The bond angle (H–O–H) in a water molecule is approximately:

The correct answer is approximately 105° (more precisely 104.5°).

In a water molecule (H₂O), the central oxygen atom is sp³-hybridised. Four electron pairs surround it: two bonding pairs (to the two hydrogen atoms) and two lone pairs.

If all four pairs were identical, the geometry would be a perfect tetrahedron with bond angles of 109.5°. However, lone pairs occupy more space and repel more strongly than bonding pairs (lone pair–lone pair > lone pair–bond pair > bond pair–bond pair).

This extra repulsion from the two lone pairs squeezes the two O–H bonds closer together, reducing the H–O–H angle from 109.5° to about 104.5°, giving water its characteristic bent (V-shaped) structure.

Therefore, the bond angle in a water molecule is approximately 105°.