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Coordination complexes caused great fascination among chemists during the nineteenth century, although it wasn’t until the century’s end that Alfred Werner determined the fundamental principles of their structure. A fuller understanding of their structures was developed over the first half of the twentieth century. This understanding developed in stages, with newer theories and models building on previous ones rather than replacing them completely.
Crystal Field Theory
In the early 1930’s, American physicists John Hasbrouck van Vleck and Hans Bethe developed a theoretical treatment of crystalline transition metal compounds that explained their magnetic and optical properties. This treatment did not consider bonding in any way, but it did concern itself with the geometry of ligands, or counterions, around a central metal ion. Electrostatic considerations formed the basis of the theory, with attention to repulsive forces between ligand electrons and \(d\) electrons on the metal ion. The model gave good predictions about the effect of ligand geometry on the electronic structure of the metal ion.
Ligand Field Theory
During the 1950’s, British chemists John Stanley Griffith and Leslie Orgel combined some aspects of crystal field theory with molecular orbital theory to produce a bonding model for coordination complexes. Rather than just considering electrostatic interactions between metal and ligand, this approach used orbital interactions to model the role of covalency in determining the electronic structure of the complex.
Angular Overlap Model
This model is a specific approach to considering how ligand and metal orbitals combine to produce molecular orbitals in a coordination complex. It uses the symmetry and orientation of atomic orbitals to determine what combinations of these orbitals should be considered.
Computational Chemistry
The three "pencil-and-paper models" described above for bonding in coordination complexes have been greatly augmented by the power of modern computational chemistry. Although based on the same principles, computational methods offer a much higher level of quantitative information about electronic energy and allow relatively easy assessment of possible structural features such as coordination geometry. Pioneering methods included extended Hückel theory, originally developed by Roald Hoffman and Robert Burns Woodward to study pericyclic reactions. The method was later adapted by Hoffman and others to investigate inorganic and organometallic complexes. More recently, developments such as density functional theory have greatly facilitated computational work on transition metal systems because of their simplified treatment of many-electron systems. These approaches and others like them allow for the prediction of coordination complex properties through a computer interface.
10.02: Bonding Theories
As the name implies, crystal field theory was developed as a way of explaining phenomena in ionic crystalline solids. Bethe and van Vleck wished to provide a rationale for the magnetic properties of these materials, as well as their colors. The latter resulted from the wavelength of light absorbed, revealing something about differences in electronic energy levels in the ions of the crystalline solid. The magnetic properties also depended on electronic energy levels, because magnetism depended upon whether electrons were paired, and whether or not electrons are paired depends on the presence or absence of orbitals at the same energy level.
They first considered a transition metal ion in an octahedral hole formed between layers of counterions packed in a crystalline lattice. The metal ion would have six near neighbors: three above and three below. Of course, the ions would be held together by electrostatics: the negative charges on the counterions would be attracted to the positive charge on the metal ion.
But in addition to that positive charge, the metal ion also has electrons of its own. How did the electrons of those neighboring counterions affect the energy of those metal electrons? "...If we’re going to work this out carefully, we need to consider three different situations", thought the physicists. "We need to start by considering the energy of the metal valence orbitals in the absence of counterions. Since these are transition metal ions, we will pay close attention to the five $d$ orbitals. What happens when those $d$ orbitals are placed in a field of surrounding electrons, if the field is evenly distributed around them? In other words, what happens to them in a spherical field of negative charge? And then, what happens if those electrons are not evenly distributed? What happens if the surrounding electrons approach only along the Cartesian axes?"
The Cartesian axes are important here because, in an octahedron, the six ligands are ninety degrees ($90^{\circ}$) apart from each other in space. We can think of two of those ligand electrons as lying in either direction from the metal ion along the z axis; two of them along the x axis; and the last two along the y axis.
Of course, when these metal electrons are placed in a field of negative charge, repulsion results, and the electrons on the metal go up in energy. They all increase in energy by the same amount.
The octahedral field is a different question, however, because the five $d$ orbitals have different spatial distributions. Two of them (the $d_{z^2}$ and $d_{x^2-y^2}$) lie along the axes. These orbitals are referred to as having $e_g$ symmetry. The other three (the $d_{xy}, d_{xz}$, and $d_{yz}$) lie in between the axes. These three are referred to as having $t_{2g}$ symmetry.
Those two orbitals along the axes experience repulsion from the neighboring anions. Because all of the negative charge is focused in six positions rather than being uniformly distributed, the repulsion experienced by these orbitals is greater than it would be in a spherical field. On the other hand, the three orbitals between the axes experience very little repulsion; their energy is actually much lower than it would be in a spherical field of negative charge.
These two sets of orbitals are therefore found at two different energy levels. The energy difference between them is called the octahedral field splitting, $\Delta_o$. Because the total amount of repulsion is the same in the octahedral field and the spherical field – it is just distributed differently – the average energy level of the orbitals should be the same in the two fields. This average energy of the orbitals is called the barycenter, a term borrowed from astronomy. Because two of the orbitals are raised above the barycenter in the octahedral field and three are lowered below the barycenter, then the $e_g$ set must be $\frac{3}{5}\Delta_o$ ($=0.6\Delta_o$) above the barycenter and the $t_{2g}$ set must be $\frac{2}{5} \Delta_o$ ($=0.4\Delta_o$) below the barycenter. That means the average energy lies at the barycenter.
A parameter called the crystal field stabilization energy (CFSE) was developed to assess the energy difference between the metal ion in an octahedral coordination geometry and a spherical field. To determine CFSE, we simply add up the energy of all the electrons relative to the energy level of electrons in the spherical field, in units of $\Delta_o$. For an octahedral case:
$\text{CFSE}= \left[ \frac{2}{5} \left( \text{# of electrons in } t_{2g}\right) + \frac{3}{5} \left( \text{# of electrons in } e_g \right) \right] \times \Delta_o \nonumber$
Problems
Exercise $1$
Determine the crystal field stabilization energy (CFSE) in the following octahedral ions:
a) $\ce{V^3+}$ b) $\ce{Ni^2+}$ c) $\ce{Cr^3+}$ d) $\ce{Zn^2+}$
Answer a
$\ce{V^3+}$ is $d^2$, with two electrons in the $t_{2g}$ and zero electrons in $e_g$.
CFSE = $[2(-0.4) + 0(0.6)]\times \Delta_o = -0.8\Delta_o$
Answer b
$\ce{Ni^2+}$ is $d^8$, with six electrons in $t_{2g}$ and two electrons in $e_g$.
CFSE = $[6(-0.4) + 2(0.6)]\times \Delta_o = -1.2\Delta_o$
Answer c
$\ce{Cr^3+}$ is $d^3$, with three electrons in $t_{2g}$ and zero electrons in $e_g$.
CFSE = $[3(-0.4) + 0(0.6)]\times \Delta_o = -1.2\Delta_o$
Answer d
$\ce{Zn^2+}$ is $d^{10}$, with six electrons in $t_{2g}$ and four electrons in $e_g$.
CFSE = $[6(-0.4) + 4(0.6)]\times \Delta_o = 0\Delta_o$. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.02%3A_Bonding_Theories/10.2.01%3A_Crystal_Field_Theory.txt |
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 dxy, dxz, and dyz 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 dxy, dyz, and dxz. 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.
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 dxy, dxz, and dyz, 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 π symmetry. Instead of a lone pair, this acceptor takes the form of a π* orbital on the ligand. That means there has to be a π 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 π donation has the effect of breaking a π bond in the ligand because of the fact that electrons are donated into a ligand π* 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 π donates into a carbon monoxide, populating the CO π* orbital, the CO bond gets weaker.
Because we are still looking at π bond formation, the orbital pictures look somewhat similar to the ones for ligand-to-metal π donation, but with π* 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 π* 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.1
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 (= C42) 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 (= C42) 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 (= C42) i 6S4 8S6 3σh 3σd
A1g 1 1 1 1 1 1 1 1 1 1 x2 + y2 + z2
A2g 1 1 -1 -1 1 1 -1 1 1 -1
Eg 2 -1 0 0 2 2 0 -1 2 0 (2z2 – x2 – y2, x2 - y2)
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 = A1g + T1u + Eg. 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 A1g representation corresponds to the metal s orbital. The T1u representation corresponds to the three metal p orbitals. The Eg representation corresponds to two of the d orbitals: 2z2 – x2 – y2 (usually abbreviated as z2) and x2 – y2. The other three d orbitals (dxy, dxz, and dyz) have T1u 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.
References
(1) Point Groups https://chem.libretexts.org/@go/page/269925 (accessed Jul 2, 2021).
Example $1$
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
A1g: 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
A2g: 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
Eg: 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
T1g: 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
T2g: 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
A1u: 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
A2u: 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
Eu: 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
T1u: 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
T2u: 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 = A1g + T1u + Eg.
Example $1$
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.
σh: 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.
σd: An off-axis reflection, so all vectors have moved. The character is 0.
Oh E 8C3 6C2 6C4 3C2 (= C42) i 6S4 8S6 3σh 3σd
Γ 12 0 0 0 -4 0 0 0 0 0 | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.03%3A_Ligand_Field_Theory/10.3.01%3A__Ligand_Field_Theory_-_Molecular_Orbitals_for_an_Octahedral_Complex.txt |
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 t2g 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 t2g 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 eg orbitals. It would be at a higher energy level, but it would avoid that repulsive interaction with the other electron in the t2g orbital. That would be the high-spin case. The fourth electron has gone into the higher possible orbital rather than the lower one.
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 t2g and eg levels (the octahedral field splitting, Δo) 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 Δo. First, comparison between Δo 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 Δo becomes. These examples are illustrated in Table 1.
Table 1. Comparison of Octahedral Field Splitting Between M(II) and M(III) Ions1
Complex Δo (cm-1) Complex Δo (cm-1)
Cr(OH2)62+ 14,000 Cr(OH2)63+ 17,600
Mn(OH2)62+ 7,500 Mn(OH2)63+ 21,000
Fe(OH2)62+ 10,000 Fe(OH2)63+ 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, Δo is much larger for second-row than for first-row metals of the same group, as shown in Table 2. The value of Δo 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 Δo 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 2. Comparison of Octahedral Field Splitting Between Ions from Different Periods of the Periodic Table2
Complex Δo (cm-1) Complex Δo (cm-1)
CoCl63- Not available Co(NH3)63+ 23,000
RhCl63- 20,300 Rh(NH3)63+ 33,900
IrCl63- 24,900 Ir(NH3)63+ 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 Δo, whereas σ-donors would give a smaller Δo 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 3. Comparison of Octahedral Field Splitting Between Ions from Different Periods of the Periodic Table2
Complex Δo (cm-1) Complex Δo (cm-1)
CrBr63- Not available NiBr62- 7,000
CrCl63- 13,600 NiCl62- 7,300
Cr(OH2)63+ 17,400 Ni(OH2)62+ 8,500
Cr(NH3)63+ 21,600 Ni(NH3)62+ 10,800
Cr(CN)63- 26,300 Ni(CN)62- Not available
Among the chromium complexes, the π-accepting cyanide gives a much larger value of Δo 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 (pKa = -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. Πc 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 Πc is a positive unit of energy.
The second component involves quantum mechanical exchange between like spins: Πe is a stabilizing factor. When two electrons of like spin can exchange with each other, energy decreases: Πe is a negative unit of energy. For example, in a d2 octahedral metal, the two electrons are both at the t2g 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 4. Comparison of Total Pairing Energies Between M(II) and M(III) Ions3
Complex Π (cm-1) Complex Π (cm-1)
Mn(OH2)62+ 25,500 Mn(OH2)63+ 28,000
Fe(OH2)62+ 17,600 Fe(OH2)63+ 30,000
Co(OH2)62+ 22,500 Co(OH2)63+ 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 Δo 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 eg level. The pairing energies are not that large compared to Δo 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
1. Demonstrate the exchanges possible in the following configurations.
2. Determine the difference in Πc between the high spin and low spin configuration in each of the following cases:
a) d4 b) d5 c) d6
3. Determine the difference in Πe between the high spin and low spin configuration in each of the following cases:
a) d5 b) d6 c) d7
4. Given the value of Δo and Π in each case, 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+o = 14,000 cm-1)4
d) [Co(OH2)6]3+o = 19,000 cm-1)4
Solutions
1.
2. Determine the difference in Πc between the high spin and low spin configuration in each of the following cases:
a) d4
The difference is: ΔE = ls – hs = Πc – 0 = Πc.
b) d5
The difference is: ΔE = ls – hs = 2Πc – 0 = 2Πc.
c) d6
The difference is: ΔE = ls – hs = 2Πc – 0 = 2Πc.
3. Determine the difference in Πe between the high spin and low spin configuration in each of the following cases:
a) d5
The difference is: ΔE = ls – hs = 4Πe - 4Πe = 0
b) d6
The difference is: ΔE = ls – hs = 6Πe - 4Πe = 2Πe
c) d7
The difference is: ΔE = ls – hs = 6Πe - 5Πe = Πe
4. Given the value of Δo and Π in each case, predict whether the complex will be high spin or low spin.
a) [Mn(OH2)6]2+
Δo = 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+
Δo = 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+o = 10,000 cm-1)4
Δo = 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+o = 19,000 cm-1)4
Δo = 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 | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.03%3A_Ligand_Field_Theory/10.3.02%3A_Orbital_Splitting_and_Electron_Spin.txt |
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 eg level is destabilized by 0.6Δo compared to undifferentiated d orbitals in a spherical field, whereas the t2g level is 0.4Δo 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 t2g level and the last one goes into the eg 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.
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.
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 interelectron repulsion in complexes can be lower than in free ions based on the degree of covalency in the complex.
Example $1$
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.
Example $2$
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)]Δo = -2.0Δo
2. High-spin d5: LFSE = [0.6 (2) – 0.4 (3)]Δo = [1.2 - 1.2]Δo = 0Δo
3. Low-spin d7: LFSE = [0.6 (1) – 0.4 (6)]Δo = [0.6 - 2.4]Δo = -1.8Δo
4. High-spin d7: LFSE = [0.6 (2) – 0.4 (5)]Δo = [1.2 - 2.0]Δo = -0.8Δo
Example $3$
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 = (-2.0Δo + 2Πc + 4Πe) - (0Δo + 4Πe) = -2Δo
1. d7: ΔE low spin – high spin = (-1.8Δo + 3Πc + 6Πe) - (-0.8Δo + 2Πc + 5Πe) = -Δo + Πc + Πe | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.03%3A_Ligand_Field_Theory/10.3.03%3A_Ligand_Field_Stabilization_Energy.txt |
Tetrahedral geometry is even more common in chemistry than square planar geometry. 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 (dxy, dxz, dyz) whereas the others point at the faces of the cube (dx2-y2, dz2). 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.
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, Td, shown in the table.
Td E 8C3 6C2 6S4 6σd
A1 1 1 1 1 1 x2 + y2 + z2
A2 1 1 1 -1 -1
E 2 -1 2 0 0 (2z2 - x2 - y2, x2 - y2)
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, Γσ , shown in the table. This representation reduces to the irreducible representations, A1 + T2. The d orbitals represented here, the T2 set, are indeed the dxy, dxz and dyz. The non-bonding d orbitals are the E group, corresponding to dz2 and dx2-y2.
We can go further with the group theory approach and to determine the reducible representation for pi bonding, Γπ , 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 dz2 and dx2-y2. Note, however, that they also include the dxy, dxz, and dyz 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 \(1\)
Demonstrate these symmetry operations on the drawing of the tetrahedron within a cube shown above.
1. C3
2. C2
3. S4
4. σd | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.03%3A_Ligand_Field_Theory/10.3.04%3A_Tetrahedral_Complexes.txt |
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 dz2 orbital drops in energy because it is no longer forming the antibonding combination with ligand orbitals along the z axis.
The dz2 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 dz2 orbital overlaps with these ligand orbitals to a much lesser extent than the dx2-y2 orbitals, so it drops to a level well below the dx2-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, dxy, dxz and dyz, are degenerate in an octahedral geometry but not in a square planar one. The dxy orbital is in the plane of the metal and ligands whereas the dxz and dyz 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 dxy orbital as lying at higher energy than the dxz, dyz pair because of the potential for interaction with the ligands that lie in the same plane as the dxy 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 dxz, dyz 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 dx2-y2.
Sometimes, square planar d orbital splitting diagrams show the dxy orbital above the dz2 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 D4h symmetry.
D4h E 2C4 C2 2C2' 2C2" i 2S4 σh 2σv 2σd
A1g 1 1 1 1 1 1 1 1 1 1 x2 + y2, z2
A2g 1 1 1 -1 -1 1 1 1 -1 -1 Rz
B1g 1 -1 1 1 -1 1 -1 1 1 -1 x2 - y2
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:
Γσ = A1g + B1g + Eu
Returning to the character table, we find that the matching orbitals on the metal include the dz2 and the dx2- y2, as well as the s (represented by x2+ y2), the px and py. 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 dz2 and the dx2- 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:
Γπ || = A2g + B2g + Eu
Consulting the character table, we find that the corresponding orbitals on the central atom are dxy, px and py. In reality, the interaction with the dxy is likely to be much more pronounced than with either the px or the py because of stronger overlap between the dxy and the ligand p orbital.
The second orientation is perpendicular to the plane of the complex.
This time, the irreducible representation is as follows:
Γπ ⊥ = A2g + B2g + Eu
According to the character table, this time the corresponding orbitals on the central atom are dxz, dyz, and pz. Once again, because of stronger overlap between the dxy or dyz with the ligand p orbital compared to ligand overlap with the metal pz, 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 D4h 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 dxy orbital at a higher energy level than the nominally sigma-antibonding dz2 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 dx2- 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 dx2- y2. Square planar complexes are most often observed with d7 or d8 metal ions, which avoids populating that highest-energy d orbital.
Problems
1. a) Demonstrate how to arrive at the reducible representation for sigma bonding under the D4h symmetry of a square planar complex.
b) Determine the irreducible representation.
2. a) Demonstrate how to arrive at the reducible representation for pi bonding in the plane of the complex under the D4h symmetry of a square planar complex.
b) Determine the irreducible representation.
3. a) Demonstrate how to arrive at the reducible representation for pi bonding perpendicular to the plane of the complex under the D4h symmetry of a square planar complex.
b) Determine the irreducible representation.
Solutions.
1. a) There are 4 unchanged vectors for E. For the others:
b) Γσ: Remember, ai = 1hQN∙χ(R)∙χ(R)Q
A1g: 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
A2g: 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
B1g: 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
B2g: 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
Eg: 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
A1u: 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
A2u: 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
B1u: 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
B2u: 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
Eu: 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
Γσ = A1g + B1g + Eu
2. a) There are 4 unchanged vectors for E. For the others:
b) Γπ||:
A1g: 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
A2g: 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
B1g: 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
B2g: 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
Eg: 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
A1u: 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
A2u: 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
B1u: 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
B2u: 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
Eu: 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
Γπ|| = A2g + B2g + Eu
3. a) There are 4 unchanged vectors for E. For the others:
b) Γπ⊥ :
A1g: 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
A2g: 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
B1g: 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
B2g: 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
Eg: 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
A1u: 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
A2u: 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
B1u: 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
B2u: 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
Eu: 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
Γπ⊥ = A2u + B2u + Eg | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.03%3A_Ligand_Field_Theory/10.3.05%3A_Square-Planar_Complexes.txt |
The angular overlap model is an approach to quantifying the interaction between metal and ligand orbitals in different geometries, with a focus on the metal d orbitals. Although developed in the 1970's, this approach is still used as a starting point for theoretical calculations using advanced computational chemistry methods available today.1
The core concept of the angular overlap model is that different ligand orbitals will overlap with metal d orbitals to different extents because of the variety of angles at which these orbitals would approach each other. Stronger overlap leads to greater interaction. That means both greater stabilization of the ligand-centered bonding orbital and greater destabilization of the metal-centered antibonding orbital.
Let's look at some examples. Consider the overlap of the \(d_{z^2}\) orbital with an axial ligand in octahedral geometry, with the assumption that the \(d_{z^2}\) orbital lies in the axial direction. There should be considerable overlap between the metal and ligand orbitals, leading to both significant stabilization of the ligand donor electrons and similar destabilization of an electrons in the metal \(d_{z^2}\) orbital. By comparison, an equatorial ligand would have significantly less overlap with the \(d_{z^2}\) orbital. Instead of overlapping with the substantial lobe along the z axis, the ligand would be interacting with the minimal \(d_{z^2}\) toroid in the xy plane. The bonding orbital would be stabilized to a significantly lesser extent compared to the axial ligand. The antibonding orbital would be destabilized by a correspondingly smaller amount.
In the case of a tetrahedral ligand, there is essentially no overlap with the \(d_{z^2}\) orbital because its on-axis lobe is too far away from the cubic corner positions occupied by ligands in a tetrahedral array. There is really no bonding or antibonding in this case. On the other hand, the dxz orbital reaches a little closer to that corner position, allowing for some overlap with the ligand orbital. Consequently, there is some stabilization of the bonding electrons and destabilization of the antibonding d orbital. At first glance, the amount of overlap in this case, and the amount of stabilization or destabilization, appears much more similar to the case of the equatorial ligand with the \(d_{z^2}\) orbital than the axial ligand with the \(d_{z^2}\) orbital.
We will not go into the mathematics that explore the exact extent of overlap expected in each case. Instead, we will go straight to a summary of the results in the next section.
Exercise \(1\)
Estimate the degree of interaction between the ligand and metal orbital: large, medium, or none.
Answer
a) none b) medium c) medium (although it may appear greater than in (b), and it is) d) large
10.04: Angular Overlap
The pictures we considered previously explicitly addressed sigma overlap between the ligand donor orbital and the metal d acceptor orbital. The donor electrons drop in energy into the new sigma bonding combination and any electrons in the d orbital are raised into the new sigma antibonding combination. For example, when an axial ligand interacts with a metal dz2 orbital, the in-phase, bonding combination drops in energy by a quantity that we will call eσ. At the same time, the out-of-phase, antibonding combination is raised in energy by the same quantity, eσ. A real computational treatment would reveal that this is not exactly true; the amount of energy by which the bonding orbital is stabilized is slightly different from the amount by which the antibonding orbital is destabilized. However, the distinction is pretty minor for our purposes.
From the interactions we have already looked at, we know that this specific interaction is a strong one. Other metal-ligand interactions may involve less overlap than this one and will lead to smaller changes in energy. For this reason, other metal-ligand interactions are expressed as fractions of the quantity, eσ, found in the case of the dz2 orbital interacting with an axial ligand.
Our first task is to label the positions that will be occupied by ligands in a number of different geometries. To avoid a cluttered diagram of these positions, we can use three separate illustrations to illustrate the ligand positions in some common geometries. The first drawing below includes all the geometries in which ligands are found along Cartesian coordinates. The second drawing describes tetrahedral geometry, whereas the third one describes trigonal structures.
Given those ligand positions, we can assess the interactions that would occur if ligands were purely sigma donors. For reference, the dz2 orbital is assumed to lie along the axis between positions 1 and 6 and the magnitudes of the interactions are all scaled relative to the interaction of a ligand at position 1 or 6 with the dz2 orbital. We will not address the approach that was taken to arrive at these relative numbers; we will simply use the results tabulated here.
Sigma Interactions of Ligands with Metal d Orbitals (units of eσ)
Ligand Positions dz2 dx2-y2 dxy dxz dyz
1 1 0 0 0 0
2 1/4 3/4 0 0 0
3 1/4 3/4 0 0 0
4 1/4 3/4 0 0 0
5 1/4 3/4 0 0 0
6 1 0 0 0 0
7 0 0 1/3 1/3 1/3
8 0 0 1/3 1/3 1/3
9 0 0 1/3 1/3 1/3
10 0 0 1/3 1/3 1/3
11 1/4 3/16 9/16 0 0
12 1/4 3/16 9/16 0 0
In order to quantify the interactions between ligands and metal orbitals, we simply tally up the interactions for each orbital. For example, consider a complex like hexaamminecobalt(III) chloride, [Co(NH3)6]Cl3. It has six sigma donor ligands, forming an octahedral structure. The total interactions with each orbital in this geometry would be calculated by totaling the interactions at all the ligand positions for each orbital.
Looking under the dz2 column and totaling only the values for positions 1 through 6, we find:
dz2: (1 + 1/4 + 1/4 + 1/4 + 1/4 + 1) eσ = 3eσ
A similar approach using the dx2-y2 column leads to:
dx2-y2: (0 + 3/4 + 3/4 + 3/4 + 3/4 + 0) eσ = 3eσ
However, the remaining three d orbitals have no interactions with the ligands in these six positions:
dxy: 0; dxz: 0; dyz: 0
So, we find that the dz2 and dx2-y2 orbitals are both raised in energy by 3eσ. At the same time, we know that the ligand donor orbitals are stabilized by their interaction with the metal orbitals. To see how much, we can tally up the interaction for each ligand at its position. That is, for the ligand in position 1, we would add in its interaction with each of the five d orbitals to determine the amount by which it is stabilized by bonding. We simply add up the values across that row.
Ligand in position 1: - (1 + 0 + 0 + 0 + 0) eσ = - eσ
Ligand in position 2: - (1/4 + 3/4 + 0 + 0 + 0) eσ = - eσ
Ligand in position 3: - (1/4 + 3/4 + 0 + 0 + 0) eσ = - eσ
Ligand in position 4: - (1/4 + 3/4 + 0 + 0 + 0) eσ = - eσ
Ligand in position 5: - (1/4 + 3/4 + 0 + 0 + 0) eσ = - eσ
Ligand in position 6: - (1 + 0 + 0 + 0 + 0) eσ = - eσ
We have now generated a diagram that looks very much like what we saw from ligand field theory. We see the familiar d orbital splitting diagram for octahedral geometry as well as stabilization of the ligand donor orbitals in sigma bonds.
The results from the angular overlap model are somewhat simplified compared to the results from ligand field theory. For example, they completely neglect any interaction between ligand orbitals and metal s or p orbitals. However, this model allows a very straightforward calculation of the relative d orbital energy levels.
Problems
1. Use the results of the calculation for the octahedral geometry to calculate the net stabilization due to bonding (in units of eσ) for the following complexes.
a) [Co(NH3)6]Cl3 (assume low spin) b) [Fe(NH3)6](NO3)3 (assume high spin)
c) [Ni(NH3)6]Cl2
2. Use the table of sigma interactions to calculate orbital energy stabilization or destabilization for the following geometries.
a) trigonal planar ML3 b) square planar ML4 c) trigonal bipyramidal ML5
Solutions
1.
2. a) Positions 2, 11, 12.
dz2: (1/4 + 1/4 + 1/4) eσ = 3/4 eσ
dx2-y2: (3/4 + 3/16 + 3/16) eσ = 18/16 = 9/8 eσ
dxy: (0 + 9/16 + 9/16) eσ = 18/16 = 9/8 eσ
dxz: 0
dyz: 0
Ligand in position 2: - (1/4 + 3/4 + 0 + 0 + 0) eσ = - eσ
Ligand in position 11: - (1/4 + 3/16 + 9/16 + 0 + 0) eσ = - eσ
Ligand in position 12: - (1/4 + 3/4 + 9/16 + 0 + 0) eσ = - eσ
b) Positions 2, 3, 4, 5.
dz2: (1/4 + 1/4 + 1/4 + 1/4) eσ = eσ
dx2-y2: (3/4 + 3/4 + 3/4 + 3/4) eσ = 3 eσ
dxy: 0
dxz: 0
dyz: 0
Ligand in position 2: - (1/4 + 3/4 + 0 + 0 + 0) eσ = - eσ
Ligand in position 3: - (1/4 + 3/4 + 0 + 0 + 0) eσ = - eσ
Ligand in position 4: - (1/4 + 3/4 + 0 + 0 + 0) eσ = - eσ
Ligand in position 5: - (1/4 + 3/4 + 0 + 0 + 0) eσ = - eσ
c) Positions 1, 2, 6, 11, 12.
dz2: (1 + 1/4 + 1+ 1/4 + 1/4) eσ = 11/4 eσ
dx2-y2: (0 + 3/4 + 0 + 3/16 + 3/16) eσ = 18/16 = 9/8 eσ
dxy: (0 + 0 + 0 + 9/16 + 9/16) eσ = 18/16 = 9/8 eσ
dxz: 0
dyz: 0
Ligand in position 1: - (1 + 0 + 0 + 0 + 0) eσ = - eσ
Ligand in position 2: - (1/4 + 3/4 + 0 + 0 + 0) eσ = - eσ
Ligand in position 6: - (1 + 0 + 0 + 0 + 0) eσ = - eσ
Ligand in position 11: - (1/4 + 3/16 + 9/16 + 0 + 0) eσ = - eσ
Ligand in position 12: - (1/4 + 3/4 + 9/16 + 0 + 0) eσ = - eσ | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.04%3A_Angular_Overlap/10.4.01%3A_Sigma_Bonding_in_the_Angular_Overlap_Model.txt |
So far, we have considered only the effects of sigma donation on the d orbital splitting diagram using the angular overlap model. When we looked at ligand field theory, we saw that pi-donor and pi-acceptor effects produced significant changes in these diagrams. We generally see these effects in any ligands that have orbitals that can accept electron density from the metal (back-bonding). The paradigm of a pi acceptor is carbon monoxide, of course. Similar effects can be found in related ligands in which the donor atoms participate in pi bonding with another atom in the ligand, thus making a pi* orbital available for back-bonding. In addition, back-bonding is a feature of phosphine ligands and some N-heterocyclic carbenes.
We usually think about this interaction as illustrated below. The empty ligand orbital approaches so that it is perpendicular to the bond axis, allowing overlap with the filled metal d orbital. The interaction lowers the energy of the d electrons, which become bonding in nature, and raises the energy of the empty ligand orbital.
We can use the same ligand positions for pi acceptors that we already used for sigma donors. The orbitals will be oriented differently than the sigma donor orbitals but they will approach from the same directions..
As before, we can use the results of calculations of the strengths of these interactions based on the amount of overlap; we don't need to know exactly how the numbers in the table below came about. This time, the maximum overlap occurs between a dxz orbital and a p orbital approaching in position 1, perpendicular to the bond axis. Several other combinations will be equally strong. This time, the stabilization is expressed in terms of eπ rather that eσ. The amount of energy in this case is somewhat smaller than in sigma donation because of a lesser degree of metal-ligand orbital overlap.
Sigma Interactions of Ligands with Metal d Orbitals (units of eπ)
Ligand Positions dz2 dx2-y2 dxy dxz dyz
1 0 0 0 1 1
2 0 0 1 1 0
3 0 0 1 0 1
4 0 0 1 1 0
5 0 0 1 0 1
6 0 0 0 1 1
7 2/3 2/3 2/9 2/9 2/9
8 2/3 2/3 2/9 2/9 2/9
9 2/3 2/3 2/9 2/9 2/9
10 2/3 2/3 2/9 2/9 2/9
11 0 3/4 1/4 1/4 3/4
12 0 3/4 1/4 1/4 3/4
These interactions modify the picture we built previously for simple sigma donors. In the new interaction diagram, a second set of ligand p orbitals is destabilized by the pi interaction. At the same time, some of the d orbitals are stabilized by the additional interaction. This modification is illustrated below for octahedral geometry.
Problems
1. Use the table of pi interactions to calculate orbital energy stabilization or destabilization for the following geometries.
a) trigonal planar ML3 b) square planar ML4 c) trigonal bipyramidal ML5
Solutions
1. a) Positions 2, 11, 12.
dz2: 0
dx2-y2: - (0 + 3/4 + 3/4) eπ = 6/4 = - 3/2 eπ
dxy: - (1 + 1/4 + 1/4) eσ = 6/4 = - 3/2 eπ
dxz: - (1 + 1/4 + 1/4) eσ = 6/4 = - 3/2 eπ
dyz: - (0 + 3/4 + 3/4) eσ = 6/4 = - 3/2 eπ
Ligand in position 2: (0 + 0 + 0 + 1 + 1) eπ = 2eπ
Ligand in position 11: (0 + 3/4 + 1/4 + 1/4 + 3/4) eπ = 2eπ
Ligand in position 12: (0 + 3/4 + 1/4 + 1/4 + 3/4) eπ = 2eπ
b) Positions 2, 3, 4, 5.
dz2: 0
dx2-y2: 0
dxy: - (1 + 1 + 1 + 1) = -4 eπ
dxz: - (1 + 0 + 1 + 0) = -2 eπ
dyz: - (0 + 1 + 0 + 1) = -2 eπ
Ligand in position 2: (0 + 0 + 1 + 1 + 0) eπ = 2 eπ
Ligand in position 3: (0 + 0 + 1 + 0 + 1) eπ = 2 eπ
Ligand in position 4: (0 + 0 + 1 + 1 + 0) eπ = 2 eπ
Ligand in position 5: (0 + 0 + 1 + 0 + 1) eπ = 2 eπ
c) Positions 1, 2, 6, 11, 12.
dz2: 0
dx2-y2: - (0 + 0 + 0 + 3/4 + 3/4) eπ = -3/2 eπ
dxy: - (0 + 1 + 0 + 1/4 + 1/4) eπ = -3/2 eπ
dxz: - (1 + 1 + 1 + 1/4 + 1/4) eπ = -7/2 eπ
dyz: - (1 + 0 + 1 + 3/4 + 4/4) eπ = -7/2 eπ
Ligand in position 1: (0 + 0 + 0 + 1 + 1) eπ = 2 eπ
Ligand in position 2: (0 + 0 + 1 + 1 + 0) eπ = 2 eπ
Ligand in position 6: (0 + 0 + 0 + 1 + 1) eπ = 2 eπ
Ligand in position 11: (0 + 3/4 + 1/4 + 1/4 + 3/4) eπ = 2 eπ
Ligand in position 12: (0 + 3/4 + 1/4 + 1/4 + 3/4) eπ = 2 eπ | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.04%3A_Angular_Overlap/10.4.02%3A_Pi_Acceptors_in_the_Angular_Overlap_Model.txt |
Ligands in which the donor atom has more than one lone pair are capable of pi donation to the metal. Common examples include halide and alkoxide ligands. The overlap question in this case is exactly the same as in the case of pi acceptors; the same table of interactions applies here. However, since the ligand electrons are forming the pi bond in this case, the ligand electrons are stabilized and the d orbitals with which they overlap are destabilized.
In octahedral geometry, for example, we see the off-axis d orbitals raised modestly in energy in the presence of pi donor ligands. The angular overlap model allows us to quantify the relative changes in energy because of these ligand interactions.
Example \(1\)
Confirm the stabilization and destabilization of orbitals due to pi donation in the octahedral interaction diagram shown above.
Solution
Positions 1, 2, 3, 4, 5, 6.
• dz2: 0
• dx2-y2: 0
• dxy: (0 + 1 + 1 + 1 + 1 + 0) = 4 eπ
• dxz: (1 + 1 + 0 + 1 + 0 + 1) = 4 eπ
• dyz: (1 + 0 + 1 + 0 + 1 + 1) = 4 eπ
• Ligand in position 1: (0 + 0 + 0 + 1 + 1) eπ = 2 eπ
• Ligand in position 2: (0 + 0 + 1 + 1 + 0) eπ = 2 eπ
• Ligand in position 3: (0 + 0 + 1 + 0 + 1) eπ = 2 eπ
• Ligand in position 4: (0 + 0 + 1 + 1 + 0) eπ = 2 eπ
• Ligand in position 5: (0 + 0 + 1 + 0 + 1) eπ = 2 eπ
• Ligand in position 6: (0 + 0 + 0 + 1 + 1) eπ = 2 eπ
10.4.04: The Spectrochemical Series
The spectrochemical series was determined through an examination of the absorption spectra of a series of octahedral \(\ce{Co(III)}\) complexes by Tsuchida in the 1930's. The position of the d-d absorption band is influenced by the field strength of the ligand, which leads to greater or lesser values of \(Δ_o\). That interaction depends on both the relative energies of the metal and ligand orbitals and the degree of overlap between these orbitals. The closer in energy the two orbitals are to each other, the greater the interaction. Also, the greater the overlap between the two orbitals, the greater the interaction.
Recall that sigma donors simply donate a lone pair to the metal but do not have additional metal-ligand interactions. The classic example is ammonia. The nitrogen in ammonia has only one lone pair to donate and is a simple sigma donor. Ethylenediamine, \(\ce{NH2CH2CH2NH2}\) (en), is also a sigma donor because it has a single lone pair on each nitrogen atom. The en ligand is a slightly stronger sigma donor than ammonia. The difference is explained by the slightly stronger basicity of en compared to ammonia; the nitrogen lone pairs in en are better donors to both protons and metal ions.
en > NH3 (basicity)
The common pi donors include the halides and some oxygen donors. In these cases, the donor atom has an additional lone pair (or more) which may engage in formation of a pi bond by donation to a metal d orbital. Among the halides, fluoride produces the largest field splitting and iodide the smallest. This trend is also explained in terms of the relative basicity of these halides; fluoride is more basic than bromide. In addition, there is also understood to be better overlap between fluorine's p orbital and the d acceptor of a positively charged metal ion, compared to larger halides such as iodine.
F- > Cl- > Br- > I- (basicity)
Oxygen donors do not follow this basicity trend. Hydroxide is thought to act as a better pi donor than water, partly because of its negative charge. A carboxylate ligand is not quite as strong a pi donor as hydroxide, probably because its lone pair is delocalized into the carbonyl group.
H20 > RCO2- > HO- (pi donation anion vs neutral)
The pi acceptors include familiar examples such as carbon monoxide (carbonyl ligand) as well as the aromatic phenanthroline.
CO > -CN > phenanthroline > NO2- > SCN-
The phenanthroline may be an unfamiliar ligand. It is an aromatic amine, like pyridine. Other aromatic ligands, if they donate through a lone pair such as a phenyl, \(\ce{C6H5^{-}}\), can also be considered pi acceptors. These compounds feature a pi bond that includes the donor atom, so there is a pi* orbital at that position capable of undergoing back-donation from the metal.
Again, there are subtle variations among the field strengths exhibited by these pi acceptors, and there may be several factors contributing to those differences. For example, the cyanide is only slightly weaker field than the isoelectronic carbon monoxide. That small difference may be due to the negative charge on the cyanide rendering it a weaker electron acceptor. On the other hand, phenanthroline is neutral, but it is still a weaker donor than carbon monoxide. That weaker field may be due to the smaller lobe on the nitrogen atom in the pi* orbital in phenanthroline. Because nitrogen is more electronegative than carbon, it makes a greater contribution to the C=N pi bonding orbital than carbon; the opposite is true in the antibonding orbital. Carbon monoxide, in contrast, features a larger lobe on carbon in its pi* orbital. The result is better metal-ligand pi overlap with carbon monoxide than with phenanthroline.
Phosphines are also commonly considered to be pi acceptors, for subtle reasons that have been subject to some debate.1
All three of these series can be collated to arrive at a combined list, which is a more complete spectrochemical series. Note that there may be some overlap between the different types of donors. Nevertheless, the overall trend holds: pi acceptors lead to the greatest field splitting on average, whereas di donors lead to the smallest.
Example \(1\)
Some additional common ligands are displayed according to field strength, below. Classify these ligands as pi donors, sigma donors, or pi acceptors.
Solution
Ph3, based on its structure, appears to be a sigma donor only because it has just one lone pair on the donor atom. However, phosphines tend to behave as pi acceptors; that is, they appear to withdraw some electron density from the metal as indicated by the response of reporter ligands such as CO.
Bpy, py and CH3CN (acetonitrile) all appear to be pi acceptors. Like phen, they behave as weak pi acceptors.
Acac, oxide, oxalate and sulfide are all pi donors. The donor atom in each case has a lone pair in addition to the sigma donor pair. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.04%3A_Angular_Overlap/10.4.03%3A_Pi_Donors_in_the_Angular_Overlap_Model.txt |
In the angular overlap model (AOM), the field splitting, $Δ$, can be shown to result from a combination of the interaction parameters eσ and eπ. For example, in octahedral geometry,
$Δ = 3eσ - 4eπ \label{eq1}$
in which the value of eπ is positive for a pi donor but negative for a pi acceptor.
The magnitude of these parameters varies from one complex to another. As we have seen previously in examination of ligand field theory, both the metal ion and the identity of the ligand play a role in determining the magnitude of the field splitting. As outlined in the previous section on ligand field theory, the identity of the metal influences the d orbital splitting. Even for the same element, an increase in charge translates into a significant increase in the splitting parameter. For this reason, aqueous $\ce{Co(II)}$ ion is high spin, whereas the aqueous $\ce{Co(III)}$ ion is low spin.
The spectrochemical series tells us that the ligands also play a major role in determining the magnitude of the field splitting, with pi donors producing a smaller splitting than pi acceptors. The splitting produced by an aquo ligand is much smaller than the produced by a cyano ligand, for example.
Spectroscopic data can be used to determine the parameters eσ and eπ. A series of examples for different ligands coordinated to the $\ce{Cr^{3+}}$ ion are outlines in Table $1$.
Table $1$: Angular Overlap Parameters in Octahedral Cr(III) Complexes1
Ligand Δ (cm-1)
π acceptors
-CN 7,530 -930 26,310
pyridine 6,150 -330 19,770
σ donors
en 7,260 assumed 0 21,780
NH3 7,180 assumed 0 21,540
π donors
HO- 8,600 2,150 17,200
H2O 7,550 1,850 15,250
F- 8,200 2,000 16,600
Cl- 5,700 980 13,180
Br- 5,380 950 12,430
I- 4,100 670 9,620
It is worth noting that the parameters observed for Cr(III), although mostly consistent with the spectrochemical series, are not identical to those seen in Co(III). In particular, the places of water and hydroxide are switched in the series. Once again, the identity of the metal plays a role in the strength of interaction with the ligand, and certain ligands may be observed to interact more strongly with some metals than with others.
Example $1$
Calculate Δ for tetrahedral $\ce{Ni^{2+}}$ when coordinated with the following ligands, given the parameters eσ and eπ.1
1. $\ce{PPh3}$ with $eσ = 5,000 \, \text{cm}^{-1}$ and $eπ = -1,750 \, \text{cm}^{-1}$
2. $\ce{Cl^{-}}$ with $eσ = 3,900\, \text{cm}^{-1}$ and $eπ = 1,500 \, \text{cm}^{-1}$
3. $\ce{Br^{-}}$ with $eσ = 3,600\, \text{cm}^{-1}$ and $eπ = -1,000 \, \text{cm}^{-1}$
Solution
This question is a direct application of Equation \ref{eq1}:
1. $\ce{PPh3}$: \begin{align*} Δ &= 3eσ - 4eπ \[4pt] &= 3(5,000) - 4(-1,750) cm^{-1} \[4pt] &= 22,000\, \text{cm}^{-1}.\end{align*} \nonumber
2. $\ce{Cl^{-}}$: \begin{align*}Δ &= 3(3,900) - 4(1,500) \[4pt] &= 5,700 \, \text{cm}^{-1}.\end{align*} \nonumber
3. $\ce{Br^{-}}$: \begin{align*}Δ &= 3(3,600) - 4(1,000) \[4pt] &= 6,800 \, \text{cm}^{-1}.\end{align*} \nonumber
Example $2$
AOM parameters eσ were calculated from spectroscopic data for a series of bidentate nickel complexes with octahedral geometry, $\ce{Ni(en')2(NCS)2}$.2
• en' = $\ce{H2NCH2CH2NH2}$ with eσ = 4,010 cm-1
• en' = $\ce{(CH3)2NCH2CH2N(CH3)2}$ with eσ = 3,165 cm-1
• en' = $\ce{(CH3CH2)2NCH2CH2NH2}$ with eσ = 2,485 cm-1 ($\ce{-NEt2}$); eσ = 4,650 cm-1 ($\ce{-NH2}$)
Explain the reasons for the differences in the eσ values.
Solution
Greater steric hindrance appears to decrease the interaction between the sigma donor and the metal. Thus, the magnitude of eσ decreases from the least hindered $\ce{-NH2}$ group to the most hindered $\ce{NEt2}$ group. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.04%3A_Angular_Overlap/10.4.05%3A_The_Magnitude_of_Parameters_e_e_and.txt |
The relative strengths of metal-ligand binding interactions distilled into the spectrochemical series depend on an inference drawn from the difference between the ground and excited states. In contrast, the magnetochemical series offers similar information based solely on observation of the ground state. These observations are possible in iron(III) porphyrin complexes because of a subtle change in spin state that occurs upon substitution of the axial position with different ligands. The result is a series analogous to the spectrochemical series that is called the magnetochemical series. This series was developed chiefly by the lab of Christopher Reed at University of Southern California and University of California, Riverside.1
There are two unusual things that happen with these porphyrin complexes that allow measurement of metal-ligand interactions in this way. A distortion that occurs in the weak field case of these square pyramidal complexes results in spin pairing in the weak field configuration rather than the high field case. Also, in these porphyrin complexes, a quantum mechanical admixture occurs in which the 5/2 and 3/2 states exist in superposition with each other. As a result, the spin state in these complexes is often intermediate between these two cases.
One additional feature makes these porphyrin complexes quite useful in measuring a magnetochemical series. Paramagnetic complexes produce dramatic shifts in nuclear magnetic resonance spectroscopy. In this case, the hydrogen atoms of the 5-membered pyrrole rings in the porphyrin system shift from about -60 ppm in the S = 3/2 case to about +80 ppm in the S = 5/2 case. In the case of a quantum admixture, the shift ranges in between these two limiting values. This NMR shift can be used to compare the field strength of the axial ligand.
An array of experiments eventually lead to a magnetochemical series. Example ligands from this series are shown in order here, from strong field to weak field:
$\ce{NO+ = CO > R3Sn- > -CH3 > RS- > F- > -OPh > N3- = -OAc > NCS- > Cl- = -OH > Br- > I-} \nonumber$
In addition, this method has allowed the inclusion of several very weakly bound ligands. These can be appended to the series as follows:
$\ce{I- > ReO4- > BF4- > CF3SO3- > ClO4- > H2O > SbF6- > CB11H12-} \nonumber$
Note the unusual position of water, which shows up as a much lower field ligand than in the Co(III) based spectrochemcial series for octahedral complexes. The difference is thought to come from the low spin Co(III) vs. the usually high spin Fe(III). Pi bonding is less favourable in the latter case and so the order in the iron porphyrin complexes is more strongly reflective of sigma donating effects. The anionic hydroxide is a better sigma donor than water because of greater electrostatic attraction to the metal.
Example $1$
Organometallic ligands such as $\ce{-CH3}$ are not typically included in the spectrochemical series.
1. Characterize $\ce{-CH3}$ in terms of ligand type (pi donor, sigma donor, pi acceptor).
2. Explain why it appears so high in the magnetochemical series compared to other anions such as F-.
Solution
1. -CH3 is a sigma donor.
2. The -CH3 anion would be extraordinarily basic. The pKa of CH4 is estimated to be approximately 50, compared to a pKa of approximately 4 for HF. The -CH3 anion is a very strong sigma donor. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.04%3A_Angular_Overlap/10.4.06%3A_The_Magnetochemical_Series.txt |
Certain metal ions frequently display distortions from ideal geometry, such that the symmetry of the compound is lowered. In tetragonal distortions of octahedral complexes, for example, the metal-ligand bond distances of two axial ligands may be significantly longer than the distances of the equatorial ligands. Alternatively, two of the metal-ligand distances may be compressed compared to others.
Sometimes, these distortions can be experimentally observed. For example, EXAFS (X-ray absorption fine structure) data for [Cu(OH2)6](BrO3)2, an octahedral complex in which all six Cu-O bonds ought to be the same length, show two different Cu-O distances of 1.96(1) and 2.32(2) Å (the digit in parentheses is used to convey the degree of uncertainty in X-ray measurements).1 That's an approximately 20% difference between the same kind of metal-ligand bond distances within the same complex.
These distortions, often called Jahn-Teller distortions, are observed in complexes of transition metals with specific electron configurations: high spin d4, low spin d7, and d9 (for which there is neither a low nor high spin case). These configurations give rise to unequally occupied degenerate orbitals. Degenerate orbitals are multiple orbitals at the same energy. The three p orbitals in an atomic shell are degenerate, for example. In contrast, non-degenerate orbitals occur at different energy levels. These two cases are illustrated below with the simplest conceivable case, a pair of orbitals.
A set of degenerate orbitals might be equally or unequally occupied depending on the number of available electrons. For example, in the simple case of two degenerate orbitals, equal occupancy would occur if there were two electrons or four electrons because those even numbers of electrons could be divided evenly between the two orbitals. Either one or three electrons would lead to unequally occupied orbitals; there is no way to put the same number of electrons in each orbital in this case.
These cases of unequally occupied degenerate orbitals can lead to opportunities for distortion. When the symmetry of a complex drops, two otherwise degenerate orbitals might no longer exist at the same energy as each other. That means, for example, that instead of having three electrons at the same energy, we would have two electrons at lower energy and one at higher. There would be a net decrease in energy in the non-symmetric, distorted case.
In the context of an octahedral d9 complex, elongation of the bonds to the axial ligands leads to a very slight lowering of the d orbitals that interact with ligands along the z axis: the dz2 and the dxz and dyz. The other orbitals are shown increasing slightly according to the principle that the average energy of the orbitals is preserved at the orbital barycenter (as opposed to the energy of the electrons, which decreases in this case).
Not all cases of unequally occupied degenerate orbitals produce measurable Jahn Teller distortions. In octahedral geometry, examples of strong Jahn Teller effects are seen when the unequally occupied orbitals are in the eg set rather than the t2g set. The eg set is the one that is linked to sigma bonding, after all, so the orbitals in that level are strongly influenced by the strength of that sigma bonding interaction, and hence the bond length.
Changes in bond length are not the only observable result of the Jahn Teller effect. The absorption spectra of some octahedral complexes show two distinct d-d bands rather than the one band absorbed for pure octahedral symmetry. This observation is a consequence of having additional orbital levels giving rise to distinct electronic transitions. This phenomenon is also seen if the excited state, rather than the ground state, experiences Jahn Teller distortion.
The Jahn Teller effect also plays a role in reactivity. For example, it is thought to promote the lability of octahedral Cu(II) complexes by weakening the bonds to axial ligands via the tetragonal distortion. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.05%3A_The_Jahn-Teller_Effect.txt |
Students sometimes want to be able to predict the coordination geometry that will be adopted by a particular combination of metal ion and ligands. In reality, there are enough variables involved in both the ligand and the metal that such predictions are not possible with any certainty.
Nevertheless, there are some broad trends that can be highlighted by looking at what the angular overlap model says about geometric preferences. One of the most common questions is whether a four coordinate complex will adopt a tetrahedral or square planar geometry. Probably the most important factor in this case is the steric bulk of the ligands. The ligand-metal-ligand bond angles in tetrahedral geometry are ideally 109 degrees, compared to 90 degrees in square planar geometry. As a result, ligands are spaced further apart in tetrahedral geometry than in square planar geometry, so bulkier ligands favor tetrahedral complexes.
On the other hand, the angular overlap model (as well as other approaches) suggests some advantages of square planar geometry. A comparison of eσ at a range of d electron configurations for both high spin and low spin cases is plotted below.
From the outset, we should note that low spin tetrahedral complexes are almost unheard of, so we should really compare square planar to the high spin tetrahedral case. In the comparison between high spin tetrahedral and high spin square planar geometry, square planar is favored only for d3, d4, d8 and d9 configuration. However, as we will see shortly, square planar complexes of d3 and d4 metals are not very common because metals with those electron configurations are more likely to adopt higher coordination geometries, such as five or six coordination. All of the other cases are more likely to adopt a tetrahedral geometry because if electronic stabilization factors are equal then steric factors (relief of ligand crowding) will prevail.
If we are comparing low spin square planar to high spin tetrahedral cases, there is an increased preference for the square planar geometry for all but the d1, d2 and d10 configurations. That suggests square planar geometry is usually favored unless there are steric factors that push the geometry to tetrahedral. Of course, the comparison we have used ignores eπ. Because the magnitude of eσ and eπ vary with different ligands and metals, we can't easily include both factors in one graph. However, we do know that π donors will make it more likely that we are dealing with a high spin square planar case, whereas π acceptors will lead to a low spin square planar case. These two cases have a dramatic effect on the d orbital splitting diagram. The angular overlap model suggests a dramatic reordering of the square planar d orbital splitting diagram in these cases.
