A B 3 LYP study on electronic structures of [ ( X ) mMn ( μ-oxo ) 2 Mn ( Y ) n ] q + ( X , Y = H 2 O , OH and O ) as a Mn cluster model of OEC

Electronic and molecular structures of [(X)mMn (μ-oxo)2Mn(Y)n] (X, Y = H2O, OH and O), which are Mn cluster models at catalytic sites of OEC, were studied by broken-symmetry unrestricted B3LYP method. Two paths from the S0 to S3 states of Kok cycle were investigated. One is a path starting from [Mn(II) (μ-oxo)2Mn(III)] at the S0 state, and another is from [Mn(III) (μ-oxo)2Mn(III)] at S0. Results found in this study are summarized as, 1) In [Mn(II), Mn(III)], it is not possible that H2O molecules coordinate to the Mn atoms with retaining the octahedral configuration. 2) The OH anion selectively coordinates to Mn(IV) rather than Mn(III). 3) When the oxo atom directly bind to the Mn atom, the Mn atom must be a Mn(IV). From these results, the catalytic mechanism for four-electron oxidation of two H2O molecules in OEC is proposed. 1) The Mn4(II, III, IV, IV) at S0 is ruled out. 2) For Mn4(III, III, IV, IV) at S1, the Mn atom coordinated by OH anion is a Mn(IV) not Mn(III). 3) Only Mn(III) ion which is coordinated by a H2O molecule at S0 plays crucial roles for the oxidation.


INTRODUCTION
Manganese atoms play important roles in the oxygenevolving complex (OEC) of photosystem II, the catalytic site for the four-electron oxidation of water to molecular oxygen such as 2H 2 O → O 2 + 4e -+ 4H + .Kok and co-workers have proposed, known as Kok cycle, that the oxidation proceeds through five states denoted as from S 0 to S 4 under four times of irradiation of light and remove of four protons [1].Recently, the structures of OEC have been reported by the X-ray crystallographic studies [2][3][4][5].
Four manganese atoms and calcium ion are present at the catalytic site.The manganese atoms and calcium ion are arranged in cubanelike geometry with Mn 2 O 2 face of the di-μ-oxo-bridged [Mn(μ-O) 2 Mn].The Mn-Mn distances were observed to be about 2.7 for Mn 2 O 2 face [6][7][8][9].
As shown in Figure 1, three electrons and two protons are released from the catalytic site during the oxidation from S 0 to S 3 .The molecular oxygen is formed from S 3 through unstable S 4 by releasing an electron and two protons.The oxidation states of the four Mn atoms at the S 0 state have been suggested as Mn 4 (II, III, IV, IV) or Mn 4 (III, III, III, IV) [10][11][12][13][14][15][16].Even if the oxidation states of four manganese atoms at S 0 are different, the oxidation states change as Mn 4 (III, III, IV, IV) at the S 1 state → Mn 4 (III, IV, IV, IV) at S 2 → Mn 4 (III, IV, IV, IV) at S 3 → Mn 4 (III, IV, IV, IV) at S 4 .However, the oxidation states of four manganese atoms and ligand state at S 3 are still under debate.Two possibilities have been mostly discussed [17][18][19][20][21][22]: 1) Mn-centered oxidation; 2) oxidation of water molecule forming Mn-O or Mn-OH bonds as a direct Mn-ligand.
[(H 2 O) 4 Mn(μ-O) 2 Mn(H 2 O) 4 ] complex is a most fundamental model of the S 0 state of catalytic site [49,50].This model has the possibility to present the intermediate S 1 , S 2 and S 3 states where the proton(s) is(are) released from the H 2 O molecule.Figure 1 shows schematically the expected changes of the oxidation states for two Mn atoms and (total charge and spin multiplicity) of the Mn 2 O 2 core through from S 0 to S 3 states in the Kok cycle.From the proposals that the oxidation states of four Mn atoms are assigned as Mn 4 (II, III, IV, IV) or Mn 4 (III, III, III, IV) at the S 0 state, it is reasonable to assume that the oxidation states of two Mn atoms are Mn 2 (II, III) or Mn 2 (III, III) at S 0 .If we assume that the oxidation proceeds through the release of an electron from Mn atoms at each step, for the high spin state of Mn 2 (II, III) at the S 0 state, the electronic state changes as (1, 10) at S 0 → (1, 9) at S 1 → (2, 8) at S 2 .For Mn 2 (III, III), it changes as (2, 9) at S 0 → (2, 8) at S 1 → (3, 7) at S 2 .It is apparent that the S 2 state of Mn 2 (II, III) series is perfectly same as the S 1 state of Mn 2 (III, III) series.For the S 2 → S 3 transition, (2, 9) and (2, 7) are possible at S 3 for the Mn 2 (II, III) series, while (3,8) and (3,6) are possible for the Mn 2 (III, III) series.At this stage, the spin state of Mn 2 depends on the electronic structure of the formed Mn-O and Mn-OH bonds.To our knowledge, the experimental and theoretical studies of [Mn(μ-O) 2 Mn] complexes with direct Mn-OH bonds are not found.

