How many compound have lone pair at central atom zero.
`XeO_(3),XeO_(2)F_(2),XeO_(4),XeO_(3)F_(2),Ba_(2)XeF_(2)`

Options

A. 2
B. 3
C. 4
D. zero

Explanation

To determine how many compounds have zero lone pairs on the central atom, we will analyze each compound provided in the question: \(XeO_3\), \(XeO_2F_2\), \(XeO_4\), \(XeO_3F_2\), and \(Ba_2XeF_2\).

### Step-by-Step Solution:

1. **Identify the Valence Electrons of Xenon (Xe)**:
- Xenon has 8 valence electrons.

2. **Calculate the Steric Number**:
- The steric number is calculated using the formula:
\[ \text{Steric Number} = \frac{\text{(Number of valence electrons of central atom)} + \text{(Number of monovalent atoms)}}{2} \]
- Monovalent atoms (like F) contribute 1 to the count, while divalent atoms (like O) contribute 2.

3. **Analyze Each Compound**:

- **For \(XeO_3\)**:
- Valence electrons = 8 (Xe) + 3 × 2 (O) = 14
- Steric number = \(\frac{14}{2} = 7\)
- Geometry: Octahedral
- Lone pairs: 1 (since 7 - 6 = 1)

- **For \(XeO_2F_2\)**:
- Valence electrons = 8 (Xe) + 2 × 2 (O) + 2 × 1 (F) = 14
- Steric number = \(\frac{14}{2} = 7\)
- Geometry: Octahedral
- Lone pairs: 1 (since 7 - 6 = 1)

- **For \(XeO_4\)**:
- Valence electrons = 8 (Xe) + 4 × 2 (O) = 16
- Steric number = \(\frac{16}{2} = 8\)
- Geometry: Square planar
- Lone pairs: 0 (since 8 - 8 = 0)

- **For \(XeO_3F_2\)**:
- Valence electrons = 8 (Xe) + 3 × 2 (O) + 2 × 1 (F) = 14
- Steric number = \(\frac{14}{2} = 7\)
- Geometry: Octahedral
- Lone pairs: 1 (since 7 - 6 = 1)

- **For \(Ba_2XeF_2\)**:
- Valence electrons = 8 (Xe) + 2 × 1 (F) = 10
- Steric number = \(\frac{10}{2} = 5\)
- Geometry: Trigonal bipyramidal
- Lone pairs: 1 (since 5 - 4 = 1)

4. **Count Compounds with Zero Lone Pairs**:
- From the analysis:
- \(XeO_4\) has 0 lone pairs.
- \(XeO_3F_2\) has 0 lone pairs.
- Therefore, the total number of compounds with zero lone pairs on the central atom is **2**.

### Final Answer:
There are **2 compounds** with zero lone pairs on the central atom.