Recall that paramagnetic means it contains at least one unpaired electron and diamagnetic is the lack thereof.
#"O"_2# is paramagnetic, with one electron each in its #pi_(2p_x)^"*"# and #pi_(2p_y)^"*"# antibonding molecular orbitals.
When we go back over to #"N"_2#, since #"N"# has one less electron than #"O"# in its atomic orbitals, #"N"_2# has two less electrons than #"O"_2# in its molecular orbitals.
Furthermore, going from #"O"_2# to #"N"_2# doesn't change the energy ordering for the #pi_(2p_x)^"*"# and #pi_(2p_y)^"*"# relative to the #sigma_(2p_z)# or #pi_(2p_x)# or #pi_(2p_y)#, so this assumption is valid.
That means #"N"_2# is diamagnetic, with no unpaired electrons.
In fact, its highest energy occupied molecular orbital (HOMO) is its #sigma_(2p_z)# bonding orbital, which currently contains two electrons.
(Due to orbital mixing effects between the #sigma_(2s)# and #sigma_(2p_z)# from #"Li"_2# to #"N"_2#, the #sigma_(2p_z)# is higher in energy relative to if the effects were not present, and the #sigma_(2s)# is lower in energy instead. #"N"_2# and #"O"_2# mark the border-crossing for when these effects are not significant; that is, when the #sigma_(2s)# and #sigma_(2p_z)# are too far in energy to interact.)
#"N"_2^(+)# therefore involves the removal of one #sigma_(2p_z)# electron.
Thus, #"N"_2^(+)# has a paramagnetic configuration due to the unpaired #sigma_(2p_z)# electron.
Think about it; if you lose an electron in a bonding molecular orbital, does the bond get weaker or stronger? #"N"_2^(+)# has less bonding character than #"N"_2#, which means it is less thermodynamically stable.
Does that mean it's a weaker bond? You should be able to figure this out from here.