COMPUTATIONAL DETAILS
The complexes including single OH ligand have axial and equatorial configurations of the Mn-OH bonds for the [Mn(μ-O) 2 Mn] face, as shown in Figure 2. The first character in the symbol (a, •) shows configuration of a ligand on the left side of two manganese atoms, while second one shows a ligand on the right side.Here we assign the left and right sides of two Mn atoms as Mn a and Mn b .Then, the symbol (a, •) means that OH binds to the Mn a atom in axial configuration and the Mn b atom has no OH ligand and has only H The geometries of these configurations were fully optimized without any geometry constraints for the LS and HS states with all possible total charges shown in Figure 1.For the stationary geometries, the vibrational frequency analyses were carried out to confirm that the stationary point is a local minimum on potential energy surface.The usual unrestricted Hartree-Fock (UHF) method leads to poor estimations for the binuclear systems because of lack of the electron correlation effects in the transition metals.The hybrid exchange-correlation functional B3LYP method [51,52] was most widely used for the transitionmetal systems.Since the B3LYP method contains the moderate static correlation effects, we employed the B3LYP method to estimate the electronic structures.The usual unrestricted method was used for the HS states, while the broken symmetry (BS) method was used for the LS states.The Ahlrichs DZV [53] and 6-31G * [54] basis sets were employed for Mn atom and for O and H atoms, respectively.After the optimized geometries were determined, the single point calculations were carried out using the Ahlrichs TZV [55] basis set for the Mn atom.Mulliken population analysis was used to estimate the distributions of spin densities.In this report, energies and spin densities estimated by TZV are summarized in Tables.All calculations were performed using the Gaussian 03 program package [56].
The transformation of the BS solution to first-order density matrix ( (1)   ) gives the natural orbitals (NOs: with their occupation numbers of electrons ( i n  ), which provide a powerful tool to analyze the electronic structures of the LS states of binuclear systems and to determine the oxidation states of two Mn atoms   ˆN where 2 and 1 2 N  and N  are numbers of up-and down-spins, respectively.Although the BS solution has the spin contamination 2 SC to the spin angular momentum, it can be related to the occupation numbers of electrons ( Here, present the distribution of the electron spin localized on Mn atom.Thus, from 2 SC and i  Ŝ  , the oxidation states of Mn atoms and characteristic of spin coupling between two Mn atoms can be easily determined.The MMOs i   are essentially same as the corresponding orbitals proposed by Amos and Hall [57]. Many procedures are defined for the (bi-)radical characters in the LS states.As an alternative way, the index of the (bi-)radical character for the antiferromagnetically coupled i   and i   can be defined by using the Shannon entropy theory [58,59].For the BS solution, Although the BS solution suffers from the spin contamination, the projection to the pure spin state can be easily carried out by using the paired MMOs This equation can be rewritten by using the paired NOs The spin projected first-order density matrix can be constructed by using Eq.8 and the index of the spin projected (bi-)radical character is defined as

RESULTS AND DISCUSSION
Computational results for all optimized geometries are summarized in Tables S1, S2 and S3 as supporting materials.Table S1 summarizes the oxidation states, spin angular momentums and relative energies for all optimized geometries starting from Mn 2 (III, II) at the S 0 state, while Table S2 summarizes the results for all optimized geometries starting from Mn 2 (III, III) at the S 0 state.Table S3 summarizes the effective exchange integrals of the lowest LS states in energy.In text, low-lying configurations of them are discussed as well as the previous ones with the Mn = O bond [49].

