## Chemistry and Chemical Reactivity (9th Edition)

Molecular orbital configuration: $Li_2:\ [core]\ (\sigma_{2s})^2$ $B_2:\ [core]\ (\sigma_{2s})^2\ (\sigma_{2s}^*)^2\ (\pi_{2p})^2$ $C_2:\ [core]\ (\sigma_{2s})^2\ (\sigma_{2s}^*)^2\ (\pi_{2p})^4$ $N_2:\ [core]\ (\sigma_{2s})^2\ (\sigma_{2s}^*)^2\ (\sigma_{2p})^2\ (\pi_{2p})^4$ $O_2:\ [core]\ (\sigma_{2s})^2\ (\sigma_{2s}^*)^2\ (\sigma_{2p})^2\ (\pi_{2p})^4\ (\pi_{2p}^*)^2$ Since $B.O.=(N.\ Bonding -N.\ Antibonding)/2$, the bond orders are, respectively: 1,1,2,3,2, thus the shortest bond is in $N_2$ because it has the highest bond order.