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?*