CRYSTAL FIELD THEORY CHAPTER OVERVIEW
CRYSTAL FIELD THEORY One of the most striking characteristics of transition-metal complexes is the wide range of colors they exhibit. Crystal field theory (CFT) is a bonding model that explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. The central assumption of CFT is that metal–ligand interactions are purely electrostatic in nature.
COLORS OF COORDINATION COMPLEXES The color for a coordination complex can be predicted using the Crystal Field Theory (CFT). Knowing the color can have a number of useful applications, such as the creation of pigments for dyes in the textile industry. The tendency for coordination complexes to display such a wide array of colors is merely coincidental; their absorption energies happen to fall within range of the visible light spectrum.
CRYSTAL FIELD STABILIZATION ENERGY A consequence of Crystal Field Theory is that the distribution of electrons in the d orbitals may lead to net stabilization (decrease in energy) of some complexes depending on the specific ligand field geometry and metal d-electron configurations. It is a simple matter to calculate this stabilization since all that is needed is the electron configuration and knowledge of the splitting patterns.
CRYSTAL FIELD THEORY Crystal field theory (CFT) describes the breaking of orbital degeneracy in transition metal complexes due to the presence of ligands. CFT qualitatively describes the strength of the metal-ligand bonds. Based on the strength of the metal-ligand bonds, the energy of the system is altered. This may lead to a change in magnetic properties as well as color. This theory was developed by Hans Bethe and John Hasbrouck van Vleck.
INTRODUCTION TO CRYSTAL FIELD THEORY One of the most striking characteristics of transition-metal complexes is the wide range of colors they exhibit. In this section, we describe crystal field theory (CFT), a bonding model that explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. The central assumption of CFT is that metal–ligand interactions are purely electrostatic in nature.
MAGNETIC MOMENTS OF TRANSITION METALS Magnetic moments are often used in conjunction with electronic spectra to gain information about the oxidation number and stereochemistry of the central metal ion in coordination complexes. A common laboratory procedure for the determination of the magnetic moment for a complex is the Gouy method which involves weighing a sample of the complex in the presence and absence of a magnetic field and observing the difference in weight. A template is provided for the calculations involved.
MAGNETISM Movement of an electrical charge (which is the basis of electric currents) generates a magnetic field in a material. Magnetism is therefore a characteristic property of all materials that contain electrically charged particles and for most purposes can be considered to be entirely of electronic origin.
METALS, TETRAHEDRAL AND OCTAHEDRAL NON-OCTAHEDRAL COMPLEXES OCTAHEDRAL VS. TETRAHEDRAL GEOMETRIES A consequence of Crystal Field Theory is that the distribution of electrons in the d orbitals can lead to stabilization for some electron configurations. It is a simple matter to calculate this stabilization since all that is needed is the electron configuration.
ORGEL DIAGRAMS Orgel diagrams are useful for showing the energy levels of both high spin octahedral and tetrahedral transition metal ions.
QUALITATIVE ORGEL DIAGRAMS TANABE-SUGANO DIAGRAMS Tanabe Sugano diagrams are used to predict the transition energies for both spin-allowed and spin-forbidden transitions, as well as for both strong field (low spin), and weak field (high spin) complexes. In this method the energy of the electronic states are given on the vertical axis and the ligand field strength increases on the horizontal axis from left to right.
1 9/20/2021 TETRAHEDRAL VS. SQUARE PLANAR COMPLEXES High spin and low spin are two possible classifications of spin states that occur in coordination compounds. These classifications come from either the ligand field theory, which accounts for the energy differences between the orbitals for each respective geometry, or the crystal field theory, which accounts for the breaking of degenerate orbital states, compared to the pairing energy.
THERMODYNAMICS AND STRUCTURAL CONSEQUENCES OF D-ORBITAL SPLITTING The energy level splitting of the d-orbitals due to their interaction with the ligands in a complex has important structural and thermodynamic effects on the chemistry of transition-metal complexes. Although these two types of effects are interrelated, they are considered separately here.
2 9/20/2021 Colors of Coordination Complexes The color for a coordination complex can be predicted using the Crystal Field Theory (CFT). Knowing the color can have a number of useful applications, such as the creation of pigments for dyes in the textile industry. The tendency for coordination complexes to display such a wide array of colors is merely coincidental; their absorption energies happen to fall within range of the visible light spectrum. Chemists and physicists often study the color of a substance not to understand its sheer appearance, but because color is an indicator of a chemical's physical proprieties on the atomic level.
The Electromagnetic Spectrum The electromagnetic spectrum (EM) spectrum is made up of photons of different wavelengths. Photons, unique in displaying the properties of both waves and particles, create visible light and colors in a small portion of the EM spectrum. This visible light portion has wavelengths in approximately the 400-700 nanometer range (a nanometer, “nm,” is 10-9 meters). Each specific wavelength corresponds to a different color (Figure 1), and when all the wavelengths are present, it appears as white light.
Figure 1: A linear representation of the visible light spectrum. (Public Domain; Gringer via Wikipedia) The wavelength and frequency of a wave are inversely proportional: as one increases, the other decreases; this is a consequence of all light traveling at the same speed. λ ∝ ν −1 (1)
Because of this relationship, blue light has a much higher frequency and more energy than red light.
Perceiving Color Color is perceived in two ways, through additive mixing, where different colors are made by combining different colors of light, and through subtractive mixing, where different wavelengths of light are taken out so that the light is no longer pure white. For colors of coordination complexes, subtractive mixing is considered. As shown in Figure 2, the idea behind subtractive mixing is that white light (which is made from all the colors mixed together) interacts with an object. The object absorbs some of the light, and then reflects or transmits (or both, depending on the object) the rest of the light, which contacts the eye. The object is perceived as whichever color is not absorbed. In Figure 2, white light (simplified as green, red, and blue bands) is shone through a solution. The solution absorbs the red and green wavelengths; however, the blue light is reflected and passes through, so the solution appears blue. This procedure takes place whenever an object displays visible color. If none of the light is absorbed, and all is reflected back off, the object appears white; if all of the light is absorbed, and there is none left to reflect or transmit through, the object appears black.
Figure 2: Subtractive mixing Colors of Coordination Complexes: Crystal Field Splitting When ligands attach to a transition metal to form a coordination complex, electrons in the d orbital split into high energy and low energy orbitals. The difference in energy of the two levels is denoted as ∆, and it is a characteristic a property both of the metal and the ligands. This is illustrated in Figure 3; the "o" subscript on the ∆ indicates that the complex has octahedral geometry.
1 9/4/2021 https://chem.libretexts.org/@go/page/528 Figure 3: d-Orbitals Splitting in an octahedral ligand field. (CC BY-SA-NC; anonymous by request)
If ∆o is large, and much energy is required to promote electrons into the high energy orbitals, the electrons will instead pair in the lower energy orbitals, resulting in a "low spin" complex (Figure 4A); however, if ∆o is small, and it takes little energy to occupy the higher orbitals, the electrons will do so, and remain unpaired (until there are more than five electrons), resulting in a “high spin” complex (Figure 4B). Different ligands are associated with either high or low spin —a "strong field" ligand results in a large ∆o and a low spin configuration, while a "weak field" ligand results in a small ∆o and a high spin configuration. For more details, see the Crystal Field Theory (CFT) page.
Figure 4: (a) Low Spin, Strong Field (Δo > P ). (b): High Spin, Weak Field (Δo < P ). (CC BY-SA-NC; anonymous)
A photon equal to the energy difference ∆o can be absorbed, promoting an electron to the higher energy level. As certain wavelengths are absorbed in this process, subtractive color mixing occurs and the coordination complex solution becomes colored. If the ions have a noble gas configuration, and have no unpaired electrons, the solutions appear colorless; in reality, they still have a measured energy and absorb certain wavelengths of light, but these wavelengths are not in the visible portion of the EM spectrum and no color is perceived by the eye.
In general, a larger Δo indicates that higher energy photons are absorbed, and the solution appears further to the left on the EM spectrum shown in Figure 1. This relationship is described in the equation
Δo = hc/λ (2) where h and c are constants, and λ is the wavelength of light absorbed. Using a color wheel can be useful for determining what color a solution will appear based on what wavelengths it absorbs (Figure 6). If a complex absorbs a particular color, it will have the appearance of whatever color is directly opposite it on the wheel. For example, if a complex is known to absorb photons in the orange range, it can be concluded that the solution will look blue. This concept can be used in reverse to determine ∆ for a complex from the color of its solution.
2 9/4/2021 https://chem.libretexts.org/@go/page/528 Figure 5: Color wheel with wavelengths marked (CC BY-SA 4.0 International; Tem5psu via source)
Relating the Colors of Coordination Complexes to the Spectrochemical Series According to the Crystal Field Theory, ligands that have smaller Δ) values are considered "weak field" and will absorb lower- energy light with longer λ values (ie a "red shift"). Ligands that have larger Δ) values are considered "strong field" and will absorb higher-energy light with shorter λ values (ie a "blue shift"). This relates to the colors seen in a coordination complex. Weaker-field ligands induce the absorption of longer wavelength (lower frequency=lower energy) light than stronger- field ligands since their respective Δo values are smaller than the electron pairing energy.
The energy difference, Δo, determines the color of the coordination complex. According to the spectrochemical series, the high spin ligands are considered "weak field," and absorb longer wavelengths of light (weak Δo), while complexes with low spin ligands absorb light of greater frequency (high Δo). The color seen is the complementary color of the color associated with the absorbed wavelength. To predict which possible colors and their corresponding wavelengths are absorbed, the spectrochemical series can be used:
(Strong field/large Δ0/low spin) (weak field/ small Δ0/high spin) - - - - 4- - - 2------CO , NO , CN >NO2 >en>py≈NH3>EDTA >NCS >H2O>ONO >ox >OH >F >SCN >Cl >Br >I
Example 1 If a solution with a dissolved octahedral complex appears yellow to the eye, what wavelength of light does it absorb? Is this complex expected to be low spin or high spin? Solution A solution that looks yellow absorbs light that is violet, which is roughly 410 nm from the color wheel. Since it absorbs high energy, the electrons must be raised to a higher level, and Δo is high, so the complex is likely to be low spin.
Example 2
An octahedral metal complex absorbs light with wavelength 535 nm. What is the crystal field splitting Δo for the complex? What color is it to the eye? Solution To solve this question, we need to use the equation hc Δ = o λ with −34 h is Planck’s constant and is 6.625 ×10 J ⋅ s and c is the speed of light and is 2.998 ×108 m/s. It is also important to remember that 1 nm is equal to 1 ×10−9 meters. With all this information, the final equation looks like this:
3 9/4/2021 https://chem.libretexts.org/@go/page/528 (6.625 ×10−34 J ⋅ s)(2.998 ×108 m/s) Δ = = 3.712 J/molecule o 1 m (535nm) ( ) 1 ×109 nm
It is not necessary to use any equations to solve the second part of the problem. Light that is 535 nm is green, and because green light is absorbed, the complex appears red (refer to Figures 1 and 6 for this information). Note: the fact that the complex is octahedral makes no impact when solving this problem. Although the splitting is different for complexes of different structures, the mechanics of solving the problem are identical.
Example 3
There are two solutions, one orange and one blue. Both solutions are known to be made up of a cobalt complex; however, one has chloride ions as ligands, while the other has ammonia ligands. Which solution is expected to be orange? Solution In order to solve this problem, it is necessary to know the relative strengths of the ligands involved. A sample ligand strength list is given here, but see Crystal Field Splitting for a more complete list: - - - - CN > en > NH3 > H2O > F >SCN > Cl - From this information, it is clear that NH3 is a stronger ligand than Cl , which means that the complex involving NH3 has a greater ∆, and the complex will be low spin. Because of the larger ∆, the electrons absorb higher energy photons, and the solution will have the appearance of a lower energy color. Since orange light is less energetic than blue light, the NH3 containing solution is predicted to be orange
References 1. Cox, P. A. Instant Notes Inorganic Chemistry. Second ed. Grand Rapids: Garland, Incorporated, 2004. 2. Nassau, Kurt. The Physics and Chemistry of Color : The Fifteen Causes of Color. Second ed. New York: Wiley- Interscience, 2001. 3. Petrucci, Ralph H., William S. Harwood, and Geoff E. Herring. General Chemistry : Principles and Modern Applications. Ninth ed. Upper Saddle River: Prentice Hall PTR, 2006. 4. Petrucci, Ralph H., Carey Bissonnette, F. Geoffrey Herring, Jeffrey D. Madura. General Chemistry: Principles and Modern Applications. Tenth Ed. Upper Saddle River: Pearson Education, Inc. 2011.
Problems 1. What color will a complex be that absorbs light that is 600 nm be? −19 2. What color will a complex an octahedral complex appear if it has a Δo of 3.75 ×10 J? 3. Would you expect a violet solution to be high spin or low spin? What about a red solution? 3+ - 3- 4. There are two solutions, one which is yellow and another which is violet. The solutions are [Co(H2O)6] and [Co(CN )6] . What are the colors of each solution?
Answers 1. Blue. The color absorbed is orange. 2. It is red. Using Δ=hc/λ, h=6.626*10-34J*s, c=2.998*108m/s, wavelength would equal 530 nm. So green is absorbed, and the complementary color of green is red, so red is the color of the complex. 3. It would be high spin. The complementary color of violet is yellow, which has a wavelength of 570 nm. For a red solution, the complementary color absorbed is green, with a wavelength of 530 nm, so it would be considered low spin. 3+ - 3- 4. [Co(H2O)6] is violet and [Co(CN )6] is yellow. Looking at the spectrochemical series, H2O is a weak field ligand, so it absorbs colors of long wavelengths—in this case, the longer wavelength is yellow, so the color reflected is violet. CN- is a strong field ligand, so it absorbs colors of shorter wavelengths-in this case, the shorter wavelength is violet, so the color reflected is yellow.
4 9/4/2021 https://chem.libretexts.org/@go/page/528 Contributors and Attributions Deyu Wang (UCD)
5 9/4/2021 https://chem.libretexts.org/@go/page/528 Crystal Field Stabilization Energy A consequence of Crystal Field Theory is that the distribution of electrons in the d orbitals may lead to net stabilization (decrease in energy) of some complexes depending on the specific ligand field geometry and metal d-electron configurations. It is a simple matter to calculate this stabilization since all that is needed is the electron configuration and knowledge of the splitting patterns.
Definition: Crystal Field Stabilization Energy
The Crystal Field Stabilization Energy is defined as the energy of the electron configuration in the ligand field minus the energy of the electronic configuration in the isotropic field.
