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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?

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The $\ce{NR2}$ groups also act as pi donors with their formally "lone" electron pairs coupling to the molybdenum $4d$ and $5p$ orbitals (compare with the mesomeric effect of such groups in electrophilic aromatic substitution). Therefore the complex has fifteen electrons accessing the molybdenum valence orbitals. Under the threefold symmetry of the complex the three remaining vacant states split into an $a_1$ state and an $e$ pair, matching the symmetries of the bonding orbitals (apart from a center of inversion) in $\ce{N2}$.

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