Answer
See the answer below.
Work Step by Step
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.
