Valence Bond Theory

Metal-ligand bonds are regarded as coordinate covalent bonds in which electron pairs from the donating ligand enter vacant electron orbitals on the central atom.

The example above shows Be (starting with the neutral atom).  Be loses 2 electrons to form the Be²⁺ ion.  The complex ion [Be(H₂O)₄]²⁺ is described by an electronic configuration in which 8 electrons (2 from each water molecule) are donated into the vacant orbitals of Be.  Provided that each ligand donates a pair of electrons, the number of ligands will equal the number of vacant orbitals that will be filled on the central atom.  Hybridization is invoked since the coordinate bonds that are formed are indistiguishable from each other.  In this particular case, sp³ hybridization is suggested.  Following this line of thinking, one may state that the number of hybridized orbitals will be determined by the number of ligands.

---

Transition metals have low-lying vacant d orbitals.  As a result, the availability of these low-lying d orbitals leads to new structures - octahedral, trigonal bipyramid and square planar geometries.

---

Here is the valence bond picture for a transition metal.

First, with transition metals, as the ion is formed, the valence s and p orbitals (4s and 4p, in the case of a first row transition metal such as Co) become higher in energy compared to the valence d orbital (3d in the case of Co).  The six valence electrons of Co³⁺ can be assigned to the 3d orbitals, following Hund's rule.  Co³⁺ can coordinate to six ligands (each ligand donating a pair of electrons) which leads to an octahedral complex.  Two examples are shown above, the fluoro complex is an example of a complex anion while the ammonia (ammine) complex is considered a complex cation.  The above diagram also shows two possibilities for the electronic configuration.  The fluoro complex keeps the Co electrons in the same configuration as the isolated Co³⁺ ion.  The electrons from the fluoro ligands simply occupy the upper vacant orbitals on Co.  With hybridization, this will be a set of d²sp³ hybrids.  The same type of hybridization can be imagined in the hexaammine complex.  In this case, however, the electrons originally with Co³⁺ are forced to pair first and the electrons from the ligands will also occupy the orbitals vacated due to the pairing of the Co electrons.  This happens because ammonia is known to be a "strong" ligand.

From the above example, a drawback of the valence bond theory becomes evident.  One needs to know first if the ligand is strong enough to cause electrons on the metal to pair first.

Crystal Field Theory

With this theory, the bonding between the central metal atom and the ligands is completely ignored.  The ligands are viewed simply as mere point charges.  One focuses on the valence d orbitals of the central transition metal atom and examines how the relative energies of the d orbitals change upon introduction of external negative point charges (the ligands).  Thus, it is necessary to go back and examine the various d orbitals.

Octahedral geometry

In the figure below, an octahedral field created by six point charges is assumed:

If the distribution of the point charges is spherical then all d orbitals will be affected in the same manner.  All of the d orbitals will rise in energy.  In an octahedral arrangement, the point charges approach the metal atom along the Cartesian x, y and z axes.  Thus, the orbitals that lie along these axes (d_z2 and d_x2-y2) should rise in energy by a greater amount compared to the orbitals that do not lie exactly along the x, y and z axes (d_xy d_xz d_yz).  The d orbitals in the presence of an octahedral field can be described by the following energy diagram:

With the above diagram in mind, one can therefore assign electronic configurations for various octahedral complexes:

It is straightforward as long as the number of d electrons is less than or equal to 3 (or greater than or equal to 8).
With 4 to 7 d electrons, two options are possible for octahedral complexes.
An example is Co³⁺ in an octahedral complex:
Hexafluorocobaltate(III), [CoF₆]^3-, is characterized by a smaller crystal field splitting (the energy difference between the two sets of d orbitals).  As a result, electrons choose to occupy the higher d orbitals instead of pairing with another electron (pairing has an energy penalty since electrons repel each other).  On the other hand, hexaamminecobalt(III), [Co(NH₃)₆]³⁺, has a much larger crystal field splitting.  In this case, the d electrons are forced to pair.  The result is a configuration that is diamagnetic.  High spin corresponds to a higher number of unpaired electrons while low spin corresponds to the opposite.

