Bond angle (H-O-H) in H₂O is

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104°30'

107°18'

109°28'

90°

answer is B.

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Detailed Solution

Molecule	Hybrisation of central atom	B.A	Geometry
H₂O	sp³	104°31'	V-shape
CO₂	sp	180°	Linear
NH₃	sp³	107°	Pyramidal
CH₄	sp³	109°28'	Tetrahedral

The molecule in which if the central atom undergoes sp³ hybridisation, then the bond angle between adjacent covalent bonds should be 109°28' [Regular BA] & the geometry of the molecule should be tetrahedral. [Regular geometry].

The above said statement is only applicable to the molecules without L.P's but if the molecule posses L.P along with BP's, then the geometry is said to be irregular geometry (Distorted geometry) & the bond angle (B.A) is said to be irregular B.A.

In case of H₂O 2-L.P's are present instead of 2-B.P's. L.P's occupies more space than B.P's and hence L.P's on the central atom causes strong repulsion on the O-H bond pair electrons and hence the two O-H bonds moves closer to each other, therefore bond angle decreases from 109°28' to 104°31'.

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