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.
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