S 0 -S 2 States
Figure 3 shows optimized geometries of LS states from S 0 to S 2 states derived from Mn 2 (II, III) and Mn 2 (III, III) at S 0 .The corresponding HS states have similar geometries to the LS states.Table 1 summarizes the oxidation states of Mn atoms, spin contaminations, relative energies and spin coupling constants for the LS and HS states.Tables 2 and 3 summarize the selected interatomic distances and spin densities of selected atoms, respectively.Two LS states 1 and 3 are assumed to be Mn 2 (III, II) and Mn 2 (III, III) at the S 0 state, respectively.For the optimized state 1 in Table 1, 2 SC = 3.9923 indicates that the doublet state 1 has four antiferromagnetic pairs of spins localized on two Mn atoms and the natural orbital analysis indicates that unpaired single spin is localized on the Mn b atom.Accordingly, the oxidation state of Mn atoms in the doublet state 1 is apparently assigned as Mn 2 (III, II).The oxidation state of Mn 2 (III, II) can be also confirmed by the spin densities of two Mn atoms, -3.783 and 4.715 e shown in Table 3.The Mn-Mn distance is given by 2.724 Å, comparable with 2.7 Å in the crystal structure of the active site of OEC.However, as can be seen from Figure 3, the coordinations of water molecules are remarkably deformed from octahedral configuration of Mn atoms.Two μ-oxo atoms shift to the Mn(III) ion.Differences of bond distances of Mn(III)-μO and Mn(II)-μO are given by 0.235 and 0.264 Å (Table 2).The decet state 2, which is 2.3 kcal mol -1 above the doublet state 1, has similar spin densities of 3. 788 and 4.761 e on the Mn a and Mn b atoms to the state 1, respectively, showing that the decet state 2 is obviously the HS state corresponding to the LS doublet state 1.   2 and 3 summarize the selected interatomic distances and spin densities of selected atoms, respectively.on two Mn atoms.Thus, the oxidation states of two Mn atoms in the state 3 are apparently assigned by Mn 2 (III, III).From Figure 3, the Mn-Mn distance is given by 2.652 Å, slightly shorter than 2.7 Å estimated by X-ray crystallographic study.However, coordination of water molecules is pseudo-octahedral configuration with high symmetry, in contrast with the doublet state 1 of Mn 2 (III, II).The corresponding HS state 4 is 6.9 kcal•mol -1 higher than the state 3.
It can be predicted from the crystal structure of OEC that four Mn atoms are coordinated by pseudo-octahedral configurations.It is, therefore, expected that the oxidation states of Mn atoms at the S 0 state is Mn 2 (III, III) rather than Mn 2 (III, II).In other word, it is thought that Mn 4 (II,III,IV,IV) at the S 0 state in OEC can be ruled out.
When single OH -anion coordinates to Mn atom at the S 1 and S 2 states shown in Figure 1, axial and equatorial configurations are possible.As shown in Figure 1, the expected oxidation states of two Mn atoms are Mn 2 (III, III), Mn 2 (IV, III) and Mn 2 (IV, IV).Obtained was the LS singlet state 5 shown in Figure 3, which is thought to correspond to the S 1 state with Mn 2 (III, III) oxidation states.The ligand OH -anion is definitely deformed from axial configuration, although we set up an initial geometry in axial configuration before optimizing the geometry.Two Mn atoms are not strictly coordinated by H 2 O molecules and OH -anion in octahedral configuration.The OH - anion is located at middle position between axial and equatorial configurations.The Mn-Mn distance is given by 2.639 Å shorter than 2.7 Å of OEC.The BS solution of 5 has 2 SC = 4.0029, indicating that the oxidation state of Mn atoms is Mn 2 (III, III), consistent with spin densities of 3.791 and -3.796 e shown in Table 3.We tried to obtain the optimized equatorial geometry of the OH -anion by starting from equatorial configuration.However, the equatorial geometry was not obtained by approaching to the geometry 5. Accordingly, it could be con-sidered from these results that the oxidation state of Mn atom coordinated by the OH -anion is not Mn(III), even the OH -anion is generated as an intermediate through the oxidation of water molecules in OEC.Reversely, the ligand coordinating to Mn(III) is not a OH -anion.