Crystal Field Theory indeed becomes attractive when one has spectroscopy as a tool.  The crystal field splitting is a measurable quantity.  From various experiments, one can establish an empirical scale that ranks ligands according to increasing crystal field splittings:

Cl^- <  F^- < H₂O < NH₃ < ethylenediamine (en) < NO₂^- < CN^-, CO

Unfortunately, the above ranking raises serious questions against the crystal field theory since a neutral ligand, CO, is found to be one of the strongest ligands.

Square planar geometry

This should be similar to the octahedral case.  In fact, one can imagine slowly removing the point charges along the z axis until they are completely gone.  The figure below shows the correlation diagram:
Based on the energies of the d orbitals for a square planar complex, one would predict that this arrangement would be quite popular for d⁸ species.

 

Tetrahedral geometry

Imagine the metal atom inside a cube with its d_z2 and d_x2-y2 orbitals pointing at the center of the cube faces.  In a tetrahedral arrangement, the point charges approach via the four opposite corners of the cube.  The orbitals d_xy, d_xz, and d_yz are pointing towards the center of the cube edges (These are closer to the corner).  Therefore,
The splitting in a tetrahedral field is much smaller than the splitting in an octahedral case.  Thus, tetrahedral complexes, if given an option, always assume a high spin configuration.

Molecular Orbital Theory (Quantum mechanics and symmetry)

The following is a crude and approximate molecular orbital diagram for an octahedral complex:

By the way, the symbols you see beside the complex orbitals are called Mulliken symbols.
They describe the symmetry of the orbital:
Uppercase letter - the number of orbitals, A, B for one, E for two, T for three
A is symmetric with respect to highest rotation axis C_n, B is antisymmetric.
The subscripts 1 and 2 refer to symmetric and antisymmetric, respectively, with respect to a C₂ axis in which the C₂ axis is perpendicular to C_n.
The subscripts u and g are with respect to an inversion center, in which g is symmetric.

Chelation

In some cases, the ligand has two or more atoms capable of donating a pair of electrons for forming a coordinate covalent bond to a metal atom.  If the spatial distance between such atoms is not too close or too far, these atoms may coordinate to the same atom.  The first example is the carbonate ion.  This ion is an example of a case in which the distance between the two potential donors is too short.

The carbonato complex is not stable since a 4-membered ring is highly strained.
5-membered and 6-membered rings are less strained as illustrated by the example below.

The phenomenom of ring formation by a ligand in a complex is called chelation.

 

---

Other examples of chelating agents

Forming 5-membered rings:

---

Forming 6-membered rings:

The above examples are all bidentate ligands.  There are ligands that are capable of
attaching to a metal atom via more than two atoms.

diethylenetriamine is an example of a tridentate ligand

porphyrin (found in heme) is a tetradentate ligand

ethylenediaminetetracetate ion (EDTA) is a hexadentate ligand

---

You should be able to explain the chelation phenomenon
using arguments involving entropy.

Molecular Geometry and Isomerism

Linkage Isomerism - This happens when a particular ligand is capable of coordinating to a metal
in two different and distinct ways.  Here is an example:

---

Coordination - Sphere Isomers - Examples:

[Cr(H₂O)₅Cl] Cl₂^.H₂O
[Cr(H₂O)₆]Cl₃
[Cr(H₂O)₄Cl₂]Cl^.(H₂O)₂

 

---

Stereoisomers (Same chemical bonds but different three dimensional arrangement)

Geometrical isomers
Enantiomers (nonsuperimposable mirror images)

 

---

Molecular geometry and stereoisomerism are intimately related

Square planar - Since this arrangement is planar (there is a mirror plane of symmetry), these complexes do not have enantiomers.  Geometric isomerism, however, is possible. Here is an example.

Tetrahedral - In a tetrahedral arrangement, ligands occupy corners of a tetrahedron.  These corners are all adjacent to each other, therefore, tetrahedral complexes are not capable of geometric isomerism.  Enantiomers, however, are possible, when all four ligands are different.

Octahedral - Both geometric isomers and enantiomers are possible.  Below are some examples.

The above is true only if the like ligands are all cis to one another.  If two like ligands are trans then there will be a mirror plane of symmetry normal to the line connecting the ligands that are trans to each other.

The above is an example of stereoisomerism with chelating ligands.

source:internet