`XeF_(6)+H_(2)O overset(excess)to` Product.
How many hybrid orbitals will be present in Xe in product?

Answer

To solve the question regarding the reaction of \( \text{XeF}_6 \) with excess water and to determine the number of hybrid orbitals present in xenon (\( \text{Xe} \)) in the product, we can follow these steps:

### Step 1: Identify the Reaction
The reaction of \( \text{XeF}_6 \) with excess water leads to hydrolysis, forming \( \text{XeO}_3 \) and \( \text{HF} \). The balanced reaction can be represented as:
\[ \text{XeF}_6 + 3 \text{H}_2\text{O} \rightarrow \text{XeO}_3 + 6 \text{HF} \]

### Step 2: Determine the Hybridization of Xenon in \( \text{XeO}_3 \)
To find the hybridization of xenon in \( \text{XeO}_3 \), we need to calculate the steric number (hybrid number) using the formula:
\[ \text{Hybrid number} = \frac{1}{2} \left( \text{Number of valence electrons} + \text{Number of monovalent atoms} - \text{Cationic charge} + \text{Anionic charge} \right) \]

### Step 3: Calculate the Valence Electrons
- Xenon (\( \text{Xe} \)) is in group 18 and has 8 valence electrons.
- In \( \text{XeO}_3 \), there are no monovalent atoms attached to xenon (only oxygen atoms).
- There are no cationic or anionic charges in this case.

### Step 4: Substitute Values into the Formula
Now, substituting the values into the hybrid number formula:
\[ \text{Hybrid number} = \frac{1}{2} \left( 8 + 0 - 0 + 0 \right) = \frac{1}{2} \times 8 = 4 \]

### Step 5: Determine the Hybridization Type
A hybrid number of 4 corresponds to \( \text{sp}^3 \) hybridization. Therefore, xenon in \( \text{XeO}_3 \) has 4 hybrid orbitals.

### Final Answer
The number of hybrid orbitals present in xenon in the product \( \text{XeO}_3 \) is **4**.

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