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 $N_2$ can either be lower or higher than the dissociation energy of $N_2^{+}$
• C. dissociation energy of $N_2$ > dissociation energy of $N_2^{+}$
• D. dissociation energy of $N_2^{+}$ > dissociation energy of $N_2$

Answer 1

The correct option is C dissociation energy of $N_2$ > dissociation energy of $N_2^{+}$

$Bondorder\alpha bondenergy$

So the bond order of $N_2$ and $N_2^{+}$ is as follows

Bond order of $N_2=3$,

Bond order of $N_2^{+}=2.5$,

the bond order of $N_2$ is more than $N_2^{+}$,

Hence, dissociation energy of $N_2$ > dissociation energy of $N_2^{+}$

So the correct option is [C].