The Ligand Field

Ions of the same d configuration show important similarities independent of the identity of the element. This means that Co(III) is closer in many properties to (f Fe(II) than to (f Co(II). The variable valency of the transition metals leads to many cases of isoconfigurational ions, and this idea helps us predict new complexes from the existence of isoconfigurational analogs. Numerous andogies of this type have been established for the pair Ir(III) and Ru(II), for example. 

FIGURE 1.6 Molecular orbital, or ligand field picture, of M-L bonding in an octahedral ML6 complex. The box contains the d orbitals that are filled with n electrons to give the d electron configuration. The star denotes antibonding. 

The L lone pairs in the free ligand become bonding pairs shared between L and M when the M-L a bonds are formed these are the six lowest orbitals in Fig. 1.6 and are always completely filled with 12. Each M-L a-bonding MO is formed by the combination of the ligand lone pair, L(a), with and has both M and L character, but L((j) 

Ligands are generally nucleophilic because they have high-lying lone pair electrons available, while a metal ion is electrophilic because it has low-lying empty d orbitals available. A nucleophilic ligand, a lone-pair donor, can thus attack an electrophilic metal, a lone pair acceptor, to give a metal complex. Metal ions can accept multiple lone pairs so that the complex formed is ML (n = 2-9). 

Hypervalenqr, the ability of an element to exceed the valence electron count normally appropriate for the orbitals that are available, is best established in the main-group elements, such as sulfur, where an octet of eight valence electrons is appropriate for its single s and three p orbitals. In hypervalent SFe, for example, six electrons come from S and one each from the six F atoms for a total of 12 valence electrons, greatly exceeding the expected octet. The modem theory of hypervalency avoids the earlier idea, now exploded, that empty d orbitals (3d orbitals for S) allow the atom to house the excess electrons. 

FIGURE 1.4 Crystal field splitting patterns for the common coordination geometries tetrahedral and square planar. For the square planar arrangement, the z axis is conventionally taken to be perpendicular to the square plane. 

FIGURE 1.5 The molecular orbital, or ligand field picture of metal ligand bonding 

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The effect of configurational mixing of higher-lying s orbitals into the ligand field d-orbital basis set is also likely to favour elongation rather than contraction. ... 

Two symmetry parameterizations of the angular overlap model of the ligand field. Relation to the crystal field model. C. E. Schaffer, Struct. Bonding (Berlin), 1973,14, 69-110 (33). 

The ligand field photosubstitution reactions of d6 hexacoordinate metal complexes. P. C. Ford, Coord. Chem. Rev., 1982, 44, 61-82 (44). 

The complex Ti(H20)6]3+ absorbs light of wavelength 510. nm. What is the ligand field splitting in the complex in kilojoules per mole (kj-mol ) ... 

What is the value (in kilojoules per mole) of the ligand field splitting ... 

In octahedral complexes, the e -orbitals (dz< and dx2 -yi) lie higher in energy than the t2 -orbitals (dxy, dyz, and dzx). The opposite is true in a tetrahedral complex, for which the ligand field splitting is smaller. 

FIGURE 16.28 Tbe energy levels of the d-orbitals in a tetrahedral complex with the ligand field splitting A,. Each box (that is, orbital) can hold two electrons. The subscript g is not used to label the orbitals in a tetrahedral complex. 

Transitions between d-orbitals or between the ligands and the metal atom in complexes give rise to color the wavelength of d-d transitions can be correlated with the magnitude of the ligand field splitting. 

J 6 Determine the ligand field splitting from the wavelength of light absorbed by a complex (Example 16.3). 

Is there a correlation between the ligand field strength of the halide ions F, Cl, Br, and 1 and the electronegativity of the halogen If so, can this correlation be explained by ligand field theory Justify your answer. 

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