CF SE = ΔE = Eligand field −Eisotropic field (1)
The CSFE will depend on multiple factors including: Geometry (which changes the d-orbital splitting patterns) Number of d-electrons Spin Pairing Energy Ligand character (via Spectrochemical Series)
For an octahedral complex, an electron in the more stable t2g subset is treated as contributing −2/5Δo whereas an electron in the higher energy eg subset contributes to a destabilization of +3/5Δo. The final answer is then expressed as a multiple of the crystal field splitting parameter Δo. If any electrons are paired within a single orbital, then the term P is used to represent the spin pairing energy.
Example 1: CFSE for a high Spin d7 complex
What is the Crystal Field Stabilization Energy for a high spin d7 octahedral complex? Solution The splitting pattern and electron configuration for both isotropic and octahedral ligand fields are compared below.
The energy of the isotropic field (Eisotropic field) is
Eisotropic field = 7 ×0 +2P = 2P
The energy of the octahedral ligand field Eligand field is
Eligand field = (5 ×−2/5Δo) +(2 ×3/5Δo) +2P = −4/5Δo +2P
So via Equation 1, the CFSE is
CF SE = Eligand field −Eisotropic field
= (−4/5Δo +2P ) −2P
= −4/5Δo
Notice that the Spin pairing Energy falls out in this case (and will when calculating the CFSE of high spin complexes) since the number of paired electrons in the ligand field is the same as that in isotropic field of the free metal ion.
1 9/7/2021 https://chem.libretexts.org/@go/page/15736 Example 2: CFSE for a Low Spin d7 complex
What is the Crystal Field Stabilization Energy for a low spin d7 octahedral complex? Solution The splitting pattern and electron configuration for both isotropic and octahedral ligand fields are compared below.
The energy of the isotropic field is the same as calculated for the high spin configuration in Example 1:
Eisotropic field = 7 ×0 +2P = 2P
The energy of the octahedral ligand\) field Eligand field is
Eligand field = (6 ×−2/5Δo) +(1 ×3/5Δo) +3P
= −9/5Δo +3P
So via Equation 1, the CFSE is
CF SE = Eligand field −Eisotropic field
= (−9/5Δo +3P ) −2P
= −9/5Δo +P
Adding in the pairing energy since it will require extra energy to pair up one extra group of electrons. This appears more a more stable configuration than the high spin d7 configuration in Example 1, but we have then to take into consideration the Pairing energy P to know definitely, which varies between 200 −400 kJ mol−1 depending on the metal.
Table 1: Crystal Field Stabilization Energies (CFSE) for high and low spin octahedral complexes Total d- Crystal Field Stabilization Isotropic Field Octahedral Complex electrons Energy
High Spin Low Spin
Eisotropic field Configuration Eligand field Configuration Eligand field High Spin Low Spin
d0 0 t2g 0eg 0 0 t2g 0eg 0 0 0 0
d1 0 t2g 1eg 0 -2/5 Δo t2g 1eg 0 -2/5 Δo -2/5 Δo -2/5 Δo
d2 0 t2g 2eg 0 -4/5 Δo t2g 2eg 0 -4/5 Δo -4/5 Δo -4/5 Δo
d3 0 t2g 3eg 0 -6/5 Δo t2g 3eg 0 -6/5 Δo -6/5 Δo -6/5 Δo
d4 0 t2g 3eg 1 -3/5 Δo t2g 4eg 0 -8/5 Δo + P -3/5 Δo -8/5 Δo + P
d5 0 t2g 3eg 2 0 Δo t2g 5eg 0 -10/5 Δo + 2P 0 Δo -10/5 Δo + 2P
d6 P t2g 4eg 2 -2/5 Δo + P t2g 6eg 0 -12/5 Δo + 3P -2/5 Δo -12/5 Δo + P
d7 2P t2g 5eg 2 -4/5 Δo + 2P t2g 6eg 1 -9/5 Δo + 3P -4/5 Δo -9/5 Δo + P
d8 3P t2g 6eg 2 -6/5 Δo + 3P t2g 6eg 2 -6/5 Δo + 3P -6/5 Δo -6/5 Δo
d9 4P t2g 6eg 3 -3/5 Δo + 4P t2g 6eg 3 -3/5 Δo + 4P -3/5 Δo -3/5 Δo
d10 5P t2g 6eg 4 0 Δo + 5P t2g 6eg 4 0 Δo + 5P 0 0
P is the spin pairing energy and represents the energy required to pair up electrons within the same orbital. For a given metal ion P (pairing energy) is constant, but it does not vary with ligand and oxidation state of the metal ion).
2 9/7/2021 https://chem.libretexts.org/@go/page/15736 Octahedral Preference Similar CFSE values can be constructed for non-octahedral ligand field geometries once the knowledge of the d-orbital splitting is known and the electron configuration within those orbitals known, e.g., the tetrahedral complexes in Table 2. These energies geoemtries can then be contrasted to the octahedral CFSE to calculate a thermodynamic preference (Enthalpy-wise) for a metal-ligand combination to favor the octahedral geometry. This is quantified via a Octahedral Site Preference Energy defined below.
Definition: Octahedral Site Preference Energies
The Octahedral Site Preference Energy (OSPE) is defined as the difference of CFSE energies for a non-octahedral complex and the octahedral complex. For comparing the preference of forming an octahedral ligand field vs. a tetrahedral ligand field, the OSPE is thus:
OSP E = CF SE(oct) −CF SE(tet) (2)
The OSPE quantifies the preference of a complex to exhibit an octahedral geometry vs. a tetrahedral geometry.
Note: the conversion between Δo and Δt used for these calculations is: 4 Δ ≈ Δ (3) t 9 o which is applicable for comparing octahedral and tetrahedral complexes that involve same ligands only. Table 2: Octahedral Site Preference Energies (OSPE) OSPE (for high spin Total d-electrons CFSE(Octahedral) CFSE(Tetrahedral) complexes)**
High Spin Low Spin Configuration Always High Spin*
0 0 d 0 Δo 0 Δo e 0 Δt 0 Δo 1 1 d -2/5 Δo -2/5 Δo e -3/5 Δt -6/45 Δo 2 2 d -4/5 Δo -4/5 Δo e -6/5 Δt -12/45 Δo 3 2 1 d -6/5 Δo -6/5 Δo e t2 -4/5 Δt -38/45 Δo 4 2 2 d -3/5 Δo -8/5 Δo + P e t2 -2/5 Δt -19/45 Δo 5 2 3 d 0 Δo -10/5 Δo + 2P e t2 0 Δt 0 Δo 6 3 3 d -2/5 Δo -12/5 Δo + P e t2 -3/5 Δt -6/45 Δo 7 4 3 d -4/5 Δo -9/5 Δo + P e t2 -6/5 Δt -12/45 Δo 8 4 4 d -6/5 Δo -6/5 Δo e t2 -4/5 Δt -38/45 Δo 9 4 5 d -3/5 Δo -3/5 Δo e t2 -2/5 Δt -19/45 Δo 10 4 6 d 0 0 e t2 0 Δt 0 Δo
P is the spin pairing energy and represents the energy required to pair up electrons within the same orbital.
Tetrahedral complexes are always high spin since the splitting is appreciably smaller than P (Equation 3).
3 9/7/2021 https://chem.libretexts.org/@go/page/15736 After conversion with Equation 3. The data in Tables 1 and 2 are represented graphically by the curves in Figure 1 below for the high spin complexes only. The low spin complexes require knowledge of P to graph.
Figure 1: Crystal Field Stabilization Energies for both octahedral fields (CF SEoct) and tetrahedral fields (CF SEtet ). Octahedral Site Preference Energies (OSPE) are in yellow. This is for high spin complexes. From a simple inspection of Figure 1, the following observations can be made: The OSPE is small in d1, d2, d5, d6, d7 complexes and other factors influence the stability of the complexes including steric factors The OSPE is large in d3 and d8 complexes which strongly favor octahedral geometries
Applications The "double-humped" curve in Figure 1 is found for various properties of the first-row transition metals, including Hydration and Lattice energies of the M(II) ions, ionic radii as well as the stability of M(II) complexes. This suggests that these properties are somehow related to Crystal Field effects. In the case of Hydration Energies describing the complexation of water ligands to a bare metal ion: 2+ 2+ M (g) +H2O → [M(OH2)6] (aq) (4)
Table 3 and Figure 1 shows this type of curve. Note that in any series of this type not all the data are available since a number of ions are not very stable in the M(II) state. Table 3: Hydration energies of M 2+ ions M ΔH°/kJmol-1 M ΔH°/kJmol-1
Ca -2469 Fe -2840 Sc no stable 2+ ion Co -2910 Ti -2729 Ni -2993 >V -2777 Cu -2996 Cr -2792 Zn -2928 Mn -2733
Graphically the data in Table 2 can be represented by:
4 9/7/2021 https://chem.libretexts.org/@go/page/15736 Figure 2: hydration energies of M 2+ ions
Contributors and Attributions Prof. Robert J. Lancashire (The Department of Chemistry, University of the West Indies)
5 9/7/2021 https://chem.libretexts.org/@go/page/15736 Crystal Field Theory Crystal field theory (CFT) describes the breaking of orbital degeneracy in transition metal complexes due to the presence of ligands. CFT qualitatively describes the strength of the metal-ligand bonds. Based on the strength of the metal-ligand bonds, the energy of the system is altered. This may lead to a change in magnetic properties as well as color. This theory was developed by Hans Bethe and John Hasbrouck van Vleck.
Basic Concept In Crystal Field Theory, it is assumed that the ions are simple point charges (a simplification). When applied to alkali metal ions containing a symmetric sphere of charge, calculations of bond energies are generally quite successful. The approach taken uses classical potential energy equations that take into account the attractive and repulsive interactions between charged particles (that is, Coulomb's Law interactions). q q E ∝ 1 2 (1) r with E the bond energy between the charges and q1 and q2 are the charges of the interacting ions and r is the distance separating them. This approach leads to the correct prediction that large cations of low charge, such as K + and Na+ , should form few coordination compounds. For transition metal cations that contain varying numbers of d electrons in orbitals that are NOT spherically symmetric, however, the situation is quite different. The shapes and occupations of these d-orbitals then become important in building an accurate description of the bond energy and properties of the transition metal compound. When examining a single transition metal ion, the five d-orbitals have the same energy (Figure 1). When ligands approach the metal ion, some experience more opposition from the d-orbital electrons than others based on the geometric structure of the molecule. Since ligands approach from different directions, not all d-orbitals interact directly. These interactions, however, create a splitting due to the electrostatic environment.
Figure 1: Distributing a charge of −6 uniformly over a spherical surface surrounding a metal ion causes the energy of all five d orbitals to increase due to electrostatic repulsions, but the five d orbitals remain degenerate. Placing a charge of −1 at each vertex of an octahedron causes the d orbitals to split into two groups with different energies: the dx2−y2 and dz2 orbitals increase in energy, while the, dxy, dxz, and dyz orbitals decrease in energy. The average energy of the five d orbitals is the same as for a spherical distribution of a −6 charge, however. Attractive electrostatic interactions between the negatively charged ligands and the positively charged metal ion (far right) cause all five d orbitals to decrease in energy but does not affect the splittings of the orbitals. The two eg orbitals point directly at the six negatively charged ligands, which increases their energy compared with a spherical distribution of negative charge. In contrast, the three t2g orbitals point between the negatively charged ligands, which decreases their energy compared with a spherical distribution of charge. For example, consider a molecule with octahedral geometry. Ligands approach the metal ion along the x, y, and z axes. Therefore, the electrons in the dz2 and dx2−y2 orbitals (which lie along these axes) experience greater repulsion. It requires more energy to have an electron in these orbitals than it would to put an electron in one of the other orbitals. This causes a splitting in the energy levels of the d-orbitals. This is known as crystal field splitting. For octahedral complexes, crystal field splitting is denoted by Δo (or Δoct). The energies of the dz2 and dx2−y2 orbitals increase due to greater interactions with the ligands. The dxy , dxz, and dyz orbitals decrease with respect to this normal energy level and become more stable.
Electrons in Orbitals According to the Aufbau principle, electrons are filled from lower to higher energy orbitals (Figure 1). For the octahedral case above, this corresponds to the dxy, dxz, and dyz orbitals. Following Hund's rule, electrons are filled in order to have the highest number of unpaired electrons. For example, if one had a d3 complex, there would be three unpaired electrons. If one were to add an electron, however, it has the ability to fill a higher energy orbital ( dz² or dx²-y²) or pair with an electron residing in the dxy, dxz, or dyz orbitals. This pairing of the electrons requires energy (spin pairing energy). If the pairing energy is less than the crystal field splitting energy, ∆₀, then the next electron will go into the dxy, dxz, or dyz orbitals
1 9/3/2021 https://chem.libretexts.org/@go/page/529 due to stability. This situation allows for the least amount of unpaired electrons, and is known as low spin. If the pairing energy is greater than ∆₀, then the next electron will go into the dz² or dx²-y² orbitals as an unpaired electron. This situation allows for the most number of unpaired electrons, and is known as high spin. Ligands that cause a transition metal to have a small crystal field splitting, which leads to high spin, are called weak-field ligands. Ligands that produce a large crystal field splitting, which leads to low spin, are called strong field ligands.
4 Figure 2: Low Spin, Strong Field (∆o˃P) High Spin, Weak Field (∆o˂P)Splitting for a d complex under a strong field (left) and a weak field (right). The strong field is a low spin complex, while the weak field is a high spin complex. As mentioned above, CFT is based primarily on symmetry of ligands around a central metal/ion and how this anisotropic (properties depending on direction) ligand field affects the metal's atomic orbitals; the energies of which may increase, decrease or not be affected at all. Once the ligands' electrons interact with the electrons of the d-orbitals, the electrostatic interactions cause the energy levels of the d-orbital to fluctuate depending on the orientation and the nature of the ligands. For example, the oxidation state and the strength of the ligands determine splitting; the higher the oxidation state or the stronger the ligand, the larger the splitting. Ligands are classified as strong or weak based on the spectrochemical series: ------2- - - 4- - - I < Br < Cl < SCN < F < OH < ox < ONO < H2O < SCN < EDTA < NH3 < en < NO2 < CN - Note that SCN and NO2- ligands are represented twice in the above spectrochemical series since there are two different Lewis base sites (e.g., free electron pairs to share) on each ligand (e.g., for the SCN- ligand, the electron pair on the sulfur or the nitrogen can form the coordinate covalent bond to a metal). The specific atom that binds in such ligands is underlined. In addition to octahedral complexes, two common geometries observed are that of tetrahedral and square planar. These complexes differ from the octahedral complexes in that the orbital levels are raised in energy due to the interference with electrons from ligands. For the tetrahedral complex, the dxy, dxz, and dyz orbitals are raised in energy while the dz², dx²-y² orbitals are lowered. For the square planar complexes, there is greatest interaction with the dx²-y² orbital and therefore it has higher energy. The next orbital with the greatest interaction is dxy, followed below by dz². The orbitals with the lowest energy are the dxz and dyz orbitals. There is a large energy separation between the dz² orbital and the dxz and dyz orbitals, meaning that the crystal field splitting energy is large. We find that the square planar complexes have the greatest crystal field splitting energy compared to all the other complexes. This means that most square planar complexes are low spin, strong field ligands.
