VBT, CFT & LFT Theories - 1 | Chemistry Optional Notes for UPSC PDF Download

Bonding in Coordination Compounds

• This chapter is devoted to bonding theories for coordination compounds. Let us first think about, what a good theory should be able to do in general. The answer is, that it should be able to make many correct explanations for experimental observations based on a few, sensible, assumptions. In addition, it should be able to predict experimental observations. The more the theory can explain and predict, and the fewer the necessary assumptions, the better the theory. What does this mean for a bonding theory? What would a good bonding theory for coordination compounds be able to do? It should certainly be able to explain and predict the number of bonds and the shape of a molecule. 
• In addition, it should be able to explain the magnetism of molecules, in particular dia- and paramagnetism. Remember, a molecule is diamagnetic when it has no unpaired electrons. It is paramagnetic when there are unpaired electrons. A diamagnetic molecule is repelled by an external magnetic field. A paramagnetic molecule is attracted by an external magnetic field. It should further be able to explain the stability and reactivity of complexes, as well as the optical properties of complexes. Optical properties of compounds are linked to bonding because they are related to electronic states.

Valence Bond Theory

• There are essentially three bonding concepts that are used to describe the bonding in coordination compounds. The first one is the valence bond theory. The valence bond concept was introduced by Linus Pauling in 1931 to explain covalent bonding in molecules of main group elements.
  
• The basic idea is to overlap half-filled valence orbitals to form covalent bonds in which the two electrons are shared between the bonding partners (Fig. 7.1.1). These orbitals can either be atomic orbitals, or hybridized atomic orbitals. The concept works very well to explain the shapes of molecules of main group elements. The valence bond concept in its original form assumes that each bonding partner contributes one electron to the covalent bond. This is not consistent with the dative bonding in coordination compounds where it is assumed that one partner donates an electron pair and the other partner accepts it. 
• To adapt valence bond theory to suit coordination compounds, Pauling suggested that a dative bond is formed via the overlap of a full valence orbital of the donor and an empty valence orbital of the acceptor. We will see that this concept can explain the shapes of coordination compounds in some cases, but overall it does not work very well. We will also see that valence bond theory can explain magnetism in some cases, but also here the valence bond theory has significant deficits. By its nature, valence bond theory cannot explain optical properties. Overall, valence bond theory is far more suitable for main group element molecules compared to transition metal complexes.

Crystal Field Theory

• The second major theory is the crystal field theory. It is actually not a bonding theory because it is based on repulsive electrostatic interactions. It was originally developed to explain color in ionic crystals. Later, it was found that it can also explain colors in molecular coordination compounds, and is suitable to explain shapes and magnetism of complexes. 
• However, because it is based on repulsive electrostatic interactions it cannot actually explain what holds the atoms in a molecule together. However, the crystal field theory is quite simple and convenient to use, and there is a lot of practicality to it.

Ligand Field Theory

• The third theory is the ligand field theory. It is the most powerful theory, but also the most complicated one. Basically, it is molecular orbital theory applied to coordination compounds. 
• It can make detailed statements about the number of bonds and shapes of molecules, and can explain the magnetism and optical properties of coordination compounds.

Valence Bond Theory for Coordination Compounds

Octahedral Complexes

• Let us have a closer look at the valence bond theory, and assess valence bond theory for complexes by a number of examples.
  
