In water molecule, the bond angle of `104.5^(@)` around oxygen is accounted due to

Options

A. lp-lp repulsion is more than lp-bp repulsion.
B. lp-bp repulsion is more than lp-lp repulsion.
C. lp-lp repulsion is equal to lp-bp repulsion.
D. Presence of lone pair does not affect the bond angle.

Explanation

To solve the question regarding the bond angle of 104.5° in a water molecule, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Molecular Geometry**:
- Water (H₂O) has a bent molecular geometry. The central atom is oxygen, which is bonded to two hydrogen atoms.

2. **Determine the Hybridization of Oxygen**:
- Oxygen in water has two bonding pairs (from the O-H bonds) and two lone pairs of electrons. This leads to an sp³ hybridization.

3. **Expected Bond Angle**:
- In an ideal tetrahedral arrangement (which corresponds to sp³ hybridization), the bond angle is approximately 109.5°.

4. **Consider the Effect of Lone Pairs**:
- The presence of lone pairs affects the bond angle. Lone pairs occupy more space than bonding pairs due to their higher electron-electron repulsion.

5. **Analyze the Repulsion**:
- Lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion, which is in turn greater than bond pair-bond pair repulsion. This means that the lone pairs push the bonding pairs closer together.

6. **Calculate the Adjusted Bond Angle**:
- Due to the increased repulsion from the two lone pairs, the bond angle is reduced from the ideal tetrahedral angle of 109.5° to approximately 104.5°.

7. **Conclusion**:
- The bond angle of 104.5° in the water molecule is primarily due to the greater repulsion exerted by the lone pairs compared to the bonding pairs.

### Final Answer
The bond angle of 104.5° around oxygen in a water molecule is accounted for by the greater lone pair-lone pair repulsion compared to lone pair-bond pair and bond pair-bond pair repulsions.

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