7.4: Ligand Field Theory

Ligand Field Theory

Crystal field theory is successful at providing some general insights into the differing energy levels of d orbitals in coordination complexes. That knowledge can help us to understand some of the magnetic properties of these compounds, which are determined by the number of unpaired electrons. It can also help us to understand some of the absorbed wavelengths of light in the UV-visible spectrum, which in some cases depends on the energy difference between different sets of d orbitals.

Ligand field theory picks up where crystal field theory left off by taking into consideration the importance of orbital overlap for covalent bonding. That factor adds another level of detail to our model of metal complex behavior, and it provides additional nuance to our understanding of why two related complexes may have very different properties.

We can start by taking another look at octahedral coordination geometry. That’s probably the most common coordination geometry. It’s also the one we looked at when we considered crystal field theory. In this geometry, all six ligands are located along the x, y, or z axes.

Once again, only two of the valence d orbitals are located along those axes. In crystal field theory, we considered that orientation as an opportunity for electron-electron repulsion between ligand lone pairs (or charges on neighboring anions) and the d electrons. This time, we’ll think about the opportunity for covalent bonding that exists between ligand orbitals and the d orbitals along these axes.

In addition to the d orbitals, we should also think about the potential for bonding with the metal’s valence s and p orbitals. All three p orbitals are located along the axes where we find the ligands. They are ideally located for bonding with ligand orbitals along those axes. The s orbital is different: it is non-directional. No matter where a ligand is located, if it has sigma bonding, it will have the same overlap with that spherical s orbital. That means the s orbital is a good candidate for forming a covalent sigma bond with any ligand, no matter what the geometry.

That leaves us with six possible metal orbitals that could potentially overlap with the six ligands. Each metal orbital could potentially interact with any one of a number of these ligands. In the illustration below, we’ve made the assumption that the ligand orbitals are p orbitals. That’s not a bad bet given that we are generally considering p-block elements as donor atoms in most ligands. The s orbital could presumably interact with up to six ligands. The metal p orbitals could interact with ligand orbitals in either direction along the axis. By inspection, it appears that the d orbitals could interact with two or four orbitals.

That ratio of six ligand orbitals to six ligand orbitals is coincidental. A one-to-one correspondence between metal orbitals and ligand orbitals isn’t strictly required, but it’s easier to think about that way. Let’s begin by assuming that the s orbital interacts with just one donor orbital. Just like in MO construction for main group diatomics, if we start with two atomic orbitals and combine them, we get two molecular orbitals. One of them is the in-phase combination and the other one is the out-of-phase combination. The number of molecular orbitals we obtain is the same as the number of atomic orbitals we start with.

We can think of these two combinations as the addition of two orbitals together, with a positive coefficient if they are in phase and a negative coefficient if they are out of phase.

ψM-L = aψM(s) + bψL(p)

ψ*M-L = cψM(s) - dψL(p)

Notice that the molecular orbital interaction diagram is asymmetric. The ligand orbital is at lower energy than the metal orbital. The donor atom is a p block element; it’s to the right of the transition metals in the periodic table. The donor atom is more electronegative than the transition metal, so its electrons are at lower energy. In a case like that, the coefficients in the orbital combination follow a pattern. In the in-phase combination, the more electronegative element gets a larger coefficient than the less electronegative element (b > a). The opposite is true in the out-of-phase combination (c > d).

The molecular orbital more closely resembles the atomic orbital to which it is closest in energy, both spatially and energetically. Consequently, we often talk about the bonding orbital, σ, as though it is still a ligand p orbital, and the antibonding orbital, σ, as though it is still a metal s orbital. The bonding orbital is ligand-centered and the antibonding orbital is metal-centered.

If we consider the three metal p orbitals and their overlap with three of the ligand donor orbitals, we find a very similar outcome. The three σ* orbitals are metal-centered, whereas the three σ orbitals are ligand-centered.

The d orbitals, on the other hand, present a significant variation. Only two of them appear to be capable of overlap with ligand p orbitals. If we pair those two d orbitals off with the remaining two ligand donor orbitals, we have three leftover metal d orbitals. Those three metal orbitals do not overlap with the ligand and so they are non-bonding.

We can superimpose those three pictures to get a look at the full molecular orbital interaction diagram. It is still somewhat simplified. We are ignoring any core orbitals on the metal and we are also ignoring any ligand orbitals that aren’t involved in sigma bonding to the metal. Maybe we have ammine ligands but we are ignoring all of those N-H bonds and focusing solely on the M-N bonds.

The six donor atoms are each donating a pair of electrons to the metal. Upon covalent bond formation, each of those lone pairs slides down into an energy well; they are stabilized by formation of the metal-ligand bond.

So far, we have ignored the electrons on the metal because we are looking at a general case. There could be anywhere from zero to ten electrons in the atomic d orbitals, depending on the metal and its oxidation state. These electrons will be crucial to many aspects of the behavior of the complex because they are found at the frontier. Frontier orbitals are often the key to understanding reactivity as well as physical properties such as magnetism and interaction with light.

We often consider only these frontier orbitals when we consider the properties of transition metal complexes. We ignore what seems like the most important part of the picture: the bonding orbitals. We do that because those bonding orbitals are not frontier orbitals. Furthermore, the stability of those bonding orbitals is reflected in the antibonding level. The lower the bonding electrons sink, the higher the antibonding levels rise. One immediate consequence of that connection is that the bond strength may control the gap between nonbonding and antibonding d orbitals. Strong sigma donors provide a large d-d gap, whereas weak sigma donors give a smaller d-d gap. Understanding donor strength can be a difficult task, but remember for example that more basic ligands are often stronger donors.

If we think about a specific case of metal ion, we can add some d electrons to the picture. Suppose we have a d⁴ metal ion. Maybe it’s a Cr²⁺ ion. This ion can take on two different electron configurations in an octahedral environment. With stronger donors, the d-d gap is potentially too large to promote a fourth electron into the next energy level, so the complex is left with two unpaired electrons. With weaker donors, the relatively small d-d gap may be smaller than the energy it takes to put two electrons into the same orbital, so there are four unpaired electrons. We’ll return to the consequences of these situations on an upcoming page.

Pi Donor Ligands

The sigma donor strength of the ligand can have an appreciable effect on the d orbital splitting in the complex, and that might influence the properties of the complex. That’s not something that we considered in crystal field theory, but thinking about bonding interactions has provided another layer of information. What other variations in ligand donation might play a role in the d orbital splitting?

Suppose a donor atom had more than one lone pair. Could it donate a second? On paper, that makes another bond between the ligand and the metal. Bond formation is energy-releasing and stabilizing, although sometimes we get in trouble when we draw too many bonds, because we’ve run out of orbitals to interact with each other.

When we draw a pair of bonds between two atoms, we usually think of the first bond as a sigma bond and the second one as a pi bond. The question here is, can bromine form a pi bond with a transition metal? When we learn about pi bonds, we start by looking at two nitrogen atoms or two carbon atoms using parallel p orbitals to bond with each other. The resulting pi bond avoids the sigma bond because it is above and below the bond axis. We have already established that the metal can use its p orbitals for bonding. On paper, a metal p orbital in an octahedral complex could pi bond with four different ligands.

The trouble with that scheme is that we are already using these metal p orbitals for sigma bonding. We have other metal orbitals that could overlap with these ligand p orbitals, however, and we aren’t using them for sigma bonding. These are the d_xy, d_xz, and d_yz orbitals. This set looks even more promising for pi bonding; the d orbitals reach out towards the ligand p orbitals, maximizing overlap. Once again, each d orbital could potentially form pi bonds with up to four ligands.