Consequently, square planar geometry is expected to be much more common in cases with π acceptor ligands, whereas tetrahedral geometry should be more common in cases with π donor ligands.
It is worth noting that the first reported group of low spin tetrahedral complexes, the crystallographically characterized tetrakis(norbornyl)cobalt(IV), (nor)4Co, as well as the cationic and anionic (nor)4Co+ and (nor)4Co-, are d5, d4 and d6 complexes, respectively.1 AOM predicts very similar stabilization in low spin square planar and low spin tetrahedral geometries in these cases. However, the tetrahedral geometry is largely enforced by the bulky norbornyl ligands.
As mentioned above, there is a general preference for six coordination over four coordination, especially at lower d electron counts. A simple AOM comparison indicates octahedral geometry is favored for the low spin case between d0 and d7. Even high spin octahedral geometry is favored electronically over low spin square planar up to d4, with only a modest preference for square planar between d5 and d7. It isn't surprising that octahedral geometry is so common. In very simple terms, this preference comes from the energetically favored formation of six metal-ligand bonds rather than only four metal-ligand bonds.
Example \(1\)
Confirm the stabilization values from the angular overlap model for (a) high spin and (b) low spin tetrahedral geometry.
Solution
Stabilization (σ only) ΔE = 0(# e electrons) - 43(# t2 electrons) - 4(2 electrons per ligand) eσ
a) For high spin:
• d0: ΔE = 0(0) + 43(0) - 8 eσ = -8 eσ
• d1: ΔE = 0(1) + 43 (0) - 8 eσ = -8 eσ
• d2: ΔE = 0(2) + 43 (0) - 8 eσ = -8 eσ
• d3: ΔE = 0(2) + 43 (1) - 8 eσ = -623 eσ
• d4: ΔE = 0(2) - 43(2) - 8 eσ = -513 eσ
• d5: ΔE = 0(2) - 43 (3) - 8 eσ = -4 eσ
• d6: ΔE = 0(3) - 43 (3) - 8 eσ = -4 eσ
• d7: ΔE = 0(4) - 43 (3) - 8 eσ = -4 eσ
• d8: ΔE = 0(4) - 43 (4) - 8 eσ = -223 eσ
• d9: ΔE = 0(4) - 43 (5) - 8 eσ = -113 eσ
• d10: ΔE = 0(4) - 43 (6) - 8 eσ = 0 eσ
b) For low spin:
• d0: ΔE = 0(0) + 43 (0) - 8 eσ = -8 eσ
• d1: ΔE = 0(1) + 43 (0) - 8 eσ = -8 eσ
• d2: ΔE = 0(2) + 43 (0) - 8 eσ = -8 eσ
• d3: ΔE = 0(3) + 43 (0) - 8 eσ = -8 eσ
• d4: ΔE = 0(4) -43 (0) - 8 eσ = -8 eσ
• d5: ΔE = 0(4) - 43 (1) - 8 eσ = -623 eσ
• d6: ΔE = 0(4) - 43 (2) - 8 eσ = -513 eσ
• d7: ΔE = 0(4) - 43 (3) - 8 eσ = -4 eσ
• d8: ΔE = 0(4) - 43 (4) - 8 eσ = -223 eσ
• d9: ΔE = 0(4) - 43 (5) - 8 eσ = -113 eσ
• d10: ΔE = 0(4) - 43 (6) - 8 eσ = 0 eσ
Example \(2\)
Calculate the preference for square planar geometry for each d electron count in (a) the case of high spin square planar vs. high spin tetrahedral, and (b) the case of low spin square planar vs. high spin tetrahedral.
Solution
Preference = ΔΔE = ΔE(square planar) - ΔE(tetrahedral)
a) For high spin:
• d0: ΔΔE = - 8 -(- 8) eσ = 0 eσ
• d1: ΔΔE = - 8 -(- 8) eσ = 0 eσ
• d2: ΔΔE = - 8 -(- 8) eσ = 0 eσ
• d3: ΔΔE = - 8 - (-623 )eσ = -113eσ
• d4: ΔΔE = -7 - (-513) eσ = -123eσ
• d5: ΔΔE = -4 - (4) eσ = 0 eσ
• d6: ΔΔE = -4 - (4) eσ = 0 eσ
• d7: ΔΔE = -4 - (4) eσ = 0 eσ
• d8: ΔΔE = -4 - (223) eσ = -113 eσ
• d9: ΔE = -3 - (113) eσ = -123 eσ
• d10: ΔE = 0 - (0) eσ = 0 eσ
b) For low spin:
• d0: ΔΔE = - 8 -(- 8) eσ = 0 eσ
• d1: ΔΔE = - 8 -(- 8) eσ = 0 eσ
• d2: ΔΔE = - 8 -(- 8) eσ = 0 eσ
• d3: ΔΔE = - 8 - (-623 )eσ = -113eσ
• d4: ΔΔE = -7 - (-513) eσ = -123eσ
• d5: ΔΔE = -7 - (4) eσ = -3 eσ
• d6: ΔΔE = -7 - (4) eσ = -3 eσ
• d7: ΔΔE = -7 - (4) eσ = -3 eσ
• d8: ΔΔE = -6 - (223) eσ = -313 eσ
• d9: ΔE = -3 - (113) eσ = -123 eσ
• d10: ΔE = 0 - (0) eσ = 0 eσ | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.06%3A_Four-_and_Six-Coordinate_Preferences.txt |
Although four- and six-coordinate geometries are probably the most commonly observed in inorganic chemistry, a range of other shapes have also been reported.1 The other most common coordination number is five. Geometries for five-coordination include both trigonal bipyramidal and square pyramidal. Both are common, and distorted geometries in between these two limiting cases are even more commonly observed. The fractional parameter, τ (tau), is usually reported with structural analyses of five-coordinate compounds to convey where the structure falls on this continuum. A value of τ = 1 corresponds to perfect trigonal bipyramidal and a value of τ = 0 corresponds to perfect square pyramidal. Most reported complexes have values solidly in between these extremes. However, there are some ligands, such as porphyrins, in which a rigid ring of four basal donors enforces square pyramidal geometry more closely.
Lower coordination numbers are less common but are reported in certain cases. Two-coordinate compounds are sometimes observed in complexes of the coinage metals. They include a range of ligands, including σ donors in [Ag(NH3)2]+ ion, π donors in [CuCl2]- ion and π acceptors in [Au(CN)2]- ion, for example. Three coordinate complexes have been reported from across the periodic table, especially with sterically demanding ligands such as Bradley and Chisholm's [(Me3Si)2N]3M (Sc, Ti, V, Cr, Fe) or Wolczanski's (t-Bu3SiO)3Ta.2,3
Coordination numbers higher than six are sometimes observed. These cases often involve early transition metals in high oxidation states that are less electronically saturated. Examples are also seen in the lanthanides and actinides owing to their large size. There is even a reported sixteen-coordinate complex, although it is of an alkali metal ion and not a transition metal ion.
References
1. Gispert, J. R. Coordination Chemistry, Wiley-VCH: Weinheim, Germany, 2008, pp. 59-80.
2. Bradley, D. C.; Copperthwaite, R. G.; Extine, M. W.; Reichert, W. W.; Chisholm, M. H. (1978). "Transition Metal Complexes of Bis(Trimethyl-silyl)Amine (1,1,1,3,3,3-Hexamethyldisilazane)" Inorganic Syntheses. 1978, 18. p. 112.
3. Neithamer, D. R.; LaPointe, R. E.; Wheeler, R. A.; Richeson, D. S.; Van Duyne, G. D.; Wolczanski, P. T. "Carbon monoxide cleavage by (silox)3Ta (silox = tert-Bu3SiO-): physical, theoretical, and mechanistic investigations", J. Am. Chem. Soc. 1989, 111, 25, 9056-9072.
4. Pollak, D.; Goddard, R.; Pörschke, K.-R. "Cs[H2NB2(C6F5)6] Featuring an Unequivocal 16-Coordinate Cation". J. Am. Chem. Soc. 2016, 138, 30, 9444-9451.
Problems
1. Use the angular overlap model to
i) calculate d orbital energy destabilization and
ii) construct d orbital splitting diagrams for linear (two-coordinate) geometry with the following ligand types.
a) sigma donor b) pi acceptor c) pi donor
2. Use the angular overlap model with sigma-only interactions to calculate d orbital energy destabilization in square pyramidal geometry.
3. Use the angular overlap model with sigma-only interactions to construct d orbital splitting diagrams for the following geometries.
a) trigonal planar b) trigonal pyramidal c) square bipyramidal
Solutions
1. i) a) Positions 1, 6.
dz2: (1 + 1) eσ = 2 eσ
dx2-y2: (0 + 0) eσ = 0
dxy: (0 + 0) eσ = 0
dxz: 0
dyz: 0
b) Positions 1, 6.
dz2: 0
dx2-y2: - (0 + 0) eπ = 0
dxy: - (0 + 0 ) eπ = 0
dxz: - (1 + 1) eπ = -2 eπ
dyz: - (1 + 1) eπ = -2 eπ
c) Positions 1, 6.
dz2: 0
dx2-y2: (0 + 0) eπ = 0
dxy: (0 + 0 ) eπ = 0
dxz: (1 + 1) eπ = 2 eπ
dyz: (1 + 1) eπ = 2 eπ
ii)
2. Positions 1, 2, 3, 4, 5:
dz2: (1 + 1/4 + 1/4 + 1/4 + 1/4) eσ = 2eσ
dx2-y2: (0 + 3/4 + 3/4 + 3/4 + 3/4) eσ = 3eσ
dxy: 0
dxz: 0
dyz: 0
3. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/10%3A_Coordination_Chemistry_II_-_Bonding/10.07%3A_Other_Shapes.txt |
Introduction
The d-orbital splitting in coordination complexes results in a gap ($\Delta$) that happens to be just the right magnitude to absorb visible light. Because metal complexes can absorb visible light, they display an array of colors. Not only is the color attractive to the eye, it is an indication of the chemical and physical properties of the metal complex. The color (like the magnitude of $\Delta$) depends on the identity of the metal ion, the coordination geometry, and the ligand identity. Chemists don't just "look" at color, though - we measure it using electronic absorption spectroscopy. This is usually done in a lab using a UV-visible spectrophotometer.
An example of such a measurement is shown below in Figure $1$ for a Cu(II) complex. The sample appears a pink color to the eye, and when it is measured using a UV-visible spectrometer, it is shown to absorb visible light at approximately 530 nm. The absorption spectrum can indicate the oxidation state of Cu, the ligands bound to the Cu(II) ion, and the coordination geometry. The color of the solution in Figure $1$ is a shade of pink.
We observe the complementary color of light absorbed
The absorption spectrum shown above in Figure $1$ is a simple case in which only one absorption band is observed in the visible region of the spectrum. In a simple case like this, the color of a complex can be predicted as the complementary color of the light absorbed by the solution. When a solution or object absorbs a certain wavelength, we see the complementary color; or the color opposite to the absorbed wavelength on the color wheel in Figure $2$. In the case of the Cu(II) complex spectrum shown in Figure $1$, the color of the light absorbed at 530 nm is green, and the predicted color observed is pink.
The table below lists the approximate colors of absorption corresponding to wavelengths of light absorbed, and gives similar information to that deduced from Figure $2$.
Table $1$: Approximate wavelengths of absorption and their complementary colors observed. * This prediction is limited to simple cases where there is only one absorption band in the visible spectrum.
$\lambda$ absorbed E (cm$^{-1}$) E (eV) Approximate color absorbed Predicted color observed (by eye)
$> 700$ nm $< 14,000 \frac{1}{\text{cm}}$ $< 1.77$ eV Infrared not observable
$\approx 700-635$ nm $\approx 14,300- 16,000\frac{1}{\text{cm}}$ $\approx 1.77-1.95$ eV Red Green
$\approx 635-590$ nm $\approx 15,700- 16,900 \frac{1}{\text{cm}}$ $\approx 1.95-2.10$ eV Orange Blue
$\approx 590-560$ nm $\approx 16,900-17,900 \frac{1}{\text{cm}}$ $\approx 2.10-2.21$ eV Yellow Violet
$\approx 560-520$ nm $\approx 17,900- 19,200 \frac{1}{\text{cm}}$ $\approx 2.21-2.38$ eV Green Red
$\approx 520-490$ nm $\approx 19,200- 20,400 \frac{1}{\text{cm}}$ $\approx 2.38-2.53$ eV Cyan Red-Orange
$\approx 490-450$ nm $\approx 20,400- 22,200 \frac{1}{\text{cm}}$ $\approx 2.53-2.76$ eV Blue Orange
$400-450$ nm $\approx 22,200- 25,000 \frac{1}{\text{cm}}$ $\approx 2.76-3.10$ eV Violet Yellow
<400 nm $> 25,000 \frac{1}{\text{cm}}$ $> 3.10$ eV Ultraviolet (UV) not observable
Energy of electronic absorption
The absorption spectrum of a metal complex can be used to calculate the splitting energy, $\Delta$, when the absorption corresponds to a $d \rightarrow d$ transition. Let's use the $d^9$ Cu(II) complex (discussed above) as an example. A $d^9$ metal ion has only one visible-light $d \rightarrow d$ transition. Let's assume that the coordination geometry is approximately octahedral (although it is actually a Jahn-Teller distorted octahedron, and more like a square plane). If we assume it's octahedral, then the $d$-orbital splitting diagram (see Figure $1$) leads us to expect one electronic transition: an electron is excited from the $t_{2g}$ to $e_g$. The energy absorbed is equal to the energy of the $\Delta$.
Many cases are not as simple as a $d^9$ octahedral case because there are multiple possible electronic transitions, and also multiple absorption bands in the UV-vis spectrum. In these more complex cases, the actual energy of the transition are affected by differences in electron-electron repulsion energies in the ground state and the excited states. We will learn how to account for multiple possible excited states and electron-electron repulsions using Tanabe-Sugano diagrams later in this chapter.
11.01: Absorption of Light
When light passes through a solution that absorbs light, it enters the solution with an initial intensity ($I_o$) at a given wavelength, and it emerges with an intensity, $I$.
The Beer-Lambert Law defines the relationship between absorbance at a given wavelength and the concentration of the solution.
$\log \left( \frac{I_{o}}{I} \right)=A=\varepsilon l c$
The absorbance (A) is a unitless number because $\frac{I_{o}}{I}$ is unitless. The absorbance depends on the concentration ($c$) and the path length ($l$). The concentration of the sample solution is measured in molarity (M) and the length of the light path in centimeters (cm). The Greek letter epsilon ($\varepsilon$) in these equations is called the molar absorptivity (also called the molar absorption coefficient). The units of $\varepsilon$ are $\frac{L}{mol \times cm}$ or $L \times mol^{-1}\times cm^{-1}$.
Chemists most often measure and report absorbed light in terms of wavelength ($\lambda$) in units of nanometers (nm). But the wavelength scale is inconvenient for measuring energy because it is inversely proportional to both frequency and energy. In other disciplines, like physics for example, absorption spectra are more often reported in terms of frequency ($\nu$) using units of inverse centimeters ($cm^{-1}$). The relationships between energy (E), $\nu$, and $\lambda$ are given by the equation below:
$E=h v=\frac{h c}{\lambda}=h c\left(\frac{1}{\lambda}\right)=h c \bar{v}$ | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/11%3A_Coordination_Chemistry_III_-_Electronic_Spectra/11.01%3A_Absorption_of_Light/11.1.01%3A_Beer-Lambert_Absorption_Law.txt |
Note
The quantum numbers of atoms (and ions) correspond to quantized energy states, called microstates, that depend on the electron configuration. Electrons within a subshell or orbital can adopt different configurations of their electrons, with different individual quantum numbers. These different electron configurations are microstates, and they can be grouped into terms and ordered according to their relative energies. Since electronic transitions occur between terms of different energies, knowledge of the terms for a given atom or ion can aid in the interpretation of electronic spectra.
Introduction
Transition metal ions can give rise to a spectrum of beautifully colored colored complexes. The colors are often caused by absorption of visible light due to electronic transitions involving metal d-orbitals. These electronic transitions are not only attractive to the eye, they are useful spectroscopic signals because the transitions occur between quantized electronic energy states, called microstates. Spectroscopy, coupled with knowledge of the possible microstates and their energies can yield clues about molecular structure.
To learn what a microstate is, let's use the simple example of a carbon atom with a $1s^2 2s^2 2p^2$ electron configuration. The $s$ subshell is full, and there is only one way to put two electrons into an $s$ subshell. Both electrons must occupy the orbital having $m_l=0$, and each must have opposite spins, ($m_s =+\frac{1}{2}$), and ($m_s =-\frac{1}{2}$). On the other hand, the $p$ subshell of carbon has two electrons that can occupy any of three orbitals (Figure $1$). This leaves room for different orientations of the electrons within the $p$ subshell. Each different electron configuration is a microstate, and each has an energy that can be distinguished using electronic spectroscopy (UV-vis for example). In the case of a carbon atom with a two $p$ electrons, there are 15 different possible microstates, as illustrated in Figure $1$.
Just as individual electrons have quantum numbers ($n, l, m_l, m_s$), the electronic states of atoms have quantum numbers ($L, M_l, M_s, S, J$). Each individual microstate in Figure $1$ can be described by quantum numbers $M_l$ and $M_s$. The values of total angular momentum ($L$), and total intrinsic spin ($S$) arise from sets of microstates, and are related to quantum numbers $M_l$ and $M_s$ for individual microstates. The relative energies of the terms for 3d metals and other light atoms can be predicted (roughly) by Hund's Rules, according to values of S and L. These atomic quantum numbers and Hund's rules are described below.
A closer look at electronic spectra
Let us take a closer look at optical absorption spectra, also called electronic spectra, of coordination compounds. We have previously argued that ligand field theory can predict and explain the electronic spectra. However, ligand field theory (LFT) is sufficient to explain the spectra in only a few cases. For example the $\ce{[Ni(H2O)6]^2+}$ ion is an octahedral $d^8$-complex ion. According to LFT, the metal $d$-orbitals in an octahedral field are the $t_{2g}$ and the $e_g$–orbitals (Figure $2$). Six electrons are in the $t_{2g}$ orbitals, and two electrons are in the $e_{g}$ orbitals (Figure $2$). Ligand field theory (LFT) would predict that there is one electron transition possible, namely the promotion of an electron from a $t_{2g}$ into an $e_{g}$ orbital. This process would be triggered by the absorption of light whereby the wavelength of the light would depend on the $\Delta_o$ between the $t_{2g}$ and the $e_{g}$ orbitals. Overall, this should lead to a single absorption band in the absorption spectrum of the complex. We can check this prediction by experimentally recording the absorption spectrum of the complex (Figure $2$).
What we find is that the absorption spectrum is far more complex than expected. Instead of just a single absorption band there are multiple ones. Obviously, LFT is unable to explain this spectrum. The question is: why? The answer is: LFT assumes that there are no electron-electron interactions. However, in reality there is repulsion between electron in d-orbitals and this has an effect on their energy. And, electrons within the d-subshell, for example, repel each other to different extents depending on the relative orientations of their orbitals in space.
To illustrate how electrons in different orbitals might have different interactions, let's consider the case of a $d^2$ excited state in an octahedral ligand field (Figure $3$).
According to LFT, both electrons would be in the $t_{2g}$ orbitals in the ground state. For instance, they could be in the microstate where one electron is in $d_{xy}$, and the other is in the $d_{xz}$ orbital (not pictured). This configuration is called a microstate of a state because there are other combinations of orbitals possible. For example, the ground state would also be realized by a microstate in which the electrons were in the $d_{xz}$ and the $d_{yz}$ orbitals. Upon absorption of light, the electron in the $d_{xy}$ orbital could be excited into one of the higher-energy $e_{g}$ orbitals. That excited electron could occupy either the $d_{z^2}$ or the $d_{x^2-y^2}$ orbitals. Those two possibilities reflect two different microstates associated with the excited state. For a moment, let's assume that the excited electron goes into the $d_{z^2}$ orbital. In this microstate one electron would be in the $d_{xz}$ orbital and the other one in the $d_{z^2}$ orbital (Figure $3$, right). There is another possibility for how to excite an electron from the ground state. We could assume that instead of the $d_{xy}$ electron being promoted, the $d_{xz}$ electron gets promoted. In this case, we would realize a microstate in which one electron in the $d_{xy}$ orbital and the other one in the $d_{z^2}$ orbital (Figure $3$, left).
Now let us compare the two cases shown in Figure $3$. Ligand field theory would argue that both excited microstates have the same energy. However, in fact they do not. Why? It is because the electrons in the first excited microstate interact differently than those of the second excited microstate. This difference becomes plausible when considering the orbital shapes and orientations (Figure $3$, bottom). The $d_{xz}$ orbital has electron density on the z-axis, while the $d_{xy}$ orbital is perpendicular to $d_{z^2}$. The different relative orientations of $d_{xy}$ and $d_{xz}$ with respect to $d_{z^2}$ cause electrons in the $d_{z^2}$ orbital to interact differently with an electron in a $d_{xy}$ than in a $d_{xz}$. As a result, the two excited microstates do not have the same energy. In other words, to achieve either of these different excited microstates from the ground state, we need different amounts of energy. Thus, the complex would absorb light with different energies (or wavelengths). This is in contrast to what LFT predicts.
If LFT cannot predict the number of electronic transitions, then how can we correctly predict how many absorption bands we get? The answer is, we must find all possible microstates for the $d^2$ electron configuration and group together those with the same energy. A group of microstates with the same energy is called a term. The number of electron transitions can then be predicted from the number of terms.
Quantum Numbers
The table below summarizes the quantum numbers of atoms (individual microstates and sets of microstates). You may wish to review the quantum numbers for individual electrons and then refer to this table as you read this and the next sections. Recall the meanings of the quantum numbers for individual electrons ($n, l, m_l, m_s$ that were described in a previous section (Section 2.2.2).
Symbol Name Allowed Range Comment
$L$ Total Orbital Angular Momentum $|l_1 + l_2 | , |l_1 + l_2 ‐ 1| , \ldots, |l_1 ‐ l_2 |$ Total orbital angular momentum of a collection of microstates, designated as S, P, D, F..etc
$S$ Total Intrinsic Spin $|{m_s}_1 + {m_s}_2 |, |{m_s}_1 +{m_s}_2 ‐ 1| , ... ,| {m_s}_1 ‐ {m_s}_2 |$ Total spin of a collection of microstates
$M_l$ Magnetic Quantum Number $L, L-1, L-2, ..., -L$ Direction of total angular momentum for an individual microstate
$M_s$ Spin Magnetic Quantum Number $S, S-1, S-2, ...-S$ Total spin of electrons for an individual microstate
$J$ Total Angular Momentum $L+S,..., | L-S |$ Total angular momentum
Quantum Numbers S and L for sets of microstates
The quantum numbers $S$ and $L$ represent sets of microstates. Once these values are found, they can be ordered in terms of relative energies, first according to the value of $S$, then $L$.
Total Orbital Angular Momentum Quantum Number: L
$L$ gives the total sum of orbital angular momentum vectors in a multielectron atom. The possible values of $L$ can be found from the values of $l$ for individual electrons in the system. For example, in a system with two electrons with values of $m_{l_1}$ and $m_{l_2}$, the possible values of $L$ are:
$L = \sum_i^n m_{l_i} = |m_{l_1} + m_{l_2} | , |m_{l_1} + m_{l_2} ‐ 1| , \ldots, |m_{l_1} ‐ m_{l_2} |$
The absolute value symbols indicate that $L$ cannot be a negative value. The values of $L$ correspond to different energy levels for groups of microstates. Microstates with values of $L=0, 1, 2, 3...$ correspond to symbols $S, P, D, F...$ respectively. This is analogous to the relationship between the electron quantum number, $l$, and the $s, p, d, f..$ orbital subshells. However, the capital letters used for microstates do not indicate orbital or subshell assignments for the electrons.
The Total Intrinsic Spin Quantum Number: $S$
The sum total of the spin vectors of all of the electrons is called $S$. The values of $S$ are computed from $m_s$ in a manner very similar to how $L$ is computed from $m_l$. Because $S$ measures the magnitude of a vector, it cannot be negative. For a system with two electrons, each with spin ${m_s}_1$, and ${m_s}_2$, the possible values of $S$ are given below.
$S =\sum_i^n {m_s}_i = |{m_s}_1 + {m_s}_2 |, |{m_s}_1 +{m_s}_2 ‐ 1| , ... ,| {m_s}_1 ‐ {m_s}_2 | \label{spin}$
The possible values of $S$ fall into a series that depend on whether there are an odd or even number of electrons.
• Odd number of electrons: $S=\frac{1}{2}, \frac{3}{2}, \frac{5}{2},...$
• Even number of electrons: $S = 0, 1, 2, 3,...$
Example 11.2.1 : Total Orbital Angular Momentum of Carbon
What are the possible $L$ values for the electrons in the $1s^2 2s^2 2p^2$ configuration of carbon?
Solution
Both electrons (i.e., the 2p electrons) are $l = 1$. The possible combinations are 0, 1, and 2 corresponding to symbols S, P, and D, respectively.
Example 11.2.2 : Total Orbital Angular Momentum of Unknown Species
What are the possible $L$ values for the electrons in the $[Xe]6s^2 4f^1 5d^1$ ?
Solution
We can ignore the electrons in the $[Xe]$ core and the electrons in the $6s$ block. So all we have to consider is the lone $f$ electron ($l=3$) and the lone $d$ electron ($l=2$).
The two extremes for possible $L$ values are $3+2 = 5$ and $3 ‐ 2 = 1$.
Thus, possible values of $L$ for this Xe atom are 5(H), 4(G), 3(F), 2(D), and 1(P).
Example 11.2.5 : The Hydrogen Ground State
Find $S$ for $1s^1$.
Solution
$S$ must be $\frac{1}{2}$, since that’s the spin of a single electron and there’s only one electron.
Example 11.2.6 : The Beryllium Excited State
Find S for $1s^2 2s^1 2p^1$.
Solution
$S = 1, 0$
Example 11.2.7 : The Carbon Ground State
Find $S$ for carbon atoms with the $1s^2 2s^2 2p^2$ electron configuration.
Solution
$S=1,0$ This is the same as the previous problem. Notice that $S$ is not affected by which orbitals are occupied by electrons. $S$ depends only on the number of unpaired electrons. These are usually the electrons in partially-filled subshells (i.e., unpaired electrons in open shells).
Example 11.2.8 : The Nitrogen Ground State
Find $S$ for nitrogen atoms with the $1s^2 2s^2 2p^3$ electron configuration.
Solution
$S$ can be $S=\frac{3}{2}, \frac{1}{2}$.
Quantum numbers $M_l$ and $M_s$ for individual microstates
Once S and L are found, the allowed values of $M_l$ and $M_s$ can be calculated.
The Total Magnetic Quantum Number: $M_l$
The Total Magnetic Quantum Number $M_l$ is the total $z$-component of all of the relevant electrons’ orbital momentum. While $L$ describes the total angular momentum in the system, $M_l$ tells you which direction it is pointing. $L$ can be assigned to a collection of microstates, but $M_l$ is unique to a specific microstate in that group. Unlike $L$, $M_l$ is allowed to have negative values. The possible values of $M_l$ are integer values ranging from the largest positive sum to the most negative sum of possible $m_l$ values:
$M_l = L, L-1, L-2, ..., -L$
For the $p^2$ case, the $m_l$ values of $p$ orbitals are $-1, 0, +1$. The largest value of $M_l$ would come from a state where both electrons occupy $m_l=+1$, thus the maximum is $M_l=L=2$. Likewise, the minimum value of $M_l$ comes from both electrons occupying $m_l=-1$ to give $M_l=-L=-2$. The possible $M_l$ values for $p^2$ are the series of integer values $+2, +1, 0, -1, -2$.
It is worth noting that there are values of $M_l$ that are forbidden due to the Pauli exclusion principle. For example, the value of $M_l=+3$ could only come from three electrons occupying the $m_l=+1$ orbital. That is impossible because more than one electron would possess the same set of electron quantum numbers.
The Total Spin Magnetic Quantum Number: $M_s$
$M_s$ is the sum total of the z-components of the electrons’ inherent spin in an individual microstate. The difference between $S$ and $M_s$ is subtle, but important. $M_s$ indicates the total z-component of the electrons’ spins, while $S$ indicates the entire resultant vector. It is also distinct from $M_l$, which is the sum total of the z-component of the orbital angular momentum. $M_s$ can be computed from $S$, as shown below. Note that while $S$ must be positive, $M_s$ can have negative values.
$M_s = S, S-1, S-2, ...-S \label{Ms}$
For the $p^2$ case, the possible values of $M_s$ are $M_s = +1, 0, -1$. The value $M_s = +1$ comes from the sum of two electrons with spin "up", $m_s=+\frac{1}{2}$. The value $M_s = -1$ comes from the sum of two electrons with spin "down", $m_s=-\frac{1}{2}$. And the value $M_s = 0$ comes from the sum of one electron with $m_s=+\frac{1}{2}$ and the other with $m_s=-\frac{1}{2}$.
There are some values of $M_s$ that will be forbidden, but not in the case of $p^2$. However in the case of a $p^4$, for example, the value $M_s$=2 is forbidden because there is no way to put four "up" electrons in three orbitals without violating the Pauli exclusion principle.
Example 11.2.3 : Total Magnetic Quantum Number of the Zirconium Ground State
What are the possible values $M_l$ of a zirconium atom with the $[Kr] 5s^2 4d^2$ electron configuration?
Solution
Both open-shell electrons (i.e., the 4d electrons) are $l=2$, so the values are 4, 3, 2, 1, 0, -1, -2, -3, -4.
Example 11.2.4 : Total Spin Magnetic Quantum Number of the Carbon Ground State
What are the $M_s$ values for $1s^2 2s^2 2p^2$ ?
Solution
This system is paramagnetic, with two unpaired electrons in the $2p$ orbital. The three $p$ orbitals have values of $m_s=-1,0,1$. The maximum value of $M_s=+1$ would come from the two electrons occupying $m_{s_1}=+1$ and $m_{s_2}=0$. The minimum value, $M_s=-1$ would come from the two electrons occupying $m_{s_1}=-1$ and $m_{s_2}=0$. Thus, the possible values for the total spin magnetic quantum number are $M_s = 1, 0, ‐1$, where the value of $M_s=0$ comes from a configuration where, for example, $m_{s_1}=+1$ and $m_{s_2}=-1$.
Example 11.2.4 : Total Spin Magnetic Quantum Number of the Nitrogen Ground State
What are the $M_s$ values for $1s^2 2s^2 2p^3$ ?
Solution
$M_s = +\frac{3}{2}, \; +\frac{1}{2}, \; -\frac{1}{2}, \; -\frac{3}{2}$
11.02: Quantum Numbers of Multielectron Atoms
Constructing "Free Ion" Term Symbols and their Relative Energies
The different electronic states of an atom or ion are described by $L$ and $S$ in the form of a term symbol (Figure $1$). Keep in mind that $M_l$ is related to $L$, and $M_s$ is related to $S$, and in some cases their values are equal.
One term symbol can describe one or more microstates with the same energy. The free ion terms are useful in the interpretation of electronic spectra. These term symbols are also commonly called free ion terms (or Russell-Saunders terms) because they describe the atomic or ionic state, free of any ligands. The term symbol indicates the total orbital angular momentum of the atom, ($L$), and the multiplicity, $(2S+1)$, of the state. The multiplicity can be found with the formula, $(2S+1)$, but it is also equal to the number of unpaired electrons plus one.
$\text{Multiplicity } = \left( 2S+1 \right) = \left( \text{# of unpaired electrons }+1 \right)$
The generic form of a free ion term symbol is shown in Figure $1$. Recall that $L=0,1,2, 3...$ is denoted by a capital letter $S, P, D, F...$, respectively. The latter designation for $L$ is used in the term symbol.
Example 11.2.1.1 : Hydrogen Ground State
What is the free ion term symbol of an H atom in the ground state ($1s^1$)? How many microstates are there for the ground state hydrogen atom? Comment on the degeneracy and multiplicity of the state.
Solution
There is only one electron in a $s$ orbital. That electron must have $l=0$, and $s$ could be either $+\frac{1}{2}$ or $-\frac{1}{2}$. This is two possible microstates, one with spin "up" and one with spin "down". In either case there is only one value of the total intrinsic spin quantum number, $S=+\frac{1}{2}$. Therefore, the only possible value of spin multiplicity is Multiplicity $=2\times (+\frac{1}{2})+1 = 2$. The $s$ orbital corresponds to $l =0$, so $L=0$, or $S$. Thus the term symbol is $^2S$.
There are two possible microstates of the ground state (both represented by the ground state term). One microstate has $m_s=+\frac{1}{2}$ and the other has $m_s=-\frac{1}{2}$.
The question about degeneracy specifically asks about degeneracy of the state. Let's give the most complete answer possible. In terms of energy levels (the ground state in this case), there are two microstates with approximately equal (degenerate) energy, thus the degeneracy of the state is two (2) and is the same as the multiplicity ($2S+1=2$). However, sometimes we also refer to the "degeneracy" of electron configurations. It is important to recognize that ground state electron configuration with $M_s=m_s=+\frac{1}{2}$ is singly degenerate, and the same is true for $M_s=m_s=-\frac{1}{2}$. Recognising degeneracy of electron configurations will be useful later in this chapter because the degeneracy of an electron configuration can be helpful for interpreting term symbols in an octahedral field.
Exercise $1$
Find the term symbols for the following cases:
1. $L=0, \ S=0$
2. $L=0, \ S=\frac{1}{2}$
3. $L=1, \ S=1$
4. $L=2, \ S=\frac{3}{2}$
5. $L=3, \ S=2$
Answer
(a) $^1S$
(b) $^2S$
(c) $^3P$
(d) $^4D$
(e) $^5F$
Finding all possible microstates for an atom or ion
What are the possible microstates and the overall number of microstates for a free ion with $d^2$-electron configuration? We can express this in a so-called microstate table ($2$).
You can see the microstate table for the $d^2$ electron configuration depicted above. Each column represents a possible value $M_l$. For the $d^2$ configuration, $M_l$ can adopt values from -4 to +4, hence there there are 9 columns. Each row represents possible $M_s$ value. For the $d^2$ electron configuration, $M_s$ can vary between +1 and -1. $M_s=+1$ means both electrons have the spin $+\frac{1}{2}$, $M_s=0$ means one electron has the spin $+\frac{1}{2}$ and the other one has the spin $-\frac{1}{2}$. The $M_s=-1$ value is adopted when both electrons have the spin $-\frac{1}{2}$. Now we can combine each $M_s$ value with each $M_l$ value, which defines a particular field in the table. You can see that some fields are empty due to the Pauli principle. For example, the field for $M_l=-4$ and $M_s=-1$ is not filled because in this case both electrons would have the same quantum numbers, namely $m_l=2$, and $m_s=-\frac{1}{2}$. We can further see that for some fields only one combination of electrons is possible, while for others there are multiple possible microstates. For example, for the field with $M_l=4$ and $M_s=0$ there is only one combination of electrons possible. You can see the notation “$2^- 2^+$” in this field. This means that the first electron has an $m_l$ value of 2 and a spin of $-\frac{1}{2}$ and the second electron also has an $m_l$ value of 2 and an $m_s$ value of $+\frac{1}{2}$. The most populated field is the field for $M_l=0$ and $M_s=0$; there are overall five microstates with that combination of $M_l$ and $M_s$ values. If we count the overall number of microstates in the table then we arrive at the number 45. This is consistent with what we would expect according to the following formula. $\text{#microstates} = (2L+1)(2S+1)$
In our example L=4 and S = 1, and thus the number of microstates is: $((2x4)+1)((2x1)+1)=45$. In general for a $d^n$ electron configuration with $n$ d-electrons, the number of microstates is $(10!)/((10-n)!n!)$.
One useful property of a microstate table is that we can derive all possible terms and term symbols for a particular electron configuration from it. For simplification's sake, we indicate a possible microstate in the table in Figure $5$ just by an “x”. Then we draw the largest possible rectangular box into the microstate table that contains only fields with at least one possible electron combination (Figure $5$).
You can see that the largest possible box is the red box drawn (Figure $5$). This box contains the microstates that belong to the first term of the electron configuration $d^2$. The number of microstates is equal to the number of fields within the red box. That makes $7\times 3=21$ microstates. What is the term symbol for this term? In order to answer this question we need to find the microstate in the red box that has the highest $M_l$ and the highest $M_s$ value. You can see that it is the one with $M_l=3$ and $M_s=2$. Thus, L=3 and S=1. This defines the term as a 3F term because 2S+1 = 3, and L=3 corresponds to the term symbol F. This term is a triplet term. Note that a triplet term also includes microstates with opposite spins. Overall seven of the 21 microstates have $M_s=0$.
Next, we look what is the next-largest rectangular box we can draw, and which contains microstates we did not consider yet. We can see that the blue box contains nine fields equaling nine microstates. The microstate within this box that has the highest $M_l$ and $M_s$ values is the one with $M_l=4$ and $M_s=0$. This defines a 1G term. This is a singlet term. Note that a singlet term ONLY has microstates with paired spins (Ms=0).
There is another rectangular box which also contains nine microstates. It is the green one. The microstate with the highest $M_l$ and $M_s$ values is the one with $M_l=1$ and $M_s=1$. Thus, $L =1$, and $2S+1=3$, which defines a $^3P$ term. Note that despite this being a triplet term, which indicates two unpaired electrons (\S=1\), the term also contains three microstates that have opposite spins with a total $M_s=0$.
The next-largest box is the purple one containing five fields and five microstates. What is the term with the highest $M_l$ and $M_s$ values? It is the one with $M_l=2$ and $M_s=0$. This means that L=2 and S=0, which defines a 1D term, a singlet term with only microstates with paired spins. There is one microstate left that we did not consider thus far, the one with $M_l=0$ and $M_s=0$. These are the highest $M_l$ and $M_s$ values because there is no other microstate. Thus, this term contains only one microstate and the term symbol is $^1S$. Now we have found all the terms and term symbols.
Example 11.2.1.2 : Boron
What are the possible term symbols for $1s^2 2s^2 2p^1$ electronic configuration of boron? Which is the ground state?
Solution
We only need to consider the one electron in the $2p$ orbital, since that is the only open shell electron. For one electron, there is only one value of $S = ½$, so multiplicity must be $2S+1=2$. The possible values of $L$ for one electron in a $p$ orbital are $L=0,1$ (or S, and P terms). This gives the free ion term symbols $^2P$ and $^2S$. Both of these terms have multiplicity of 2, so the ground state would be the one with the highest value for $L$. The ground state term is $^2P$.
Example 11.2.1.3 : Beryllium Excited State
What are the term symbols for the $1s^2 2s^1 2p^1$ excited-state electronic configuration of Beryllium?
Solution
The term symbols will be of the form $^1P$ and $^3P$. For the $^1 P$ state, $L=1$ and $S=0$, so the total angular momentum $J=1$. For the second state, $L=1$ and $S=1$, so the possible values of $J=2,1,0$. There are four microstates for this configuration with term symbols of $^1 P_1$ and $^3 P_2$, $^3P_1$, and $^3P_0$.
Finding The Ground State Term
...from a list of microstates
Each term symbol represents an atomic state with an associated energy level. If you have found all the possible microstates using a microstate table, and have identified their terms, then the term with the lowest energy (ie the ground state) can be found using Hund's first two rules. In abbreviated form, Hund's rules state...
1. The lowest energy electron configuration has the maximum multiplicity (the maximum number of unpaired electrons).
2. For a given multiplicity, the configuration with the largest total orbital angular momentum ($L$) value has the lowest energy.
In the case of the $p^2$ carbon atom (see Figure $1$), the lowest energy state would have a multiplicity of 3 (two unpaired electrons). Of the states with a multiplicity of 3, the largest value of $L$ is 1 (and thus, $P$). Thus, the ground state term for carbon is $^3P$. These rules are reliable for identifying the ground state term for any atom, but have limited use for ordering relative energies outside of the ground state for 3d metal ions and other light atoms. For heavier atoms (e.g., 4d metals) another type of interaction, called jj coupling must be considered, but this is outside the scope of this discussion. Tip: There is a deep symmetry that connects different electronic configurations. It turns out that a $p^1$ configuration has the same term symbols as a $p^5$. Similarly, $p^2 = p^4$. A similar relationship can be used to figure out high electron number term symbols for the $d$ and $f$ orbitals.
...if you just want the find Ground State term
To assign transitions in an electronic spectrum, we should know the ground state term, as well as all other terms and their relative energies for a given atom or ion. However, the identification of all possible terms becomes tedious with more than one or two electrons in the $d$ orbitals. Fortunately, there are tools (like correlation diagrams) that can help us find all terms for an atom or ion, as long as the ground state term can be identified. Here is a quick shortcut for finding the ground state term for transition metal complex with partially-filled $d$ shell.
Quickly find the ground state free ion tern
In this case, we'll focus on transition metal d-orbitals. To begin, you'll need to know the metal ion identity and its oxidation state. These instructions assume that the $d$ orbital is the only shell that is partially filled.
1. Determine the number of d-electrons.
2. Then fill these electrons into an electron orbital diagram of five degenerate orbitals, putting electrons into the highest value of $m_l$ first, and filling orbitals with single electrons before pairing them (just as you may have done in an introductory chemistry course).
3. Calcuate the total intrinsic spin (\S\) and spin multiplicity using the diagram drawn in #2.
4. Calculate $L$ using the diagram drawn from #2.
5. Write the term symbol for the ground state.
Let's use the example of the $\ce{[Cr(NH3)6]^3+}$ complex ion, and determine its ground state term following the shortcut described above.
1. Determine the number of $d$-electrons.
$\ce{[Cr(NH3)6]^3+}$ contains $\ce{Cr^3+}$, which is a $d^3$ metal ion. There are three $d$ electrons.
2. Draw a diagram of five degenerate $d$ orbitals, with $m_l$ values labeled. Fill in the electrons to maximize $m_s$ and $m_l$.
3. Calculate spin multiplicity (should be the maximum possible value!).
$S= m_{s_1} + m_{s_2} + m_{s_3} = +\frac{1}{2} + +\frac{1}{2} + +\frac{1}{2} = \frac{3}{2}$
Spin multiplicity = $2S+1 = 2 \frac{3}{2} + 1 = 4$
4. Calculate $L$ (should be the maximum possible value after spin is maximized.
$L = m_{l_1}+ m_{l_2} + m_{l_3} = 2 + 1 + 0 = 3$
5. Write the ground state free ion term:
$^{(2S+1)}L = \; ^4F$
Exercise $2$
Earlier on this page, you found all the terms for a $d^2$ metal ion. Use both of the methods (Hund's rule, and the shortcut method) described here to identify the ground state term for $d^2$.
Answer
Using Hund's rule: The terms identified for $d^2$ are $^3F, \; ^1G, \; ^3P, \; ^1D,$ and $^1S$. Hund's first rule tells us that either $^3F$ or $^3P$ could be the ground state because these two terms have the highest multiplicity. Hund's second rule tells us that $^3F$ must be the ground state because it has the largest value of $L$.
Using the shortcut: The drawing is shown below, with one electron in $m_l=+2$ and the second electron in $m_l=+1$. Both electrons are unpaired with $m_s=+\frac{1}{2}$. This gives $L=3$ or an F term. And, $S=1$ with a multiplicity of 3. The ground state term is $^3F$.
Exercise $3$
Find the ground state free ion terms for the metal ions in the following cases:
(a) $\ce{[Ti(H2O)6]Cl2}$
(b) $\ce{[Ti(NH3)6]Cl3}$
(c) $\ce{[Cu(H2O)6]Cl2}$
Answer
(a) $\ce{Ti^2+}$ is a $d^2$ metal ion, $^3F$
(b) $\ce{Ti^3+}$ is a $d^1$ metal ion, $^2D$
(c) $\ce{Cu^1+}$ is a $d^9$ metal ion, $^2D$. Note that $d^9$ and $d^1$ are related by the symmetry of their electronic states.
11.2.02: Spin-Orbit Coupling
The magnetic fields created by $S$ and $L$ are not isolated from one another; they interact through spin-orbit coupling (aka Russell-Saunders coupling). We will consider only this simple form of coupling in this text. Its application is limited to the elements with $z<40$, including the first row of the transition elements. In the case of heavier elements, we must also consider $jj$ coupling; however, we will not discuss the latter here.
In the Russell-Saunders spin-orbit coupling scheme, the interaction between $S$ and $L$ is expressed by an additional quantum number, the total angular momentum quantum number ($J$). The possible values of $J$ are values between $L+S$ and $|L-S|$.
$J = L+S, \; L+S-1, \; L+S-2, \; ..., \; |L-S|$
A value of $J$ must be positive or zero for a multielectron system. J values can fall into series $\frac{1}{2}, \frac{3}{2}, \frac{5}{2},...$ or $0, 1, 2,...$. The quantum number $J$ is added to the term symbol as a subscript to the right of the letter describing the term. A full term symbol is as follows:
$^{(2S+1)}L_J$The result of spin-orbit coupling is that a term for the free ion is split into states of different energies. For example, a $^3P$ state of a carbon atom with a $p^2$ electron configuration would be split into three different energy states (according to the three possible J values 0, 1, and 2): $^3P_0, ^3P_1, ^3P_2$.
The relative energies of the states can be predicted from Hund's Third Rule.
Hund's third Rule:
• For subshells that are less than half-filled, the lowest energy state has the lowest $J$ value.
• For subshells that are exactly half-filled, there is only one J value, thus it is the lowest energy.
• For subshells that are more than half-filled, the lowest energy state has the largest $J$ value.
Thus, in this case where the $p$ subshell is less than half full, the lowest energy state from the $^3P$ free ion term would be that with $J=0$, $^3P_0$, followed by $J=1$ and $J=2$. The splitting and relative energies are depicted in Figure $1$. Spin-orbit coupling and the splitting of the free ion terms have important implications for electronic spectra because they affect the energies of electronic transitions. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/11%3A_Coordination_Chemistry_III_-_Electronic_Spectra/11.02%3A_Quantum_Numbers_of_Multielectron_Atoms/11.2.01%3A_Finding_Microstates_and_Term_Symbols.txt |
Types of transitions related to the metal ion:
1. d-d transitions: d-d transitions are electronic transitions that occur between the molecular orbitals (MOs) that are mostly metal in character: specifically, the orbitals that we think of as the d-orbitals of a transition metal complex. These transitions are useful in determining the energy of splitting and can be used to indicate coordination chemistry (geometry and ligand sets). In octahedral complexes, d-d transitions occur between the $t_{2g}$ and $e_g$ orbitals (across $\Delta$). These transitions cannot occur in metal complexes where the d-orbital is completely empty ($d^0$) or completely full ($d^{10}$). In other words, a d-d transition is only possible in $d^1 - d^9$ metal ions. In a UV-visible absorption spectrum, d-d transitions appear as relatively weak absorptions with extinction coefficients ($\varepsilon$) less than 1,000.
2. Charge transfer (CT) transitions: Charge transfer transitions occur between MOs that are mostly metal in character and those that are mostly ligand in character. These transitions depend on the type of ligand: they occur only when the metal is bound to ligands that are $\pi$-donors or $\pi$-acceptors. And there are two types of CT transitions. If the metal is bound to a $\pi$-donor ligand, electrons from lower-energy MO's that are mostly ligand in character can become excited to MO's that are mostly metal in character. These are ligand to metal charge transfers (LMCT) transitions (Figure $1$, left diagram). If the the metal is bound to ligands that are $\pi$-acceptors, electrons from the MO's that are mostly metal in character can become excited to higher-energy orbitals that are mostly ligand in character. These are metal to ligand charge transfer (MLCT) transitions (Figure $1$, right diagram). In a UV-visible absorption spectrum, CT transitions appear as relatively intense absorptions with extinction coefficients ($\varepsilon$) much greater than 1,000.
11.03: Electronic Spectra of Coordination Compounds
Electronic transitions between states of different energies give rise to electronic spectra. However, some transitions are more probable, and thus more intense, than others. In UV-vis spectroscopy, for example, the transitions that are "allowed" can give rise to absorption bands that are much more intense than transitions that are "forbidden". To be clear, "forbidden" transitions are still possible, and are sometimes observable, but are less intense because they happen less frequently than "allowed" transition when the molecules are exposed to electromagnetic radiation.
Selection Rules that govern Electronic Transitions
The Selection Rules governing transitions between electronic energy levels (including microstates and terms) are:
1. The Spin Selection Rule, $\Delta S = 0$.
2. The Laporte (or orbital) Selection Rule, for centrosymmetric molecules $g \rightarrow u$ or $u \rightarrow g,$ or $\Delta l = \pm 1$.
Selection rules that govern electronic transitions
Spin selection rule
The Spin Selection Rule forbids transitions between states with different total spin, and thus different spin multiplicity. This rule allows transitions only between states with the same total intrinsic spin ($\Delta S = 0$), and thus the same spin multiplicity value in the term symbol. In other words, the direction of the promoted electron's spin should not change. Just in case it is not obvious yet, the spin multiplicity is the left superscript in the term symbol, so this rule allows transitions between terms with the same superscript. For example, this rule would allow a transition from a $^3T$ term to a $^3A$ term, but not from $^3T$ to $^2D$.
Laporte (orbital) Selection Rule
The Laporte Selection Rule applies to molecules that have a center of symmetry (aka center of inversion, centrosymmetric). This rule forbids transitions between states with the same parity (symmetry) with respect to an inversion center ($i$). Parity is indicated on molecular orbitals and on term symbols with subscripts $g$ (gerade, or even) and $u$ (ungerade, or uneven). Transitions between $u$ and $g$ terms are allowed (eg $T_{2g} \rightarrow T_{1u}$ is allowed), but those between two $g$ or two $u$ terms are forbidden (eg $T_{2g} \rightarrow T_{1g}$ is forbidden). Because different types of orbitals have different symmetries with respect to $i$, this rule is sometimes referred to as the "orbital selection rule". It forbids transitions within one type of orbital subshell. For example, $p$ orbitals are antisymmetric with respect to $i$, while both $s$ and $d$ orbitals are symmetric with respect to $i$. This rule forbids $s \rightarrow s$, $p \rightarrow p$, $d \rightarrow d$, and $d \rightarrow s$ transitions, but allows transition between $d \rightarrow p$ and $d \rightarrow s$ orbitals. This rule is important in transition metal complexes because it fobids $d-d$ transitions in centrosymmetric geometries, including octahedral and linear coordination geometries.
Despite these selection rules, $d \rightarrow d$ transitions are a hallmark feature of octahedral transition metal complexes, and are often responsible for their brilliant colors. These $d \rightarrow d$ transtions, and other forbidden transitions may still occur, primarily through relaxation of these rules in specific cases.
Breaking the rules!
Relaxation of the Laporte and Spin Selection Rules rules can occur through:
• Vibronic coupling: the bonds of metal complexes vibrate and may cause temporary distortions in molecular symmetry. These distortions can cause temporary loss in symmetry (as well as some orbital mixing), and thus allow $d \rightarrow d$ transitions in those moments of distortion. Despite being forbidden by the Laporte Selection rule, the $d \rightarrow d$ transitions in octahedral complexes appear, but are weak (of low intensity) with molar absorptivities $\le 100 \; M^{-1} cm^{-1}$.
• Orbital Mixing: In the case of octahedral complexes, $\pi$-acceptor and $\pi$-donor ligands can mix with the d-orbitals so that transitions are no longer purely $d \rightarrow d$ (but these are usually considered "charge transfer" transitions, not $d \rightarrow d$). In the case of tetrahedral complexes, the molecule has no center of symmetry (thus no $u$ or $g$ subscripts on the terms). In the valence orbital model, a tetrahedral molecule is said to have $sp^3$ and $sd^3$ hybridized orbitals, while MO theory predicts MO's with mixtures of some s, p, and d character - this mixing of orbital types can allow transitions between "mixed" orbitals that would otherwise be forbidden in "pure" orbitals. A similar phenomenon would occur in octahedrons with significant distortion, or in other coordination geometries that provide potential for orbital mixing.
• Spin-Orbit coupling: This gives rise to spin-forbidden bands of low intensity, usually with very weak molar absorptivities approximately $\le 1 M^{-1} cm^{-1}$. This phenomenon is usually more important for transition metals of the second row $4d$ and beyond.
