Fifth group elements form hydrides to type `AH_(3)`. The hydrides have a lone pair of electrons. The hydries are reducing in nature and the reducing power is related to the stability of A-H bonds. The hydrides are covalent and low boiling. Their boiling points depends on their ability to from hydrogen bond and their molecular size which decide the intermolcular forces in the hydrides .
The H-M-H bond angle of `V` group hydrides decrease from `107^(circ)` to `90^(circ)` for `NH_(3)` to `SbH_(3)` , this is due to:

Options

A. `SbH_(3) lt AsH_(3) lt PH_(3) lt NH_(3)`
B. `NH_(3) lt PH_(3) lt AsH_(3) lt SbH_(3)`
C. `NH_(3) lt SbH_(3) lt AsH_(3) lt PH_(3)`
D. `NH_(3) lt PH_(3) lt SbH_(3) lt AsH_(3)`

Answer

A

Explanation

In all these hydrides, the central atom 'M' is sphybridized forming four sp-hybrid orbitals. Three of these hybrid orbitals form three sigma `(sigma)` bonds with three hydrogen atoms, whereas the fourth hybrid orbital contains a lone pair. As a result, these hydrides have trigonal pyramidal geometry (or structure).

The H-M-H bond angle decreases from `NH_3" to "SbH_3.` Now, as we go down the group from nitrogen to bismuth, the size of central atom goes on increasing and its electronegativity goes on decreasing. Consequently, the position of bond pairs of electrons shifts more and more away from the central atom on moving from `NH_3" to "BiH_3.` For example, the bond pair in N-H bond is more closer to nitrogen than the bond pair in P- Hbond. As a result, the force of repulsion between the bonded pair of electrons in `NH_3` is more than that in `PH_3`. In general, the force of repulsion of between bonded pairs of electrons decreases as we move from `NH_3" to "BiH_3` and, therefore, the bond angle also decreases in the same order.