• The first example is the hexaammine chromium (3+) cation (Fig. 7.1.2). From experiment we know that it has an octahedral shape, with six dative Cr-N bonds. Can valence bond theory explain the six bonds and the octahedral shape satisfactorily? In order to explain the six dative Cr-N bonds we would need to overlap six empty chromium valence orbitals with six filled valence orbitals of N. We can see that the six ammine ligands have one electron lone pair each that can serve as the valence orbitals. Does chromium have six empty valence orbitals? In order to assess this, we first need to know the oxidation state of the chromium. It is +3 because the ligands are all neutral when the bonds are cleaved heteroleptically, and the complex cation has a 3+ charge. Therefore, the chromium is a Cr³⁺ cation.
• Next, we need to know the electron configuration of the Cr³⁺. A neutral Cr atom has the electron configuration 4s¹3d⁵. When a transition metal loses electrons to form a cation, it always loses its two valence electrons first, and then its d electrons. For chromium this means that we must remove the one 4s electron, and two of the five 3d-electrons. The three remaining 3d electrons are expected to be spin up in three different d orbitals according to Hund’s rule. How many empty valence orbitals remain? These would be two 3d and the 4s orbitals. 
• In addition, it would also be justified to consider the three 4p orbitals as valence orbitals because the 4p orbitals are energetically only slightly higher than the 4s orbital. That means that we would have the six valence orbitals that we would need to explain the six bonds. There is, however, a complication. The six bonds in the complex are not distinguishable, but the six valance orbitals in the Cr³⁺ ion are distinguishable, for example, the 3d orbitals have different shape and energy than the 4s orbital, which is different from the 4p orbitals. Therefore, if we overlapped these orbitals with the electron lone pairs at N, the bonds would not be equivalent, or indistinguishable. We would have difficulty to explain the highly symmetric octahedral shape of the molecule. To go around this issue, valence bond theory uses the concept of hybridization. 
• In this concept we mathematically mix the wave functions of the valence orbitals to form hybridized orbitals. In our example we would mix the two empty d-orbitals, the 4s orbital, and the three 4p orbitals to form six so-called d²sp³ hybridized orbitals. They have the same shape and size, and their lobes point toward the corners of an octahedron. Therefore, we can now create overlap between these six orbitals, and the six electron lone pairs at N to form six equivalent, indistinguishable Cr-N bonds. We conclude that we have now satisfactorily explained the bonding and the shape of the complex.
• Can we also explain its magnetism? From experiment we know that the complex is paramagnetic, and that there are three unpaired electrons. Does valence bond theory predict the same? Yes, it does. There are three unpaired electrons in the three half-filled d-orbitals.

Tetrahedral Complexes

• Our next is example is a tetrahedral complex, the tetrahydroxo zincate (2-) complex anion, Fig. 7.1.3.
  
• When viewing it as a Lewis-acid base complex with dative bonds it can be thought as an adduct of a Zn2+ and four hydroxide anions. One of the three electron lone pairs at the hydroxide ions would donate its electrons into empty Zn valence orbitals. That means we would need overall four empty Zn valence orbitals to explain the four Zn-O bonds. A neutral zinc atom has the electron configuration 4s² 3d¹⁰. We can derive this from the fact that zinc is in group 12 of period 4 in the periodic table. A Zn²⁺ ion has two electrons less. Because we must remove s electrons before we remove d electrons, the Zn²⁺ has the electron configuration 3d¹⁰. Like in the previous example we can justifiably consider the 4p orbitals as additional valence orbitals. 
• We can see that we have four empty orbitals available to make the four bonds, namely the 4s and the 4p orbitals, but these orbitals are not equivalent, and do not have the correct orientation to explain the tetrahedral shape of the complex. There is a 90° angle between the p-orbitals which is smaller than the 109.5° tetrahedral bond angle in the molecule. However, we can solve this problem by hybridizing the 4s and the three 4p orbitals to form four sp³-hybridized orbitals. These hybrid orbitals have the property that their lobes point toward the corners of a tetrahedron. Thus, they are suitable to explain the tetrahedral shape of the molecule. We can place the ligands around the Zn²⁺ ion and approach the ligands on the bond axes to create orbital overlap between the empty sp³-hybridized orbitals and one electron lone pair at the oxygen atom. This produces the tetrahedral tetrahydroxo zincate (2-) anion.
• Can we also explain the magnetism of the molecule? What magnetism would valence bond theory predict? We can see that there are no unpaired electrons in any of the metal valence orbitals. Thus, the complex should be diamagnetic. This is also what we find experimentally. Thus, valence bond theory is able to explain the magnetism of this complex anion.

Square Planar Complex

• Now let us see if the valence bond theory can also explain a square planar complex such as tetracyanonickelate (2-).
  