Exercise $1$
Consider the $d^2$ case presented in a previous section. The free ion terms for $d^2$ were found to be $^3F, \; ^1G, \; ^3P, \; ^1D,\; ^1S$. First, identify the ground state term. Then identify the Spin-Allowed excitations starting from the ground state.
Answer
The ground state is $^3F$. There is only one spin-allowed transition: $^3F \rightarrow ^3P$.
The spin-forbidden transitions from $^3F$ include $^3F \rightarrow ^1D$, $^3F \rightarrow ^1G$, and $^3F \rightarrow ^1S$. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/11%3A_Coordination_Chemistry_III_-_Electronic_Spectra/11.03%3A_Electronic_Spectra_of_Coordination_Compounds/11.3.01%3A_Selection_Rules.txt |
Splitting of Terms in an Octahedral Field
Thus far we have only considered free ion terms, which means terms without the presence of a ligand field. Let us think next about the influence of an octahedral field on a term. Terms are wavefunctions, just like orbitals, and therefore they behave like orbitals in a ligand field. For example, the $d$-orbitals split into $t_{2g}$and $e_g$ orbitals in an octahedral ligand field. A $D$-term behaves similarly. It splits into $T_{2g}$ and $E_g$ terms (note the capital letters that describe term, and the lower-case letters that describe orbitals!). \The $p$-orbitals are triple-degenerate, having $T_{1g}$ symmetry in the $O_h$ (octahedral) point group, and do not split in energy: they give the $t_{1g}$ orbitals under an octahedral field. Similarly, the $P$-terms have the same symmetry and also do not split. The $P$-term becomes a $T_{1g}$ term in an octahedral field.
Following analogous arguments, $S$ terms become $A_{1g}$ terms. $F$ terms do split in energy like $f$-orbitals and become $T_{1g}$, $T_{2g}$, and $A_{2g}$ terms. Overall, the presence of the octahedral field increases the number of terms from four to seven (Figure $1$). It is easy to see that the ligand field leads to many states, and many potential electron transitions. Thus, we would expect quite complicated spectra. For other ligand fields, the terms also behave analogously to orbitals. For, instance in a tetrahedral field D-terms split into E and T2 terms, and so forth.
Term Splitting for octahedral d2 metal complexes
Now let us think about how the term energies of our free d2-ion changes when placed in an octahedral ligand field, depending on the ligand field strength. We can express this by a correlation diagram (Figure $2$). In a correlation diagram, we plot the energies of the terms relative to the field strength.
On the left side we plot the terms without any field according to their energies. In the case of a d2 ion, the energies are
$\ce{^{3}F < ^{1}D < ^{3}P < ^{1}G < ^{1}S}. \nonumber$
Next, we plot the relative energies in a weak octahedral ligand field, and label the terms according to their symmetry. We can see that the D, F, and G terms split in energy, while the S and P terms do not. Because of the weak field, energy differences are very small. Now let us increase the ligand field strength continuously, until we have reached a very strong ligand field. We see that some of the terms move up in energy, while other terms move down as the ligand field increases. For example, two of the three terms resulting from the F-terms increase in energy while one decreases. It is also possible that a term does not change its energy. For example, the 3T1g term from the 3P term does not change its energy. In very strong ligand field there are three groups of terms that have similar energy. In the hypothetical case of an infinitely strong ligand field, the terms that belong to a particular group become identical in energy. In this case, there are only three states for the electrons possible. The field is considered so strong that the energy associated with electron-electron interactions become negligible compared to the energy of the field. The electrons behave as though there were no electron-electron interactions. Therefore, we can call the lowest energy state the t2g2 state. It is equivalent to the state of the two electrons being in the t2g orbitals. The second state is called the t2geg state. It is equivalent to the state of one electron being in the t2g and one electron being in the eg orbitals. The third state is called the eg2 state, which is equivalent to the state with both electrons in the eg2-orbitals. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/11%3A_Coordination_Chemistry_III_-_Electronic_Spectra/11.03%3A_Electronic_Spectra_of_Coordination_Compounds/11.3.02%3A_Correlation_Diagrams.txt |
Tanabe-Sugano diagrams are modified forms of correlation diagrams. The modification is simply that the energies of terms are plotted in terms of transition energy and $\Delta$ divided by a constant, $B$, called the Racha parameter. Terms are plotted relative to the lowest-energy term; in other words, the ground state term is plotted as the horizontal axis ($x$-axis). The Tanabe-Sugano diagram of a $d^2$ transition metal complex is shown below in Figure $1$. Tanabe-Sugano diagrams for other d-electron counts are available in the Reference Section of LibreTexts.
Tanabe-Sugano diagram of a d2 octahedral complex
The difference between the diagram shown in Figure $1$ and the previously-discussed correlation diagram for a $d^2$ metal ion is that the energy of the ground state is plotted horizontally, and the energy of all other terms are plotted relative to that. In the case of the $d^2$ electron configuration, the $^3T_{1g}$ term is the ground term and is plotted as a horizontal line. We can see that the ligand field strength on the x-axis is given in units of B (specifically as $\frac{\Delta}{B}$), and the energy of the terms is also given in units of B (specifically $\frac{E}{B}$. $B$ is a so-called Racah parameter, which is a quantum-mechanical energy unit for the electromagnetic interactions between the electrons. It is chosen because it provides “handy” numbers.
You can see that some lines in Figure $1$ are bent (eg the two $^1 A_{1g}$ terms), and some are straight. Bending of lines occurs when two terms interact with each other because they are close in energy and have the same symmetry. This is again an analogy to orbitals. Just as orbitals interact when they have the same symmetry type and similar energy, terms also interact when they have the same symmetry and similar energy. Without taking their interactions into account, their energies can cross when the energy of term A declines and the energy of term B increases with increasing field strength (Figure $2$). The closer the terms come to the point where they cross, the stronger their interactions, because their energies become more and more similar. These interactions lead to the "bending away" of the terms from each other, leading to bent curves. This means that curves for two terms of the same symmetry type will bend away in a Tanabe-Sugano diagram and never cross. For example the terms for the two $^1 A_{1g}$ terms bend away from each other and do not cross.
Next, let us think about which electron transitions would be allowed by considering spin selection and the Laporte rule. First, notice that all the terms in Figure $1$ have "gerade" symmetry and are labeled with a "g" subscript. In fact, this is the case for the terms of any octahedral complex. What does this mean for the allowance of electron transitions? It means that no electron transition would be allowed by the Laporte Rule, and that would imply that the complex could not absorb light. The Laporte selection rule, however, does not hold strictly. It only says that the probability of the electron-transition is reduced - but not forbidden. This means that an absorption band that disobeys the Laporte rule will have lower intensity compared to one that follows the Laporte rule, but it can still be observed. The spin-selection rule, however, holds strictly, and transitions between terms of different spin multiplicity are strictly forbidden, meaning that they have near zero probability to occur. Overall, we can therefore excite an electron from the $^3T_1$ ground state to other triplet terms, namely the $^3T_2$ term, and the $^3A_2$ term (Figure $1$).
Tanabe-Sugano diagram of d3 octahedral complexes
Now, let us have a look at the Tanabe-Sugano diagram of a d3 ion in an octahedral ligand field (Figure $3$). What is the ground term? We can see that the term designation on the horizontal line reads "$^4A_{2g}$”, therefore this term is the ground term. How many electron transitions from the ground state should we expect? To answer this question we need to count the number of other quartet terms. There is the $^4T_{2g}$, the $^4T_{1g}$, and another $^4T_{1g}$. Thus, there are overall three electron transitions possible.
Can we understand why the ground state is a quartet term? It helps to consider how we would fill the electrons into the d-orbitals for the electron configuration d3. All three electrons would be filled spin-up in the t2g orbital following Hund’s rule (Figure $4$). Because each electron has the spin +1/2 the total spin of all three electrons is 3x1/2=3/2. Thus, the spin multiplicity is ((2x3/2)+1)=4. Note that the microstate we have drawn is actually only one of the (2L+1)(2S+1) microstates. S=3x1/2=3/2, but what is L? You can see on the left side of the diagram that the 4A2 term originated from a 4F term. This means L=3, and (2L+1)((2S+1)=7x4=28. This means that there are actually 27 other microstates that have the same energy as the microstate that we drew. Why did we draw this microstate in favor of the others? This is because this microstate is the state with the maximum ML (=L) and Ms (=S) values determining the term symbol.
Tanabe-Sugano diagram of d4 octahedral complexes
Now let us look at the Tanabe-Sugano diagram of a $d^4$ octahedral complex (Figure $5$). You can see that this diagram is separated into two parts separated by a vertical line. The line indicates the ligand field strength at which the complex changes from a high spin complex to a low spin complex. At lower ligand field strengths, the ground term is a $^5E_g$ term (solid turquoise line). At higher field strength the ground term is a $^3T_{1g}$ term (dashed purple line).
We can rationalize this again by drawing the orbital box representation of the d-orbitals in the octahedral ligand field. In the high spin state, there are four unpaired electrons, thus S=4x1/2=2, and 2S+1=5. In the low spin state, there are two unpaired electrons, and thus S=2x1/2=1, and 2S+1=3. This explains the quintet and the triplet nature of the high and low spin ground terms. Note again, that the two microstates represented by the orbital box diagrams ($6$) are not the only microstates that have the respective energy. They are only the "representative" microstates because they have the maximum ML and MS values.
How many electron transitions are possible from the ground term? For a high spin complex there is only one because the is only one other quintet term, namely the $^5T_{2g}$ term. For the low spin complex, there are five transitions because there are five other triplet terms.
Tananbe-Sugano diagram of d5 octahedral complexes
The Tanabe-Sugano diagram of a d5 octahedral complex is also divided into two parts separated by a vertical line (Figure $7$). The left part reflects the high spin and the right part the low spin complex.
The high spin ground state is a sextet term, and the low spin ground state is a doublet term. We can understand the sextet and doublet nature of the terms when considering that the associated electron box diagrams have five and one unpaired electrons respectively. S=5x1/2=5/2 and 2S+1=6 for the high spin term, and S=1x1/2=1/2 and 2S+1=2 for the low spin term. What are the possible electron transitions from the ground state? For the high-spin complex there is no other sextet term, meaning that there is no electron transition possible. Hence, high-spin octahedral d5-complexes are colorless. An example is the hexaaqua manganese (2+) complex. A solution of this complex is near colorless, only very slightly pinkish. The slight color is because also spin-forbidden transitions can occur, albeit at a very low probability. For a d5-low spin complex there are three additional doublet states, and thus there are three electron transitions possible.
Tanabe-Sugano diagram of d6 octahedral complexes
The next diagram is the one for the d6 electron configuration (Fig. 8.2.13). Again, the diagram is separated into parts for high and low spin complexes. The dashed lines in the diagram indicate the terms that have a different spin multiplicity than the ground term. This way we can more easily see how many electron transitions are allowed.
The ground term for the high-spin complex is the quintet 5T2 term. It is a quintet term because four electrons in the d-orbitals are unpaired, and two are paired. The value for S is thus 4x1/2=2, and the spin multiplicity is 2S+1=5. The ground term for the low spin complex is a 1A1 term. It is a singlet term because all electrons are paired, and thus S=0, and 2S+1=1. How many electron transitions are there for the high-spin complex? There is only one because the 5E term is the only other quintet term. There are five transitions possible for the low-spin case because there are five additional singlet terms.
Tanabe-Sugano diagram of d7 octahedral complexes
Next, let us look at the Tanabe-Sugano diagram of a d7 octahedral complex (Fig. 8.2.15). In this case, the high spin complex has a 4T1 ground term, and the low spin complex has a 2E ground term.
The microstate that "represents" the high spin ground term has three unpaired electrons, hence the spin quantum number S=3/2 and the spin multiplicity is 2S+1=4. The microstate that represents the low spin ground term has one unpaired electron, an S value of ½, and a spin multiplicity of 2. There are three other quartet terms, and four other doublet terms, hence there are three electron transitions for the high-spin complex, and four for the low spin complex.
Tanabe-Sugano diagram of d8 octahedral complexes
Now let us look at an octahedral complex with d8 electron configuration. For this electron configuration, there are no high and low spin complexes possible, therefore, the Tanabe-Sugano diagram is no longer divided into two parts (Fig. 8.2.17).
There is a single ground term of the type 3A2. It is a triplet state because the microstate representing the term has two unpaired electrons in the eg orbitals (Fig. 8.2.18). Thus, S=2x1/2=1, and the spin multiplicity is 2S+1=2. How many electron transitions would you expect? There are three other triplet states, namely the 3T2 and two 3T1 terms. Therefore, there are three electron transitions possible.
Unnecessary diagrams for $d^1, \; d^9, \; d^10$
We could also ask: Are there Tanabe-Sugano diagrams for d1, d9, and d10? For, d1 there are no electron-electron interactions, thus the simple orbital picture is sufficient. The 2D term splits into T2g and Eg terms, and there is only one electron transition possible. The d9 electron configuration is the “hole-analog” of the d1 electron configuration. It has also just one 2D term which splits into a T2g and an Eg term in the octahedral ligand field. Therefore, also in this case there is only one electron transition from the T2g into the Eg term possible. In the case of d10 all microstates are filled with orbitals, and there is only the 1S term which does not split in an octahedral ligand field. Therefore, there are no electron transitions in this case.
Jahn-Teller distortions and other geometries
Some of the electron configurations discussed above are ones that are particularly susceptible to Jahn-Teller distortion. The Jahn-Teller theorem predicts that electron configurations with asymmetrically populated orbitals (i.e., those that have spin multiplicity of a doublet or greater), will distort. Distortions affect the electronic spectra of coordination complexes, and in practice Jahn-Teller distortions are significant only in the cases where the $e_g$ orbitals are asymmetrically populated (e.g., octahedral $d^9$ and high spin $d^4$).
Finally, it should be mentioned that it is also possible to construct Tanabe-Sugano diagrams for other shapes such as the tetrahedral shape, but we will not discuss these further here.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/11%3A_Coordination_Chemistry_III_-_Electronic_Spectra/11.03%3A_Electronic_Spectra_of_Coordination_Compounds/11.3.03%3A_Tanabe-Sugano_Diagrams.txt |
While the free ion terms symbols are derived using Russell-Saunders terms, we only briefly discussed how the term symbols may be found from a character table upon ligand-field splitting. The symmetry labels (called Mulliken Labels) that are used in character tables to describe the symmetry of wavefunctions were described in a previous section (Section 4.3.3). Just as orbitals are wavefunctions that are described by the irreducible representations that are part of a character table, so are the terms. When we are discussing the terms of an octahedral complex, we can apply what we already know about octahedral symmetry and symmetry labels to understand how the terms labels are assigned.
Upon close inspection of any of the Tanabe-Sugano diagrams shown previously, or in the Resources Section, you may notice that each free ion term (listed at the right of the diagram) splits into a term or set of terms that retain the same multiplicity (the left superscript). Under an octahedral field, all terms have symbols $A, B, E$ or $T$ because they are singly, doubly, or triply degenerate. While some of these terms have subscripts "1" or "2", all the terms also have "$g$" subscripts due to the symmetry of the $d$ orbitals. Some of the symbols might also look familiar, as they are similar to the labels you know for splitting of $d$ orbitals, like $E_g$ and $T_{2g}$.
It is useful to know that under an octahedral field, the $D, F, G, H, I$ terms split, but the terms $S, P$ do not split. From inspection of the $O_h$ character table, we find that free ion terms will split as follows:
$\begin{array}{|l|l|} \hline \textbf{Free Ion Term} & \textbf{Terms under } O_h\ \hline \text{S} & A_{1g} \ P & T_{1g} \ D & T_{2g}+E_g \ F & A_{2g}+T_{2g}+T_{1g} \ G & A_{1g}+E_g+T_{2g}+T_{1g} \ H & E_g+T_{1g}+T_{1g}+T_{2g} \ I & A_{1g}+A_{2g}+E_g+T_{1g}+T_{2g}+T_{2g} \ \hline \end{array} \nonumber$
The term symbols under an octahedral field have symmetry that matches symmetry of other elements within the molecule, including the $d$-orbitals. A reminder of their symmetry meanings is below (see also Section 4.3.3).
Review of Mulliken Labels
Each wavefunction (including terms) can be described by a Mulliken label (symmetry label). The labels used for terms in an octahedral ligand field are listed in the table below.
$\begin{array}{l|l} \hline \textbf{Mulliken Labels} & \textbf{meaning}\ \hline \text{A or B} & \text{singly degenerate} \ E & \text{doubly degenerate} \ T & \text{triply degenerate} \ \hline \textbf{Subscripts} & \textbf{meaning} \ \hline 1 & \text{symmetric to } \sigma_v \text{ or perpendicular } C_2 \ 2 & \text{anti-symmetric to } \sigma_v \text{ or perpendicular } C_2 \ g & \text{symmetric to inversion center} \ u & \text{anti-symmetric to inversion center} \ \hline \end{array}$
We can use our knowledge of symmetry to derive the term labels for a given electron configuration under an octahedral field. Let's focus only on the capital letter that is assigned as the term symbol. Assignment of the octahedral term subscripts will not be described here, but are similar to assignment of these same subscripts to orbitals with matching symmetry.
For example, the $d^1$ electron configuration has a free ion ground state term of $^2D$. According to the table shown above, these terms will split into $E_g$ and $T_{2g}$ terms. Which would be its ground state term under an octahedral field? And which is the excited state term? To answer these questions we need to look at the electron configurations of the ground and excited states in an octahedral field. It is convenient to know that the term symbol under an octahedral field indicates the degeneracies of the related electron configurations.
Let's take a look at the electron configurations of the split $d$-orbitals to determine which term is the ground state. In the ground state of a $d^1$ under an octahedral ligand field, there would be one electron in the $t_{2g}$ orbitals, thus the electron configuration is $t_{2g}^1e_g^0$. For simplicity we can consider only the cases where $m_s$ values are $+\frac{1}{2}$. There are three ways to put the electron in $t_{2g}$ therefore, the ground state is triply degenerate, with a symbol "$T$". This is illustrated in the left panel of Figure $1$.
In the excited state of a $d^1$ under an octahedral ligand field, there would be one electron in the $e_{g}$ orbitals, or a configuration of $t_2g^0e_g^1$. There are two ways to arrange the electron with $m_s=+\frac{1}{2}$ in the $e_{g}$ orbitals (Figure $1$, left). Therefore, the excited state is doubly degenerate, with a symbol "$E$".
In summary, the $^2D$ free ion term for $d^1$ splits into a lower-energy $^2T_{2g}$ and a higher-energy $^2E_g$ term under an octahedral field (Figure $1$, left).
We could use a similar treatment to derive the relative energies and terms of a $d^9$ ion in an octahedral field. The ground state term of $d^9$ is the same as for $d^1$: it is $^2D$. If we look at the ground state and excited state electron configurations of a $d^9$ ion, we find that the ground state configuration is $t_{2g}^6 e_g^3$, which is doubly degenerate. On the other hand, the excited state has an electron configuration of $t_{2g}^5 e_g^4$, and has triple degeneracy. In summary, in the case of $d^9$, the $^2D$ free ion term splits into a lower energy $^2E_g$ term and a higher-energy $^2T_{2g}$ term. This is exactly the opposite of $d^1$!
The $d^1$ and $d^9$ cases are opposite because they are related by what is said to be a "positive hole" concept. The $d^9$ configuration could be derived from a $d^{10}$ configuration with one positive "hole" (an electron removed). The positive hole is similar to the lone electron in $d^1$ except that in the ground state, the hole is in $e_g$, while the electron in $d^1$ is in $t_{2g}$; thus the two situations are exact opposites.
Just as $d^1$ and $d^9$ are related by the hole concept, so are $d^4$ and $d^6$. If you consider that $d^1$ is the addition of one electron to the totally symmetric $d^0$ case, then perhaps you can see how $d^6$ is similar in that it is the addition of one electron to the totally symmetric $d^5$ case. Thus, $d^1$ and $d^6$ have similar splittings under the octahedral field. Similarly, $d^9$ is created by removing one electron from the totally symmetric case of $d^{10}$ and $d^4$ is created by removing one electron from $d^5$. Thus $d^4$ and $d^9$ have identical splittings for their ground state free ion terms, which in these cases are a $D$ term.
Due to the relationships between $d^1, d^4, d^6,$ and $d^9$, the same Orgel Diagram is used to show term splitting for these cases (Figure $2$). Orgel diagrams are yet another form of correlation diagrams.
Just as there are relationships between $d^1,d^4,d^6,$ and $d^9$, there are similar relationships between $d^2,d^3,d^7,$ and $d^8$, and there is an Orgel diagram that can represent the splitting of their relevant free-ion terms.
Tetrahedral splitting
Because tetrahedral field splitting has the opposite pattern to octahedral field splitting of the $d$-orbitals, there are relationships between tetrahedral and octahedral terms. In general, for a tetrahedral field with $d^n$ electrons, the Tanabe-Sugano diagram of an octahedral diagram for $d^{10-n}$ can be used to interpret the tetrahedral case. The Orgel diagrams are labeled in general terms for this reason. They can be applied to both tetrahedral and octahedral cases.
Jahn-Teller distortions
We can also see what would happen to the terms in the case of Jahn-Teller distortion. In the case of a Jahn-Teller distortion (most common for $d^9$ and high-spin $d^4$ where the $e_g$ orbitals are asymmetrically occupied), we can use the degeneracy of orbital electron configurations to identify term labels. Let's walk through an example using the most simple case: a $d^1$ metal ion (Figure $3$).
Jahn-teller usually causes a tetragonal distortion of an octahedron. This would be a change from $O_h$ to $D_{4h}$ symmetry.
The $t_{2g}$ orbitals (the set $d_{xy}, d_{xz}, d_{yz}$ are split into $e_g$ ($d_{xz},d_{yz}$) and $b_{2g}$ ($d_{xy}$) by this symmetry change. Thus, we can expect any $T_{2g}$ terms to be split similarly into $E_g$ and $B_{2g}$ terms by Jahn-Teller distortion. The $e_g$ orbitals under $O_h$ are split into $a_1g$ ($d^{z^2}$) and $b_1g$ ($d_{x^2-y^2}$) under $D_{4h}$. Thus, we can expect $E_g$ terms to be split into $B_{1g}$ and $A_{1g}$ terms by Jahn-Teller distortion (Figure $4$).
Just as $d^1$ and $d^9$ gave opposite order of their terms under an octahedral field, they also give opposite order of terms in the Jahn-Teller distortion. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/11%3A_Coordination_Chemistry_III_-_Electronic_Spectra/11.03%3A_Electronic_Spectra_of_Coordination_Compounds/11.3.04%3A_Symmetry_labels_for_split_terms.txt |
The Tanabe-Sugano diagrams can be used to interpret absorption spectra and gain insight into the properties of a coordination complex. For example, you could use the appropriate diagram to predict the number of transitions, assign the identity of a specific transition, or calculate the value of $\Delta$ for a specific metal complex.
How to use the Tanabe-Sugano Diagrams
1. Determine the $d$-electron count of the metal ion of interest.
2. Choose the appropriate Tanabe-Sugano diagram: this is the one matching the $d$-electron count of the metal ion. There is a full list of Tanabe-Sugano diagrams in the Resources Section.
3. Acquire an electronic spectrum of the metal complex and identify $\lambda_{max}$ for spin-allowed (strong intensity) and spin forbidden (weak intensity) transitions.
4. Convert wavelength ($\lambda_{max}$) to energy (E) in wavenumbers ($cm^{-1}$) and generate energy ratios relative to the lowest-energy allowed transition. (i.e. $\frac{E_2}{E_1}$ and$\frac{E_3}{E_1}$).
5. Using a ruler, slide it across the printed Tanabe-Sugano 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, $Delta$ (Sometimes this is labeled as $10D_q$, and it is useful to know that $\Delta=10D_q$).
Example $1$: Chromium Splitting
A Cr3+ metal complex has strong transitions and $\lambda_{max}$ at
• 431.03 nm,
• 781.25 nm, and
• 1,250 nm.
Determine the $Δ_{oct}$ for this complex.
Solution
1. Cr has 6 electrons. Cr3+ has three electrons, so it has a d-configuration of d3
2. Locate the d3 Tanabe-Sugano diagram
3. Convert to wavenumbers:
$\dfrac{10^7(nm/cm)}{1250\; nm}= 8,000\; cm^{-1}$
$\dfrac{10^7(nm/cm)}{781.25\; nm}= 13,600\; cm^{-1}$
$\dfrac{10^7(nm/cm)}{431.03\; nm}= 23,200\; cm^{-1}$
1. Allowed transitions are $\ce{^4T_{1g}} \leftarrow \ce{ ^4_{\,}A_{2g}}$, $\ce{^4T_{1g} \leftarrow ^4_{\,}A_{2g}}$ and $\ce{^4T_{2g}\leftarrow ^4_{\,}A_{2g}}$.
Transition Energy cm-1 Ratios to lowest
$\ce{^4T_{1g}} \leftarrow \ce{ ^4_{\,}A_{2g}}$ 23,200 2.9
$\ce{^4T_{1g} \leftarrow ^4_{\,}A_{2g}}$ 13,600 1.7
$\ce{^4T_{2g}\leftarrow ^4_{\,}A_{2g}}$ 8,000 1
1. 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
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
1. Based on the two tables above it should be assessed that the Δoct/B value is 10. B is found by dividing E by the height.
Energy cm-1 Height B
23,200 29 800
13,600 17 800
8,000 10 800
1. Next multiply Δoct/B by B to yield the Δoct energy. $10 \times 800 = 8000\; cm^{-1}=Δ_{oct}$
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.
11.3.06: Tetrahedral Complexes
Transitions in tetrahedral complexes are Laporte-allowed
Tetrahedral metal complexes often have more intense electronic transitions than their octahedral counterparts. This is due to the fact that the $d-d$ transitions in a tetrahedron are allowed by the Laporte selection rule, while $d-d$ transitions in an octahedral complex are Laporte-forbidden. Recall that the Laporte selection rule applies to centrosymmetric complexes only. The Laporte rule applies to octahedral complexes but not to tetrahedral complexes because a tetrahedron does not have a center of inversion. Notice that the terms (and orbital labels) in a tetrahedron do not include the $g$ subscripts that are present under octahedral symmetry (Figure $1$). The splitting pattern of a tetrahedral complex is exactly opposite to the octahedral case. In the case of a tetrahedron, however, the "$g$" subscripts are inappropriate because of the tetrahedron's lack of a center of inversion, and transitions between the terms in a tetrahedron do not violate the Laporte Rule.
Another way to explain this in terms of electron transitions between orbitals is through the orbital mixing required to form a tetrahedral complex. Orbital types (i.e., $s,p,d$) must mix to form the moleculare orbitals of a tetrahedral transition metal complex. The mixing of $s$ and $p$ orbitals with the $d$ orbitals allows transitions that are forbidden in the case of pure d-orbitals.
It is also worth noting that $\Delta_t = \frac{4}{9} \Delta_o$. The smaller $\Delta$ for transition metals means that the tetrahedral complexes can absorb at a lower energy and longer wavelength relative to an analogous octahedron.
Tanabe-Sugano diagrams for tetrahedral complexes
Due to the opposite splitting pattern, the transitions for a $d^n$ tetrahedral complex are sufficiently represented by the $d^{10-n}$ Tanabe-Sugano diagram (just drop the $g$ subscripts from the diagrams). For example, the electronic spectrum of a $d^8$ tetrahedral complex (e.g., $\ce{[Ni(H2O)6]^2+}$) can be interpreted using the $d^2$ Tanabe-Sugano diagram. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/11%3A_Coordination_Chemistry_III_-_Electronic_Spectra/11.03%3A_Electronic_Spectra_of_Coordination_Compounds/11.3.05%3A_Applications_of_Tanabe-Sugano_Diagrams.txt |
Charge Transfer Transitions
We are still not done with our electronic spectra. Thus far, we have only considered transitions of d-electrons between d-orbitals, and their terms. These are called d-d transitions. However, there are also so-called charge transfer transitions possible, which are not d-d transitions. We can easily see that there must be transitions other than d-d transitions when we look at the colors of d10 and d0 ions. For those, there are no d-d transitions possible. Therefore, they all should be colorless. However, that is not always true. Some of these ions are indeed colorless, but some are not (Figure \(2\)). For example, Zn2+, a d10 ion, is colorless in complexes, but not Cu(I), which is also d10. While tetrakis(acetonitrile)copper(+) is colorless, bis(phenanthrene) copper(+) is dark orange. d0 ions have similar properties: While TiF4 and TiCl4 are colorless, TiBr4 is orange, and TiI4 is brown. Some d0 species are even extremely colorful, for example, permanganate with Mn7+, which is extremely purple, and dichromate with Cr(VI), which is bright orange.
The explanation for these phenomena is charge-transfer transitions (\(2\)). There are two types of charge-transfer transitions: ligand-to-metal (LMCT) and metal-to-ligand (MLCT) charge transfer transitions. For the ligand-to-metal transitions, electrons from bonding σ and π-orbitals get excited into metal d-orbitals in the ligand field, for example the t2g and the eg orbitals in an octahedral complex. If the energy difference between the σ/π-orbitals and the d-orbitals is small enough, then this electron transition is associated with the absorption of visible light. The transition is called a ligand-to-metal transition because the ligand σ/π-orbitals are mostly located at the ligands, while the metal-d-orbitals in a ligand field are mostly located at the metal. Vice versa, the metal-to-ligand transition involves the transition of an electron from metal d-orbitals in a ligand field to ligand π*-orbitals. This essentially moves electron density from the metal to the ligand, hence the name ligand-to-metal-charge transfer transition. If the energy-difference between the ligand π* and the metal orbitals is small enough, then the absorption occurs in the visible range. Charge-transfer transitions are usually both spin- and Laporte allowed; hence, if they occur, the color is often very intense. How can we distinguish between d-d and charge transfer transitions? Charge transfer transitions often change in energy as the solvent polarity is varied (solvatochromic), as there is a change in polarity of the complex associated with the charge transfer transition. This can be used to distinguish between d-d transitions and charge-transfer bands.
LMCT Transitions
Can we predict when the energy windows between the bonding molecular orbitals and the metal d-orbitals are small enough for LMCT transitions in the visible to take place? Generally, it would be desirable if the energy of the metal orbitals were as low as possible and the energy of the bonding ligand orbitals were as high as possible. The energy of metal d-orbitals decreases with increasing positive charge at the metal because the effective nuclear charge on the metal increases. This means that very high metal oxidation states favor LMCT transitions. The d-orbitals should have few or no electrons, so that electrons can be promoted into the orbitals, and orbital energy decreases because electron-electron repulsion is minimized. Examples are Mn(VII), Cr(VI), and Ti(IV). The energy of MOs from bonding ligand orbitals increases when the ligand orbitals have high energy; this is typically the case for π-donor ligands with a negative charge (Fig. 8.2.21).
Examples of ligands are oxo- and halo ligands. This explains, for example, the LMCT transitions in permanganate. The Mn is in the very high oxidation state +7, and the ligands are oxo-ligands, which are π-donors with a 2- negative charge. The transitions are both Laporte and spin-allowed, leading to a very high intensity of light absorption, and thus color (Fig. 8.2.21).
MLCT Transitions
What are favorable metal ion and ligand properties for a metal-to-ligand transition, then? In this case we would like to keep the energy of the metal orbitals as high as possible so that the energy difference between a metal d-orbital and a π*-orbital is minimized. This is accomplished when the positive charge at the metal ion is small, and there are many d-electrons that can repel each other, thereby increasing orbital energies, for example Cu(I), Fig. 8.2.22.
The ligand should be a π-acceptor with low-lying π*-orbitals - for example, phenanthroline, CN-, SCN-, and CO. The bis(phenanthroline) copper(+) ion, for instance, is dark orange and has an MLCT absorption band at 458 nm. This MLCT transfer is both spin and Laporte-allowed.
It should be mentioned that some complexes allow for both metal-to-ligand and ligand-to-metal transitions. For example, in the Cr(CO)6 complex, the σ-orbitals are high enough and the π*-orbitals are low enough in energy to allow for light absorption in the visible range. Finally, intraligand bands are also possible when the ligand is a chromophore.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/11%3A_Coordination_Chemistry_III_-_Electronic_Spectra/11.03%3A_Electronic_Spectra_of_Coordination_Compounds/11.3.07%3A_Charge-Transfer_Spectra.txt |
Transition metal complexes with strong $\pi$-accepting ligands have proven quite useful across many applications in industry, materials science, and medicine due to their intense absorption of light, photostability, and unique geometric, redox, and photodynamic properties. These complexes usually have low-valent metal ions (low oxidation state) with $\pi$-accepting ligands constructed of conjugated poly-aromatic rings. Intense light absorption occurs through metal-to-ligand charge transfer (MLCT), causing a formal oxidation of the metal ion and reduction of the ligand. These compounds are used as sensors, and their excited states can be used as indispensable reactants. Some examples of their applications are discussed briefly below.
Ruthenium bypyridine derivatives
Let's use the example of a simple and interesting metal complex: tris(bypyridine)ruthenium(II), or $\ce{[Ru(bpy)_3]^2+}$. This molecule is the most generic form of a family of derivatives that have this core structure in common. $\ce{[Ru(bpy)_3]^2+}$ absorbs ultraviolet light at 452 nm due to an intense MLCT. It also absorbs light at shorter wavelenghts due to ligand-centered $\pi * \leftarrow \pi$, metal-centered $d \leftarrow d$, and LMCT transitions. The ground state is a singlet state (all paired electrons), and the spin-selection rule allows formation of a singlet excited state. However, in the case of high-atomic-number $d^6$ metal ions, like Ru, there is intense spin-orbit coupling that favors intersystem crossing. Intersystem crossing is the phenomenon by which a system's spin multiplicity can change by the transition of an electron's spin without energy cost. In the case of $\ce{[Ru(bpy)_3]^2+}$, the excited singlet state transitions to an isoenergetic (or degenerate) triplet excited state. $\ce{[Ru(bpy)_3]^2+}$ has a long-lived excited state due to the fact that the transition between the triplet excited state and the singlet ground state is spin-forbidden. The excited state can relax to the singlet ground state radiatively (releasing a photon) or non-radiatively (as phonons/heat).
Bioimaging
The $\ce{[Ru(bpy)_3]^2+}$ complex is fluorescent, with a quantum yeld of approximately 2%. Derivatives of this core structure offer a variety of interesting properties, including a tunable color of emission. Other modifications have been made to make Ru bipyridine derivatives more functional for cellular imaging. For example, these metal complexes have been modified to bind to carbohydrates, specific organelles, or specific organs for targeted biological imaging applications.
Energy and catalysis
Due to the long lifetime of the $\ce{[Ru(bpy)_3]^2+}*$ excited triplet state, this molecule has potential as a photosensitizer for the oxidation and reduction of water (water splitting) for energy production. The MLCT excitation is essentially a removal of an electron from Ru (an oxidation of Ru) and a placement of an electron onto a bpy ligand (a reduction of the ligand). The resultant $\ce{[Ru^3+]}$ center is a powerful oxidant, capable of oxidizing water, while the unpaired electron on reduced bpy is a powerful reductant. The power of this complex for catalyzing the splitting of water has been demonstrated using an electron mediator, methyl viologin, a Pt catalyst, and a sacrificial reductant. Once the excited $\ce{[Ru(bpy)_3]^2+}*$ has formed, the excited electron can be captured by methyl viologen and passed to a Pt catalyst which reduces hydrogen ion ($\ce{H^+}$) to $\ce{H_2}$. The oxidized $\ce{[Ru(bpy)_3]^3+}$ complex is then reduced by reaction with a sacrificial reductant like EDTA or triethanolamine to re-form $\ce{[Ru(bpy)_3]^2+}$.
$\ce{[Ru(bpy)_3]^2+}$ is also used as a photosensitizer for organic synthesis. Many analogues of $\ce{[Ru(bpy)_3]^2+}$ are employed as well. These transformations exploit the redox properties of excited $\ce{[Ru(bpy)_3]^2+}*$ and its reductively quenched derivative $\ce{[Ru(bpy)_3]^+}$. (from Wikipedia)
Other applications
Derivatives of $\ce{[Ru(bpy)_3]^2+}$ are numerous. Such complexes are widely discussed for applications in biodiagnostics, photovoltaics, and organic light-emitting diodes, but no derivative has been commercialized. Application of $\ce{[Ru(bpy)_3]^2+}$ and its derivatives to fabrication of optical chemical sensors is arguably one of the most successful areas so far. (from Wikipedia) | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/11%3A_Coordination_Chemistry_III_-_Electronic_Spectra/11.03%3A_Electronic_Spectra_of_Coordination_Compounds/11.3.08%3A_Applications_of_Charge-Transfer.txt |
Although there are a vast array of reactions that occur at transition metal centers, we will touch on only a subset of the possibilities. This chapter will focus on the reactions of octahedral and square planar metal complexes, and the reactions will be divided into three types:
1. Substitution reactions at the metal center: these are reactions in which one ligand is replaced with another ligand within the metal's inner coordination sphere.
2. Redox: these are reactions in which the metal oxidation state formally changes by gaining or loosing electrons.
3. Ligand reactions: these are reactions that occur primarily on the ligands that are bound to metal ions.
Review of reaction coordinate diagrams and terms
Before we begin our discussion of reactions, let's review some of the vocabulary and models that chemists use to understand reactions.
A chemical reaction can be represented with a reaction coordinate diagram in which the total potential energy of a state is plotted in the vertical $y$ axis, and the "reaction coordinate" (sometimes also referred to as "reaction progress", or "extent of reaction") is plotted on the horizontal $x$ axis. The reaction coordinate is often drawn as the progress of a reaction from reactants (R) on the left to products (P) on the right.* Examples of reaction coordinate diagrams for a one-step and a two-step reaction are shown in Figure $1$. Notice the difference in the reaction profile that depends on the number of reaction steps. A one-step reaction has one "hill" shape, while a two step reaction takes on a "saddle" shape.
Thermodynamics
The difference in energy between reactants (R) and products (P) indicates whether reactants or products will dominate at thermodynamic equilibrium. At standard conditions, this difference is the standard Gibbs Free Energy of the reaction ($\Delta G^{\circ}_{rxn}$ or $\Delta G^{\circ}$). The Gibbs free-energy change is a combination of enthalpy change ($\Delta H^{\circ}$) and entropy change ($\Delta S^{\circ}$):
$\Delta G^{\circ} = \Delta H^o - T\Delta S^o \nonumber$
where
• $T$ is the temperature in Kelvin (recall that the Kelvin temperature is simply the Celsius temperature plus 273.15).
• Enthalpy change ($\Delta H^{\circ}$) is the heat released or absorbed by the reaction.
• Entropy change ($\Delta S^{\circ}$) is the change in disorder from reactants to products. In a reaction in which one molecule cleaves into two smaller molecules, for example, disorder increases, so $\Delta S^{\circ}$ is positive.
The standard Gibbs free energy change for a reaction can be related to the reaction's equilibrium constant $K_{eq}$ by the equation:
$\Delta G^{\circ} = -RT \ln K_{eq} \nonumber$
where $R$ is the gas constant (8.314 J/mol×K) and $T$ is the temperature in Kelvin (K).
A negative value for $\ce{\Delta G_{rnx}^{\circ}}$ (an exergonic reaction) corresponds to $\ce{K_{eq}}$ being greater than 1, an equilibrium constant which favors product formation. Conversely, an endergonic reaction has a positive value of $\ce{\Delta G ^{\circ}}$, and a $\ce{K_{eq}}$ between 0 and 1.
Kinetics
The highest point between two energy minima is a transition state (TS). The difference in energies of the reactant and the highest-energy TS is the standard free energy of activation ($\ce{\Delta G^{\circ \ddagger}}$). The activation energy, in combination with the temperature at which the reaction is being run, determines the rate of a reaction: the higher the activation energy, the slower the reaction. The rate constant ($k$) is the proportionality constant relating the rate of the reaction to the concentrations of reactants. The relationship between activation energy and the rate constant, $k$, is given by the Arrhenius equation:
$k = Ae^{-\frac{E_a}{RT}} \quad \quad \text{or} \quad \quad \ln k=\ln A-\frac{E_{A}}{R T} \nonumber$
A rate law is an expression showing the relationship of the reaction rate to the concentrations of each reactant. The rate is defined as the change in reactants or products over time. For example, in a reaction of two reactants ($R_1$ and $R_2$) to form product (P), a rate law could be written as follows:
$\frac{d\ce{[P]}}{dT}=\ce{k[R_1][R_2]} \nonumber$
The rate law above is second order overall, and first order with respect to both $R_1$ and $R_2$. The order indcates the power of reactant concentrations for individual reactants or overall for the reaction.
In a one-step reaction, the product is formed in one step. There is an energy barrier to overcome so that this process can happen. At the height of that energy barrier is the TS. In a one-step reaction, there is one transition state. In contrast, in a two-step reaction, there are two transition states and an intermediate (denoted by the letter I). The first step from R to I, passes over transition state $TS_1$, is endergonic. The second step passing through transition state $TS_2$ is exergonic. The intermediate (I) is thus depicted as an energy 'valley' (a local energy minimum) situated between the two energy maxima; $TS_1$ and $TS_2$.
In Figure $1$B, notice that the energy barrier for the first step is larger than the energy barrier for the second step. This means that the first step is slower than the second. In a multi-step reaction, the slowest step - the step with the largest energy barrier - is referred to as the rate-limiting or rate-determining step. The rate-determining step can be thought of as the 'bottleneck' of the reaction: a factor which affects the rate-determining step will affect the overall rate of the reaction. Conversely, a factor which affects only a much faster step will not significantly affect the rate of the overall reaction.
Summary and Key Terms
Reaction coordinate diagrams convey some very important ideas about how the reaction can proceed, and about the thermodynamics and kinetics of the reaction. In the sections that follow, the following key terms will be used:
Description of Key Terms
Reactants (R): The molecular species that are the starting materials for a chemical reaction. These are often written at the left side of a reaction coordinate diagram.
Products (P): The molecular species that result from a chemical reaction. These are often written at the right side of a reaction coordinate diagram.
Transition State (TS): a transition state is the transient species that exists at an energy maximum. In other words, TS's are highest-energy structures along a reaction pathway. Hammond's postulate suggests that the structure of a transition state should resemble the lower-energy species (R, P, or I) adjacent to it along the reaction coordinate, and most closely resembles the structure of the species closest in energy.
Intermediate (I): an intermediate is a high-energy structure at a local minimum. Intermediates are sometimes thought of as long-lived transition states. Unlike transition states, intermediates sometimes exist long enough and in high enough concentration to be detectable experimentally.
Microscopic reversibility: if a reaction can proceed in the forward direction from R to P, then in principle, it can also proceed in the reverse direction from P to R. There may be many possibilities for how a reaction could happen, but the lowest-energy pathway in the forward direction is also the lowest energy pathway in the reverse direction. In other words, the reactions passes through the same lowest-energy I and TS in the forward and reverse directions.
Steady-State Approximation: the concentration of an intermediate is assumed to react as quickly as it forms. This approximation allows us to assume that concentration of an intermediate remains "steady" throughout the reaction progress. This approximation is useful for analysis of reaction kinetics.
Order: The order of a reaction indicates the power of total reactant concentration in a kinetic rate law. The order of a reactant indicates that reactant's power in the rate law. For example, is a rate law that is $\ce{Rate = k [A]^2[B]}$, the reaction is second order in A, first order in B and third order overall. If there were also a reactant, C, that was not part of the rate law, the reaction would be zero order with respect to C.
Exercise $1$
Figure $1$ is replicated below for convenience. This figure shows a reaction coordinate diagram for a two-step reaction.
a. ($\Delta G^{\circ}$) and $\Delta G^{\circ \ddagger}$) are labeled on Figure $1A$, but not on Figure $1B$. Add the appropriate labels to diagram B. (copied below for convenience).
b. Consider the energy barriers for the first and second steps and explain why the steady-state approximation is a valid assumption.
Answer
a) The activation energy for this reaction is the difference between R and $TS_2$ because $TS_2$ is the highest energy along the reaction pathway.
b) The activation energy for step 2 is much smaller than for step 1. This tells us that step 2 occurs faster than step 1. So, the concentration of intermediate should not increase over the course of the reaction, and it should react at least as fast as it is formed. The fact that the activation barrier for step 2 is less than for step one allows us to assume that [I] does not change over the course of the reaction. This should generally be true for all intermediates.
Exercise $1$
Use the reaction coordinate diagram below to answer the questions.
1. Is the overall reaction endergonic or exergonic in the forward (A to D) direction?
2. How many steps does the reaction mechanism have?
3. How many intermediates does the reaction mechanism have?
4. Redraw the diagram if necessary. Add a label showing the energy barrier for the rate-determining step of the forward reaction.
5. Add a label showing $\ce{\Delta G^{\circ}_{rxn}}$ for the reverse reaction (D to A).
6. What is the fastest reaction step, considering both the forward and reverse directions?
Answer
1. Is the overall reaction endergonic or exergonic in the forward (A to D) direction?
2. How many steps does the reaction mechanism have?
3. How many intermediates does the reaction mechanism have?
4. The first step is rate determining since it has the largest activation barrier.
5. $\ce{\Delta G^{\circ}_{rxn}}$ for the reverse reaction (D to A) is the same as for the froward direction, with a revers of its sign.
6. The fastest step has the lowest barriers: this would be the step to/from B and C.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
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Ligand substitution refers to the replacement of one ligand in a coordination complex with another ligand.
Remember, a ligand in coordination chemistry is just a Lewis base that binds to a metal atom or ion. It does so by donating a lone pair (or other pair of electrons). Generallly, this donation is reversible. The donor can always take its electrons back. Typically, there may be some balance between the metal's need for more electrons and the donor's attraction for its own electrons; donor atoms are frequently more electronegative than the metal.
Even though the reaction is pretty simple, it can occur in different ways.
That is, the elementary steps involved in the reaction can occur in different orders. The elementary reactions are the individual bond-making or bond-breaking events that lead to an overall change. Sometimes the order of steps is referred to as the mechanism or the mechanistic pathway.
• The mechanism is the order of elementary reaction steps.
• Elementary reaction steps are individual bond-making and breaking steps.
You may have seen reaction mechanisms before. For example, carbonyl addition chemistry can involve lengthy mechanisms, in which a number of proton transfers and other bond-making and bond-breaking steps must occur to get from one state to another. Because ligand substitution is simpler than that, it is a good place to study mechanism in a little more depth, without getting overwhelmed by the details.
The sequence of steps in the mechanism influences how different factors will impact the reaction. For example, changing concentrations of different components in a reaction mixture can affect the time it takes for a reaction to finish.
• The mechanism can have a dramatic impact on the outcome of the reaction under different circumstances.
These kinds of considerations have a dramatic impact on industrial processes such as pharmaceutical production. In that setting, chemical engineers need to make decisions about how much of each reactant must be admitted to a reaction mixture and how long they should be allowed to react together. If they allow the reaction to proceed for too, long, there may be “side-reactions” that start to occur, interfering with the quality of the product, and they will waste valuable time in the production pipeline. If they don’t allow it to react long enough, the reaction may not finish, and the product will be contaminated with leftover starting materials.
In this chapter, we will look at how this simple reaction can occur in different ways. We will see some different methods that are used to tell which way the reaction occurs (i.e. evidence of what is really happening). We will also look at some different factors that may influence whether the reaction is likely to occur one way or the other (i.e. reasons it is happening that way, or reasons we expect it will happen that way).
Exercise \(1\)
Some kind of substitution occurs in each of the following reactions: an atom or group replaces another. In each case, identify what is being replaced, and what replaces it.
12.2.02: Inert and Labile Complexes
Kinetics of Substitution Reactions
Kinetics is a branch of chemistry that is concerned with the rates of chemical reactions. In this section, we will discuss the rates of metal-ligand (M-L) substitution reactions.
Let's start with some examples. In the table below are three examples of ligand substitution reactions of hexaaquo metal complexes to form hexaammine complexes. These reactions are nearly identical with the exception of the metal ion. The products are thermodynamically favored in all cases.
Reaction Rate constant ($k$) Labile or Inert
$\ce{[Ni(OH2)]^{2+} + 6 NH3 <=>> [Ni(NH3)6]^{2+}} \quad$ $k = 10^{4} s^{-1}$ (1 ns) Labile (happens in < 1 min)
$\ce{[Cr(OH2)]^{3+} + 6 NH3 <=>> [Cr(NH3)6]^{3+}} \quad$ $k = 10^{-3} s^{-1}$ (6 days) Inert (slow, takes hours)
$\ce{[Cu(OH2)]^{2+} + 6 NH3 <=>> [Cu(NH3)6]^{2+}}\quad$ $k = 10^{8} s^{-1}$ Very Labile (happens in seconds)
Definitions
• Kinetically Labile - Metal complexes that undergo "kinetically fast" substitution reactions are kinetically labile. Taube suggested that these are reactions in which half of the reactant is consumed in one minute or less ($t_{1/2}$ <1 min).
• Kinetically Inert - Metal complexes that undergo "kinetically slow" substitution reactions are kinetically inert or non-labile. Taube suggested that these are reactions in which $t_{1/2}$ >1 min.
A common pitfall is to confuse the meaning of kinetic terms, like labile and inert, with thermodynamic terms, like stable and unstable. It is important to distinguish between kinetics and thermodynamics. For example, the complex $[Fe(H_2O)Cl]^{2+}$ has a large formation constant and is thermodynamically stable; yet it is also labile. On the other hand, the complex $[Co(NH_3)_6]^{3+}$ is unstable in acidic aqueous solution and decomposes spontaneously to $[Co(H_2O)_6]^{3+}$; yet it decomposes slowly because it is inert. It is good practice to use clear terms such as "kinetically labile" or "kinetically inert" and "thermodynamically stable" and "thermodynamically unstable".
Exercise $1$
Draw the reaction coordinate diagrams for a reaction of the form $[ML_6]^{n+} + X \rightleftharpoons [ML_5X]^{n+} + L$ in the following scenarios:
1. $[ML_6]^{n+}$ is theromdynamically stable and kinetically inert.
2. $[ML_6]^{n+}$ is theromdynamically unstable and kinetically inert.
3. $[ML_6]^{n+}$ is theromdynamically stable and kinetically labile.
4. $[ML_6]^{n+}$ is theromdynamically unstable and kinetically labile.
Answer a)
Answer b)
Answer c)
Answer d)
Factors that affect rates of substitution reactions:
Some of the factors that affect the kinetic rates of ligand substitution are the same factors that affect thermodynamic stability (see Chapter 10). The same factors that make a complex stable can also make it more inert. Why are kinetic factors related to thermodynamic stability? It is because the structure and stability of the reactant complex is related to the structure and stability of the transition state. The reactant complex must change its geometry to form an intermediate or transition state. When the reactant is particularly stable, it can result in a higher activation energy associated with moving away from the stable configuration. However, it is incorrect to assume that stability is always correlated with reaction rates. Kinetic and thermodynamic factors are related, yet separate.
There are three important factors that influence kinetic rate of substitution:
1. Ligand Field Stabilization Energy (LFSE): Electron configurations that place electrons in higher-energy orbitals (particularly antibonding orbitals) result in more labile complexes. As long as there are not electrons in higher-energy orbitals, the lability correlates roughly with LFSE. The more negative the LFSE, the more inert.
2. Coulombic attraction between the metal and ligand: In general, higher charge density on the metal ion or on the ligand(s) leads to stronger electrostatic attraction between metal and ligand. Stronger Coulombic attraction generally leads to slower dissociation steps and faster association steps. The effect that these have on the reaction rate depends on elementary steps that take part in the rate-determine step(s).
3. Denticity: Multi-dentate ligands create particularly inert complexes as a result of the kinetic chelate effect.
All three considerations are described in more detail below.
Exercise $2$
In which compound from each pair would you expect the strongest ionic bonds? Why?
a) LiF vs KBr
b) CaCl2 vs. KCl
Answer a
The ions in LiF are both smaller than in KBr, so the force of attraction between the ions in LiF is greater because of the smaller separation between the charges.
Answer b
Calcium has a 2+ charge in CaCl2, whereas potassium has only a + charge, so the chloride ions are more strongly attracted to the calcium than to the potassium.
Taube's observations of metal complex substitution rates
Henry Taube (Nobel Prize, 1983) tried to understand kinetic lability by comparing the factors that govern bond strengths to observations about the rates of reaction of coordination complexes. He saw some things that were unsurprising. He also drew some new conclusions based on ligand field theory.
