Question

Nitrogen forms \[{{\text{N}}_{\text{2}}}\] but phosphorus is converted into \[{{\text{P}}_4}\] from \[{{\text{P}}_2}\]. The reason or this is:
A.Triple bond is present between phosphorus atoms.
B.${{p\pi - p\pi }}$ bonding is weak
C.${{p\pi - p\pi }}$ bonding is strong
D.Multiple bonds are formed easily.

Answer 1

Hint: The nitrogen belongs to the second period with the second shell being the valence shell, while the phosphorus belongs to the third period with the third shell as the valence shell. It has a 3d orbital.

Complete Stepwise Solution:
As per their group numbers, both nitrogen and phosphorous have five electrons in their valence shell, out of which two electrons are there in the s-subshell while the other three are present in the three ‘p’ orbitals are lone or single unpaired electrons as per the Hund’s rule. As there are no more orbitals in the nitrogen atom so the two nitrogen atoms share their three unpaired electrons to form three bonds.
In the case of phosphorus, the case is different. The 3rd orbit has s, p, and d subshells. So the phosphorus atom can accommodate more electrons in the 3d levels that the nitrogen atom cannot do. In the \[{{\text{P}}_2}\] structure, the large orbitals of the 3rd level are inefficient in bonding to the ${{p\pi - p\pi }}$ bonds. So they are unable to triple bond. Hence the \[{{\text{P}}_4}\] structure is preferred over the \[{{\text{P}}_2}\] structure.

Hence, the correct option is B.

Notes: The triple bond of the nitrogen molecule, in which there are two ${{p\pi - p\pi }}$ bonds and a sigma bond, is very stable and is difficult to break. Hence the bond dissociation energy of the nitrogen molecule is very high.