A) \[{{109}^{o}}28'\]
B) \[{{107}^{o}}10\]
C) \[{{104.5}^{o}}\]
D) \[{{92}^{o}}\]
Correct Answer: C
Solution :
The \[H-O-H\] bond angle in water is\[{{104}^{o}}28'\], in accordance with the VSEPR prediction of slightly less than tetrahedral due to the presence of lone pairs of electrons on oxygen atom. Thus the orbitals used for bonding by \[O\] are close to \[s{{p}^{3}}\] hybrids.
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