S
As shown in Figure 1, the respective S 2 and S 1 states derived from Mn 2 (III, II) and Mn 2 (III, III) at S 0 are expected to be common Mn 2 (IV, III).For both LS and HS states, axial and equatorial geometries were optimized as  values of 7 and 9 are, respectively, given by 3.0141 and 2.9857, indicating that three spins localized on both Mn atoms are antiferromagnetically coupled each other.From the spin densities shown in Table 3, single unpaired spin is localized on Mn b atom.Thus, the oxidation states of two Mn atoms are apparently Mn 2 (IV, III).The OH -anion coordinates to Mn(IV) not Mn(III).

S
In the geometry of the LS doublet state 7, the H-atom of the axial H 2 O on Mn(III) ion faces to the O-atom of the OH ion, indicating the possibility that the H-atom of H 2 O transfers to the OH bond to yield the Mn(III)-OH.Figure 4 shows the energy change for the H-transfer from state 7.The total energy increases and any local minimum or transition state was not found on the potential energy surface.The spin densities of both Mn atoms do not change to maintain the oxidation states of Mn(IV) and Mn(III).At R OH = 1.6 Å, H 2 O and OH are formed on the Mn(IV) and Mn(III), respectively.We could not find the optimized geometries in which the OH -anion coordinates to Mn(III).In other word, the OH -anion selectively coordinates to the Mn(IV) ion in the case of Mn 2 (III, IV).
The optimized states 11 -14, which correspond to S 2 derived form Mn 2 (III,III) at S 0 , are summarized in Table 1.The geometries of the LS singlet states 11 and 13 shown in Figure 3 are similar to 7 and 9. Mn-Mn distances, 2.710 and 2.733 Å, close to 2.7 Å of the X-ray structure, are slightly longer than 2.649 and 2.674 Å of 7 and 9. From the 2 SC -values and spin densities (Table 3), it is apparent that the oxidation state of Mn atoms are Mn 2 (IV,IV) with three antiferromagnetic spin couplings.Similar to stabilities of 7 -10, the LS singlet state of axial configuration is more stable than that of equatorial configuration and the LS states are more stable than the corresponding HS state.

S
In summary, OH -anion can coordinate to the Mn(IV) ion rather than the Mn(III) ion with retaining the octahedral geometry.It is reasonable that this result will be retained in the OEC.Thus, It is expected that the OH -anion produced through the oxidation of water molecules coordinates to the Mn(IV) ion in the [Mn(μ-O) 2 Mn] plane of the OEC in axial configuration.

