• # question_answer 105)   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.

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.
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