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 $Ne_{2}$ no bond.

(N/A) Formation of $N_{2}$ molecule:
Electronic configuration of $N$-atom: ${ }_{7} N = 1s^{2}, 2s^{2}, 2p_{x}^{1}, 2p_{y}^{1}, 2p_{z}^{1}$
$N_{2}$ molecule: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \pi 2p_{x}^{2}, \pi 2p_{y}^{2}, \sigma 2p_{z}^{2}$
Bond order = $\frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (10 - 4) = 3$. $A$ bond order of $3$ indicates a triple bond.
Formation of $F_{2}$ molecule:
Electronic configuration of $F$-atom: ${ }_{9} F = 1s^{2}, 2s^{2}, 2p_{x}^{2}, 2p_{y}^{2}, 2p_{z}^{1}$
$F_{2}$ molecule: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, \pi 2p_{x}^{2} = \pi 2p_{y}^{2}, \pi^{*} 2p_{x}^{2} = \pi^{*} 2p_{y}^{2}$
Bond order = $\frac{1}{2} (10 - 8) = 1$. $A$ bond order of $1$ indicates a single bond.
Formation of $Ne_{2}$ molecule:
Electronic configuration of $Ne$-atom: ${ }_{10} Ne = 1s^{2}, 2s^{2}, 2p_{x}^{2}, 2p_{y}^{2}, 2p_{z}^{2}$
$Ne_{2}$ molecule: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, \pi 2p_{x}^{2} = \pi 2p_{y}^{2}, \pi^{*} 2p_{x}^{2} = \pi^{*} 2p_{y}^{2}, \sigma^{*} 2p_{z}^{2}$
Bond order = $\frac{1}{2} (10 - 10) = 0$. $A$ bond order of $0$ indicates no bond exists.