Q.

Bond angle (H-O-H) in H2O is

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a

90°

b

104°30'

c

107°18'

d

109°28'

answer is B.

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Detailed Solution

MoleculeHybrisation of central atomB.AGeometry
H2Osp3104°31'V-shape
CO2sp180°Linear
NH3sp3107°Pyramidal
CH4sp3109°28'Tetrahedral
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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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