The bond angle in `XeO_3` molecule decreases from `109^@28^1` to `130^@` due to 

Options

A. greater repulsions among lone pairs
B. greater repulsions among lone pair and bond pair
C. greater repulsions among bond pairs
D. steric effect

Answer

To solve the question regarding the bond angle in the `XeO_3` molecule, we will analyze the factors affecting the bond angle and the molecular geometry.

### Step-by-Step Solution:

1. **Identify the Molecular Geometry**:
- The molecular formula `XeO_3` indicates that xenon (Xe) is bonded to three oxygen (O) atoms.
- Xenon is a noble gas and can form compounds due to its ability to expand its octet.

2. **Determine the Hybridization**:
- In `XeO_3`, xenon undergoes `sp^3` hybridization. This is because it has four regions of electron density (three bonding pairs with oxygen and one lone pair).

3. **Understand the Bond Angle**:
- In a perfect tetrahedral geometry (which corresponds to `sp^3` hybridization), the bond angle is `109.5°`.
- However, the presence of lone pairs affects the bond angles due to repulsion.

4. **Consider the Effect of Lone Pairs**:
- The lone pair occupies more space than bonding pairs, leading to greater repulsion between the lone pair and the bonding pairs.
- This repulsion pushes the bonding pairs closer together, resulting in a decrease in the bond angle.

5. **Calculate the Actual Bond Angle**:
- In the case of `XeO_3`, the bond angle is observed to be approximately `103°`, which is less than the ideal tetrahedral angle of `109.5°`.
- The bond angle decreases further to `130°` due to increased repulsion effects as more lone pairs or other factors come into play.

6. **Conclusion**:
- The bond angle in `XeO_3` decreases from `109° 28'` to `130°` primarily due to greater repulsions among the lone pairs and bond pairs.

### Final Answer:
The bond angle in `XeO_3` decreases from `109° 28'` to `130°` due to greater repulsions among the lone pairs and bond pairs.