Let’s consider the consequences of pi bond formation using the molecular orbital interaction diagram. This time we’re just considering lone pairs on three ligands that could donate to the three different d orbitals of appropriate symmetry: the d_xy, d_yz, and d_xz. We get a bonding and an antibonding combination, but the three ligand pairs all go down in energy. As long as there are fewer than six d orbitals there is a net decrease in energy. Just like before, when these orbitals combine, the bonding combination has more ligand character whereas the antibonding combination has more metal character. The key change in terms of the frontier orbitals is that the d orbital splitting becomes smaller than in the case of the simple sigma bond donor.

Pi Acceptor Ligands

Ligand-to-metal donation is just one way to make a metal-ligand double bond. In those cases in which the metal has d electrons, the double bond could also be formed by donation from the metal to the ligand. The d electrons are most likely to occupy the d_xy, d_xz, and d_yz, as those orbitals are the lowest available ones. Those are the same orbitals that were involved in π bonding in ligand-to-metal donation. In addition to d electrons, there must also be an acceptor orbital on the ligand. That is to say, there must be an empty orbital on the ligand with \(\pi\) symmetry. Instead of a lone pair, this acceptor takes the form of a \(\pi\)* orbital on the ligand. That means there has to be a \(\pi\) bond between the donor atom and a second atom within the ligand. Carbon monoxide is the classic example of such a ligand.

Metal-to-ligand \(\pi\) donation has the effect of weakening a \(\pi\) bond in the ligand because of the fact that electrons are donated into a ligand \(\pi\)* orbital. In terms of Lewis structures, this looks like an interaction that would strengthen the metal-ligand bond but weaken bonds within the ligand itself. When a metal \(\pi\) donates into a carbon monoxide, populating the CO \(\pi\)* orbital, the CO bond gets weaker.

Because we are still looking at \(\pi\) bond formation, the orbital pictures look somewhat similar to the ones for ligand-to-metal π donation, but with \(\pi\)* molecular orbitals instead of p atomic orbitals.

This time, the relevant ligand orbitals are higher in energy than the metal orbitals because they are antibonding. Antibonding orbitals are typically higher in energy than atomic orbitals. That means that when we construct orbital combinations from these pairs, the metal-ligand pi bonding orbital has more metal character. Conversely, the metal-ligand pi antibonding orbital has more ligand character. If there are any d electrons, they will drop into a lower energy well through formation of the pi bond. The ligand \(\pi\)* molecular orbitals are raised in energy, but that change has no energetic consequences since there aren’t any electrons there.

In contrast to ligand-to-metal pi bond donation, metal-to-ligand pi bond formation has the effect of increasing the d orbital splitting. That gives us three different magnitudes of d orbital splitting for three different categories of ligand. The splitting follows the order pi donors < sigma donors < pi acceptors, at least as a rough trend; individual ligands may deviate from this trend for other reasons, such as the strength of sigma donation.

Group Theory as a Tool in Ligand Field Theory

What if the orbital combinations are not obvious? What if you can’t decide by inspection which ligand orbital would overlap with which metal orbital? There is a more general method of evaluating these things using group theory. We should use that approach for the octahedral geometry, because we already have a concrete example of what that should look like. We can use our knowledge of that outcome to build confidence in our results from group theory. Group theory may be a newer approach to us, and so it will be helpful to validate the results.

To get started, we are going to need a character table for an octahedral geometry, and we will need to consider how the bonding orbitals behave under the symmetry operations of this group. You can brush up on point groups and symmetry here.

The top row of the character table for the octahedral point group organizes the symmetry operations and tells us how many of each operation we can find.

Oh

E

8C3

6C2

6C4

3C2 (= C4^2)

i

6S4

8S6

3σh

3σd

Apart from the identity element, there are ten additional symmetry elements in this point group. The C₃ axis passes through the four pairs of opposite faces of the octahedron. We can rotate by either 120 degrees or 240 degrees, making a total of eight operations. Six different C₂ axes pass through opposite edges of the octahedron. Additional C₂ axes are coincident with the C₄ axis passing through opposite corners of the octahedron (because C₂ = C₄², but not C₄¹ or C₄³). There is an inversion center at the position of the metal ion. That inversion element makes possible an S₄ axis coincident with the C₄ axis and an S₆ axis coincident with the C₃ axis. Finally, there are two sets of mirror planes, one set in the equatorial planes and one set bisecting the faces of the octahedron.

We can use the character table to determine appropriate orbitals for bonding with the ligands. A set of vectors is often used for this purpose but p orbitals can be used just as well. We consider how each of the p orbitals will change under a particular symmetry element.

Has a p orbital remained in place, unchanged? That counts as 1. Has it moved into an entirely different position? That counts as 0. Has it remained in place, but with the opposite orientation? That counts as -1.

Looking at each of the symmetry elements gives us a reducible representation.

Oh	E	8C3	6C2	6C4	3C2 (= C4^2)	i	6S4	8S6	3σh	3σd
Γ	6	0	0	2	2	0	0	0	4	2

If we look at the complete character table, we can find the irreducible representations that this reducible representation is composed of.

Oh	E	8C3	6C2	6C4	3C2 (= C4^2)	i	6S4	8S6	3σh	3σd
A1g	1	1	1	1	1	1	1	1	1	1		x^2 + y^2 + z^2
A2g	1	1	-1	-1	1	1	-1	1	1	-1
Eg	2	-1	0	0	2	2	0	-1	2	0		(2z^2 – x^2 – y^2, x^2 - y^2)
T1g	3	0	-1	1	-1	3	1	0	-1	-1	(Rx, Ry, Rz)
T2g	3	0	1	-1	-1	3	-1	0	-1	1		(xz, yz, xy)
A1u	1	1	1	1	1	-1	-1	-1	-1	-1
A2u	1	1	-1	-1	1	-1	1	-1	-1	1
Eu	2	-1	0	0	2	-1	0	1	-2	0
T1u	3	0	-1	1	-1	-3	-1	0	1	1	(x, y, z)
T2u	3	0	1	-1	-1	-3	1	0	1	-1

We find that Γ = A_1g + T_1u + E_g. If we add together the characters for each of the symmetry elements in those three representations, we get the characters in our reducible representation for the ligand p orbitals.

The A_1g representation corresponds to the metal s orbital. The T_1u representation corresponds to the three metal p orbitals. The E_g representation corresponds to two of the d orbitals: 2z² – x² – y² (usually abbreviated as z²) and x² – y². The other three d orbitals (d_xy, d_xz, and d_yz) have T_1u symmetry and are not a match for the set of ligand orbitals that we examined. This is exactly the outcome that we established by looking at metal orbitals and ligand orbitals and deciding which ones would overlap. The match between the two approaches should provide some confidence about the use of character tables to make decisions about symmetry-appropriate bonding.

Example \(\PageIndex{1}\)

The formula for decomposing a reducible representation is based on taking the dot product and normalizing:

\[ai = 1h QN∙ χ(R)∙χ(R)Q \nonumber \]

in which ai is the number of times the irreducible representation appears in the reducible representation; h is the order of the point group (the total number of symmetry operations); N is the number of operations for a given symmetry element, Q; χ(R)is the character of the reducible representation and χ(R)is the character of the irreducible representation.