Taube observed that many M+1 ions (M = metal) are more labile than many M+3 ions, in general. That isn't too surprising, since metal ions function as electrophiles (Lewis acids) and ligands function as nucleophiles (Lewis bases) in forming coordination complexes. In other words, metals with higher charges ought to be stronger Lewis acids, and so they should bind ligands more tightly. However, there were exceptions to that general rule. For example, Taube also observed that $\ce{Mo+5}$ compounds are more labile than $\ce{Mo+3}$ compounds. So, there must be more going on here than just the effects of electrostatic attraction.
Another factor that governs ionic bond strengths is the size of the ion. Typically, ions with smaller atomic radii form stronger bonds than ions with larger radii. Taube observed that Al3+, V3+, Fe3+ and Ga3+ ions are all about the same size. All these ions exchange ligands at about the same rate. That isn't surprising, because they have the same charge and the same radius. However, Cr3+ is also about the same size as those ions and it also has the same charge, but it is much less labile. Once again, there are exceptions to our regular expectations based on simple electrostatic considerations. Furthermore, 4d and 5d transition metals (Y$\rightarrow$Cd, and Ac$\rightarrow$Hg) are much more inert than 3d transition metals (Sc$\rightarrow$Zn). This is unexpected when we consider size; the 4d and 5d metals are much larger than the 3d metals. This unexpected behavior tells us that electrostatics alone cannot predict lability.
Taube came up with a hypothesis that could explain the seeming contradictory oservations described above: kinetic lability must be affected by d-electron configuration. This idea forms the basis of Taube's rules about lability.
For example, metals like Ni2+ and Cu2+ are very labile. The d orbital splitting diagrams for those compounds would have d electrons in the eg set. Remember, the eg set arises from interaction with the ligand donor orbitals; this set corresponds to a $\sigma$ antibonding level.
By comparison, V2+ is rather inert. The d orbital splitting diagram in this case has electrons in the t2g set, but none in the eg set.
So, having electrons in the higher energy, antibonding eg level weakens the bond to the ligand, so the ligand can be replaced more easily. In the absence of those higher energy electrons, the bond to the ligand is stronger, and the ligand isn't replaced as easily.
On the other hand, metals like Ca2+, Sc3+ and Ti4+ are pretty labile. The d orbital splitting diagrams in those cases are pretty simple: there are no d-electrons at all in these ions.
That means having no electrons in these mostly non-bonding levels leaves the complex susceptible to ligand replacement. But it's hard to see why population of an orbital that is mostly non-bonding would have an effect on ligand bond strength.
Instead, this factor probably has something to do with the part of ligand substitution that we have ignored so far. Not only does one ligand need to leave, but a second one needs to bond in its place. So, having an empty orbital for the ligand to donate electrons into (or, put another way, not having electrons in the way that may complicate donation from the ligand) makes that part of the reaction easier.
Exercise $3$
Consider all possible electron configurations for octahedral complexes ($d^0$ to $d^{10}$, high spin and low spin cases): predict whether each case would be inert, intermediate, or labile.
Answer
Kinetically Inert (Slow) Kinetically Labile (Fast)
$d^n$ electron count and ligand field strength
Octahedral configurations with empty $e_g$ orbitals:
• $d^3$
• low-spin $d^4, d^5, d^6$
Strong-field $d^8$ (square planar)
Octahedral configurations with occupied $e_g$ orbitals:
• $d^1,d^2, d^7,d^9,d^10$
• high-spin $d^4,d^5, d^6$
Weak-field $d^8$ (usually tetrahedral)
Exercise $4$
Put the metal ions in order of decreasing reaction rate (from labile to inert):
a) $Al^{3+}, Na^+, Mg^{2+}$
b) $Ca^{2+}, Mg^{2+}, Sr^{2+}$
Answer (a)
Most Labile to most inert: $Na^+ > Mg^{2+} > Al^{3+}$
These are metal ions with similar size and varying charge. They are in order of increasing charge and increasing density from left to right.
Answer (b)
Most labile to most inert: $Sr^{2+} > Ca^{2+} > Mg^{2+}$
These metal ions have the same charge, and vary in size. They are in order of decreasing charge and increasing charge density from left to right.
Exercise $5$
Some metals, like Mn2+, can be either labile or inert, depending on whether they are high spin or low spin. Explain why using d orbital splitting diagrams.
Answer
Add texts here. Do not delete this text first.
Exercise $6$
Predict whether the following metals, in octahedral complexes, are labile or not.
a) Co3+ (high spin)
b) Co3+ (low spin)
c) Fe2+ (low spin)
d) Fe2+ (high spin)
e) Zn2+
Answer
Answer a
labile (electrons in higher energy d orbital set)
Answer b
not labile (all electrons in lower energy d orbitals)
Answer c
not labile (all electrons in lower energy d orbitals)
Answer d
labile (electrons in higher energy d orbital set)
Answer e
labile (electrons in higher energy d orbital set)
Generalizations for how metal ion affects kinetics:
These are some generalizations about how the kinetics of substitution is affected by the metal ion identity and the $d^n$ electron configuration.
1. s-block metals are very labile, except for those with very high charge density (eg. $Mg^{2+}\ is\: inert$)
2. $d^{10}$ metals are labile (eg: $Zn^{2+}, Cu^+, Hg^{2+}$)
3. Other ions with a full shell are labile (eg: $Ln^{3+}$ of f-block)
4. 3d $M^{2+}$, when high spin, are generally labile (eg. $Cu^{2+} is\: very\: labile$)
5. 4d and 5d are usually inert due to higher LFSE (low spin, high LFSE)
6. $M^{2+}$ is more labile than the same metal as $M^{3+}$
7. $d^3$ and low spin $d^6$ are inert (eg. $Cr^{3+}, Co^{3+},$ low spin $Fe^{2+}$)
Chelate Complexes
In addition to considering factors related to the metal ion and charge or covalent character of metal-ligand bonds, an important consideration is the effect a chelate ligand on reaction kinetics. This effect is also discussed in more detail later in this chapter (Section 12.4.5).
Recall the Thermodynamic Chelate Effect
Chelating complexes tend to be more stable than complexes with monodentate ligands. This is called the “thermodynamic chelate effect”. The effect deserves an explanation. The explanation is the increase of entropy that occurs when two or more monodentate ligands are replaced by a chelating ligand. The entropy increases because the overall number of particles increases as the substitution takes place.
For example, the substitution of six ammine ligands in the hexaammine nickel (2+) complex by three ethylenediamine chelating ligands increases the number of molecules from four to seven, and hence the entropy increases, in this case by 88 J K-1 and mol-1 (Figure $3$)
The Kinetic Chelate Effect
In addition to the thermodynamic chelate effect, there is the kinetic chelate effect. Chelate complexes are frequently more inert than complexes with monodentate ligands. Chelate complexes are more inert for two reasons (Figure $4$).
Firstly, the whole ligands needs to rotate and bend in order to cleave the first metal-ligand bond. This requires time and slows the kinetics of the bond cleavage. The second reason is that the detached donor atom cannot leave the proximity of the complex because the ligand is still attached via the other donor atom. This increases the probability of the re-formation of the metal-ligand bond which decreases the probability of both bonds being cleaved.
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A mechanism is the sequence of elementary steps by which a reaction proceeds. There are two aspects that describe the mechanism of a substitution reaction. One is called the stoichiometric mechanism, and the other one is called the intimate mechanism. The stoichiometric part of the mechanism is defined by the identity of the intermediate, or lack thereof. The intimate part of the mechanism is defined by the rate limiting step, and how the rate is dependent on the identity of the incoming ligand.
Stoichiometric Mechanisms
The stoichiometric part of the mechanism is defined by the identity of the intermediate, or lack thereof. There are three classes of stoichiometric mechanisms based on the type of intermediates that can be characterized. If the intermediate can be isolated it is either an Associative (higher-coordination number intermediate) or Dissociative (lower-coordination number intermediate) reaction. On the other hand, if the intermediate cannot be isolated because it is short-lived (or does not exist) the reaction is classified as Interchange.
Associative (A) Mechanism
Associative (A) mechanisms involve a first step where the incoming ligand bonds to the metal ion, creating an intermediate with a higher coordination number. The leaving group leaves in a second step to form the product. The generic reaction below illustrates the steps of an associative mechanism. Note that the intermediate, $\color{red}{\ce{ML_{n}XY}}$, has a higher coordination number than that of the reactant, $\ce{ML_{n}X}$.
$\begin{array}{rc} \text{Overall Reaction:} & \ce{ML_{n}X + Y <=> ML_{n}Y + X} \ \hline \text{Step 1 (Association):} & \ce{ML_{n}X + Y <=>} \color{red}{\ce{ML_{n}XY}}\ \text{Step 2 (Dissociation):} & \textcolor{red}{\ce{ML_{n}XY}} \ce{<=> ML_{n}Y + X} \end{array} \nonumber$
Associative mechanisms are typical of square planar, $d^8$ complexes. The important features that distinguish this mechanism from others is that the intermediate is long-lived enough that it is detectable, and that the intermediate has a higher coordination number than the reactant complex.
Dissociative (D) Mechanism
Dissociative (D) mechanisms involve a first step where a bond between the metal ion and the leaving group breaks, creating an intermediate with a lower coordination number. The entering group enters in a second step to form the product. The generic reaction below illustrates the steps of a dissociative mechanism. Note that the intermediate, $\color{blue}{\ce{ML_{n}}}$, has a lower coordination number than that of the reactant, $\ce{ML_{n}X}$.
$\begin{array}{rc} \text{Overall Reaction:} & \ce{ML_{n}X + Y <=> ML_{n}Y + X} \ \hline \text{Step 1 (Dissociation):} & \ce{ML_{n}X <=>}\textcolor{blue}{\ce{ML_{n}}}\ce{ + X}\ \text{Step 2 (Association):} & \textcolor{blue}{\ce{ML_{n}}}\ce{ + Y <=> ML_{n}Y} \end{array} \nonumber$
The important features that distinguish this mechanism from others is that the intermediate is long-lived enough that it is detectable, and that the intermediate has a lower coordination number than the reactant complex.
Interchange (I) Mechanism
Interchange (I) Mechanisms take place in one concerted step where the entering group enters as the leaving group leaves. Bond formation and bond breaking occur simultaneously. In the case of an interchange mechanism, no intermediate is detectable. This means that either there is no intermediate, or that the intermediate is too high-energy and short-lived to be detected. The generic reaction below illustrates the single step of an interchange mechanism. Note that the species, $\color{green}{\ce{[Y\bond{...}ML_{n}\bond{...}X]^{\ddagger}}}$, can be defined as either a transition state or a very short-lived intermediate. The distinction between these two possibilities is difficult; when the intermediate is undetectable it is not considered a true intermediate.
$\begin{array}{rc} \text{Overall Reaction:} & \ce{ML_{n}X + Y <=> ML_{n}Y + X} \ \hline \text{Step 1 (Concerted):} & \ce{ML_{n}X + Y <=>}\textcolor{green}{ \ce{[Y\bond{...}ML_{n}\bond{...}X]^{\ddagger}}} \ce{+ <=> ML_{n}Y + X} \end{array} \nonumber$
The interchange mechanism is common for many six-coordinate (octahedral) metal complexes. The hallmark feature that distinguishes the interchange mechanism from other possible mechanisms is the absence of a detectable intermediate. If an intermediate is detected, the mechanism is considered associative (A). Another piece of evidence that can indicate an interchange mechanism is stereochemical changes from reactant to product. If specific sterochemistry (ie cis or trans relationships) is changed, it may be taken as evidence that an intermediate exists long enough to allow rearrangement to occur.
Distinguishing A, D, and I Mechanisms
Figure $1$ shows examples of possible reaction profiles for each of the stoichiometric mechanisms described above. The profile of A and D mechanisms can be similar; they each require an intermediate, and the primary difference between A and D is the sequence of steps. An I mechanism lacks a true intermediate. These three cases can be difficult to distinguish experimentally, and especially if characterization of an intermediate is difficult. Notice that the reaction coordinate diagram for A and D pathways, shown in Figure $1$, are similar. The only necessary difference between them is in the identity of the intermediate species.
Intimate Mechanisms
The intimate mechanism is defined by characteristics of the rate-limiting step. The reaction is classified as associatively-activated (using a subscript, $a$) if the rate changes by more than 10-fold as the entering group is varied. The reaction is classified as dissociatively-activated (using a subscript, $d$) if the rate changes by less than 10-fold as the entering group is varied.
Associatively-Activated Mechanisms
The rate-limiting step of an associatively activated pathway is formation of a bond between the entering group and the central metal ion. These reactions also require formation of an encounter complex in a pre-equilibrium step prior to bond forming or bond breaking steps discussed here (see Eigen-Wilkins Mechanism).
Associatively-activated A and I ($\ce{A_{a}}$ and $\ce{I_{a}}$)
In the case of an associatively-activated associative pathway ($\bf\ce{A_{a}}$) or an associatively-activated interchange pathway ($\bf\ce{I_{a}}$), the rate-limiting step is the association of the entering group (Y) with the reactant, $\ce{ML_{n}X}$. In the case of the $\ce{A_{a}}$ pathway, it is the first step (bolded) below.
$\begin{array}{rcc} & \ce{A_{a}} \text{ Mechanism} & \ \hline \textbf{Step 1 (Association):} & \bf \ce{ML_{n}X + Y <=>[k_1][k_{-1}] ML_{n}XY} & \Longleftarrow \textbf{RATE LIMITING ASSOCIATION}\ \text{Step 2 (Dissociation):} & \ce{ML_{n}XY <=>[k_2][k_{-2}] ML_{n}Y + X} \end{array} \nonumber$
In the case of $\ce{A_{a}}$, the first step shown above is slower than the second step ($k_1 < k_2$). In this case, an intermediate may not be detected because its steady state concentration is close to zero (ie it reacts as soon as it forms). Thus, distinguishing between $\ce{A_{a}}$ and $\ce{I_{a}}$ mechanisms is often impossible.
Associatively-activated D ($\ce{D_{a}}$)
In the case of an associatively-activated dissociative reaction ($\bf\ce{D_{a}}$), the rate limiting step is the association of the entering group (Y) with the intermediate $\ce{ML_{n}}$; the second step (bolded) in the mechanism shown below:
$\begin{array}{rcc} & \ce{D_{a}} \text{ Mechanism} & \ \hline \text{Step 1 (Dissociation):} & \ce{ML_{n}X <=>[k_1][k_{-1}] ML_{n} + X} &\ \textbf{Step 2 (Association):} & \bf \ce{ML_{n}X + Y <=>[k_2][k_{-2}] ML_{n}Y} & \Longleftarrow \textbf{RATE LIMITING ASSOCIATION} \end{array} \nonumber$
In this case, the intermediate concentration should be measurable since the second step is slower than the first ($k_1 > k_2$).
Exercise $1$
Draw the reaction coordinate diagrams (aka reaction profiles) of reactions that proceed through $\ce{A_{a}}$, $\ce{I_{a}}$, and $\ce{D_{a}}$ pathways. Use these diagrams to explain why it is impossible to distinguish between $\ce{A_{a}}$ and $\ce{I_{a}}$.
Hint
Notice that the reaction coordinate diagrams for A and D pathways, shown in Figure $1$, are similar. The only necessary difference between them is the identity of the intermediate. These diagrams can be modified to represent associatively-activated pathways by altering the relative energy of the transition state for the associative step; the energy barrier for the associative step must be greater than that of the dissociative step. In other words, the difference between $\ce{A_a}$, $\ce{D_a}$ is in the relative energies of the first and second transition states. In the case of $\ce{I_a}$, the diagram could be similar to the I profile shown in Figure $1$, with the exception that the energy barrier for making the M-Y bond is greater than the energy barrier for breaking the M-X bond.
Answer
The three reaction coordinate diagrams shown in the figure below represent $\ce{A_{a}}$, $\ce{I_{a}}$, and $\ce{D_{a}}$ mechanisms respectively. If your diagram does not match this perfectly, that may be OK. The important features are:
• $\ce{A_{a}}$: This reaction profile should show an intermediate with higher coordination number than the reactant. The intermediate should be relatively stable. The energy barrier for formation of this intermediate is the associative step, and therefore it should also be the rate-limiting step for an $\ce{A_{a}}$ mechanism; in other words, the first step should have the highest energy barrier.
• $I_a$: This reaction profile should lack a stable intermediate. The energy barrier for creating of the M-Y bond must be rater-limiting. The diagram shown below is an acceptable solution, but a diagram like shown in Panel C of Figure $1$ is also correct. It is not clear from the diagram alone whether bond breaking or bond making are rate-limiting; this nuance would lie primarily in the identity of the transition state, and whether the M-Y bond is stronger or weaker than the M-X bond in the transition state.
• $\ce{D_{a}}$: This reaction profile should show an intermediate with lower coordination number than the reactant. The intermediate should be relatively stable. The energy barrier for association of the entering group with the intermediate (the second step) must be rate-limiting in a $\ce{D_{a}}$ mechanism. In other words, the second step should have the highest energy barrier.
The $\ce{A_{a}}$ and $I_a$ mechanisms have different reaction profiles while still the rate-limiting step is the initial association of the incoming ligand to the metal center. These two types of mechanisms are difficult to distinguish because in the case of $\ce{A_{a}}$, there is a relatively small energy barrier for reaction of the intermediate to form product; thus this intermediate is not long-lived and would be difficult to detect. Experimental distinction between $\ce{A_{a}}$ and $I_a$ lies in the detection and characterization of this intermediate, and since it may be elusive in $\ce{A_{a}}$, it is difficult to unambiguously determine whether an associatively-activated mechanism is $\ce{A_{a}}$ or $I_a$.
Dissociatively-Activated Mechanisms
When the rate-determining step is the breaking of a bond between the central metal ion and the leaving group, it is considered a dissociatively-activated (subscript $d$) mechanism. The rates of these reactions are largely independent of the identity of the incoming ligand.
Dissociatively-activated A ($\ce{A_{d}}$)
In the case of a dissociatively-activated associative pathway ($\bf\ce{A_{d}}$), the rate-limiting step is the dissociation of the leaving group (X) from the intermediate, $\ce{ML_{n}XY}$; the second step (bolded) below.
$\begin{array}{rcc} & \ce{A_{d}} \text{ Mechanism} & \ \hline \text{Step 1 (Association):} & \ce{ML_{n}X + Y <=>[k_1][k_{-1}] ML_{n}XY} & \ \textbf{Step 2 (Dissociation):} & \bf \ce{ML_{n}XY <=>[k_2][k_{-2}] ML_{n}Y + X} & \Longleftarrow \textbf{RATE LIMITING DISSOCIATION} \end{array} \nonumber$
In this case, the intermediate concentration should be measurable since the second step is slower than the first ($k_1 > k_2$.
Dissociatively-activated D and I ($\ce{D_{a}}$ and $\ce{I_{a}}$)
In the case of a dissociatively-activated dissociative reaction ($\bf\ce{D_{d}}$), or a dissociatively-activated interchange pathway ($\bf\ce{I_{d}}$), the rate limiting step is the dissociation of the leaving group (X) from the reactant, $\ce{ML_{n}X}$. For $\ce{D_{d}}$ the rate limiting step is the first step (bolded) in the mechanism shown below:
$\begin{array}{rcc} & \ce{D_{d}} \text{ Mechanism} & \ \hline \textbf{Step 1 (Dissociation):} & \bf \ce{ML_{n}X <=>[k_1][k_{-1}] ML_{n} + X} & \Longleftarrow \textbf{RATE LIMITING DISSOCIATION} \ \text{Step 2 (Association):} & \ce{ML_{n}X + Y <=>[k_2][k_{-2}] ML_{n}Y} & \end{array} \nonumber$
In the case of $\ce{D_{d}}$, the first step shown above is slower than the second step ($k_1 < k_2$ and an intermediate may not be detected because its steady state concentration is close to zero (ie it reacts as soon as it forms). Thus, distinguishing between $\ce{D_{d}}$ and $\ce{I_{d}}$ mechanisms is often impossible.
Exercise $2$
You should complete Excercise $1$ before attempting this excercise.
Draw the reaction coordinate diagrams (aka reaction profiles) of reactions that proceed through $\ce{A_{d}}$, $\ce{I_{d}}$, and $\ce{D_{d}}$ pathways.
Answer
Curated or created by Kathryn Haas | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.02%3A_Substitutions_Reactions/12.2.03%3A_Mechanistic_Possibilities.txt |
Since the different mechanisms have different rate-limiting steps, there are kinetic consequences for each type of mechanism. We can used kinetic data, therefore, to help distinguish the mechanism of a reaction. However, the information gleaned from rate laws alone can be ambiguous. For example, what does it mean if a reaction depends on on the concentration of incoming ligand? How can we distinguish between an associative mechanism (\(A\)), and an associatively-activated dissociative mechanism (\(D_a\)?
12.03: Kinetics Hint at the Reaction Mechanism
Derrivation of the Rate Law
A Dissociative (D) mechanism is two steps in which the first step is the dissociation of a ligand. The intermediate, $\color{blue}{\ce{ML_{n}}}$, has a lower coordination number than that of the reactant, $\ce{ML_{n}X}$.
$\begin{array}{rc} \text{Step 1 (Dissociation):} & \ce{ML_{n}X <=>[k_1][k_{-1}]}\textcolor{blue}{\ce{ML_{n}}}\ce{ + X}\ \text{Step 2 (Association):} & \textcolor{blue}{\ce{ML_{n}}}\ce{ + Y <=>[k_2][k_{-2}] ML_{n}Y} \ \hline \text{Overall Reaction:} & \ce{ML_{n}X + Y <=> ML_{n}Y + X} \end{array} \nonumber$
If we assume that $k_{-2}<<k_2$, the rate law for the formation of products is the rate law of the second step.
$\frac{d\left[\mathrm{ML}_{n}\right]}{d t}=k_{2} \textcolor{blue}{\ce{[ML_{n}]}} \ce{[Y]} \nonumber$
We can assume that the concentration of the intermediate is small enough that it would be difficult or impossible to measure its concentration. Here, we can evoke the steady-state approximation and assume that the concentration, $\color{blue}{\ce{[ML_{n}]}}$, is approximately unchanging over the course of reaction. The $\color{blue}{\ce{[ML_{n}]}}$ depends on $k_1, k_{-1}$ and $k_2$. These assumption are expressed mathematically as follows.
$\frac{d\left[\mathrm{ML}_{n}\right]}{d t}=k_{1}\left[\mathrm{ML}_{n} \mathrm{X}\right]-k_{-1}\left[\mathrm{ML}_{n}\right][\mathrm{X}]-k_{2}\left[\mathrm{ML}_{n}\right][\mathrm{Y}]=0 \nonumber$
We can then use the equation above to solve for $\color{blue}{\ce{[ML_{n}]}}$.
$\textcolor{blue}{\ce{ML_{n}}}=\frac{k_{1}\left[\mathrm{ML}_{n} \mathrm{X}\right]}{k_{-1}[\mathrm{X}]+k_{2}[\mathrm{Y}]} \nonumber$
Then substitution this into the rate law given above for the second step of the mechanism. This would yield a rate law that depends only on species with measurable concentrations.
$\frac{d\left[\mathrm{ML}_{n} \mathrm{Y}\right]}{d t}=\frac{k_{2} k_{1}\left[\mathrm{ML}_{n} \mathrm{X}\right][\mathrm{Y}]}{k_{-1}[\mathrm{X}]+k_{2}[\mathrm{Y}]} \nonumber$
The rate law above suggests that the rate of product formation has an inverse relationship to the concentration of the outgoing ligand, X. This is one piece of evidence that can be used to distinguish a Dissociative mechanism; that is if the rate of reaction decreases as [X] in solution is increased it is evidence of a D mechanism. This rate law also suggests that ther is a complicated dependence of the reaction rate on the concentration of incoming ligand, Y. The importance of [Y] compared to [X] depends mathmatically on the weight of the two values in the demoninator of the rate law. If [X] >> [Y], then a valid approximation is to assume that the $k_2\ce{[Y]}$ term is relatively small and simplify the rate law by dropping it out completely.
$\text{If X >> Y, then simplify: } \frac{d\left[\mathrm{ML}_{n} \mathrm{Y}\right]}{d t}=\frac{k_{2} k_{1}\left[\mathrm{ML}_{n} \mathrm{X}\right][\mathrm{Y}]}{k_{-1}[\mathrm{X}]} \nonumber$
The resulting rate law (above) has a simpler dependence on [Y]; that is when [X] is large, the reaction rate is directly related to [Y].
The opposite extreme is under conditions of high [Y]. This yields a rate law that is first order in its dependence on the reactant metal complex. When [Y] is large, and the $k_{-1}\ce{[X]}$ term is relatively insignificant, the rate law can be simplified to the pseudo-first order rate law as follows.
$\text{If Y >> X, then simplify: } \frac{d\left[\mathrm{ML}_{n} \mathrm{Y}\right]}{d t}= k_{1}\ce{[ML_{n}X]} \nonumber$
The best type of experimental evidence to determine a dissociative mechanism would be a systematic study of the dependence of rate on both [X] and [Y]. However, the dissociative reaction is unique in that it displays this type of saturation kinetics at high concnetration of the incoming ligand, and so the study of how a reaction rate behaves under high [Y] is common.
First-order reactions
Under conditions of high [Y], the a reaction following a D mechanism is said to follow "first order" kinetics. This type of reaction is sometimes called a first order reaction. That means the rate law depends on only one concentration term.
Why does the dissociative mechanism depend on concentrations in this specific way?
The rate depends on one molecule losing a ligand. Once it does so, a second ligand can replace the one that left. However, losing a ligand may be harder to do than gaining a new one. To lose a ligand, a bond must be broken, which costs energy. To gain a new ligand, a bond is made, releasing energy. That first step is harder to do, so it takes longer. It is a bottleneck that slows the reaction down. It is called the rate-determining step.
• The rate-determining step is the slow step of the reaction.
• The rate-determining step controls the rate of the overall reaction; everything else has to wait for that step to happen.
• Once the rate-determining step has occurred, everything else follows very quickly.
No collision is necessary for the metal complex to lose a ligand. Instead, a bond in the metal complex has to break. That just takes time and energy. As a result, concentration of the incoming ligand matters very little.
We should think a little more about energy requirements, available energy and reaction rate. It takes a certain amount of energy to break a bond. Over any given period of time, a specific amount of energy is available in the surroundings to use. That energy is not available uniformly. Some molecules will get more energy from their surroundings and others will get less. There will be a statistical distribution, like a bell curve, of energy available in different molecules. That means bond-breaking events are governed by statistics.
Figure $2$: The relationship between temperature, energy available, and energy barrier. The black line represents energy needed to start the reaction, also called the energy barrier or the activation barrier. The blue curve is the distribution of available energy in a group of molecules at a cooler temperature. The yellow curve is for a group of molecules that is a little warmer, and the red curve even warmer.
In figure $2$, most of the molecules at the low temperature (blue) do not have enough energy to begin the reaction. A small portion do, and so the reaction will proceed, but very slowly. In the yellow curve, there is more energy available, and so a large fraction of molecules have the energy necessary to begin the reaction. In the red curve, the vast majority have sufficient energy to react. Thus, one of the factors governing how quickly a reaction will happen is the energy needed, or activation barrier. A second factor is the energy available, as indicated by the temperature.
Of course, even if there is enough energy for the reaction, the reaction might not occur yet. Energy is necessary but not sufficient to start a reaction. There are also statistical factors in terms of whether a molecule has its energy allotted into the right places, or in some cases, whether two molecules that need to react together are oriented properly.
Suppose at a given temperature it takes a specific amount of time for half the molecules to gain enough energy so that they can undergo the reaction. That amount of time is called the half life of the reaction. After one half life, half the molecules have reacted and half remain. After a second half life, half the remaining molecules (another quarter, for three quarters of the original material in all) have also reacted, and a quarter still remain. After a third half life, half the remaining ones (another eighth, making it seven eighths reacted in total) will have reacted, leaving an eighth of the original material behind.
• Exponential decay is based on a statistical distribution of energy availability.
• The concept of half life is related to exponential decay.
• It takes a fixed period of time for a half of the metal complex obtain enough energy to dissociate.
Thus, the time it takes for the reaction to happen does not really depend on the concentration of anything.
However, the change in concentration over time -- the quantity that we can usually measure most easily -- depends on the original concentration, and for that reason the concentration of the metal complex appears in the rate law.
Figure $3$: The reactions in the top row and bottom row are proceeding with the same half-life as we move from left to right. However, the top row starts out more concentrated than the bottom row. As a result, the concentrations in the top row are changing more quickly than in the bottom row.
Suppose the half life for a particular case of ligand substitution is one second. After a half life, a 1 M solution becomes 0.5 M, so the rate of change in concentration per time is 0.5M/s. But after the same half life, a 0.5 M solution becomes 0.25 M, so the change in concentration is 0.25 M/s.
Exercise $1$
If a first order reaction has a half-life of 120 seconds, how much of the original material is left after
a) four minutes? b) six minutes? c) eight minutes? d) ten minutes?
Answer a
4 minutes = 240 seconds = 2 x 120 second = 2 half lives.
Material left = 50% x 50% = 0.5 x 0.5 = 0.25 = 25% left
Answer b
6 minutes = 360 seconds = 3 x 120 second = 3 half lives.
Material left = 0.5 x 0.5 x 0.5 = 0.125 = 12.5% left
Answer c
8 minutes = 480 seconds = 4 x 120 second = 4 half lives.
Material left = 0.5 x 0.5 x 0.5 x 0.5 = 0.0625 = 6.25% left
Answer d
10 minutes = 600 seconds = 5 x 120 second = 5 half lives.
Material left = 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 0.03125 = 3.125% left
Exercise $2$
Given the first-order dissociative rate law above, what would happen to the reaction rate for substitution in each of the following cases?
1. the concentration of ligand is doubled, and the concentration of metal complex is doubled
2. the concentration of ligand is tripled, and the concentration of metal complex is halved
3. the concentration of ligand is doubled, and the concentration of metal complex is tripled
4. the concentration of ligand is halved, and the concentration of metal complex is halved
Answer a
Dissociative Rate Law: Rate = [MLn], if MLn is the complex. There is no dependence on [X], if X is the new ligand.
Rate will double: Rate = 2 x [MLn]0, if [MLn]0 is the original concentration.
Answer b
Rate will be halved: Rate = 0.5 x [MLn]0.
Answer c
Rate will triple: Rate = 3 x [MLn]0.
Answer d
Rate will be halved: Rate = 0.5 x [MLn]0.
Exercise $3$
Plot graphs of initial rate vs concentration to show what you would see in dissociative substitution.
a) The concentration of metal complex, [MLn], is held constant at 0.1 mol/L and the concentration of ligand is changed from 0.5 mol/L to 1 mol/L and then to 1.5 mol/L.
b) The concentration of new ligand, [X], is held constant at 0.1 mol/L and the concentration of metal complex is changed from 0.5 mol/L to 1 mol/L and then to 1.5 mol/L
Exercise $4$
Given the following sets of initial rate data, determine whether each case represents a dissociative substitution. [MLn] = concentration of the coordination complex; [X] = concentration of incoming ligand.
a)
b)
Exercise $5$
In the following data, the concentration of the metal complex and the incoming ligand were held constant, but more of the departing ligand was added to the solution.
a) Explain what the data says about rate dependence on this concentration.
b) Explain this rate dependence in terms of the reaction.
Answer a
The rate of the reaction is depressed when the concentration of the departing ligand is increased.
Answer b
This dependence could indicate an equilibrium in the dissociative step. The more departing ligand is added, the more the equilibrium is pushed back towards the original metal complex. With less dissociated metal complex around, the entering ligand cannot form the new complex as quickly.
Exercise $6$
In certain solvents, such as THF, acetonitrile and pyridine, the rate law for substitution often appears to be Rate = k1[MLn] + k2[MLn][X], in which X is the incoming ligand and MLn is the metal complex.
a) What do these solvents have in common?
b) What is a possible explanation for this rate law?
c) This rate law has been shown to be consistent with an entirely associative mechanism. How is that possible?
Answer a
These solvents all have lone pairs. They could be Lewis bases or nucleophiles.
Answer b
It looks like two competing mechanisms. On term suggests a dissociative mechanism, whereas the other term suggests a dissociative mechanism. They could be happening in competition with each other.
Answer c
On the other hand, it could be that there is one mechanism with two different nucleophiles. If the incoming ligand is the nucleophile, the term on the right shows up in the rate law. If the solvent is the nucleophile, forming a third complex, the term on the left shows up in the rate law. That's because we would typically change the amount of metal complex and the amount of ligand that we add to the solution in order to determine the rate law, but we wouldn't normally be able to change the concentration of the solvent, so it would be a constant. (How could you confirm this explanation in an experiment?) | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.03%3A_Kinetics_Hint_at_the_Reaction_Mechanism/12.3.01%3A_Rate_Law_for_Dissociative_Mechanisms.txt |
Interchange (I) Mechanisms take place in one concerted step where there is no intermediate, or the intermediate is elusive. The reactio nessentially occurs in once step in which bond forming and bond breaking is concerted, as shown below. The species, $\color{green}{\ce{[Y\bond{...}ML_{n}\bond{...}X]^{\ddagger}}}$, can be defined as either a transition state or a very short-lived intermediate.
$\begin{array}{rc} \text{Step 1 (Concerted):} & \ce{ML_{n}X + Y <=>}\textcolor{green}{ \ce{[Y\bond{...}ML_{n}\bond{...}X]^{\ddagger}}} \ce{+ <=>[k_1][k_{-1}] ML_{n}Y + X} \end{array} \nonumber$
The rate law of this one-step reaction depends on whether the relative rates of the forward and reverse reactions. When the reaction is practically irreversible, then the rate depends only on the formation of product, and not its reaction to re-form product. In this case, the rate law is a simple second-order rate law.
$\frac{d\ce{[ML_{n}Y]}}{dt}=k_1 \ce{[ML_{n}X][Y]} \nonumber$
On the other hand, if the rate of re-formation of reactant is significant, then the rates of forward and reverse reactions must be considered in the rate law.
$\frac{d\ce{[ML_{n}Y]}}{dt}=k_1 \ce{[ML_{n}X][Y]} - k_{-1}\ce{[ML_{n}X][X]} \nonumber$
In the latter case where there is a dependence of rate on both [X] and [Y], an effective strategy is to employ the condition where solution concentration of both [X] and [Y] are simultaneously high so that the reaction is effectively a competition of two pseudo-first order reactions that depend primarily on the metal complex concentrations. In such a case as this, another helpful piece of information is the equilibrium constant for the reaction ($K=\frac{k_1}{k_{-1}}$). The rate law is written below.
$\text{When [X] and [Y] are very high: } \frac{d\ce{[ML_{n}Y]}}{dt}=k_1 \ce{[ML_{n}Y]} - k_{-1}\ce{[ML_{n}X]} = -\frac{d\ce{[ML_{n}X]}}{dt} \nonumber$
One distinguishing feature of an I mechanism compared to a D mechanism can be determined from such an experiment under high concentration of X and Y. In the case of a D mechanism and sufficiently high concentration of incoming ligand, [Y], the reaction demonstrations saturation kinetics to give a pseudo-first order rate law depending only on the concentration of reactant metal complex. In the case of I, the reaction rate is not a simple first order rate law because even under high [Y] and [X], there is a dependence of rate on concentration of the product complex.
Reactions that are $I_a$ or $I_d$ are distinguished by the relative strength of the M-X and M-Y bond in the transition state. These reactions can be distinguished by varying the Identities of X and Y to determine how the reaction rate depends on identity of X and Y. If there is a large dependence on the incoming ligand, and less so on the outgoing ligand, it could be classified as an $I_a$ mechanisms. The opposite is also true.
12.3.03: Rate Law for Associative Mechanisms
Associative (A) mechanisms involve a first step where the incoming ligand bonds to the metal ion, creating an intermediate, $\color{red}{\ce{ML_{n}XY}}$, with a higher coordination number than that of the reactant, $\ce{ML_{n}X}$.
$\begin{array}{rc} \text{Step 1 (Association):} & \ce{ML_{n}X + Y <=>[k_1][k_{-1}]} \color{red}{\ce{ML_{n}XY}}\ \text{Step 2 (Dissociation):} & \textcolor{red}{\ce{ML_{n}XY}} \ce{<=>[k_2][k_{-2}] ML_{n}Y + X} \ \hline \text{Overall Reaction:} & \ce{ML_{n}X + Y <=> ML_{n}Y + X} \end{array} \nonumber$
To determine the rate law, we can use a similar process as we followed for D mechanisms, which are also two steps. Assuming that $k_2 >> k_{-2}$, the rate of formation of product from the second step is: $\frac{d\ce{[ML_{n}Y]}}{dt} = k_2 \textcolor{red}{\ce{[ML_{n}XY]}} \nonumber$
The steady state approximation allows us to assume that the intermediate is in low concentration (too low to be reliably measured) and that it is unchanging. We can use the rate law for the first step to find $\color{red}{\ce{[ML_{n}XY]}}$ in terms of species that can be reliably measured.
$\frac{d\textcolor{red}{\ce{[ML_{n}XY]}}}{dt} = k_1 \ce{[ML_{n}X][Y]}-k_{-1}\textcolor{red}{\ce{[ML_{n}XY]}} - k_2\textcolor{red}{\ce{[ML_{n}XY]}}=0 \nonumber$
...solving this first-step rate law for $\textcolor{red}{\ce{[ML_{n}XY]}}$ gives $\textcolor{red}{\ce{[ML_{n}XY]}} = \dfrac{k_1\ce{[ML_{n}X][Y]}}{k_{-1}+k_2}$
And then substitution $\textcolor{red}{\ce{[ML_{n}XY]}}$ into the rate law for the second step, and let $k=\frac{k_1 k_2}{k_{-1}+k_2}$ to gives the overall rate law below.
$\frac{d\ce{[ML_{n}Y]}}{dt} = \frac{k_1 k_2 \ce{[ML_{n}X][Y]}}{k_{-1}+k_2} = k \ce{[ML_{n}X][Y]} \nonumber$
This overall rate law is second order, and first order with respect to each of the reactants. This means that there is a direct (linear) relationship between concentration of either reactant and the rate of reaction.
Why does the associative mechanism depend on concentrations in this specific way?
This is a case of two molecules coming together. If both compounds are dissolved in solution, they must “swim around” or travel through the solution until they bump into each other and react. The more concentrated the solution is, or the more crowded it is with molecules, the more likely are the reactants to bump into each other. If we double the amount of new ligand in solution, an encounter between ligand and complex becomes twice as likely. If we double the amount of metal complex in solution, an encounter also becomes twice as likely.
Exercise $1$
Given the associative rate law above, what would happen to the reaction rate for an associative substitution in the following cases?
1. the concentration of ligand is doubled, and the concentration of metal complex is doubled
2. the concentration of ligand is tripled, and the concentration of metal complex is doubled
3. the concentration of ligand is tripled, and the concentration of metal complex is tripled
4. the concentration of ligand is halved, and the concentration of metal complex is doubled
Answer a
Associative Rate Law: $Rate = [ML_{n}][X]$, if MLn is the complex and X is the new ligand.
Rate will quadruple: $Rate = (2 \times [ML_{n}]_{0}) \times (2 \times [X]_{0})= 4 \times [ML_{n}]_{0}[X]_{0}$, if [X]0 and [MLn]0 are the original concentrations.
Answer b
Rate will sextuple: $Rate = (3 \times [ML_{n}]_{0}) \times (2 \times [X]_{0}) = 6 \times [ML_{n}]_{0}[X]_{0}$.
Answer c
Rate will nonuple: $Rate = (3 \times [ML_{n}]_{0}) \times (3 \times [X]_{0}) = 9 \times [ML_{n}]_{0}[X]_{0}$.
Answer d
Rate will stay the same: $Rate = (0.5 \times [ML_{n}]_{0}) \times (2 \times [X]_{0}) = 1 \times [ML_{n}]_{0}[X]_{0}$.
Exercise $2$
Plot graphs of initial rate vs. concentration to show what you would see in associative substitution.
a) The concentration of new ligand, [X], is held constant at 0.1 mol/L and the concentration of metal complex is changed from 0.5 mol/L to 1 mol/L and then to 1.5 mol/L.
b) The concentration of metal complex, [MLn], is held constant at 0.1 mol/L and the concentration of ligand is changed from 0.5 mol/L to 1 mol/L and then to 1.5 mol/L.
Answer
Exercise $3$
In the previous problem, the experiment was run in a particular way for particular reasons.
a) Why was one concentration held constant while the other one was changed? Why not change both?
b) Why does the graph report "initial rate" -- just the rate at the very beginning of the reaction?
Answer a
Changing both concentrations at once would leave some doubt about whether one concentration had affected the rate, or the other concentration, or both. In practice, one concentration is usually held constant while the other is kept in excess and varied.
Answer b
Rate changes over time because the concentrations of reactants change as they are consumed. By reporting only the initial rate (usually meaning less than 5% or 10% complete, but possibly even less than that if a lot of data can be gathered very quickly), the concentrations are still about what you started with. That means you can report a rate that corresponds to a given concentration with confidence.
Exercise $4$
Given the following sets of initial rate data (rates measured at the beginning of a reaction), determine whether each case represents an associative substitution.
a)
b)
c)
d)
Exercise $5$
What information can be gained from the slopes of lines in Exercise $4$ ?
Answer
Because $Rate = k[ML_{n}][L]$ is held constant while [L] is varied, then the slope of the line is k [MLn]. Since you would know the value of [MLn], you could obtain the rate constant from the quantity (slope/ [MLn]).
Note that the rate law for an A mechanism, and a reversible I mechanism are of the same form: they are each first order with respect to each reactant. Thus, an unambiguous determination between the two is difficult because it would require detection of the intermediate, which exists only at a very low concentration. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.03%3A_Kinetics_Hint_at_the_Reaction_Mechanism/12.3.02%3A_Rate_Laws_for_Interchange_Mechanisms.txt |
Most octahedral complexes react through either an associative or dissociative interchange mechanism ($I_a$ or $I_d$). Although the rate laws should be different for these two cases, it is difficult to distinguish between them. The difficulty lies in the seemingly conflicting observations from experiments performed under limiting conditions of the incoming ligand. For example, in the case of the reaction of hexaaquochromium(III) ( $\ce{[Cr(H2O)6]}$ ) with ammonia ( $\ce{NH3}$ ), the rate laws determined at high and low concentrations of the incoming ligand give different apparent rate laws. Under most conditions, the rate law appears to indicate a dissociative mechanism (ie rate is independent of the $\ce{[NH3]}$). But, at very high $\ce{[NH3]}$, the rate law appears to indicate an associative mechanism (ie the rate depends on $\ce{[NH3]}$).
This information might seem contradictory. However, it can be explained by the formation of a transient ion pair, usually called an encounter complex, in a step prerequisite to the rate-determining step(s). The Eigen-Wilkins Mechanism is based on this idea.
The Eigen-Wilkins mechanism
The Eigen-Wilkins mechanism is also a rate law, and it governs the reactions of octahedral metal complexes. This mechanism does not define the rate-limiting step; rather, it defines the existence of a pre-equilibrium step (ie an initial step that is not rate-determining) that results in formation of an encounter complex. The encounter complex is a short-lived ion pair of the metal complex and the incoming ligand; it is an intermediate formed through Coulomb interactions. For the conversion of a generic metal complex, where X is the leaving group and Y is the entering group, the overall reaction, pre-equilibrium step, and formation of products from the encounter complex are shown below:
$\begin{array}{rcc} & \text{CHEMICAL EQUATION} & \text{EQUILIBRIUM EXPRESSION} \ \hline \text{Formation of Encounter Complex} & \ce{ML5X + Y<=>[{k_1}][{k_{-1}}] (ML5X*Y)} &\ce{K_E = \ce{\frac{[(ML5X*Y)]}{[ML5X][Y]}}} \ \text{Formation of Products:} & \ce{(ML5X*Y) ->[{k_2}] ML5Y + X} &\ce{K_2 = \ce{\frac{[ML5Y][X]}{[(ML5X*Y)]}}}\ \hline \text{Overall Reaction:} & \ce{ML5X + Y <=> ML5Y + X} & \ce{K = \ce{\frac{[ML5Y][X]}{[ML5X][Y]}}} \ \end{array} \nonumber$
Fast pre-association
One possibility (that is most common) is that $k_1$ and $k{-1}$ are much larger than $k_2$. In this case, the encounter complex forms quickly, and once it forms, it also can quickly fall apart to re-form the reactant complex. When $k_2$ is relatively small then reaction of the encounter complex to form the product is the rate-determining step, and sometimes the concentration of encounter complex can be experimentally determined. However, it is rare that this is the case. A second, more common scenario iis that $k_2$ is also relatively large. Assuming that the formation of the encounter complex is still a fast step, then the rate still depends on the reaction of that encounter complex to form product. In either case, one path for deriving the rate laws is given below. The rate of formation of the product is defined by the second step:
$\text{Rate}=\frac{d\ce{[ML5Y]}}{dt}=k_2 \ce{[(ML5X*Y)]} \nonumber$
Although the encounter complex may exist in high enough concentration to be detected, that is rare and technically difficult. Rather, the concentration of the encounter complex can be determined by manipulation of the equilibrium constant expression and some helpful assumptions. The equilibrium constant for formation of the encounter complex is:
$K_E = \ce{\frac{[(ML5X*Y)]}{[ML5X][Y]}} \nonumber$
This can be rearranged to solve for concentration of the encounter complex.
$\ce{[(ML5X*Y)]} = K_E\ce{[ML5X][Y]} \label{encounter}$
...and we can re-write the rate experssion to remove the encounter complex from the expression. But by doing so, we remove the problem of the unmeasurable encounter complex being in the expression, but add the problem of the reactant metal complex being part of the expression:
$\text{Rate}=k_2 K_E\ce{[ML5X][Y]} \nonumber$
The problem is that we don't precisely know what the value of $\ce{[ML5X]}$. We only know the total amount we initially added. But, we assume that $k_1$ is large, and so the initial concentration of the metal complex may change immediately to produce some of the encounter complex. As long as $k_2$ is relatively small, we can assume that the initial total concentration of metal complex at the start of reaction consists primarily of the reactant complex plus the encounter complex:
$\ce{[M]_{total}}=\ce{ [ML5X] +[(ML5X*Y)]} \nonumber$
We can substitute XXX into XXX to remove the encounter complex from the expression, then rearrange to solve for the reactant metal ion concentration, $\ce{[ML5X]}$:
$\begin{array}{rl} \ce{[M]_{total}}&=\ce{ [ML5X]} +K_E \ce{[ML5X][Y]} \ &=\ce{ [ML5X]}(1 +K_E \ce{[Y]}) \ \ce{[ML5X]} & = \frac{\ce{[M]_{total}}}{1 +K_E \ce{[Y]}} \end{array} \nonumber$
Now this allows us to do one more substitution into the rate expression to get the rate into terms of constants and solution concentrations that are known:
$\text{Rate}=k_2 K_E\frac{\ce{[M]_{total}[Y]}}{1 +K_E \ce{[Y]}} \nonumber$
This rate expression is complex, but can be simplified under extremely high or low concentrations of the incoming ligand, [Y].
High [Y]: If the recation proceeds at high concentration of Y, then the denominator, $1 +K_E \ce{[Y]}$ becomes approximately $K_E \ce{[Y]}$. The expression is simplified to a first-order rate law under high [Y]:
$\text{Rate}=k_2 K_E\frac{\ce{[M]_{total}[Y]}}{K_E \ce{[Y]}}=k_2 \ce{[M]_{total}} \nonumber$
This would allow determination of $k_2$ because the rate could be directly observed by varying the concentration of $\ce{[M]_{total}}$ at high [Y].
Low [Y]: If the reaction is run at very low [Y], then the denominator, $1 +K_E \ce{[Y]}$ becomes approximately 1. We can let the observed rate constant, $k_{obs}=k_2K_E$ to get a simplified second-order rate law under very low [Y]:
$\text{Rate}=k_2 K_E\frac{\ce{[M]_{total}[Y]}}{K_E \ce{[Y]}}=\frac{d\ce{[ML5Y]}}{dt}=k_{obs} \ce{[M]_{total}}[Y] \nonumber$
The observed rate constant can be measured experimentally, and $K_E$ can be determined theoretically ($k_{obs}=k_2K_E$) | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.03%3A_Kinetics_Hint_at_the_Reaction_Mechanism/12.3.04%3A_Preassociation_Complexes.txt |
The rate law shows how the rate of a reaction depends on concentrations of different species in solution. The proportionality constant, k, is called the rate constant. It contains other information about the energetic requirements of the reaction.
All reactions must overcome activation barriers in order to occur. The activation barrier is the sum of the energy that must be expended to get the reaction going. An activation barrier is often thought of, cartoonishly, as a hill the molecule has to climb over during the reaction. Once, there, it can just slide down the other side of the hill to become products. At the top of the hill, the molecule exists in what is called the "transition state". At the transition state, the structure is somewhere between its original form and the structure of the products.
The type of diagram shown in figure LS6.1 is sometimes called a "reaction progress diagram". It shows energy changes in the system as a reaction proceeds. One or more activation barriers may occur along the reaction pathways, as various elementary steps occur in the reaction. In the above case, it is easy to imagine the source of the energy barrier, because some energy must be expended to break the bond to ligand C.
However, after that barrier is passed, energy is lowered again. This can happen for several reasons. Once C has separated from the metal complex, it is free to vibrate, tumble, roll and zip around all on its own. That means it can put its energy into any of those modes, independently of the metal complex. As a result, the entropy of the system increases. That lowers the overall "free energy" of the system. In addition, there may be some relief of crowding as the molecule changes from a four-coordinate complex to a three-coordinate complex, so strain energy is also lowered.
Exercise $1$
Make drawings depicting the relationship between reaction progress and energy for the following cases:
a) a new ligand binds to a four-coordinate complex, forming a five coordinate complex.
b) a two-step process in which a new ligand binds to a four-coordinate complex, forming a five coordinate complex, and then an old ligand dissociates to form a new, four-coordinate complex.
The rate constant gives direct insight into what is happening at the transition state, because it gives us the energy difference between the reactants and the transition state. Based on that information, we get some ideas of what is happening on the way to the transition state.
The rate constant can be broken down into pieces. Mathematically, it is often expressed as
$k = (\frac{RT}{Nh}) e^{-\frac{\Delta G \ddagger}{RT}} \nonumber$
In which R = the ideal gas constant, T = temperature, N = Avogadro's number, h = Planck's constant and Δ G= the free energy of activation.
The ideal gas constant, Planck's constant and Avogadro's number are all typical constants used in modeling the behaviour of molecules or large groups of molecules. The free energy of activation is essentially the energy requirement to get a molecule (or a mole of them) to undergo the reaction.
Note that k depends on just two variables:
• Δ G or the energy required for the reaction
• T or the temperature of the surroundings, which is an index of the available energy
The ratio of activation free energy to temperature compares the energy needs to the energy available. The more energy available compared to the energy needed, the lower this ratio becomes. As a result, the exponential part of the function becomes larger (since the power has a minus sign). That makes the rate constant bigger, and the reaction becomes faster.
The activation free energy is constant for a given reaction. It can be broken down in turn to:
$\Delta G^{\ddagger} = \Delta H^{\ddagger} -T \Delta S^{\ddagger} \nonumber$
in which Δ H = activation enthalpy and Δ S = activation entropy.
The activation enthalpy is the energy required for the reaction. The activation entropy deals with how the energy within the molecule must be redistributed for the reaction to occur. These two parameters can be useful in understanding events leading to the transition state.
For example, in ligand substitution, an associative pathway is marked by low enthalpy of activation but a negative entropy of activation. The low enthalpy of activation results because bonds don't need to be broken before the transition state, so it doesn't cost much to get there. That's favorable and makes the reaction easier. However, a decrease in entropy means that energy must be partitioned into fewer states. That's not favorable and makes the reaction harder. The reason the energy must be redistributed this way is that two molecules (the metal complex and the new ligand) are coming together to make one bigger molecule. They can no longer move independently of each other, and all of their combined energy must be reapportioned together, with a more limited range of vibrational, rotational and translational states to use for that purpose.
