Yves Jean - Molecular Orbitals of Transition Metal Complexes (793957), страница 10
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(1) [Cr(CO)6 ];(2) [W(CO)5 ]; (3) [Mn(CO)5 Cl]; (4) [TiCl4 ]; (5) [Co(CO)3 (Et)];(6) [Re(PR3 )(CO)4 Cl]; (7) [Fe(CO)4 (H)2 ]; (8) [Fe(CO)4 (η2 -H2 )];(9) [ReH9 ]2− ; (10) [ReH5 (PR3 )2 (SiR3 )2 ]; (11) [Ni(CO)4 ];(12) [Cu(SR)3 ]2− ; (13) [Ni(CN)5 ]3− ; (14) [RhI3 (CO)2 (Me)]− ;(15) [RhI3 (CO)(COMe)]− ; (16) [MoF4 (O)2 ]2− ; (17) [Re(NR3 )4 (O)2 ]+ ;(18) [Mn(η6 -C6 H6 )(CO)3 ]+ ;(19) [Zr(η5 -C5 H5 )2 (H)(Cl)];5(20) [Nb(η -C5 H5 )2 (Me)3 ]; (21) [Os(η5 -C5 H5 )(CO)2 Cl]; (22) [WCl6 ];(23) [Fe(CO)4 (η2 -C2 H4 )]; (24) [Zn(η5 -C5 H5 )(Me)].1.3Wilkinson’s catalyst [RhL3 Cl], where L = PPh3 , is used for thehydrogenation of alkenes.
The reaction proceeds as follows:H2C 2 H4[RhL3 Cl] −→[RhL3 Cl(H)2 ] −→ [RhL2 Cl(H2 )] + L −→ [RhL2 Cl(H)2 (η2 -C2 H4 )] + L↓[RhL3 Cl] + CH3 -CH3 ←− [RhL3 Cl(H)(CH2 CH3 )] ←− [RhL2 Cl(H)(CH2 CH3 )] + LFor each step, give the oxidation state of the metal (no), the electronic configuration of the complex (dn ), and the total number ofelectrons (Nt ).1.41. Give the charge on the ligands CO, Cl, Et, PR3 , H, H2 , SiR3 , SR,CN, I, Me, COMe, F, O, NR3 , C2 H4 , C6 H6 , C5 H5 if the ionicmodel is adopted (§ 1.2).Setting the scene2.
Formulate the complexes given in § 1.1.2.2 if this model is adopted,and hence deduce the oxidation state of the metal.1.51. Give the oxidation state of the metal no, the electronicconfiguration dn , and the total number of electrons Nt for thecomplexes [Cr(η6 -C6 H6 )2 ] and [Ru(η6 -C6 H6 )(η4 -C6 H6 )].2. How can one explain the change in the coordination mode η6 →η4 of one of the ligands when chromium in the complex is replacedby ruthenium?3.
Using the same argument, predict the hapticity x fora cyclopentadienyl ligand in the following complexes:(i) [Mn(CO)3 (ηx -C5 H5 )]; (ii) [W(CO)2 (η5 -C5 H5 )(ηx -C5 H5 )];(iii) [Fe(CO)2 (η5 -C5 H5 )(ηx -C5 H5 )].1.6The borohydride ligand (BH−4 in the ionic model) can bind in theη1 , η2 , or η3 modes, depending on whether one, two, or three B-Hbonds interact with the metal centre.MHHHBH Hh1H HBMHHBMHHh2h3HRationalize the coordination mode in the following complexes:[Cu(PR3 )3 (η1 -BH4 )], [Cu(PR3 )2 (η2 -BH4 )], and [Ti(CO)4 (η3 -BH4 )]− .1.7What is the oxidation state of the metal centres inthe following binuclear complexes: (1) [Re(CO)5 ]2 ; (2)3[ReCl4 (H2 O)]2−2 ; (3) [MoCl2 (PR3 )2 ]2 ; (4) [Pd(η -C3 H5 )(µ-Cl)]2 ; (5)5[Mo(η -C5 H5 )(CO)2 (µ-SR)]2 .1.8Consider the interaction between a metal atom and a ligand with ans orbital.Exercises1.
Show that there can always be an interaction between the ligandorbital and the s orbital of the metal. Sketch the shapes of theresulting bonding and antibonding MO.2. What position must the ligand adopt to give (i) maximal and(ii) minimal overlap with the px orbital of the metal? What is itsvalue in the latter case?3. Repeat question 2 for the xy and z2 metal orbitals.This page intentionally left blank Principal ligand fields: σinteractions1Those ligands for which it is necessary toconsider π -type interactions with the metalare studied in Chapter 3.In this chapter, we shall construct the molecular orbitals (MO) of monometallic complexes MLℓ and thereby deduce their electronic structureby distributing the electrons in these orbitals.
