The bond order of H2O is 2, and that of NH3 is 3, which makes sense, when considering the number of bonds they have.
However, if using basic group theory, you get the MO diagrams as presented in this paper:
Upon examination, we see that in the case of H2O, there are two orbitals that are strongly bonding, one weakly bonding orbital, and one non-bonding orbital. If we then tried to apply the bonding order equation to this MO diagram, would we not end up with a bond order of 3 - despite the a1 orbital being only weakly bonding, its occupation still stabilizes the molecule relative to its constituent atoms. Is it the case that the concept of bonding order breaks down using these kinds of MO diagrams?
A further example is given using the NH3 MO diagram. Here, they calculate the bond order as 3, ignoring the fact that the NH3 a1 orbital is weakly bonding. If we were to in theory calculate the MO diagram for NH3 in a trigonal planar geometry (same as for BH3), we would also get the bond order as 3. Of course, the occupied a1 orbital is weakly bonding in the trigonal pyrimidal NH3 geometry, leading to a preference over the trigonal planar geometry, but why is this not reflected in the bond order?