Question

Given reasons for the following:
(i) When R is an alkyl group, $R_{3}P=O$ exist but $R_{3}N=O$ doesn't.
(ii) $PbC{l}_4$ is more covalent than $PbC{l}_2$.
(iii) $N_2$ is much less reactive at room temperature.

Answer 1

(i) Nitrogen lacks the d orbital so nitrogen cannot expand its coordination number beyond four so $R_{3}N=O$ does not exist.

(ii) According to Fazan's rule, a higher charge on cation or anion makes compound more covalnet, +4+4 state is more stable then +2+2 state. Hence PbCl4$PbC{l}_4$ is more covalent than$PbC{l}_2$

(iii) Dinitrogen is formed by sharing three electron pairs between two nitrogen atoms.The two nitrogen atoms are joined by triple bond(N≡N). The nitrogen atom is very small in size ,therefore the bond length is also quite small(109.8 pm) & as a result the bond dissociation energy is quite high(946Kj / mol).This reason leads dinitrogen to be very less reactive at room temperature.