Description of d-Orbitals To understand CFT, one must understand the description of the lobes:
dxy: lobes lie in-between the x and the y axes. dxz: lobes lie in-between the x and the z axes. dyz: lobes lie in-between the y and the z axes. dx2-y2: lobes lie on the x and y axes. dz2: there are two lobes on the z axes and there is a donut shape ring that lies on the xy plane around the other two lobes.
Figure 3: Spatial arrangement of ligands in the an octahedral ligand field with respect to the five d-orbitals.
Octahedral Complexes In an octahedral complex, there are six ligands attached to the central transition metal. The d-orbital splits into two different levels (Figure 4). The bottom three energy levels are named dxy , dxz, and dyz (collectively referred to as t2g). The two upper energy levels are named dx2−y2 , and dz2 (collectively referred to as eg).
Figure 4.
2 9/3/2021 https://chem.libretexts.org/@go/page/529 The reason they split is because of the electrostatic interactions between the electrons of the ligand and the lobes of the d-orbital. In an octahedral, the electrons are attracted to the axes. Any orbital that has a lobe on the axes moves to a higher energy level. This means that in an octahedral, the energy levels of eg are higher (0.6∆o) while t2g is lower (0.4∆o). The distance that the electrons have to move from t2g from eg and it dictates the energy that the complex will absorb from white light, which will determine the color. Whether the complex is paramagnetic or diamagnetic will be determined by the spin state. If there are unpaired electrons, the complex is paramagnetic; if all electrons are paired, the complex is diamagnetic.
Tetrahedral Complexes In a tetrahedral complex, there are four ligands attached to the central metal. The d orbitals also split into two different energy levels. The top three consist of the dxy , dxz, and dyz orbitals. The bottom two consist of the dx2−y2 and dz2 orbitals. The reason for this is due to poor orbital overlap between the metal and the ligand orbitals. The orbitals are directed on the axes, while the ligands are not.
Figure 5: (a) Tetraheral ligand field surrounding a central transition metal (blue sphere). (b) Splitting of the degenerate d-orbitals (without a ligand field) due to an octahedral ligand field (left diagram) and the tetrahedral field (right diagram).
The difference in the splitting energy is tetrahedral splitting constant (Δt), which less than (Δo) for the same ligands:
Δt = 0.44 Δo (2)
Consequentially, Δt is typically smaller than the spin pairing energy, so tetrahedral complexes are usually high spin.
Square Planar Complexes In a square planar, there are four ligands as well. However, the difference is that the electrons of the ligands are only attracted to the xy plane. Any orbital in the xy plane has a higher energy level (Figure 6). There are four different energy levels for the square planar (from the highest energy level to the lowest energy level): dx2-y2, dxy, dz2, and both dxz and dyz.
Figure 6: Splitting of the degenerate d-orbitals (without a ligand field) due to an square planar ligand field.
The splitting energy (from highest orbital to lowest orbital) is Δsp and tends to be larger then Δo
Δsp = 1.74 Δo (3)
Moreover, Δsp is also larger than the pairing energy, so the square planar complexes are usually low spin complexes.
Example 1 3- For the complex ion [Fe(Cl)6] determine the number of d electrons for Fe, sketch the d-orbital energy levels and the distribution of d electrons among them, list the number of lone electrons, and label whether the complex is paramagnetic or diamagnetic. Solution Step 1: Determine the oxidation state of Fe. Here it is Fe3+. Based on its electron configuration, Fe3+ has 5 d-electrons. Step 2: Determine the geometry of the ion. Here it is an octahedral which means the energy splitting should look like:
3 9/3/2021 https://chem.libretexts.org/@go/page/529 Step 3: Determine whether the ligand induces is a strong or weak field spin by looking at the spectrochemical series. Cl- is a weak field ligand (i.e., it induces high spin complexes). Therefore, electrons fill all orbitals before being paired.
Step four: Count the number of lone electrons. Here, there are 5 electrons. Step five: The five unpaired electrons means this complex ion is paramagnetic (and strongly so).
Example 2
A tetrahedral complex absorbs at 545 nm. What is the respective octahedral crystal field splitting (Δo)? What is the color of the complex? Solution hc Δ = t λ (6.626 ×10−34J ⋅ s)(3 ×108m/s) = 545 ×10−9 m = 3.65 ×10−19 J
However, the tetrahedral splitting (Δt) is ~4/9 that of the octahedral splitting (Δo).
Δt = 0.44Δo Δ Δ = t o 0.44 3.65 ×10−19J = 0.44 = 8.30 ×10−18J
This is the energy needed to promote one electron in one complex. Often the crystal field splitting is given per mole, which requires this number to be multiplied by Avogadro's Number (6.022 ×1023). This complex appears red, since it absorbs in the complementary green color (determined via the color wheel).
4 9/3/2021 https://chem.libretexts.org/@go/page/529 Electronic Structure of Coordination Complexes
Video:
Problems For each of the following, sketch the d-orbital energy levels and the distribution of d electrons among them, state the geometry, list the number of d- electrons, list the number of lone electrons, and label whether they are paramagnetic or dimagnetic: 2+ 1. [Ti(H2O)6] 2- 2. [NiCl4] 3- 3. [CoF6] (also state whether this is low or high spin) 3+ 4. [Co(NH3)6] (also state whether this is low or high spin) 5. True or False: Square Planer complex compounds are usually low spin.
Answers 1. octahedral, 2, 2, paramagnetic
2. tetrahedral, 8, 2, paramagnetic (see Octahedral vs. Tetrahedral Geometries)
3. octahedral, 6, 4, paramagnetic, high spin
4. octahedral, 6, 0, diamagnetic, low spin
5 9/3/2021 https://chem.libretexts.org/@go/page/529 5. True
Contributors and Attributions Asadullah Awan (UCD), Hong Truong (UCD) Prof. Robert J. Lancashire (The Department of Chemistry, University of the West Indies)
6 9/3/2021 https://chem.libretexts.org/@go/page/529 Introduction to Crystal Field Theory
Learning Objectives To understand how crystal field theory explains the electronic structures and colors of metal complexes.
One of the most striking characteristics of transition-metal complexes is the wide range of colors they exhibit. In this section, we describe crystal field theory (CFT), a bonding model that explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. The central assumption of CFT is that metal– ligand interactions are purely electrostatic in nature. Even though this assumption is clearly not valid for many complexes, such as those that contain neutral ligands like CO, CFT enables chemists to explain many of the properties of transition-metal complexes with a reasonable degree of accuracy. The Learning Objective of this Module is to understand how crystal field theory explains the electronic structures and colors of metal complexes. d-Orbital Splittings CFT focuses on the interaction of the five (n − 1)d orbitals with ligands arranged in a regular array around a transition-metal ion. We will focus on the application of CFT to octahedral complexes, which are by far the most common and the easiest to visualize. Other common structures, such as square planar complexes, can be treated as a distortion of the octahedral model. According to CFT, an octahedral metal complex forms because of the electrostatic interaction of a positively charged metal ion with six negatively charged ligands or with the negative ends of dipoles associated with the six ligands. In addition, the ligands interact with one other electrostatically. As you learned in our discussion of the valence-shell electron-pair repulsion (VSEPR) model, the lowest-energy arrangement of six identical negative charges is an octahedron, which minimizes repulsive interactions between the ligands. We begin by considering how the energies of the d orbitals of a transition-metal ion are affected by an octahedral arrangement of six negative charges. Recall that the five d orbitals are initially degenerate (have the same energy). If we distribute six negative charges uniformly over the surface of a sphere, the d orbitals remain degenerate, but their energy will be higher due to repulsive electrostatic interactions between the spherical shell of negative charge and electrons in the d orbitals (Figure 1a). Placing the six negative charges at the vertices of an octahedron does not change the average energy of the d orbitals, but it does remove their degeneracy: the five d orbitals split into two groups whose energies depend on their orientations. As shown in Figure 1b, the dz2 and dx2−y2 orbitals point directly at the six negative charges located on the x, y, and z axes. Consequently, the energy of an electron in these two orbitals (collectively labeled the eg orbitals) will be greater than it will be for a spherical distribution of negative charge because of increased electrostatic repulsions. In contrast, the other three d orbitals (dxy, dxz, and dyz, collectively called the t2g orbitals) are all oriented at a 45° angle to the coordinate axes, so they point between the six negative charges. The energy of an electron in any of these three orbitals is lower than the energy for a spherical distribution of negative charge.
1 9/8/2021 https://chem.libretexts.org/@go/page/6520 Figure 1: An Octahedral Arrangement of Six Negative Charges around a Metal Ion Causes the Five d Orbitals to Split into Two Sets with Different Energies. (a) Distributing a charge of −6 uniformly over a spherical surface surrounding a metal ion causes the energy of all five d orbitals to increase due to electrostatic repulsions, but the five d orbitals remain degenerate. Placing a charge of −1 at each vertex of an octahedron causes the d orbitals to split into two groups with different energies: the dx2−y2 and dz2 orbitals increase in energy, while the dxy, dxz, and dyz orbitals decrease in energy. The average energy of the five d orbitals is the same as for a spherical distribution of a −6 charge, however. Attractive electrostatic interactions between the negatively charged ligands and the positively charged metal ion (far right) cause all five d orbitals to decrease in energy but does not affect the splittings of the orbitals. (b) The two eg orbitals (left) point directly at the six negatively charged ligands, which increases their energy compared with a spherical distribution of negative charge. In contrast, the three t2g orbitals (right) point between the negatively charged ligands, which decreases their energy compared with a spherical distribution of charge. (CC BY-SA-NC; anonymous by request)
The difference in energy between the two sets of d orbitals is called the crystal field splitting energy (Δo), where the subscript o stands for octahedral. As we shall see, the magnitude of the splitting depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. (Crystal field splitting energy also applies to tetrahedral complexes: Δt.) It is important to note that the splitting of the d orbitals in a crystal field does not change the total energy of the five d orbitals: the two eg orbitals increase in energy by 0.6Δo, whereas the three t2g orbitals decrease in energy by 0.4Δo. Thus the total change in energy is
2(0.6Δo) +3(−0.4Δo) = 0. (1) Crystal field splitting does not change the total energy of the d orbitals. Thus far, we have considered only the effect of repulsive electrostatic interactions between electrons in the d orbitals and the six negatively charged ligands, which increases the total energy of the system and splits the d orbitals. Interactions between the positively charged metal ion and the ligands results in a net stabilization of the system, which decreases the energy of all five d orbitals without affecting their splitting (as shown at the far right in Figure 1a).
2 9/8/2021 https://chem.libretexts.org/@go/page/6520 Electronic Structures of Metal Complexes We can use the d-orbital energy-level diagram in Figure 1 to predict electronic structures and some of the properties of transition-metal complexes. We start with the Ti3+ ion, which contains a single d electron, and proceed across the first row of the transition metals by adding a single electron at a time. We place additional electrons in the lowest-energy orbital available, 2 1 3 3+ while keeping their spins parallel as required by Hund’s rule. As shown in Figure , for d –d systems—such as [Ti(H2O)6] , 3+ 3+ [V(H2O)6] , and [Cr(H2O)6] , respectively—the electrons successively occupy the three degenerate t2g orbitals with their spins parallel, giving one, two, and three unpaired electrons, respectively. We can summarize this for the complex 3+ 3 3+ 3 [Cr(H2O)6] , for example, by saying that the chromium ion has a d electron configuration or, more succinctly, Cr is a d ion.
Figure 2: The Possible Electron Configurations for Octahedral dn Transition-Metal Complexes (n = 1–10). Two different 4 5 6 7 configurations are possible for octahedral complexes of metals with d , d , d , and d configurations; the magnitude of Δo determines which configuration is observed. (CC BY-SA-NC; anonymous by request) When we reach the d4 configuration, there are two possible choices for the fourth electron: it can occupy either one of the empty eg orbitals or one of the singly occupied t2g orbitals. Recall that placing an electron in an already occupied orbital results in electrostatic repulsions that increase the energy of the system; this increase in energy is called the spin-pairing energy (P). If
Δo is less than P, then the lowest-energy arrangement has the fourth electron in one of the empty eg orbitals. Because this arrangement results in four unpaired electrons, it is called a high-spin configuration, and a complex with this electron 2+ configuration, such as the [Cr(H2O)6] ion, is called a high-spin complex. Conversely, if Δo is greater than P, then the lowest- energy arrangement has the fourth electron in one of the occupied t2g orbitals. Because this arrangement results in only two unpaired electrons, it is called a low-spin configuration, and a complex with this electron configuration, such as the 3− 5 6 7 [Mn(CN)6] ion, is called a low-spin complex. Similarly, metal ions with the d , d , or d electron configurations can be either high spin or low spin, depending on the magnitude of Δo. In contrast, only one arrangement of d electrons is possible for metal ions with d8–d10 electron configurations. For example, 2+ 8 2+ 9 the [Ni(H2O)6] ion is d with two unpaired electrons, the [Cu(H2O)6] ion is d with one unpaired electron, and the 2+ 10 [Zn(H2O)6] ion is d with no unpaired electrons.
If Δo is less than the spin-pairing energy, a high-spin configuration results. Conversely, if Δo is greater, a low-spin configuration forms.
Factors That Affect the Magnitude of Δo
The magnitude of Δo dictates whether a complex with four, five, six, or seven d electrons is high spin or low spin, which affects its magnetic properties, structure, and reactivity. Large values of Δo (i.e., Δo > P) yield a low-spin complex, whereas small values of Δo (i.e., Δo < P) produce a high-spin complex. As we noted, the magnitude of Δo depends on three factors: the charge on the metal ion, the principal quantum number of the metal (and thus its location in the periodic table), and the nature of the ligand. Values of Δo for some representative transition-metal complexes are given in Table 1.
