The shape of a molecule is determined by electron-pair repulsions in the valence shell. A lp occupies a larger space than a bp because it is not shared by two nuclei. Thus, the lp-lp repulsion is greater than the lp - bp repulsion, which in turn is greater than the bp - bp repulsion. The presence of lp causes distortion of bond angles, hence, a deviation from an ideal shape. The extent of distortion depends upon the orientation of the Ip's around the central atom. In a trigonal bipyramid, the lp's occupy equatorial positions than the axial ones. In AB type molecules, as the EN of A increases, the bp's come closer and the repulsion between them increases. On the other hand, as EN of B increases, the bp's get farther and repulsion decreases.
Which of the following statements is true?

Options

A. F-N-F angle in `NF_3` is greater than H-N-H angle in `NH_3` .
B. F-N-F angle in `NF_3` is smaller than H-N-H angle in `NH_3`
C. H-O-H angle in `H_2O` is greater than H-N-H angle in `NH_3`
D. F-O-F angle in `F_2O` is greater than H-N-H angle in `H_2O`

Answer

B

Explanation

EN of F > EN of H. Hence, bond angle of `NH_3 gt NF_3`