In `XeO_(3)F_(2)`, d-orbital which is not used in bonding is :

Options

A. `d_(x^(2)-y^(2))`
B. `d_(z^(2))`
C. `d_(xy)`
D. `d_(yz)`

Explanation

To determine which d-orbital is not used in bonding in the compound \( \text{XeO}_3\text{F}_2 \), we can follow these steps:

### Step 1: Determine the Valence Electrons
- **Xenon (Xe)** has 8 valence electrons.
- **Oxygen (O)** has 6 valence electrons, and since there are 3 oxygen atoms, that contributes \( 3 \times 6 = 18 \) electrons.
- **Fluorine (F)** has 7 valence electrons, and since there are 2 fluorine atoms, that contributes \( 2 \times 7 = 14 \) electrons.

**Total valence electrons = 8 (Xe) + 18 (O) + 14 (F) = 40 electrons.**

### Step 2: Identify the Hybridization
- To find the hybridization, we need to determine the number of bond pairs and lone pairs.
- In \( \text{XeO}_3\text{F}_2 \), there are 5 bonding pairs (3 from oxygen and 2 from fluorine) and no lone pairs on xenon.

**Hybridization = Number of bond pairs + Number of lone pairs = 5 + 0 = 5.**

- A hybridization of 5 corresponds to \( \text{sp}^3\text{d} \).

### Step 3: Determine the Shape
- The molecular geometry for \( \text{sp}^3\text{d} \) hybridization is trigonal bipyramidal.

### Step 4: Identify the Involved Orbitals
- In \( \text{XeO}_3\text{F}_2 \):
- The \( \text{d} \) orbitals involved in bonding are \( d_{z^2} \) (for sigma bonds) and \( d_{xy}, d_{yz}, d_{xz} \) (for pi bonds).
- The \( s \) orbital and \( p \) orbitals (specifically \( p_x, p_y, p_z \)) are also involved in bonding.

### Step 5: Identify the Non-bonding d-Orbital
- The \( d_{x^2-y^2} \) orbital is not involved in any bonding interactions in this compound.

### Conclusion
The d-orbital that is not used in bonding in \( \text{XeO}_3\text{F}_2 \) is the **\( d_{x^2-y^2} \)** orbital.

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