In Nitrogen family the H-M-H angle in the hydrides `MH_(3)` gradually becomes closer to `90^(@)` on going from N to Sb. This due to

Options

A. The basic strength of hydrides increases
B. Almost pure p-orbitals are used for M  H bonding
C. The bond energies of M  H bond increase
D. The bond pairs of electrons become nearer to the central atom

Answer

B

Explanation

Similarly all the hydrides have trigonal pyramidal shape with a lone pair of electrons.
Assume that no hybridization occurs and all these central atoms are using pure P - orbitals. Then the bond angle H - M - H should be `90^@`. This is because each M - H bond uses one P - orbital and each P - orbital is oriented `90^@` from the other `(P_x , P_y and P_z )`.
Thus the degree of hybridization decreases down the group and the angles become closer to `90^@`. In `NH_3` the bond pair – bond pair repulsion is highest due to small size of N atom. So the H - N - H angle opens up a bit wider and the bond angle is maximum.