Use the formula to confirm the finding that, for sigma bonding in an octahedral geometry:

\[ Γ = A_{1g} + T_{1u} + E_g. \nonumber \]

Solution

ai = 1h QN∙ χ(R)∙χ(R)Q

A_1g: ai = 148 [1∙6∙1+8∙0∙1+6∙0∙1+6∙2∙1+3∙2∙1+ 1∙0∙1+6∙0∙1 + 8∙0∙1+3∙4∙1+6∙2∙1] = 148 (6+12+ 6+12+12)= 148∙48=1

A_2g: ai = 148 [1∙6∙1+8∙0∙1+6∙0∙(-1)+6∙2∙(-1)+3∙2∙1+ 1∙0∙1+6∙0∙(-1) + 8∙0∙1+3∙4∙1+6∙2∙(-1)] = 148 (6-12+ 6+12-12)= 148∙0=0

E_g: ai = 148 [1∙6∙2+8∙0∙(-1)+6∙0∙0+6∙2∙0+3∙2∙2+ 1∙0∙2+6∙0∙0 + 8∙0∙(-1)+3∙4∙2+6∙2∙0] = 148 (12+12+ 24)= 148∙48=1

T_1g: ai = 148 [1∙6∙3+8∙0∙0+6∙0∙(-1)+6∙2∙1+3∙2∙(-1)+ 1∙0∙3+6∙0∙1 + 8∙0∙0+3∙4∙(-1)+6∙2∙(-1)] = 148 (18+12- 6-12-12)= 148∙0=0

T_2g: ai = 148 [1∙6∙3+8∙0∙0+6∙0∙1+6∙2∙(-1)+3∙2∙(-1)+ 1∙0∙3+6∙0∙(-1) + 8∙0∙0+3∙4∙(-1)+6∙2∙1] = 148 (18-12- 6-12+12)= 148∙0=0

A_1u: ai = 148 [1∙6∙1+8∙0∙1+6∙0∙1+6∙2∙1+3∙2∙1+ 1∙0∙(-1)+6∙0∙(-1) + 8∙0∙(-1)+3∙4∙(-1)+6∙2∙(-1)] = 148 (6+12+ 6-12-12)= 148∙0=0

A_2u: ai = 148 [1∙6∙1+8∙0∙1+6∙0∙(-1)+6∙2∙(-1)+3∙2∙1+ 1∙0∙(-1)+6∙0∙1 + 8∙0∙(-1)+3∙4∙(-1)+6∙2∙1] = 148 (6-12+ 6-12+12)= 148∙0=0

E_u: ai = 148 [1∙6∙2+8∙0∙(-1)+6∙0∙0+6∙2∙0+3∙2∙2+ 1∙0∙(-2)+6∙0∙0 + 8∙0∙1+3∙4∙(-2)+6∙2∙0] = 148 (12+12- 24)= 148∙0=0

T_1u: ai = 148 [1∙6∙3+8∙0∙0+6∙0∙(-1)+6∙2∙1+3∙2∙(-1)+ 1∙0∙(-3)+6∙0∙(-1) + 8∙0∙0+3∙4∙1+6∙2∙1] = 148 (18+12- 6+12+12)= 148∙48=1

T_2u: ai = 148 [1∙6∙3+8∙0∙0+6∙0∙1+6∙2∙(-1)+3∙2∙(-1)+ 1∙0∙(-3)+6∙0∙1 + 8∙0∙0+3∙4∙1+6∙2∙(-1)] = 148 (18-12- 6+12-12)= 148∙0=0

Γ = A_1g + T_1u + E_g.

Example \(\PageIndex{2}\)

Determine the reducible representation for the symmetry of the pi-bonding orbitals in an octahedral geometry. You can use simple vectors to represent the bias of the p orbitals.

Solution

E: All 12 vectors remain in same place in same orientation. The character is 12.

C₃: An off-axis rotation, so all vectors have moved. The character is 0.

C₂: An off-axis rotation, so all vectors have moved. The character is 0.

C₄: An on-axis rotation, but the vectors are off-axis. Note how the labeled vectors make the movement out of position clear. The character is 0.

C₂ (= C₄²): An on-axis rotation, but this time the four vectors located along the rotational axis just switch bias. The others move out of position. The character is -4.

i: All 12 vectors move to the opposite side of the structure. The character is 0.

S₄: An on-axis rotation, followed by inversion. Note how the labeled vectors make the movement out of position clear. The character is 0.

S₆: An off-axis rotation, so all vectors have moved. The character is 0.

σ_h: Four vectors change position (character is 0); four vectors remain in position and keep original bias (character is 4); four vectors remain in position and switch bias (character is -4); net character is 0.

σ_d: An off-axis reflection, so all vectors have moved. The character is 0.

Oh	E	8C3	6C2	6C4	3C2 (= C4^2)	i	6S4	8S6	3σh	3σd
Γ	12	0	0	0	-4	0	0	0	0	0

Spin Multiplicity

Once we have a d orbital splitting diagram for a particular geometry of a complex, we can populate the diagram with the known number of d electrons for a specific metal ion. If the complex is octahedral and the metal ion has 1, 2, or 3 d electrons, then the electrons will simply go in the lower level, the t_2g orbitals. For the d³ case, one electron will occupy each orbital, with parallel spins.

What about a fourth electron? If the metal ion has a d⁴ configuration, we could imagine two situations. The electron may also occupy one of the t_2g orbitals, which lie at lower energy. To do so, the fourth electron must be spin-paired with the other occupant of that orbital.

That’s the low-spin configuration. The fourth electron has gone into the lower possible orbital rather than the higher possible one. Alternatively, the fourth electron could occupy one of the e_g orbitals. It would be at a higher energy level, but it would avoid that repulsive interaction with the other electron in the t_2g orbital. That would be the high-spin case. The fourth electron has gone into the higher possible orbital rather than the lower one.

High Spin vs Low Spin

The terms “high-spin” and “low-spin” really refer to the net spin of the atom. Each electron has a spin of a certain magnitude, but spin is a vector quantity. If two spins are pointing in the same direction, they add together, so the overall spin of the atom increases. If two spins are pointing in opposite directions, they cancel out, so the overall spin of the atom decreases. Having electrons paired in the same orbital leads to a lower spin for the atom.

The electron configuration of a d⁴ metal ion in an octahedral complex depends broadly on two factors: the difference in energy between the t_2g and e_g levels (the octahedral field splitting, Δ_o) and the energy associated with pairing two electrons in the same orbital. We have already looked at some of the factors that influence the field splitting, so let’s start by looking at that factor.

It is useful to be aware of some general trends in Δ_o. First, comparison between Δ_o measured for +3 cations and for +2 cations of the first-row transition metals manganese, iron, and cobalt shows that charge exerts a significant influence. The higher the charge on an ion, the larger Δ_o becomes. These examples are illustrated in Table 1.

Table \(\PageIndex {1}\). Comparison of Octahedral Field Splitting Between M(II) and M(III) Ions¹

Complex	Δo (cm^-1)	Complex	Δo (cm^-1)
Cr(OH2)6^2+	14,000	Cr(OH2)6^3+	17,600
Mn(OH2)6^2+	7,500	Mn(OH2)6^3+	21,000
Fe(OH2)6^2+	10,000	Fe(OH2)6^3+	14,000

In each case, the M(III) ion has an octahedral field splitting that is significantly larger than the corresponding value in the M(II) case. However, comparisons of the field splitting between metals in different columns is complicated, with no simple trend.

Second, Δ_o is much larger for second-row than for first-row metals of the same group, as shown in Table 2. The value of Δ_o is even larger for third-row transition metals than second-row transition metals. As a result, transition metal ions from the second and third rows are usually low-spin. First-row transition metal ions, with their smaller Δ_o values, are often high spin, but they can also be low spin, and charge is an important factor in determining which case will occur.

