question_answer 100)

  How will you account for \[104.5{}^\circ \]bond angle in water?

Answer:

  In \[{{H}_{2}}O\] molecule the oxygen is \[s{{p}^{3}}\text{ }-hybridized\]and thus, tetrahedral configuration comes into existence. Two positions are occupied by H atoms by forming sigma bonds with two hybrid orbitals and two positions are occupied by lone pairs. The expected bond angle should be\[109.5{}^\circ \], but the actual angle is \[104.5{}^\circ \]. The lone pair-lone pair repulsions are greater than bond pair-bond pair repulsions. As a result, bond angle in water is reduced from \[{{109.5}^{o}}\] to \[{{104.5}^{o}}\].