In a Molybdenum complex of the form Mo(NR2)3, the Mo is in the oxidation state of +3, leaving it with 3 d electrons. When combined with an additional 2*3 electrons from the ligands, this leaves it with 9 electrons surrounding the metal. This is 9 less than the full 18, and so it could be said to have 9 frontier orbitals?
Why then, does is this complex considered isolobal with the nitrogen atom in N2, which has a full 8 valence electrons and so 0 frontier orbitals?