Yves Jean - Molecular Orbitals of Transition Metal Complexes (793957), страница 20
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Forexample, xy is placed in the a2 representationwhereas we indicated that it has b2 symmetry.These differences arise because the axissystem previously used for the octahedron isnot the one conventionally adopted for theC2v group. For example, the C2 axis should bethe z-axis, rather than the y-axis. In otherwords, the orbitals sketched on the right-handside of Figure 2.15 do indeed correspond tothe symmetries indicated, and only theirnames would be changed if one adopted theaxis system appropriate for the C2v group.We have already established the d orbitals of an octahedral complexin the orientation chosen above, with four ligands placed along thebisectors of the x- and y-axes (§ 2.1.2.5, Figure 2.4); they are shown inthe left-hand side of Figure 2.15.Removal of the ligands L5 and L6 has no effect on either the shapeor the energy of the three nonbonding t2g orbitals of the octahedron(zero coefficients on these ligands). These orbitals (a2 (xz), b1 (yz), and1a1 (x 2 −y2 )) therefore remain degenerate, not, as in the octahedron, bysymmetry, but ‘accidentally’ in the lower-symmetry point group C2v .
Inthe eg block, the z2 orbital is only very slightly stabilized by the removalof the two weak antibonding interactions in the xy plane. It thereforestays as a strongly antibonding orbital (3a1 ). In contrast, the xy orbitalis strongly stabilized by the removal of two of the four antibondinginteractions with the ligands located in the xy plane (b2 ).13 This stabilization is enhanced by mixing with the px orbital, which polarizes it ineg3a12a1b21a1t2gFigure 2.15. Derivation of the d-block orbitalsfor a ‘butterfly’ ML4 complex from the dorbitals of an octahedral ML6 complex. Thehybrid s–p orbital (2a1 ) placed between the b2and 3a1 orbitals of the d block is also indicated.b1a2MMOther complexes or MLn fragmentsthe direction opposite to the ligands L1 and L2 , in a way which we havealready seen several times (2-92).+=2-92There are therefore four nonbonding or weakly antibonding d orbitals.
If these orbitals are doubly occupied, the result, taking the four bondsinto account, will be a 16-electron complex. As in the examples that wehave already met, this result is linked to the existence of a nonbondingorbital on the metal which is not of d type. In the ML4 butterfly complex,this is an s–p hybrid orbital of a1 symmetry (2-93) which points along theC2 symmetry axis of the complex, that is, between the two ligands whichwere removed in the original octahedral complex.
Its energy, which isnot very high due to the participation of the s orbital, is lower than thatof the most antibonding orbital in the d block, as shown in Figure 2.15.2-932.8.4. Bent ML2 complexesR 3PNiR 3P2-94ML2 complexes usually adopt a linear (or essentially linear) geometry (see § 2.7). Knowledge of the orbital structure for stronglybent ML2 species is therefore useful mainly when one wishes toconsider them as fragments in larger complexes. For example, the [(η2 ethylene)Ni(PR 3 )2 ] complex (2-94) can be described in terms of a bentML2 entity interacting with a molecule of ethylene.Starting from the butterfly ML4 fragment previously studied, a bentML2 fragment may be obtained by removing the ligands L3 and L4 (2-95).L2L3zxyL2ML1L1ML42-95The d orbitals of this fragment are readily obtained from those ofthe ML4 fragment (Figure 2.16).
The shape and energy of the threenonbonding orbitals (a2 , b1 , and 1a1 ), and of the antibonding b2 orbital,are all unchanged (zero coefficients on L3 and L4 ). However, the twomain antibonding interactions in the z2 orbital are eliminated, so thatPrincipal ligand fields: σ interactions3a12b12a13a1b2b22a1Figure 2.16.
Derivation of the d-block orbitalsfor a bent ML2 complex from the d orbitals ofa ‘butterfly’ ML4 complex. The hybrid s–porbital (2a1 ) and the metal p orbital (lyingabove the d block) are also indicated.b11a11a1b1a2a2MMthis orbital becomes nearly nonbonding (2a1 ). The five orbitals in the dblock are therefore nonbonding or weakly antibonding. Once the fourelectrons associated with the two M–L bonds are taken into account, ad10 electronic configuration leads to a 14-electron complex.This lack of four electrons compared to the 18-electron rule arisesbecause there are two nonbonding orbitals on the metal, which are notof d-type (2-96).
