Question

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) $PbC{l}_4$ is more covalent than $PbC{l}_2.$
(iii) At room temperature, $N_2$ is not very reactive.

Answer 1

(i) P due presence of vacant d-orbitals forms $p\pi -d\pi$ multiple bonds and hence can expand its covalency beyond four. Therefore, $R_{3}P=O$ exists in which covalency of P is 5. In $R_{3}N=O,$ the covalency of N is 5 but due to absence of d-orbitals N can not form $p\pi -d\pi$ bonds and thus cannot expand its covalency beyond 4. Hence, $R_{3}N=O$ does not exist.

(ii) O.S of Pb in $PbC{l}_4$ is +4 while in $PbC{l}_2$ it is +2. Due to large amount of energy needed to lose all the four valence electrons, $P{b}^{4+}$ cannot be formed and $PbC{l}_4$ (formed by sharing of electrons) will be thus more covalent than $PbC{l}_2.$

(iii) In $N_2$ (which is a gas at room temperature), $N\equiv N$ is present. Due to its high bond dissociation energy $\left(941.410kJmo{l}^{-1}\right)$, it is unreactive at room temperature.