• In the valence bond picture we view the Ni-CN bonds as dative bonds, and the complex is considered an adduct of Ni2+ and CN-. To explain the four bonds, the Ni2+ ion would need to have four empty valence orbitals. Ni is a group 10 metal and a neutral Ni atom has the electron configuration 4s23d8. To create a Ni2+ ion we must remove the two 4s electrons, and thus the Ni2+ has the electron configuration 3d8. Do we have four empty orbitals available? Yes, the 4s and the three 4p orbitals are empty but again they are not equivalent and thus not suitable to explain four equivalent Ni-C bonds. Can we hybridize these orbitals? Yes, we can, but the resulting four sp3 hybridized orbitals would not be suitable to explain the square planar shape, only the tetrahedral shape. 
• What valence bond theory suggests in this case is to reverse the spin of one of the unpaired d electrons and move it into the other half-filled d-orbital. This produces an empty d-orbital that we can now hybridize with the 4s and two of the 2p orbitals to four dsp2-hybridized orbitals. These four orbitals have the property that their lobes point toward the vertices of a square, thus they are suitable to explain the square-planar shape. We can approach the ligands now on the bond axes to create orbital overlap between the empty dsp2 Ni and the electron lone pairs of the ligands. We can also say that the ligands donate their electron lone pairs into the hybridized metal orbitals. This produces the four covalent bonds that we need and yields a molecule of a square planar shape.
• We can see that the valence bond theory can still explain the square planar shape, but only with the help of the additional assumption that one of the d-electrons gets spin-reversed and moves into another d-orbital. An assumption a theory makes should always be reasonable, so let us critique how reasonable this assumption is. Firstly, is the spin-reversal reasonable? Spin-reversal is a quantum-mechanically forbidden process, and thus it is questionable to assume that it happens. Secondly, there is no good explanation for why the electron moves. The energy of two spin-paired electrons in the same orbital is actually higher than that of two spin-paired electrons in different orbitals. So overall, we see that valence bond theory has difficulties to explain the square planar shape. It must make assumptions that are not very plausible.

Octahedral d⁵ High and Low Spin Orbital Complex

• The valence bond theory has also difficulties to explain so-called high spin and low spin octahedral complexes. For example, it is known from magnetic measurements for 3d⁵ transition metal ions that they can make octahedral complexes with either one unpaired electron or five unpaired electrons, depending on the ligand. In the first case, the number of paired electrons in the d-orbitals is maximized, and we have a low-spin complex, in the other case the number of unpaired electrons is maximized, and we have a high spin complex. What approach does valence bond theory take to explain this phenomenon?
  
• In the case of a low spin-complex, valence bond theory assumes a so-called inner orbital complex. Like in the square planar complex it is assumed that unpaired electrons reverse their spins and move into other half-filled d-orbitals so that spin-pairing is maximized. In the case of a d5 ion, two electrons reverse their spin, and move into two other half-filled orbitals. This leaves one unpaired electron. We see that due to the movement of the two electrons two 3d-orbitals are empty now, and so are the 4s and the 4p orbitals. The six empty orbitals can now be combined to form d²sp³-hybridized orbitals that can explain the octahedral shapes. Approaching the ligands overlaps the electron lone pair at the ligand with the empty hybrid orbitals to form a dative, covalent bond. We can also say the ligands donate electron lone pairs to form six covalent bonds. We can again criticize that spin-reversal is forbidden and spin-pairing is energetically unfavorable making the approach valence bond theory takes to explain the low-spin complex unsatisfactory.
• What about the 3d5 high spin complex (Fig. 7.1.6)? In this case we cannot pair spins to create empty d-orbitals because we need to explain five unpaired electrons. Now, valence bond theory makes another new assumption. It assumes that the outer 4d orbitals get involved in the bonding. These orbitals are empty and available for hybridization. We can therefore hybridize two 4d, the 4s, and the three 4p orbitals to form d²sp³ hybridized orbitals. In the last step we can approach the ligands, and the ligands can donate their electron lone pairs into the transition metal d-orbitals. Now we have explained the six bonds, the octahedral shape, and the five unpaired electrons.
• We can again critique the valence bond approach. What justification is there to assume that the 4d orbitals are involved. The answer is: Very little. These orbitals are just too high in energy to be considered valence orbitals. It is not reasonable to assume that they are involved in the bonding. Therefore, again, we see that valence bond theory has difficulties to explain the properties of a complex. Valence bond theory also does not explain distortions of octahedral complexes due to the Jahn-Teller effect.

