• # question_answer44) What is meant by the term bond order? Calculate the bond order of ${{N}_{2}},{{O}_{2}},O_{2}^{+}$ and $O_{2}^{-}$.

Bond order is defined as half of the difference between the number of electrons present in the bonding and antibonding orbitals, i.e., Bond order = $\frac{{{N}_{b}}-{{N}_{a}}}{2}$ ${{N}_{2}}:KK\sigma {{(2s)}^{2}}\overset{*}{\mathop{\sigma }}\,{{(2s)}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}\pi {{(2{{p}_{z}})}^{2}}$ Bond order = $\frac{8-2}{2}=2$ ${{O}_{2}}:KK\sigma {{(2s)}^{2}}\overset{*}{\mathop{\sigma }}\,{{(2s)}^{2}}\sigma {{(2{{p}_{z}})}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}$ $\overset{*}{\mathop{\pi }}\,{{(2{{p}_{x}})}^{1}}\overset{*}{\mathop{\pi }}\,{{(2{{p}_{y}})}^{1}}$ Bond order = $\frac{8-4}{2}=2$ $O_{2}^{+}:KK\sigma {{(2s)}^{2}}\overset{*}{\mathop{\sigma }}\,{{(2s)}^{2}}\sigma {{(2{{p}_{z}})}^{2}}$ $\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}\overset{*}{\mathop{\pi }}\,{{(2{{p}_{x}})}^{1}}$ Bond order $=\frac{8-3}{2}=2.5$ $O_{2}^{-}:KK\sigma {{(2s)}^{2}}\overset{*}{\mathop{\sigma }}\,{{(2s)}^{2}}\sigma {{(2{{p}_{z}})}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}$ $\overset{*}{\mathop{\pi }}\,{{(2{{p}_{x}})}^{1}}\overset{*}{\mathop{\pi }}\,{{(2{{p}_{y}})}^{1}}$ Bond order $=\frac{8-5}{2}=1.5$