Give reasons for the following : (CH3)3 P = O exists but (CH3)3 N = O does not.
N atom cannot expand its covalency beyond four due to absence of low lying vacant d- orbitals, whereas P atom possesses low lying vacant d- orbitals. As a result, (CH3)3 P = O exists but (CH3)3 N = O does not.
N due to the absence of d – orbitals, cannot form p Π –d Π multiple bonds. As a result, N cannot expand its covalency beyond four but in R3N=O, N has a covalency of 5. Therefore, the compound R3N=O does not exist. In contrst, P due to the presence of d- orbitals forms p Π – d Π multiple bonds and hence can expand its covalency beyond 4. Therefore, P forms R3P=O I n which the covalency of P is 5.