Octahedral d⁷ High and Low Spin Orbital Complex

• High-spin and low-spin complexes are not only observed for octahedral complexes of d5-ions, but for example also for octahedral d7 ion complexes. A low-spin complex has three unpaired electrons and a high-spin complex has one unpaired electron. We will see that valence bond theory has even greater troubles to explain these compounds.
  
• In a d⁷ ion there are four paired and three unpaired electrons according to Hund’s rule (Fig. 7.1.7). We can reverse the spin of one unpaired electron and pair it with an unpaired electron in another half-filled orbital to reduce the number of unpaired electrons to one. However, this gives us only one empty 3d orbital available for d²sp³-hybridization. In this case we cannot produce more empty 3d-orbitals by reversing the spin. Therefore, we must make again the questionable assumption that outer orbitals are involved in the bonding such as the 4d orbitals. Valence bond theory now suggests to move the unpaired electron from the 3d to the 4d orbital. This is simply done to create another empty 3d orbital that we need for d²sp³-hybridization. However, why would the 3d electron just go into another orbital of much higher energy? If we make this questionable assumption though, we have indeed six orbitals available for hybridization, and we can let the ligands donate an electron pair into the empty hybrid orbitals.
  
• Finally, let us discuss an octahedral d7 high-spin outer orbital complex (Fig. 7.1.8). In this case we cannot pair any spins in the 3d orbitals. Therefore we assume again that the 4d orbitals get involved in the bonding, and hybridize two of them with the 4s and the three 4p orbitals. The six ligands can then donate six electron pairs into the orbitals thereby creating six bonds and explaining the octahedral shape.
• Overall, we see that in the valence bond theory we move around electrons as we please in order to explain shapes and magnetism of complexes without good justification. Therefore, the valence bond theory, while extremely valuable for main group compounds, is only of limited use for transition metal complexes.

Crystal Field Theory

• Now let us discuss the second bonding theory for coordination compounds, the crystal field theory. It is actually not a bonding theory because it is based on repulsive electrostatic interactions. Nonetheless, it has many features of a bonding theory in the sense that it can explain many phenomena that a bonding can explain, in particular molecular shape, magnetism, and optical properties.
• What are the principles of crystal field theory? Crystal field theory assumes that the electrons in the metal d-orbitals are surrounded by an electric field which is caused by the ligand electrons. This electric field is called the crystal field. The name crystal field comes from the fact that this principle was first applied to transition metal ions surrounded by anions in crystals, and was only later extended to transition metal ions surrounded by ligands in molecular coordination compounds. The assumption that ligands surrounding a transition metal ion produce an electric field makes sense because the ligands contain electrons that are associated with an electric field. It is further assumed that the crystal field raises the energy of the metal-d-orbitals because of electrostatic repulsion between the ligand electrons and the metal electrons.
  
