What Is the Bond Order of B2?

The bond order of B2 is 1. To find the bond order of a diatomic molecule such as B2, a chemistry student starts by writing out the electronic configuration of a single atom of boron to find out which electrons are located in bonding orbitals and anti-bonding orbitals.

In mapping out the electronic structure of boron, the student should find that the one-sigma and the two-sigma orbitals are nullified by the one-sigma-star and the two-sigma-star orbitals, meaning that these two orbitals’ electrons don’t count toward the bonding or anti-orbitals. The student should find that only the one-pi orbital, which contains two bonding electrons, counts toward the bond order of B2. Next, the student defers to the formula: Bond Order = 0.5 (Na -Nb), with Na being the total number of electrons in the bonding orbitals and Nb being the total number of electrons in the anti-bonding orbitals.

Applying the formula, the bond order is calculated: Bond Order = 0.5 (6 – 4) = 1. An interesting physical property that manifests in boron as a result of this bond order is paramagnetism, a slight attraction to a magnetic field. This results from the two bonding electrons being unpaired, allowing them to respond to a magnetic field by assuming different spins and angular momenta.