• Associative pathway: more bond making than bond breaking; lower enthalpy needs
• Associative pathway: two molecules must be aligned and come together; fewer degrees of freedom for energy distribution; decrease in entropy
On the other hand, the dissociative pathway is marked by a higher enthalpy of activation but a positive entropy of activation. The higher enthalpy of activation results because a bond must be broken in the rate determining step. That's not favorable. However, the molecule breaks into two molecules in the rate determining step. these two molecules have more degrees of freedom in which to partition their energy than they did as one molecule. That's favorable.
• Dissociative pathway: more bond breaking in rate determining step, higher enthalpy needs
• Dissociative pathway: one molecule converts to two molecules in rate determining step, greater degrees of freedom in two independently moving molecules, entropy increases
Thus, looking at the activation parameters can reveal a lot about what is going on in the transition state.
Exercise $2$
What factor(s) other than entropy might raise the free energy of the transition state going into an associative step between a metal complex and an incoming ligand? (What factor might make the first, associative step slower than the second, dissociative step?)
Answer
The metal centre is becoming more crowded as the new ligand arrives, so an increase in energy owing to steric hindrance may also play a role in the transition state energetics.
Exercise $3$
Other mechanisms for ligand substitution are also possible. The following case is referred to as an associative interchange (IA).
a) Describe in words what happens in an associative interchange.
b) Predict the rate law for the reaction.
c) Qualitatively predict the activation entropy and enthalpy, compared with
i) an associative mechanism and
ii) a dissociative mechanism.
Answer a
The new ligand, B, is arriving at the same time as the old ligand, A, is departing. We might also describe it as new ligand B pushing old ligand A out of the complex.
Answer b
$Rate = k[ML_{5}A][B]$, which looks like an associative rate law.
Answer c
This is a thought-provoking question without a definite answer. Associative mechanisms typically have lower activation enthalpy than dissociative mechanisms, because there has also been some bond-making prior to the bond-breaking in the rate determining step. The associative interchange would be a little more like the associative mechanism than dissociative. The mix of bond-making and bond-breaking at the transition state would make the enthalpy of activation relatively low.
Associative mechanisms have negative activation entropies, whereas dissociative mechanisms have positive activation entropies. The associative interchange could be in between the two, given that the elementary step would be close to entropically neutral overall. What happens at the transition state is a little harder to imagine, but it might reflect the small changes in entropy through the course of the reaction, producing a small entropy of activation. On the other hand, if the incoming ligand is forced to adopt some specific approach as it comes into the molecule (to stay out of the way of the departing ligand, for example) then that restriction could show up as a small negative activation entropy.
Exercise $4$
For the following mechanism:
a) Describe in words what is happening.
b) Predict the rate determining step.
c) Predict the rate law for the reaction.
d) Qualitatively predict the activation entropy and enthalpy, compared with
i) both an associative mechanism and
ii) a dissociative mechanism.
e) Suggest some ligands that may be able to make this mechanism occur.
Answer a
The lone pair donation from one ligand appears to push another ligand out.
Answer b
The first step is probably rate determining, because of the bond breaking involved.
Answer c
If the first step is rate determining, $Rate = k [ML_{5}A]$.
Answer d
Another question without a very clear answer. Compared with an associative mechanism, the activation entropy is probably much more positive, because additional degrees of freedom are being gained as the molecule heads over the activation barrier and one of the ligands separates to be on its own. However, the activation entropy may be less positive than in a regular dissociation, because in this case the breaking of one bond has to be coordinated with the formation of another.
The enthalpy of activation has both a bond-making and bond-breaking component, a little like in an associative mechanism. However, the amount of bond making here is probably less important, because pi bonds are typically not as strong as sigma bonds. The activation enthalpy is probably higher than an associative pathway but not as high as a dissociative one.
Answer e
The donor ligand must have a lone pair. Oxygen donors would be good candidates, because even if one lone pair is already donating in a sigma bond, an additional lone pair may be available for pi donation. The same thing is true for halogen donors. It would also be true for anionic nitrogen donors but not for neutral nitrogen donors, because a neutral nitrogen has only one lone pair. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.03%3A_Kinetics_Hint_at_the_Reaction_Mechanism/12.3.05%3A_Activation_Parameters.txt |
We usually look for physical reasons why a given compound might undergo a reaction via one mechanism and not another. That ability adds to our understanding of chemistry. If we can take information and give it predictive value, then we may be able to make educated decisions about what is probably happening with new reactions.
Why might a reaction undergo a dissociative reaction rather than an associative one? What factors might prevent an associative pathway?
One reason may be that there is not enough room. In an associative step, an additional ligand comes in and binds to the metal. If it is already crowded, that may be difficult.
Figure \(1\): The role of steric crowding in ligand substitution. In one of these cases, the associative mechanism is less favored because of crowding that will occur in the transition state.
• Steric crowding may lead to a dissociative, rather than associative, mechanism.
Another reason has to do with electronics. Maybe the compound cannot easily accept an additional bonding pair. That may be the case if the compound already has eighteen electrons.
Figure \(2\): The role of electron count in ligand substitution. In one of these cases, the associative mechanism is less favored because the metal is already electronically saturated.
• Electronic saturation may lead to a dissociative, rather than associative, mechanism.
However, if there is less crowding, and more electrons can be accommodated, an associative pathway may result.
Exercise \(1\)
i) Draw structures for the following reactions. Pay attention to geometry.
ii) Predict whether each of the substitutions would occur through associative or dissociative mechanisms.
a) AuCl3py + Na N3 → Na+ [AuCl3N3]- + py
b) Rh(C2H4)2(acac) + C2D4 → Rh(C2D4)2(acac) + C2H4
c) [Co(NH3)5Cl]2+ + H2O → [Co(NH3)5(OH2)]2+ + Cl-
d) trans-(Et3P)2PtCl2 + -SCN → trans-(Et3P)2PtCl(SCN) + Cl-
e) Rh(NH3)4Cl2+ + -CN → Rh(NH3)4Cl(CN)+ + Cl-
f) trans-[Cr(en)2Br2 ]+ + -SCN → cis- and trans-[Cr(en)2Br(SCN)]+ + Br-
Answer
Exercise \(2\)
Maurice Brookhart of UNC, Chapel Hill, makes organometallic compounds in order to study fundamental questions about reactivity. In this case, he has reported making a new compound capable of "C-H activation", a reaction in which unreactive C-H bonds can be forced to break. This process holds the future promise of converting coal and natural gas into important commodities currently obtained from petroleum.
a) Draw, with curved arrows, a mechanism for the ligand substitution in the synthesis of this C-H activating complex.
b) Explain your reasons for your choice of reaction mechanism.
Answer
Exercise \(3\)
Bob Grubbs of Cal Tech was awarded the Nobel Prize in chemistry for his development of catalysts for olefin metathesis. Olefin metathesis is important both in the reforming of petroleum and in the synthesis of important commodities such as pharmaceuticals. In the following study, he replaced chlorides on a "Grubbs Generation I catalyst" to study the effect on the olefin metathesis reaction.
a) Draw, with curved arrows, a mechanism for the ligand substitution in this complex.
b) Explain your reasons for your choice of reaction mechanism.
c) What factor(s) do you think Grubbs hoped to study by making this substitution in the catalyst?
Answer
Exercise \(4\)
Sometimes, kinetic studies can give insight into a reaction if controlled changes in the reaction produce measurable results.
a) Draw, with curved arrows, a mechanism for the ligand substitution in this reaction.
b) Explain your reasons for your choice of reaction mechanism.
c) Explain the following kinetic data.
Exercise \(5\)
Several different structures were proposed for Ni(cysteine)22-. Kinetic studies of substitution in this complex showed the rate was dependent in the concentration of both the metal complex and the incoming ligand. Which structure do you think is correct? Why? | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.03%3A_Kinetics_Hint_at_the_Reaction_Mechanism/12.3.06%3A_Some_Reasons_for_Differing_Mechanisms.txt |
This section will review some of the common experiments that can provide evidence supporting different mechanisms, with a focus on the reactions of octahedral complexes.
12.04: Experimental Evidence in Octahedral Substitutions
Consider what must happen for the simplest case of a purely dissociative (D) reaction mechanism for an octahedral metal complex. The first step, and the rate-limiting step must be dissociation of one ligand to form an intermediate with a lower coordination number. Such a situation, where an octahedral complex looses a ligand, X, to become a square pyramidal intermediate is shown below in Figure $1$.
If the dissociation of a ligand is rate-determining, factors that lower the energy of the transition state or intermediate structures will speed the reaction rate. Here we could consider the differences in ligand-field stabilization energy (LFSE). The difference between the LFSE of the octahedral reactant and that of the intermediate is defined as the ligand field activation energy (LFAE). The LFAE can be calculated for a given $d$-electron count for an octahedral reactant and a square pyramidal intermediate using the information provided in Figure $1$. Metal ions with particularly low (more negative) LFAE values tend to be more labile, while those with particularly high (less negative) values tend to be more inert. However, most octahedral complexes are expected to react through dissociative mechanisms due to the unfavorable steric crowding that must occur during an associative pathway.
The ease by which the M-X bond is broken will also influence rate of a dissociative reaction, and this can be probed experimentally. The following experiments can give evidence that support the argument that a reaction occurs by a dissociative pathway, or that it is dissociatively activated.
Experimental evidence that supports dissociative mechanism
1. Identity of the entering ligand has little effect on reaction rate.
If dissociation is the rate-determining step, the identity of the entering ligand, Y, should have little effect on the reaction rate. Therefore, an experiment that varies Y while holding X constant can indicate whether or not Y is involved in the rate-determining step. If the identity of Y has little effect on rate, then it must not be involved in the rate-determining step. Such experimental evidence can be taken as evidence of a dissociative mechanism. For example, the rate constants for reaction of $\ce{[Cr(NH3)5H2O}]^3+}$ with different incoming ligands, shown below, has rate constants of similar magnitude (they are approximately within one order of magnitude), indicating a dissociatively activated mechanism. In this case, other lines of evidence suggest this is a reaction with a $I_d$ mechanism.
$\ce{[Cr(NH3)5H2O]^3+ + Y^{-} <=> [Cr(NH3)5Y]^2+ + H2O} \nonumber$
Table $1$: Rate Constants for Exchange of Aquo Ligand of pentaammineaquo chromium(III) ion by incoming ligand, Y (chemical equation shown above). There is little effect of the identity of the incoming ligand on the rate constant. This data was sourced from Miessler, Fischer and Tarr's Inorganic Chemistry, 5th edition, pg 449.
Entering Ligand, $\ce{Y^-}$ Rate Constant, $k_1 \; (10^{-4}M^{-1}s^{-1})$
$\ce{NCS^-}$ 4.2
$\ce{Cl^-}$ 0.7
$\ce{Br^-}$ 3.7
$\ce{CF3CO2^-}$ 1.4
2. Steric Crowding of the Metal Complex
Steric crowding of the metal complex would inhibit a dissociative pathway through steric crowding of the entering ligand. However, steric crowding can increase the stability of the intermediate after dissociation of the ligand from the crowded octahedral complex. If the reaction rate increases with steric crowding of the reactant complex, this is taken as evidence to support a dissociative mechanism.
3. Volume of Activation
One molecule generally takes up less space than two sperate molecules. This usually holds true in solution. Measurement of a rate constant's dependence on pressure can be used to determine the volume of activation ($\Delta V_{act}$) and give insight into the mechanism of that reaction. In a dissociation mechanism, the dissociation causes one metal complex to become two separate molecules. Thus, the intermediate should, in principle, take up more space than the reactant complex. In a situation like this, we expect that the rate of reaction should decrease with increasing pressure, giving a positive value for $\Delta V_{act}$. However, solvation of ions can influence volume, and this should be considered in interpretation of pressure-dependence data.
4. Increased coulombic attraction with the leaving group decreases reaction rate.
Couloubic (electrostatic) attraction between the dissociating ligand, X, and the metal complex should slow ligand dissociation. Coulombic attraction increases with increased magnitude of charges and with decreased distance. If the reaction rate slows with increased charge of the metal ion or the ligand itself, this is taken as evidence of a dissociative mechanism. Likewise, if rate decreases as the radius of the central metal ion is decreased, this is evidence to support a dissociative mechanism.
5. Increased strength of the M-X bond decreases reaction rate.
Free energy relationships between the formation constant of M-X and the kinetic rate constant of the reaction can provide evidence to support a dissociative mechanism (Discussed in the next section). | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.04%3A_Experimental_Evidence_in_Octahedral_Substitutions/12.4.01%3A_Dissociation.txt |
The Linear Free Energy Relation (LFER) is a tool that can indicate the importance of bond breakage or bond formation in the rate-determining step. In the case of a dissociative mechanism, for example, bond breaking is critical for reaction progress. Therefore, the strength of the M-X bond influences not only the extent to which the reaction will happen, but also the rate. Although thermodynamic stability of a complex does not necessarily indicate the kinetic rate at which it will react, there is a relationship between bond stability and the rate of reaction if the breaking or formation of that bond is involved in the rate-determining step. The relationship can be stated as follows:
$ln \; k = ln \; K +c \nonumber$
And this relationship can be justified using the Arrhenius equation and equation for the temperature-dependence of the equilibrium constant.
$\ln k=\ln A-\frac{E_{A}}{R T} \quad \text { and } \quad \ln K=\frac{-\Delta H^{\circ}}{R T}+\frac{\Delta S^{\circ}}{R} \nonumber$
In an experiment designed to test the Linear Free Energy Relationship, several reactions would be conducted under nearly identical conditions, where only the identity of the leaving group, X, is varied. In these experiment, the temperature would be constant and the same in each case. If we assume that the pre-exponential factor, $A$ is the same for each reaction condition (and it should be at identical temperatures for analogous reactions), then the $\ln K$ term of the Arrhenius equation becomes proportional to the activation energy, $E_a$ (and proportional to Gibb's energy of activation, $\Delta G^{\ddagger}$, and Gibb's energy $\Delta G$). And since the Gibb's energy is related to the equilibrium constant, K, by $\Delta G = RT \ln K$, the value of $K$ and $k$ should be correlated. Likewise, if $T$ is constant over the reaction conditions and $\Delta S^{\circ}$ is similar across the analogous reactions (which it should be), then $\ln K$ becomes proportional to enthalpy, $\Delta H^{\circ}$, which is proportional to Gibb's free energy of reaction $\Delta G^{\circ}$.
If a Linear Free Energy Relation plot of $\ln K$ vs $\ln k$ is linear, but with a slope less than one, it indicates a dissociative mechanism with some degree of associative character (for example, an $I_d$ mechanism or a mechanism involving the formation of preassociation complex). If the plot of $\log K$ vs $\log k$ for the reactions is linear, this is evidence that the reaction is dissociative. An example for a linear free energy plot is shown in Figure $1$ for the hydrolysis of $\ce{[Co(NH3)5X]^+2}$ (this is the substitution of $\ce{X^-}$ with $\ce{H2O}$).
When the slope is approximately 1, as it is in the case of $\ce{[Co(NH3)5X]^+2}$ in Figure $1$, it indicates that the variation of the $\ce{X^-}$ leaving group has a similar effect on the magnitude of both $\Delta G$ (thermodynamic, related to $K$) and $\Delta G^{\ddagger}$ (kintetic, related to $k$). In other words, a slope close to 1 indicates a purely dissociative pathway (D). This data shown in Figure $1$, for example, is evidence of a dissociative mechanism for $\ce{[Co(NH3)5X]^+2}$.
The argument that a reaction is dissociative can also be supported by evidence showing little effect of the incoming ligand on reaction rate. For example, data supporting the argument that $\ce{[Co(NH3)5H2O]^+3}$ reacts by a dissociative pathway is shown below. The data show that variation of the identity of the incoming ligand, $\ce{Y^-}$, has little effect on the rate constant (they are approximately within one order of magnitude).
$\text{For the reaction: }\ce{[Co(NH3)5H2O]^3+ + Y^{m-} <=> [Co(NH3)5Y]^{(3-m)+} + H2O} \nonumber$
Table $1$: Rate Constants for Exchange of Aquo Ligand of pentaammineaquo cobalt(III) ion by incoming ligand, Y (chemical equation shown above). There is little effect of the identity of the incoming ligand on the rate constant. This data was sourced from Miessler, Fischer and Tarr's Inorganic Chemistry, 5th edition, pg 448.
Entering Ligand, $\ce{Y^-}$ Rate Constant, $k_1 \; (10^{-6}s^{-1})$
(pseudo-first order with excess $\ce{[Y^{-}]}$)
$\ce{H2O}$ 100
$\ce{N3^-}$ 100
$\ce{SO4^2-}$ 24
$\ce{Cl^-}$ 21
$\ce{NCS^-}$ 16
12.4.03: Associative Mechanisms
Associative mechanisms tend to be less common for octahedral complexes as a result of steric hindrance of the metal ion by six ligands. However, reactions of octahedral complexes with associative character are more likely if the ligands have low steric bulk and/or the central metal ion is larger with longer bonds (as in the case of $4d$ and $5d$ metal ions). Reactions of octahedrons have also been observed in cases of low $d$-electron count or low electron density around the central ion, presumably making nucleophilic reaction more favorable. If an octahdral complex is determined to have associative character, in most cases it is associatively-active interchange ($I_a$).
The evidence that supports arguments for associative mechanisms is somewhat the opposite as that which would support dissociative character. One typical piece of evidence that supports an associative pathway is a large effect of the incoming ligand on the reaction rate constant. For example, in the reaction of hexaaquo chromium (III) ion ($\ce{[Cr(H2O)6}]^3+}$) with various incoming ligands (shown below), rate constants vary by several orders of magnitude. This data is evidence of an associative mechanism ($I_a$). In contrast, the reaction of pentaamineaquo chromium (III) ion ($\ce{[Cr(NH3)5H2O}]^3+}$) occurs by an associative mechanism (data shown in Section 12.4.1). The difference in reaction mechanisms for different chromium (III) complexes can be justified by the different electron densities around their metal ion center. The ammine ligand is a stronger sigma donor and thus the metal center is more electron rich in the case of $\ce{[Cr(NH3)5H2O}]^3+}$. Nucleophilic reaction by an incoming ligand is thus more likely in the case of $\ce{[Cr(H2O)6}]^3+}$, which has a more electrophilic center.
$\ce{[Cr(H2O)6]^3+ + Y^{-} <=> [Cr(H2O)5Y]^2+ + H2O} \nonumber$
Table $1$: Rate Constants for Exchange of Aquo Ligand of hexaaquo chromium(III) ion by incoming ligand, Y (chemical equation shown above). There is little effect of the identity of the incoming ligand on the rate constant. This data was sourced from Miessler, Fischer and Tarr's Inorganic Chemistry, 5th edition, pg 449.
Entering Ligand, $\ce{Y^-}$ Rate Constant, $k_1 \; (10^{-4}M^{-1}s^{-1})$
$\ce{NCS^-}$ 180
$\ce{NO3^-}$ 73
$\ce{Cl^-}$ 2.9
$\ce{Br^-}$ 0.9
$\ce{I^-}$ 0.08
Another piece of evidence that can support associative mechanisms is the temperature-dependence of the rate law, which in turn can yield the entropy of activation (represented as $\Delta S^{\ddagger}$ or $\Delta S_{act}$). When an incoming ligand associates with a metal complex and two molecules beome one, there would be a negative change in the entropi (entropy decreases). A negative value for the entropy of activation indicates an associative pathway. An example of this is in the substitution of the aquo ligand in $\ce{[Ru(EDTA)(H2O)]^-}$, which has a negative value of $\Delta S^{\ddagger}$, indicating an $I_a$ mechanism. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.04%3A_Experimental_Evidence_in_Octahedral_Substitutions/12.4.02%3A_Linear_Free_Energy_Relationships.txt |
The substitution reaction of acidic octahedral complexes (with ligands that can donate a proton) can be catalyzed in the presence of hydroxide ion ($\ce{OH^-}$. Through an initial acid-base reaction, the metal complex can be deprotonated to its conjugate base. The deprotonation increases electron density on the metal center, facilitating the loss of the leaving group ligand, especially when the leaving ligand is trans to the deprotonation site. This mechanism is referred to as the conjugate-base mechanism or the ${\mathrm{S}}_{\mathrm{N}}\mathrm{1CB}$ (Substitution, Nucleophilic, first-order in the Conjugate Base mechanism) mechanism. Although it involves a dissociative rate-limiting step, the rate law is consistent with the second-order kinetics because the rate also depends on the initial reaction of the base with the metal complex. In this mechanism, the first step is formation of the conjugate base of the metal complex by deprotonation of one of its ligands. This is a case where numerous lines of evidence are necessary to establish the mechanism.
Cobalt pentammine complexes, $\ce{Co(NH3)5X^{n+}}$, have been studied extensively and are found to react through the conjugate-base mechanism. The way in which the actual mechanism was determined, and how other possible mechanism were ruled out are described below.
Evidence under acidic aqueous conditions
In acidic aqueous solutions, the reaction
$\ce{ [Co(NH3)_5X]^{2+} + Y^{-} -> [Co(NH3)_5Y]^{2+} + X^{-}} \nonumber$
was found to proceed through an aquo complex intermediate, $\ce{[Co(NH3)_5{OH_2}]^{3+}}$, which is also the dominant product despite the identity of the incoming ligand. In other words, to form the product of the reaction above, there are two substitution steps necessary (shown below).
\begin{align*} \ce{[Co(NH3)_5X]^{2+} + H2O} &\rightleftharpoons \ce{[Co(NH3)_5(OH2)]^{3+} + X^{-}} \tag{step 1, form aquo intermediate}\[4pt] \ce{[Co(NH3)5(OH2)]^{3+} + Y^{-}} &\rightleftharpoons \ce{[Co(NH3)5Y]^{2+} + H2O } \tag{step 2, form final product} \end{align*}
The first step in the reaction is the breaking of a $\ce{Co \bond{-} X}$ bond and the formation of a $\ce{Co \bond{-} OH2}$ bond (step 1). Subsequently, $\ce{Y^{-}}$ can replace the aquo group (step 2).
In aqueous solution, water is always present at a much higher concentration than the various possible entering groups $\ce{Y^{-}}$, so it is reasonable that the aquo complex should be favored in the competition to form the new bond to $\ce{Co(III)}$. Nevertheless, the strength of the $\ce{Co \bond{-} Y}$ bond should depend on the nucleophilicity of $\ce{Y^{-}}$ in these substitution reactions. Thus, the amount of $\ce{[Co(NH3)_5Y]^{2+}}$ product should increase with increasing nucleophilicity of $\ce{Y^{-}}$. The fact that the aquo complex is the predominant reaction product despite the identity of $\ce{Y^{-}}$ strongly suggests that the the $\ce{Co \bond{-} X}$ bond breaking is more important in the reaction outcome than formation of the new bond to the incoming $\ce{Y^{-}}$.
This evidence indicates dissociative character in the reaction. But the nature of the substitution mechanism is still unclear. For example, is the original $\ce{Co \bond{-}X}$ bond completely broken before the new $\ce{Co \bond{-} OH2}$ bond has begun to form so that $\ce{Co(NH3)_5^{3+}}$ is a true intermediate? Does the mechanism involve a transitory intermediate where bonds are breaking and forming in a concerted fashion?
Evidence under basic aqueous conditions
The evidence of reaction mechanism under basic conditions is seemingly contradictory to what is described above. When the entering group, $Y^{-}$, is the hydroxide ion, the reaction is
$\ce{ Co(NH3)_5X^{n+} + OH^{-} \to Co(NH3)_5OH^{2+} + X^{q-}} \nonumber$
Substitution by hydroxide ion is kinetically faster than by the aquo ligand in acidic solutions, and the rate is law found to be second-order:
$\frac{d\left[\ce{Co(NH3)_5OH^{2+}}\right]}{dt}=k\ \left[\ce{Co(NH3)_5X^{n+}}\right]\left[\ce{OH^{-}}\right] \nonumber$
This rate law is consistent with an associative reaction involving nucleophilic attack by the hydroxide ion at the cobalt center. In an associative interchange mechanism, the $\ce{Co \bond{-} OH}$ bond formation occurs simultaneously with breaking of the $\ce{Co \bond{-} X}$ bond. However, just as in the case above, the hydroxo complex dominates even in the presence of $\ce{Y^{-}}$. Thus, the hydroxide ion seems to be a uniquely effective nucleophile toward cobalt(III). However, nucleophilic displacements have been investigated on many other electrophile, and in general, hydroxide is not a particularly effective nucleophile toward other electrophilic centers. So, assignment of an associative mechanism to this reaction is reasonable only if we can explain why hydroxide is uniquely reactive in this case and not in others.
The conjugate-base mechanism
An alternative mechanism, referred to as the conjugate-base mechanism or the ${\mathrm{S}}_{\mathrm{N}}\mathrm{1CB}$ (Substitution, Nucleophilic, first-order in the Conjugate Base mechanism) mechanism, is also consistent with the second-order rate law. In this mechanism, the first step is formation of the conjugate base of the metal complex by deprotonation of one of its ligands. An important requirement of this mechanism is the presence of a metal-bound ligand with acidic protons (including amine, ammine, or aquo ligands). Under basic conditions the mecahnism is catalyzed by hydroxide ion.
In the case of $\ce{[Co(NH3)_5{OH_2}]^{3+}}$ reacting under basic conditions, the hydroxide removes a proton from one of the ammine ligands to give a six-coordinate conjugate base intermediate (step 1 below). This conjugate base intermediate contains an amido ($\ce{NH^{-}_2}$) ligand which is a $\pi$-donor ligand, allowing facilitated dissociation of one of the other ligands (particulary the ligand in the trans position). This intermediate loses the leaving group $X^{-}$ in the rate determining step to form a five-coordinate intermediate, $\ce{Co(NH3)4NH^{2+}}$ (step 2 below). This five-coordinate intermediate quickly picks up a water molecule from the bulk solution to give the aquo complex (step 3 below). In a series of proton transfers to (step 4) and from (step 5) the aqueous solvent, the aquo complex rearranges to the final product. The mechanism is shown below for $\ce{[Co(NH3)_5{OH_2}]^{3+}}$ reacting with $\ce{Cl^{-}}$:
\begin{align*} \ce{[Co(NH3)_5{Cl}^{2+}+OH]^{-}} &\rightleftharpoons \ce{[Co(NH3)_4(NH2){Cl}]^{+} + H_2O} \tag{step 1, conjugate base}\[4pt] \ce{[Co(NH3)_4(NH2){Cl}]^{+}} &\rightleftharpoons \ce{[Co(NH3)_4{(NH2)}]^{2+} + {Cl}^{-}} \tag{step 2, dissociation} \[4pt] \ce{Co(NH3)_4{(NH2)}^{2+} + H_2O} &\rightleftharpoons \ce{Co(NH3)_4{(NH2)(OH2)}^{2+}} \tag{step 3, association} \[4pt] \ce{Co(NH3)_4{(NH2)(OH2)}^{2+} + OH^{-}} &\rightleftharpoons \ce{Co(NH3)_4{(NH2)OH}^{+} + H_2O} \tag{step 4, proton transfer} \[4pt] \ce{Co(NH3)_4{(NH2)OH}^{+} + H_2O} &\rightleftharpoons \ce{Co(NH3)_5{OH}^{2+} + OH^{-}} \tag{step 5, proton transfer} \end{align*}
The evidence in favor of the ${\mathrm{S}}_{\mathrm{N}}\mathrm{1CB}$ mechanism is persuasive. It requires that the ammine protons be acidic, so that they can undergo the acid–base reaction in the first step. That this reaction occurs is demonstrated by proton-exchange experiments. In basic $\ce{D_2O}$, the ammine protons undergo $\ce{H-D}$ exchange according to
$\ce{[Co(NH3)5Cl]^{2+} + D2O \rightleftharpoons [Co(ND3)5Cl]^{2+} + HDO} \nonumber$
The ammine protons are also necessary; the reaction does not occur for similar compounds, like $\ce{Co(2,2^'-bipyridine)_2(O_2CCH_3)^{+}_2}$, which lack acidic protons on the nitrogen atoms that are bound to cobalt (i.e., there are no $\ce{H-N-Co}$ moieties).
The evidence that $\ce{Co(NH3)_4{(NH2)}^{2+}}$ is an intermediate is also persuasive. When the base hydrolysis reaction is carried out in the presence of other possible entering groups, $Y^{p-}$, the rate at which $\ce{[Co(NH3)_5X]^{n+}}$ is consumed is unchanged, but the product is a mixture of $\ce{[Co(NH3)_5{OH}]^{2+}}$ and $\ce{[Co(NH3)5Y]^{n+}}$. If this experiment is done with a variety of leaving groups, $X^{q-}$, the proportions of $\ce{[Co(NH3)_5{OH}^{2+}]}$ and $\ce{[Co(NH3)5Y]^{n+}}$ are constant despite which leaving group exists in the reactant complex. These observations are consistent with the hypothesis that all reactants, $\ce{[Co(NH3)_5X]^{n+}}$, give the same intermediate, $\ce{[Co(NH3)_4{(NH2)}]^{2+}}$. The product distribution is always the same, because it is always the same species undergoing the product-forming reaction.
12.4.05: The Kinetic Chelate Effect
We discussed the increased thermodynamic stability of chelating ligands in a previous section on the chelate effect (Section 10.1.1). This thermodynamic benefit is related to a change in the rate of binding and dissociation events called the kinetic chelate effect. This effect was also mentioned earlier in this chapter (Section 12.2.2).
Recall that there is a relationship between the equilibrium constant and rate constants of any chemical reaction. The relationship is demonstrated below for a general reaction.
$aA+bB \rightleftharpoons cC+dD \nonumber$
$\text{forward rate} = k_f[A]^a [B]^b \nonumber$
$\text{reverse rate} = k_r[C]^c [D]^d \nonumber$
At equilibrium, the rate of the forward reactions is equal to the rate of the reverse reaction. Therefore:
$\text{forward rate} = \text{reverse rate} \nonumber$
$k_f[A]^a [B]^b = k_r[C]^c [D]^d \nonumber$
Rearrangement of the equation above gives:
$\dfrac{k_f}{k_r}=\dfrac{[C]^c [D]^d }{[A]^a [B]^b } = K_{eq} \nonumber$
Because chelating ligands have larger values of $K_{eq}$ than analogous monodentate ligands, there must also be a change in the relative rates of the forward (binding) and reverse (dissociation) reactions. In fact, there is a kinetic benefit that makes a metal bound to a polydentate ligand more inert than the analogous complex with monodentate ligands: this is called the kinetic chelate effect.
The kinetic chelate effect is a result of a slower first dissociation step and faster re-association step relative to that of a monodentate ligand. The complete dissociation of a bidentate ligand, for example, would require two dissociation steps (see first reaction in figure below). It is the first dissociation step that is slower in a bidentate ligand ($k_1 < k'_1$ in the figure below). This first dissociation step is slower in part because the chelate would have to rotate to move the free ligand away from the open coordinate site on the metal ion. For the same reason, the reverse of this step (association) is faster than the association of a monodentate ligand ($k_{-1} > k'_{-1}$).
(1) Carter, M. J.; Beattie, J. K. Kinetic Chelate Effect. Chelation of Ethylenediamine on Platinum(II). Inorg. Chem.1970, 9 (5), 1233–1238. https://doi.org/10.1021/ic50087a044.
Curated or created by Kathryn Haas | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.04%3A_Experimental_Evidence_in_Octahedral_Substitutions/12.4.04%3A_The_conjugate_base_mechanism.txt |
The stereochemical outcomes of a reaction can depend on numerous factors. For simplicity, the discussion in this section will be limited to octahedral complexes that react through dissociative mechanisms, with emphasis on those that occur through the limiting case of a dissociative (\(D\)) mechanism. In light of this focus, the reactions of Co(III) complexes, particularly those with ethylenediamine ligands, provide appropriate examples of well-characterized systems.
In the purely dissociative case, the octahedral metal complex completely loses one ligand to become a five-coordinate intermediate. Therefore, the structure of the product depends primarily on the structure of the intermediate and its interaction with the entering ligand. The outcome of a \(D\) reaction should not be influenced by the identity of the leaving group. However, the identity of the entering group and the reaction conditions are appropriate considerations. For example, under basic aqueous conditions, the conjugate base mechanism can alter the stereochemical outcome compared to the same reaction under more acidic conditions.
Structure of the intermediate
When an octahedral metal complex reacts through a \(D\) mechanism, a six-coordinate metal complex forms a five-coordinate intermediate, which goes on to react with the entering group. The structure of that 5-coordinate intermediate has a major influence on the stereochemistry of the product. Recall from a previous section (Section 9.4) that 5-coordinate metal complexes can adopt either a square pyramidal or trigonal bipyramidal geometry. * In general, if the intermediate adopts a square pyramidal structure, the incoming ligand will enter at the same site where the leaving group was lost, and stereochemistry is retained from reactant to product. On the other hand, if the intermediate structure is a trigonal pyramid, at least some of the product will not retain the original configuration.
* Note: Recall from a previous section (Section 9.4) that square pyramid and trigonal bipyramid geometries typically have similar energies and readily interconvert through pseudorotation. However, during a D reaction, the intermediate exists for such a short moment in time, that interconversion is assumed not to occur unless there is an unusually long-lived intermediate.
Cis vs trans reactants
This section will present generalizations for the stereochemical consequences of substitution in trans-bisethylenediamine (trans-en) and cis-bisethylenediamine (cis-en) octahedral Co(III) complexes. Cis-en complexes tend to retain a square pyramidal intermediate structure and retain the cis stereochemistry of the product, except when the conjugate base mechanism promotes isomerization. On the other hand, trans-en complexes are more likely to undergo stereochemical rearrangments through a trigonal pyramidal intermediate. However, especially in the case of trans-en complexes, the product ratios are a result of mixtures of intermediates.
12.05: Stereochemistry of Octahedral Reactions
The reaction of metal complexes of the form $\ce{[M(en)2LX]^{n+}}$ have multiple possible pathways in terms of the structure of the intermediate.
When the X ligand departs from a trans complex, the most likely case is that the L ligand will remain in the axial position. In that case, there is one primary structure for the square pyramidal intermediate. Reaction of that intermediate with the incoming ligand, Y, will result in retention of trans stereochemistry.
An alternative possibility is that the intermediate adopts a trigonal bipyramidal structure. In that case, the outcome of the reaction depends on which of the possible trigonal bipyramidal structures exist and their relative amounts. There are two chemically distinguishable positions on the trigonal bipyramid; the axial and equatorial positions. The remaining L ligand can occupy either the axial position, which has a slightly longer bond, or the equatorial positions, which are slightly closer to the metal and more sterically hindered in space. To adopt either of the trigonal bipyramidal structures, rearrangement of the complex is necessary. It is more likely that the rearrangement will result in the trans ligand being in the equatorial position, since that requires less distortion of the original positions of each ligand (this is not obvious from the structures in Figure $1$). A trigonal structure with axial L requores one of the en ligand nitrogens to move by 90 degrees, while a structure with equitoral L requires only 30 degree movement from any ligand. As a result, the trigonal bipyramid with equitorial L (bottom of Figure $1$) is the dominant trigonal bipyramidal structure. That structure can produce three optical isomers from the three different angles of ligand Y entering along the equitorial plane. When Y enters trans to L, it produces the trans product. When Y enters cis to L, it produces either of the two cis optical isomers. The cis-$\Lamda$ and $cis-\Delta$ isomers are equally likely, and are produced as a racemic miture from an optically inactive starting trans reactant. The expected ratios of cis to trans are 2:1.
Regarding the trigonal bipyramidal intermediates, the identity of L influences its preference for the axial or equatorial positions. For example, $\pi$-donor ligands prefer equatorial positions due to the shorter bond lengths involved in $\pi$-donor interactions. Thus, when L is a strong $\pi$-donor, reaction products are more likely to come from the trigonal intermediate with an equatorial L. This last point is relevant to reactions where the conjugate base mechanisms is at play. The deprotonated ligand that results from the conjugate base mechanism is a stronger $\pi$-donor, and has preference for occupation of the equatorial positions of a trigonal bipyramidal intermediate. Even in cases where the square pyramid is typically preferred, the conjugate base mechanism can result in preference for a trigonal intermediate.
The actual proportion of reaction products of a given stereochemistry depends on the extent to which each of the possible intermediates leads to the products. However, the expected percentages of cis-trans from any single intermediate is rarely found in experimental data (See Table $1$).
Table $1$ gives the % of trans stereochemistry retained in the displacement of $\ce{Cl^-}$ by either $\ce{H2O}$ or $\ce{OH^-}$in a series of $\ce{trans-[Co(en)2LCl]^+}$ complexes where the identity of the trans ligand, $\ce{L^-}$, is varied.
The reactions run under acidic conditions is shown below, and its data is shown in the middle column in Table $1$ :
$\ce{trans-[Co(en)2LCl]^+ + H2O <==> [Co(en)2L(H2O)]^2+ + Cl^-} \nonumber$
The reactions run under basic conditions is shown below, and its data is shown in the right column in Table $1$ :
$\ce{trans-[Co(en)2LCl]^+ + OH^- <==> [Co(en)2L(OH)]^+ + Cl^-} \nonumber$
Table $1$: % trans product under different conditions
$\ce{L^-}$ % trans product under acidic conditions
(stereochemistry of reactant is retained)
% trans product under basic conditions
(stereochemistry of reactant is retained)
$\ce{OH^-}$ 25 6
$\ce{NCS^-}$ 30-50 24
$\ce{Br^-}$ 50 100
$\ce{Cl^-}$ 65 95
$\ce{NH3^-}$ 100 24
$\ce{NO2^-}$ 100 94
Data from Miessler, G. L.; Fischer, P. J.; Tarr, D. A. Inorganic Chemistry, 5th Ed. Pearson: Boston, 2014, p. 459. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.05%3A_Stereochemistry_of_Octahedral_Reactions/12.5.01%3A_Substitution_in_trans-en_octahedral_complexes.txt |
The $D$ substitution reactions of cis complexes occur through the same intermediate structures described above for trans complexes. For example, the reaction,
$\ce{cis{-}[M(en)2LX]^+ + Y^- -> [Co(en)2LY]^2+ + X^-} \nonumber$
can proceed through square planar or trigonal bipyramidal intermediates. Complete the exercise below.
Exercise $1$
Follow the example of Figure 12.5.1.1 to sketch the possible reaction intermediates and products for $\ce{cis{-}[M(en)2LX]^+ + Y^-}$. Which intermediates are most likely? In the case of each trigonal bipyramid, what would be the ratios of products if ligand attack from the three angles along the trigonal plane are equally likely?
Answer
Add texts here. Do not delete this text first.
Table $1$ gives the % of cis stereochemistry in the displacement of $\ce{Cl^-}$ by either $\ce{H2O}$ or $\ce{OH^-}$ in a series of $\ce{cis-[Co(en)2LCl]^+}$ complexes where the identity of the trans ligand, $\ce{L^-}$, is varied.
The reactions run under acidic conditions is shown below, and its data is shown in the middle column in Table $1$ :
$\pm \ce{cis-[Co(en)2LCl]^+ + H2O <==> [Co(en)2L(H2O)]^2+ + Cl^-} \nonumber$
The reactions run under basic conditions is shown below, and its data is shown in the right column in Table $1$. Note that in all but one case, the reactions run under basic conditions were done so using optically pure starting material, and gave an optically active product.
$\Delta-\ce{cis-[Co(en)2LCl]^+ + OH^- <==> [Co(en)2L(OH)]^+ + Cl^-} \nonumber$
Table $1$: % cis product under different conditions
$\ce{L^-}$ % cis product under acidic conditions from racemic cis reactant
(stereochemistry of reactant is retained)
% cis and trans product under basic conditions from $\Delta-cis$ starting material
(stereochemistry of reactant is retained)
$\ce{OH^-}$ 100 61% $\Delta-cis$ | 36% $\Lambda-cis$ | 3% trans
$\ce{NCS^-}$ 100 56% $\Delta-cis$ | 24% $\Lambda-cis$ | 20% trans
$\ce{Br^-}$ 100 -
$\ce{Cl^-}$ 100 21% $\Delta-cis$ | 16% $\Lambda-cis$ | 63% trans
$\ce{NH3^-}$ - 60% $\Delta-cis$ | 24% $\Lambda-cis$ | 16% trans
$\ce{NO2^-}$ 100 46% $\Delta-cis$ | 20% $\Lambda-cis$ | 34% trans
Data from Miessler, G. L.; Fischer, P. J.; Tarr, D. A. Inorganic Chemistry, 5th Ed. Pearson: Boston, 2014, p. 459.
Under acidic conditions, and when the conjugate base mechanism is unlikely, the product specifically retains the stereochemistry of the starting material. This data (middle column above) indicates that the $D$ reaction of the cis isomer, in contrast to trans, reacts exclusively through the square pyramidal intermediate under acidic conditions. This is because in every case, the ligand trans to the X leaving group is an amine from the en ligand, which is not a $\pi$ donor. The ligand cis to the leaving group (L) has no influence on the reaction outcome under acidic conditions.
Under basic conditions, where the conjugate base mechanism is possible, cis stereochemistry is not entirely retained, and the cis products are produced in approximately a 2:1 ration of $Delta:\Lambda$ from a $Delta$ reactant.
Exercise $1$
Consider the data in the right colum of Table $1$. What does the presence of trans product and a 2:1 ratio of the two cis isomers indicate about the intermediates of reaction under basic conditions?
Answer
Add texts here. Do not delete this text first.
12.5.03: Isomerization of Chelate Rings
Recall that the stereoisomers of octahedral complexes with two and three bidentate ligands were discussed previously (Section 9.3). This page will discuss the interconversion of stereoisomers, which can occur through two primary mechanisms involving either (1) bond breaking and bond making steps, and (2) twisting.
Isomerization through dissociative substitution
One way to convert one steroisomer to another is through bond breaking and bond re-making steps. This type of structural rearrangement is essentially a substitution reaction, as described previously in this chapter, except that the leaving group and entering group are the same ligand. Evidence for this type of mechanism comes from the study of isotopically labeled amibidentate ligands (those that have different modes of coordination). For example, an acetyl group with a labeled \(\ce{CD3}\) can be added as an "outside" group (adjacent to the coordinating groups) in a tris(acac)cobalt(III) complex. The labeled group moves to the "inside" (directly coordinated to the metal ion) during isomerization from an optically pure solution to a racemic mixture. This change can only occur through bond breaking and re-forming steps.
Isomerism through twisting
The second pathway of isomerism is through twisting; it does not involve bond breaking or bond forming. Twisting that causes interconversion of octahedral isomers was also discussed in Chapter 9. A figure from that chapter is re-posted here for convenience.
An octahedroal coordination sphere is just a trigonal antiprism in which all edge lengths are identical. Rotation of one triangular face relative to its opposite until the two are eclipsed gives a triganal prismatic geometry. In fact, since continuation of this rotation gives another octahedral complex the trigonal prismatic geometry is an intermediate in isomerization reactions involving octahedral complexes. In tris- and bis-chelates such isomerizations are said to occur by a Bailar twist or a Ray-Dutt twist, which differ only in the relationship between the chelate rings and the faces twisted. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.05%3A_Stereochemistry_of_Octahedral_Reactions/12.5.02%3A_Substitution_in_cis-en_octahedral_complexes.txt |
Substitutions in square planar complexes tend to occur through mechanisms with associative character, although the degree of associative character may vary (eg \(A, \; I_a, \; I_d\). The outcome of square planar substitutions depend on the identities of the leaving group, the ligand trans to the leaving group, and the entering group. The reactions of Pt(II) are well studied and will serve as the primary examples in this section. Pt(II) complexes are of particular interest because of their role as effective chemotherapeutics in the treatment of several types of solid tumor cancers.
The generic chemical equation for substitution in a square planar complex is shown below, where X is the leaving group, Y is the entering group, and \(\ce{[ML3X]}\) is the reactant complex:
12.06: Substitutions in Square Planar Complexes
Kinetics
In general, the reaction kinetics of square-planar associative substitution takes the form of the following rate law, which has two terms:
$\frac {d \ce{[ML3Y]}}{dT} =k_1 \ce{[ML3X]} + k_2 \ce{[ML3X][Y]} \nonumber$
Each of the terms corresponds to one of the two possible reaction pathways. The term, $k_2 \ce{[ML3X][Y]}$, is second-order, and corresponds to a simple associative pathway. The $k_1 \ce{[ML3X]}$ term is first-order, although it also corresponds to an associative pathway. The reason it appears to be first order is because square planar complexes can react through a solvent-assisted pathway where bulk solvent replaces the leaving group, and then is replaced by the entering group (Figure $2$). Because solvent concentration does not appreciably change as the reaction proceeds, it is not part of the rate law term even though it is involved in the rate-determining association. The step in which solvent is substituted by the entering group must happen faster than the solvent-assisted replacement of X.
The kinetic rate constats in square planar substitution can depend on the identities of the leaving group, the ligand trans to the leaving group, and the entering group. The ligands that are in the $cis$ positions with respect to the leaving group do not have a significant affect on the rate, and these are referred to as spectator ligands.
Stereochemistry
Square planar complexes react to give products in which the stereochemistry of the reactant is retained. In other words, a trans-reactant will give a trans-product, and a cis-reactant will give a cis-product (Figure $3$). The identity of the trans ligand has particular importance in the outcome of square planar substitutions (discussed further in the next sections). We will designate the trans ligand generally with the symbol T to indicate its position (Figure $3$), the leaving group as X, and the entering group as Y. The spectator ligands (ligands that are $cis$ to X) will be desginated L and L'. For the discussion that follows, the plane of the metal complex will be designated as the $xy$ plane, while the $z$ axis is perpendicular to the plane (Figure $3$). The $x$ and $y$ axes will run colinear to the bond axes, with the $x$ axis running colinear to the T-M-X bonds (Figure $3$).
The retention of stereochemistry, and the fact that X, Y, and T can influence the rate of reaction provide hints at the nature of intermediates and/or transition states involved in these reactions. The existence of five-coordinate square pyramidal structures, like those of $\ce{[Ni(CN)5]^3-}$ and $\ce{[Pt(SnCl3)5]^3-}$, suggests that five-coordinate intermediates are possible. However the substitutions of square planar complexes are generally classified as associative interchange mechanisms ($I_a$) because isolation of a true intermediate is rarely accomplished. In any case, the retention of stereochemistry is explained by a five-coordinate trigonal bipyramidal transition state in which X, Y, and T are each in the trigonal plane (along with the metal center). As the M-X bond is breaking, the M-Y bond is forming and Y replaces X in the position trans to T. Because T, Y and X are in the trigonal plane in the transition state, all three ligands are able to interact with the same $d$ orbitals of the central metal ion. Thus the identities of X, Y, and T should influence the ability of the M-X bond to break and the M-Y bond to form. On the other hand, L and L' are in a perpendicular plane, and would interact with different orbitals than X and Y. This explains why spectator ligands L and L' have little effect on the reaction rate, as they are interacting with different orbitals than those involved in bond breaking and bond making.
The solvent-assisted pathway can occur through similar transition states and mechanistic steps, except that there are additional steps involving solvent association and displacement.
Although there is general consensus that square planar complexes react through associative pathways, the characterization of intermediates and/or transition states has remained elusive, and there is still debate as to the intricate details of the reaction mechanisms involving square planar complexes. For example, there are arguments that a six-coordinate transition state may exist in the solvent-assistant pathway.
12.6.02: Evidence for Associative Reactions
The reactions of Pt(II) complexes have been thoroughly investigated, and there are several lines of experimental evidence that have lead to the general conclusion that square planar complexes react through associative mechanisms. The experimental evidence is presented here.
The entering group
In the reaction of $\ce{trans-[PtL2Cl2]}$, the rate of reaction depends on the identity of the incoming ligand Y.
For example, the rate of the reaction shown above depends heavily on the ligand, Y, as follows:
$\text{most kinetically labile }\ce{PR_3 > CN^- > SCN^- > I^- > Br^- > N3^- > NO2^- > py > NH3 ~ Cl^- > CH3OH } \text{ most kinetically inert} \nonumber$
In general, ligands that would form the strongest bonds with Pt(II) due to their soft character or $\pi$-binding ability react fastest. The rate constants in the above series range by nearly ten orders of magnitude. This huge range of rate constants as a factor of Y demonstrates suggests that M-Y bond formation is part of the rate limiting step. This is evidence for rate-determining association $A$ or $I_a$.
The leaving group
Since the leaving group occupies a similar position in the transition state as the entering group, we should expect the leaving group to have a similar effect on the rate constants as was found for Y. In fact, this is true. For the reaction of $\ce{[Pt(diene)X]^+}$ with pyridine, the follow trend was observed by varying the leaving group, X:
$\text{most kinetically labile }\ce{NO3^- > Cl^- > Br^- > I^- > N3^- > SCN^- > NO2^- > CN^-} \text{ most kinetically inert} \nonumber$
The rate of reaction follows an order that is nearly opposite as that observed for Y: The softer the ligand, the slower the dissociation. The rate constants in this series span a range of approximately five orders of magnitude, illustrating that breaking of the M-X bond is part of the rate-determining step. This is evidence for a mechanism involving associative interchange character ($I_a$ or $I_d$. The importance of the leaving group somewhat depends on the identity of the entering group, and vice-versus.
The trans ligand
The trans ligand also occupies a similar position to X and Y in the transition state. So, as you might expect, it too has influence on the rate of reaction. In general, the rate of substitution of X in Pt(II) complexes general follows the order given below when T is varied as:
$\text{M-X most kinetically labile }\ce{CN^- ~ CO ~ C2H4 > PH3 ~ SH2 > NO2^- > I^- > Br^- > py ~ NH3 > OH^- > H2O } \text{ M-X most kinetically inert} \nonumber$
The effect of the trans ligand deserves a more through discussion, and will be addressed in the next section. It is mentioned here because it is evidence for the trigonal bipyramidal intermediate that was described earlier for associative substitution in square planar metal complexes. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.06%3A_Substitutions_in_Square_Planar_Complexes/12.6.01%3A_Kinetics_and_Stereochemistry_of_Square_Planar_Rea.txt |
Occasionally in ligand substitution there is a situation in which there are two identical ligands that could be replaced, but two different products would result depending on which ligand left. This situation often happens in square planar complexes, for example. Replacement of one ligand would lead to a cis product. Replacement of the other one would lead to a trans product.
An important example of this issue is in the synthesis of cis-platin, an antitumour medication frequently used to treat ovarian and testicular cancer.
Cis-platin could be made from treatment of tetraammineplatinum(II) with chloride salts. The chloride ion could replace two of the ammonia ligands.
But that doesn't work. That synthesis results in the formation of trans-platin, a compound that has all of the nasty side effects of the cis isomer but with none of the therapeautic benefit.
If instead you were to start with tetrachloroplatinate salts and treat them with ammonia, you could replace two of the chloride ligands. That works really well, and it provides cis-platin, not trans-platin.
Exercise $1$
What do you think is the mechanism of substitution of the two reactions above? Why?
Answer
This is probably an associative mechanism because of the square planar geometry.
This reaction, if run under these conditions, is clearly not always under thermodynamic control. Two different products result, depending on how the reaction is done. One of those isomers is probably more stable than the other; if thermodynamics were in charge, it would form the same thing both times.
Instead, there may be an element of kinetic control for at least one of the pathways. A given product might be made, not because it is more stable, but simply because it forms more quickly than the other one.
Take another look at those two reactions. One of the things that they have in common is that the ligand that gets replaced is trans to a chloride. It is not trans to an ammonia. Maybe the other ligands in the complex can influence how quickly one ligand can leave.
Specifically, the "trans effect" is the role of trans-ligands in influencing substitution rates in square planar complexes.
The following kinetic data were obtained for substitutions on square planar platinum complexes, in the reaction:
$trans-(PEt_{3})_{2}PtLCl + py \rightarrow trans-(PEt_{3})_{2}PtLpy^{+} + Cl^{-} \nonumber$
L kobs (s-1) T, °C
PMe3 0.20 0
H- 0.047 0
PEt3 0.041 0
CH3- 6.0 x 10-4 25
C6H5- 1.2 x 10-4 25
Cl- 3.5 x 10-6 25
Ref: Cooper Langford & Harry Gray, Ligand Substitution Processes, W.A. Benjamin, NY, 1965, p. 25.
Exercise $2$
Draw structures for each of the complexes listed in the table.
Answer
Exercise $3$
Provide a mechanism with arrows for the reaction studied in the table.