S 3 States from Mn 2 (III, II) at S 0
The geometries were fully optimized for all possible configurations of the coordination of two OH -anions and all possible spin states.All optimized states of the S 3 states derived from Mn 2 (III, II) and Mn 2 (III, III) at S 0 are, respectively, summarized in Tables S1 and S2.The states with the oxo-ligand, which are previously discussed in ref. 49, are also listed in both Tables.The selected configurations are summarized in Table 4. Two lowest configurations in energy and oxo-ligation are selected for the S 3 state from Mn 2 (III, II) at S 0 and three lowest configurations for the S 3 state from Mn 2 (III, III) at S 0 are selected.Tables 5 and 6 summarize the selected interatomic distances and spin densities, respectively.
As shown in Table 4, the LS singlet state 15 with t-(a, a) configuration of two OH -anions is most stable in energy.The corresponding HS septet state 16 is higher by only 0.8 kcal•mol -1 , consistent with small J-value of -42 cm -1 shown in Table 4.As can be seen from Figure 5, the optimized geometry has pseudo-octahedral configuration with the Mn-Mn distance of 2.654 Å slightly shorter than the observed 2.7 Å in OEC.It is found from 2 SC = 3.0719 shown in Table 4 that three antiferromagnetic spin couplings exist in the LS singlet state 15.From Table 6, the spin densities of Mn atoms are 2.897 e and -2.877 e, indicating that the oxidation states of Mn atoms are Mn 2 (IV, IV).Since the charge densities of two OHs are respectively given by -0.351 and -0.343 e with the spin densities of -0.002 and 0.015 e, two OHs are apparently OH -anions.The Mn-OH distances of 1.788 and 1.798 Å are comparable with 1.795 Å of 11 with the Mn 2 (IV, IV) spin configuration (see Tables 2 and 5).

S
The second lowest LS singlet state 17 (Table 4 and Figure 5) has the c-(a, a) configuration with 4.7 kcal•mol -1 higher than 15.The singlet state 17 is a Mn 2 (IV, IV)    Subsequently, one electron is removed from the Mn(III) ion to give a Mn(IV)-OH -bond.This is reasonably consistent with fact that the OH -anion does not coordinate to the Mn(III) ion as found for the states 7 -10, in Table 1.When one proton is removed from Mn(IV)-OH -bond, the geometry 37 with Mn(IV) = O 2-bond is formed.The subsequent remove of one electron from another Mn(III) ion rather than the Mn(IV) ion occurs to give the singlet state 37.As discussed above, this 37 is unstable and a transient state on the potential energy surface.

S 3 States from Mn 2 (III, III) at S 0
The S 3 states under consideration corresponds to those in which one electron is removed from the S 3 states derived from Mn 2 (III, II) at S 0 which are discussed in the preceding section.However the stabilities and electronic structures are significantly different from the preceding S 3 states.The most stable state is a LS doublet state 19 with axial configuration of oxo ligand in contrast with the t-(a, a) configuration of two OH -anions in the singlet state 15.The states, 19 -22, are perfectly same as the states, 7 -10, discussed in ref. 49.The oxidation states of Mn atoms in all states are Mn 2 (IV, IV) and the Mn-O bond character is formally presented as Mn(IV)-O -. Figure 6 shows the magnetic orbitals MMO 3 corresponding to the Mn(IV)-O -bond.These orbitals are composed of the in-phase and out-phase combinations of the bonding and antibonding orbitals of 3d yz (Mn) and 2p y (O).The bonding characters of both orbitals MMO 3 are due to the larger weight of the bonding orbital.Table 7 summarizes the biradical indexes of the antiferromagnatically coupled orbitals.The biradical index of the MMO 3 is estimated to be 0.366, showing the definitely small bira-dical character, consistent with the bonding character of the MMO 3.
    The second higher states in energy are states, 23 -25, in which two OHs coordinate to the Mn atoms with t-(a, a) configuration.The LS doublet state 23 is most stable among the t-(a, a) configurations with 5.5 kcal mol -1 higher than 19.The  spin coupling exist in the system.From the spin densities of two Mn atoms (2.715 and -2.910 e in Table 6), it is apparent that the oxidation state of Mn atoms is Mn 2 (IV, IV).
In changing from S 2 to S 3 states through removes of one proton and one electron, it is natural to consider that each OH group is an anion and one electron is removed from Mn atom to yield Mn 2 (IV, V).However, our results show that the oxidation state on Mn atoms is Mn 2 (IV, IV) and one electron must be removed from OH -anion or oxo atoms binding two Mn atoms.As can be confirmed from Table 6, the spin densities of oxo and OH a anion are given to be 0.715 and 0.478 e, respectively.Their summation is equal to 1.193 e close to unity.Figure 6 shows the MMO 2 that correspond to the antiferromagnetic spin coupling including contribution from oxo and OH a anion.The orbital MMO + 2, which is occupied by a down-spin, is composed of the antibonding combi-nation between the p z orbital on oxo and lone pair orbital on OH a .The MMO-2, which is occupied by a up-spin, distributes over the d yz orbital on the Mn b atom.The pure radical orbital (n = 1) is localized on another Mn a atom.Thus, in the oxidation state of Mn 2 (IV, IV) of the LS doublet state 23, two spins on the Mn atoms are coupled antiferro-magnetically each other.The biradical index of the coupled orbitals MMO 2 is estimated to be 0.890 close to unity, showing the strong biradical character contrast with weak biradical character of the Mn(IV)-O - bond.