Table \(\PageIndex {2}\). Comparison of Octahedral Field Splitting Between Ions from Different Periods of the Periodic Table²

Complex	Δo (cm^-1)	Complex	Δo (cm^-1)
CoCl6^3^-	Not available	Co(NH3)6^3+	23,000
RhCl6^3^-	20,300	Rh(NH3)6^3+	33,900
IrCl6^3^-	24,900	Ir(NH3)6^3+	Not available

Although the available data is limited here, second-row rhodium has a larger octahedral field splitting than first-row cobalt, and third-row iridium displays a larger field splitting than second-row rhodium. Note that these examples are compared using complexes with identical ligands.

The configuration of first-row transition metals is also strongly influenced by the ligands in the complex, indicated in Table 3. For a given ion, we would expect π-acceptors to give relatively large values of Δ_o, whereas σ-donors would give a smaller Δ_o and π-donors would give a smaller value still. There may be some overlap between these groups, because there are stronger and weaker π-acceptors, for example, or stronger and weaker σ-donors, but that is the general trend that we would expect.

Table \(\PageIndex {3}\). Comparison of Octahedral Field Splitting Between Ions from Different Periods of the Periodic Table²

Complex	Δo (cm^-1)	Complex	Δo (cm^-1)
CrBr6^3^-	Not available	NiBr6^2^-	7,000
CrCl6^3^-	13,600	NiCl6^2^-	7,300
Cr(OH2)6^3+	17,400	Ni(OH2)6^2+	8,500
Cr(NH3)6^3+	21,600	Ni(NH3)6^2+	10,800
Cr(CN)6^3-	26,300	Ni(CN)6^2-	Not available

Among the chromium complexes, the π-accepting cyanide gives a much larger value of Δ_o than the π-donating chloride, as expected from molecular orbital considerations. In both the chromium and nickel cases, the ammine ligand, a simple σ-donor, provides a larger splitting that the aquo ligand, which is a π-donor. We may not be able to make predictions about electron configuration based solely on the type of ligand, but generally we would expect that a complex with π-accepting ligandswould be more likely to be low-spin than a complex with π-donating ligands.

In addition to comparing ligands of different classes, we may also want to look at differences between ligands of the same class. Frequently, among σ-donors and π-donors, basicity of the ligand plays a valuable predictive role. For example, when coordinated to Ni(II), chloride results in a larger field splitting than bromide. Chloride is more basic than bromide, as shown by the pKa of the conjugate acids (pK_a = -6 for HCl vs. -9 for HBr), leading to stronger coordination in the case of chloride; that means a lower energy for the donor electrons but a higher energy for the σ* orbital compared to bromide coordination.

Let’s take a look at the other factor that plays a role in determining the electron configuration: the pairing energy. We refer generally to the pairing energy as Π, but it actually has two components. Π_c is the coulombic portion of this energy. It arises from repulsion between two electrons that occupy the same region of space (the same orbital). Repulsion between the electrons causes energy to increase, so Π_c is a positive unit of energy.

The second component involves quantum mechanical exchange between like spins: Π_e is a stabilizing factor. When two electrons of like spin can exchange with each other, energy decreases: Π_e is a negative unit of energy. For example, in a d² octahedral metal, the two electrons are both at the t_2g level and have the same spin; that situation allows them to freely exchange with each other, which results in a decrease in energy.

In a d³ system, the addition of just one more electron leads to a significant lowering of energy because the amount of exchange triples. Now, the electron in the first orbital can exchange with either the electron in the second or the third orbital. The electrons in the second and third orbitals can also exchange with each other. That makes a total of three possible exchanges.

These two factors, repulsion and exchange, contribute to an overall or total pairing energy. Some values of total pairing energy are shown in Table 4.

Table \(\PageIndex {4}\). Comparison of Total Pairing Energies Between M(II) and M(III) Ions³

Complex	Π (cm^-1)	Complex	Π (cm^-1)
Mn(OH2)6^2+	25,500	Mn(OH2)6^3+	28,000
Fe(OH2)6^2+	17,600	Fe(OH2)6^3+	30,000
Co(OH2)6^2+	22,500	Co(OH2)6^3+	21,000

Like the field splitting, pairing energy is somewhat complex and is governed by more than one factor. However, it is worth pointing out that pairing energy is frequently (but not always) larger for more highly charged ions. For example, the pairing energy for both Mn(III) and Fe(III) are greater than the respective values for Mn(II) and Fe(II). That fact reflects the contraction of the more highly-charged ions. Confined to a smaller volume, repulsion between the electrons becomes greater.

In general, notice that the magnitudes of these pairing energies are pretty large compared to many of the octahedral field splitting values in the previous tables. In particular, they are larger than most of the values of Δ_o for first-row transition metals. That’s why many first-row ions are high spin: the energy required to pair electrons in a low-spin configuration is often greater than the energy required to place an electron in the e_g level. The pairing energies are not that large compared to Δ_o for second- and third-row transition metals, so ions of those metals are more likely to be low spin.

So, if we are comparing the energy difference between two configurations, we need to take into account both the energy difference between the two orbital energy levels and the energy difference resulting from spin pairing. That could include both differences in electron-electron repulsion and differences in energy because of exchange.

Problems

Exercise \(\PageIndex{1}\)

Demonstrate the exchanges possible in the following configurations.

Answer

Exercise \(\PageIndex{2}\)

Determine the difference in Π_c between the high spin and low spin configuration in each of the following cases:

a) d⁴

b) d⁵

c) d⁶

Answer

a) d⁴

The difference is: ΔE = ls – hs = Π_c – 0 = Π_c.

b) d⁵

The difference is: ΔE = ls – hs = 2Π_c – 0 = 2Π_c.

c) d⁶

The difference is: ΔE = ls – hs = 2Π_c – 0 = 2Π_c.

Exercise \(\PageIndex{3}\)

Determine the difference in Π_e between the high spin and low spin configuration in each of the following cases:

a) d⁵

b) d⁶

c) d⁷

Answer

a) d⁵

The difference is: ΔE = ls – hs = 4Π_e - 4Π_e = 0

b) d₆

The difference is: ΔE = ls – hs = 6Π_e - 4Π_e = 2Π_e

c) d⁷

The difference is: ΔE = ls – hs = 6Π_e - 5Π_e = Π_e

Exercise \(\PageIndex{4}\)

4. Given the value of Δ_o and Π from the above tables, predict whether the complex will be high spin or low spin.

a) [Mn(OH₂)₆]²⁺

b) [Mn(OH₂)₆]³⁺

c) [Co(OH₂)₆]²⁺ (Δ_o = 14,000 cm^-1)⁴

d) [Co(OH₂)₆]³⁺ (Δ_o = 19,000 cm^-1)⁴

Answer

a) [Mn(OH₂)₆]²⁺

Δ_o = 7,500 cm^-1 and Π = 25,500 cm^-1; because it costs more to pair electrons than to promote one, this complex will be high spin.

b) [Mn(OH₂)₆]³⁺

Δ_o = 21,000 cm^-1 and Π = 28,000 cm^-1; because it costs more to pair electrons than to promote one, this complex will be high spin.

c) [Co(OH₂)₆]²⁺

Δ_o = 10,000 cm^-1 and Π = 22,500 cm^-1; because it costs more to pair electrons than to promote one, this complex will be high spin.

d) [Co(OH₂)₆]³⁺

Δ_o = 19,000 cm^-1 and Π = 21,000 cm^-1; because it costs more to pair electrons than to promote one, this complex will be high spin, but note how similar the values are. Many Co(III) complexes are low spin.

