question_answer

The bond angle of water \[{{104}^{o}}31\] due to:

A)  repulsion between \[1p\] and bond pair

B)  \[s{{p}^{3}}\]-hybridization of O

C)  H-bonding

D)  higher electronegativity of O

Correct Answer: A

Solution :

In \[{{H}_{2}}O\] molecule, although central O atom is \[s{{p}^{3}}\]hybridized, \[H-O-H\]bond angle should be  \[{{109}^{o}}\,28,\]but the angle is\[{{104}^{o}}\text{ }31\]. The decrease is bond angle is due to the repulsion between lone pairs present on O atom and bond pairs.