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) $P b C l_{4}$ is more covalent than $P b C l_{2}$ .
(iii) $N_{2}$ is much less reactive at room temperature.

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Solution

(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 state is more stable then +2 state. Hence $P b C l_{4}$ is more covalent than $P b C l_{2}$

(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.

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