We shall study differenttypes of ‘ligand field’, each one being characterized by the number ofligands and by their geometrical arrangement around the metal centre(octahedral complexes ML6 , tetrahedral, or square-planar ML4 , etc.).We shall always begin the analysis by establishing the molecularorbitals of the associated model complex, in which all the ligands (i) areidentical and (ii) only have σ -type interactions with the metal (Chapter 1,§ 1.5.1). The resulting orbital scheme is characteristic of the ligand fieldbeing studied (octahedral, tetrahedral, square-planar, etc.).
In a firstapproximation, it is applicable to all complexes of this type, even if thetwo conditions specified above are not met exactly. The main reason forthis is that σ interactions exist in all complexes, and they are strongerthan π interactions when the latter are present. Therefore, if π-typeinteractions are added in a ‘real’ complex to σ interactions, while theresults are different from those obtained for the model complex, theyare not completely transformed. The π effects will therefore be treatedsubsequently as perturbations to be added to the orbital scheme established for the model complex.1 In the same way, different σ interactionsassociated with non-equivalent ligands produce changes compared tothe model complex that can initially be neglected.There are several ways of representing the ligand orbital that isinvolved in the σ interaction with the metal: as an s orbital (2-1a), as a porbital (2-1b), or as a hybrid (s–p) orbital directed towards the metalliccentre (2-1c).
We shall use this last representation, since, except forhydride ligands which possess only a single valence orbital 1s and veryelectronegative monoatomic ligands such as F in which the contributionof the p orbital dominates, it is the most appropriate for all other ligands.The orbital on ligand Li will be written σi , due to the nature of the bondwhich it can form with the metal.MMM2-1a2-1b2-1cPrincipal ligand fields: σ interactionsTo construct the molecular orbitals of a complex, we have todetermine their shape, by which we mean the contributions of themetal and ligand orbitals to each one, as well as their relative energies.The ‘fragment method’ is widely used for the construction of the MO ofa complex MLℓ ; we allow the atomic orbitals (AO) of the central metalto interact with those on the ligands.
Knowledge of the symmetry properties of these orbitals allows us to simplify the construction of thesediagrams very considerably. If the two orbitals have the same symmetryproperties, their overlap is non-zero and an interaction can occur. Onthe other hand, two orbitals of different symmetry have zero overlap(‘by symmetry’) and do not interact at all (Chapter 1, § 1.3.3). In thiscontext, we shall sometimes use basic ideas from group theory, as wellas several results established in Chapter 6 which the reader may consultwhen necessary.2.1. Octahedral ML6 complexesConsider a complex in which the metal centre is surrounded by sixidentical ligands (L1 –L6 ) placed at the vertices of an octahedron (2-2).The metal is placed at the origin, and the ligands on the axes x (L2 , L4 ),y (L1 , L3 ), and z (L5 , L6 ).
Each ligand Li possesses an orbital σi directedtowards the metal (2-3).6L6L1L24zL4M L3xy1M325L52-22-32.1.1. Initial analysis of the metal–ligand orbitalinteractions1s2-4 (S = 0)Our first task is to examine how each of the orbitals on the ligands caninteract with the s, p, and d orbitals on the metal centre.As the s orbital has spherical symmetry, its overlap with any one ofthe σi orbitals is non-zero (σ1 in 2-4).
And as the six ligands are equivalent, this overlap is the same no matter which orbital σi is considered,so the interactions between s orbital and each of the ligand orbitals areidentical.We therefore find a bonding MO between the metal and the ligands (2-5) and a corresponding antibonding MO (2-6), where eachOctahedral ML6 complexes2-5(bonding)2-6(antibonding)63py2-7 (S = 0)2-9(bonding)2-11(bonding)py2-8 (S = 0)2-10(antibonding)2-12(antibonding)σi contribution is equal in both orbitals.
Since the ligands are moreelectronegative than the metal, the bonding orbital is mainly concentrated on the ligands but the antibonding orbital on the metal (Chapter 1,§ 1.6.1).We now turn to the p-type orbitals on the central atom. The pyorbital overlaps with the orbitals on the ligands L1 and L3 located onthe y-axis (σ3 in 2-7). But the four other ligands L2 , L4 –L6 , are placedin the nodal plane of py (xz), which is a symmetry element of the complex. The σi orbitals associated with these ligands are symmetric withrespect to this plane, whereas the py orbital is antisymmetric: the overlaps between py and each of σ2 , σ4 –σ6 are therefore zero ‘by symmetry’.In the example shown in 2-8, it is clear that the σ6 orbital has a positiveoverlap with the grey lobe of the py orbital but an equivalent negativeoverlap with the white lobe, so the total overlap is equal to zero.
Therefore, there cannot be any interaction between the py orbital and any ofthese four ligand orbitals.The py orbital on the metal therefore combines with the σ1 and σ3orbitals to form a bonding MO, mainly based on the ligands (2-9), andan antibonding MO mainly based on the metal (2-10).This analysis is readily extended to the px and pz orbitals, which cancombine only with σ2 and σ4 for px (2-11 and 2-12) or σ5 and σ6 for pz(2-13 and 2-14). The interactions involving the p orbitals of the metaltherefore lead to the formation of three bonding and three antibondingMO.
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