Question

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 1

Molecular orbital energy level diagram of $N_2$

$N_2\left(14\right):\sigma 1S^2\sigma \ast 1S^2\sigma 2S^2\sigma \ast 2S^2\pi 2P_x^2=pi2P_y^2\sigma P_z^2$




Bond Order=$\frac{1}{2}$(electrons in BMO-electrons in ABMO)

Bond Order $\left(N_2\right)=\frac{1}{2}(10-4)=3$
So, the bond order is 3. hence, it would have a triple bond.

Molecular orbital energy level diagram of $F_2$

$F_2\left(18\right)=\sigma 1s^2\sigma \ast 1s^2\sigma 2s^2\sigma \ast 2s^2\sigma 2s^2\sigma \ast 2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^2={\pi }^{\ast }2p_y^2$


Bond Order $\left(F_2\right)=\frac{1}{2}(10-8)=1$
So, the bond order is 1. Hence, it would have a single bond.

Molecular orbital energy level diagram of $N{e}_2$

$N{e}_2\left(20\right):\sigma 1s^2\sigma s^2\sigma 2s^2\sigma \ast 2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^2={\pi }^{\ast }2p_y^2{\sigma }^{\ast }2p_z^2$



Bond Order $\left(N{e}_2\right)=\frac{1}{2}(10-10)=0$
No bond formed in $N{e}_2$ molecule. So, $N{e}_2$ molecule would not exist.