A) \[C{{H}_{4}}\]
B) \[N{{H}_{3}}\]
C) \[{{H}_{2}}O\]
D) \[{{H}_{2}}{{O}_{2}}\]
Correct Answer: A
Solution :
\[\text{C}{{\text{H}}_{\text{4}}}\text{,N}{{\text{H}}_{\text{3}}}\]and \[{{\text{H}}_{\text{2}}}\text{O}\]all the three molecules shows \[\text{s}{{\text{p}}^{3}}-\]hybridisation. In these molecules \[\text{C}{{\text{H}}_{\text{4}}}\]has bond angle \[\text{10}{{\text{9}}^{\text{o}}}\] 28 and have tetrahedral structure. But \[\text{N}{{\text{H}}_{3}}\]and \[{{\text{H}}_{\text{2}}}\text{O}\]has one and two lone pair of electrons respectively. Due to the presence of lone pair of electrons \[\text{N}{{\text{H}}_{3}}\]and \[{{\text{H}}_{\text{2}}}\text{O}\]their bond angle is decreased as \[\text{10}{{\text{7}}^{\text{o}}}\] and \[\text{10}{{\text{4}}^{\text{o}}}\] respectively and have pyramidal and V shape respectively. Hence, the angle is maximum in\[\text{C}{{\text{H}}_{4}}.\]
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