question_answer

The correct order of decreasing H-C-H angle in the following molecule is

A)  I > II > III

B)  II > I > III

C)  III > II > I

D)  I > III > II

Correct Answer: B

Solution :

 H-C-H bond angle is about \[120{}^\circ \], when C is bonded through a double bond but in case of \[\text{C}{{\text{H}}_{\text{4}}}\text{,}\]the H-C-H bond angle is only \[\text{10}{{\text{9}}^{o}}\,28.\] Cyclic compounds have a H-C-H bond angle greater than \[\text{10}{{\text{9}}^{o}}\,28\] due to hindrance. Thus, the correct order of H- C-H bond angle is \[\text{II}\,\text{}\,\text{I}\,\text{}\,\text{II}\]