question_answer 1)

  Use the molecular orbital energy level diagram to show that \[{{N}_{2}}\] would be expected to have a triple bond, \[{{F}_{2}}\], a single bond and \[N{{e}_{2}}\], no bond.

Answer:

    Molecular orbital energy level diagram shows the distribution of electrons present in the molecule or ion in various molecular orbitals. This helps to determine bond order. \[\text{Bond order =}\frac{\begin{array}{*{35}{l}}    \text{No}\text{. of electrons in bonding orbitals -}  \\    \text{No}\text{. of electrons in antibonding orbitals}  \\ \end{array}}{\text{2}}\]\[{{N}_{2}}:14\] \[KK\sigma {{(2s)}^{2}}{{\sigma }^{*}}{{(2s)}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}\sigma {{(2{{p}_{z}})}^{2}}\] Bond order = \[\frac{8-2}{2}=3,\] i.e., \[{{N}_{2}}\]has triple bond between two nitrogen atoms. \[{{F}_{2}}:18\]\[KK\sigma {{(2s)}^{2}}{{\sigma }^{*}}{{(2s)}^{2}}\sigma {{(2{{p}_{z}})}^{2}}\pi {{(2{{p}_{x}})}^{2}}\]                                 \[\pi {{(2{{p}_{y}})}^{2}}{{\pi }^{*}}{{(2{{p}_{x}})}^{2}}{{\pi }^{*}}{{(2{{p}_{y}})}^{2}}\] Bond order = \[\frac{8-6}{2}=1,\]  i.e., \[{{F}_{2}}\] has single bond between two fluorine atoms. \[N{{e}_{2}}:\]\[20KK\sigma {{(2s)}^{2}}{{\sigma }^{*}}{{(2s)}^{2}}\sigma {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}\]                         \[{{\pi }^{*}}{{(2{{p}_{x}})}^{2}}{{\pi }^{*}}{{(2{{p}_{y}})}^{2}}{{\sigma }^{*}}{{(2{{p}_{z}})}^{2}}\] Bond order = \[\frac{8-8}{2}=0,\]    i.e., \[N{{O}_{2}}\] has no bond.