Question

The electronic configuration of valence shell of nitrogen molecule in ground state is $ (\sigma 2{{s}^{2}}),({{\sigma }^{*}}2{{s}^{2}}),(\pi 2{{p}^{4}}),(\sigma 2{{p}^{2}}) $ . Hence, the bond order in nitrogen molecule is

• A. 3
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (A)

Bond order $ =\frac{{{N}_{b}}-{{N}_{a}}}{2} $
where, $ {{N}_{b}}= $ Number of bonding electrons
$ {{N}_{a}}= $ Number of antibonding electrons
$ {{N}_{2}}(14)=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},\pi 2p_{y}^{2} $
$ \approx \pi 2p_{z}^{2},\sigma 2p_{x}^{2} $
Bond order of $ {{N}_{2}}=\frac{10-4}{2}=3 $