The H-O-H bond angle in water is approximately 104°. This angle is determined by the arrangement of the atoms in the water molecule and is influenced by the repulsion between electron pairs in the molecule.

Explanation:
1. Molecular Structure of Water:
- Water (H2O) is a bent or V-shaped molecule, with two hydrogen atoms bonded to a central oxygen atom.
- The oxygen atom is located at the center of the molecule, and the two hydrogen atoms are attached to it.
- The oxygen atom has six valence electrons, and each hydrogen atom contributes one electron, giving a total of eight valence electrons in the molecule.

2. Valence Shell Electron Pair Repulsion (VSEPR) Theory:
- The VSEPR theory states that the arrangement of electron pairs around a central atom is determined by minimizing their repulsion.
- According to this theory, the electron pairs in the valence shell of an atom, whether they are bonding or non-bonding pairs, will arrange themselves in a way that maximizes their separation.

3. Electron Pair Repulsion in Water:
- In water, the oxygen atom has two lone pairs of electrons in addition to the two bonding pairs formed with the hydrogen atoms.
- The presence of these lone pairs creates repulsion between them and the bonding pairs.
- The lone pairs exert a greater repulsive force compared to the bonding pairs due to their closer proximity to the oxygen atom.

4. Effect on Bond Angle:
- The repulsion between the lone pairs and bonding pairs causes the H-O-H bond angle to be less than the ideal tetrahedral angle of 109.5°.
- The greater repulsion from the lone pairs pushes the bonding pairs closer together, resulting in a smaller bond angle.
- As a result, the H-O-H bond angle in water is reduced to approximately 104°.

Therefore, the correct answer is option 'C' (1040).