Table 1: Crystal Field Splitting Energies for Some Octahedral (Δo)* and Tetrahedral (Δt) Transition-Metal Complexes Octahedral Octahedral Tetrahedral Δ (cm−1) Δ (cm−1) Δ (cm−1) Complexes o Complexes o Complexes t
3 9/8/2021 https://chem.libretexts.org/@go/page/6520 Octahedral Octahedral Tetrahedral Δ (cm−1) Δ (cm−1) Δ (cm−1) Complexes o Complexes o Complexes t
3+ 4− [Ti(H2O)6] 20,300 [Fe(CN)6] 32,800 VCl4 9010 2+ 3− 2− [V(H2O)6] 12,600 [Fe(CN)6] 35,000 [CoCl4] 3300 3+ 3− 2− [V(H2O)6] 18,900 [CoF6] 13,000 [CoBr4] 2900 3− 2+ 2− [CrCl6] 13,000 [Co(H2O)6] 9300 [CoI4] 2700 2+ 3+ [Cr(H2O)6] 13,900 [Co(H2O)6] 27,000 3+ 3+ [Cr(H2O)6] 17,400 [Co(NH3)6] 22,900 3+ 3− [Cr(NH3)6] 21,500 [Co(CN)6] 34,800 3− 2+ [Cr(CN)6] 26,600 [Ni(H2O)6] 8500 2+ Cr(CO)6 34,150 [Ni(NH3)6] 10,800 4− 3− [MnCl6] 7500 [RhCl6] 20,400 2+ 3+ [Mn(H2O)6] 8500 [Rh(H2O)6] 27,000 3− 3+ [MnCl6] 20,000 [Rh(NH3)6] 34,000 3+ 3− [Mn(H2O)6] 21,000 [Rh(CN)6] 45,500 2+ 3− [Fe(H2O)6] 10,400 [IrCl6] 25,000 3+ 3+ [Fe(H2O)6] 14,300 [Ir(NH3)6] 41,000 *Energies obtained by spectroscopic measurements are often given in units of wave numbers (cm−1); the wave number is the reciprocal of the wavelength of the corresponding electromagnetic radiation expressed in centimeters: 1 cm−1 = 11.96 J/mol.
Source of data: Duward F. Shriver, Peter W. Atkins, and Cooper H. Langford, Inorganic Chemistry, 2nd ed. (New York: W. H. Freeman and Company, 1994).
Factor 1: Charge on the Metal Ion Increasing the charge on a metal ion has two effects: the radius of the metal ion decreases, and negatively charged ligands are more strongly attracted to it. Both factors decrease the metal–ligand distance, which in turn causes the negatively charged ligands to interact more strongly with the d orbitals. Consequently, the magnitude of Δo increases as the charge on the metal ion increases. Typically, Δo for a tripositive ion is about 50% greater than for the dipositive ion of the same metal; for example, 2+ −1 3+ −1 for [V(H2O)6] , Δo = 11,800 cm ; for [V(H2O)6] , Δo = 17,850 cm . Factor 2: Principal Quantum Number of the Metal For a series of complexes of metals from the same group in the periodic table with the same charge and the same ligands, the magnitude of Δo increases with increasing principal quantum number: Δo (3d) < Δo (4d) < Δo (5d). The data for hexaammine complexes of the trivalent Group 9 metals illustrate this point: 3+ −1 [Co(NH3)6] : Δo = 22,900 cm 3+ −1 [Rh(NH3)6] : Δo = 34,100 cm 3+ −1 [Ir(NH3)6] : Δo = 40,000 cm
The increase in Δo with increasing principal quantum number is due to the larger radius of valence orbitals down a column. In addition, repulsive ligand–ligand interactions are most important for smaller metal ions. Relatively speaking, this results in shorter M–L distances and stronger d orbital–ligand interactions.
Factor 3: The Nature of the Ligands
Experimentally, it is found that the Δo observed for a series of complexes of the same metal ion depends strongly on the nature of the ligands. For a series of chemically similar ligands, the magnitude of Δo decreases as the size of the donor atom − − − increases. For example, Δo values for halide complexes generally decrease in the order F > Cl > Br > I− because smaller, more localized charges, such as we see for F−, interact more strongly with the d orbitals of the metal ion. In addition, a small neutral ligand with a highly localized lone pair, such as NH3, results in significantly larger Δo values than might be expected.
4 9/8/2021 https://chem.libretexts.org/@go/page/6520 Because the lone pair points directly at the metal ion, the electron density along the M–L axis is greater than for a spherical anion such as F−. The experimentally observed order of the crystal field splitting energies produced by different ligands is called the spectrochemical series, shown here in order of decreasing Δo: − − − 2− − − − − − CO ≈ CN > NO2 > en > NH3 > SCN > H2O > oxalate > OH > F > acetate > Cl > Br > I strong-field ligands intermediate-field ligands weak-field ligands
The values of Δo listed in Table 1 illustrate the effects of the charge on the metal ion, the principal quantum number of the metal, and the nature of the ligand.
The largest Δo splittings are found in complexes of metal ions from the third row of the transition metals with charges of at least +3 and ligands with localized lone pairs of electrons.
Example 1
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present. 3− a. [CoF6] − b. [Rh(CO)2Cl2] Given: complexes Asked for: structure, high spin versus low spin, and the number of unpaired electrons Strategy: a. From the number of ligands, determine the coordination number of the compound. b. Classify the ligands as either strong field or weak field and determine the electron configuration of the metal ion. c. Predict the relative magnitude of Δo and decide whether the compound is high spin or low spin. d. Place the appropriate number of electrons in the d orbitals and determine the number of unpaired electrons. Solution a. A With six ligands, we expect this complex to be octahedral. B The fluoride ion is a small anion with a concentrated negative charge, but compared with ligands with localized lone pairs of electrons, it is weak field. The charge on the metal ion is +3, giving a d6 electron configuration.
C Because of the weak-field ligands, we expect a relatively small Δo, making the compound high spin. D In a high-spin octahedral d6 complex, the first five electrons are placed individually in each of the d orbitals with their spins parallel, and the sixth electron is paired in one of the t2g orbitals, giving four unpaired electrons. b. A This complex has four ligands, so it is either square planar or tetrahedral. B C Because rhodium is a second-row transition metal ion with a d8 electron configuration and CO is a strong-field ligand, the complex is likely to be square planar with a large Δo, making it low spin. Because the strongest d-orbital interactions are along the x and y axes, the orbital energies increase in the order dz2dyz, and dxz (these are degenerate); dxy; and dx2−y2.
D The eight electrons occupy the first four of these orbitals, leaving the dx2−y2. orbital empty. Thus there are no unpaired electrons.
Exercise 1
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present. 2+ a. [Mn(H2O)6] 2− b. [PtCl4]
Answer a octahedral; high spin; five
5 9/8/2021 https://chem.libretexts.org/@go/page/6520 Answer b square planar; low spin; no unpaired electrons
Crystal Field Stabilization Energies Recall that stable molecules contain more electrons in the lower-energy (bonding) molecular orbitals in a molecular orbital diagram than in the higher-energy (antibonding) molecular orbitals. If the lower-energy set of d orbitals (the t2g orbitals) is selectively populated by electrons, then the stability of the complex increases. For example, the single d electron in a d1 3+ complex such as [Ti(H2O)6] is located in one of the t2g orbitals. Consequently, this complex will be more stable than expected on purely electrostatic grounds by 0.4Δo. The additional stabilization of a metal complex by selective population of the lower-energy d orbitals is called its crystal field stabilization energy (CFSE). The CFSE of a complex can be calculated by multiplying the number of electrons in t2g orbitals by the energy of those orbitals (−0.4Δo), multiplying the number of electrons in eg orbitals by the energy of those orbitals (+0.6Δo), and summing the two. Table 2 gives CFSE values for octahedral complexes with different d electron configurations. The CFSE is highest for low-spin d6 complexes, which accounts in part for the extraordinarily large number of Co(III) complexes known. The other low-spin configurations also have high CFSEs, as does the d3 configuration. Table 2: CFSEs for Octahedral Complexes with Different Electron Configurations (in Units of Δo)
High Spin CFSE (Δo) Low Spin CFSE (Δo)
t2g eg t2g eg d 0 0 d 1 ↿ 0.4 d 2 ↿ ↿ 0.8 d 3 ↿ ↿ ↿ 1.2 d 4 ↿ ↿ ↿ ↿ 0.6 ↿⇂ ↿ ↿ 1.6 d 5 ↿ ↿ ↿ ↿ ↿ 0.0 ↿⇂ ↿⇂ ↿ 2.0 d 6 ↿⇂ ↿ ↿ ↿ ↿ 0.4 ↿⇂ ↿⇂ ↿⇂ 2.4 d 7 ↿⇂ ↿⇂ ↿ ↿ ↿ 0.8 ↿⇂ ↿⇂ ↿⇂ ↿ 1.8 d 8 ↿⇂ ↿⇂ ↿⇂ ↿ ↿ 1.2 d 9 ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿ 0.6 d 10 ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ 0.0
CFSEs are important for two reasons. First, the existence of CFSE nicely accounts for the difference between experimentally measured values for bond energies in metal complexes and values calculated based solely on electrostatic interactions. Second, CFSEs represent relatively large amounts of energy (up to several hundred kilojoules per mole), which has important chemical consequences. Octahedral d3 and d8 complexes and low-spin d6, d5, d7, and d4 complexes exhibit large CFSEs.
Summary Crystal field theory, which assumes that metal–ligand interactions are only electrostatic in nature, explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. Crystal field theory (CFT) is a bonding model that explains many properties of transition metals that cannot be explained using valence bond theory. In CFT, complex formation is assumed to be due to electrostatic interactions between a central metal ion and a set of negatively charged ligands or ligand dipoles arranged around the metal ion. Depending on the arrangement of the ligands, the d orbitals split into sets of orbitals with different energies. The difference between the energy levels in an octahedral complex is called the crystal field splitting energy (Δo), whose magnitude depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. The spin-pairing energy (P) is the increase in energy that
6 9/8/2021 https://chem.libretexts.org/@go/page/6520 occurs when an electron is added to an already occupied orbital. A high-spin configuration occurs when the Δo is less than P, which produces complexes with the maximum number of unpaired electrons possible. Conversely, a low-spin configuration occurs when the Δo is greater than P, which produces complexes with the minimum number of unpaired electrons possible. Strong-field ligands interact strongly with the d orbitals of the metal ions and give a large Δo, whereas weak-field ligands interact more weakly and give a smaller Δo. The colors of transition-metal complexes depend on the environment of the metal ion and can be explained by CFT.
7 9/8/2021 https://chem.libretexts.org/@go/page/6520 Magnetic Moments of Transition Metals Magnetic moments are often used in conjunction with electronic spectra to gain information about the oxidation number and stereochemistry of the central metal ion in coordination complexes. A common laboratory procedure for the determination of the magnetic moment for a complex is the Gouy method which involves weighing a sample of the complex in the presence and absence of a magnetic field and observing the difference in weight. A template is provided for the calculations involved. For first row transition metal ions in the free ion state, i.e. isolated ions in a vacuum, all 5 of the 3d orbitals are degenerate. A simple crystal field theory approach to the bonding in these ions assumes that when they form octahedral complexes, the energy of the d orbitals are no longer degenerate but are split such that two orbitals, the dx2-y2 and the dz2 (eg subset) are at higher energy than the dxy, dxz, dyz orbitals (the t2g subset). For octahedral ions with between 4 and 7 d electrons, this gives rise to 2 possible arrangements called either high spin/weak field or low spin/strong field respectively. The energy gap is dependent on the position of the coordinated ligands in the SPECTROCHEMICAL SERIES.
Note
6 6 A good starting point is to assume that all Co(III), d complexes are octahedral and LOW spin, i.e. t2g .
In tetrahedral complexes, the energy levels of the orbitals are again split, such that the energy of two orbitals, the dx2−y2 and the dz2 (e subset) are now at lower energy (more favored) than the remaining three dxy , dxz, dyz (the t2 subset) which are destabilized. Tetrahedral complexes are ALL high spin since the difference between the 2 subsets of energies of the orbitals is much smaller than is found in octahedral complexes. The usual relationship quoted between them is: 4 Δ ≈ Δ (1) tet 9 oct Square planar complexes are less common than tetrahedral and d8 e.g. Ni(II), Pd(II), Pt(II), etc, have a strong propensity to form square planar complexes. As with octahedral complexes, the energy gap between the dxy and dx2−y2 is Δoct and these are considered strong field / low spin hence they are all diamagnetic, μ=0 Bohr Magneton (B.M.) The formula used to calculate the spin-only magnetic moment can be written in two forms; the first based on the number of unpaired electrons, n, and the second based on the electron spin quantum number, S. Since for each unpaired electron, n = 1 and S = 1/2 then the two formulae are clearly related and the answer obtained must be identical. −−−−−−− μso = √n(n +2) (2) −−−−−−−− μso = √4S(S +1) (3)
Comparison of calculated spin-only magnetic moments with experimental data for some octahedral complexes
Ion Config μso / B.M. μobs / B.M.
1 1 Ti(III) d (t2g ) √3 = 1.73 1.6-1.7 2 2 V(III) d (t2g ) √8 = 2.83 2.7-2.9 3 3 Cr(III) d (t2g ) √15 = 3.88 3.7-3.9 4 3 1 Cr(II) d high spin (t2g eg ) √24 = 4.90 4.7-4.9 4 4 Cr(II) d low spin (t2g ) √8 = 2.83 3.2-3.3 5 3 2 Mn(II)/ Fe(III) d high spin (t2g eg ) √35 = 5.92 5.6-6.1 5 5 Mn(II)/ Fe(III) d low spin (t2g ) √3 = 1.73 1.8-2.1 6 4 2 Fe(II) d high spin (t2g eg ) √24 = 4.90 5.1-5.7
1 9/8/2021 https://chem.libretexts.org/@go/page/19707 Ion Config μso / B.M. μobs / B.M.
6 6 Co(III) d low spin (t2g ) 0 0 7 5 2 Co(II) d high spin (t2g eg ) √15 = 3.88 4.3-5.2 7 6 1 Co(II) d low spin (t2g eg ) √3 = 1.73 1.8 8 6 2 Ni(II) d (t2g eg ) √8 = 2.83 2.9-3.3 9 6 3 Cu(II) d (t2g eg ) √3 = 1.73 1.7-2.2
Comparison of calculated spin-only magnetic moments with experimental data for some tetahedral complexes
Ion Config μso / B.M. μobs / B.M. Cr(V) d1 (e1) √3 = 1.73 1.7-1.8
2 Cr(IV) / Mn(V) d2 (e ) √8 = 2.83 2.6 - 2.8 3 2 1 Fe(V) d (e t2 ) √15 = 3.88 3.6-3.7 4 2 2 - d (e t2 ) √24 = 4.90 - 5 2 3 Mn(II) d (e t2 ) √35 = 5.92 5.9-6.2 6 3 3 Fe(II) d (e t2 ) √24 = 4.90 5.3-5.5 7 4 3 Co(II) d (e t2 ) √15 = 3.88 4.2-4.8 8 4 4 Ni(II) d (e t2 ) √8 = 2.83 3.7-4.0 9 4 5 Cu(II) d (e t2 ) √3 = 1.73 -
Contributors and Attributions Prof. Robert J. Lancashire (The Department of Chemistry, University of the West Indies)
2 9/8/2021 https://chem.libretexts.org/@go/page/19707 Magnetism Movement of an electrical charge (which is the basis of electric currents) generates a magnetic field in a material. Magnetism is therefore a characteristic property of all materials that contain electrically charged particles and for most purposes can be considered to be entirely of electronic origin.