As found in Table 4, the next higher energy state is 26 with the c-(a, a) configuration, which is only 1.2 kcal•mol -1 higher than the t-(a, a) configuration, 23.However, the state 26 is a sextet state, in contrast with the doublet state 23.The spin contamination <S 2 > SC of the septet state 26 is given by 0.9821, indicating that one pair of the antiferromagnetic spin coupling exists in the system.From Table 6, the spin densities of 2.977 and 2.979 e of two Mn atoms shows that three spins are localized on each Mn atom.The spin densities of -0.511 and -0.528 e of two OH anions shows that single down-spin is delocalized over two OHs.It is expected that this down-spin is antiferromagnetically coupled with the up-spin of the Mn atoms.
The Mn-Mn distance of the sextet state 26 is estimated to be 2.670 Å longer by 0.029 Å than the singlet state 17 with same cis configuration and Mn 2 (IV, IV).The Mn-OH distances are also longer than those of 17.Although one OH forms a hydrogen bond to another OH in 17, the O-O (in OH -anions) distance of 26 is given by shorter 2.071 Å and two OHs make the face-to-face conformation.Then the state 26 has a slightly deformed pseudo-octahedral conformation, as can be seen from Figure 5.
Figure 6 shows the orbitals MMO 1 corresponding to the antiferromagnetic spin coupling, which were obtained by linear combination of the natural orbitals using the occupation numbers of 1.433 and 0.567.The MMO-1 is delocalized over two OHs in antibonding manner of the vertical lone pair orbitals of OH.The corresponding MMO + 1 is composed of the antibonding character of the d xz orbitals on two Mn atoms.The lone pair orbital on OH is combined with the d xz orbital on Mn atom in bonding character.As found from Table 7, the biradical index is estimated to be 0.572, showing the small biradical character contrast with the strong biradical index of 0.868 of the sextet state 25 with the t-(a, a) configuration.This small biradical character is consistent with bonding character of the lone pair orbital and d xz orbital.In changing from S 2 to S 3 state, one electron is removed from the formed two OHs with retaining the oxidation state Mn 2 (IV, IV), not from the Mn atoms.


The doublet state 27 with the c-(a, a) configuration is 1.4 kcal mol -1 higher that the sextet state 26.This state S 0 : Mn 4 (III, III, III, IV)