• Let us first assume the hypothetical case that the ligand electrons surround the metal d-orbitals exactly spherically (Fig. 7.1.9, bottom left). In this case the electric field is completely isotropic, and this means that the energy of all five metal d-orbitals increases to the same extent. Now let us consider the practical case that the ligands surround the metal in octahedrally, which is the case in an octahedral complex. We can say we have an octahedral crystal field. The electric field will now not be spherical any more, it will be the strongest where the ligands are, namely on the vertices of the octahedron, and less strong elsewhere. The vertices of the octahedron lie on the x, y, and z axes of the coordinate system. Thus, the crystal field is the strongest on the axes, and less strong elsewhere. What consequence does this have on the metal d-orbital energies relative to the spherical crystal field? Orbitals that have their electron density mostly on the axes will experience a greater electrostatic repulsion from the crystal field, and therefore will be higher in energy. Orbitals that will have their electron density mostly elsewhere, meaning not on the axes, will experience smaller repulsion, and thus the energy will be smaller compared to the spherical crystal field. Which are the orbitals that have their electron density mostly on the axes? These are the d_z² and the d_x²-_y² orbitals. The d_z² orbital has its energy density mostly on the z-axis, the d_x²- _y²– orbital has its energy mostly on the x and the y axes. The energy of both orbitals is increased by exactly the same amount. This is not obvious given that the orbitals have very different shapes. To understand this, it helps to remember that when orbitals are symmetrically degenerate, they also must be energetically degenerate.
• An octahedral complex belongs to the point group O_h and in the point group O_h the d_z² and the d_x²-_y²-orbitals are degenerate and have eg symmetry. Therefore, the d_x²-_y² and the d_z²-orbitals in an octahedral crystal field are also often just called the eg-orbitals. The remaining d-orbitals, the dxy, the d_yz, and the d_xz orbitals have their electron density mostly in between the axes, therefore their energy is lower compared to the spherical crystal field. The energy of all three orbitals is reduced by exactly the same amount. We can again understand this when considering that these orbitals are triple-degenerate in the point group Oh and have the symmetry type t2g. For that reason the d_xy, the d_yz, and the d_xz in an octahedral crystal field are also often called the t_2g-orbitals. The energy difference between the t_2g and the eg orbitals is called Δ_o. The energy of the t_2g orbitals is decreased by 2/5 Δ_o relative to the spherical crystal field, and the energy of the eg-orbitals is increased by 3/5 Δ_o relative to the spherical crystal field. Where do the factors 2/5 and 3/5 come from? They are due to the law of the conservation of energy. The overall energy reduction due to the energy decrease of the t_2g-orbitals must be equal to the overall energy increase due to the energy increase of the eg orbitals: ΣE(t_2g)=-Σ(E(eg). Because there are three t_2g orbitals but only two eg orbitals this equation only holds when the energy of the eg orbitals is increased by 3/5 Δ_o and the energy of the t_2g orbitals is decreased by 2/5Δ_o, or 3 x 2/5 Δ_o = 2 x 3/5 Δ_o. Note that the energy of all orbitals will be greater in comparison to the case of no electrical field existing, but the energy is increased to a greater extent for the eg-orbitals compared to the t_2g orbitals.

The Tetrahedral Crystal Field

• What about a tetrahedral complex with a tetrahedral crystal field?
  
• In this case, the ligands do not approach on the axes (Fig. 7.1.10). We can understand this when we consider that we can inscribe a tetrahedron in a cube. If we connect every other corner of a cube then we obtain a tetrahedron. We can define the coordinate system so that the three axes go through the centers of the square faces of the cube. We can see that the axes do not point toward the ligands, thus the ligands do not approach on the axes. Therefore, the orbitals that have their electron density mostly on the axes have a decreased energy relative to the spherical crystal field. These are the d_z² and the d_x²-_y² orbitals. Their energy is the same despite the fact that they have quite different shape. We can explain this again with symmetry arguments. In the point group Td the d_x²-_y² and the d_z² – orbitals are double-degenerate and have the symmetry type e. Because they are symmetrically degenerate, they are also energetically degenerate. The energy of the d_xy, the d_yz, and the d_xz orbitals have most of their energy density in between the axes. Therefore, their energies are increased relative to the spherical crystal field. 
• They are increased by the same amount because the orbitals are triply degenerated in the point group Td and have the symmetry type t₂. The energy difference between the e and the t₂ orbitals is called Δ_t. The energy of the t₂ orbitals is decreased by 2/5 Δ_t. The energy of the e orbitals is increased by 3/5 Δ_t. This is again the due to the law of the conservation of energy. The total amount of decreased energy must equal the total amount of increased energy. The tetrahedral crystal field energy is smaller than that of the octahedral field because the octahedral field interacts more strongly with the d-orbitals compared to the tetrahedral field. One can calculate that it is actually just 4/9 of the octahedral field. This is mainly because the lobes of the t₂ orbitals to not point exactly toward the vertices of the tetrahedron, while the lobes of the e_g orbitals do point exactly toward the vertices of the octahedron.

Tetragonally Distorted Octahedral and Square Planar Crystal Fields

• One nice feature of crystal field theory is that can can readily explain distortions such as the tetragonal distortion of octahedral complexes (Fig. 7.1.11).
  