Figure 1: The Right Hand Rule for an induced magnetic field In an atom, the magnetic field is due to the coupled spin and orbital magnetic moments associated with the motion of electrons. The spin magnetic moment is due to the precession of the electrons about their own axes whereas the orbital magnetic moment is due to the motion of electrons around the nucleus. The resultant combination of the spin and orbital magnetic moments of the constituent atoms of a material gives rise to the observed magnetic properties. Historically, magnetism has been recognized for thousands of years. An account, that is probably apochryphal, tells of a shepherd called Magnes in Crete who around 900 B.C discovered the naturally occurring magnet lodestone (a form of the the spinel magnetite, Fe3O4) in a region later named Magnesia. Supposedly while he was walking over a deposit, the lodestone pulled the nails out of his sandals and the metal tip from his staff.
The Classical Theory of Magnetism The classical theory of magnetism was well developed before quantum mechanics. Lenz's Law states that when a substance is placed within a magnetic field, H, the field within the substance, B, differs from H by the induced field, 4πI, which is proportional to the intensity of magnetization, I. That is; B = H +4πI (1) where B is the magnetic field within the substance and H is the applied magnetic field and I is the intensity of magnetization
Lenz's Law (1834)
Lenz's Law can also be written as B 4πI = 1 + (2) H H or B = 1 +4πκ (3) H where B/H is called the magnetic permeability of the material and κ is the magnetic susceptibility per unit volume, (I/H)
By definition, κ in a vacuum is zero, so under those conditions the equation would reduce to B = H . It is usually more convenient to measure mass than volume and the mass susceptibility, χg, is related to the volume susceptibility, κ, through the density. κ χ = (4) g ρ
1 9/8/2021 https://chem.libretexts.org/@go/page/19711 where ρ is the density.
Finally to get our measured quantity on a basis that can be related to atomic properties, we convert to molar susceptibility
χm = χg ×RMM (5)
Since this value includes the underlying diamagnetism of paired electrons, it is necessary to correct for the diamagnetic portion of χm to get a corrected paramagnetic susceptibility. ′ χm = χm +χdia (6)
Examples of these corrections are tabulated below. Table 1: Table of Diamagnetic Corrections (Pascal's constants, 10-6 c.g.s. units) Ion DC Ion DC
Na+ 6.8 Co2+ 12.8 K+ 14.9 Co3+ 12.8
+ 2+ NH4 13.3 Ni 12.8 Hg2+ 40 VO2+ 12.5 Fe2+ 12.8 Mn3+ 12.5 Fe3+ 12.8 Cr3+ 12.5 Cu2+ 12.8 Cl- 23.4
- 2- Br 34.6 SO4 40.1 I- 50.6 OH- 12
- 2- NO3 18.9 C2O4 34 - - ClO4 32 OAc 31.5 - IO4 51.9 pyr 49.2 CN- 13 Me-pyr 60 NCS- 26.2 Acac- 62.5
H2O 13 en 46.3 EDTA4- ~150 urea 33.4 these can be converted to S.I units of m3 mol-1 by multiplying by 4 π x 10-7 There are numerous methods for measuring magnetic susceptibilities, including, the Gouy, Evans and Faraday methods. These all depend on measuring the force exerted upon a sample when it is placed in a magnetic field. The more paramagnetic the sample, the more strongly it will be drawn toward the more intense part of the field.
Determination of Magnetic Susceptibility The Gouy Method: The underlying theory of the Gouy method is described here and a form for calculating the magnetic moment from the collected data is available as well. The Evans method: The Evans balance measures the change in current required to keep a pair of suspended magnets in place or balanced after the interaction of the magnetic field with the sample. The Evans balance differs from that of the Gouy in that, in the former the permanent magnets are suspended and the position of the sample is kept constant while in the latter the position of the magnet is constant and the sample is suspended between the magnets.
Orbital contribution to magnetic moments From a quantum mechanics viewpoint, the magnetic moment is dependent on both spin and orbital angular momentum contributions. The spin-only formula used last year was given as: −−−−−−−− μs.o. = √4S(S +1) (7)
2 9/8/2021 https://chem.libretexts.org/@go/page/19711 and this can be modified to include the orbital angular momentum −−−−−−−−−−−−−−−−− μS+L = √ 4S(S +1) +L(L +1) (8)
An orbital angular momentum contribution is expected when the ground term is triply degenerate (i.e. a triplet state). These show temperature dependence as well. In order for an electron to contribute to the orbital angular momentum the orbital in which it resides must be able to transform into an exactly identical and degenerate orbital by a simple rotation (it is the rotation of the electrons that induces the orbital contribution). For example, in an octahedral complex the degenerate t2g set of orbitals (dxz,dyx,dyz) can be interconverted by a o 90 rotation. However the orbitals in the eg subset (dz2,dx2-y2) cannot be interconverted by rotation about any axis as the orbital shapes are different; therefore an electron in the eg set does not contribute to the orbital angular momentum and is said to be quenched. In the free ion case the electrons can be transformed between any of the orbitals as they are all degenerate, but there will still be partial orbital quenching as the orbitals are not identical. 3 3 Electrons in the t2g set do not always contribute to the orbital angular moment. For example in the d , t2g case, an electron in the dxz orbital cannot by rotation be placed in the dyz orbital as the orbital already has an electron of the same spin. This process is also called quenching. Tetrahedral complexes can be treated in a similar way with the exception that we fill the e orbitals first, and the electrons in these do not contribute to the orbital angular momentum. The tables in the links below give a list of all d1 to d9 configurations including high and low spin complexes and a statement of whether or not a direct orbital contribution is expected. Octahedral complexes Tetrahedral complexes
A and E ground terms
The configurations corresponding to the A1 (free ion S term), E (free ion D term), or A2 (from F term) do not have a direct contribute to the orbital angular momentum. For the A2 and E terms there is always a higher T term of the same multiplicity as the ground term which can affect the magnetic moment (usually by a only small amount).
μeff = μs.o.(1 −αλ/Δ) (9) where α is a constant (2 for an E term, 4 for an A2 term) and λ is the spin-orbit coupling constant which is generally only available for the free ion but this does give important information since the sign of the value varies depending on the orbital occupancy. Table 1: Some spin-orbit coupling constants for 1st row TM ions metal ion Ti(III) V(III) Cr(III) Mn(III) Fe(II) Co(II) Ni(II) Cu(II)
d 1 2 3 4 6 7 8 9 configuration
λ / cm-1 155 105 90 88 -102 -172 -315 -830
1 4 6 9 For d to d the value is positive hence μeff is less than μso and for d to d the value is negative hence μeff is greater than μso . Δ is the crystal field splitting factor which again is often not available for complexes. 2- For the tetrahedral Co(II) ion, CoCl4 , the observed experimental magnetic moment, μobs = 4.59 Bohr Magneton (B.M.) The spin-only magnetic moment, μs.o. = 3.88 B.M. which is not in good agreement. How can we improve the analysis?
4 4 Since the ground term in the tetrahedral field is split from a F to a A2 term then we can apply the Equation 9. For an A term the constant α = 4. The spin-orbit coupling constant, λ for the free ion is -172 cm-1 which we can use as an approximation and Δ= 3100 cm-1. Hence
μeff = 3.88 ×(1 −(4 ∗ −172)/3100) (10) which comes out at μeff = 4.73 B.M. This gives a much better fit than the spin-only formula. In the case of the series;
3 9/8/2021 https://chem.libretexts.org/@go/page/19711 2- 2- 2- 2- CoI4 , CoBr4 , CoCl4 , Co(NCS)4 the magnetic moments have been recorded as 4.77, 4.65, 4.59, 4.40 BM assuming that λ is roughly a constant, then this variation shows the inverse effect of the spectrochemical series on the magnetic moment, since Δ is expected to increase from I- to NCS-.
T ground terms
The configurations corresponding to the T2 term (from D) or a T1 term (from an F term) are those where there is a direct contribution to orbital angular momentum expected. The magnetic moments of complexes with T terms are often found to show considerable temperature dependence. This is as a result of spin-orbit coupling that produces levels whose energy differences are frequently of the order kT, so as a result, temperature will have a direct effect on the population of the levels arising in the magnetic field.
In a Kotani plot μeff is plotted against kT/λ and when this corresponds to a value of 1 then μ equals the "spin-only" value. If this is extrapolated to infinity then the value corresponds to μS+L.
Measuring the magnetic moment at 80 K and 300 K often shows up this variation with temperature.
Example 1:
Account for the magnetic moments of the complex, (Et4N)2[NiCl4] recorded at 80, 99 and 300 K.
80 K 99 K 300 K
3.25 B.M. 3.43 B.M. 3.89 B.M.
Ni2+ is a d8 metal ion. The formula suggests a 4 coordinate complex and we can assume that the complex is tetrahedral with a d electron 4 4 configuration of e t2 therefore the spin-only magnetic moment can be calculated as 2.83 BM.
Why did we ignore the possibility of it being square-planar? The free ion Russell-Saunders ground term is 3F (L=3 and S=1) which will give rise to a lowest energy T term in a tetrahedral field and hence the resultant magnetic moment is expected to be temperature dependent and have a direct orbital contribution. The observed values may be quite different then to the calculated spin only magnetic moment. The value of μS+L can be calculated as: −−−−−−−−−−−−−−−−− muS+L = √ 4S(S +1) +L(L +1)
or −−−−− μS+L = √8 +12
or −− μS+L = √20 = 4.472 B. M.
4 9/8/2021 https://chem.libretexts.org/@go/page/19711 8 2+ From the observed values it can be seen that the magnetic moment of the d Ni complex is intermediate between the μso and μS+L values (probably due to partial quenching of the orbital angular momentum contribution) and is dependent on temperature. Further worked examples and some selected magnetic data are available.
Contributors and Attributions Prof. Robert J. Lancashire (The Department of Chemistry, University of the West Indies)
5 9/8/2021 https://chem.libretexts.org/@go/page/19711 Metals, Tetrahedral and Octahedral
Figure 1: The dz² (left) and dx²−y² (right) orbitals. Images used with permission from Wikipedia.
Let us continue to consider an octahedral complex. The remaining d orbitals, dxy, dxz and dyz see their energy increase to a lesser extent. We refer to the dxy, dxz and dyz orbitals collectively as the t2g d orbitals. Hybrid orbital theory can be used to describe how metals bond to ligands. When metals bond to ligands, magnetic data shows that some electrons are paired when there is no obvious reason for them to be paired. Molecular orbitals begin to account for this phenomenon by allowing wave functions to interfere in a constructive, low energy (bonding) or destructive, or high energy (antibonding) manner. Thus, the electrons can fill the lowest energy molecular orbitals available to them. However, the electron pairing may be different if the electrons were allowed to fill the lowest energy atomic orbitals available to them.
This diagram shows the field splitting of a metal with ligands in an octahedral configuration. The thick horizontal lines represent atomic orbitals of the metal (left) and ligands (right). The colors correspond to s (black), p (green) and d (red) orbitals. The middle column of horizontal lines represents molecular orbitals made of bonding (lower energy) and antibonding (higher energy) components. If the ligands are oriented on the cartesian coordinate axes, the metal will still own 3 d orbitals, xy, xz, and xz, which do not intersect (and therefore do not interact) with the ligands. These are considered "nonbonding" orbitals, and are represented by dotted lines in the diagram.
References 1. Jean, Yves; "molecular orbitals of transition metal complexes"; Oxford University Press, 2005
1 9/8/2021 https://chem.libretexts.org/@go/page/532 Non-octahedral Complexes
Learning Objectives Understand the d-orbital degeneracies of square planar and tetrahedral metal complexes.
Tetragonal and Square Planar Complexes + If two trans- ligands in an octahedral complex are either chemically different from the other four (as in trans-[Co(NH3)4Cl2] ), or at a different distance from the metal than the other four, the result is a tetragonally distorted octahedral complex. The electronic structures of such complexes are best viewed as the result of distorting an octahedral complex. Consider, for 3+ example, an octahedral complex such as [Co(NH3)6] : two trans- NH3 molecules are slowly removed from the metal along the ±z axes, as shown in the top half of Figure 1. As the two axial Co–N distances increase simultaneously, the d-orbitals that interact most strongly with the two axial ligands decrease in energy due to a decrease in electrostatic repulsions between the electrons in these orbitals and the negative ends of the ligand dipoles. The affected d orbitals are those with a component along the ±z axes—dz2, dxz, and dyz . These orbitals are not affected equally, however: because the dz2 orbital points directly at the two ligands being removed, its energy will decrease much more rapidly than the degenerate energies of the dxz and dyz, as shown in the bottom half of Figure 1. In addition, the effective positive charge on the metal increases somewhat as the axial ligands are removed, increasing the attraction between the four remaining ligands and the metal. This increases the extent of their interactions with the other two d orbitals and increases their energies. Again, the two d orbitals are not affected equally: because the dx2−y2 orbital points directly at the four in-plane ligands, its energy increases to a greater extent than the energy of the dxy orbital, which points between the in-plane ligands. If the two axial ligands are moved infinitely far away from the metal, a square planar complex is formed. The energy of the dxy orbital actually surpasses that of the dz2 orbital in the process. The largest orbital splitting in a square planar complex, between the dx2−y2 and dxy energy levels, is identical in magnitude to Δo.
Figure 1 d-Orbital Splittings for Tetragonal and Square Planar Complexes. (CC BY-SA-NC; anonymous by request) Moving the two axial ligands away from the metal ion along the z axis initially generates an elongated octahedral complex (the center compound of Figure 1) and eventually produces a square planar complex (right). As shown below the structures, an
d 2 2 d axial elongation causes the dz2 dxz and dyz orbitals to decrease in energy and the x −y and xy orbitals to increase in energy. The change in energy is not the same for all five d orbitals. The dz2 orbital has a smaller xy component than does the dxy orbital, so it reaches a lower energy level; thus, the order of these orbitals is reversed.
