A) dissociation energy of \[N_{2}^{+}\] > dissociation energy of \[N_{2}^{+}\]
B) dissociation energy of \[{{N}_{2}}=\] dissociation energy of \[N_{2}^{+}\]
C) dissociation 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 \[N_{2}^{+}\]
Correct Answer: C
Solution :
The dissociation energy will be more when the bond order will be greater or Bond order \[\propto \] Dissociation energy Molecular orbital configuration of \[{{N}_{2}}(14)=\] \[\sigma 1\,{{s}^{2}},\,\sigma *\,1{{s}^{2}},\sigma 2{{s}^{2}},\,\sigma \,2\,{{s}^{2}},\,\pi 2\,P_{y}^{2},\pi 2p_{z}^{2},\,\sigma 2p_{x}^{2}\] So bond order of \[{{N}_{2}}=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-4}{2}=3\] and bond order of \[N_{2}^{+}=\frac{9-4}{2}=2.5\] As the bond order of \[{{N}_{2}}\] is greater than \[N_{2}^{+}\] So, the dissociation energy of \[{{N}_{2}}\] will be greater than \[N_{2}^{+}\] .
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