• In a tetragonally distorted complex there is a tetragonally distorted crystal field. In an elongated octahedron two ligands are further away from the metal than the four others. Let us assume these two ligands are on the z-axis. Then, the crystal field is weaker on the z-axis. To keep the overall crystal field constant we must bring the other four ligands closer to the metal center. That means that we compress along the x and the y-axis. What will happen to the energy of the orbitals as the octahedron distorts? Because we elongate in the z-direction, the d_z² orbital, that has most electron density on the z-axis goes down in energy. Because we compress on the x and the y-axis, the energy of the d_x²-_y² orbital increases. What about the t₂^g orbitals? The dxy orbital goes up in energy because it has electron density in the xy plane, but not along the z-axis. 
• The d_xz and d_yz decrease in energy because they have a significant electron density in z-direction, and the electron density in z-direction is the same for both orbitals. We can now also think about, if the crystal field theory can explain why tetragonal distortion occurs preferentially for certain electron configurations of the metal. For example, it is known that metal ions with d₉ electron configuration often make octahedral complexes with tetragonal distortions. For instance, the hexaaqua copper (2+) complex is an example of a tetragonally distorted complex with a Cu²⁺ ion that has a d⁹ electron configuration. We can understand that the tetragonal distortion occurs when comparing the energy of the electrons in the undistorted vs the distorted octahedron. In the undistorted octahedron we have three electrons in the degenerated eg orbitals. As we distort, we can move two electrons in the energetically lower d_z²-orbital and fill the third one into the energetically higher d_x²-_y²-orbital. Thus, overall the electrons have a lower energy explaining the distortion. This is an example of the Jahn-Teller effect. In general, the Jahn-Teller effect can occur when there are partially occupied degenerate orbitals. In this case a molecule can lower its energy through distortion. 
• Note that not only partial occupation of the eg-orbitals, but also partial occupation of the t₂^g-orbitals can cause the Jahn-Teller effect, although the effect is typically smaller. For example in complexes with metal ions the d4-electron configuration, all four electrons can be stabilized through Jahn-Teller distortion. It should also be noted that in addition to an elongation, there is also the possibility of compressing the octahedron along the z-axis. In this case the order of energy of the d_z² and the d_x²-_y² orbital reverses, and the order of energy of the dxy, as well as the dxz and dyz reverses as well.
• Finally, let us look at the square planar crystal field in square planar complexes. To understand the square planar crystal field it helps to understand the square planar shape as an infinitely elongated octahedron. If we move the two ligands along the z-axis infinitely far away from the metal ion, then we have created a square planar structure. How will the orbital energies change compared to an elongated octahedron? The d_x²-_y² orbital will have an even higher energy due to the necessity to compensate for the decreased field associated with the ligands on the z-axis by further compressing along the x and the y-axis. The d_z²-orbital is even further decreased in energy because the ligands along the z-axis are now completely gone. The dxy orbital is increased in energy because of the enhanced field in the xy-plane. It is now higher than the d_z²-orbital. The dxz and the dyz orbitals are further decreased in energy because they have significant electron density in z-direction.
• Viewing the square planar shape as an extreme case of a tetragonally elongated octahedron also lets us understand why the square planar shape is so often adopted by d8-metal complexes. We can see that the stabilization energy, and thus the tendency to distort is the greatest when two electrons are in the metal eg orbitals. In this case two electrons lower their energy through distortion and no electron has an increased energy. Thus, we would understand that the distortion becomes so great, so that the octahedral complex eventually loses two ligands and adopts the square planar shape. This is a nice example how crystal field theory can explain shapes and the number of bonds in a complex without actually being a bonding theory.