Summary for OEC
The molecular and electronic structures of the model system, [Mn(μ-O) 2 Mn], corresponding to the states from S 0 to S 3 along the Kok cycle, have been investigated using the density functional method.Several rules can be enumerated from our results in this study.
1) In the combination of Mn(II)-Mn(III), the Mn atoms do not keep the octahedral structure as can be seen from Figure 3. Accordingly, it is thought that the Mn(II) ion does not exist at the S 0 state in OEC.
2) In the combination of Mn(III)-Mn(III) coordinated by OH -anion, the octahedral structure is remarkably deformed.In the Mn(III)-Mn(IV), the OH -anion selectively coordinates to Mn(IV).Accordingly, it is thought that the OH -anion coordinates to the Mn(IV) ion in OEC.
3) When the oxo atom directly bind to the Mn atom, the Mn atom must be a Mn(IV) ion.
It is possible from these results to propose the reaction mechanism of oxidation of two H 2 O molecules from the S 0 to S 3 states in the OEC, as shown in Figure 7.It is known that the spin state of the S 2 state is a low spin state with S = 1/2 [60,61].Following the reverse path from S 2 to S 0 , it is considered that the spin state of the S 0 state is S = 1/2.Although it has been suggested that the oxidation states of the Mn atoms at the S 0 state are Mn 4 (III, III, III, IV) and Mn 4 (II, III, IV, IV), both have the spin state of S = 1/2 as can be seen form Figure 7.However, from above rule 1) the H 2 O molecules coordinate to the Mn 2 (III, IV) rather than the Mn 2 (II, III) in the case of the Mn 4 (II, III, IV, IV).Accordingly the Mn(II) ion cannot be considered to exist in OEC, indicating that the Mn 4 (II, III, IV, IV) can be ruled out.In the Ferreira's structure [2], each Mn atom has rooms where the H 2 O molecules occupy in octahedral configuration.Thus, the oxidation state at the S 0 state must be Mn 4 (III, III, III, IV).It is considered that the H 2 O molecule coordinated to Mn 2 (III, III) plays an important role of the oxidation in OEC.Even if the Mn 4 (II, III, IV, IV) exists in OEC, the H 2 O molecules must coordinate to the Mn(III) and Mn(IV) ions of the Mn 4 (II, III, IV, IV).
At the S 1 state, the OH -anion is formed by removing a proton from the H 2 O molecule.Since the formed OH - anion selectively coordinates to the Mn(IV) ion from above rule 2) one electron is removed to yield the Mn(IV) ion from the Mn(III) ion coordinated by the H 2 O molecule from which the proton is released.As shown in Figure 4, the reversible reaction, Mn(III)-OH 2 + Mn(IV)-OH - Mn(III)-OH -+ Mn(IV)-OH 2 , does not definitely occur.The spin state of the S 1 state is given by S = 0, removing the up-spin from the Mn(III) ion.


The S 2 state is formed by removing one electron from the S 1 state.Judging from the change from 7 to 11, one electron is not removed to yield the Mn(V)-OH -from the Mn atom of the Mn(IV)-OH -, but the Mn(IV)-OH 2 is formed by removing one electron from the Mn atom of Mn(III)-OH 2 .Thus, the spin state changes from the singlet (S = 0) to doublet (S = 1/2) state.
Further releases of one electron and one proton lead to the S 3 state which is modeled by 19, 23 and 26 in this text.The doublet state 19, which is formed by change of Mn(IV)-OH -→ Mn(IV)-O -, is more stable than the doublet 23 and sextet 26 states which have two OH -anions.Since the doublet state 23 has trans-configuration of two OH, it is impossible that this conformation is put in the cubic Mn cluster of OEC.Although the sextet state 26 with the cis-configuration can be put in OEC, it is 6.9 kcal•mol -1 higher than 19.Further problem is that the sextet state 26 has three spins on the Mn atoms in parallel alignment.The change from the S 2 to S 3 states does not continuously link due to the spin flipping.The doublet state 27 does not have the above problems, but the 27 has further high energy than the 26.Accordingly it is thought that the Mn(IV)-O -exists in the S 3 state.
In the changes from the S 0 to S 3 states, one of four Mn atoms plays important roles for four-electron oxidation of H 2 O molecules in OEC.The Mn(IV)-O -bond is formed by removing two protons and two electrons from the Mn(III)-OH 2 .The possible spin state of the S 3 state is singlet or triplet.Considering that the O 2 molecule in the triplet state is produced, the spin state of the S 3 state would be triplet.Additionally the oxidation state of Mn(V) cannot be formed through the oxidation of H 2 O molecules in OEC.
The problem which Mn atom of four atoms takes part in the oxidation has been left.In the Ferreira's structure [2], the Mn atom which is most close to Asp61 and linked by the hydrogen-bonding to Asp61 is the branching Mn atom of Mn-μO-Mn.Presumably, these two Mn atoms will be reactive, as shown in Figure 7.This year, the X-ray structure at resolution of 1.9 Å has been published [5].Four Mn atoms are coordinated by ligands in octahedral configuration, while only one of four is coordinated by two H 2 O molecules.The Ca ion is also coordinated by two H 2 O molecules.One of two H 2 O molecules on the Mn atom is hydrogen-bonded to one of two H 2 O molecules on the Ca ion.These two hydrogenbonded H 2 O molecules might be substrates of formation of O 2 molecule.From the above rules, the oxidation state of four Mn atoms must be Mn 4 (III, III, III, IV) at the S 0 state.It is expected that H 2 O molecule on Mn atom is oxidized to yield the Mn(IV)-O -bond at the S 3 state.This proposal for OEC at 1.9 Å resolution consistently matches with the results of our trial calculation using the structure at 1.9 Å resolution [62].
Another problem is a Ca ion in the cubic form of the catalytic site.It was proposed that it is involved in the oxidation mechanism of two H 2 O molecules.Even if the Ca ion is involved in the oxidation mechanism, the H 2 O molecule coordinated to the Ca ion might be oxidized at the stage of the S 4 state.However, the proposal for the Ca ion has not been established yet.