References

1. Dunn, T. M.; McClure, D. S.; Pearson, R. G. Some Aspects of Crystal Field Theory. Harper & Row: New York, 1965, p.82.
2. Sienko, M. A.; Plane, R. A. Physical Inorganic Chemistry. W. A. Benjamin: New York, 1963, p. 56.
3. Miessler, G. L.; Fischer, P. J.; Tarr, D. A. Inorganic Chemistry, 5th Ed. Pearson: Boston, 2014, p. 374.
4. Figgis, B. N.; Hitchman, M. A, Ligand Field Theory and Its Applications. Wiley-VCH: Brisbane, 2000, p. 215

Ligand Field Stabilization Energy

If we want to compare the stability of a particular electron configuration compared to the imaginary d electron configuration in a spherical electric field, we can calculate the ligand field stabilization energy. Remember, in crystal field theory, we compared the d electrons in a spherical field to the situation in which ligands approached in an octahedral geometry. The e_g level is destabilized by 0.6Δ_o compared to undifferentiated d orbitals in a spherical field, whereas the t_2g level is 0.4Δ_o lower in energy than the d orbitals in a spherical field.

Let’s look at the case of a d⁴ complex. That’s an interesting case, because as we fill in the d electrons one by one, it is the first example in which there is the possibility of either a high spin or a low spin configuration. If we first consider the high-spin case, then we see that three of the electrons drop into the t_2g level and the last one goes into the e_g level.

The ligand field stabilization energy is the difference between energy in a spherical field and in an octahedral field. In the high spin d⁴ case, that means three electrons are lower in energy and one is higher in an octahedral environment.

• High spin d⁴: \[ \begin{align*} \text{LFSE} &= [0.6 (1) – 0.4 (3)]Δ_o \\[4pt] &= [0.6 – 1.2]Δ_o \\[4pt] &= -0.6Δ_o \end{align*} \nonumber \]

For comparison, in the low spin case, all four of the electrons are lower in energy in the presence of the octahedral field.

• Low spin d⁴: \[ \begin{align*} \text{LFSE} &= [0.6 (0) – 0.4 (4)]Δ_o \\[4pt] &= -1.6Δ_o \end{align*} \nonumber \]

Of course, the ligand field stabilization energy isn’t the only contributor to energetic differences between possible electron configurations. Pairing energy will also play a role, including Coulombic or repulsive terms as well as exchange terms. The total energy difference between a high spin and low spin configuration will compare those energies as well. For the d⁴ case:

\[\begin{align*} ΔE (\text{low spin – high spin}) &= (-1.6Δ_o + Π_c + 3 Π_e) - (-0.6Δ_o + 3 Π_e) \\[4pt] &= -Δ_o + Π_c \end{align*} \nonumber \]

One application of ligand field stabilization energy is found in the hydration energies of metal ions. As a first approximation, we might expect that the energy released when a bond is formed between a ligand and a metal ion would be related to Coulomb’s Law. If we look at values for the hydration of gas-phase ions, we expect the energies of reaction to become more negative as we move across the transition metals from left to right. That increase in magnitude of the exothermicity of hydration reflects the periodic trend in sizes of the ions. Ions of the same charge get smaller as we go to the right because of the increasing number of protons in the nucleus. As the radius of the atom decreases, the energy released upon binding a ligand increases.

The following graph illustrates this general phenomenon for a series of \(\ce{M^{2+}}\) ions. There are gaps in the graph where data was unavailable. Some of the early transition metals are rarely observed as divalent ions.

Figure \(\PageIndex{1}\): Enthalpy of hydration for M²⁺ ions. Dashed line shows the general (liner) trend. Data from J. Chem. Educ. 1977, 54 (9), 540. https://doi.org/10.1021/ed054p540. (CC BY-NC-SA; Catherine McCusker)

Overall, we can see a general progression towards more negative heats of hydration as we move across the series. That observation is consistent with Coulomb’s Law. However, if we look carefully, we can see that some of the data points are a little higher compared to the rest (or a little lower, depending on your perspective). The higher data points occur at d⁰, d⁵ and d¹⁰ (\(\ce{Ca^{2+}}\), \(\ce{Mn^{2+}}\) and \(\ce{Zn^{2+}}\)). Assuming we are dealing with high spin configurations, which is often the case for the first row of transition metals, then these are exactly the cases in which we expect ligand field stabilization energy to be absent. The table below illustrates that point.

Table 3.3.1. Ligand Field Stabilization Energy for High Spin Configuration

Electron count (dn)	LFSE (Δo)
0	0
1	-0.2
2	-0.4
3	-0.6
4	-0.3
5	0
6	-0.2
7	-0.4
8	-0.6
9	-0.3
10	0

We can confirm that what we are seeing is related to the d electron count, rather than some intrinsic property of individual metals, by looking at a similar series of M³⁺ ions. Once again, we observe the overall agreement with Coulomb’s Law, and we see that the d⁰ cases display a lower heat of hydration than the others because of a lack of ligand field stabilization energy.

Figure \(\PageIndex{2}\): Enthalpy of hydration for M³⁺ ions. Dashed line shows the general (liner) trend. Data from J. Chem. Educ. 1977, 54 (9), 540. https://doi.org/10.1021/ed054p540. (CC BY-NC-SA; Catherine McCusker)

This time, \(\ce{Fe^{3+}}\) is an outlier because it is d⁵, rather than the d⁴ \(\ce{Mn^{3+}}\). Again, some of the data is missing here because some of the later transition metals are rarely found as trivalent ions. Conversely, the ions that exhibit ligand field stabilization energies are depressed from the overall trend, displaying additional stabilization in an octahedral coordination environment.

Neither the hydration energies for the \(\ce{M^{2+}}\) ions nor the hydration energies for the \(\ce{M^{3+}}\) ions track perfectly with ligand field stabilization energies. There are a number of other factors that also impact these observed hydration energies, but they are beyond the scope of the current discussion. These factors include the nephelauxetic (“cloud-expanding”) effect, in which inter-electron repulsion in complexes can be lower than in free ions based on the degree of covalency in the complex.

Example \(\PageIndex{3}\)

Draw a diagram showing the energetic differences between d⁷ metal ions in a spherical field, a weak field octahedral environment, and a strong field octahedral environment.

Solution

Example \(\PageIndex{4}\)

Show how you would calculate the ligand field stabilization energy in the following cases:

1. Low-spin d⁵
2. High-spin d⁵
3. Low-spin d⁷
4. High-spin d⁷

Solution

1. Low-spin d⁵: LFSE = [0.6 (0) – 0.4 (5)]Δ_o = -2.0Δ_o
2. High-spin d⁵: LFSE = [0.6 (2) – 0.4 (3)]Δ_o = [1.2 - 1.2]Δ_o = 0Δ_o
3. Low-spin d⁷: LFSE = [0.6 (1) – 0.4 (6)]Δ_o = [0.6 - 2.4]Δ_o = -1.8Δ_o
4. High-spin d⁷: LFSE = [0.6 (2) – 0.4 (5)]Δ_o = [1.2 - 2.0]Δ_o = -0.8Δ_o

Example \(\PageIndex{5}\)

Calculate the difference between low spin and high spin electronic configurations in the following cases:

1. d⁵
2. d⁷

Solution

1. d⁵: ΔE low spin – high spin = (-2.0Δ_o + 2Π_c + 4Π_e) - (0Δ_o + 4Π_e) = -2Δ_o

2. d⁷: ΔE low spin – high spin = (-1.8Δ_o + 3Π_c + 6Π_e) - (-0.8Δ_o + 2Π_c + 5Π_e) = -Δ_o + Π_c + Π_e

Tetrahedral Geometry

Tetrahedral geometry is one of the two possible geometry options for a 4 coordinate metal complex. Assessing the orbital interactions in tetrahedral geometry is somewhat more complicated, however, and it is common to proceed directly to a group theory approach. Nevertheless, let's take a look at this geometry and see what we can determine through simple observation before we see the results from a more rigorous approach. To begin, it helps to know that a tetrahedral geometry is defined as having four atoms arranged at alternating corners of a cube around a central atom.

