It depends on the MO diagram.
ORBITAL MIXING EFFECTS
#"B"_2# will experience orbital mixing, just like #"N"_2# and every homonuclear diatomic molecule on period 2 to the left of #"N"_2#, where the #sigma_(2p_z)# MO increases in energy and the #sigma_(2s)# decreases in energy relative to if there is NOT orbital mixing.
So, for #"B"_2#, the #sigma_(2p_z)# MO will be higher in energy than the #pi_(2p_x)# and #pi_(2p_y)# MOs, NOT lower!
As predicted, the #sigma_(2p_z)# MO (labeled #sigma_(g)(2p)#) is higher in energy than the #pi_(2p_x)# and #pi_(2p_y)# MOs (labeled as #pi_(u)(2p)#).
Notice the trend in how the #sigma_(g)(2p)# MO has been moving upwards as we move from #"Ne"_2# to #"Li"_2#; that is indicating that the orbital mixing is greater towards #"Li"_2#, and becomes less than significant for #"O"_2#, #"F"_2#, and #"Ne"_2#.
RATIONALIZING THE ELECTRON CONFIGURATION OF #\mathbf(B_2)#
Since boron contributes three valence electrons, two borons give a total of six valence electrons. So, the #sigma# and #pi# MOs generated by the #2s# and #2p# atomic orbitals in forming #"B"_2# will be filled with six valence electrons total.
Two fill the #sigma_(2s)# (labeled #sigma_(g)(2s)#) and two fill the #sigma_(2s)^"*"# (labeled #sigma_(u)^"*"(2s)#). Finally, one electron goes in the #pi_(2p_x)# and the other goes into the #pi_(2p_y)#.
That accounts for all six valence electrons.
CONCLUSION
Therefore, #"B"_2# contains two unpaired electrons and is paramagnetic.
CHALLENGE: Had the #pi_(u)(2p)# MOs been higher in energy than the #sigma_(g)(2p)# MO, #B_2# would have been diamagnetic. Why?
