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.

Explanation

Formation of `N_(2)` molecules.
Electrons configuration of N-atom `._(7)N= 1s^(2), 2s^(2), 2p_(x)^(1), 2p_(y)^(1), 2p_(z)^(1), N_(2)` molecules `=sigma1s^(2), sigma^(star)1s^(2), sigma2s^(2), pi2p_(x)^(2)=pi2p_(y)^(2), pi2p_(z)^(2)`

Bond order =`1/2[N_(b)-N_(a)]=1/2(10-4)=3`
Bond order value of 3 means that `N_(2)` contains a triple bond.
Formation of `F_(2)` molecule, `._(9)F=1s^(2), 2s^(2), 2p_(x)^(2), sigma^(star)2s^(2), sigma2p_(z)^(2), pi2p_(x)^(2)=pi2p_(y)^(2), pi^(star)2p_(z)^(2)=pi^(star)2p_(y)^(2)`

Bond order =`1/2[N_(b)-N_(a)]=1/2(10-8)=1`
Bond order value 1 means that `F_(2)` contains single bond.
Formation of `Ne_(2)` molecule `10Ne=1s^(2),2s^(2),2p_(x)^(2), 2p_(y)^(2), 2p_(z)^(2)`
`Ne_(2)` molecule = `sigma1s^(2), sigma^(star)1s^(2), sigma2s^(2), sigma^(star)2s^(2), sigma2p_(z)^(2), pi2p_(x)^(2)=pi2p_(y)^(2), pi^(star)2p_(x)^(2)=pi^(star)2p_(y)^(2), sigma^(star)2p_(z)^(2)`

Bond order =`1/2[N_(b)-N_(a)] = 1/2(10-10)=0`
Bond order value zero means that there is no formation of bond between two Ne-atoms. Hence, `Ne_(2)` molecule does not exist.