11th Class Chemistry Chemical Bonding and Molecular Structure

  • question_answer 44)
    What is meant by the term bond order? Calculate the bond order of \[{{N}_{2}},{{O}_{2}},O_{2}^{+}\] and \[O_{2}^{-}\].

    Answer:

    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 \[=\frac{8-5}{2}=1.5\]


You need to login to perform this action.
You will be redirected in 3 sec spinner