VSEPR theory suggests that bond length and bond angle depend upon the presence or absence of lone pairs. If the bond angle of `H-O-H` in `H_(2)O` is `105^(@)` (approx.), then the bond angle of `F-O-F` in `F_(2)O` will be

Options

A. greater than `105^(@)`
B. lesser than `105^(@)`
C. equal to `105^(@)`
D. equal to `180^(@)`

Answer

To solve the question regarding the bond angle of F-O-F in F₂O based on the VSEPR theory, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Molecular Geometry**:
- Both H₂O (water) and F₂O (difluorooxide) have a central oxygen atom with lone pairs. In H₂O, there are two hydrogen atoms bonded to oxygen, while in F₂O, there are two fluorine atoms bonded to oxygen.

2. **Identifying Lone Pairs**:
- In H₂O, the bond angle is approximately 105° due to the presence of two lone pairs on the oxygen atom which repel the bonding pairs of electrons.
- In F₂O, the oxygen also has two lone pairs.

3. **Electronegativity Consideration**:
- Fluorine is more electronegative than hydrogen. This means that in F₂O, the bonding electrons are drawn closer to the fluorine atoms compared to how they are distributed in H₂O.

4. **Effect of Electronegativity on Bond Angles**:
- The greater electronegativity of fluorine compared to hydrogen causes the bond pairs in F₂O to be pulled away from the central oxygen atom. This results in a slight increase in the bond angle compared to H₂O.

5. **Comparing Bond Angles**:
- Since the bond pairs in F₂O are displaced more than in H₂O due to the higher electronegativity of fluorine, the bond angle in F₂O will be less than that in H₂O.
- Therefore, we conclude that the bond angle of F-O-F in F₂O will be less than 105°.

6. **Final Answer**:
- Based on the analysis, the bond angle of F-O-F in F₂O is less than 105°.