Question

B2 molecule has less energy and bond order than compared to n2 molecule why. Explain?

Answer 1

Answer:

both [N2]+ and [N2]– have a formal bond order of 2.5, less than the triple bond in N2. If you found that surprising, that's actually a good thing, because for a molecule to lose bond strength upon both oxidation *and* reduction is not necessarily an intuitive result, and it's not the case for most diatomic molecules.  

As to the reason for it, if you've correctly deduced the result, you should be able to work through the underlying rationale. I assume you have a molecular orbital diagram for a dinitrogen species, and that you know which orbitals are bonding and antibonding in character. So it should be apparent that the HOMO (highest occupied) of N2 is bonding in character, while the LUMO (lowest unoccupied) is antibonding. Removing one electron from the bonding HOMO weakens the bond. Adding one electron to the antibonding LUMO *also* weakens the bond. Either way, the bond order is reduced by half-a-bond's worth.  

Note that the number of *unpaired* electrons is irrelevant. Unpaired electrons can (sometimes) make a molecule highly reactive, but that in and of itself says nothing about bond strengths or stability. O2 has two unpaired electrons, but [O2]–, which has only one, is less stable and has a weaker bond.