Question

Bond order of $ {{N}_{2}} $ molecule is

• A. $ 3 $
• B. $ 2 $
• C. $ 1 $
• D. $ 0 $

Answer 1

Answer: (A)

The molecular configuration of $ {{N}_{2}} $ is $ \sigma 1{{s}^{2}},\,\overset{*}{\mathop{\sigma }}\,1{{s}^{2}}\sigma 2{{s}^{2}}\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\left\{ \begin{matrix} \pi 2p_{y}^{2} \\ \pi 2p_{x}^{2} \\ \end{matrix} \right.\overset{*}{\mathop{\sigma }}\,2{{p}_{z}}^{2} $ $ \left\{ \begin{matrix} \overset{*}{\mathop{\pi }}\,2{{p}_{y}} \\ \overset{*}{\mathop{\pi }}\,2{{p}_{x}} \\ \end{matrix} \right.\overset{*}{\mathop{\sigma }}\,2{{p}_{z}} $ number of bonding electrons $ Bond\,\,order=\frac{-number\,\,of\,\,antibonding\,\,electrons}{2} $ $ =\frac{10-4}{2} $ $ \therefore $ $ =3 $