A) 2.0
B) 2.5
C) 1.5
D) 3.0
Correct Answer: D
Solution :
The bond order in \[{{\text{N}}_{\text{2}}}\]molecule is 3. \[{{N}_{2}}=14{{e}^{-}}\]configuration is \[\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2p{{x}^{2}},\] \[\pi 2p{{y}^{2}},\pi 2{{p}^{2}}.\] \[\text{Bond}\,\text{order = }\frac{{{N}_{b}}-{{N}_{a}}}{2}\] \[=\frac{10-4}{2}\] \[=\frac{6}{2}=3\]
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