The first is an s–p hybrid orbital (3a1 ), identical to thatfound for the ‘butterfly’ ML4 fragment (Figure 2.16, 2a1 → 3a1 ). Thesecond is the p orbital perpendicular to the plane of the fragment (2b1 ),which is strictly nonbonding. It is higher in energy than the 3a1 orbital,mainly because the s contribution in this latter hybrid orbital lowers itsenergy (εs < εp ). Notice that these two orbitals have the same shapeas the two nonbonding MO in bent AH2 molecules, where A is a maingroup element (Chapter 1, Figure 1.5), and that their energetic order isthe same (ε(a1 ) < ε(b1 )).a1b12.8.5. ML complexes2-96COCuW2-972-98A coordination number of 1 is rare for stable complexes. The examplesthat are known involve a very bulky ligand such as the aryl radical,2, 4, 6−Ph3 C6 H2 , which forms d10 ML complexes with copper and silver(2-97).
The M–L unit can also be considered as a fragment in a complexwith a higher coordination number, such as W–CO in the complex[(η2 -acetylene)3 W(CO)] (2-98).The orbitals of this fragment can be obtained in different ways, forexample, by removing a ligand from a linear ML2 complex (2-99) whoseorbitals have already been derived (§ 2.7.1).The four strictly nonbonding d orbitals of the ML2 complex (πg andδg ) remain unchanged.
The z2 orbital is stabilized by the elimination ofExerciseszyL1L1xMML22-99u22ggFigure 2.17. Derivation of the d-block orbitalsfor an ML complex from the d orbitals of alinear ML2 complex. The nonbonding orbitals(s–p hybrid and pure p), located above the dblock, are also shown (2σ and 2π ).LLLg11an antibonding interaction (Figure 2.17). This stabilization is increasedby mixing with the pz orbital which polarizes z2 in the direction oppositeto the ligand (as happens for the z2 orbital in an ML5 complex (SBP),2-44). Notice that the removal of the inversion centre (D∞h → C∞v )leads to the disappearance of the g and u subscripts in the symmetrylabels (Chapter 6, § 6.2.5.2).The d block of the M–L fragment therefore contains five low-energyorbitals that are all likely to be occupied.
The rare stable complexes doindeed have a d10 electronic configuration (2-97). They are therefore12-electron complexes with three empty nonbonding orbitals on themetal, an s–p hybrid orbital (σ symmetry) and two pure p orbitals of πsymmetry that are higher in energy (2-100).2-100Exercises2.11. Give the dn electronic configuration of [M(H2 O)6 ]2+ octahedralcomplexes from the first transition series (M=Sc, Ti, V, Cr, Mn,Fe, Co, Ni, Cu, Zn).Principal ligand fields: σ interactions2. If these are high-spin complexes, give their ground-state electronicconfiguration (just for the d block).3. How many unpaired electrons are there in each case?2.21.
Give the dn electronic configuration of the complex [Cr(CO)6 ].2. Is this a high-spin or a low-spin complex? Deduce its ground-stateelectronic configuration (just for the d block).3. Reduction of this complex leads to dissociation into [Cr(CO)5 ]−and CO, rather than to the formation of [Cr(CO)6 ]− . Suggest anexplanation.2.3Making use of the results given in Figure 2.3, give the approximateshapes and relative energies of the bonding MO for an octahedralcomplex whose ligands are located as shown in 2-29.2.4Give the shapes and relative energies of the d-block MO for a squareplanar ML4 complex whose ligands are located on the bisectors ofthe x- and y-axes.2.51. How many electrons are there around the metal in the[Rh(PPh3 )3 ]+ complex whose geometry is given in 2-82?2. Link this result to the existence of nonbonding MO, lying above thed block, in a ‘T-shaped’ ML3 complex (see Figure 2.14, right-handside, for this d block).3.
Give the shapes of these MO.2.61. Construct the d-orbital correlation diagram linking a squareplanar (θ = 90◦ ) to a pyramidal ML4 complex (θ = 125.5◦ ,nodal cone angle for the z2 orbital).zθ = 90°zxyzθ = 125°2. What happens to the shape and the energy of the nonbondingmetal orbital that is not a d orbital?ExercisesApical and basal bonds in ML5 complexes (SBP)2.7LaLbLbMzLbLb1. Give the shapes and relative energies of the five d-block orbitalsfor an ML5 complex with an SBP geometry, if the metal is in thebasal plane.In what follows, we shall only consider this geometry, even ifthe metal in some cases is above the basal plane.2.
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