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Use the molecular orbital energy level diagram to show that $\mathrm{N}_{2}$ would be expected to have a triple bond, $\mathrm{F}_{2}$ a single bond and $\mathrm{Ne}_{2}$ no bond.
Solution
Formation of $\mathrm{N}_{2}$ molecule :
Electronic configuration of $\mathrm{N}$-atom ${ }_{7} \mathrm{~N}=1 s^{2}, 2 s^{2}, 2 p_{x^{\prime}}^{1}, 2 p_{y}^{1}, 2 p_{z}^{1}$
$\mathrm{N}_{2}$ molecule $=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \pi 2 p_{x}^{2}$
$=\pi 2 p_{y}^{2}, \sigma 2 p_{z}^{2}$
Bond order value of $3$ means that $\mathrm{N}_{2}$ contains a triple bond.
Formation of $\mathrm{N}_{2}$ molecule :
Electronic configuration of $\mathrm{N}$-atom ${ }_{7} \mathrm{~N}=1 s^{2}, 2 s^{2}, 2 p_{x}^{1}, 2 p_{y}^{1}, 2 p_{z}^{1}$
$\mathrm{N}_{2}$ molecule $=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \pi 2 p_{x}^{2}$
$=\pi 2 p_{y}^{2}, \sigma 2 p_{z}^{2}$