Anonymous 1 9/5/2021 https://chem.libretexts.org/@go/page/11223 Tetrahedral Complexes In a tetrahedral arrangement of four ligands around a metal ion, none of the ligands lies on any of the three coordinate axes (illustrated in part (a) in Figure 2); consequently, none of the five d orbitals points directly at the ligands. Nonetheless, the dxy, dxz, and dyz orbitals interact more strongly with the ligands than do dx2−y2 and dz2 ; this again results in a splitting of the five d orbitals into two groups. The splitting of the energies of the orbitals in a tetrahedral complex (Δt) is much smaller than that for an octahedral complex (Δo), however, for two reasons: first, the d orbitals interact less strongly with the ligands in a tetrahedral arrangement; second, there are only four negatively-charged regions rather than six, which decreases the electrostatic interactions by one-third if all other factors are equal. It can be shown that for complexes of the same metal ion with the same Δ = 4 Δ charge, the same ligands, and the same M–L distance, t 9 o . The relationship between the splitting of the five d orbitals in octahedral and tetrahedral crystal fields imposed by the same ligands is shown schematically in part (b) in Figure 2.
Figure 2: d-Orbital Splittings for a Tetrahedral Complex. (a) In a tetrahedral complex, none of the five d orbitals points directly at or between the ligands. (b) Because the dxy, dxz, and dyzorbitals (the t2g orbitals) interact more strongly with the ligands than do the dx2−y2 and dz2 orbitals (the eg orbitals), the order of orbital energies in a tetrahedral complex is the opposite of the order in an octahedral complex. (CC BY-SA-NC; anonymous by request)
Δt < Δo because of weaker d-orbital–ligand interactions and decreased electrostatic interactions.
Example 1: Predicting Structure
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present. 2+ 1. [Cu(NH3)4] 2− 2. [Ni(CN)4] Solution
Because Δo is so large for the second- and third-row transition metals, all four-coordinate complexes of these metals are square planar due to the much higher crystal field stabilization energy (CFSE) for square planar versus tetrahedral structures. The only exception is for d10 metal ions such as Cd2+, which have zero CFSE and are therefore tetrahedral as predicted by the VSEPR model. Four-coordinate complexes of the first-row transition metals can be either square planar or tetrahedral; the former is favored by strong-field ligands, whereas the latter is favored by weak-field ligands. For example, 2− 2− the [Ni(CN)4] ion is square planar, while the [NiCl4] ion is tetrahedral. 1. The copper in this complex is a d9 ion and it has a coordination number of 4. So it is probably either square planar or tetrahedral. To estimate which, we need to fill in the CFT splitting diagrams for each with 9 electrons and ask which has a lower energy. Comparing the square planer (Figure 1) splitting diagram with tetrahedral (Figure 2), suggests that 9 electrons will have a net lower total energy for square planar (since the dx2−y2 orbital is high in energy, the others are lower). For the square planar structure, neither high nor low spin states are possible (only one state) with a single unpaired electron.
Anonymous 2 9/5/2021 https://chem.libretexts.org/@go/page/11223 2. The nickle in this complex is a d8 ion and it has a coordination number of 4. So it is probably either square planar or tetrahedral. To estimate which, we need to fill in the CFT splitting diagrams for each with 8 electrons and ask which has a lower energy. Comparing the square planer (Figure 1) splitting diagram with tetrahedral (Figure 2), suggests that 8 electrons will have a net lower total energy for square planar (since the dx2−y2 orbital is high in energy, the others are lower).
For the square planar structure, it is a low spin complex since a high spin requires a lot of energy to promote to the dx2−y2 orbital. Hence, there are no unpaired electrons
Exercise 1
What are the geometries of the following two complexes? − 1. [AlCl4] + 2. [Ag(NH3)2]
Answer 1 tetrahedral Answer 2 linear
Summary Distorting an octahedral complex by moving opposite ligands away from the metal produces a tetragonal or square planar arrangement, in which interactions with equatorial ligands become stronger. Because none of the d orbitals points directly at the ligands in a tetrahedral complex, these complexes have smaller values of the crystal field splitting energy Δt. In tetrahedral molecular geometry, a central atom is located at the center of four substituents, which form the corners of a tetrahedron. Tetrahedral geometry is common for complexes where the metal has d0 or d10 electron configuration. The CFT diagram for tetrahedral complexes has dx2−y2 and dz2 orbitals equally low in energy because they are between the ligand axis and experience little repulsion. In square planar molecular geometry, a central atom is surrounded by constituent atoms, which form the corners of a square on the same plane. The square planar geometry is prevalent for transition metal complexes with d8 configuration. The CFT diagram for square planar complexes can be derived from octahedral complexes yet the dx2−y2 level is the most destabilized and is left unfilled.
Anonymous 3 9/5/2021 https://chem.libretexts.org/@go/page/11223 Octahedral vs. Tetrahedral Geometries How do we tell whether a particular complex is octahedral, tetrahedral, or square planar? Obviously if we know the formula, we can make an educated guess: something of the type ML6 will almost always be octahedral (there is an alternative geometry for 6-coordinate complexes, called trigonal prismatic, but it's pretty rare), whereas something of formula ML4 will usually be tetrahedral unless the metal atom has the d8 electron configuration, in which case it will probably be square planar. But what if we take a particular metal ion and a particular ligand? Can we predict whether it will form an octahedral or a tetrahedral complex, for example? To an extent, the answer is yes... we can certainly say what factors will encourage the formation of tetrahedral complexes instead of the more usual octahedral. The Crystal Field Stabilization Energy (CFSE) is the additional stabilization gained by the splitting of the orbitals according to the crystal field theory, against the energy of the original five degenerate d orbitals. So, for example, in a d1situation such as 3+ [Ti(OH2)6] , putting the electron into one of the orbitals of the t2g level gains -0.4 Δo of CFSE. Generally speaking, octahedral complexes will be favored over tetrahedral ones because: It is more (energetically) favorable to form six bonds rather than four
The CFSE is usually greater for octahedral than tetrahedral complexes. Remember that Δo is bigger than Δtet (in fact, Δtet is approximately 4/9 Δo).
If we make the assumption that Δtet = 4/9 Δo, we can calculate the difference in stabilization energy between octahedral and tetrahedral geometries by referencing everything in terms of Δo.
Example 1: d3 Stabilized Structures
Which is the preferred configuration for a d3 metal: tetrahedral or octahedral? Solution To answer this, the Crystal Field Stabilization Energy has to be calculated for a (d3 metal in both configurations. The geometry with the greater stabilization will be the preferred geometry. For a d3 octahedral configuration, the Crystal Field Stabilization Energy is
3 ×−0.4Δo = −1.2Δo (1) For a d3 tetrahedral configuration (assuming high spin), the Crystal Field Stabilization Energy is
−0.8Δtet (2)
Remember that because Δtet is less than half the size of Δo, tetrahedral complexes are often high spin. We can now put this in terms of Δo (we can make this comparison because we're considering the same metal ion and the same ligand: all that's changing is the geometry) So for tetrahedral d3, the Crystal Field Stabilization Energy is:
CFSE = -0.8 x 4/9 Δo = -0.355 Δo. And the difference in Crystal Field Stabilization Energy between the two geometries will be:
1.2 - 0.355 = 0.845 Δo.
If we do a similar calculation for the other configurations, we can construct a Table of Δo, Δtet and the difference between them (we'll ignore their signs since we're looking for the difference between them). Table 1: Crystal Field Stabilization Energies (not splitting parameters). This table compares the values of the CFSE for octahedral and tetrahedral geometries, assuming high spin configurations. The units are Δo, and we're assuming that Δtet = 4/9 Δo. Octahedral Tetrahedral Difference
d0, d5, d10 0 0 0 d1, d6 0.4 0.27 0.13 d2, d7 0.8 0.53 0.27
1 9/8/2021 https://chem.libretexts.org/@go/page/11274 Octahedral Tetrahedral Difference
d3, d8 1.2 0.36 0.84 d4, d9 0.6 0.18 0.42
These values can be plotted:.
CFSE
Δtet Δoct
1.2Δo
x 0 1 2 3 4 5 6 7 8 9 10
Figure 1: Crystal Field Stabilization Energy as a function of d-electrons for a hypothetical molecule in the octahedral (green curve) and tetrahedral (blue curve) geometries. (CC BY-NC; Ümit Kaya) Notice that the Crystal Field Stabilization Energy almost always favors octahedral over tetrahedral in most cases, but the degree of favorability varies with the electronic configuration. In other words, for d1 there's only a small gap between the oct and tet lines, whereas at d3 and d8 there's a big gap. However, for d0, d5 high spin and d10, there is no CFSE difference between octahedral and tetrahedral. The ordering of favorability of octahedral over tetrahedral is: d3, d8 > d4, d9> d2, d7 > d1, d6 > d0, d5, d10
The units of the graph are Δo. So if we have strong field ligands present, Δo will be bigger anyway (according to the spectrochemical series), and any energy difference between the oct and tet lines will be all the greater for it. A bigger Δo might also push the complexes over to low spin. Similarly, as we saw previously, high oxidation states and metals from the 2nd and
3rd rows of the transition series will also push up Δo. On the other hand, if large or highly charged ligands are present, they may suffer large interligand repulsions and thus prefer a lower coordination number (4 instead of 6). Consequently if you set out to make something that would have a tetrahedral geometry, you would use large, negatively charged, weak field ligands, and use a metal atom with a d0, d5 or d10 configuration from the first row of the transition series (though of course having weak field ligands doesn't matter in these three configurations because the difference between oct and tet is 0 Δo). As Table 2 shows, you can find tetrahedral complexes for most configurations, but there are very few for d3 and d8. Table 2: Tetrahedral complexes of different d electron counts 0 - 5 2- d MnO4 d MnCl4 1 - 6 2- d TiCl4 d FeCl4 2 7 2- d Cr(OR)4 d CoCl4
Contributors and Attributions Dr Mike Morris, March 2001
2 9/8/2021 https://chem.libretexts.org/@go/page/11274 Orgel Diagrams Orgel diagrams are useful for showing the energy levels of both high spin octahedral and tetrahedral transition metal ions. They ONLY show the spin-allowed transitions. For complexes with D ground terms only one electronic transition is expected and the transition energy corresponds directly to D. Hence, the following high spin configurations are dealt with: d1, d4, d6 and d9.
D Orgel diagram On the left hand side d1, d6 tetrahedral and d4, d9 octahedral complexes are covered and on the right hand side d4, d9 tetrahedral and d1, d6 octahedral. For simplicity, the g subscripts required for the octahedral complexes are not shown.
For complexes with F ground terms, three electronic transitions are expected and D may not correspond directly to a transition energy. The following configurations are dealt with: d2, d3, high spin d7 and d8.
F Orgel diagram On the left hand side, d2, d7 tetrahedral and d3, d8 octahedral complexes are covered and on the right hand side d3, d8 tetrahedral and d2 and high spin d7 octahedral. Again for simplicity, the g subscripts required for the octahedral complexes are not shown.
On the left hand side, the first transition corresponds to D, the equation to calculate the second contains expressions with both D and C.I. (the configuration interaction from repulsion of like terms) and the third has expressions which contain D, C.I. and the Racah parameter B. 4 4 T2g <--- A2g transition energy = D 4 4 T1g(F) <--- A2g transition energy = 9/5 *D - C.I. 4 4 T1g(P) <--- A2g transition energy = 6/5 *D + 15B' + C.I. On the right hand side, the first transition can be unambiguously assigned as: 3 3 T2g <--- T1g transition energy = 4/5 *D + C.I. But, depending on the size of the ligand field (D) the second transition may be due to: 3 3 A2g <--- T1g transition energy = 9/5 *D + C.I. for a weak field or
Robert J. Lancashire 1 9/5/2021 https://chem.libretexts.org/@go/page/11252 3 3 T1g(P) <--- T1g transition energy = 3/5 *D + 15B' + 2 * C.I. for a strong field.
Tanabe-Sugano Diagrams An alternative method is to use Tanabe Sugano diagrams, which are able to predict the transition energies for both spin- allowed and spin-forbidden transitions, as well as for both strong field (low spin), and weak field (high spin) complexes. In this method the energy of the electronic states are given on the vertical axis and the ligand field strength increases on the horizontal axis from left to right. Linear lines are found when there are no other terms of the same type and curved lines are found when 2 or more terms are repeated. This is as a result of the "non-crossing rule". The baseline in the Tanabe-Sugano diagram represents the lowest energy or ground term state.
case (not many examples documented) 3+ The electronic spectrum of the V ion, where V(III) is doped into alumina (Al2O3), shows three major peaks with -1 -1 -1 frequencies of: ν1 =17400 cm , ν2 =25400 cm and ν3 =34500 cm . These have been assigned to the following spin-allowed transitions. 3 3 T2g<--- T1g 3 3 T1g(P)<--- T1g 3 3 A2g<--- T1g
ν1 The ratio between the first two transitions is calculated as ν2 which is equal to 25400 / 17400 = 1.448.
To calculate the Racah parameter, B, the position on the horizontal axis where the ratio between the lines representing ν2 and ν1 is equal to 1.448, has to be determined. On the diagram below, this occurs at D/B=30.9. Having found this value, a vertical line is drawn at this position.
Figure 1: Tanabe-Sugano diagram for d2 octahedral complexes On moving up the line from the ground term to where lines from the other terms cross it, we are able to identify both the spin-forbidden and spin-allowed transition and hence the total number of transitions that are possible in the electronic spectrum. Next, find the values on the vertical axis that correspond to the spin-allowed transitions so as to determine the values of n1/B, n2/B and n3/B. From the diagram above these are 28.78, 41.67 and 59.68 respectively.
Robert J. Lancashire 2 9/5/2021 https://chem.libretexts.org/@go/page/11252 Knowing the values of n1, n2 and n3, we can now calculate the value of B. Since n1/B=28.78 and n1 is equal to 17,400 cm-1, then n 17400 B = 1 = (1) 28.78 28.78 or B = 604.5 cm−1 (2)
Then it is possible to calculate the value of D. Since D/B=30.9, then: D=B*30.9 and hence: D = 604.5 * 30.9 = 18680 cm- 1 case 3+ 3+ -1 -1 -1 Calculate the value of B and D for the Cr ion in [Cr(H2O)6)] if n1=17000 cm , n2=24000 cm and n3=37000 cm .
