Yves Jean - Molecular Orbitals of Transition Metal Complexes (793957), страница 33
Текст из файла (страница 33)
These three MO are all doubly occupied,since we are considering a complex with a d6 electronic configuration.For each conformation, there is a d-block orbital that is stabilized by abonding interaction with the π ∗ orbital on ethylene and a nonbondingorbital. The preference for the eclipsed conformation arises from thethird occupied orbital, which is nonbonding in the eclipsed structurebut destabilized in the staggered case (SS). This destabilization shouldbe linked to the analysis illustrated in Figure 4.2, which shows a repulsiveinteraction between SS orbitals in the staggered conformation.In conclusion, we note that the highest-occupied molecular orbital(HOMO) ‘rule’, according to which the more stable of two structures ofa given species is the one whose HOMO is at lower energy, is perfectlyapplicable to this conformational problem, and to the one treated in thepreceding section (see Figure 4.1).4.1.3.
d6 -[ML4 (η2 -C2 H4 )2 ] complexes: coupling of twoπ-acceptor ligandsA new conformational problem arises when there are two ligands forwhich several orientations are possible, instead of just one: does the orientation of one of them have an influence on the other? Steric repulsionsmay, of course, be significant, but here we shall be interested in purelyelectronic factors that may induce a coupling between the orientationsof the two ligands.Octahedral complexes of the type d6 -[ML4 (η2 -C2 H4 )2 ], in whichthere are two ethylene ligands in trans positions, provide a characteristicexample to illustrate this point.
The problem of the orientation of eachC==C bond with respect to the rest of the complex has been treatedin the preceding section, where it was shown that in the most stable−L bonds. If thisconformation, these bonds eclipse the neighbouring M−result is accepted, there is still the question of the relative orientationof the two ethylene ligands.
Two limiting structures can be imagined:the coplanar conformation (4-10a), which has D2h symmetry, in whichthe two C==C bonds are located in the same plane (yz); and the perpendicular conformation (4-10b), with D2d symmetry, in which one ofthe ethylene molecules has turned by 90◦ to be in the xz plane. Experimentally, this second orientation is observed in this class of compounds,for example, in the complexes trans-[Mo(PMe3 )4 (C2 H4 )2 ] and trans[Mo(diphos)2 (C2 H4 )2 ] (diphos == Ph2 PCH2 CH2 PPh2 ). The barrier torotation about the metal–olefin bond has been estimated to be about15 kcal/mol in this latter compound. Since the two ethylene moleculesare well separated, by about 4 Å, it seems unlikely that steric factors arethe origin of this conformational preference.
It is therefore natural toApplicationslook for an electronic factor, by analysing the interactions between theligands and the residual metallic fragment.zMyMx4-10a (coplanar)4-10b (perpendicular)We shall consider the π and π ∗ orbitals on the ethylene ligandsin each conformation, as they interact with the orbitals of the squareplanar d6 -ML4 fragment in which the three nonbonding d orbitals, xy,xz, and yz (Chapter 2, § 2.2.1) are doubly occupied.
Two other orbitals will also be considered in the following analysis, since they pointtowards the two ethylene molecules (4-11): the empty z2 orbital whichis weakly antibonding, and pz which is nonbonding but higher in energy(Chapter 2, § 2.2.2).pzz2xyyzxz4-11We consider first the interactions of the occupied π orbitals. Thesituation is identical to that described in the preceding section for theeclipsed confirmation of a d6 [ML5 (η2 -C2 H4 )] complex: the overlapbetween a π orbital and any of the occupied nonbonding orbitals on themetallic fragment is zero. However, there are two interactions with theempty z2 and pz orbitals, as shown in 4-12 for the coplanar conformation.The in-phase combination of the π1 and π2 orbitals (π (+) ) interacts withz2 (4-12a), and the out-of-phase combination (π (−) ) with pz (4-12b).These are two-electron interactions (ligand → metal donation), andafter the interaction, only the bonding MO, mainly concentrated on theligands, are doubly occupied.
Due to the cylindrical symmetry of thez2 and pz orbitals, there is no change to the overlaps involved if one ofConformational problemsthe two ethylene molecules is rotated by 90◦ to give the perpendicularconformation. The interaction between the two π orbitals is changed,but the distance between the ligands is too large for this change togive any energetically significant consequences. The donation interactions do not, therefore, give rise to any pronounced conformationalpreference.11(+)z2(–)pz224-12a4-12bWe now turn our attention to the interactions that involve the emptyorbitals. As in the monoethylene complex (§ 4.1.2), the overlap foreach of the π ∗ orbitals with z2 is zero by symmetry.
The same is truefor the pz orbital, which, like z2 , has cylindrical symmetry with respectto the z-axis. We must still consider the interactions with the nonbonding and doubly occupied d orbitals (4-11), that is, the metal → ligandsback-donation interactions. Following the method developed in § 3.3 ofthe preceding chapter, we shall examine the overlap of the π ∗ orbitalswith each of the three nonbonding d orbitals, remembering that a nonzero overlap leads to a bonding interaction d ↔ π ∗ that stabilizes thed orbital (π-acceptor character of the ligand).
