S States

2H2O->O2 + 4H+ Em = 0.9 VeIn the initial step, P680 accepts exciton energy from the antenna pigments, creating an excited state, P680* . This excited state decays via the reduction of an adjacent pheophytin molecule (pheo). The resulting P680+ cation has a reduction potential of 1.2 V, which is sufficient for the oxidation of the redox active tyrosine cofactor, YZ, which has a reduction potential of approximately 1.0 V [8]. Substantial rearrangements of the cluster and the energetic penalties of burying an additional positive charge within a hydrophobic membrane protein represent kinetic and thermodynamic hurdles. In order to avoid the buildup of excess charge, it is widely accepted that protons are released to achieve charge balance, keeping the oxidation potential of the OEC in the various S states within a range accessible to oxidation by YZ [68] .

In Section 5.2.6, a series of quantum mechanical studies targeting the structural properties of the OEC were discussed. In addition, Siegbahn and coworkers have used DFT to examine models for the various S States. The relative free energy for each S state led the researchers to divide the Kok cycle (Figure 5.1) into two energetically favorable transitions for S0-S1 and S1-S2, and two energetically unfavorable steps in S2-S3 and S3-S4 which are compensated for by the large favorability ofthe S4-S0 step [69]. For the S4 state, the researchers propose an oxyl radical and not a Mn=O oxo species, which is in agreement with results that will be discussed in Section 5.3.2. However, a potential problem with the studies of Siegbahn and colleagues is their use of the highest electronic spin state, the pure ferromagnetically coupled state, for their calculations. As discussed in Section 5.2.1, this state is quite different from the antiferromagnetically coupled ground state. More recent studies have used the "broken symmetry" approach that more accurately reflects certain properties the ground state of the system [65].

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