Solution These values have been assigned to the following spin-allowed transitions. 4 4 T2g ← A2g (3) 4 4 T1g ← A2g (4)
4 4 T1g(P ) ← A2g (5)
From the information given, the ratio \(n_2 / n_1 = 24000 / 17000 = 1.412\). Using a Tanabe-Sugano diagram for a d3 system this ratio is found at D/B=24.00
Figure 2: Tanabe-Sugano diagram for d3 octahedral complexes Interpolation of the graph to find the Y-axis values for the spin-allowed transitions gives: \(\dfrac{n_1}{B}=24.00\) \(\dfrac{n_2}{B}=33.90\) \(\dfrac{n_3}{B}=53.11\) −1 Recall that n1 = 17000 cm . Therefore for the first spin-allowed transition, 17000 /B =24.00 from which B can be obtained, B=17000 / 24.00 or B=708.3 cm-1. This information is then used to calculate D. Since D / B=24.00 then D = B*24.00 = 708.3 * 24.00 = 17000 cm-1.
Robert J. Lancashire 3 9/5/2021 https://chem.libretexts.org/@go/page/11252 It is observed that the value of Racah parameter B in the complex is 708.3 cm-1, while the value of B in the free Cr3+ ion is 1030 cm-1. This shows a 31% reduction in the Racah parameter indicating a strong Nephelauxetic effect. The Nephelauxetic Series is as follows: - 2- - - - - 2- - F >H2O>urea>NH3 >en~C2O4 >NCS >Cl ~CN >Br >S ~I . Ionic ligands such as F-give small reduction in B, while covalently bonded ligands such as I- give a large reduction in B.
The original paper by Tanabe and Sugano[10] had the d5 and d6 diagrams each missing a T term from excited I states. These diagrams were reproduced in the often quoted text by Figgis[12(a)] and so the errors have been perpetuated. An exception is the text by Purcell and Kotz[15] where the missing T terms have been included, however in their case they have ignored lower lying terms from excited D, F, G and H states which for d5 are the main transitions seen in the spin forbidden spectra of Mn(II) complexes.
A set of qualitative diagrams have been drawn for each configuration (which include the missing T terms) and along with the newest release of "Ligand Field Theory and its applications" by Figgis and Hitchman [12(b)] represent the only examples of Tanabe-Sugano diagrams that provide a comprehensive set of terms for spectral interpretation
References 1. Basic Inorganic Chemistry, F.A.Cotton, G.Wilkinson and P.L.Gaus, 3rd edition, John Wiley and Sons, Inc. New York, 1995. 2. Physical Inorganic Chemistry, S.F.A.Kettle, Oxford University Press, New York, 1998. 3. Complexes and First-Row Transition Elements, D.Nicholls, Macmillan Press Ltd, London 1971. 4. The Chemistry of the Elements, N.N.Greenwood and A.Earnshaw, Pergamon Press, Oxford, 1984. 5. Concepts and Models of Inorganic Chemistry, B.E.Douglas, D.H.McDaniel and J.J.Alexander 2nd edition, John Wiley & Sons, New York, 1983. 6. Inorganic Chemistry, J.A.Huheey, 3rd edition, Harper & Row, New York, 1983. 7. Inorganic Chemistry, G.L.Meissler and D.A.Tarr, 2nd edition, Prentice Hall, New Jersey, 1998. 8. Inorganic Chemistry, D.F.Shriver and P.W.Atkins, 3rd edition, W.H.Freeman, New York, 1999. 9. Basic Principles of Ligand Field Theory, H.L.Schlafer and G.Gliemann, Wiley-Interscience, New York, 1969. 10. Y.Tanabe and S.Sugano, J. Phys. Soc. Japan, 9, 1954, 753 and 766. 11. a). Inorganic Electronic Spectroscopy, A.B.P.Lever, 2nd Edition, Elsevier Publishing Co., Amsterdam, 1984. 12. (b). A.B.P.Lever in Werner Centennial, Adv. in Chem Series, 62, 1967, Chapter 29, 430. 13. (a). Introduction to Ligand Fields, B.N.Figgis, Wiley, New York, 1966. 14. (b). Ligand Field Theory and its applications, B.N. Figgis and M.A. Hitchman, Wiley-VCH, New York, 2000. 15. E.Konig, Structure and Bonding, 9, 1971, 175. 16. Y. Dou, J. Chem. Educ, 67, 1990, 134. 17. Inorganic Chemistry, K.F. Purcell and J.C. Kotz, W.B. Saunders Company, Philadelphia, USA, 1977.
Contributors and Attributions Prof. Robert J. Lancashire (The Department of Chemistry, University of the West Indies)
Robert J. Lancashire 4 9/5/2021 https://chem.libretexts.org/@go/page/11252 Qualitative Orgel Diagrams A set of qualitative diagrams have been drawn for each configuration (which include the missing T terms) and along with the newest release of "Ligand Field Theory and its applications" by Figgis and Hitchman [12(b)] represent the only examples of Tanabe-Sugano diagrams that provide a comprehensive set of terms for spectral interpretation.
1 9/8/2021 https://chem.libretexts.org/@go/page/97478 2 9/8/2021 https://chem.libretexts.org/@go/page/97478 3 9/8/2021 https://chem.libretexts.org/@go/page/97478 4 9/8/2021 https://chem.libretexts.org/@go/page/97478 Tanabe-Sugano Diagrams Tanabe-Sugano diagrams are used in coordination chemistry to predict electromagnetic absorptions of metal coordination compounds of tetrahedral and octahedral complexes. The analysis derived from the diagrams can also be compared to experimental spectroscopic data. Armed with spectroscopic data, an approximation to the crystal field splitting energy (10Dq), generated by ligands attached to a metal center, can be found.
Crystal Field Splitting Energy Within Crystal Field Theory, the interaction of the metal and ligand arise from the positive charge of the metal and negative charge on the ligands. The theory is developed by looking at the five degenerate d-orbitals and how the energies are changed on being surrounded by the negative point charges of the ligands. As the ligands are moved closer to the metal the repulsion between the electrons of the metal and ligands break the degeneracy of the d-orbitals. In the case of an octahedral complex 6 ligands surround a metal center with a single pair on each axis. This raises the energies of dx^2-y^2, dz^2 relative to those of dxy 1 dxz dyz. This energy split is called Δoct. The tetrahedral energy split is about 4/9Δoct. Racah Parameters Racah parameters were generated as a means to describe the effects of electron-electron repulsion within the metal complexes. The Racah parameters are A, B and C. In the case of Tanabe-Sugano diagrams each electron configuration split has an energy that can be related by the B value. A is ignored because it is roughly the same for any metal center and C generally approximated as being 1/4B. What B represents is an approximation of the bond strength between the ligand and metal.1 Comparisons between tabulated free ion B and B of a coordination complex is called the nephelauxetic ratio (the effect of reducing electron-electron repulsion via ligands). This effect is what gives rise to the spectrochemical series of ligands described later. β β = complex (1) βfreeion
Parameters
The x-axis in a Tanabe-Sugano diagram is in terms of the crystal field splitting parameter, 10Dq, or Δoct, scaled by the B Racah Parameter. The y-axis is in terms of energy of a electronic transition, E, scaled by B. Each line represents the energy of an electronic state while varying the strength of octahedral ligand field. And while only a few electronic states are spin allowed the spin forbidden electron transitions are included since spin forbidden transitions can appear in spectrum. Each term symbol is created from the splitting of term symbols from spherical to octahedral symmetry. With the relative energy ordering of the states are determined via Hund's rules. Diagrams for d4, d5, d6, and d7 metal ions have a discontinuity in energies as the ligand field is varied. The discontinuity, shown with the vertical line, represents complexes changing from high-spin to low-spin complexes. At the line, the spin pairing energy is equal to the crystal field splitting energy. To the left of the line metal complexes are high-spin as the spin pairing energy is greater than that of the ligand field splitting. To the right of the line metal complexes are low-spin as the spin pairing energy is less than that of the ligand field splitting energy.
Diagrams
d2 Tanabe-Sugano diagram d3 Tanabe-Sugano diagram d4 Tanabe-Sugano diagram d5 Tanabe-Sugano diagram Note: Each of the images is a thumbnail clicking on one will expand the image.
1 9/8/2021 https://chem.libretexts.org/@go/page/533 d6 Tanabe-Sugano diagram d7 Tanabe-Sugano diagram d8 Tanabe-Sugano diagram
How to use the Diagrams 1. Before looking at the diagrams the d-configuration of the metal ion must be determined. 2. Choose the appropriate Tanabe-Sugano diagram matching the d-configuration (http://chemistry.bd.psu.edu/jircitano/TSdiagram.pdf has full page diagrams necessary for measurements). 3. Take a spectrum of the complex and identify λmax for spin-allowed (strong intensity) and spin forbidden (weak intensity) transitions. λ 4. Convert max to wavenumbers and generate energy ratios relative to the lowest allowed transition. (i.e. E2/E1and E3/E1) 5. Using a ruler, slide it across the printed diagram until the E/B ratios between lines is equivalent to the ratios found in step 4.
6. Solve for B using the E/B values (y-axis, step 4) and Δoct/B (x-axis, step 5) to yield the ligand field splitting energy 10Dq.
Example 1: Chromium Splitting
3+ A Cr metal complex has strong transitions and λmax at 431.03 nm, 781.25 nm, and 1,250 nm.
Determine the Δoct for this complex. Solution 1. Cr has 6 electrons. Cr3+ has three electrons so its has a d-configuration of d3 2. Locate the d3 Tanabe-Sugano diagram 3. Convert to wavenumbers: 107(nm/cm) = 8, 000 cm−1 (1) 1250 nm 107(nm/cm) = 13, 600 cm−1 (2) 781.25 nm 107(nm/cm) = 23, 200 cm−1 (3) 431.03 nm 4 4 4 4 4 4 4. Allowed transitions are T1 g ← A2 g , T1 g ← A2 g and T2 g ← A2 g .
Transition Energy cm-1 Ratios to lowest
4 4 T1 g ← A2 g 23,200 2.9 4 4 T1 g ← A2 g 13,600 1.7 4 4 T2 g ← A2 g 8,000 1
5. Sliding the ruler perpendicular to the x-axis of the d3 diagram yields the following values:
Δoct/B 10 20 30 40
Height E(ν3)/B 29 45 64 84
Height E(ν2)/B 17 30 40 51
2 9/8/2021 https://chem.libretexts.org/@go/page/533 Δoct/B 10 20 30 40
Height E(ν1)/B 10 20 30 40
Ratio E(ν3)/E(ν1) 2.9 2.25 2.13 2.1
Ratio E(ν2)/E(ν1) 1.7 1.5 1.33 1.275
6. Based on the two tables above it should be assessed that the Δoct/B value is 10. B is found by finding the dividing E by the height.
Energy cm-1 Height B
23,200 29 800 13,600 17 800 8,000 10 800
7. Next multiply Δoct/B by B to yield the Δoct energy. −1 10 ×800 = 8000 cm = Δoct (4)
Each problem is of varying complexity as several steps may be needed to find the correct Δoct/B values that yield the proper energy ratios.
Nephelauxetic Effect 3+ 2 Imagine you had an abundance of V(H2O)6 (d ) which has two absorptions. If you had no other available metal centers, but an abundance of ligands, the complex's absorption spectrum (therefore its color) could be changed via application of the spectrochemical series: − − 2− − − − − − − 2− − I < Br < S < SCN < Cl < NO3 < N3 < F < OH < C2O4 ≈ H2O < NCS < CH3CN < py (pyridine) < NH3 < en − − (ethylenediamine) < bipy (2,2'-bipyridine) < phen (1,10-phenanthroline) < NO2 < PPh3 < CN ≈ CO
If you wanted higher energy absorptions (shift toward purple colors) you use ligands to the right of H2O and if you wanted redder colors attach ligands to the left of H2O. Here you could measure the nephelauxetic ratio of the ligands to compare how each ligand modifies the B Racah parameter. The series of complexes would also serve as a very good demonstration for a classroom. The reason behind modifying metal ligands has implications for commercial products. Unique colors in ceramics and glass products can be traced back to many metal compounds.
References 1. H.A.O. Hill, P. Day. Physical Methods in Advanced Inorganic Chemistry. John Wiley & Sons, Inc. 1968. 2. Tanabe, Yukito; Sugano, Satoru (1954). "On the absorption spectra of complex ions I". Journal of the Physical Society of Japan 9 (5): 753–766. 3. Tanabe, Yukito; Sugano, Satoru (1954). "On the absorption spectra of complex ions II". Journal of the Physical Society of Japan 9 (5): 766–779 4. Tanabe, Yukito; Sugano, Satoru (1956). "On the absorption spectra of complex ions III". Journal of the Physical Society of Japan 11 (8): 864–877.
Problems 7 1. For a d metal ion determine the energy ratios for allowed transitions at Δoct/B of 20. 6 -1 2. For a d metal ion of Δoct/B = 30 and B=530 cm what would the energies of the 5 allowed transitions be? How many are in the UV-Vis range? How many are in the IR range? 3. Write out the allowed transitions for a d5 metal ion in a E/B> 28 ligand field. 4. A d4 complex exhibits absorptions at 5500 cm-1 (strong) and 31350 cm-1 (weak). What are the transitions that are being
exhibited in the complex? What is the corresponding Δoct for the complex? 7 5. A spectrum of d metal complex seemingly exhibits only two intense transitions. What is the Δoct/B that this situation occurs? Please use reference to specific transitions and energy splitting.
3 9/8/2021 https://chem.libretexts.org/@go/page/533 Answers
1. Δoct/B of 20 yields E/B values of 38, 32, 18. Ratios then are 2.11 and 1.78 -1 2. Δoct/B = 30 yields E/B heights of 27, 40, 57, 65, 85. Energies are then 14310, 21200, 30210, 34450 and 45050 cm . All are in the UV-Vis range. *note you need to infer the E/B value for the last transition as the diagram does not extend that far up. 2 2 2 2 2 2 2 2 3. A2g<- T2g, T1g<- T2g, Eg<- T2g, and A1g<- T2g. -1 4. 31,350/5,500 gives a ratio of 5.7/1. The only Δoct/B value that matches is at 10. B value is then 550 cm . Δoctequals 5500 cm-1. 4 4 5. Three transitions are generated at low Δoct/B. However, at about a value of Δoct/B = 13 the transitions A2g<- T1g, and 4 4 T1g<- T1g have the same energies which results in the appearance of only two absorptions.
Contributors and Attributions Evan Sarina, UC Davis
4 9/8/2021 https://chem.libretexts.org/@go/page/533 Tetrahedral vs. Square Planar Complexes
Learning Objectives Discuss the d-orbital degeneracy of square planar and tetrahedral metal complexes.