If we consider the orientation of the d orbitals, we find that they fall into two different groups. Although all five orbitals lie off-axis with respect to the ligands, some of them are pointed directly at the edges of the cube (d_xy, d_xz, d_yz) whereas the others point at the faces of the cube (d_x²-y², d_z²). The edge-touching orbitals lie a little closer to the ligands; we'll define this distance as r, which in this case is half the edge length of the cube. The face-touching orbitals are slightly farther away: based on the Pythagorean theorem, they are r²+ r² = 2 r² = r² away from the ligands.

Based on that simple observation, we might start to think about the dxy, dxz and dyz group as forming the antibonding orbitals upon interaction with the ligands. The dz2 and dx2-y2 would be left as non-bonding orbitals. This result would be exactly the opposite of the octahedral case. We would therefore expect an orbital splitting diagram that is exactly the inverse of the octahedral one. Maybe the splitting between orbital levels would be a little smaller, though, because of the lack of direct overlap between the ligands and the metal orbitals. After all, even the closest set of metal orbitals don't point directly at the ligands like in the octahedral case.

Group Theory Treatment of Teterhedral Complexes

This supposition is confirmed through a group theory approach. We can use vectors pointing along the ligand-metal axes to examine sigma bonding, as shown below. We would subject these vectors to the symmetry transformations in the tetrahedral space group, T_d, shown in the table.

Td	E	8C3	6C2	6S4	6σd
A1	1	1	1	1	1		x^2 + y^2 + z^2
A2	1	1	1	-1	-1
E	2	-1	2	0	0		(2z^2 - x^2 - y^2, x^2- y^2)
T1	3	0	-1	1	-1	(Rx, Ry, Rz)
T2	3	0	-1	-1	1	(x, y, z)	(xy, xz, yz)
Γσ	4	1	0	0	2		A1 + T2
Γπ	8	-1	0	0	0		E + T1 + T2

That analysis leads to the reducible representation for sigma bonding, Γ_σ , shown in the table. This representation reduces to the irreducible representations, A₁ + T₂. The d orbitals represented here, the T₂ set, are indeed the d_xy, d_xz and d_yz. The non-bonding d orbitals are the E group, corresponding to d_z² and d_x²-y².

We can go further with the group theory approach and to determine the reducible representation for pi bonding, Γ_π , also shown in the table. Pi bonding is otherwise even more difficult to assess via simple inspection than was sigma bonding. The resulting representation reduces to E + T₁ + T₂. The d orbitals represented here include the expected d_z² and d_x²-y². Note, however, that they also include the d_xy, d_xz, and d_yz orbitals. That means that in the presence of a pi-donor ligand, the latter set are antibonding with respect to both sigma and pi bonding.

Example \(\PageIndex{6}\)

Demonstrate these symmetry operations on the drawing of the tetrahedron within a cube shown above.

1. C₃
2. C₂
3. S₄
4. σ_d

Solution

Square Planar Geometry

The other geometry option for a 4-coordinate metal complex is square planar. Square planar geometry is much less common than octahedral, but square planar complexes assert their importance through their frequent appearance in key catalytic processes and other settings. Furthermore, having learned something about bonding in octahedral complexes, we can make some educated guesses about metal-orbital interactions in square planar complexes. Both geometries are nicely described by Cartesian coordinates and so it is relatively easy to draw comparisons between the two.

We can imagine how we might arrive at a square planar geometry simply by taking an octahedral geometry and removing two axial ligands. The four remaining equatorial ligands form a square planar complex.

Since we already know something about the d orbital splitting diagram in an octahedral case, we can draw some rational conclusions about the consequence of this change. The axially-oriented d_z² orbital drops in energy because it is no longer forming the antibonding combination with ligand orbitals along the z axis.

The d_z² orbital does not drop all the way to the non-bonding level, however, because the toroid (the donut around the central node of the orbital) is still in plane with ligand orbitals along the x and y axes. Nevertheless, the d_z² orbital overlaps with these ligand orbitals to a much lesser extent than the d_x²-y² orbitals, so it drops to a level well below the d_x²-y² orbital.

The d orbital splitting diagram shown above is not the one you will normally see for a square planar complex. The three non-bonding orbitals, d_xy, d_xz and d_yz, are degenerate in an octahedral geometry but not in a square planar one. The d_xy orbital is in the plane of the metal and ligands whereas the d_xz and d_yz are above and below that plane.

We therefore might not expect all three of these orbitals to be at the exact same energy level in the square planar coordination environment. We usually think of the d_xy orbital as lying at higher energy than the d_xz, d_yz pair because of the potential for interaction with the ligands that lie in the same plane as the d_xy orbital. The typical drawing of a d orbital splitting diagram reflects that subtle difference, showing the five metal d orbitals lying at four different energy levels with just one degenerate set (the d_xz, d_yz pair).

Note that there is a large splitting and two smaller splittings between the d orbitals, rather than the single splitting observed in an octahedral environment. As a result, when we talk about possible high spin and low spin electron population in the square planar environment, we are generally concerned with whether the electrons can surmount the large splitting and occupy the top orbital, the d_x²-y².

Sometimes, square planar d orbital splitting diagrams show the d_xy orbital above the d_z² orbital and sometimes vice versa; the exact order varies with the ligands involved. The reasons for these differences are somewhat complicated. For example, this order can reflect the importance of pi bonding in a particular complex, as we will see later.

Group Theory Treatment of Square Planar Complexes

In octahedral coordination, we were able to use group theory to confirm the bonding picture we had arrived at through simple observation. We can do the same thing in the square planar case. This time, we need to use a character table for D_4h symmetry.

D4h	E	2C4	C2	2C2'	2C2"	i	2S4	σh	2σv	2σd
A1g	1	1	1	1	1	1	1	1	1	1		x^2+ y^2, z^2
A2g	1	1	1	-1	-1	1	1	1	-1	-1	Rz
B1g	1	-1	1	1	-1	1	-1	1	1	-1		x^2 - y^2
B2g	1	-1	1	-1	1	1	-1	1	-1	1		xy
Eg	2	0	-2	0	0	2	0	-2	0	0	(Rx, Ry)	(xz, yz)
A1u	1	1	1	1	1	-1	-1	-1	-1	-1
A2u	1	1	1	-1	-1	-1	-1	-1	1	1	z
B1u	1	-1	1	1	-1	-1	1	-1	-1	1
B2u	1	-1	1	-1	1	-1	1	-1	1	-1
Eu	2	0	-2	0	0	-1	0	2	0	0	(x, y)

The ten symmetry elements listed in this table may be easier to grasp than the ones in the higher-symmetry octahedral point group. In this case, we see several two- or four-fold axes and some mirror planes. They are illustrated below.

If we consider only sigma bonding from the ligands, which was the initial consideration we thought about earlier, then we could look at how this picture operates when transformed by these symmetry elements:

In that case, we obtain a reducible representation that can be reduced to the following:

Γ_σ = A_1g + B_1g + Eu

Returning to the character table, we find that the matching orbitals on the metal include the d_z² and the d_{x²- y²}, as well as the s (represented by x²+ y²), the p_x and p_y. If we are just interested in the d orbital splitting diagram, that gives us the picture that we had obtained before. Two d orbitals display some antibonding character whereas the other three are non-bonding. Of course, this treatment does not take into account the subtly different interactions of the ligands with the d_z² and the d_{x²- y²} orbitals. Although they are of like symmetry, these orbitals overlap with the ligands to different extents.

