If we consider the molecule B2H6 (diborane Figure 1), there are 12 valence electrons at our disposal for chemical bonding (B has 3, and H has 1, so 2xB + 6xH =12).
Each terminal B–H bond is a standard vanilla two-electron bond, and there are four of these, thus accounting for a total of eight electrons. This leaves a total of four electrons to share between the two bridging H atoms and the two B atoms. Consequently, two B–H–B bridging bonds are formed, each of which consists of two electrons (Figure 2), forming what are called three-center-two-electron bonds (i.e., 3 atoms share 2 electrons) – sometimes called ‘banana’ bonds, as they are not linear but curved.
Each B atom is, approximately, sp3 hybridized (a hybridization is just a mathematical tool, so you can just as easily have s1.05p2.95 hybridized orbitals!), and if we consider just one of the B atoms, two of the four sp3 hybrid orbitals form s bonds to the terminal H atoms (1s orbitals). That leaves two B sp3 hybrid orbitals, one of which contains an electron, one of which is empty. For each bridge, therefore, one sp3 orbital from each of the B atoms combines (Figure 3) with the 1s orbital of the bridging H atom to form three new molecular orbitals (MOs) – as always, n atomic orbitals (AO) form n MOs. One B atom gives its remaining valence electron to one bridge, and the other B atom gives to the other. Each bridge, therefore, has two electrons, which fill our new MO scheme starting with the lowest energy bonding MO.