In the planar conformation, the two π ∗ orbitals, concentrated in the yz plane, overlap withthe yz orbital which is therefore stabilized by two bonding interactions(4-13a). In other words, yz is stabilized by the out-of-phase combination(π ∗(−) ) of the orbitals π1∗ and π2∗ . But the xz and xy orbitals, whichare antisymmetric with respect to the yz plane, have zero overlap withthe π ∗ orbitals, which are symmetric with respect to this plane.
Theirshapes and energies therefore stay unchanged (4-13b and c).π∗1*yz*(–)xzxy*24-13a (yz + 2π ∗ )4-13b (xz)4-13c (xy)ApplicationsIn the perpendicular conformation, the π1∗ orbital still interactswith the orbital yz (4-14a), but π2∗ now overlaps with the orbital xz(4-14b), and xy remains as a pure d orbital (4-14c). In this conformation,two d orbitals are therefore stabilized to an equivalent extent (bondinginteractions), but the third is unchanged.
1*xzyzxy *24-14a (yz + π ∗ )4-14b (xz + π ∗ )4-14c (xy)These results are given as an interaction diagram in Figure 4.4;symmetry labels from the D2h point group are used for the coplanarconformation (on the left), and from the D2d point group for the perpendicular conformation (on the right). The π ∗(+) and π ∗(−) combinationsin the coplanar structure have b2u and b3g symmetry, respectively,whereas the π ∗ orbitals in the perpendicular structure form a degenerate pair of orbitals of e symmetry. Due to the large spatial separationbetween the ethylene ligands, the energies of the π ∗(+) and π ∗(−) combinations are very similar, and only slightly different from the energyof the degenerate π ∗ orbitals in the perpendicular conformation.
Thethree nonbonding d orbitals on the metal, whose symmetries are b3g(yz), b2g (xz), and b1g (xy) in the D2h point group, and e (yz, xz) andb2 (xy) in the D2d point group, respectively, are placed in the centre ofthe figure. Interaction between orbitals of the same symmetry leads,as we have shown above, to the stabilization of one d orbital in thecoplanar conformation (4-13a), whereas two of them are stabilized inthe perpendicular conformation (4-14a and b).b 3gFigure 4.4. Diagram for the interactionbetween the nonbonding d orbitals (in thecentre) and the π ∗ orbitals on the ethyleneligands in the coplanar (D2h , on the left) andperpendicular (D2d , on the right)conformations of an octahedral trans[ML4 (η2 -C2 H4 )2 ] complex with a d6electronic configuration.eb 2ub 1gb 2gb 3g∆Ecb2e∆EpConformational problemsCommentIn each conformation, the three doubly occupied MO are the orbitalsderived from the t2g block of a regular octahedral complex with a d6electronic configuration.From the energetic point of view, four electrons are stabilized inthe perpendicular conformation, but only two in the coplanar case.However, the stabilization of the orbital is larger for the coplanar conformation (Ec > Ep ), since there are two bonding interactions(4-13a), as compared to only one for the degenerate orbitals (4-14aand b).
In order to deduce the conformational preference, it is thereforenecessary to know whether Ec is smaller than, equal to, or more thantwice as large as Ep .Since these are interactions between orbitals with different energies, the stabilizations E are proportional to S2 /ε, where S is theoverlap between the orbitals and ε the energy difference betweenthem (Chapter 1, § 1.3.2). The term ε is essentially the same for theconformations, since the energies of the π ∗(−) (coplanar conformation) and π ∗ (perpendicular conformation) orbitals are almost identical.What can be said about the S2 term? If Sdπ ∗ is the bonding overlapbetween a d orbital and a π ∗ orbital, the overlap between the fragment orbitals in the perpendicular conformation is Sp = Sdπ ∗ , andthe total stabilization due to the four electrons (Figure 4.4, right-hand2 . In the coplanar conformation, the overlapside) is proportional to 4Sdπ∗takes place between a d orbital and the π ∗(−) orbital, which is writ√ten as (1/ 2)(π1∗ − π2∗ ) in normalised form.
The overlap involved,√√Sc , is therefore (1/ 2)(Sdπ ∗ + Sdπ ∗ ) = 2Sdπ ∗ , and the stabilization of the two electrons (Figure 4.4, left-hand side) is proportional to√2 . We therefore come to exactly the same result2( 2Sdπ ∗ )2 = 4Sdπ∗for the two conformations, so we can apparently deduce that there willbe essentially free rotation for an ethylene ligand, but this conclusionis contradicted by the experimental results that were presented brieflyat the beginning of this section, which indicate a barrier to rotation ofabout 15 kcal mol−1 .However, we must remember that the expression showing that theenergetic stabilization is proportional to S2 /ε holds only when thereis a large energy difference ε between the two interacting orbitals. Ifthe energy gap is not large, then the stabilization is proportional to S(Chapter 1, § 1.3.2, Note 8). In the case under discussion, the energyseparation d − π ∗ may be as little as about 2 eV for a molybdenumcomplex (an estimate obtained from calculations that use the extended Hückel method).
It is therefore worthwhile to examine the otherlimiting case (small ε, stabilization proportional to S) to comparethe two conformations. If we reuse the overlaps calculated above, we√find that the stabilization is proportional to 2 2Sdπ ∗ in the coplanarApplicationsconformation, but to 4Sdπ ∗ in the perpendicular conformation. The latter is now clearly favoured. This analysis shows us that we can anticipatea marked conformational preference for the perpendicular structure ifthe d − π ∗ energy separation is not too large.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