Pi Bonding

If we also want to include pi bonding in our understanding of these complexes, we have to think about two different orientations of the ligand p orbitals, which are not symmetrically equivalent in this case. The first orientation is parallel to the plane of the metal-ligand complex. It looks like this:

Treatment of those vectors with the symmetry elements leads to a reducible representation that can be represented by this one:

Γ_π || = A_2g + B_2g + E_u

Consulting the character table, we find that the corresponding orbitals on the central atom are d_xy, p_x and p_y. In reality, the interaction with the d_xy is likely to be much more pronounced than with either the p_x or the p_y because of stronger overlap between the d_xy and the ligand p orbital.

The second orientation is perpendicular to the plane of the complex.

This time, the irreducible representation is as follows:

Γ_π ⊥ = A_2g + B_2g + E_u

According to the character table, this time the corresponding orbitals on the central atom are d_xz, d_yz, and p_z. Once again, because of stronger overlap between the d_xy or d_yz with the ligand p orbital compared to ligand overlap with the metal p_z, the former case is likely to be much more important than the latter.

Thus, we see that the d orbitals that were not originally involved in sigma bonding have the potential to be involved in pi bonding. Which ones will actually be involved depends on the orientations of the ligands that are capable of pi bonding with these orbitals. That may be all of them in a more symmetric case (such as a homoleptic complex, in which all four ligands are the same as each other). It may be fewer in a complex with lower symmetry, in which the overall D_4h symmetry is broken by different ligands.

The nature of the ligands is probably of greater significance in terms of the magnitude of splittings in the d orbital diagram. If the metal forms a pi bond with the ligand via interaction with a p orbital on the ligand, then the resulting pi bond will be closer in both energy and character to the lower-energy ligand p orbital. We still think of that orbital as largely based on the more electronegative ligand. That means that the corresponding antibonding combination is more like the metal orbital in energy and character. It is still mostly a d orbital, for example. That results in a decrease in the splittings between d orbitals as the otherwise non-bonding set is pushed up in energy. We might even see the d_xy orbital at a higher energy level than the nominally sigma-antibonding d_z² orbital, given a strong enough pi-bonding interaction.

On the other hand, if the metal orbital interacts with the empty π* orbital of a ligand such as cyanide or carbon monoxide, this situation will be reversed. Because of its antibonding nature, the ligand π* orbital lies above the metal d orbital in energy. When the two orbitals combine, the ligand π* orbital becomes the metal-ligand π* orbital. The metal d orbital drops in energy to form the metal-ligand bonding combination. Consequently, the metal orbitals involved in pi bonding to a pi acceptor drop in energy and splittings get larger. In particular, the gap between the pi bonding metal orbitals and the purely sigma bonding metal orbitals grows wider in this case.

The difference in overall splitting in these cases can be quite significant. For example, the differences between energy levels, denoted Δ1 and Δ2 below, are about 50% greater with the strongly pi-accepting cyanide ligand than with the pi-donating chloride in the corresponding homoleptic palladium complexes.¹

In all of these cases, the splitting between the highest-lying d_{x²- y²} orbital and the next highest is much larger than the other splittings. That factor leads to square planar complexes generally adopting a low-spin configuration, which in this case means the lower orbitals are all occupied before the d_{x²- y²}. Square planar complexes are most often observed with d⁷ or d⁸ metal ions, which avoids populating that highest-energy d orbital.

Problems

Exercise \(\PageIndex{5}\)

Demonstrate how to arrive at the reducible representation for sigma bonding under the D_4h symmetry of a square planar complex.

Determine the irreducible representations.

Answer

There are 4 unchanged vectors for E. For the others:

Γ_σ: Remember, ai = 1hQN∙χ(R)∙χ(R)Q

A_1g: ai = 1/16 [1∙4∙1 + 2∙0∙1 + 1∙0∙1 + 2∙2∙1 + 2∙0∙1 + 1∙0∙1 + 2∙0∙1 + 1∙4∙1 + 2∙2∙1 + 2∙0∙1] = 1/16 [4 + 4 + 4 + 4] = 1/16(16) = 1

A_2g: ai = 1/16 [1∙4∙1 + 2∙0∙1 + 1∙0∙1 + 2∙2∙(-1) + 2∙0∙(-1) + 1∙0∙1 + 2∙0∙1 + 1∙4∙1 + 2∙2∙(-1) + 2∙0∙(-1)] = 1/16 [4 - 4 + 4 - 4] = 1/16(0) = 0

B_1g: ai = 1/16 [1∙4∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙2∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙0∙(-1) + 1∙4∙1 + 2∙2∙1 + 2∙0∙(-1)] = 1/16 [4 + 4 + 4 + 4] = 1/16(16) = 1

B_2g: ai = 1/16 [1∙4∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙2∙(-1) + 2∙0∙1 + 1∙0∙1 + 2∙0∙(-1) + 1∙4∙1 + 2∙2∙(-1) + 2∙0∙1] = 1/16 [4 - 4 + 4 - 4] = 1/16(0) = 0

E_g: ai = 1/16 [1∙4∙2 + 2∙0∙0 + 1∙0∙(-2) + 2∙2∙0 + 2∙0∙0 + 1∙0∙2 + 2∙0∙0 + 1∙4∙(-2) + 2∙2∙0 + 2∙0∙0] = 1/16 [8 - 8] = 1/16(0) = 0

A_1u: ai = 1/16 [1∙4∙1 + 2∙0∙1 + 1∙0∙1 + 2∙2∙1 + 2∙0∙1 + 1∙0∙(-1) + 2∙0∙(-1) + 1∙4∙(-1) + 2∙2∙(-1) + 2∙0∙(-1)] = 1/16 [4 + 4 - 4 - 4] = 1/16(0) = 0

A_2u: ai = 1/16 [1∙4∙1 + 2∙0∙1 + 1∙0∙1 + 2∙2∙(-1) + 2∙0∙(-1) + 1∙0∙(-1) + 2∙0∙(-1) + 1∙4∙(-1) + 2∙2∙1 + 2∙0∙1] = 1/16 [4 - 4 - 4 + 4] = 1/16(0) = 0

B_1u: ai = 1/16 [1∙4∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙2∙1 + 2∙0∙(-1) + 1∙0∙(-1) + 2∙0∙1 + 1∙4∙(-1) + 2∙2∙(-1) + 2∙0∙1] = 1/16 [4 + 4 - 4 - 4] = 1/16(0) = 0

B_2u: ai = 1/16 [1∙4∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙2∙(-1) + 2∙0∙1 + 1∙0∙(-1) + 2∙0∙1 + 1∙4∙(-1) + 2∙2∙1 + 2∙0∙(-1)] = 1/16 [4 - 4 - 4 + 4] = 1/16(0) = 0

E_u: ai = 1/16 [1∙4∙2 + 2∙0∙0 + 1∙0∙(-2) + 2∙2∙0 + 2∙0∙0 + 1∙0∙(-2) + 2∙0∙0 + 1∙4∙2 + 2∙2∙0 + 2∙0∙0] = 1/16 [8 + 8] = 1/16(16) = 1

Γ_σ = A_1g + B_1g + E_u

Exercise \(\PageIndex{6}\)

Demonstrate how to arrive at the reducible representation for pi bonding in the plane of the complex under the D_4h symmetry of a square planar complex.

Determine the irreducible representations.

