Question

why the bond angle in water less than the normal tetrahedral value
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Answer 1

Answer:

The H—C—H bond angle in methane is the tetrahedral angle, 109.5°. This angle is obtained when all four pairs of outer electrons repel each other equally. The bond angles in ammonia and in water are less than 109.5° because of the stronger repulsion by their lone pairs of electrons.

Answer 2

Answer:

An ideal tetrahedral geometry have each angle of 109.28'. But in H2O the bond angle is found to be of 104.5.

But we must know that only an ideal geometry shows the accurate bond angles in between the bonds which were as mentioned, and an ideal geometry is shown by such molecules or compounds which have all the surrounding atoms or molecules identical to each other.

And in the case of water, the central atom i.e. oxygen is having 2 lone pairs, so we can't consider water for ideal tetrahedral geometry, and as we knows that the lone pairs repel bond pair more in comparison to bond pairs.

Therefore, the lone pairs of oxygen repel the bond pairs, and thus the bond angle is reduced to 104.5 from 109.28.