Abstract
From ab initio quality calculations on model systems, we conclude that in unliganded Fe-porphyrin the FE lies in the plane for both the high-spin (q) and intermediate-spin (t) states. Thus, the high-spin d6 Fe is not too big to fit into the porphyrin plane (as often suggested). We find the q state lower for a porphyrin hole radius greater than 1.94 A and the t state lower for smaller sizes. For the five-coordinate complex including an axial nitrogenous ligand [a model for myoglobin (Mb) and hemoglobin (Hb)], we find the ground state to be q with the Fe 0.3 A out of the plane (recent x-ray data on deoxy Mb suggests about 0.4 A). The origin of this out-of-plane displacement is the nonbonded repulsions between the axial ligand and porphyrin nitrogen orbitals. Pushing the Fe of the five-coordinate complex into the plane does not lead to a stable low-spin state (as usually suggested), the q and t states being the low-lying states. Bonding the O2 to form the six-coordinate complex stabilizes the t form of the Mb model, leading to a singlet state of MbO2 with Fe in the plane. (It has often been suggested that the Fe of MbO2 and HbO2 is low-spin Fe2+; however, we find this not to be the case.) The bonding in the MbO2 model confirms the ozone model of the bonding, leading to a structure consistent with the Pauling model (our calculated FeOO bond angle is 119 degrees). The total charge transfer to the O2 is 0.10 electron, in disagreement with the Weiss model. Molecular orbital calculations (Hartree-Fock) incorrectly lead to septet ground state (S = 3) for the MbO2 model. The implications for the cooperative O2 binding in hemoglobin and protein modifications of the energetics of the active site are considered. Use of our calculated force constants for displacement of Fe perpendicular to the heme plane suggests that the movement of the Fe upon a change in the quaternary structure from the T to the R form is only about 0.04 A toward the heme plane.
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Selected References
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