Tetrahedral Geometry Tetrahedral geometry is a bit harder to visualize than square planar geometry. Tetrahedral geometry is analogous to a pyramid, where each of corners of the pyramid corresponds to a ligand, and the central molecule is in the middle of the pyramid. This geometry also has a coordination number of 4 because it has 4 ligands bound to it. Finally, the bond angle between the ligands o is 109.5 . An example of the tetrahedral molecule CH4 , or methane.
In a tetrahedral complex, Δt is relatively small even with strong-field ligands as there are fewer ligands to bond with. It is rare for the Δt of tetrahedral complexes to exceed the pairing energy. Usually, electrons will move up to the higher energy orbitals rather than pair. Because of this, most tetrahedral complexes are high spin.
Square Planar Complexes In square planar molecular geometry, a central atom is surrounded by constituent atoms, which form the corners of a square on the same plane. The geometry is prevalent for transition metal complexes with d8 configuration. This includes Rh(I), Ir(I), Pd(II), Pt(II), and Au(III). Notable examples include the anticancer drugs cisplatin (PtCl2(NH3)2). A square planar complex also has a coordination number of 4. The structure of the complex differs from tetrahedral because the ligands form a simple square on the x and y axes. Because of this, the crystal field splitting is also different (Figure 1). Since there are no ligands along the z-axis in a square planar complex, the repulsion of electrons in the dxz, dyz, and the dz2 orbitals are considerably lower than that of the octahedral complex (the dz2 orbital is slightly higher in energy to the "doughnut" that lies on the x,y axis). The dx2−y2 orbital has the most energy, followed by the dxy orbital, which is followed by the remaining orbtails (although dz2 has slightly more energy than the dxz and dyz orbital). This pattern of orbital splitting remains constant throughout all geometries. Whichever orbitals come in direct contact with the ligand fields will have higher energies than orbitals that slide past the ligand field and have more of indirect contact with the ligand fields. So when confused about which geometry leads to which splitting, think about the way the ligand fields interact with the electron orbitals of the central atom.
z
dx² - y²
large Δ low spin dxy x y
dz² d x² - y² dxz dyz
Figure 1: In square planar complexes Δ will almost always be large, even with a weak-field ligand. Electrons tend to be paired rather than unpaired because paring energy is usually much less than Δ. Therefore, square planar complexes are usually low spin. (CC BY-SA; Ümit Kaya) In square planar complexes Δ will almost always be large (Figure 1), even with a weak-field ligand. Electrons tend to be paired rather than unpaired because paring energy is usually much less than Δ. Therefore, square planar complexes are usually low spin.
1 9/8/2021 https://chem.libretexts.org/@go/page/531 [PdCl ]2 − [NiCl ]2 − Advanced: 4 is square planar and 4 is tetrahedral
2 − The molecule [PdCl4] is diamagnetic, which indicates a square planar geometry as all eight d-electrons are paired in [NiCl ]2 − 8 the lower-energy orbitals. However, 4 is also d but has two unpaired electrons, indicating a tetrahedral 2 − − geometry. Why is [PdCl4] square planar if Cl is not a strong-field ligand? Solution 2 − 2 − The geometry of the complex changes going from [NiCl4] to [PdCl4] . Clearly this cannot be due to any change in the ligand since it is the same in both cases. It is the other factor, the metal, that leads to the difference. Consider the splitting of the d-orbitals in a generic d8 complex. If it were to adopt a square planar geometry, the electrons will be stabilized (with respect to a tetrahedral complex) as they are placed in orbitals of lower energy. However, this comes at a cost: two of the electrons, which were originally unpaired in the tetrahedral structure, are now paired in the square-planer structure:
We can label these two factors as ΔE (stabilization derived from occupation of lower-energy orbitals) and P (spin pairing energy) respectively. One can see that: If ΔE > P , then the complex will be square planar If ΔE < P , then the complex will be tetrahedral. This is analogous to deciding whether an octahedral complex adopts a high- or low-spin configuration; where the crystal field splitting parameter Δo ΔE does above. Unfortunately, unlike Δo in octahedral complexes, there is no simple graphical way to represent ΔE on the diagram above since multiple orbitals are changed in energy between the two geometries. Interpreting the origin of metal-dependent stabilization energies can be tricky. However, we know experimentally that 2 + 2 + Pd has a larger splitting of the d-orbitals and hence a larger ΔE than Ni (moreover P is also smaller). 8 Practically all 4d and 5d d ML4 complexes adopt a square planar geometry, irrespective if the ligands are strong-field 2 − − ligand or not. Other examples of such square planar complexes are [PtCl4] and [AuCl4] .
Summary In tetrahedral molecular geometry, a central atom is located at the center of four substituents, which form the corners of a tetrahedron. Tetrahedral geometry is common for complexes where the metal has d0 or d10electron configuration. 2 2 2 The CFT diagram for tetrahedral complexes has dx −y and dz orbitals equally low in energy because they are between the ligand axis and experience little repulsion. In square planar molecular geometry, a central atom is surrounded by constituent atoms, which form the corners of a square on the same plane. The square planar geometry is prevalent for transition metal complexes with d8 configuration. The CFT diagram for square planar complexes can be derived from octahedral complexes yet the dx2-y2 level is the most destabilized and is left unfilled.
2 9/8/2021 https://chem.libretexts.org/@go/page/531 Contributors and Attributions Angad Oberoi (UCD), Justin Nuckles (UCD) StackExchange (orthocresol)
3 9/8/2021 https://chem.libretexts.org/@go/page/531 Thermodynamics and Structural Consequences of d-Orbital Splitting The energy level splitting of the d-orbitals due to their interaction with the ligands in a complex has important structural and thermodynamic effects on the chemistry of transition-metal complexes. Although these two types of effects are interrelated, they are considered separately here.
Structural Effects There are two major kinds of structural effects: effects on the ionic radii of metal ions with regular octahedral or tetrahedral geometries, and structural distortions observed for specific electron configurations.
Ionic Radii Figure 1 is a plot of the ionic radii of the divalent fourth-period metal ions versus atomic number. The dashed line represents the behavior predicted based on the effects of screening and variation in effective nuclear charge (Zeff), assuming a spherical distribution of the 3d electrons. Because these radii are based on the structures of octahedral complexes and Cr2+ and Cu2+ do not form truly octahedral complexes, the points for these ions are shown as open circles.
Figure 1: The Effect of d-Orbital Splittings on the Radii of the Divalent Ions of the Fourth-Period Metals. Only Ca2+(d0), Mn2+ (high-spin d5), and Zn2+ (d10) lie on the smooth dashed curve. All other divalent ions fall below this curve because they have asymmetrical distributions of d electrons. To explain why an asymmetrical distribution of d electrons makes a metal ion 2+ 2 smaller than expected, consider the Ti ion, which has a d configuration with both electrons in the t2g orbitals. Because the t2g orbitals are directed between the ligands, the two d-electrons are unable to shield the ligands from the nuclear charge. Consequently, the ligands experience a higher effective nuclear charge than predicted, the metal–ligand distance is unusually short, and the ionic radius is smaller than expected. If instead the two electrons were distributed uniformly over all five d orbitals, they would be much more effective at screening the ligands from the nuclear charge, making the metal–ligand distances longer and giving the metal a larger ionic radius. 2+ 3 A similar effect is observed for the V ion, which has a d configuration. Because the three electrons in the t2g orbitals provide essentially no shielding of the ligands from the metal, the ligands experience the full increase of +1 in nuclear charge that occurs from Ti2+ to V2+. Consequently, the observed ionic radius of the V2+ ion is significantly smaller than that of the Ti2+ ion. Skipping the Cr2+ ion for the moment, consider the d5 Mn2+ ion. Because the nuclear charge increases by +2 from V2+ to 2+ 2+ 2+ 2+ 2+ Mn , Mn might be expected to be smaller than V . The two electrons added from V to Mn occupy the eg orbitals, however, which are oriented directly toward the six ligands. Because these electrons are localized directly between the metal ion and the ligands, they are effective at screening the ligands from the increased nuclear charge. As a result, the ionic radius actually increases significantly from V2+ to Mn2+, despite the higher nuclear charge of the latter. The same effects are observed in the second half of the first-row transition metals. In the Fe2+, Co2+, and Ni2+ ions, the extra electrons are added successively to the t2g orbitals, resulting in increasingly poor shielding of the ligands from the nuclei and in abnormally small ionic radii. Skipping over Cu2+, adding the last two electrons causes a significant increase in the ionic radius of Zn2+, despite its greater nuclear charge.
1 9/7/2021 https://chem.libretexts.org/@go/page/11224 The Jahn–Teller Effect Because simple octahedral complexes are not observed for the Cr2+ and Cu2+ ions, only estimated values for their radii are 1 2+ 2+ shown in Figure . Since both Cr and Cu ions have electron configurations with an odd number of electrons in the eg orbitals. Because the single electron (in the case of Cr2+) or the third electron (in the case of Cu2+) can occupy either one of two degenerate eg orbitals, both systems have degenerate ground states. The Jahn–Teller theorem states that such non-linear systems are not stable; they undergo a distortion that makes the complex less symmetrical and splits the degenerate states, which decreases the energy of the system. The distortion and resulting decrease in energy are collectively referred to as the Jahn–Teller effect. Neither the nature of the distortion nor its magnitude is specified, and in fact, they are difficult to predict. In principle, Jahn–Teller distortions are possible for many transition-metal ions; in practice, however, they are observed only for 2+ 2+ systems with an odd number of electrons in the eg orbitals, such as the Cr and Cu ions. 2+ 2+ Consider an octahedral Cu complex, [Cu(H2O)6] , which is elongated along the z axis. As indicated in Figure 2, this kind of distortion splits both the eg and t2g sets of orbitals. Because the axial ligands interact most strongly with the dz2 orbital, the splitting of the eg set (δ1) is significantly larger than the splitting of the t2g set (δ2), but both δ1 and δ2 are much, much smaller than the Δo. This splitting does not change the centerpoint of the energy within each set, so a Jahn–Teller distortion results in 4 no net change in energy for a filled or half-filled set of orbitals. If, however, the eg set contains one electron (as in the d ions, Cr2+ and Mn3+) or three electrons (as in the d9 ion, Cu2+), the distortion decreases the energy of the system. For Cu2+, for 2+ example, the change in energy after distortion is 2(−δ1/2) + 1(δ1/2) = −δ1/2. For Cu complexes, the observed distortion is always an elongation along the z axis by as much as 50 pm; in fact, many Cu2+ complexes are distorted to the extent that they are effectively square planar. In contrast, the distortion observed for most Cr2+ complexes is a compression along the z axis. In both cases, however, the net effect is the same: the distorted system is more stable than the undistorted system. Jahn–Teller distortions are most important for d9 and high-spin d4 complexes; the distorted system is more stable than the undistorted one.
Figure 2: The Jahn–Teller Effect Increasing the axial metal–ligand distances in an octahedral d9 complex is an example of a Jahn–Teller distortion, which causes the degenerate pair of eg orbitals to split in energy by an amount δ1; δ1 and δ2 are much smaller than Δo. As a result, the distorted system is more stable (lower in energy) than the undistorted complex by δ1/2.
Thermodynamic Effects As previously noted, crystal field splitting energies (CFSEs) can be as large as several hundred kilojoules per mole, which is the same magnitude as the strength of many chemical bonds or the energy change in most chemical reactions. Consequently, CFSEs are important factors in determining the magnitude of hydration energies, lattice energies, and other thermodynamic properties of the transition metals.
2 9/7/2021 https://chem.libretexts.org/@go/page/11224 Hydration Energies The hydration energy of a metal ion is defined as the change in enthalpy for the following reaction: M 2+ +H O → M 2+ (1) (g) 2 (l) (aq)
Although hydration energies cannot be measured directly, they can be calculated from experimentally measured quantities using thermochemical cycles. In Figure 3a, a plot of the hydration energies of the fourth-period metal dications versus atomic number forms a curve with two valleys. Note the relationship between the plot in Figure 3a and the plot of ionic radii in Figure 1 the overall shapes are essentially identical, and only the three cations with spherically symmetrical distributions of d electrons (Ca2+, Mn2+, and Zn2+) lie on the dashed lines. In Figure 3a, the dashed line corresponds to hydration energies 2+ calculated based solely on electrostatic interactions. Subtracting the CFSE values for the [M(H2O)6] ions from the experimentally determined hydration energies gives the points shown as open circles, which lie very near the calculated curve. Therefore, CFSEs are primarily responsible for the differences between the measured and calculated values of hydration energies.
Figure 3: Thermochemical Effects of d-Orbital Splittings. (a) A plot of the hydration energies of the divalent fourth-period metal ions versus atomic number (solid circles) shows large deviations from the smooth curve calculated, assuming a spherical distribution of d electrons (dashed line). Correcting for CFSE gives the points shown as open circles, which, except for Ti2+ and Cr2+, are close to the calculated values. The apparent deviations for these ions are caused by the fact that solutions of the Ti2+ ion in water are not stable, and Cr2+ does not form truly octahedral complexes. (b) A plot of the lattice energies for the fourth-period metal dichlorides versus atomic number shows similar deviations from the smooth curve calculated, assuming a spherical distribution of d electrons (dashed lines), again illustrating the importance of CFSEs.
Lattice Energies Values of the lattice energies for the fourth-period metal dichlorides are plotted against atomic number in part (b) of Figure 3. Recall that the lattice energy is defined as the negative of the enthalpy change for the reaction below. Like hydration energies, lattice energies are determined indirectly from a thermochemical cycle. 2+ − M (g) +2Cl (g) → MCl2(s) (2)
The shape of the lattice-energy curve is essentially the mirror image of the hydration-energy curve in part (a) of Figure 3, with only Ca2+, Mn2+, and Zn2+ lying on the smooth curve. It is not surprising that the explanation for the deviations from the curve is exactly the same as for the hydration energy data: all the transition-metal dichlorides, except MnCl2 and ZnCl2, are more stable than expected due to CFSE.
Summary Distorting an octahedral complex by moving opposite ligands away from the metal produces a tetragonal or square planar arrangement, in which interactions with equatorial ligands become stronger. Because none of the d orbitals points directly at the ligands in a tetrahedral complex, these complexes have smaller values of the crystal field splitting energy Δt. The crystal field stabilization energy (CFSE) is the additional stabilization of a complex due to placing electrons in the lower-energy set of d orbitals. CFSE explains the unusual curves seen in plots of ionic radii, hydration energies, and lattice energies versus atomic
3 9/7/2021 https://chem.libretexts.org/@go/page/11224 number. The Jahn–Teller theorem states that a non-linear molecule with a spatially degenerate electronic ground state undergoes a geometrical distortion to remove the degeneracy and lower the overall energy of the system.
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