Question
The relationship between the dissociation energy of $${N_2}$$ and $$N_2^ + $$ is
A.
dissociation energy of $$N_2^ + > \,$$ dissociation energy of $${N_2}$$
B.
dissociation energy of $${N_2} = \,$$ dissociation energy of $$N_2^ + $$
C.
issociation energy of $${N_2} > $$ dissociation energy of $$N_2^ + $$
D.
dissociation energy of $${N_2}$$ can either be lower
or higher than the dissociation energy of $$N_2^ + $$
Answer :
issociation energy of $${N_2} > $$ dissociation energy of $$N_2^ + $$
Solution :
The dissociation energy will be more when the bond order will be greater and bond order $$ \propto $$ dissociation energy
$${\text{Molecular orbital configuration of}}$$
$${N_2}\left( {14} \right) = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 1{s^2},$$ $$\mathop \sigma \limits^* 2{s^2},\pi 2p_y^2 \approx \pi 2p_z^2,\sigma 2p_x^2$$
$$\eqalign{
& {\text{So, bond order of}} \cr
& {N_2} = \frac{{{N_b} - {N_a}}}{2} \cr
& \,\,\,\,\,\,\,\, = \frac{{10 - 4}}{2} \cr
& \,\,\,\,\,\,\,\, = 3 \cr
& {\text{and bond order of}} \cr
& N_2^ + = \frac{{9 - 4}}{2} \cr
& \,\,\,\,\,\,\,\,\,\, = 2.5 \cr} $$
As the bond order of $${N_2}$$ is greater than $$N_2^ + $$ so, the dissociation energy of $${N_2}$$ will be greater than $$N_2^ + $$ .