Question

Explain the formation of the following molecules using valence bond theory
(a) ${N_2}$ molecules
(b) ${O_2}$ molecules

Answer 1

Hint: Valence bond theory explains formation of molecule. If in a system, force of attraction between electrons of an atom and nucleus of another atom increases, energy of the system decreases and then the possibility of chemical bonding exists.

Complete step by step answer:
We can explain the formation of ${O_2}$ and ${N_2}$ molecules on the basis of valence bond theory. A covalent bond is formed by partial overlay of two half-filled atomic orbitals containing electrons with opposite spin.
Formation of ${O_2}$molecule:
Electronic configuration of oxygen is, $1{s^2}2{s^2}2p_x^22p_y^12p_2^1.$
When an atom of oxygen approaches another atom unpaired electrons present in $2{p_y}$ and $2{p_z}$ orbital overlap and form covalent bonds.
Two atoms are held by one $\sigma $-bond and one $\pi $-bond.
 
Formation of ${N_2}$ molecule:
Electronic configuration of nitrogen is, $1{s^2}2{s^2}2p_x^12p_y^12p_2^1.$
When one atom of Nitrogen approaches another atom of $N$,then$2{p_x},2{p_y}$ and $2{p_z}$ atomic orbitals overlap and form a covalent bond. As orbitals approach each other the energy of the system gets lower and forms stable molecules both nitrogen atoms held together by one $\sigma $ and two $\pi $ bonds.
 
Sigma bonds form by overlap of atomic orbitals of two atoms along the internuclear axis.
 
P1 $(\pi )$ bond formed by lateral overlapping of p-orbitals.

Note: Oxygen molecule has double bond between two O-atoms one bond is sigma bond and other bond is $\pi $ bond. Nitrogen molecules have a Triple bond between two N-atoms of which one bond is a sigma bond and two bonds or $\pi $-bonds.
Triple bond is stronger than double bond.