The bond order in `N_(2) ^(-)` is

Explanation

The electronic configurations should be written for the 4 molecule according to the buildup principle.
`{:(N_(2) ,, "(K electrons)" sigma_(s)^(2) sigma_(s)^(**2) pi_(x, y)^(4) sigma_(p_(z))^(2)) , (N_(2)^(+) ,, "(K electrons)" sigma_(s)^(2) sigma_(s)^(**2)pi _(x, y)^(4) sigma_(p_(z))^(1)), (NO ,, "(K electrons)" sigma_(s)^(2) sigma_(s)^(**2) pi_(x,y)^(4) sigma_(p_(z))^(2) pi_(x , y)^(** 1)), (NO^(+) ,, "(K electrons)" sigma_(s)^(2) sigma_(s)^(** 2) pi_(x , y)^(4) sigma_(p_(z))^(2)):}`
The computed bond orders for `N_2 and N_2^(+)` are 3 and `2 1/2`, respectively. `H_2` therfore has the stronger bond and should have the shorter bond length. The computed bond orders for NO and `NO^(+)` are `2 1/2` and 3, respectivley. `NO^(+)` has the stronger bond and should have the shorter bond length. As opposed to ionization of `N_2, which involves the loss of an electron in a bonding orbital, ionization of NO involves the loss of an electron in an antibonding orbital.