Give reasons for the following : (i) Where R is an alkyl group, `R_(3)P=O" exists but "R_(3)N=O` does not. (ii) `PbCl_(4)` is more covalent than `PbCl

(i) N due to the absence of d - orbitals, cannot form pp-dp multiple bonds. Thus, N cannot expand its covalency beyond four but in `R_(3)N = 0,` N has a covalency of 5. So, the compound `R_(3)N = 0` does not exit. On the other hand, P due to the presence of d - orbitals forms pp - dp multiple bonds and hence can expand its covalency beyond 4. Therefore, P forms `R_(3)P = O` in which the covalency of P is 5.
(ii) `PbCl_(4)` is more covalent than `PbCl_(2),` due to inert pair effect the higher oxidation state is stable. Therefore 4+ state is stable and so 2+ state losses 2 electrons and gets converted to 4+ state. This process is called oxidation `PbCl_(2)` oxides itself. Therefore its a reducing agent. So this is also the reason for `PbCl_(4)` not being an oxidising agent.
(iii) `N_(2)` is less reactive at room temperature because of strong `p pi-p pi` overlap resulting into the triple bond, `(N equiv N),` consequently high bond enthalpy.