(a) Draw the structures of the following :
(i) `PCl_(5)`(s)
(ii) `SO_(3)^(2-)`
Explain the following observations :
(i) Ammonia has a higher boiling point than phosphin.
(ii) Helium does not form any chemical compound.
(iii) Bi (V) is a stronger oxidising agent than Sb (V).

Explanation

(a) (i) Structure of `PCl_(5)` (s) : In solid state `PCl_(5)` exists as an ionic solid `[PCl_(4)]^(+)[PCl_(6)]^(-)` in which the cation , `[PCl_(4)]^(+)` is tetrahedral while the anion `[PCl_(6)]^(-)` is octahedral.

(b) (i) Due to the bigger size of P than N, P- H bond is weaker than N-H bond . Thus , `NH_(3)` (ammonia) has a higher boiling point than `PH_(3)` (phosphine)
(ii) Bacause the ionization energy of Helium is very high and the empty d- orbitals are also absent in it.
(iii) On moving down the group , the stability of +5 oxidation state decrreases while +3 oxidation state increases due to intert -pair effect . thus , +5 oxidation state of Bi is less satble than +5 oxidation state of Sb. Therefore , Bi (V) is stronger oxidising agent than Sb (V)