Figure 1 .
Figure 1.Schematic representation from S 0 to S 3 states corresponding to Kok cycle.Values in parenthesis denote total charge and spin multiplicity for high spin states.

SFor singlet state 3, 2 SC= 3 .
9978  shows that the state 3 has four antiferromagnetic pairs of spins localized

Figure 3 .
Figure 3. Optimized geometries of LS states from S 0 to S 2 states derived from Mn 2 (II,III) and Mn 2 (III,III) at S 0 .Mn-Mn and Mn-O(OH) distances are shown in unit of Å.states.Tables2 and 3summarize the selected interatomic distances and spin densities of selected atoms, respectively.

7 Figure 4 .
Figure 4. Energy change of H-transfer from H 2 O coordinating to Mn b to OH coordinating to Mn a in the LS doublet state 7. Relative energies are estimated at B3LYP/DZV level.

23 Mn 26 Mn 2 (Figure 5 .
Figure 5. Optimized geometries of LS states in S 3 states derived from Mn 2 (II,III) and Mn 2 (III,III) at S 0 .Mn-Mn and Mn-O(OH) distances are shown in unit of Å.

2 : 2 S = 1 / 2 S = 0 S = 0 or 1 Figure 7 .
Figure 7. Proposal of reaction mechanism of oxidation of H 2 O molecules from S 0 to S 3 states catalyzed by Mn cluster in OEC.
2 O ligands.The complexes including two OH ligands have a large number of com- binations of OH configurations.The pre-character t and c of t-(a,a) and c-(a,a) shown in Figure 2 means trans and cis configurations of two OHs binding to different Mn atoms, respectively.For the [(H 2 O) 3 (HO)Mn-(μ-O)
2 O molecules as ligands; b) Configuration of OH -anion; c) Total charges of system; d) In unit of kcal•mol -1 ; e) In unit of cm -1 .

Table 1 .
Figure3shows the LS doublet states 7 and 9.The corresponding HS octet states 8 and 10 have similar geometries to the LS states 7 and 9.The optimized geometries retain the pseudo-octahedral configuration with Mn-Mn distances of 2.649 and 2.674 Å for 7 and 9, respectively.The states 7 and 9 have similar Mn-O(OH -) distances of 1.812 and 1.815 Å.From

Table 1 ,
the LS states are more stable by 4.4 and 0.3 kcal•mol -1 than the HS states for the axial and equatorial configurations, respectively.The axial state 7 is 13.7 kcal•mol -1 lower than the equatorial state 9.The 2 SC
a) Symbol • means H 2 O molecules as ligands; b) Configuration of OH -anions and oxo atom; c) Total charges of system; d) In unit of kcal•mol -1 ; e) Same as 15 shown in ref. 48; f) Same as 1 and 2 shown in ref. 48; g) Same as 7, 8, 9 and 10 shown in ref. 48; h) In unit of cm -1 .

Table 7 .
Biradical index and electron occupation numbers of antiferromagnetically coupled orbitals.