Answer

There are 4 unchanged vectors for E. For the others:

b) Γ_π||:

A_1g: ai = 1/16 [1∙4∙1 + 2∙0∙1 + 1∙0∙1 + 2∙(-2)∙1 + 2∙0∙1 + 1∙0∙1 + 2∙0∙1 + 1∙4∙1 + 2∙(-2)∙1 + 2∙0∙1] = 1/16 [4 - 4 + 4 - 4] = 1/16(0) = 0

A_2g: ai = 1/16 [1∙4∙1 + 2∙0∙1 + 1∙0∙1 + 2∙(-2)∙(-1) + 2∙0∙(-1) + 1∙0∙1 + 2∙0∙1 + 1∙4∙1 + 2∙(-2)∙(-1) + 2∙0∙(-1)] = 1/16 [4 + 4 + 4 + 4] = 1/16(16) = 1

B_1g: ai = 1/16 [1∙4∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙(-2)∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙0∙(-1) + 1∙4∙1 + 2∙(-2)∙1 + 2∙0∙(-1)] = 1/16 [4 - 4 + 4 - 4] = 1/16(0) = 0

B_2g: ai = 1/16 [1∙4∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙(-2)∙(-1) + 2∙0∙1 + 1∙0∙1 + 2∙0∙(-1) + 1∙4∙1 + 2∙(-2)∙(-1) + 2∙0∙1] = 1/16 [4 - 4 + 4 - 4] = 1/16(16) = 1

E_g: ai = 1/16 [1∙4∙2 + 2∙0∙0 + 1∙0∙(-2) + 2∙(-2)∙0 + 2∙0∙0 + 1∙0∙2 + 2∙0∙0 + 1∙4∙(-2) + 2∙(-2)∙0 + 2∙0∙0] = 1/16 [8 - 8] = 1/16(0) = 0

A_1u: ai = 1/16 [1∙4∙1 + 2∙0∙1 + 1∙0∙1 + 2∙(-2)∙1 + 2∙0∙1 + 1∙0∙(-1) + 2∙0∙(-1) + 1∙4∙(-1) + 2∙(-2)∙(-1) + 2∙0∙(-1)] = 1/16 [4 - 4 - 4 + 4] = 1/16(0) = 0

A_2u: ai = 1/16 [1∙4∙1 + 2∙0∙1 + 1∙0∙1 + 2∙(-2)∙(-1) + 2∙0∙(-1) + 1∙0∙(-1) + 2∙0∙(-1) + 1∙4∙(-1) + 2∙(-2)∙1 + 2∙0∙1] = 1/16 [4 + 4 - 4 - 4] = 1/16(0) = 0

B_1u: ai = 1/16 [1∙4∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙(-2)∙1 + 2∙0∙(-1) + 1∙0∙(-1) + 2∙0∙1 + 1∙4∙(-1) + 2∙(-2)∙(-1) + 2∙0∙1] = 1/16 [4 - 4 - 4 + 4] = 1/16(0) = 0

B_2u: ai = 1/16 [1∙4∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙(-2)∙(-1) + 2∙0∙1 + 1∙0∙(-1) + 2∙0∙1 + 1∙4∙(-1) + 2∙(-2)∙1 + 2∙0∙(-1)] = 1/16 [4 + 4 - 4 - 4] = 1/16(0) = 0

E_u: ai = 1/16 [1∙4∙2 + 2∙0∙0 + 1∙0∙(-2) + 2∙(-2)∙0 + 2∙0∙0 + 1∙0∙(-2) + 2∙0∙0 + 1∙4∙2 + 2∙(-2)∙0 + 2∙0∙0] = 1/16 [8 + 8] = 1/16(16) = 1

Γ_π|| = A_2g + B_2g + E_u

Exercise \(\PageIndex{7}\)

Demonstrate how to arrive at the reducible representation for pi bonding perpendicular to the plane of the complex under the D_4h symmetry of a square planar complex.

Determine the irreducible representations.

Answer

There are 4 unchanged vectors for E. For the others:

b) Γ_π⊥ :

A_1g: ai = 1/16 [1∙4∙1 + 2∙0∙1 + 1∙0∙1 + 2∙(-2)∙1 + 2∙0∙1 + 1∙0∙1 + 2∙0∙1 + 1∙(-4)∙1 + 2∙2∙1 + 2∙0∙1] = 1/16 [4 - 4 - 4 + 4] = 1/16(0) = 0

A_2g: ai = 1/16 [1∙4∙1 + 2∙0∙1 + 1∙0∙1 + 2∙(-2)∙(-1) + 2∙0∙(-1) + 1∙0∙1 + 2∙0∙1 + 1∙(-4)∙1 + 2∙2∙(-1) + 2∙0∙(-1)] = 1/16 [4 + 4 - 4 - 4] = 1/16(0) = 0

B_1g: ai = 1/16 [1∙4∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙(-2)∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙0∙(-1) + 1∙(-4)∙1 + 2∙2∙1 + 2∙0∙(-1)] = 1/16 [4 - 4 - 4 + 4] = 1/16(0) = 0

B_2g: ai = 1/16 [1∙4∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙(-2)∙(-1) + 2∙0∙1 + 1∙0∙1 + 2∙0∙(-1) + 1∙(-4)∙1 + 2∙2∙(-1) + 2∙0∙1] = 1/16 [4 + 4 - 4 - 4] = 1/16(0) = 0

E_g: ai = 1/16 [1∙4∙2 + 2∙0∙0 + 1∙0∙(-2) + 2∙(-2)∙0 + 2∙0∙0 + 1∙0∙2 + 2∙0∙0 + 1∙(-4)∙(-2) + 2∙2∙0 + 2∙0∙0] = 1/16 [8 + 8] = 1/16(16) = 1

A_1u: ai = 1/16 [1∙4∙1 + 2∙0∙1 + 1∙0∙1 + 2∙(-2)∙1 + 2∙0∙1 + 1∙0∙(-1) + 2∙0∙(-1) + 1∙(-4)∙(-1) + 2∙2∙(-1) + 2∙0∙(-1)] = 1/16 [4 - 4 + 4 - 4] = 1/16(0) = 0

A_2u: ai = 1/16 [1∙4∙1 + 2∙0∙1 + 1∙0∙1 + 2∙(-2)∙(-1) + 2∙0∙(-1) + 1∙0∙(-1) + 2∙0∙(-1) + 1∙(-4)∙(-1) + 2∙2∙1 + 2∙0∙1] = 1/16 [4 + 4 + 4 + 4] = 1/16(16) = 1

B_1u: ai = 1/16 [1∙4∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙(-2)∙1 + 2∙0∙(-1) + 1∙0∙(-1) + 2∙0∙1 + 1∙(-4)∙(-1) + 2∙2∙(-1) + 2∙0∙1] = 1/16 [4 - 4 + 4 - 4] = 1/16(0) = 0

B_2u: ai = 1/16 [1∙4∙1 + 2∙0∙(-1) + 1∙0∙1 + 2∙(-2)∙(-1) + 2∙0∙1 + 1∙0∙(-1) + 2∙0∙1 + 1∙(-4)∙(-1) + 2∙2∙1 + 2∙0∙(-1)] = 1/16 [4 + 4 + 4 + 4] = 1/16(16) = 1

E_u: ai = 1/16 [1∙4∙2 + 2∙0∙0 + 1∙0∙(-2) + 2∙(-2)∙0 + 2∙0∙0 + 1∙0∙(-2) + 2∙0∙0 + 1∙(-4)∙2 + 2∙2∙0 + 2∙0∙0] = 1/16 [8 - 8] = 1/16(0) = 0

Γ_π⊥ = A_2u + B_2u + E_g

References

1. Gray, H. B.; Ballhausen, C. J. J. Am. Chem. Soc. 1963, 85, 260-264.