High Spin and Low Spin Complexes

• One of the greatest strength of crystal field theory is that it can explain high-spin and low spin octahedral complexes in a simple way. The basis for that is the assumption that different ligands produce crystal fields of different strengths and that the differently strong crystal fields produce different Δ_o values. This assumption is plausible because it can be expected that different ligands interact differently with a metal ion, for example, the bond length or the bond strength may be different. If the ligands produce a large crystal field then we would expect a large Δ_o, when the crystal field is small, then we would expect a small Δ_o. The size of the Δ_o determines if we get a high spin or a low spin complex. If the Δ_o is larger than the spin pairing energy, then a low spin complex is favored, if Δ_o is smaller than the spin-pairing energy, then the high spin complex is favored.
• For example in a d⁴-metal complex with small Δ_o all electrons are unpaired (Fig. 7.1.12, left). Three of them are in the t_2g-orbitals, and the fourth one is in the eg-orbital. Now let us assume a different ligand that can produce a Δ_o that is large enough to overcome the spin pairing energy. In this case the fourth electron would pair the spin of one of the three electrons in the t_2g-orbitals (Fig. 7.1.12, right). We would obtain a low spin d⁴-complex.
• In the case of a d⁵-metal high spin complex there are five unpaired electrons, and all orbitals are half-filled. If the the ligand produces a crystal field large enough, the spin pairing energy is overcome and there are two paired and one unpaired electron in the t_2g orbitals (Fig. 7.1.13).
  
• For a d⁶-metal high spin complex with a weak crystal field there are two unpaired electrons in the t_2g and eg orbitals respectively. In the case of a strong field ligand and a low spin complex all electrons are in the t²_g orbitals and all spins are paired (Fig. 7.1.14).
• In the case of a d⁷-high spin complex there are two electron pairs and one unpaired electron in the t_2g orbitals, and there are two unpaired electrons in the eg orbitals. For the low-spin complex the spin pairing energy is overcome. However, one electron must remain unpaired in the eg-orbitals because the t_2g-orbitals are fully occupied with six electrons. The number of unpaired electrons in high and low spin complexes predicted by crystal field theory is what is experimentally observed. Therefore we can state that crystal field theory can quite elegantly explain high and low spin complexes.
• We can also understand why there are no d¹, d², d³, d⁸, d⁹, and d¹⁰ low and high spin complexes. In the electron configurations d¹-d³ all electrons are unpaired in the t_2g orbitals. Therefore, there is no electron that could be moved from an eg to a t_2g orbital. For the configurations d⁸-d¹⁰ all t_2g orbitals are necessarily full. Therefore, there is no possibility to move an electron from an eg to a t_2g orbital regardless the crystal field strength.
• Further, we can explain why there are no low spin tetrahedral complexes. We have learned previously that a tetrahedral crystal field has only 4/9 of the strength of an octahedral field. Because it is so much weaker, no ligand is able to produce a field strong enough to overcome the spin pairing energy. High spin and low spin complexes are possible though for square planar complexes.

Strong and Weak Field Ligands: The Spectrochemical Series

• Crystal field theory cannot only explain magnetism well, it can also make statements about the optical properties of a coordination compound. A complex can absorb light when an electron is excited from a d orbital of lower energy to a d orbital of higher energy. The larger the Δ, the smaller the wavelength of the absorbed light. In the case of an octahedral complex, absorption of light would excite an electron from a t_2g to an e_g orbital.
  
• For example, an octahedral [Cr(H₂O)]₆²⁺ complex has a smaller Δ_o compared to an octahedral [Cr(CN)₆]^4- complex (Fig. 7.1.16), and thus the aqua complex absorbs light of longer wavelength compared to the cyano complex. Vice versa, measuring the absorption spectrum of the complexes, allows us to make statements about the relative crystal field strength of the ligands. By measuring the absorption spectrum of many complexes with a variety of ligands we can develop a so-called spectrochemical series that orders ligands according to their field strength.
  
• You can see such a series containing a non-exhaustive number of ligands in Fig. 7.1.17. You can see that the iodo ligands is on the very left side, and is the weakest field ligand, the carbonyl ligand is on the very right hand side, and is the strongest field ligand. All others are in between. We can see for example, that an aqua ligand is a stronger field ligand compared to a fluoro ligand, but a weaker ligand than an ammine ligand. Generally, ligands at the lower end of the series produce weaker fields, smaller Δs, and absorb light light with longer wavelengths. Ligands at the upper end of the series produce stronger fields, create larger Δs, and absorb light of shorter wavelengths. However, crystal filed theory cannot explain different ligand strength. Why does one ligand produce a stronger field than another? To answer this question, we need the ligand field theory.