\).