Question

Given reasons for the following:
(i) When \[R\] is an alkyl group, \[{R_3}P = O\] exists 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

Hint: We need to know the electronic configuration of $N$ and $P$ . Electronic configuration of N is $1{s^2}2{s^2}2{p^3}$ and that of P is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^3}$ . Clearly,Nitrogen does not have vacant d-orbitals hence its co-valency cannot exceed 5.

Complete step by step answer:
Now we discuss about first option as,
i) When \[R\] is an alkyl group, \[{R_3}P = O\] exist but \[{R_3}N = O\] doesn't
Since nitrogen does not have vacant d-orbitals therefore \[{R_3}N = O\] is not possible.
However,\[P\]belonging to group 15 of the periodic table has vacant d orbitals hence can exist as \[{R_3}P = O\]
An element can have a vacant d orbital if its previous orbit has at most two electrons less in the outermost shell of the p orbital. In case of phosphorus its Px, Py, Pz subshells have at least one electron. They are not empty. Hence, it can have a vacant d orbital.
(ii) \[PbC{l_4}\] is more covalent than \[PbC{l_2}\] ​.
We need to know the concept of covalent compounds and what makes a compound covalent. A covalent compound is a molecule formed by covalent bonds, in which the atoms share one or more pairs of valence electrons.The covalent nature of compounds is explained by the postulates of Fajan’s rules formulated by Kazimierz Fajan’s in 1923.
Fajan’s rule states that the size of cation, the larger the size of the anion, greater is the covalent character of the ionic bond.
Also, Greater the charge of cation, greater is the covalent character of the ionic bond.
 The +4 state of \[PbC{l_4}\] is more than +2 state of \[PbC{l_2}\] . Hence \[PbC{l_4}\] is more covalent than \[PbC{l_2}\] ​.
(iii) \[{N_2}\] is much less reactive at room temperature.
We need to the properties of dinitrogen, \[{N_2}\] .Dinitrogen is formed by sharing three electron pairs between the two atoms and hence joined by 3 covalent bond and hence a triple bond exists between the two.It is very small in size and its bond length is approximately 109.8pm and therefore the energy required to break each of the 3 bonds is quite high.This energy is known as Bond Dissociation Energy which is as high as 946KJ/mol.
Dinitrogen is bonded by triple bonds.
It has a small bond length.
The energy to break the triple bonds to destabilize it is very high.
This is the reason why it is much less reactive at room temperature, i.e, acts as an inert molecule.

Note:
It must be noted that to give the reasons for the behaviour of a particular molecule or compound,the knowledge of its properties is the key.As in the first case,the knowledge of electronic configuration of elements is needed along with knowledge of orbitals and subshells.In the second case,one must understand the postulates of Fajan’s rules and must be able to calculate the charge of the cation or anion.In the third case,the bonding properties between the given elements must be noted.