Answer
Clearly that trans ligand has a dramatic effect on how quickly the chloride can be substituted in the above study. Additional studies like this one have led to some general trends. Below, the ligands on the left have strong trans-effects. Ligands trans to them are substituted very quickly. The ligands on the right have very modest trans effects. Ligands trans to them are substituted only slowly.
Exercise $4$
Look for empirical trends in the series of ligands above. Without trying to explain exactly why, find chemically relevant factors that may be responsible for these reactivity trends.
Answer
There is an electronegativity trend: the less electronegative, the greater the trans effect (see the halogens, as well as the series O,N,C and also the orders within several pairings: S,P; O,S and N,P).
Alternatively, some of the above could be described by a polarizability trend: more polarizable atom, greater trans effect (for example, the halogens).
Most of the ligands containing π-bonds have strong trans influence (but not all).
Most of the π-donors have a weaker trans influence. However, these ligand cover a very broad range in this series.
In general, explanations of the trans effect have focused on two separate types of ligands. These are strong sigma donors and strong pi acceptors.
Strong sigma donors donate electrons very effectively to the metal via a sigma bond. Because the ligand trans to this donor would be bonding via donation to the same metal p orbital, there is a competition. The metal p orbital bonds more favorably with the strong sigma donor, and the ligand trans to it is left with a weaker bond.
The strong sigma donor gets good overlap with the metal orbital and the resulting interaction goes down low in energy.
The weak sigma donor gets poorer overlap with the metal orbital and only weak stabilization of the donor electrons.
Place these two choices together, and the metal orbital will engage in a strong bonding interaction with the strong σ-donor. Doing so lowers the electronic energy significantly. It won't interact very much with the weak σ-donor, because doing that won't result in as much lowering of electronic energy. The result is a strong bond on one side of the metal and a weak bond on the other. That weak bond will break easily and that ligand will be replaced easily.
Exercise $5$
Which of the ligands in the trans effect series are probably strong σ-donors? Why?
Answer
The strongest σ-donors are typically those with more polarizable donor atoms (such as S, P, I) as well as those with less electronegative donor ions such as C- and H-.
Strong pi acceptors exert their trans effect in a different way. They are thought to stabilize a particular geometry of the five-coordinate intermediate in substitution of square planar complexes. We haven't worried too much about the geometry of that intermediate, but it is probably trigonal bipyramidal. It would have three ligands in an equatorial plane and two more directly opposite each other, in the axial positions.
Essentially, the incoming ligand pushes two of the ligands down from the square plane to form this trigonal bipyramid. When it comes time for a ligand to leave, it is probably going to be one of these ligands that is already on the move. They are already on a trajectory out of the square plane, anyway.
A strong pi acceptor like CO exerts its trans effect by making sure it, along with the ligand opposite it, gets into that equatorial plane. It does that by a stabilizing delocalization that happens when the π-acceptor is in the electron-rich equatorial plane. In that position, it can draw electron density via π-donation from two different donors. If it were in an axial position, it could still delocalize electrons this way, but it would draw most effectively from just one donor rather than two.
So which of those two ligands is going to keep moving and leave the complex? It certainly won't be the one that is exerting a stabilizing, delocalizing effect on the complex via its strong bonding interactions. It will be the unlucky trans ligand that got dragged along with it.
Exercise $6$
Which of the ligands in the trans-effect series are probably strong π-acceptors? Show why.
Exercise $7$
Draw an orbital picture showing the -delocalization described in the trigonal bipyramidal intermediate. Label the orbitals, assuming the z axis is along the axis of the trigonal bipyramid (i.e. the equatorial plane is the xy plane).
Answer
Exercise $8$
Substitution of trans ligands in μ-oxo-bis(μ-acetato)diruthenium complexes: Synthesis and kinetic studies. Hussain, Bhatt, Kumar, Thorat, Padhiyar and Shukla, Inorganica Chemica Acta, 2009, 362, 1101-1108.
Given the structure [Ru2O(L)6(acetate)2](PF6)2, in which L is a neutral donor,
a) Draw the structure of the counterion, PF6-.
b) Provide an account of the valence electron count in the ruthenium coordination complex.
Valence electrons on metal
Total charge on ligands
Charge on the metal
Revised count on metal
Electrons donated by ligands
Total electrons on metal in complex
Data shows that the ligands L are trans to the bridging oxo ligand are labile.
c) Use orbital cartoons and words to provide an explanation for this effect.
When water is added to an acetone solution of [Ru2O(pyridine)4(L)2(acetate)2](PF6)2, then [Ru2O(pyridine)4(H2O)2(acetate)2](PF6)2 is formed.
d) Draw the product of this reaction.
When pyridine is added to [Ru2O(pyridine)4(H2O)2(acetate)2](PF6)2, then [Ru2O(pyridine)6(acetate)2](PF6)2 is formed. The following rate constants were observed for this reaction.
[pyridine] mol L-1 kobs s-1 (x 10-3)
0.005 1.75
0.012 2.53
0.025 9.7
0.049 16.7
0.100 35.2
e) Use the data from this table to determine the order in pyridine. Provide an explanation for your conclusion.
f) Draw a mechanism for the reaction, consistent with the data.
g) Write a rate law for the reaction.
h) The authors studies the same reaction with different ligands instead of pyridine. They observed an increase in rate constants with increasing basicity of the incoming ligands. Provide an explanation for this observation.
Answer a
Answer b
Answer c
The oxo is a strong sigma donor and hogs the orbital with the metal thus leaving very little room for orbital bonding trans to the sigma donor.
Answer d
Answer e
First order. Although there is some amount of error in the data, doubling the pyridine concentration generally results in a doubling of the rate.
Answer f
The mechanism you draw would have to involve a first associative step; because the complex is already 18 electrons, associative interchange is likely.
Answer g
$Rate = k [ML_{n}][py] \nonumber$
Answer h
Basic ligands have stronger attraction to the metal thus accelerating the reaction. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.07%3A_The_Trans_Effect.txt |
The oxidation state of a metal ion can be changed by reduction/oxidation (redox) reactions in which electrons are transferred between a metal center and another species. The other species could be another metal ion or some other redox partner. In the case of metal complexes, the electron transfer often occurs at the metal center, changing the oxidation state of the metal ion(s) involved. This section will focus on the redox reactions involving change in metal ion oxidation state.
There are two general mechanisms by which electrons can be transferred to or from a metal center. The definition of these two mechanisms depends on whether the redox partner is bound within the metal ion's inner sphere (bound directly to the metal ion), or whether it is just nearby the metal complex (in the metal ion's outer sphere). If the redox partner is bound to the metal ion itself, it is called an inner-sphere electron transfer. Electrons can also be transferred between two species that are close in proximity, but not bound to one another. When electrons are transferred to or from a metal through close proximity, but not through direct bonds to the metal center, it is an outer-sphere electron transfer.
12.08: Redox Mechanisms
How does an electron get from one metal to another? This might be a more difficult task than it seems. In biochemistry, an electron may need to be transfered a considerable distance. Often, when the transfer occurs between two metals, the metal ions may be constrained in particular binding sites within a protein, or even in two different proteins.
That means the electron must travel through space to reach its destination. Its ability to do so is generally limited to just a few Angstroms (remember, an Angstrom is roughly the distance of a bond). Still, it can react with something a few bond lengths away. Most things need to actually bump into a partner before they can react with it.
This long distance hop is called an outer sphere electron transfer. The two metals react without ever contacting each other, without getting into each others' coordination spheres. Of course, there are limitations to the distance involved, and the further away the metals, the less likely the reaction. But an outer sphere electron transfer seems a little magical.
Barrier to Reaction: A Qualitative Picture of Marcus Theory
So, what holds the electron back? What is the barrier to the reaction? Rudy Marcus at Caltech has developed a mathematical approach to understanding the kinetics of electron transfer, in work he did beginning in the late 1950's. We will take a very qualitative look at some of the ideas in what is referred to as "Marcus Theory". An electron is small and very fast. All those big, heavy atoms involved in the picture are lumbering and slow. The barrier to the reaction has little to do with the electron's ability to whiz around, although even that is limited by distance. Instead, it has everything to do with all of those things that are barely moving compared to the electron.
Imagine an iron(II) ion is passing an electron to an iron(III) ion. After the electron transfer, they have switched identities; the first has become an iron(III) and the second has become an iron(II) ion.
Nothing could be simpler. The trouble is, there are big differences between an iron(II) ion and an iron(III) ion. For example, in a coordination complex, they have very different bond distances. Why is that a problem? Because when the electron hops, the two iron atoms find themselves in sub-optimal coordination environments.
Exercise \(1\)
Suppose an electron is transferred from an Fe(II) to a Cu(II) ion. Describe how the bond lengths might change in each case, and why. Don't worry about what the specific ligands are.
Answer
The bonds to iron would contract because the increased charge on the iron would attract the ligand donor electrons more strongly. The bonds to copper would lengthen because of the lower charge on the copper.
Exercise \(2\)
In reality, a bond length is not static. If there is a little energy around, the bond can lengthen and shorten a little bit, or vibrate. A typical graph of molecular energy vs. bond length is shown below.
1. Why do you think energy increases when the bond gets shorter than optimal?
2. Why do you think energy increases when the bond gets longer than optimal?
3. In the following drawings, energy is being added as we go from left to right. Describe what is happening to the bond length as available energy increases.
Answer a
a) Most likely there are repulsive forces between ligands if the bonds get too short.
Answer b
b) Insufficient overlap between metal and ligand orbitals would weaken the bond and raise the energy.
Answer c
c) The range of possible bond lengths gets broader as energy is increased. The bond has more latitude, with both longer and shorter bonds allowed at higher energy.
Exercise \(3\)
The optimum C-O bond length in a carbon dioxide molecule is 1.116 Å. Draw a graph of what happens to internal energy when this bond length varies between 1.10 Å and 1.20 Å. Don't worry about quantitative labels on the energy axis.
Answer
Exercise \(4\)
The optimum O-C-O bond angle in a carbon dioxide molecule is 180 °. Draw a graph of what happens to internal energy when this bond angle varies between 170 ° and 190 °. Don't worry about quantitative labels on the energy axis.
Answer
The barrier to electron transfer has to do with reorganizations of all those big atoms before the electron makes the jump. In terms of the coordination sphere, those reorganizations involve bond vibrations, and bond vibrations cost energy. Outside the coordination sphere, solvent molecules have to reorganize, too. Remember, ion stability is highly influenced by the surrounding medium.
Exercise \(5\)
Draw a Fe(II) ion and a Cu(II) ion with three water molecules located somewhere in between them. Don't worry about the ligands on the iron or copper. Show how the water molecules might change position or orientation if an electron is transferred from iron to copper.
Answer
The water molecules may pivot toward the more highly charged Fe(III), or they may shift closer to it because of the attraction between the ion and the dipole of the water molecule.
Keep in mind that such adjustments would happen in non-polar solvents, too, although they would involve weaker IMFs such as ion - induced dipole interactions.
Thus, the energetic changes needed before electron transfer can occur involve a variety of changes, including bond lengths of several ligands, bond angles, solvent molecules, and so on. The whole system, involving both metals, has some optimum set of positions of minimum energy. Any deviations from those positions requires added energy. In the following energy diagram, the x axis no longer defines one particular parameter. Now it lumps all changes in the system onto one axis. This picture is a little more abstract than when we are just looking at one bond length or one bond angle, but the concept is similar: there is an optimum set of positions for the atoms in this system, and it would require an input of energy in order to move any of them move away from their optimum position.
It is thought that these kinds of reorganizations -- involving solvent molecules, bond lengths, coordination geometry and so on -- actually occur prior to electron transfer. They happen via random motions of the molecules involved. However, once they have happened, there is nothing to hold the electron back. Its motion is so rapid that it can immediately find itself on the other atom before anything has a chance to move again.
Consequently, the barrier to electron transfer is just the amount of energy needed for all of those heavy atoms to get to some set of coordinates that would be accessible in the first state, before the electron is transferred, but that would also be accessible in the second state, after the electron is transfered.
Exercise \(6\)
Describe some of the changes that contribute to the barrier to electron transfer in the following case.
Answer
The reactants and products are very similar in this case. However, the Fe(III) complex has shorter bonds than the Fe(II) complex because of greater electrostatic interaction between the metal ion and the ligands. These changes in bond length needed in order to get ready to change from Fe(III) to Fe(II) (or the reverse) pose a major barrier to the reaction.
In the drawing below, an electron is transferred from one metal to another metal of the same kind, so the two are just switching oxidation states. For example, it could be an iron(II) and an iron(III), as pictured in the problem above. In the blue state, one iron has the extra electron, and in the red state it is the other iron that has the extra electron. The energy of the two states are the same, and the reduction potential involved in this transfer is zero. However, there would be some atomic reorganizations needed to get the coordination and solvation environments adjusted to the electron transfer. The ligand atoms and solvent molecules have shifted in the change from one state to another, and so our energy surfaces have shifted along the x axis to reflect that reorganization.
That example isn't very interesting, because we don't form anything new on the product side. Instead, let's picture an electron transfer from one metal to a very different one. For example, maybe the electron is transferred from cytochrome c to the "copper A" center in cytochrome c oxidase, an important protein involved in respiratory electron transfer.
Exercise \(7\)
In the drawing above, some water molecules are included between the two metal centres.
1. Explain what happens to the water molecules in order to allow electron transfer to occur, and why.
2. Suppose there were a different solvent, other than water, between the complexes. How might that affect the barrier to the reaction?
Answer a
a) The drawing is an oversimplification, but in general the water molecules are shown reorienting after the electron transfer because of ion-dipole interactions. In this case, the waters are shown orienting to present their negative ends to the more positive iron atom after the electron transfer. In reality, in a protein there are lots of other charges (including charges on the ligand) that may take part in additional ion-dipole interactions.
Answer b
b) Because electron transfer is so fast, atomic and molecular reorganisations are actually thought to happen before the electron transfer. The water molecules would happen to shift into a position that would provide the greatest possible stabilisation for the ions and then the electron would be transferred. A less polar solvent than water would be less able to stabilize ions and the electron would be slower to transfer as a result. In addition, a less polar solvent than water would be a poor medium to transmit an electron, which is charged and therefore stabilized by interactions with polar solvents.
The energy diagram for the case involving two different metals is very similar, except that now there is a difference in energy between the two states. The reduction potential is no longer zero. We'll assume the reduction potential is positive, and so the free energy change is negative. Energy goes down upon electron transfer.
Compare this picture to the one for the degenerate case, when the electron is just transferred to a new metal of the same type. A positive reduction potential (or a negative free energy change) has the effect of sliding the energy surface for the red state downwards. As a result, the intersection point between the two surfaces also slides downwards. Since that is the point at which the electron can slide from one state to the other, the barrier to the reaction decreases.
What would happen if the reduction potential were even more positive? Let's see in the picture below.
The trend continues. According to this interpretation of the kinetics of electron transfer, the more exothermic the reaction, the lower its barrier will be. It isn't always the case that kinetics tracks along with thermodynamics, but this might be one of them.
But is all of this really true? We should take a look at some experimental data and see whether it truly works this way.
Oxidant k (M-1s-1) (margin of error shown in parentheses)
Co(diene)(NH3)23+ 0.12 3.0(4)
Co(diene)H2O)NCS2+ 0.38 11(1)
Co(diene)(H2O)23+ 0.53 800(100)
Co(EDTA) 0.60 6000(1000)
As the reduction potential becomes more positive, free energy gets more negative, and the rate of the reaction dramatically increases. So far, Marcus theory seems to get things right.
Exercise \(8\)
1. Plot the data in the above table.
2. How would you describe the relationship? Is it linear? Is it exponential? Is it direct? Is it inverse?
3. Plot rate constant versus free energy change. How does this graph compare to the first one?
Answer a
a) Here is a plot of the data.
Answer b
b) It doesn't look linear. If we plot the y axis on a log scale, things become a little more linear.
It looks closer to a logarithmic relationship than a linear one.
Answer c
c) Assuming one electron transfer:
The graph takes the same form but in the opposite direction along the x axis.
Marcus Inverted Region
When you look a little closer at Marcus theory, though, things get a little strange. Suppose we make one more change and see what happens when the reduction potential becomes even more positive.
So, if Marcus is correct, at some point as the reduction potential continues to get more positive, reactions start to slow down again. They don't just reach a maximum rate and hold steady at that plateau; the barrier gets higher and higher and the reactions get slower and slower. If you feel a little skeptical about that, you're in good company.
Marcus always maintained that this phenomenon was a valid aspect of the theory, and not just some aberration that should be ignored. The fact that nobody had ever actually observed such a trend didn't bother him. The reason we didn't see this kind of thing, he said, was that we just hadn't developed technology that was good enough to measure these kind of rates accurately.
But technology did catch up. Just take a look at the following data (from Miller, J. Am.Chem. Soc. 1984, 3047).
Don't worry that there are no metals involved anymore. An electron transfer is an electron transfer. Here, an electron is sent from the aromatic substructure on the right to the substructure on the left. By varying the part on the left, we can adjust the reduction potential (or the free energy change, as reported here.
Exercise \(9\)
1. Plot the data in the above table.
2. How would you describe the relationship?
Answer a
a)
Answer b
b) We can see two sides of an inverted curve. The reaction gets much faster as the free energy becomes more negative, but at some point the rate begins to decrease again.
As the reaction becomes more exergonic, the rate increases, but then it hits a maximum and decreases again. Data like this means that the "Marcus Inverted Region" is a real phenomenon. Are you convinced? So were other people. In 1992, Marcus was awarded the Nobel Prize in Chemistry for this work.
Exercise \(10\)
Take a look at the donor/acceptor molecule used in Williams' study, above. a) Why do you suppose the free energy change is pretty small for the first three compounds in the table? b) Why does the free energy change continue to get bigger over the last three compounds in the table?
Answer
The acceptor compound becomes an anion when it accepts an electron. The first three compounds do not appear to be strongly electrophilic; they can accept electrons simply because of resonance stability of the resulting anion. The last three have electron withdrawing groups (chlorines and oxygens) that would stabilize the anion even further.
Exercise \(11\)
The rates of electron transfer between cobalt complexes of the bidentate bipyridyl ligand, Co(bipy)3n+, are strongly dependent upon oxidation state in the redox pair. Electron transfer between Co(I)/Co(II) occurs with a rate constant of about 109 M-1s-1, whereas the reaction between Co(II)/Co(III) species proceeds with k = 18 M-1s-1.
1. What geometry is adopted by these complexes?
2. Are these species high spin or low spin?
3. Draw d orbital splitting diagrams for each complex.
4. Explain why electron transfer is so much more facile for the Co(I)/Co(II) pair than for the Co(II)/Co(III) pair.
Answer a
a) octahedral; bpy is a bidentate ligand.
Answer b
b) Co is first row; Co(I) and Co(II) have relatively low charge. Usually we would expect them to be high spin. Co(III) is at a cut-off point in the first row; it is just electronegative enough that it is usually low spin.
Answer c
c)
Answer d
d) In a transfer from Co(II) to Co(III), there is additional reorganization needed because the metal changes between high and low spin. Not only does one electron have to move from one metal to another metal, but additional electrons have to shuffle from one orbital to another on the same metal to accommodate the change. These reorganizations have a barrier, slowing the reaction. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.08%3A_Redox_Mechanisms/12.8.01%3A_Outer_Sphere_Electron_Transfer.txt |
In some cases, electron transfers occur much more quickly in the presence of certain ligands. For example, compare the rate constants for the following two electron transfer reactions, involving almost exactly the same complexes:
$\ce{Co(NH3)6^{3+} + Cr^{2+} -> Co^{2+} + Cr^{3+} + 6 NH3} \: \: k = 10^{-4}M^{-1}s^{-1} \nonumber$
$\ce{Co(NH3)5Cl^{2+} + Cr^{2+} -> Co^{2+} + CrCl^{2+} + 6NH3} \: \: k=6 \times 10^{5} M^{-1}s^{-1} \nonumber$
(Note: aqua ligands are omitted for simplicity. Ions, unless noted otherwise, are aqua complexes.)
Notice two things: first, when there is a chloride ligand involved, the reaction is much faster. Second, after the reaction, the chloride ligand has been transferred to the chromium ion. Possibly, those two events are part of the same phenomenon.
Similar rate enhancements have been reported for reactions in which other halide ligands are involved in the coordination sphere of one of the metals.
In the 1960’s, Henry Taube of Stanford University proposed that halides (and other ligands) may promote electron transfer via bridging effects. What he meant was that the chloride ion could use one of its additional lone pairs to bind to the chromium ion. It would then be bound to both metals at the same time, forming a bridge between them. Perhaps the chloride could act as a conduit for electron transfer. The chloride might then remain attached to the chromium, to which it had already formed a bond, leaving the cobalt behind.
Electron transfers that occur via ligands shared by the two metals undergoing oxidation and reduction are termed "inner sphere" electron transfers. Taube was awarded the Nobel Prize in chemistry in 1983; the award was based on his work on the mechanism of electron transfer reactions.
Exercise $1$
Take another look at the two electron transfer reactions involving the cobalt and chromium ion, above.
1. What geometry is adopted by these complexes?
2. Are these species high spin or low spin?
3. Draw d orbital splitting diagrams for each complex.
4. Explain why electron transfer is accompanied by loss of the ammonia ligands from the cobalt complex.
5. The chloride is lost from the cobalt complex after electron transfer. Why does it remain on the chromium?
Answer a
a) octahedral
Answer b
b) In the first row, 2+ complexes are almost always high spin. However, 3+ complexes are sometimes low spin.
Answer c
c)
Answer d
d) The Co(II) complex is high spin and labile. The ligands are easily replaced by water.
Answer e
e) The Cr(III) complex is only d3; it is inert.
Other ligands can be involved in inner sphere electron transfers. These ligands include carboxylates, oxalate, azide, thiocyanate, and pyrazine ligands. All of these ligands have additional lone pairs with which to bind a second metal ion.
Exercise $2$
Draw an example of each of the ligands listed above bridging between a cobalt(III) and chromium(II) aqua complex.
Answer
Exercise $3$
Explain, with structures and d orbital splitting diagrams, how the products are formed in the following reaction, in aqueous solution.
$\ce{Fe(OH2)6^{2+} + (SCN)Co(NH3)5^{2+} -> (NCS)Fe(OH2)5^{2+} + Co(OH2)6^{2+} + 5NH3} \nonumber$
Answer
How does the electron travel over the bridge?
Once the bridge is in place, the electron transfer may take place via either of two mechanisms. Suppose the bridging ligand is a chloride. The first step might actually involve an electron transfer from chlorine to the metal; that is, the chloride could donate one electron from one of its idle lone pairs. This electron could subsequently be replaced by an electron transfer from metal to chlorine.
Sometimes, we talk about the place where an electron used to be, describing it as a "hole". In this mechanism, the electron donated from the bridging chloride ligand leaves behind a hole. The hole is then filled with an electron donated from the other metal.
Alternatively, an electron might first be transferred from metal to chlorine, which subsequently passes an electron along to the other metal. In the case of chlorine, this idea may be unsatisfactory, because chlorine already has a full octet. Nevertheless, some of the other bridging ligands may have low-lying unoccupied molecular orbitals that could be populated by this extra electron, temporarily.
Exercise $4$
For the iron / cobalt electron transfer in problem Exercise $3$ (RO9.3.), show
1. an electron transfer mechanism via a hole migration along the bridge
2. an electron transfer mechanism via an electron migration along the bridge
Answer a
Answer b
Exercise $5$
One of the many contributions to the barrier for electron transfer between metal ions is internal electronic reorganization.
a) Draw d orbital splitting diagrams for each of the following metal ions in an octahedral environment.
Ru(II) or Os(II)
Ru(III) or Os(III)
Co(II)
Co(III)
Flash photolysis is a method in which an electron can be moved instantly “uphill” from one metal to another (e.g. from M2II to M1III, below); the electron transfer rate can then be measured as the electron “drops” back from M1II to M2III.
b) Explain the relative rates of electron transfer reaction in this system, as measured by flash photolysis in the table below.
M1II M2III kobs s-1
Os Ru > 5 x 109
Os Co 1.9 x 105
c) Does the reaction above probably occur via an inner sphere or by an outer sphere pathway? Why?
Answer a
a)
Answer b
b) The electron transfer between Os(II) and Ru(III) will not involve any electron reorganization because both are low spin to begin with. However, the electron transfer between Os(II) and Co(III) will result in cobalt changing from low spin to high spin. The need to move electrons between different d orbitals on the cobalt will add to the barrier, slowing down the reaction.
Answer c
c) The pathway is probably inner sphere because of the bridging ligand. Furthermore, the conjugation in the bridging ligand would help in conducting an electron from one end of the ligand to the other, either through an electron mechanism or a hole mechanism.
Exercise $6$
Outer sphere electron transfer rates depend on the free energy change of the reaction (ΔG°) and the distance between oxidant and reductant (d) according to the relation
Rate constant = $k = Ae^{(- \Delta G)} e^{-d}$
a) What happens to the rate of the reaction as distance increases between reactants?
One potential problem in measuring rates of intramolecular electron transfer (i.e. within a molecule) is competition from intermolecular electron transfer (between molecules).
b) What would you do in the flash photolysis experiment above to discourage intermolecular electron transfer?
c) How could you confirm whether you were successful in discouraging intermolecular reaction?
Answer a
a) The rate decreases exponentially as distance increases.
Answer b
b) You might keep the concentration low in order to increase the distance between molecules, reducing the likely hood of an outer-sphere electron transfer.
Answer c
c) If you ran the experiment at a series of dilutions, intramolecular electron transfer would be unaffected but outer sphere electron transfer would not. If the rates were the same across a number of different concentrations, the reaction would probably be intramolecular.
Exercise $7$
Stephan Isied and coworkers at Rutgers measured the following electron transfer rates between metal centers separated by a peptide. (Chem Rev 1992, 92, 381-394)
1. The proline repeating unit is crucial in ensuring a steady increase in distance between metal centers with increased repeat units, n. Why?
2. An inner sphere pathway in this case is expected to be somewhat slow because of the lack of conjugation in the polyproline bridge. Explain why.
3. Plot the data below, with logk on the y axis (range from 4-9) and d on the x axis (12-24 Angstroms).
n d (Å) kobs (s-1 )
1 12.2 5 x 108
2 14.8 1.6 x 107
3 18.1 2.3 x 105
4 21.3 5.1 x 104
5 24.1 1.8 x 104
d) A linear relationship is in agreement with Marcus theory; logk = - c x d. Is your plot linear?
Isied offers a number of possible explanations for the data, all of which involve two competing reaction pathways.
e) Suggest one explanation for the data.
Answer a
a) Rings are frequently used to introduce conformational rigidity (or decrease conformational flexibility), limiting the range of potential shapes a molecule could adopt. If the molecule can't wiggle around as much, then the distance between the ends of the molecule should be more constant.
Answer b
b) Although the ligand is bridging, it would be difficult to picture either an electron or hole mechanism of inner sphere electron transfer. There are few pi bonds or lone pairs to use as places to put electrons or temporarily remove electrons from, shuttling the electrons from place to place along the ligand. A conjugated system would be much more likely to carry out inner sphere electron transfer.
Answer c
c)
Answer d
d) The data is not linear.
Answer e
e) The data appear to show two lines that cross. That's a classic symptom of two competing mechanisms. The faster mechanism, to the left, is probably an intramolecular electron transfer. The slower mechanism, to the right, may be an intermolecular electron transfer. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.08%3A_Redox_Mechanisms/12.8.02%3A_Inner_Sphere_Electron_Transfer.txt |
So far, our coverage of the reactions of metal complexes has focused only on the reactions that occur at the metal center. However, there are also reactions that can occur to the ligands bound at the metal center. When a ligand binds to an electropositive metal center, bonds within the ligand can be polarized in new ways, leading to new or enhanced reactivity. Some of these reactions are covered in the chapters on organometallic chemistry, and there are many examples of how biology employs metal ions at enzyme active sites to catalyze biological transformations. This section will briefly describe some of the most common types of reactions that can be catalyzed by metal ion binding.
12.09: Reactions of Coordinated Ligands
Transition metal ions (but not alkali metal ions) act as Lewis acids in metal-ligand interactions. As a consequence of the metal ion interacting with the ligand, the electronic structure of the ligand is altered. In some cases, the metal-ligand interaction causes changes in the strength and reactivity of other bonds within the ligand itself. For example, when a ligand has an acidic proton, interactions with a metal ion will make that acidic proton more acidic. In other words, interaction with the metal ion can cause a decrease in the ligand's $pK_a$. The metal-ligand complex itself can act as a stronger Bronsted acid (giving off a proton) compared to the "free" ligand. We can see this type of alteration in the $pK_a$ of water when it is "free" compared to when it is bound to a metal ion.
Example: Water meets a transition metal ion1
In biology, the solvent is water. Water is also a ligand that binds to metal ions, according to the equation shown below.
$\ce{nH2O + M^{x+} <=>> [M(H2O)_{n}]^{x+}} \nonumber$
There is a LOT of water around in biological systems. Consequently, the reaction shown above is shifted far to the right (toward the aquo metal complex, $\ce{[M(H2O)_{n}]^{x+}}$) in aqueous solution. Consequently, when metals are dissolved in aqueous solution, the "free" ion does not exist, and rather the metal is in complex with water. This is generally true unless there are other ligands present that have stronger affinity for the metal ion.
Consider the two equilibrium reactions shown below. The autoionization of water is shown on the left. This reaction has a very small equilibrium constant ( $K_W = 10^{-14}$) and thus the $pK_a$ for dissociation of a proton from water is $pK_a = 14$. On the left is the dissociation of a proton from a water molecule bound to $\ce{Zn^2+}$ ion (other water ligands are not drawn for simplicity). The bond between $\ce{H2O}$ and $\ce{Zn}$ is considered to be covalent.
$\ce{H2O <=> H^+ + OH^-} \quad \quad \text{vs.} \quad \quad \ce{[Zn(H2O)]^2+ <=> H^+ + [Zn(OH)]^+} \nonumber$
Exercise $1$
Predict which reaction above is more favorable. In other words, in which case is water a better acid; when water is "free", or when it is bound to a transition metal ion like Zn2+? Defend your prediction.1
Hint
Consider these two things:
(1) In which case is the hydroxide ion more stable: when it is a free ion or when it is bound to Zn? How will this effect one case relative to the other?
(2) Consider separation of charge in each equilibrium. Is the localization of positive and negative charge more energetically favorable in the reactands or in the products in each case? How will this effect one case relative to the other?
Answer
Water is a better acid when it has a covalent bond to a metal ion. The metal-ligand interaction weakens the H-O bond and stabilizes the conjugate base that forms after deprotonation.
Exercise $2$
Water has a $pK_a$ of 14. When water is bound to a Zn2+ ion in aqueous solution, its $pK_a$ is lowered to 10. Calculate the pH of pure water and the pH of a solution of 0.1 M ZnSO4.
Hint
You will need the chemical equations above and you will need to recall the definitions of $K_a$ and $pK_a$, as well as $pH$ to complete this task. Recall that $pK_a = - \log K_a$ and $pH = - \log [H^+]$. The $K_a$ is the equilibrium constant for the chemical equation. Use an I.C.E. table to calculate the amount of hydrogen ion at equilibrium.
Answer
The pH of pure water is 7. The pH of 0.1 M $\ce{ZnSO4}$ depends only on the concentration of the Zn aquo complex that forms after dissociation of $\ce{ZnSO4}$ to $\ce{Zn^2+}$ and $\ce{SO4^2-}$. We can use the I.C.E method to determine that the concentration of hydrogen ion at equilibrium is approximately $3.2 \times 10^{-6}$ after 0.1M $\ce{Zn^2+}$ is initially added to water. This gives approximately $pH = 5.5$.
Exercise $3$
Transition metals dramatically decrease solution pH compared to pure water. On the other hand, alkali metals do not significantly affect pH.
a) Based on this information, what can you conclude about the relative covalent nature of water’s interactions with transition metals compared to alkali metals?
b) Rationalize this based on the element's position on the periodic table and their consequent electron configurations.
Answer
a) transition metals form more covalent bonds with water, while alkali metal's interactions are more electrostatic in character.
b) transition metals have partially-filled d subshells, while alkali metals have the electron configurations of the nobel gases. Perhaps the nobel gas electron configuration makes covalent bonding unfavorable, while partially filled d-orbitals enables productive bonding interactions with ligands.
Exercise $4$
When a transition metal binds to a Cys side chain, it usually binds to the thiolate anion form of this side chain. Explain why the Cys loses its proton when a metal ion binds. Use drawings to describe what happens to the $pK_a$ of the thiol hydrogen.
Hint
A cystein side chain has a thiol (R-SH) group. Recall that thiols behave similarly to alcohols, but thiols are more acidic than alcohols.
Answer
Bonding of a cystein S to a metal ion would weaken the S-H bond and increase the acidity of that proton (it's pKa would decrease). In bilogical conditions, near neutral pH, this bonding could decrease the pKa of the thiol group so much that the H is lost to give the conjugate base thiolate bound to the metal.
Biological hydrolysis2
Above you learned how metals can act as Lewis Acids to increase the acidity of a bound water molecule (or other ligand). This modulation of water's $pK_a$ can be really useful to catalyze reactions in which hydroxide ion is a reactant. The enzyme carbonic anhydrase (CA) is a prime example of how biology uses this property to catalyze an important biological acid/base reaction. CA is the enzyme that converts $\ce{CO2}$ into bicarbonate ($\ce{HCO3^-}$) very quickly. This is important in the blood to get to $\ce{CO2}$ to the lungs to be exhaled. Conversion of $\ce{CO2}$ to bicarbonate is also important in production of the aqueous fluid of eye and other secretions; and CA is an important drug target for treatment of glaucoma.
The figure shows the structure of human carbonic anhydrase IV (from PDB 1znc), with a blow up of the Zn active site where Zn is bound to 3 histidine side chains (His). The mechanism for this enzyme is shown in the chemical equations circling the structure in Figure $4$. As we discused above, binding water to a metal ion can change its $pK_a$, and make its protons more acidic by stabilizing the hydroxide ion. When hydroxide is needed to carry out hydrolysis reactions, acidic metal ions, like Zn2+ are often used to stabilize the hydroxide nucleophile in order to catalyze the reaction (lower the activation energy). The $pK_a$ of “free” water is 14. For water in [Zn(H2O)6]2+ in aqueous solution, the $pK_a$ is 10. But, in the active site of carbonic anhydrase, the $pK_a$ of water is lowered even more, to 7, by the protein environment around the Zn active site. The active site of this protein has the "power" to lower the pKa of water even more than in the case of aqueous Zn as a virtue of the hydrophobic pocket of the active site that makes charge separation more unstable compared to bulk water where charges can be better solubilize by bulk water.
Look carefully at the catalytic cycle shown in Figure $4$ and identify the step in which water looses a proton (the step labeled "water deprotonation". This step would be nearly impossible under biological conditions at pH 7.4. In other words, it would happen too infrequently to be useful in biological systems. But, when the pKa is lowered to 7, which is below the pH, that step becomes so favorable, that it would happen spontaneously in solution.
Carbonic anhydrase is just one example of how metal ions can be useful for catalyzing hydrolysis reactions. There are several more examples of catalytic enzymes and small molecule metal complexes that use metals (Cu, Co, Ni, Mn, Ca, and Mg) to catalyze hydrolysis of esters (eg hydrolysis of fats), amides (eg hydrolysis of peptide bonds with water), phosphate esters (eg hydrolysis of DNA and RNA) using mechanisms simlar to that shown above for hydrolysis of carbon dioxide.
Sources
1. Inspired by or taken directly from Metals in Acid Base Chemistry, an in-class activity created by Sheila Smith, University of Michigan- Dearborn ([email protected]) and posted on VIPEr (www.ionicviper.org) on October 17, 2009. Copyright Sheila Smith 2009. This work is licensed under the Creative Commons Attribution Non-commercial Share Alike License. To view a copy of this license visit http://creativecommons.org/about/license/.
2. From Metals in Biological Systems - Who? How? and Why?, a 5 slides about learning object created by Adam R. Johnson, Harvey Mudd College ([email protected]), Hilary Eppley, DePauw Univeristy (), and Sheila Smith, University of Michigan- Dearborn ([email protected]) and posted on VIPEr (www.ionicviper.org) on January 20, 2010. Copyright Sheila Smith 2009. This work is licensed under the Creative Commons Attribution Non-commercial Share Alike License. To view a copy of this license visit http://creativecommons.org/about/license/.
12.9.02: Template Reactions
A template reaction is one in which a metal ion acts as a template to bring reactive sites in close proximity. In the absence of the metal ion, the same reaction would proceed to form a different product. An example of the template effect is the dialykation of nickel dithiolate (Figure \(1\)). The same dialkylation in the absence of the Ni template creates a polymer.
Another example of a template reaction is in the synthesis of crown ethers from dialkylations that are templated by metal ions. Again, the same reaction carried out without the template metal would form a polymer.
12.9.03: Electrophilic Substitutions
We have seen how metal ion binding can alter the pKa of a bound ligand by polarizing bonds. In a similar fashion, binding of a ligand to a metal ion can alter the location of electron density so that positions on the ligand take on new reactivity. This can be seen in the case of acetylacetone (acac) complexes (most recently discussed in section 12.5.3). The metal coordination to the acac ligand induces enol tautomerization, placing electron density at the central alpha carbon and causing it to become a potent nucleophile. Reactions occur at that nucleophilic carbon in similar manner to the electrophilic aromatic substitution reactions that you may have learned about in an Organic Chemistry course. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms/12.09%3A_Reactions_of_Coordinated_Ligands/12.9.01%3A_Metal-catalyzed_Hydrolysis.txt |
What is Organometallic Chemistry?
This chapter will introduce a subfield of inorganic coordination chemistry; organometallic chemistry. Let’s begin with a few simple questions: what is organometallic chemistry? What, after studying organometallic chemistry, will we know about the world that we didn’t know before? Why is the subject worth studying? And what kinds of problems is the subject meant to address? The purpose of this introduction is twofold: (1) to help motivate us as we move forward (that is, to remind us that there is a point to all this!); and (2) to illustrate the kinds of problems we’ll be able to address using concepts from the field. You might be surprised by the spine-chilling power you feel after learning about the behavior of organometallic compounds and reactions!
Put most bluntly, organometallic chemistry is the study of compounds containing, and reactions involving, metal-carbon bonds. The metal-carbon bond may be transient or temporary, but if one exists during a reaction or in a compound of interest, we’re squarely in the domain of organometallic chemistry. Despite the denotational importance of the M-C bond, bonds between metals and the other common elements of organic chemistry also appear in organometallic chemistry: metal-nitrogen, metal-oxygen, metal-halogen, and even metal-hydrogen bonds all play a role. Metals cover a vast swath of the periodic table and include the alkali metals (group 1), alkali earth metals (group 2), transition metals (groups 3-12), the main group metals (groups 13-15, “under the stairs”), and the lanthanides and actinides. We will focus most prominently on the behavior of the transition metals, so called because they cover the transition between the electropositive group 2 elements and the more electron-rich main group elements.
Why is organometallic chemistry worth studying? Well, for me, it mostly comes down to synthetic flexibility. There’s a reason the “organo” comes first in “organometallic chemistry”—our goal is usually the creation of new bonds in organic compounds. The metals tend to just be along for the ride (although their influence, obviously, is essential). And the fact is that you can do things with organometallic chemistry that you cannot do using straight-up organic chemistry. Case in point:
The venerable Suzuki reaction...unthinkable without palladium!
The establishment of the bond between the phenyl rings through a means other than dumb luck seems unthinkable to the organic chemist, but it’s natural for the palladium-equipped metal-organicker. Bromobenzene looks like a potential electrophile at the bromine-bearing carbon, and if you’re familiar with hydroboration you might see phenylboronic acid as a potential nucleophile at the boron-bearing carbon. Catalytic palladium makes it all happen! Organometallic chemistry is full of these mind-bending transformations, and can expand the synthetic toolbox of the organic chemist considerably.
To throw another motive into the mix for the non-specialist (or the synthesis-spurning chemist), organometallic chemistry is full of intriguing stories of scientific inquiry and discovery. Exploring how researchers take a new organometallic reaction from “ooh pretty” to strong predictive power is instructive for anyone interested in “how science works,” in a practical sense. We’ll examine a number of classical experiments in organometallic chemistry, both for their value to the field and their contributions to the general nature of scientific inquiry.
What kinds of problems should we be able to address as we move forward? Here’s a bulleted list of the most commonly encountered types of problems in an organometallic chemistry course:
• Describe the structure of an organometallic complex…
• Predict the product of the given reaction conditions…
• Draw a reasonable mechanism based on evidence…
• Devise a synthetic route to synthesize a target organometallic compound…
• Explain the observation(s)…
• Predict the results of a series of experiments…
The first four are pretty standard organic-esque problems, but it’s the last two, more general classes that really make organometallic chemistry compelling. Just imagine putting yourself in the shoes of the pioneers and making the same predictions they did!
There you have it, a short introduction to organometallic chemistry and why it’s worth studying. The rest of this chapter will describe what organometallic chemistry really is…it will be helpful to keep these motives in mind as you study. Keep a thirst for predictive power, and it’s hard to go wrong with organometallic chemistry!
Resources for learning organometallic chemistry
For the penny-pinching student or layman, there are several good resources for organometallic chemistry on the Web. Nothing as exhaustive as Reusch’s Virtual Textbook of Organic Chemistry exists for organometallic chemistry, but the base of resources available on the Web is growing. Rob Toreki’s Organometallic HyperTextBook could use a CSS refresh, but contains some nice introductions to different organometallic concepts and reactions. Try the electron-counting quiz!
VIPER is a collection of electronic resources for teaching and learning inorganic chemistry, and includes a nice section on organometallic chemistry featuring laboratory assignments, lecture notes, and classroom activities. Awesome public lecture notes are available from Budzelaar at the University of Manitoba and Shaughnessy at Alabama (Roll Tide?). For practice problems, check out Fu’s OpenCourseWare material from MIT and Shaughnessy’s problem sets.
Historical Background and Introduction to Metallocenes
Definition: Organometallic Complex
Organometallic Complex: A complex with bonding interactions between a metal atom and one or more carbon atoms of an organic group or molecule.
An organometallic complex is defined as a complex with bonding interactions between one or more carbon atoms of an organic group or molecule and at least one metal atom. It is important to understand that just the presence of an organic ligand is not sufficient to define an organometallic compound. If there are organic ligands present, but no metal-carbon bonds, the complex is said to be "metal-organic". There must be interactions between a carbon and a metal for the complex to be considered organometallic.
Example $1$
Which of the molecules below is an organometallic complex?
$\ce{[Sn(Me)4]} \quad \quad \ce{Sn(OMe)4} \ce{cis-[PtCl2(NH3)2]}$
Solution
The tetramethyl tin ($\ce{Sn(Me)4}$) has metal-carbon bonds, and thus it is an organometallic complex.
Tetramethoxy tin ($\ce{Sn(OMe)4}$) has metal-oxygen bonds. It lacks metal-carbon bonds, but it does contain organic ligands; thus it is a "metal-organic" complex rather than organometallic.
Cisplatinum ($\ce{cis-[PtCl2(NH3)2]}$) contains no carbon and thus it is neither organometallic or metal-organic; it is just a coordination compound.
Industrial Importance of Organometallic Compounds
Organometallic compounds are a very important class of compounds in industry. For example, tens of thousand of tons of aluminum and tin alkyl compounds are produced each year for industrial purposes. The use of organometallic compounds as catalysts to produce other compounds is even more important. Organometallic catalysts are used for a range of industrial syntheses from manufacture of commodity polymers like polypropylene and polyethylene to production of simple organic molecules like acetaldehyde and acetic acid. These compounds are produced at a scale in the order of millions of tons per year.
History of Organometallic Chemistry
Let us take a historical approach to organometallic chemistry and see how the field has evolved. Arguably, the first organometallic compound was kakodyl oxide, an organo-arsenic compound (Figure $2$). It was accidentally produced by the French chemist Louis Cadet in 1760 when he was working on inks. He heated arsenic oxide and potassium acetate and obtained a red-brown oily liquid, known as Cadet’s fuming liquid. It consists mostly of cacodyl and cacodyl oxide (Figure $2$). In cacodyl, there is an As-As bond and two methyl groups attached to each As atom. In the other product, cacodyl oxide, there is an O atom bridging the two As atoms, and each As is bound to two methyl groups. The names of these two compounds come from the greek word “kakodes”, meaning "bad smell". Indeed, they have a very intense, garlic-like smell. The reaction can be used to identify arsenic in samples. For instance, if you suspected your food was poisoned with arsenic, you could heat a sample together with potassium acetate. If a bad, garlic-like smell evolved, this would indicate that there is arsenic in your sample.
Another important milestone in coordination chemistry was the discovery of Zeise’s salt by the Danish chemist William Zeise in 1827 (Figure $3$). Zeise’s salt was the first olefin complex in which an olefin was bound side-on to a metal using its π-electrons for σ-bonding. Zeise observed that when sodium hexachloroplatinate (2-) was heated in ethanol, a compound with the composition Na[PtCl3(C2H4)] could be isolated. The chemical composition of the compound suggested that there was a C2H4 organic fragment bonded to the platinum, but it was not clear how.
This question was only answered more than 100 years later in 1969 with the crystal structure analysis of Zeise’s salt. The crystal structure revealed that an ethylene molecule was bound side-on to the platinum (Figure $3$). The two carbon atoms had about the same distance to the Pt arguing that they were equally strongly involved in the bonding with Pt. The first C had a distance of 216 pm while the second carbon had a distance of 215 pm. The ethylene molecule was oriented perpendicular to the plane made of the three chloro ligands and the platinum atom. The Cl-Pt-Cl bond angles were near 90°. Overall, the structure could be described as a structure derived from a square planar structure with the ethylene as the fourth ligand that would stand perpendicular to the plane in order to minimize steric repulsion with the chloro ligands.
An important question was how to describe the bonding in the compound. It was noteworthy that the C-C bond length in Zeise’s salt was significantly longer (144 pm) than that in a free ethylene molecule (134 pm). Based on the bond-strength bond-length concept this argued that the bond was weaker than a regular C=C double bond, and that the bond order was smaller than 2. What could explain this lower bond order and the side-on coordination of the ethylene in Zeise’s salt?
The answer is that the ethylene molecule uses its π-electrons for σ-bonding with a metal d-orbital (Figure $4$). You can see above that the lobe of a dx2-y2 orbital has the correct orientation to overlap with the bonding π-orbitals of the ethylene ligand in a σ-fashion. Through this interaction electron density gets donated from the bonding ligand π-orbital into the metal-d-orbital. This results in a lower electron density for π-bonding within the ligand, and thus the bond order is decreased. In addition, there is another effect that lowers the bond order. A metal d-orbital can interact with the π*-orbitals in π-fashion. In this process, electron density can be donated from the metal d-orbital into the empty π-orbitals of the ligand. The increased electron density in the π-orbitals further decreases the bond order.
In 1890, a further milestone in organometallic chemistry was reached with the synthesis of the first carbonyl complex, the nickel tetracarbonyl. Nickel tetracarbonyl is a colorless liquid that boils at 43°C. The compound was prepared by the German chemist Ludwig Mond. Nickel tetracarbonyl forms spontaneously from nickel metal and carbon monoxide at room temperature. The reaction is the basis of the Mond-process that is used up to date in order to purify nickel (Figure $5$). Carbon monoxide gas can be streamed over impure Ni metal at temperatures of 50-60°C to form Ni(CO)4 in gaseous form. Impurities are left behind in solid form. Then, the nickel tetracarbonyl is thermally decomposed at ca. 220° to form nickel and carbon monoxide. The process can be varied to produce nickel in powder form, as spheres (Figure $5$), and as coatings. The Mond process is still used despite the very high toxicity of nickel tetracarbonyl.
\begin{align*} \ce{Ni_{(s, ~ impure)} + 4 CO_{(g)}} &\ce{ -> Ni(CO)4_{(g)}} & & 50-60^{\circ} \text{C (also spontaneous at room temp)} \[5pt] \ce{Ni(CO)4_{(g)}} &\ce{ -> Ni_{(s, ~ purified)} + 4 CO} & &220-250 ^{\circ} \text{C (Ni is purified)} \end{align*}
In 1900 the first Grignard reagents were discovered. Victor Grignard (Figure $9$) was an enthusiastic young French chemist who discovered how to make organomagnesium halides (RMgX) while working for his Ph.D.. His supervisor, Sabatier, had been trying this chemistry for some time, but Victor was the genius who solved the problem. Organomagnesium halides form from magnesium and organic halides in ethers. This discovery in 1900 changed the course of organic chemistry and won Grignard and Sabatier the Nobel Prize in 1912.
$\text{The Grignard Reaction:} \quad \ce{Mg_{(s)}} + R-X ->[ether] R-Mg-X}$
In 1917, the first lithium alkyls were prepared by Wilhelm Schlenk. Like Grignard reagents, lithium alkyls are very valuable reactants in synthetic organic chemistry. Lithium alkyls are very air-sensitive compounds, and some even self-ignite in air (for example tert-butyl lithium is explosive when exposed to air). Therefore, they need to be handled carefully under inert gas. So that the reactions could be carried out under air0free conditions, Wilhelm Schlenk made an important invention: The Schlenk-lines (Figure $6$). The Schlenk lines allows a chemist to remove all gas (air) from a reaction flasks and back-fill the flask with an inert gas. A Schlenk line has a dual manifold with several ports. One manifold is connected to a source of purified inert gas, while the other is connected to a vacuum pump. The inert-gas line is vented through an oil bubbler, while solvent vapors and gaseous reaction products are prevented from contaminating the vacuum pump by a liquid nitrogen or dry ice/acetone cold trap. Special stopcocks or Teflon taps allow vacuum or inert gas to be selected without the need for placing the sample on a separate line.
Discovery of Metallocenes
The discoveries discussed above were important milestones in the development of organometallic chemistry. But the field expanded more rapidly with the discovery of the first metallocene: ferrocene. Like often in science, ferrocene was discovered through an accident. In 1951 Peter Pauson and Thomas Kealy attempted to synthesize the organic compound fulvalene through oxidative coupling of cyclopentadienyl magnesium chloride with iron (III) chloride. However, instead of obtaining fulvalene they observed the formation hydrofulvalene together with an orange powder with “remarkable stability”. Analysis of the powder showed that the chemical compound contained two cyclopentadienyl rings and one iron atom. According to the knowledge on bonding in organometallic compounds that was known at the time Kealy and Pauson suggested a structure for the molecule in which two cyclopentadienyl group were bound to a single Fe atom via two Fe-C bonds (Figure $7$).
However, the remarkable stability of the compound was in contradiction to the proposed structure. The structure would be a 10 electron complex which would be highly coordinatively unsaturated. This would be inconsistent with the observed stability. We can briefly practice our electron counting skills again in order to prove that there are only 10 electrons (Figure $8$). (Electron counting will be discussed in more depth later in this chapter)
The compound was stable in air, and could be sublimed without decomposition. Further, the double-bonds in the cyclopentadienyl rings resisted hydrogenation. In addition, the structure was inconsistent with spectroscopic observations. Only one C-H stretch vibration was observed in the IR and only one signal was observed in the 1H NMR. Kealy and Pauson’s structure would have been consistent with three C-H stretch vibrations and three NMR resonances. For these reasons, Kealy’s and Pausons structure was soon questioned. Ernst Otto Fischer at University of Munich and Sir Geoffrey Wilkinson at Harvard University (Figure $9$) suggested an alternative structure that was very different from all known organometallic structures. The bonding in this structure was completely different from all bonding concepts considered thus far for organometallics. What structure did they propose? They proposed a sandwich structure in which the two aromatic cyclopentadienyl anions would sandwich an Fe2+ ion. What would be the bonding in this structure? The cylopentadienyl anion would donate its six π-electrons into the valence orbitals of the metal. Thus, all five carbon atoms would be equally involved in the bonding, the ligand would act as a ƞ5-ligand. Because of the donation of π-electrons in the cycopentadiene unit, the structure was called “ferrocene”. The ferrocene structure would explain the stability and low reactivity of the compound.
Electron-counting gives an 18 electron complex (Figure $10$). Let us verify this using the oxidation state method. Fe would contribute eight electrons in the neutral state. The ligands would be considered as cyclopendienyl anions with a 1- charge. This would give an oxidation state of +2 for Fe reducing the number of electrons from eight to six. The two ligands would contribute six electrons each giving twelve electrons. 12+6=18. Hence, the structure fulfills the 18 electron rule. The structure is also consistent with the spectroscopic observation of one NMR resonance and one C-H stretch vibration. All these were excellent arguments to support the ferrocene sandwich structure, however they were not an absolute proof. The proof finally came with the X-ray crystal structure determination which unambiguously confirmed the sandwich structure.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.01%3A_Introduction_to_Organometallic_Chemistry.txt |
The nomenclature of coordination compounds was described in an earlier section (Section 9.2). Organometallic compounds are named using this same system, so it may be helpful to review before proceeding. Here, we will point out some of the symbols and terminology that are used heavily in naming organometallic complexes.
Hapticity
The eta ($\eta$) symbol is used to indicate the variable hapticity of ligands with conjugated $\pi$ systems. For example, the cyclopentadiene anion (Cp, $\ce{C_5H_5}$) ligand has the capability to coordinate a metal ion in several ways due to its cyclic conjugated $\pi$ system. The hapticity (the number of atoms involved in the metal bonding interaction) must be indicated in the formula with the $\eta ^n$ notation, and in the name with the appropriate prefix. Figure $1$ illustrates three different hapticities, and the relevant formula symbols and names.
Another common ligand is the $\pi$-allyl anion ($\ce{C_3H_3}$), which can coordinate in either an $\eta^1$ or $\eta^3$ fashion, as illustrated below (Figure $2$).
Bridging ligands
Ligands that bridge two or more metal ions are indicated using the symbol, mu ($\mu$). For a ligand that bridges $n$ metal ions, the symbol is written as $\mu _n$. However the subscript is often ommitted when $n=2$. Some examples of the carbonyl ligand are given below. More specific rules for naming are found in Section 9.2.
Common Ligands and Classifications
Some common organic ligands found in organometallic compounds are listed below with their names and structures.
Name Structure CBC description
Alkyl X
Carbonyl L
Carbene (alkylidene) X2
Carbyne (alkylidyne) X3
Ethylene L
Acteylene L
Acyl X
$\pi$-allyl ($\ce{C_3H_5^-}$) X for $\eta^1$
LX for $\eta^3$
Cyclopropenyl (cyclo-$\ce{C_3H_3}$) X for $\eta^1$
LX for $\eta^3$
Cyclobutadiene (cyclo-$\ce{C_4H_4}$) L2 for $\eta^4$
Cyclopentadienyl (cyclo-$\ce{C_5H_5}$, Cp) X for $\eta^1$
LX for $\eta^3$
L2X for $\eta^5$
Benzene L3 for $\eta^6$
1,5-cyclooctadiene (1,5-COD) L2 for $\eta^4$
Covalent Bond Classification Method (CBC)
Later in this chapter, you will learn to count electrons. The way in which you count depends on what counting scheme you choose to use. But in any case, the way in which we "count" the electrons from ligands depends on the ligand type. The covalent bond classification method (CBC) has been in use since the late 1990's to classify ligands in organometallic complexes. This method classifies ligands into three main types as follows. (see Application of the Covalent Bond Classification Method for the Teaching of Inorganic Chemistry. Malcolm L. H. Green and Gerard Parkin, Journal of Chemical Education 2014 91 (6), 807-816, DOI: 10.1021/ed400504f.)
CBC Ligand Class/Type Examples Use in neutral Ligand Method of Electron Counting Use in Donor Pair Method of Electron Counting
X-type $\ce{H^-, Cl^-, H_3C^-}$ Considered neutral
Donates 1 electron to the metal
Considered anionic
Donates 2 electrons to the metal
L-type $\ce{CO, NH_3, H_2O}$ Considered neutral
Donates 2 electrons to the metal
Considered neutral
Donates 2 electrons to the metal
Z-type (rarely used)
$\ce{BR_3}$, Lewis acids
Considered neutral
Accepts 2 electrons
Considered cationic
Donates 0 electrons
13.3.01: The 18 Electron Rule
In this section we will learn how to count valence electrons in coordination compounds. Electron counting is important because the number of electrons in a complex can be used to predict the compound's stability and reactivity. There are two different methods that are commonly used, and each is described below. Each of the two methods will lead to the same result if done correctly. Which method you prefer is “personal taste”, but each method is about equally common in the literature, so it is good to be familiar with both of them.
13.03: Electron Counting in Organometallic Complexes
Method A for counting electrons: The Donor Pair Method
The donor pair method consider ligands to donate pairs of electrons to the metal. Each ligand-metal bond is considered a pair of electrons donated to the metal by the ligand. The method consists of the steps described in the box below.
How to count with the DONOR PAIR METHOD
1. Determine the oxidation state of the metal center. This step can be done by first deconstructing the complex so that each ligand "keeps" its bonded electrons (See Figure \(2\)). by considering the overall charge of the complex and the charge of each ligand, the oxidation state of the metal center can be determined. An alternative to deconstruction is to consider each ligand's class according to the Covalent Bond Classification (CBC) method discussed earlier.
2. Determine the number of metal ion valence electrons. The number of valence electrons can be determined from the metal's group on the periodic table and its oxidation state. If the oxidation state is positive we subtract electrons from the group number, if it is negative, which is rare, then we add electrons.
3. Determine the number of bonded electrons derived from ligands. Each ligand single bond provides two electrons, double bonds provide four electrons, etc.
4. Add the number of electrons from the metal ion valence (step 2) to the number of bonded electrons from the ligands to get the total valence electron count for the complex.
Below is an example of using the donor pair method to count electrons in the chemotherapy drug \(\ce{cis-[PtCl2(NH3)2]}\), also called cis-platinum (Figure \(1\)). This compound is neutral in charge.
Step 1: Deconstruction of the complex using the donor pair method would convert each metal-ligand bond to a pair of electrons on the ligand. Without consulting the Covalent Bond Classification (CBC) method, we can determine the charge on each ligand using the simple Lewis dot diagrams that result. The ligands consist of two neutral ammines and two negatively charged chlorides. The ligands account for a total negative charge of -2. Since the metal complex is neutral, there must be a +2 change on the metal. Thus, the metal oxidation state is +2.
Step 2: With the knowledge that the Pt has an oxidation state of +2, and is in group 10 of the periodic table, we can determine the number of valence electrons on Pt(II). Subtract the oxidation state from the group number to get 8 valence electrons on the Pt(II) ion.
Step 3: Each of the ligands had one bond to Pt, thus each is a 2-electron donor. The ligands combined contribute a total of 8 electrons.
Step 4: Adding the 8 electrons from Pt(II) and the 8 electrons from the ligands gives a total electron count of 16.
Method B for counting electrons: The Neutral Atom Method (aka Neutral Ligand Method)
The “neutral atom method”, is sometimes called "the neutral ligand method". In this method, each ligand is considered to donate the number of electrons that it would if the ligand were neutral. The neutral atom method is carried out according to the steps described below.
How to count using the NEUTRAL LIGAND/ATOM METHOD
1. Determine the number of valence electrons on the metal center, as if it were a neutral atom. Add those electrons to the total count.
Hint: The number valence electrons is the same as the group number of the transition metal in the periodic table (groups 3-12 for the transition metals).
2. Account for the ionic charge on the entire coordination complex. If the charge is neutral, you can ignore this step. If the complex has a positive charge, you subtract that many electrons from the total count. If the complex is an anion, you add the anionic charge to the total electron count.
3. Determine how many electrons are contributed by each ligand. (This is the most complicated step)
• Deconstruct the metal ion (break its M-L bonds) so that each resulting ligand fragment and the central metal are neutral.
• Then count the number of electrons from each ligand that contributed to the bond. Add the sum of those electrons from all the ligands to the total count.
(Remeber you've already accounted for the electrons from the neutral atom in step #1)
4. Add the metal ion valence electrons (step 1), the total ligand electrons (step 3), and accounting for the charge of the complex (step 2). The total sum is the total valence electron count for the metal complex.
Below is an example of applying the neutral atom method to count electrons in the cis-platinum complex (\(\ce{[Pt(Cl)_2(NH_3)_2]}\) (Figure \(2\))).
Step 1: The metal is platinum which is located in group 10 of the periodic table. Therefore, a neutral platinum atom has ten valence electrons.
Step 2: The complex is neutral, this we can ignore this step.
Step 3: Each Cl is considered to donate 1 electron and each ammine ligand is considered to donate two electrons. The Cl's donate a total of two electrons, and the ammines donate a total of four electrons to give a total of 6 from all ligands.
Step 4: Ten electrons from Pt and six electrons from the ligands combine to give a total electron count of 16.
There are pros and cons to each of the methods presented above. The neutral atom method has the advantage that we do not have to calculate charges of ligands or the metal oxidation states to reach the final answer. However, the disadvantage is that we must think about how to cleave bonds to create neutral fragments. We may need to cleave bonds in an way that is contrary to the real donor-acceptor nature of a coordination compound. The donor-pair method does account for the donor-acceptor nature of a coordination compound because the bonds are considered dative bonds, and the electrons are assigned to the ligands and metals accordingly. And advantage to the donor-pari method is that we do not need to think how to cleave bonds in artificial ways because we cleave the bonds always heteroleptically. However, the disadvantage is that it requires us to calcuate charges at the ligands to determine metal oxidation states, which is an additional, non-trivial step.
The 18 Electron Rule
Electron counting is a critical step in the context of an important rule in coordination chemistry: The 18 electron rule. The 18 electron rule states that transition metal complexes are normally most stable when they have a total count of 18 electrons in the valence shell. In other words, for \(d\)-block metal complexes, a valence configuration that fills all valence orbitals (\(ns^2 \; (n-1)d^{10} \; np^6\)) is most stable; it is equivalent to a nobel-gas configuration for the metal atom, and is analogous to the "octet rule" for the main-group elements. Complexes that have 18 electrons in their count are considered to be coordinatively saturated. In general, coordinatively saturated compounds are considered to be relatively stable, and when they react they usually undergo dissociation or oxidation reactions that decrease the electron count. If the there are less than 18 electrons, the species is coordinatively unsaturated, and considered less stable than a species with 18 electrons. In general, coordinatively unsaturated complexes tend to react in ways that can increase the electron count. They tend to undergo reactions involving associations to add more electrons to the system and/or tend to gain electrons through redox reactions (the metal becomes reduced). If a species has more than 18 electrons it is coordinatively oversaturated and tends to lose ligands. It is usually easily oxidized. Both loss of ligands and oxidation reduces to the number of electrons to or at least closer to 18.
Definitions:
Coordinatively saturated: when the complex has an electron count =18 electrons. These complexes are considered most stable.
Coordinatively unsaturated: when the complex has an electron count <18 electrons. These complexes tend to add more ligands, and to become reduced.
Coordinatively oversaturated: when the complex has an electron count >18 electrons. These complexes tend to lose ligands and become oxidized.
Exceptions to the 18-electron rule:
The 18 electron rule has many exceptions, and therefore needs to be applied with caution. In particular, group 3, 4, and 10 complexes deviate often from the 18 electron rule.
For illustration purposes, let us count the number of electrons of the tetrahedral tetrabenzyltitanium(0) complex by the oxidation state method. We could also use the neutral atom method, which would give the same results. This complex is a group 4 complex because titanium is in group 4. How many electrons will the titanium contribute? Because the number of electrons is always the same as the group number, it will contribute four electrons. Next, what is oxidation state of Ti? To determine it we must determine the charge at the ligands. To do that we cleave the bonds heteroleptically. This will give benzylate anions with -1 charge. There are four of these ions, and therefore there will be four negative charges overall. The complex is charge-neutral, and thus the oxidation state is +4 because -4+4=0. Therefore, we need to subtract four electrons. Because we cleaved the bond heteroleptically, each ligand contributes two electrons, giving overall eight electrons coming from the four ligands. This means that we have overall eight electrons, or an 8-electron complex. This is far, far away from 18 electrons. Nonetheless, the complex is stable. How can we explain this? The answer is that in order to achieve 18 electrons it would need to add five additional ligands if each ligand is considered a 2-electron donor. This would increase the coordination number to 9 which is too high to produce a stable complex. In order to reduce the complex to an 18 electron complex, 10 electrons would need to be added. This would produce a complex with a -10 charge which is way to high to be stable. The arguments are generalizable for group 3 and group 4 complexes. Because these elements only have a few d electrons, the ligands would need to contribute a lot of electrons to produce an 18 electron complex. This would require just too many ligands to add. The coordination numbers would get too high. If electrons are added instead of ligands, the negative charge at the complex would be to high to be stable based on electron-electron repulsion arguments.
These arguments cannot be applied for group 10 elements, because these elements have many d electrons. The explanation in this case is that these elements like to make square planar complexes when in the oxidation number is +2. Square planar complexes prefer 16 instead of 18 electrons. We will learn later, when we discuss bonding in coordination compounds, why this is. You can see that the square planar diamminedichloro palladium complex shown is square planar and has sixteen electrons. There are 10 electrons coming from Pd. If we use the neutral atom method, no electrons need to be added or subtracted due to the charge at the complex. The complex is charge-neutral. To assess how many electrons come from the ligands we need to cleave the bonds so that neutral ligands are produced. The Pd-Cl bonds need to be cleave homoleptically, the Pd-N bonds need to be cleave heteroleptically. Therefore, the two chloro ligands are 1e donors, and the two ammine ligands are 2e donors. This gives 10+4+2=16 electrons.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.02%3A_Nomenclature_Ligands_and_Classification.txt |
Organometallic complexes, which consist of centrally located metals and peripheral organic compounds called ligands, are the workhorses of organometallic chemistry. Just like organic intermediates, understanding something about the structure of these molecules tells us a great deal about their expected reactivity. Some we would expect to be stable, and others definitely not! A big part of our early explorations will involve describing, systematically, the principles that govern the stability of organometallic complexes. From the outset, I will say that these principles are not set in stone and are best applied to well controlled comparisons. Nonetheless the principles are definitely worth talking about, because they form the foundation of everything else we’ll discuss. Let’s begin by exploring the general characteristics of organometallic complexes and identifying three key classes of organic ligands.
When we think of metals we usually think of electropositive atoms or even positively charged ions, and many of the metals of OM chemistry fit this mold. In general, it is useful to imagine organic ligands as electron donors and metals as electron acceptors. When looking at a pair of electrons shared between a transition metal and main-group atom (or hydrogen), I imagine the cationic metal center and anionic main-group atom racing toward one another from oblivion like star-crossed lovers. In the opposite direction (with an important caveat that we’ll address soon), we can imagine ripping apart metal–R covalent bonds and giving both electrons of the bond to the organic atom. This heterolytic bond cleavage method reproduces the starting charges on the metal and ligand. Unsurprisingly, the metal is positive and the ligand negative.
FYI, you might see the blue bipyridine referred to as an L2 ligand elsewhere; this just means that a single bipyridine molecule possesses two L-type binding points. Ligands with multiple binding points are also known as chelating or polydentate ligands. Chelating ligands may feature mixed binding modes; for instance, the allyl ligand is of the LX-type. Chelating ligands can also bind to two different metal centers; when they act in this way, they’re called bridging ligands. But don’t let all this jargon throw you! Deconstruct complexes one binding point at a time, and you cannot go wrong.
Next, we’ll take a closer look at the metal center and expand on the purpose of the deconstruction process described here.
"Bookkeeping" through deconstruction of organometallic complexes
Now it’s time to turn our attention to the metal center, and focus on what the deconstruction process can tell us about the nature of the metal in organometallic complexes. Let’s turn our attention now to a new complex. I’ve gone ahead and deconstructed it for us.
Say hello to rhodium (Rh)! Don't fret; it's just a group 9 element.
The complex possesses one X-type and three L-type ligands, so the rhodium atom ends up with a formal charge of +1. The formal charge on the metal center after deconstruction has a very special name that you will definitely want to commit to memory: it’s called the oxidation state. It’s usually indicated with a roman numeral next to the atomic symbol of the metal (the “+” is implied). In the complex shown above, rhodium is in the Rh(I) or +1 oxidation state. Oxidation state is most useful because changes in oxidation state indicate changes in electron density at the metal center, and this can be a favorable or unfavorable occurrence depending on the other ligands around. We will see this principle in action many, many times! Get used to changes in oxidation state as everyday events in organometallic reaction mechanisms. Unlike carbon (with the exception of carbene…what’s its oxidation state?!) and other second-row elements, the transition metals commonly exhibit multiple different oxidation states. More on that later, though. For now, training yourself to rapidly identify the oxidation state of a complexed metal is most important. Please note that when a complex possesses an overall charge, the oxidation state is affected by this charge!
oxidation state = number of X-type ligands bound to metal + overall charge of complex
What of this number of d electrons concept? A very useful way to think about “number of d electrons” is as the “number of non-bonding electrons on the metal center,” and you’re probably familiar with identifying non-bonding electrons from organic chemistry. The numbers of valence electrons of each organic element are set in stone: carbon has four, nitrogen has five, et cetera. Furthermore, using this knowledge, it’s straightforward to determine the number of lone pair electrons associated with an atom by subtracting its number of covalent bonds from its total number of valence electrons. E.g., for a neutral nitrogen atom in an amine NR3, 5 – 3 = 2 lone pair electrons, typically. The extension to organometallic chemistry is natural! We can analyze complexed metal centers in the same way, but they tend to have a lot more non-bonding electrons than organic atoms, and the number depends on the metal’s oxidation state. For instance, the deconstructed rhodium atom in the figure above has 8 d electrons: 9 valence electrons minus 1 used for bonding to Cl. Dative bonds don’t affect d electron count since both electrons in the bond come from the ligand.
number of non-bonding electrons = number of d electrons = metal’s group number – oxidation state
Drawing all the non-bonding d electrons out as lone pairs would clutter things up, so they are never drawn…but we must remember that they’re around! Why? Because the number of d electrons profoundly affects a complex’s geometry. We will return to this soon, but the key idea is that the ligands muck up the energies of the d orbitals as they approach the metal (recall the “star-crossed lovers” idea), and the most favorable way to do so depends on the number of non-bonding electrons on the metal center.
Oxidation state and d electron count: two tools the OM chemist can't live without!
This post introduced us to two important bookkeeping tools, oxidation state and number of d electrons. In the final installment of the “Simplifying the Organometallic Complex” series, we’ll bring everything together and discuss total electron count. We’ll see that total electron count may be used to draw a variety of insightful conclusions about organometallic complexes.
Apply deconstruction to gain insight
So far, we’ve seen how deconstruction can reveal useful “bookkeeping” properties of organometallic complexes: number of electrons donated by ligands, coordination number, oxidation state, and d electron count (to name a few). Now, let’s bring everything together and discuss total electron count, the sum of non-bonding and bonding electrons associated with the metal center. Like oxidation state, total electron count can reveal the likely reactivity of OM complexes—in fact, it is often more powerful than oxidation state for making predictions. We’ll see that there is a definite norm for total electron count, and when a complex deviates from that norm, reactions are likely to happen.
Let’s begin with yet another new complex. This molecule features the common and important cyclopentadienyl and carbon monoxide ligands, along with an X-type ethyl ligand.
What is the total electron count of this Fe(II) complex?
The Cp or cyclopentadienyl ligand is a polydentate, six-electron L2X ligand. The two pi bonds of the free anion are dative, L-type ligands, which we’ll see again in a future post on ligands bound through pi bonds. Think of the electrons of the pi bond as the source of a dative bond to the metal. Since both electrons come from the ligand, the pi bonds are L-type binders. The anionic carbon in Cp is a fairly standard, anionic X-type binder. The carbon monoxide ligands are interesting examples of two-electron L-type ligands—notice that the free ligands are neutral, so these are considered L-type! Carbon monoxide is an intriguing ligand that can teach us a great deal about metal-ligand bonding in OM complexes…but more on that later.
After deconstruction, we see that the Fe(II) center possesses 6 non-bonding d electrons. The total electron count is just the d electron count plus the number of electrons donated by the ligands. Since the d electron count already takes overall charge into account, we need not worry about it as long as we’ve followed the deconstruction procedure correctly.
total electron count = number of d electrons + electrons donated by ligands
For the Fe(II) complex above, the total electron count is thus 6 + (6 + 2 + 2 + 2) = 18. Let’s work through another example: the complex below features an overall charge of +1. Water is a dative ligand—that “2″ is very important!
Note that the overall charge is lumped into the oxidation state and d electron count of Mo.
The oxidation state of molybdenum is +2 here…remember that the overall charge factors in to that. When everything is said and done, the total electron count is 4 + (6 + 2 + 2 + 2 + 2) = 18.
What’s up with 18?! As it turns out, 18 electrons is a very common number for stable organometallic complexes. So common that the number got its own rule—the 18-electron rule—which states that stable transition-metal complexes possess 18 or fewer electrons. The rule is analogous to organic chemistry’s octet rule. The typical explanation for the 18-electron rule points out that there are 9 valence orbitals (1 s, 3 p, 5 d) available to metals, and using all of these for bonding seems to produce the most stable complexes. Of course, as soon as the rule left the lips of some order-craving chemist, researchers set out to find counterexamples to it, and a number of counterexamples are known. Hartwig describes the rule as an “empirical guideline” with little theoretical support. In fact, theoretical studies have shown that the participation of p orbitals in complex MOs is unlikely. I know that’s not what you want to hear—but hang with me! The 18-electron rule is still a very useful guideline. It’s most interesting, in fact, when it is not satisfied.
One last example…how would you expect the complex below to react?
Cobaltocene: jonesing for chemical change.
If we assume that the 18-electron rule is true, then cobaltocene has a real problem. It possesses 7 + (6 + 6) = 19 total valence electrons! Yet, we can also reason that this complex will probably react to relieve the strain of not having 18 electrons by giving up an electron. Guess what? In practice, cobaltocene is a great one-electron reducing agent, and can be used to prepare anionic complexes through electron transfer.
\[CoCp_2 + ML_n → [CoCp_2]+[ML_n]^–\]
This post described how to calculate total electron count and introduced the power of the 18-electron rule for predicting whether a complex will donate or accept electrons. We will definitely see these ideas again! But what happens when the electron counts of two complexes we’re interested in comparing are the same? We’ll need more information. In the next post, we’ll explore the periodic trends of the transition series. Our goal will be to make meaningful comparisons between complexes of different metals. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.03%3A_Electron_Counting_in_Organometallic_Complexes/13.3.02%3A_Simplifying_the_Organometallic_Complex_by_Deconstruction.txt |
Organometallic chemistry may be taught in many ways. Some textbooks spend a significant chunk of time discussing ligands, while others forego ligand surveys to dive right in to reactions and mechanisms. I like the ligand survey approach because it allows you to get a grip on expected behaviour for each type of ligand before you see them pop up as intermediates in reaction cycles. With the general principles in hand, it becomes easier to generate explanations for observed each ligan's effect on a reactions. Instead of generalizing from complex, specific examples in the context of reaction mechanisms, we’ll look at general trends first and apply these to reaction intermediates and mechanisms later. This section will kick off with carbon monoxide, a simple but fascinating ligand.
This section begins a survey of some of the most common or most interesting ligands in organometallic complexes. We will begin with a survey of ligands that participate in "dative" bonding to metal ions in this section ($\PageIndex{}$), followed by ligands that interact with metal ions through their $\pi$ electrons, and those that have multiple metal-carbon bonds in the following sections.
13.04: Survey of Organometallic Ligands
Introduction into Carbonyl Complexes
In this chapter we will look closer at carbonyl complexes, often just called carbonyls. Why? They are interesting for a number of reasons. Firstly, they are a quite extensive class of compounds with a diverse coordination chemistry. We will learn that the CO ligand can bind to a metal in various, sometimes non-obvious ways. Further, carbonyls are frequently used as starting materials for other coordination compounds. This is because the carbonyl ligand has no charge and carbon monoxide is a gas. Because of that a ligand substitution reaction can be easily driven to the right side by purging CO out of the reaction vessel. CO is a C1 unit, and this is frequently used in catalysis with carbonyls. Carbonyls are often intermediates in reactions that add a single carbon atom to a hydrocarbon chain. A fascinating fundamental property of carbonyls is that its $\pi$-accepting properties stabilize metals in low, sometimes even negative oxidation numbers.
Bonding in CO
The carbon in the CO molecule is the more reactive end, and thus CO prefers to bind with the carbon to a metal, and not with oxygen. This is not obvious because the carbon is less electronegative than oxygen. The reason is that the HOMO of the CO molecule is an approximately non-bonding orbital which is primarily localized at the carbon (Figure $1$). In the valence bond picture it is resembled by the electron-lone pair at the carbon atom. The electron lone pair at the oxygen is resembled by the 2a1 molecular orbital which has a significantly lower energy, therefore it is rarely used in bonding. Both the $\pi$ and $\pi$*-orbitals are relatively close in energy to the HOMO, and thus these orbitals can also be used for bonding in carbonyls. This explains the diverse coordination chemistry of carbonyls.
σ-binding Modes of The Carbonyl Ligand
In the most simple case the CO uses its electron lone pair at the carbon to bind to single metal atom M (Figure $2$). In this case we call CO a terminal carbonyl ligand. In addition, it is possible that the electron lone pair is shared between two or even three metals that are interconnected via metal-metal bonds. In these cases, we say that the CO acts as a $\mu$-bridged, and $\mu_3$-brigded ligand, respectively. If two metals held together by a metal-metal bond are different, then the interaction of CO with one metal may be stronger than with the other. In this case the CO ligand acts as a semibridging ligand. In all these cases CO acts as a 2-electron donor because it donates its two electrons in the electron lone pair in the carbon atom.
In addition to the electron lone pair at the carbon the CO ligand can also use its binding $\pi$-orbitals for electron-donation into metal orbitals (Figure $3$). However, these electrons can only be used in conjunction with the electron lone pair at C because the electron pair has a higher energy than the $\pi$-electrons, and electrons of higher energy are always used first. For steric reasons the electron lone pair and the $\pi$-electrons are always donated to different metals that are held together by metal-metal bonds.
In the most simple case the CO ligand binds in $\mu_2$-$\eta_2$-mode which implies that the electron lone pair binds to one metal and two $\pi$-electrons bind to another. In this this case the CO ligand acts as a 4-electron donor. In addition, it is possible that all four $\pi$-electrons are involved involved in the bonding in addition to the electron lone pair. In this case three metal atoms are involved. The first one interacts with the electron-lone pair, the second one with the two $\pi$-electrons, and the third one with the other two $\pi$-electrons. Because three metals are bridged and both atoms of the ligand are involved in the bonding with the metal we can say that CO binds in $\mu_3$-$\eta_2$-fashion and acts as a 6-electron donor. It is also possible that the electron-lone pair is being shared between two metals, and two $\pi$-electrons are donated to the third metal. Also in this case the ligand binds in $\mu_3$-$\eta_2$-fashion, but acts as a 4-electron donor. Another way the CO ligand can act as 4-electron donor is when it acts as an isocarbonyl ligand. In this case it uses its electron pairs at both the C and the O-atoms. This is only possible when steric circumstances favor the utilization of the O-electron lone pair over the energetically higher $\pi$-electrons. This is rare.
$\pi$-binding Modes of The Carbonyl Ligand
The carbonyl ligand can use its $\pi$*-orbitals for bonding with metal d-orbitals in $\pi$-fashion. The ligand acts as a $\pi$-acceptor. There are two possibilities for the binding (Figure $4$).
The first one occurs when the CO-ligand acts as a terminal ligand binding end-on. In this case the two lobes of the $\pi$*-orbitals at the carbon interact with the the lobes of a metal d-orbital. Also the bonding $\pi$-orbitals have the right symmetry to overlap with the metal d-orbital in this mode, but their energies are much further away from those of the d-orbitals, so that these interactions can be neglected. The second possibility is that the CO binds side-on to the a metal-d-orbital. In this case one lobe at C and one lobe at O interacts with the metal d-orbital. These interactions are less strong compared to the end-on interactions because the orbital overlap is less efficient due to the unequal size of the lobes of the $\pi$*-orbital. The $\pi$*-orbital is mostly localized at the carbon-atom because the carbon atom is the more electropositive atom. Note that the bonding $\pi$-orbitals do not have the right symmetry to overlap with a metal-d orbital in $\pi$-fashion, they can only overlap in σ-fashion when the binding is side-on.
The Dewar-Chatt-Duncanson model
The σ-donor and the $\pi$-acceptor interactions in carbonyl complexes synergistically reinforce each other. This synergistic effect is called the Dewar-Chatt-Duncanson model for carbonyls. How can we understand the synergistic interactions. Let us consider a carbonyl ligand that binds end-on to a metal.
Let us first look at the σ-donor interactions. The electron lone pair at the carbon donates electron density into empty metal d-orbitals and a dative bond is formed between the metal and the carbon. The donated electron density enhances the energy of the metal d-electrons due to increased electron-electron repulsion. Because of their increased energy the d-electrons get more easily accepted by the carbonyl ligand through the $\pi$-acceptor interactions. The $\pi$-acceptor interactions increase the bond order between the metal and the carbon bond. At the same time the bond order between carbon and oxygen gets decreases because electron density has been transferred from the metal into the $\pi$*-orbitals (Figure $5$).
The strength of the $\pi$-acceptor interactions can differ significantly in carbonyls. One can draw three different structures for weak, intermediate, and strong $\pi$-acceptor interactions (Figure $6$ to $8$).
When only weak interactions are present we can represent the M-C bond as a single bond, and the C-O bond as a triple bond.
When the interactions have intermediate strength we can represent both the M-C bond and the C-O bond as a double-bonds.
When the $\pi$-acceptor interactions are strong then the M-C bond becomes a triple bond and the C-O bond becomes a single bond.
On what does the strength of the $\pi$-acceptor interactions depend? It depends mostly on the charge on the carbonyl. Carbonyl cations with a charge of +2 or higher tend to have weak $\pi$-acceptor interactions. Neutral carbonyls or carbonyls with a +1 or -1 charge have intermediate $\pi$-acceptor interactions. and those with a negative charge of 2- or higher have strong $\pi$-acceptor interactions. One can see from this that the smaller the positive charge and the higher the negative charge, the higher the metal-carbon bond-order, and the stronger the metal-carbon bond. As a consequence negative charges tend to stabilize carbonyls, while positive charges destabilize them. Therefore, carbonylate anions tend to be more stable compared to carbonyl cations. The stability of neutral carbonyls is intermediate.
Homoleptic Carbonyls of The Transition Metals
After having understood the principles of the bonding in carbonyls let us next think about what structures transition metal carbonyls make. The structures follow mostly the 18 electron rule. For the most simple homoleptic carbonyls, in which the carbonyl ligand binds end-on to the transition metal acting as a two-electron donor, we would just assemble as many ligands as needed until we have 18 electrons. This would give us the coordination number in the carbonyl from which we could deduct the structure. This works well for transition metals with an even number of valence electrons.
In this case the number of ligands x is just 18 minus the number of metal electrons divided by 2. We would therefore expect a mononuclear carbonyl of the type M(CO)x (Figure $9$). For metals with an odd number of electrons, the situation is more complicated because an odd number times two multiplied with an integer number never gives 18. In this case, as many ligands x as needed to make a 17 valence electron complex are added to the metal, and then the 17 valence electron complex dimerizes to form a dinuclear carbonyl of the composition M2(CO)x. An exception is the vanadium. It makes a 17 valence electron vanadium hexacarbonyl that does not dimerize because a coordination number of 7 is not favorable.
Homoleptic carbonyls of the 6th group (M = Cr, Mo, W)
Now let us look closer at the carbonyls with metals with an even number of electrons. We can start out with group 6 which contains the elements chromium, molybdenum, and tungsten. Our task is to determine the composition and structure of the carbonyls.
According to the 18 electron rule we need 18 electrons overall. How many are contributed by the metal? Because the metals are in group 6, they all have six electrons. The carbonyls carry no charge, so there are no electrons to be subtracted or added. How many ligand electrons do we need to get to 18? Well, that is 18-6=12 electrons. How many carbonyl ligands are needed then? Because each ligand is considered a two electron donor, we need 12/2=6 ligands. The composition of the carbonyl will therefore be M(CO)6. The carbonyl adopts octahedral shape in order to maximize the distance of the ligands from each other. The group 6 carbonyls are colorless, crystalline, and sublimable solids.
Charged Octahedral Carbonyls with 18 Electrons
Are there also other 18 e carbonyls of the type M(CO)6 with other metals but group 6 metals? Yes, but the charge at the carbonyl needs to be adjusted so that the complex has 18 electrons. For instance, we can replace Cr by V, but then the we need a 1- charge which results in a carbonylate of the formula V(CO)6-. Why? The vanadium is in group 5 and has one electron less than chromium. For that reason, we must add an electron which gives the complex a 1- charge. Also the higher homologues Nb(CO)6- and Ta(CO)6- are known. What if we go further to the left in the periodic table? Titanium is left to the vanadium, and has four valence electrons. Therefore, the 18 valence electron hexacarbonyl of titanium has a 2- charge, and the formula is Ti(CO)62-. Again, the higher homologues of Ti(CO)62- , the Zr(CO)62- and the Hf(CO)62- are also known. Can we also go to the right in the periodic table? Manganese sits to the right of the chromium. The manganese has one electron more than the chromium, therefore the 18 electron hexacarbonyl of manganese must be a carbonyl cation with a 1+ charge, and the formula is Mn(CO)62+. Of the higher homologues, both the Tc(CO)6+ and the Re(CO)6+ are known, too. To the right of the Mn, there is the Fe which has 8 valence electrons. Therefore the 18 electron iron hexacarbonyl must have a 2+ charge, and the formula is Fe(CO)62+. The higher homologues Ru(CO)62+ and Os(CO)62+ are also known. Can we go even further to the left, to the Co? The expected formula for Co carbonyl would be Co(CO)63+, but it is not known, and neither is the higher homologue, the Rh(CO)63+. Only the Ir(CO)63+ has been isolated. This reflects a general trend in carbonyl chemistry. Positive charges destabilize the carbonyl, and carbonyls with high positive charges are frequently not stable. On the other hand, negative charges stabilize a carbonyl and even carbonylates with high negative charges are stable. We can easily explain this by the fact that the bond order between the metal and the carbon decreases with increasing positive charge, and increases with increasing negative charge. Increasing bond order strengthens the stability of the carbonyl. We see from this that the $\pi$-acceptor properties are quite important for the stability of carbonyls. We also see that in carbonylates the oxidation numbers of the metals are negative. Negative oxidation numbers are quite unusual for metals. We conclude that the carbonyl ligand has the unusual property to stabilize the metal in negative oxidation states.
Is there a way to easily measure the bond order of an M-C bond? We can do this indirectly by measuring the IR-spectrum of the carbonyl. Because the bond order of the C-O bond decreases with increasing bond order for the M-C bond, the wavenumber for the C-O stretch vibration can be used as a measure for the M-C bond order. The lower the wavenumber, the lower the C-O bond order, and the higher the M-C bond order. When we look at the numbers for the different carbonyls within a period we see that as expected the wavenumbers increase with increasing group number indicating a decreasing bond order. We can also look down the groups. We see that in this case, the numbers hardly change, meaning that the bond order is hardly affected by the period of the metal.
Homoleptic Carbonyls of the 8th Group (Fe, Ru, Os)
What charge-neutral carbonyls would we expect for the group 8 elements Fe, Ru, and Os?
Overall, these carbonyl need to have 18 electrons. In this case the metals have 8 electrons. The charge is 0. This means that we would have 18-8=10 electrons that would need to come from the ligand. The number of ligands needed would therefore be 10/2=5. Thus, we would expect pentacarbonyls of the composition M(CO)5 (Figure $12$). These carbonyls adopt the trigonal-bipyramidal shape. The group 8 carbonyls are all liquids. The iron pentacarbonyl has a light-orange color while the ruthenium and the osmium carbonyls are colorless.
Charged Trigonal-Bipyramidal Carbonyls with 18 electrons
We can again ask if there are charged isosteric pentacarbonyls with 18 e of metals other than group 8 metals. The answer is yes. The manganese is left to the iron in the periodic table and has one electron less. Therefore its pentacarbonyl needs to have a 1- charge and the composition is Mn(CO)5-. Also the higher homologues, the Tc(CO)5- and the Re(CO)5- exist. The Cr has one electron less than than the Mn, therefore its pentacarbonyl has a 2- charge and the formula is Cr(CO)52-. Mo(CO)52- and W(CO)52- are also known. The vanadium has another electron less because it is in the 5th group, therefore its pentacarbonyl has a 3- charge, and the formula is V(CO)53-. Again, the higher homologues, the Nb(CO)53- and the Ta(CO)53- are also stable. Also for the pentacarbonylates high negative charges are no problem for the stability of the complexes due to the stabilizing effect of the $\pi$-acceptor interactions. What about carbonyl cations with trigonal bipyramidal shape? As we go from Fe to Co, we need a 1+ charge at the carbonyl to achieve 18 electrons. The respective Co(CO)5+ cation is stable, and so are their higher homologues, the Rh(CO)5+ and the Ir(CO)5+. However, the group 10 pentacarbonyl cations Ni(CO)52+, Pd(CO)52+, and Pt(CO)52+ are not stable. The 2+ positive charge weakens the metal-carbon bond too much. This behavior is consistent with the weakening of the M-C bond with increasing positive charge due to weaker $\pi$-acceptor effects.
Homoleptic carbonyls of group 10 (Ni, Pd, Pt)
What about the group 10 carbonyls?
In this case the metals have 10 electrons. There is no charge. The number of ligand electrons needed is 18-10 = 8 electrons. Thus, the number of ligands is 8/2=4. Therefore, the composition is M(CO)4 (Figure $14$). The shape is tetrahedral because we have an 18 and not a 16 valence electron complex. Nickel tetracarbonyl is a volatile, colorless liquid. The higher homologues, the Pd(CO)4 and the Pt(CO)4 are not stable. We can explain this when remembering that the $\pi$-orbital overlap in tetrahedral complexes is generally weak. Therefore, the M-C carbon bond is less effectively stabilized by the $\pi$-acceptor interactions compared to octahedral and trigonal-bipyramidal carbonyls. This effect is even more pronounced for the Pd and Pt carbonyls because they have larger orbitals that can overlap even less effectively with the relatively small $\pi$*-orbitals of the CO ligand. In this case the $\pi$-acceptor interactions are so weak so that the entire molecule is no longer stable.
Charged Tetrahedral Carbonyls with 18 electrons
What about charged 18 electron tetrahedral carbonyls with other metals? Again, negatively charged carbonylates are stable, even with high negative charges. Co(CO)4-, Fe(CO)42-, Mn(CO)43-, and Cr(CO)44- are all stable and so are their higher homologues. The Cu(CO))4+ which has a 1+ positive charge is also known, but its higher homologues, the Ag(CO)4+ and the Au(CO)4+ are not.
Overall the stability of carbonyls tends to decrease from group 6 to group 8 to group 10 (Figure $16$). This trend is because the octahedral shape allows for the best orbital overlap for $\pi$-acceptor interactions, followed by the trigonal bipyramidal, followed by the tetrahedral shape. The stability also tends to decline from the 4th to the 6th period. This trend can be explained by less efficient orbital overlap due to increasingly different orbital size between the ligand and the metal orbitals. These effects are sufficiently large for Pd(CO)4 and Pt(CO)4 to make them instable.
Group 4 and Group 12 Carbonyls (non-existing)
We have discussed group 6, 8, and 10 carbonyls, but not group 4 and group 12. Why? Because no carbonyls of these groups are stable. What is the reason? For group 4 carbonyls, we would need seven carbonyl ligands. This would lead to a coordination number of seven which is unacceptably high for carbonyls. For group 12 carbonyls, we would only need three carbonyl ligands. This would lead to an unacceptably low coordination number of 3.
Homoleptic carbonyls of the 7th group (M = Mn, Tc, Re)
Now let us think about what structures the homoleptic carbonyls with metals having an odd number of electrons form. Let us start with the 7th group which consists of the metals Mn, Tc, and Re. Our approach is to see how many ligands we need to produce a 17 valence electron fragment, and then dimerize that fragment.
The group 7 metals have seven valence electrons. Again, we will assume that there is no charge on the carbonyl. The number of ligand electrons to produce the 17 valence electrons fragment will therefore be 17-7 = 10. The number of ligands needed is therefore 10/2=5 assuming that the carbonyl ligand is a 2-electron donor binding end-on to the metal. This gives an M(CO)5 fragment that dimerizes to form an M2(CO)10 carbonyl with a metal-metal bond. The structure can be thought to be derived from an octahedral structure in which each metal is surrounded octahedrally by five CO ligands and an M(CO)5 fragment acting as the sixth ligand. Each M(CO)5 unit has one axial ligand and four equatorial ligands, whereby the axial ligand is co-directional with the metal-metal bond. The four equatorial ligands of the first metal are in staggered conformation relative to the four equatorial ligands of the second metal due to steric repulsion arguments.
Charged Isoelectronic Carbonyls of the Type M2(CO)10
Can we make charged isoelectronic carbonyls of the type M2(CO)10 with other metals? The answer is yes, but only carbonylate anions with low negative charge, and no carbonyl cations. The formation of carbonyl cations is prohibited because of the weakening of the $\pi$-acceptor interactions that result from the positive charge. High negative charges are not possible because too many negative charges on the metal atoms lead to electrostatic repulsion and the destabilization of the metal-metal bonds.
There are therefore only Cr2(CO)102-, and the higher homologues Mo2(CO)102- and W2(CO)102-, but no other M2CO10 type carbonyl ions.
Homoleptic carbonyls of the 9th group (M = Co, Rh, Ir)
What are the expected structures for the homoleptic carbonyls of group 9?
In this case, the metal contributes 9 electrons, and this means that 17-9=8 electrons need to come from the carbonyl ligand. We therefore need 8/2=4 CO molecules to produce an M(CO)4 17 valence electron fragment. Two of the fragments then dimerize to form a dinuclear carbonyl of the composition M2CO8. In nature, however, only the Co-carbonyl is realized, the Rh and the Ir analogs are unstable. The structure of the Co2(CO)8 carbonyl can be derived from an M(CO)4L trigonal-pyramid, whereby L is the second M(CO)4 unit (Fig. 10.2.19). According to that, the coordination number is 5. There are two axial and six equatorial ligands whereby the equatorial ligands are oriented in a staggered conformation. In solution this structure is in dynamic equilibrium with a second structure with C2v symmetry in which two CO ligands are bridging. Due to this equilibrium there is constant ligand scrambling, meaning that the CO ligands constantly move from axial to equatorial positions, and migrate from one metal atom to the other. In solid state the C2v-type structure is present exclusively indicating that it is slightly more stable.
Charged Isoelectronic Carbonyls of the Type M2(CO)8
The charged, isoelectronic carbonyls of the type M2(CO)8 behave similarly to that of the type M2(CO)10. There are no highly charged binuclear carbonylate anions, and no carbonyl cations.
There are only the weakly negatively charged [Fe2(CO)8]2- , [Ru2(CO)8]2-, and [Os2(CO)8]2-. The weakly negative charge stabilizes the metal-carbon bond without destabilizing the metal-metal bond too much. We can further see from this that a weakly negative charge is better for the stability of the species than no charge at all.
Homoleptic Carbonyls of the 11th Group (M = Cu, Ag, Au)
The group 11 metals Cu, Ag, and Au have eleven valence electrons.
Therefore, six ligand electrons are needed to achieve 17 electrons (Figure $21$). This means that 6/2=3 CO ligands are required to produce the 17 valence electron fragment. The dimer of it has the composition M2(CO)6 and its structure can be derived from a tetrahedral M(CO)3L structure in which L is the second M(CO)3 fragment. Neither the Cu, nor the Ag, nor the Au carbonyl are known due to the weak $\pi$-overlap in tetrahedral coordination. However, the weakly negatively [Ni2(CO)6]2- ion is known which may be explained by the stabilizing effect of one negative charge on the M-C bond.
Overall, we see the same stability trends for the dinuclear carbonyls as compared to the mononuclear ones. The stability decreases with the group number because $\pi$-overlap becomes smaller with smaller coordination numbers. Also, we see that the stability decreases with the period because of decreasing match of orbital sizes.
Homoleptic carbonyls of the 5th group (M = V, Nb, Ta)
The 17 valence electron fragment of the carbonyl of vanadium has the composition V(CO)6. It does not dimerize because the coordination number of 6 is much more favorable compared to the coordination number 7. V(CO)6 is a dark violet radical of octahedral shape. The radical electron is sterically inactive. V(CO)6 can be used as an oxidant because it can be reduced easily to the 18 valence electron species V(CO)6-. The higher homologues Nb(CO)6 and Ta(CO)6 are not known. This may be explained by the weaker orbital overlap between the relatively small $\pi$*-orbitals of the carbonyl ligand and the large metal d-orbitals of Nb and Ta.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
Synthesis
Metal carbonyl complexes containing only CO ligands abound, but most cannot be synthesized by the method we all wish worked, bathing the elemental metal in an atmosphere of carbon monoxide (entropy is a problem, as we already discussed for W(CO)6). This method does work for nickel(0) and iron(0) carbonyls, however.
M + n CO → M(CO)n [M = Fe, n = 5; M = Ni, n = 4]
Other metal carbonyl complexes can be prepared by reductive carbonylation, the treatment of a high-oxidation-state complex with CO. These methods usually require significant heat and pressure. One example:
WCl6 + CO + heat, pressure → W(CO)6
Still other methods employ deinsertion from organic carbonyl compounds like dimethylformamide. These methods are particularly useful for preparing mixed carbonyl complexes in the presence of reducing ligands like phosphines.
IrCl3(H2O)3 + 3 PPh3 + HCONMe2 + PhNH2 → IrCl(CO)(PPh3)2 + (Me2NH2)Cl + OPPh3 + (PhNH3)Cl + 2 H2O
The key thing to notice about the reaction above is that the CO ligand is derived from dimethylformamide (DMF). | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.04%3A_Survey_of_Organometallic_Ligands/13.4.01%3A_Carbon_Monoxide_%28Carbonyl_Complexes%29.txt |
Now let us leave the $\ce{CO}$ ligand, consider a number of related ligands, and discuss similarities and differences compared to the $\ce{CO}$ ligand.
The cyano ligand (CN-)
Let us start with the cyano ligand $\ce{CN^-}$. This ligand is isoelectronic to the $\ce{CO}$ ligand. The N atom has one electron less than O, but the negative charge at the cyanide ligand compensates for that. One question we can ask is: What is the reactive end? The answer is: In analogy to the carbonyl ligand the reactive end is the carbon atom. We can explain this by the fact that the MO diagram is similar to that of $\ce{CO}$, only the differences in atomic orbital energies are smaller due to the smaller electronegativity difference between C and N compared to C and O. Therefore, like in $\ce{CO}$, the HOMO is represented by the electron lone pair at the C, which makes C the more reactive end. Due to the smaller electronegativity difference, the difference in energy between the lone pair at C and the lone pair at N is smaller, therefore, in contrast to $\ce{CO}$, the cyano ligand acts far more often as a bridging ligand between two metals using both of its electron lone pairs.
Would we expect the cyano ligand to be a stronger or weaker $\sigma$-donor compared to the $\ce{CO}$ ligand? Think about it for a moment. The answer is: It is a stronger $\sigma$-donor because of its negative charge. The negative charge at the ligand increases electrostatic repulsion between the electrons, and this increases the orbital energies. Therefore, there is a stronger tendency to donate the electrons. Our next question is: Is the $\ce{CN^-}$ ligand a stronger or weaker $\pi$-acceptor than $\ce{CO}$? The energy of the $\pi^*$-orbitals is higher compared to $\ce{CO}$ because of the negative charge at the ligand. Because of that, electrons from the metal cannot be as easily accepted by the ligand. Therefore, the cyano ligand is a weaker $\pi$-acceptor than the carbonyl ligand. Our last question is: Are cyano complexes more stable with metals in high or low oxidation states. Because of electrostatic arguments a cyanide anion interacts more strongly with a metal cation rather than a metal in a zero or negative oxidation state. Therefore, unlike $\ce{CO}$, $\ce{CN^-}$ does not stabilize metals in low oxidation states. It prefers to make complexes with metal in high, positive oxidation numbers.
The Nitrosyl Ligand NO
The nitrosyl ligand $\ce{NO}$ is another ligand similar to $\ce{CO}$. It has one more electron that $\ce{CO}$ because N has one more electron than C. The additional electron makes $\ce{NO}$ an “odd” molecule with a radical electron. Like in $\ce{CO}$ and $\ce{CN^-}$, the more electropositive element is the reactive end. In the case of $\ce{NO}$ it is the N atom. The radical electron is the most reactive electron that can be most easily donated, however the electron lone pair at N may be donated in addition. In the former case, the $\ce{NO}$ is a 1-electron donor, in the latter it is a 3-electron donor. How can we tell if one or three electrons have been donated? When only one electron is donated, the electron lone pair at the nitrogen is sterically active and leads to a bent structure (Fig. $1$).
An example is the trans-bis-(ethylenediamine)chloronitrosyl cobalt (1+) cation. Generally, the $\ce{O-N-M}$ bond angle in nitrosyl complexes with $\ce{NO}$ as a 1e-donor can vary between 119 and 140°. We can also identify a bent nitrosyl ligand in the IR spectrum. Typical wave numbers are 1520-1729 cm-1.
When the nitrosyl ligand donates all three electrons, then it binds to the metal in a linear fashion (Fig. $2$). The electron lone pair is no longer sterically active because it is involved in the bonding. An example is the nitroprusside anion $\ce{[Fe(CN)5(NO)]^2-}$. It is used as a medication for the treatment of high blood pressure. The bond angles in complexes with $\ce{NO}$ as the 3-electron donor are often not exactly 180°, but vary between 165 and 180°. It can also be identified in the IR because it has a characteristic band between 1610-1680 cm-1.
There is a also a different view on nitrosyl complexes with linear and bent structures. In bent structures we can also consider the ligand as an $\ce{NO^-}$ anion that donates two electrons. The $\ce{NO^-}$ anion has two electron lone pairs at N. When it donates two electrons then one sterically active electron lone pair remains at the nitrogen atom. In linear structures we can regard the ligand also as an $\ce{NO^+}$ cation that donates two electrons. An $\ce{NO^+}$ cation has one electron lone pair at N, and when it donates that lone pair then there is no sterically active electron at the nitrogen left, and thus the ligand binds in a linear fashion.
We could also ask: What can neutral $\ce{NO}$ not be a 2-electron donor with the radical electron left at N? The answer is: The radical electron is the highest energy electron, and is always used first in interactions with a metal.
Phosphine Ligands
Phosphines are most notable for their remarkable electronic and steric tunability and their “innocence”—they tend to avoid participating directly in organometallic reactions, but have the ability to profoundly modulate the electronic properties of the metal center to which they’re bound. Furthermore, because the energy barrier to umbrella flipping of phosphines is quite high, “chiral-at-phosphorus” ligands can be isolated in enantioenriched form and introduced to metal centers, bringing asymmetry just about as close to the metal as it can get in chiral complexes. Phosphorus NMR is a technique that Just Works (thanks, nature). Soft phosphines match up very well with the soft low-valent transition metals. Electron-poor phosphines are even good π-acids!
Like CO, phosphines are dative, L-type ligands that formally contribute two electrons to the metal center. Unlike CO, most phosphines are not small enough to form more than four bonds to a single metal center (and for large R, the number is even smaller). Steric hindrance becomes a problem when five or more PR3 ligands try to make their way into the space around the metal. An interesting consequence of this fact is that many phosphine-containing complexes do not possess 18 valence electrons. Examples include Pt(PCy3)2, Pd[P(t-Bu)3]2, and [Rh(PPh3)3]+. Doesn’t that just drive you crazy? It drives the complexes crazy as well—and most of these coordinatively unsaturated compounds are wonderful catalysts.
Bridging by phosphines is extremely rare, but ligands containing multiple phosphine donors that bind in an Ln (n > 1) fashion to a single metal center are all over the place. These ligands are called chelating or polydentate to indicate that they latch on to metal centers through multiple binding sites. For entropic reasons, chelating ligands bind to a single metal center at multiple points if possible, instead of attaching to two different metal centers (the aptly named chelate effect). An important characteristic of chelating phosphines is bite angle, defined as the predominant P–M–P angle in known complexes of the ligand. We’ll get into the interesting effects of bite angle later, but for now, we might imagine how “unhappy” a ligand with a preferred bite angle of 120° would be in the square planar geometry. It would much rather prefer to be part of a trigonal bipyramidal complex, for instance.
The predominant orbital interaction contributing to phosphine binding is the one we expect, a lone pair on phosphorus interacting with an empty metallic d orbital. The electronic nature of the R groups influences the electron-donating ability of the phosphorus atom. For instance, alkylphosphines, which possess P–Csp3 bonds, tend to be better electron donors than arylphosphines, which possess P–Csp2 bonds. The rationale here is the greater electronegativity of the sp2 hybrid orbital versus the sp3 hybrid, which causes the phosphorus atom to hold more tightly to its lone pair when bound to an sp2 carbon. The same idea applies when electron-withdrawing and -donating groups are incorporated into R: the electron density on P is low when R contains electron-withdrawing groups and high when R contains electron-donating groups. Ligands (and associated metals) in the former class are called electron poor, while those in the latter class are electron rich.
As we add electronegative R groups, the phosphorus atom (and the metal to which it's bound) become more electron poor.
Like CO, phosphines participate in backbonding to a certain degree; however, the phenomenon here is of a fundamentally different nature than CO backbonding. For one thing, phosphines lack a π* orbital. In the days of yore, chemists attributed backbonding in phosphine complexes to an interaction between a metallic dπ orbital and an empty 3d orbital on phosphorus. However, this idea has elegantly been proven bogus, and a much more organicker-friendly explanation has taken its place (no d orbitals on P required!). In an illuminating series of experiments, M–P and P–R bond lengths were measured via crystallography for several redox pairs of complexes. I’ve chosen two illustrative examples, although the linked reference is chock full of other pairs. The question is: how do we explain the changes in bond length upon oxidation?
Upon oxidation, M–P bond lengths increase and and P–R bond lengths decrease. Why?
Oxidation decreases the ability of the metal to backbond, because it removes electron density from the metal. This explains the increases in M–P bond length—just imagine a decrease the M–P bond order due to worse backbonding. And the decrease in P–R bond length? It’s important to see that invoking only the phosphorus 3d orbitals would not explain changes in the P–R bond lengths, as the 3d atomic orbitals are most definitely localized on phosphorus. Instead, we must invoke the participation of σ*P–R orbitals in phosphine backbonding to account for the P–R length decreases. When all is said and done, the LUMO of the free phosphine has mostly P–R antibonding character, with some 3d thrown into the mix. The figure below depicts one of the interactions involved in M–P backbonding, a dπ → σ* interaction (an orthogonal dπ → σ* interaction also plays a role). As with CO, a resonance structure depicting an M=P double bond is a useful heuristic! Naturally, R groups that are better able to stabilize negative charge—that is, electron-withdrawing groups—facilitate backbonding in phosphines. Electron-rich metals help too.
Backbonding in phosphines, a sigma-bond-breaking affair.
The steric and electronic properties of phosphines vary enormously. Tolman devised some intriguing parameters that characterize the steric and electronic properties of this class of ligands. To address sterics, he developed the idea of cone angle—the apex angle of a cone formed by a point 2.28 Å from the phosphorus atom (an idealized M–P bond length), and the outermost edges of atoms in the R groups, when the R groups are folded back as much as possible. Wider cone angles, Tolman reasoned, indicate greater steric congestion around the phosphorus atom. To address electronics, Tolman used a not-so-old friend of ours—the CO stretching frequency (νCO) of mixed phosphine-carbonyl complexes. Specifically, he used Ni(CO)3L complexes, where L is a tertiary phosphine, as his standard. Can you anticipate Tolman’s logic? How should νCO change as the electron-donating ability of the phosphine ligands increases?
Tolman’s logic went as follows: more strongly electron-donating phosphines are associated with more electron-rich metals, which are better at CO backbonding (due fundamentally to higher orbital energies). Better CO backbonding corresponds to a lower νCO due to decreased C–O bond order. Thus, better donor ligands should be associated with lower νCO values (and vice versa for electron-withdrawing ligands). Was he correct? Exhibit A…
Tolman's map of the steric and electronic properties of phosphine ligands.
Notice the poor ligand trifluorophosphine stuck in the “very small, very withdrawing” corner, and its utter opposite, the gargantuan tri(tert-butyl)phosphine in the “extremely bulky, very donating” corner. Intriguing! One can learn a great deal just by studying this chart.
Bonding in Phosphine Complexes
Let us have a look at the MO diagram of the $\ce{PH3}$ molecule and compare it with the $\ce{NH3}$ molecule. As one would expect, the MOs are overall similar, but there is one important difference. While the LUMO in $\ce{NH3}$ is the anti-bonding 3a1 orbital, the LUMOs in the $\ce{PH3}$ molecule are the anti-bonding 2e orbitals. The relative energies of the 3a1 and 2e orbitals in the $\ce{PH3}$ and $\ce{NH3}$ molecules are swapped up. This can be attributed to the fact that the the P atom uses the 3s and the 3p orbitals as valence orbitals, while N uses the 2s and 2p orbitals. The 3s and the 3p orbitals are larger and overlap less effectively with the small 1s orbitals of the hydrogen. They also have a higher energy making the P-H bonds less polar than the N-H bonds. The energy of the $\ce{PH3}$ HOMO is higher than that of the $\ce{NH3}$ HOMO. Both the HOMO and the LUMO of $\ce{PH3}$ are more diffuse and polarizable than the respective orbitals in $\ce{NH3}$.
The higher energy of the HOMO in $\ce{PH3}$ makes it a better donor than $\ce{NH3}$. In addition, the $\ce{PH3}$ has $\pi$-acceptor properties because the 2e LUMO are relatively low-lying anti-bonding 2e orbitals and have suitable shape for $\pi$-overlap with metal d-orbitals. $\ce{NH3}$ does not have these $\pi$-acceptor properties because its LUMO is the 3a1 orbital, and not the 2e orbitals. The 2e orbitals in $\ce{NH3}$ are energetically too high in order to allow for significant $\pi$-acceptor interactions.
The $\sigma$-donor and $\pi$-acceptor properties of the phosphine ligand can be modified by substituting H by other groups. Generally, more electron donating groups increase the energy of the HOMO and the LUMO. This strengthens the $\sigma$-donor and reduces the $\pi$-acceptor properties. Vice versa, more electron withdrawing groups decrease the energy of the HOMO and the LUMO. As a consequence, the ligand becomes a weaker $\sigma$-donor and a stronger $\pi$-acceptor (Fig. $4$).
The table above shows the HOMO and LUMO energies of three phosphines. As expected the HOMO and LUMO energies decrease from $\ce{PMe3}$ to $\ce{PH3}$ to $\ce{PF3}$ due to the increasingly electron-withdrawing nature of the substituent. As a consequence, the $\sigma$-donor properties weaken and the $\pi$-acceptor properties strengthen from to $\ce{PMe3}$ to $\ce{PH3}$ to $\ce{PF3}$.
Not only the energies, but also the orbital overlap is important for the strength of the $\pi$-acceptor properties of the phosphine ligands. The more strongly electron-withdrawing the group is the more the anti-bonding e-type LUMO orbitals are located at the P atom. The more these orbitals are localized at P, the better they can overlap with the ligand. Vice versa the localization of the HOMO shifts toward the group as the group becomes more electron-withdrawing, thereby weakening the $\sigma$-donor properties.
Phosphine ligands with strongly electron withdrawing groups such as $\ce{PF3}$ have $\pi$-acceptor properties strong enough to stabilize metals in low oxidation numbers, similar to $\ce{CO}$. For example, the $\ce{PF3}$ ligand forms stable complexes with Pd and Pt atoms in the oxidation number 0, while the respective Pd and Pt carbonyls are not known.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.04%3A_Survey_of_Organometallic_Ligands/13.4.02%3A_Ligands_similar_to_CO.txt |
Ligands can, shockingly enough, bind through the electrons in their own $\sigma$ bonds in an L-type fashion. This binding mode depends as much on the metal center as it does on the ligand itself—to see why, we need only recognize that $\sigma$ complexes look like intermediates in concerted oxidative additions. With a slight reorganization of electrons and geometry, an L-type $\sigma$ ligand can become two X-type ligands. Why, then, are $\sigma$ complexes stable? How can we control the ratio of $\sigma$ complex to X2 complex in a given situation? How does complexation of a $\sigma$ bond change the ligand’s properties? We’ll address these questions and more in this post.
General Properties
The first thing to realize about $\sigma$ complexes is that they are highly sensitive to steric bulk. Any old $\sigma$ bond won’t do; hydrogen at one end of the binding bond or the other (or both) is necessary. The best studied $\sigma$ complexes involve dihydrogen ($\ce{H2}$), so let’s start there.
Mildly backbonding metals may bind dihydrogen “side on.” Like side-on binding in $\pi$ complexes, there are two important orbital interactions at play here: The first is a sigma bonding interaction ($\ce{ \sigma H-H \bond{->} d\sigma}$), as pictured on the left in Figure $2$; the second is a $\pi$ bonding interaction ( $\ce{d \pi \bond{->}} \sigma^* \ce{H–H}$), as pictured on the right in Figure $2$. Dihydrogen complexes can “tautomerize” to hydride $\ce{(H)2}$ isomers through cleavage of the $\ce{H–H}$ bond to the metal (this is an oxidative addition reaction, as discussed in the chapter on organometallic reactions).
$\ce{H2}$ binding in an L-type fashion massively acidifies the ligand. Changes in of over thirty pKa units have been observed upon metal binding! Analogous acidifications of X–H bonds, rarely exhibit ΔpKa > 5. What’s causing the different behavior of X–H and H–H ligands? The key is to consider the conjugate base of the ligand—in particular, how much it’s stabilized by a metal center relative to the corresponding free anion. The principle here is analogous to the famous dictum of organic chemistry: consider charged species when making acid/base comparisons. Stabilization of the unhindered anion $\ce{H^-}$ by a metal is much greater than stabilization of larger, more electronegative anions like $\ce{HO^-}$ and $\ce{NH2^-}$ by a metal. As a result, it’s more favorable to remove a proton from metal-complexed $\ce{H2}$ than from larger, more electronegative X–H ligands.
Remarkably large stabilization by an acidic metal fragment, without any counterbalancing from steric factors, explains the extreme acidification of dihydrogen upon metal binding.
The electronic nature of the metal center has two important effects on $\sigma$ complexes. The first concerns the acidity of $\ce{H2}$ upon metal binding. The principle here is consistent with what we’ve hammered into the ground so far. In the same way cationic organic acids are stronger than their neutral counterparts, $\sigma$ complexes of electron-poor metals—including (and especially) cations—are stronger acids than related complexes of electron-rich metals. The second concerns the ratio of L-type to X2-type binding. We should expect more electron-rich metal centers to favor the X2 isomer, since these should donate more strongly into the $\sigma$*H–H orbital. This idea was masterfully demonstrated in a study by Morris, in which he showed that $\ce{H2}$ complexes of π-basic metal centers show all the signs of X2 complexes, rather than L complexes. More generally, metal centers in $\sigma$ complexes need a good balance of π basicity and $\sigma$ acidity (I like to call this the “Goldilocks effect”). Because of the need for balance, $\sigma$ complexes are most common for centrally located metals (groups 6-9).
The M–H bond in hydride complexes is a good base—anyone who’s ever quenched lithium aluminum hydride can attest to this! Intriguingly, because it’s a good base, the M–H bond can participate in hydrogen bonding with an acidic X–H bond, where X is a heteroatom. This kind of bonding, called dihydrogen bonding (since two hydrogen atoms are involved), is best described as a sort of $\sigma$M–H→$\sigma$*X–H orbital interaction. Think of it as analogous to a traditional hydrogen bond, but using a $\sigma$ bond instead of a lone pair. Crazy, right?!
Other kinds of $\sigma$ complexes are known, but these are rarer than H–H complexes. One class that we’ve seen before involve agostic interactions of C–H bonds in alkyl ligands. $\sigma$ Complexes of inorganic ligands like silanes and stannanes may involve complex bonding patterns, but we won’t concern ourselves with those here.
Hydride Complexes
Metal-hydrogen bonds, also known (misleadingly) as metal hydrides, are ubiquitous X-type ligands in organometallic chemistry. There is much more than meets the eye to most M-H bonds: although they’re simple to draw, they vary enormously in polarization and pKa. They may be acidic or hydridic or both, depending on the nature of the metal center and the reaction conditions. In this post, we’ll develop some heuristics for predicting the behavior of M-H bonds so that later, we can discuss their major modes of reactivity (acidity, radical reactions, migratory insertion, etc.).
General Properties
Metal hydrides run the gamut from nucleophilic/basic to electrophilic/acidic. Then again, the same can be said of X–H bonds in organic chemistry, which may vary from mildly nucleophilic (consider Hantzsch esters and NADH) to extremely electrophilic (consider triflic acid). As hydrogen is what it is in both cases, it’s clear that the nature of the X fragment—more specifically, the stability of the charged fragments X+ and X–—dictate the character of the X–H bond. Compare the four equilibria outlined below—the stabilities of the ions dictate the position of each equilibrium. By now we shouldn’t find it surprising that the highly π-acidic W(CO)5 fragment is good at stabilizing negative charge; for a similar reason, the ZrCp2Cl fragment can stabilize positive charge.*
Metal-hydrogen bonds may be either hydridic (nucleophilic) or acidic (electrophilic). The nature of other ligands and the reaction conditions are keys to making predictions.
Let’s turn our attention now to homolytic M–H bond strength. A convenient thermodynamic cycle allows us to use the acidity of M–H and the oxidation potential of its conjugate base in order to determine bond strength. This clever method, employed by Tilset and inspired by the inimitable Bordwell, uses the cycle in the figure below. BDE values for some complexes are provided as well. From the examples provided, we can see that bond strength increases down a group in the periodic table. This trend, and the idea that bridging hydrides have larger BDEs than terminal M–H bonds, are just about the only observable trends in M–H BDE.
A clever cycle for determining BDEs from other known quantities, with select BDE values. I've left out solvation terms from the thermodynamic cycle. For more details, see the Tilset link above.
Why is knowing M-H BDEs useful? For one thing, the relative BDEs of M-C and M-H bonds determine the thermodynamics of β-hydride elimination, which results in the replacement of a covalent M-C bond with an M-H bond. Secondly, complexes containing weak M-H bonds are often good hydrogen transfer agents and may react with organic radicals and double bonds, channeling stannane and silane reductants from organic chemistry.
Hydricity refers to the tendency of a hydride ligand to depart as H–. A similar thermodynamic cycle relates the energetics of losing H– to the oxidation potentials of the conjugate base and the oxidized conjugate base; however, this method is complicated by the fact that hydride loss establishes an open coordination site. I’ve provided an abridged version of the cycle below. Hydricities are somewhat predictable from the electronic and steric properties of the metal center: inclusion of electron-donating ligands tends to increase hydricity, while electron-withdrawing or acidic ligands tend to decrease it. For five-coordinate hydrides that form 16-electron, square planar complexes upon loss of hydride, the bite angle of chelating phosphines plays an interesting role. As bite angle increases, hydricity does as well.
A thermodynamic cycle for hydricity, with some examples. Hydricity and bite angle are well correlated in five-coordinate palladium hydrides.
Bridging hydrides are an intriguing class of ligands. A question to ponder: how can a ligand associated with only two electrons possibly bridge two metal centers? How can two electrons hold three atoms together? Enter the magic of three-center, two-electron bonding. We can envision the M–H sigma bond as an electron donor itself! With this in mind, we can imagine that hydrides are able to bind end-on to one metal (like a standard X-type ligand) and side-on to another (like an L-type π system ligand, but using sigma electrons instead). Slick, no? We’ll see more side-on bonding of sigma electrons in a future post on sigma complexes.
Resonance forms of bridging hydrides, with an example. Sigma complexes like these show up in other contexts, too!
Consistent with the idea that bridging is the result of “end-on + side-on” bonding, bond angles of bridging hydrides are never 180°. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.04%3A_Survey_of_Organometallic_Ligands/13.4.03%3A_Hydrides_and_Dihydrogen_Complexes.txt |
With this section, we finally reach our first class of dative actor ligands, π systems. In contrast to the spectator L-type ligands we’ve seen so far, π systems most often play an important role in the reactivity of the OM complexes of which they are a part (since they act in reactions, they’re called “actors”). π Systems do useful chemistry, not just with the metal center, but also with other ligands and external reagents. Thus, in addition to thinking about how π systems affect the steric and electronic properties of the metal center, we need to start considering the metal’s effect on the ligand and how we might expect the ligand to behave as an active participant in reactions. To the extent that structure determines reactivity—a commonly repeated, and extremely powerful maxim in organic chemistry—we can think about possibilities for chemical change without knowing the elementary steps of organometallic chemistry in detail yet.
13.05: Bonding between Metal Atoms and Organic Pi Systems
General Properties
The π bonding orbitals of alkenes, alkynes, carbonyls, and other unsaturated compounds may overlap with dσ orbitals on metal centers. This is the classic ligand HOMO → metal LUMO interaction that we’ve beaten into the ground over the last few posts. Because of this electron donation from the π system to the metal center, coordinated π systems often act electrophilic, even if the starting alkene was nucleophilic (the Wacker oxidation is a classic example; water attacks a palladium-coordinated alkene). The π → dσ orbital interaction is central to the structure and reactivity of π-system complexes.
Then again, a theme of the last three posts has been the importance of orbital interactions with the opposite sense: metal HOMO → ligand LUMO. Like CO, phosphines, and NHCs, π systems are often subject to important backbonding interactions. We’ll focus on alkenes here, but these same ideas apply to carbonyls, alkynes, and other unsaturated ligands bound through their π clouds. For alkene ligands, the relative importance of “normal” bonding and backbonding is nicely captured by the relative importance of the two resonance structures in the figure below.
Resonance forms of alkene ligands.
Complexes of weakly backbonding metals, such as the electronegative late metals, are best represented by the traditional dative resonance structure 1. But complexes of strong backbonders, such as electropositive Ti(II), are often best drawn in the metallacyclopropane form 2. Organic hardliners may have a tough time believing that 1 and 2 are truly resonance forms, but several studies—e.g. of the Kulinkovich cyclopropanation—have shown that independent synthetic routes to metallacyclopropanes and alkene complexes containing the same atoms result in the same compound. Furthermore, bond lengths and angles in the alkene change substantially upon coordination to a strongly backbonding metal. We see an elongation of the C=C bond (consistent with decreased bond order) and some pyramidalization of the alkene carbons (consistent with a change in hybridization from sp2 to sp3). A complete orbital picture of “normal” bonding and backbonding in alkenes is shown in the figure below.
Normal bonding and backbonding in alkene complexes.
Here’s an interesting question with stereochemical implications: what is the orientation of the alkene relative to the other ligands? From what we’ve discussed so far, we can surmise that one face of the alkene must point toward the metal center. Put differently, the bonding axis must be normal to the plane of the alkene. However, this restriction says nothing about rotation about the bonding axis, which spins the alkene ligand like a pinwheel. Is a particular orientation preferred, or can we think about the alkene as a circular smudge over time? The figure below depicts two possible orientations of the alkene ligand in a trigonal planar complex. Other orientations make less sense because they would involve inefficient orbital overlap with the metal’s orthogonal d orbitals. Which one is favored?
Two limiting cases for alkene orientation in a trigonal planar complex.
First of all, we need to notice that these two complexes are diastereomeric. They have different energies as a result, so one must be favored over the other. Naive steric considerations suggest that complex 4 ought to be more stable (in most complexes, steric factors dictate alkene orientation). To dig a little deeper, let’s consider any electronic factors that may influence the preferred geometry. We’ve already seen that electronic factors can overcome steric considerations when it comes to complex geometry! To begin, we need to consider the crystal field orbitals of the complex as a whole. Verify on your own that in this d10, Pt(0) complex, the crystal-field HOMOs are the dxy and dx2–y2 orbitals. Where are these orbitals located in space? In the xy-plane! Only the alkene in 3 can engage in efficient backbonding with the metal center. In cases when the metal is electron rich and/or the alkene is electron poor, complexes like 3 can sometimes be favored in spite of sterics. The thought process here is very similar to the one developed in an earlier post on geometry. However, please note that this situation is fairly rare—steric considerations often either match or dominate electronics where alkenes are concerned. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.05%3A_Bonding_between_Metal_Atoms_and_Organic_Pi_Systems/13.5.01%3A_Linear__Systems.txt |
Arene or aromatic ligands are the subject of this post, the second in our series on π-system ligands. Arenes are dative, L-type ligands that may serve either as actors or spectators. Arenes commonly bind to metals through more than two atoms, although η2-arene ligands are known. Structurally, most η6-arenes tend to remain planar after binding to metals. Both “normal” bonding and backbonding are possible for arene ligands; however, arenes are stronger electron donors than CO and backbonding is less important for these ligands. The reactivity of arenes changes dramatically upon metal binding, along lines that we would expect for strongly electron-donating ligands. After coordinating to a transition metal, the arene usually becomes a better electrophile (particularly when the metal is electron poor). Thus, metal coordination can enable otherwise difficult nucleophilic aromatic substitution reactions.
General Properties
The coordination of an aromatic compound to a metal center through its aromatic π MOs removes electron density from the ring. I’m going to forego an in-depth orbital analysis in this post, because it’s honestly not very useful (and overly complex) for arene ligands. π → dσ (normal bonding) and dπ → π* (backbonding) orbital interactions are possible for arene ligands, with the former being much more important, typically. To simplify drawings, you often see chemists draw “toilet-bowl” arenes involving a circle and single central line to represent the π → dσ orbital interaction. Despite the single line, it is often useful to think about arenes as L3-type ligands. For instance, we think of η6-arenes as six-electron donors.
Multiple coordination modes are possible for arene ligands. When all six atoms of a benzene ring are bound to the metal (η6-mode), the ring is flat and C–C bond lengths are slightly longer than those in the free arene. The ring is bent and non-aromatic in η4-mode, so that the four atoms bound to the metal are coplanar while the other π bond is out of the plane. η4-Arene ligands show up in both stable complexes (see the figure below) and reactive intermediates that possess an open coordination site. To generate the latter, the corresponding η6-arene ligand undergoes ring slippage—one of the π bonds “slips” off of the metal to create an open coordination site. We’ll see ring slippage again in discussions of the aromatic cyclopentadienyl and indenyl ligands.
Arene ligands exhibit multiple coordination modes.
Even η2-arene ligands bound through one double bond are known. Coordination of one π bond results in dearomatization and makes η2-benzene behave more like butadiene, and furan act more like a vinyl ether. With naphthalene as ligand, there are multiple η2 isomers that could form; the isomer observed is the one that retains aromaticity in the free portion of the ligand. In fact, this result is general for polycyclic aromatic hydrocarbons: binding maximizes aromaticity in the free portion of the ligand. In the linked reference, the authors even observed the coordination of two different rhodium centers to naphthalene—a bridging arene ligand! Other bridging modes include σ, π-binding (the arene is an LX-type ligand, and one C–M bond is covalent, not dative) and L2-type bridging through two distinct π systems (as in biphenyl).
Arene ligands are usually hydrocarbons, not heterocycles. Why? Aromatic heterocycles, such as pyridine, more commonly bind using their basic lone pairs. That said, a few heterocycles form important π complexes. Thiophene is perhaps the most heavily studied, as the desulfurization of thiophene from fossil fuels is an industrially useful process.
Bonding in Ferrocene
Ferrocene was the first metallocene to be discovered and characterized. The structure of ferrocene is shown below.
Can the stability of the ferrocene structure also be explained by molecular orbital theory? Let us check! First, we need to decide on the point group. We will make the simplification that the two Cp cyclopentadienyl rings are in eclipsed formation, and then the point group is D5h. Actually, the rings are in staggered confirmation and the point group is D5d, but the energy difference is minimal, and there is a very small activation barrier between the two conformers.
\begin{array}{|c|rrrrrrrr|cc|}
\hline \bf{D_{5h}} & E & 2C_5 & 2C_5^2 & 5C_2 & \sigma _h & 2S_5 & 2S_5^2 & 5 \sigma_h & h=20 & \
\hline A_{1}’ & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2, \; z^2\A_{2}’ & 1 & 1 & 1 & -1 & 1 & 1 & 1 & -1 & R_z & \
E_{1}’ & 2 & 2cos(72 ^{\circ}) & 2cos(144 ^{\circ}) & 0 & 2 & 2cos(72 ^{\circ}) & 2cos(144 ^{\circ}) & 0 & (x, \; y) & \
E_{2}’ & 2 & 2cos(144 ^{\circ}) & 2cos(72 ^{\circ}) & 0 & 2 & 2cos(144 ^{\circ}) & 2cos(72 ^{\circ}) & 0 & & (x^2-y^2,\; xy) \
A_{1}” & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & & \
A_{2}” & 1 & 1 & 1 & -1 & -1 & -1 & -1 & 1 & z & \
E_{1}” & 2 & 2cos(72 ^{\circ}) & 2cos(144 ^{\circ}) & 0 & -2 & -2cos(72 ^{\circ}) & -2cos(144 ^{\circ}) & 0 & (R_x,\;R_y) & (xz,\; yz) \
E_{2}” & 2 & 2cos(144 ^{\circ}) & 2cos(72 ^{\circ}) & 0 & -2 & -2cos(144 ^{\circ}) & -2cos(72 ^{\circ}) & 0 & & \
\hline \end{array}
We can define the z-axis standing perpendicular to the Cp rings, and the xy plane to be coplanar with the Cp rings.
The Fe 3d, 4s and 4p orbitals will be our frontier orbitals and we can read their symmetry types from the character table of D5h. The 3dxz and 3dyz have E1’’ symmetry, the 3dx2-y2 and the 3dxy have E2’ symmetry, the 3dz2 has A1’ symmetry, the 4s has A1’ symmetry, and the 4px and the 4py orbitals have E1’ symmetry, and the 4pz has A2’’ symmetry.
Next, we need to determine the ligand group orbitals. We know that the π-ligand molecular orbitals are the ones that donate the electrons into the metal-orbitals, thus we have to have a closer look at these orbitals. The π-MOs of the ligand are made from the five 2pz orbitals of the carbon atoms that stand perpendicular to the ring. This means that there are 5 MOs to consider.
You can see the five MOs and their relative energy above. One is strongly bonding and has no node. Then, there are two double-generate weakly bonding ones that have one node, and finally there are two anti-bonding ones that have two nodes. Because the cyclopentadienyl anion has six π electrons, the bonding MO and the two weakly bonding MOs are full, the anti-bonding MOs are empty.
Because we have two Cp- anions to consider we have overall 10 MOs to combine. We would therefore expect ten ligand group orbitals (LGOs). We can determine their symmetry type by determining the reducible and irreducible representations. The results are shown below.
We can generally divide the LGOs into three groups with a different number of nodes. There are two 0-node LGOs with A1’ and A2’’ symmetry. They are made from the 0-node MOs of the ligand. There is a bonding an anti-bonding combination possible. In the bonding combination the 0–node ligand MOs have the lobes with the same algebraic sign pointing toward each other. The anti-bonding combination has the lobes with opposite algebraic sign pointing toward each other.
There are four 1-node LGOs with E1’ and E1’’ symmetry. They are constructed from the two 1-node ligand MOs. Like for the 0-node orbitals, there is a bonding and an anti-bonding combination possible.
Finally, there are four 2-node LGOs with E2’ and E2’’ symmetry. They are made from the 2-node ligand MOs and there is also a bonding and an anti-bonding combination possible.
The Molecular Orbital Diagram of Ferrocene
Now we have all the information to draw the molecular orbital diagram of ferrocene. As usual we plot the metal frontier orbitals on the left and label their symmetry. We plot the Cp LGOs on the right and also label the symmetry. We can order the LGOs according to energy with the 0-node LGOs having the lowest energy and the 2-node LGOs having the highest energy. Then, we can start to combine orbitals of the same symmetry type to form MOs. There are two metal AOs of the A1’ type and one A1’ LGO giving three MOs of this symmetry type, one bonding, one approximately non-bonding, and third one anti-bonding. We can connect AOs, LGOs, and MOs with dotted lines. Next, we can combine the A2’’ AO and the A2’’ LGO to form a bonding and an anti-bonding MO. We again connect the AOs, LGOs, and MOs with dotted lines. Then, we can produce two bonding e1’’ and two anti-bonding e1’’ MOs from the E1’’ metal AOs and the E1’’ LGOs and connect the orbitals with dotted lines. There are two E1’ LGOs and two E1’ AOs that can be combined to two bonding 1e1’ and two anti-bonding 2e1’ MOs and we again connect the orbitals with dotted lines. The two E2’ LGOs can be combined with the two E2’ d-orbitals to form a pair of bonding 1e2’ and anti-bonding 2e2’ molecular orbitals. Lastly, we notice that the E2’’ LGOs do not find a partner, and we have to write them as non-bonding with the same energy into the MO diagram.
Now we need to fill the electrons into the orbitals. The ligands have twelve electrons overall. They can be filled into the six orbitals of the lowest energy. This fills the 1a1’, the 1a2’’, the 1e1’’, the and 1e1’ orbitals. We notice that all MOs are bonding which supports the stability of the molecule. We still need to consider the metal d-orbitals. We have an Fe2+ ion and thus six metal d-electrons. They would go into the 1e2’ orbital which is bonding and the 2a1’ orbital which is weakly bonding. The 2a1 is the HOMO, and the next higher 2e1’’ is the LUMO. We can see that we can fill all metal electrons into bonding MOs. The LUMO is an anti-bonding orbital, and thus overall all bonding MOs are filled, and no non-bonding and anti-bonding orbitals need to be filled. This is the ideal situation for a stable molecule. We can also see that the MO diagram explains the 18 electron rule. All 18 electrons are in bonding MOs.
We can consider the 1e2’, the 2a1’, and the 2e1’’ metal d-orbitals in the ligand field produced by the Cp-ligands as these 5 orbitals can hold the maximum possible number of 10 d-electrons, have similar energy then the d-orbitals, and have contributions from them.
Metallocenes with other cyclic π-ligands
Is it possible to make metallocenes with other π-conjugated rings but the cylopentadienyl anion?
The answer is yes, for instance benzene is known to act as a ligand bis-(benzene) chromium (0). In this case the ligand acts as a ƞ6-ligand because all six carbon atoms are involved in the bonding. Why does chromium give stable metallocene complex with benzene? We can explain this again with the 18 electron rule. In bis-(benzene) chromium (0), chromium is in the oxidation state 0 because the benzene ligand has no charge. Thus, chromium contributes six electrons. Adding the 12 π-electrons from the two benzene ligands gives 18 electrons.
Other cyclic π-ligands
Also cyclobutadiene can act as a cylic π-ligand in complexes. The cyclobutadiene is is different from the cyclopentadienyl anion and the benzene ligand in two ways. Firstly, it has much more ring strain then the previous two, and secondly it is not an aromatic, but an anti-aromatic ligand. Remember, we have an aromatic ring when there are 4n+2 π-electrons, whereby n is an inter number. This means that rings with two (n=0), six (n=1), and ten (n=2) π-electrons are aromatic. Anti-aromatic rings are those that have 4n electrons, such as four (n=1), eight (n=2) and so forth. Cyclobutadiene has four electrons, and thus it is anti-aromatic. Anti-aromatic rings are less stable than aromatic ones because not all π-electrons are in bonding molecular orbitals. Let us illustrate this by constructing the MO diagram for the π-system of the cyclobutadiene molecule (Figure $25$)
The π-system is made of four carbon atoms contributing a half-filled 2pz orbital each (if we define the plane of the molecule as the xy plane). That makes four p-orbitals with four electrons that give four molecular orbitals. There is a bonding MO with no node, two doubly-degenerate non-bonding ones with one node, and one anti-bonding one with two nodes (Figure $25$). There are four electrons. We can fill two electrons into the bonding MO, but the other two must go into the two non-bonding ones under obedience of Hund’s rule.
Jahn-Teller distortion in cyclobutadiene
One may think that the cyclobutadiene is a square planar molecule because of complete delocalization of the π-electrons, but that is actually not the case. There are two shorter double-bonds and two longer single-bonds, and the shape of the molecules is a rectangle. This means that the two double bond are localized. We can view this effect as a Jahn-Teller distortion.
The distortion of the square to form a rectangle is energetically favorable because it lowers the energy of the two non-bonding electrons. Why? Let us look at the non-bonding orbital 1 first and elongate in x-direction, and squeeze in y-direction (Figure $26$). We can see that we bring the p-orbitals with bonding interactions further apart, and bring those with anti-bonding interactions closer together. This means that the energy of this orbital goes up and the orbital becomes anti-bonding. Let us do the same with non-bonding orbital 2. We see that the opposite happens. The bonding interactions are enhanced and the anti-bonding interactions are weakened. Therefore, this orbital becomes bonding and the energy goes down. We can now fill the two electrons into the bonding MO. We see that we have lowered the energy of the electrons through the distortion.
We could also have squeezed in x-direction and elongated in y-direction. In this case orbital 1 would have become bonding and orbital 2 anti-bonding. However, this distortion is symmetry-equivalent to the previous one, and does not produce a new molecule. Both molecules can be superposed by a simple 90° rotation.
Cyclobutadiene as ƞ4-ligand
Cyclobutadiene (Cb) in its free form is not stable because of the high ring strain and the anti-aromaticity, but complexes with cyclobutadiene as a ligand are stable. We would expect that 18 electron complexes are most stable.
What would be an 18 electron sandwich complex with two cyclobutadiene ligands? Because each cyclobutadiene contributes four electrons, ten electrons would need to come from the metal, and we would expect a nickel cyclobutadienyl complex Ni(Cb)2. This complex is not known, but derivatives are. For example a Ni complex with two tetraphenyl butadiene ligands are known (Figure $27$). Another example is the butadienyl tricarbonyl iron (0) complex. Interestingly, when cyclobutadiene acts as a ƞ4-ligand, then it is not distorted, but square planar. An explanation is that we can formally treat the cyclobutadiene ligand as an Cb2- ligand binding to a metal cation. For instance in FeCb(CO)3 the Fe would be an Fe2+, in the nickel complex, the nickel would be a Ni4+. The Cb2- anion would formally aromatic because it has 4n+2=6 electrons. The additional two electrons would be in the non-bonding orbitals. Because the two non-bonding orbitals are completely filled now, there would be no longer a driving force for the distortion. Note though that is is a formal view only, and there are arguments that speak against this view. One is that the addition of the two electrons to the ligand should further destabilize the ring because the added electrons are non-bonding. Aromaticity would rather be achieved by removing two electrons to form a Cb2+ cation that would have only two electrons in the bonding orbital.
Cyclooctetraene (cot) as a ligand
Also metallocenes with cyclooctatetraene (cot) acting as ƞ8-ligand is known. However, because of the large ring size only metals with large atomic radii, can make metallocenes with this ligand. For example uranium makes a uranocenes with two cyclooctatetraene ligands (Figure $28$).
Like cyclobutadiene, cyclooctatetraene is an anti-aromatic ligand with 4n=8 π-electrons (n=2). In the free cyclooctatetraene molecule is non-planar and the π-electrons are localized (Figure $29$).
In metallocenes however, the ring becomes planar (Figure $28$). One can again formally explain that by assuming an aromatic cot2- ligand that binds to a metal 2+ cation, however one should keep in mind again that this is a formal view and not necessarily reflects the bonding situation in the compound.
Metals having smaller atomic radius can bind to cot in ƞ2, ƞ4, and less commonly in ƞ6-mode. The 18 electron rule holds in most cases. For instance, cot can make a ƞ4-complex in tricarbonyl cyclooctatetraene ruthenium (0) (Figure $30$).
Also two metals can bind to a single cot-ligand. This is for instance realize in μ-cyclooctetraene bis(tricarbonyl ruthenium (0) (Figure $31$).
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.05%3A_Bonding_between_Metal_Atoms_and_Organic_Pi_Systems/13.5.02%3A_Cyclic__systems.txt |
Odd-numbered π systems—most notably, the allyl and cyclopentadienyl ligands—are formally LnX-type ligands bound covalently through one atom (the “odd man out”) and datively through the others. This formal description is incomplete, however, as resonance structures reveal that multiple atoms within three- and five-atom π systems can be considered as covalently bound to the metal. To illustrate the plurality of equally important resonance structures for this class of ligands, we often just draw a curved line from one end of the π system to the other. Yet, even this form is not perfect, as it obscures the possibility that the datively bound atoms may dissociate from the metal center, forming σ-allyl or ring-slipped ligands. What do the odd-numbered π systems really look like, and how do they really behave?
General Properties
Allyls are often actor ligands, most famously in allylic substitution reactions. The allyl ligand is an interesting beast because it may bind to metals in two ways. When its double bond does not become involved in binding to the metal, allyl is a simple X-type ligand bound covalently through one carbon—basically, a monodentate alkyl! Alternatively, allyl can act as a bidentate LX-type ligand, bound to the metal through all three conjugated atoms. The LX or “trihapto” form can be represented using one of two resonance forms, or (more common) the “toilet-bowl” form seen in the general figure above. I don’t like the toilet-bowl form despite its ubiquity, as it tends to obscure the important dynamic possibilities of the allyl ligand.
Can we use FMO theory to explain the wonky geometry of the allyl ligand?
The lower half of the figure above illustrates the slightly weird character of the geometry of allyl ligands. In a previous post on even-numbered π systems, we investigated the orientation of the ligand with respect to the metal and came to some logical conclusions by invoking FMO theory and backbonding. A similar treatment of the allyl ligand leads us to similar conclusions: the plane of the allyl ligand should be parallel to the xy-plane of the metal center and normal to the z-axis. In reality, the allyl plane is slightly canted to optimize orbital overlap—but we can see at the right of the figure above that π2–dxy orbital overlap is key. Also note the rotation of the anti hydrogens (anti to the central C–H, that is) toward the metal center to improve orbital overlap.
Exchange of the syn and anti substituents can occur through σ,π-isomerization followed by bond rotation and formation of the isomerized trihapto form. Notice that the configuration of the stereocenter bearing the methyl group is unaffected by the isomerization! It should be noted that 1,3-disubstituted allyl complexes almost exclusively adopt a syn,syn configuration without danger of isomerization.
The methylene and central C–H simply change places!
Upon deconstruction, the cyclopentadienyl (Cp) ligand yields the aromatic cyclopentadienyl anion, an L2X-type ligand. Cp is normally an η5-ligand, but η3 (LX) and η1 (X) forms are known in cases where the other ligands on the metal center are tightly bound. η1-Cyclopentadienyl ligands can sometimes be fluxional—the metal has the ability to “jump” from atom to atom. Variations on the Cp theme include Cp* (C5Me5) and the monomethyl version (C5H4Me). Cp may be found as a loner alongside other ligands (in “half-sandwich” or “piano-stool” complexes), or paired up with a second Cp ligand in metallocenes. The piano-stool and bent metallocene complexes are most interesting for us, since these have potential for open coordination sites—plain vanilla metallocenes tend to be relatively stable and boring.
Binding modes of Cp and general classes of Cp complexes.
Research for this post has made me appreciate the remarkable electron-donating ability of the cyclopentadienyl (Cp) ligand, which renders its associated metal center electron rich. The LUMO of Cp is high in energy, so the ligand is a weak π-acid and is, first and foremost, a σ-donor. This effect is apparent in the strong backbonding displayed by Cp carbonyl complexes. Despite its strong donating ability, Cp is rarely an actor ligand unless another ligand shakes things up—check out the nucleophilic reactivity of doubly jazzed-up ferrocene for a nice example.
It’s critical to recognize that η5-Cp is a six-electron ligand (!). Because Cp piles on electrons, the numbers of ligands bound to the metal in Cp complexes tend to be lower than we might be used to, especially in bent metallocenes containing two Cp’s. Still, the number of ligands we’d expect on the metal center in these complexes is perfectly predictable. We just need to keep in mind that the resulting complexes are likely to contain 16 or 18 total electrons.
By considering the MCp2 fragment, we can predict the nature of ancillary ligands in bent metallocenes.
Synthesis
Crabtree cleanly divides methods for the synthesis of Cp complexes into those employing Cp+ equivalents, Cp– equivalents, and the neutral diene. Naturally, the first class of reagents are appropriate for electron-rich, nucleophilic complexes, while the second class are best used in conjunction with electron-poor, electrophilic complexes. The figure below provides a few examples.
Methods for the synthesis of Cp complexes. The possibilities are exhausted by anionic, cationic, and neutral Cp equivalents!
Methods for synthesizing allyl complexes can also be classified according to whether the allyl donor is a cationic, anionic, or neutral allyl equivalent. In metal-catalyzed allylic substitution reactions, the allyl donor is usually a good electrophile bearing a leaving group displaced by the metal (an allyl cation equivalent). However, similar complexes may be obtained through oxidative addition of an allylic C–H bond to the metal, as in the synthesis of methallyl palladium chloride dimer below. Transmetalation of nucleophilic allyls from one metal to another, which we can imagine as nucleophilic attacks of an anionic allyl group, are useful when the metal is electron-poor.
Nucleophilic, electrophilic, and neutral allyl donors.
Complexes containing conjugated dienes are also viable precursors to allyl complexes. Migratory insertion of a diene into the M–H bond is a nice route to methyl-substituted allyl complexes, for example. Interestingly, analogous insertions of fulvenes do not appear to lead to Cp complexes, but some funky reductive couplings that yield ansa-metallocenes are known. External electrophilic attack (e.g., protonation) and nucleophilic attack on coordinated dienes also result in allyl complexes. In essence, one double bond of the diene becomes a part of the allyl ligand in the product; the other is used as a “handle” to establish the key covalent bond to the allyl ligand.
Dienes: brave crusaders in the quest for allyl complexes.
Reactions
Since the Cp ligand is typically a spectator, we’ll focus in this section on the reactivity of allyl ligands. As usual, the behavior of the allyl ligand depends profoundly on the nature of its coordinated metal center, which depends in turn on the metal’s ancillary ligands. Avoid “tunnel vision” as you study the examples below—depending on the issue at hand, the ancillary ligands may be as important as (or more important than) the actor ligand.
Nucleophilic allyl complexes react with electrophiles like H+, MeI, X+, and acylium ions to yield cationic alkene complexes.
Attack of electrophiles on nucleophilic allyl complexes. Notice the donating Cp ligand!
In a conceptually related process, nucleophiles can attack allyl ligands bound to electrophilic metals. This process is key in allylic substitution reactions. In some cases, the nucleophile is known to bind to the metal first, then transfer to the allyl ligand through reductive elimination. Alternatively, direct attack on the allyl ligand may occur on the face opposite the metal. The figure below illustrates both possibilities.
External and internal attack of nucleophiles on coordinated allyl ligands.
Migratory insertion of allyl ligands is known, and is analogous to insertions involving alkyl ligands (see the figure below). Movement of the allyl ligand toward the coordination site of the dative ligand is assumed. Of course if oxidative addition to form allyl complexes is possible, reductive elimination of allyl ligands with other X-type ligands is also possible. We’ll hit on these process in more detail in their dedicated posts.
Like alkyl ligands, allyls can migrate onto dative ligands like CO and π bonds.
One final post on σ-complexes will bring this series to a temporary close. In the next post, we’ll look at ligands that, rather surprisingly, bind through their σ-electrons. These “non-classical” ligands behave like other dative ligands we’ve seen before, but are important for many reactions that involve main-group single bonds, such as oxidative addition. Before their discovery, the relevance of σ-ligands for reaction mechanisms went unappreciated by organometallic chemists.
13.5.04: Fullerene Complexes
The $\pi$ systems of fullerenes (eg buckmisterfullerene, $\ce{C60}$) also act as ligands for metal complexes. Most transition metal-fullerene complexes are derived from $C_{60}$. The structure of $\ce{C60[IrCl(CO)(PMe3)2]2}$ is shown in Figure $1$.
As ligands, fullerenes behave similarly to electron-deficient alkenes, and they prefer coordination to electron-rich metal centers (metal ions with softer character like those of the 4d and 5d transition metals). They almost always coordinate in a dihapto ($\eta^2$) fashion. The $\eta^2$ metal binding most often occurs on the junction of two 6-membered rings, as shown for $\ce{[[Ph3P]2Pt]6(\eta^2-C60)}$ on the left of Figure $2$. In (\ce{Ru3(CO)9(C60)}\), the fullerene binds to the triangular face of the cluster as shown on the right in Figure $2$.
Hexahapto and pentahato binding is possible, but is less common that dihapto coordination. Modification of the fullerene with phynyl substituents makes the fullerene a more electron rich ligand so that penta- and hexahapto coordination is more favorable. For example, the pentaphenyl anion, $\ce{C60Ph5^-}$, binds to Fe in a pentahpto ($\eta^5$) fashion, similar to the interactions in ferrocene (Left, Figure $3$).
The first X-ray structure that gave insight into the spherical structure of fullerenes was derived from an oxygen adduct of osmium tetroxide (Right, Figure $3$). | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.05%3A_Bonding_between_Metal_Atoms_and_Organic_Pi_Systems/13.5.03%3A_Odd-numbered__Systems.txt |
• 13.6.1: Metal Alkyls
Metal alkyls feature a metal-carbon σ bond and are usually actor ligands, although some alkyl ligands behave as spectators. Our aim will be to understand the general dependence of the behavior of alkyl ligands on the metal center and the ligand’s substituents. Using this knowledge, we can make meaningful comparisons between related metal alkyl complexes and educated predictions about their likely behavior.
• 13.6.2: Carbenes
Fischer carbenes, Shrock carbenes, and vinylidenes are usually actor ligands, but they may be either nucleophilic or electrophilic, depending on the nature of the R groups and metal. In addition, these ligands present some interesting synthetic problems: because free carbenes are quite unstable, ligand substitution does not cut the mustard for metal carbene synthesis.
• 13.6.3: Carbyne (Alkylidyne) Complexes
• 13.6.4: Carbido and Cumulene Complexes
• 13.6.5: Carbon Wires- Polyyne and Polyene Bridges
13.06: Metal-Carbon Bonds
Part 1:
With this post we finally reach the defining ligands of organometallic chemistry, alkyls. Metal alkyls feature a metal-carbon σ bond and are usually actor ligands, although some alkyl ligands behave as spectators. Our aim will be to understand the general dependence of the behavior of alkyl ligands on the metal center and the ligand’s substituents. Using this knowledge, we can make meaningful comparisons between related metal alkyl complexes and educated predictions about their likely behavior. Because alkyl ligands are central to organometallic chemistry, I’ve decided to spread this discussion across multiple posts. We’ll deal first with the general properties of metal alkyls.
In the Simplifying the Organometallic Complex series, we decomposed the M–C bond into a positively charged metal and negatively charged carbon. This deconstruction procedure is consistent with the relative electronegativities of carbon and the transition metals. It can be very useful for us to imagine metal alkyls essentially as stabilized carbanions—but it’s also important to understand that M–C bonds run the gamut from extremely ionic and salt-like (NaCH3) to essentially covalent ([HgCH3]+). The reactivity of the alkyl ligand is inversely related to the electronegativity of the metal center.
The hybridization of the carbon atom is also important, and the trend here follows the trend in nucleophilicity as a function of hybridization in organic chemistry. sp-Hybridized ligands are the least nucleophilic, followed by sp2 and sp3 ligands respectively.
The history of transition metal alkyls is an intriguing example of an incorrect scientific paradigm. After several unsuccessful attempts to isolate stable metal alkyls, organometallic chemists in the 1920s got the idea that metal-carbon bonds were weak in general. However, later studies showed that it was kinetic instability, not thermodynamic, that was to blame for our inability to isolate metal alkyls. In other words, most metal alkyls are susceptible to decomposition pathways with low activation barriers—the instability of the M–C bond per se is not to blame. Crabtree cites typical values of 30-65 kcal/mol for M–C bond strengths.
What are the major decomposition pathways of metal alkyl complexes? β-hydride elimination is the most common. Thermodynamically, the ubiquity of β-hydride elimination makes sense—M–C bonds run 30-65 kcal/mol, while M–H bonds tend to be stronger. The figure below summarizes the accepted mechanism and requirements of β-hydride elimination. We’ll revisit this fundamental reaction of organometallic complexes in a future post.
Kinetically stable metal alkyl complexes violate one of the requirements for β-hydride elimination. Methyl and neopentyl complexes lack β-hydrogens, violating requirement 1. Tightly binding, chelating ligands may be used to prevent the formation of an empty coordination site, violating requirements 3a and 3b. Titanium complexes are known that violate requirement 4 and eliminate only very slowly—back-donation from the metal to the σ*C–H is required for rapid elimination (see below).
Reductive elimination is a second common decomposition pathway. The alkyl ligand hooks up with a second X-type ligand on the metal, and the metal is reduced by two units with a decrease in the total electron count by two units. I’ve omitted curved arrows here because different mechanisms of reductive elimination are known. We’ll discuss the requirements of reductive elimination in detail in a future post; for now, it’s important to note that the thermodynamic stability of C–X versus that of (M–X + M–C) is a critical driving force for the reaction.
When X = H, reductive elimination is nearly always thermodynamically favorable; thus, isolable alkyl hydride complexes are rare. This behavior is a feature, not a bug, when we consider that hydrogenation chemistry depends on it! On the other hand, when X = halogen reductive elimination is usually disfavored. Reductive elimination of C–C (X = C) can be favored thermodynamically, but is usually slower than the corresponding C–H elimination.
Complexes that cannot undergo β-hydride elimination are sometimes susceptible to α-elimination, a mechanistically similar process that forms a metal carbene. This process is particularly facile when the α-position is activated by an adjacent electron donor (Fischer carbenes are the result).
In some metal alkyl complexes, C–H bonds at the α, β, or even farther positions can serve as electron donors to the metal center. This idea is supported by crystallographic evidence and NMR chemical shifts (the donating hydrogens shift to high field). Such interactions are called agostic interactions, and they resemble an “interrupted” transition state for hydride elimination. Alkyl complexes that cannot undergo β-hydride elimination for electronic reasons (high oxidation state, d0 metals) and vinyl complexes commonly exhibit this phenomenon. The fact that β-hydride elimination is slow for d0 metals—agostic interactions are seen instead—suggests that back-donation from a filled metal orbital to the σ*C–H is important for β-hydride elimination. Here’s an interesting, recent-ish review of agostic interactions.
In the next post in this series, we’ll explore the synthesis of metal alkyl complexes in more detail, particularly clarifying the question: how can we conquer β-hydride elimination?
Part 2:
In this post, we’ll explore the most common synthetic methods for the synthesis of alkyl complexes. In addition to enumerating the reactions that produce alkyl complexes, this post will also describe strategies for getting around β-hydride elimination when isolable alkyl complexes are the goal. Here we go!
Properties of Stable Alkyl Complexes
Stable alkyl complexes must be resistant to β-hydride elimination. In the last post we identified four key conditions necessary for elimination to occur:
1. The β-carbon must bear a hydrogen.
2. The M–C and C–H bonds must be able to achieve a syn coplanar orientation (pointing in the same direction in parallel planes).
3. The metal must bear 16 total electrons or fewer and possess an open coordination site.
4. The metal must be at least d2.
Stable alkyl complexes must violate at least one of these conditions. For example, titanium(IV) complexes lacking d electrons β-eliminate very slowly. The complex below likely also benefits from chelation (see below).
Complexes have been devised that are unable to achieve the syn coplanar orientation required for elimination, or that lack β-hydrogens outright. A few examples are provided below—one has to admire the cleverness of the researchers who devised these complexes. The metallacyclobutane is particularly striking!
Using tightly binding, chelating ligands or a directing group on the substrate, the formation of 16-electron complexes susceptible to β-hydride elimination can be discouraged. Notice how the hyrdrogen-bonding L2 ligands in the central complex below hold the metal center in a death grip.
Finally, it’s worth noting that complexes with an open coordination site—such as 16-electron, square-planar complexes of Ni, Pd, and Pt important for cross-coupling—are susceptible to reactions with solvent or other species at the open site. Bulky alkyl ligands help prevent these side reactions. In the example below, the methyl groups of the o-tolyl ligands extend into the space above and below the square plane, discouraging the approach of solvent molecules perpendicular to the plane.
Many transition metal complexes catalyze (E)/(Z) isomerization and the isomerization of terminal alkenes (α-olefins) to internal isomers via β-hydride elimination. This is a testament to the importance of this process for alkyl complexes. Of course, transient alkyl complexes may appear to be susceptible to β-hydride elimination, but if other processes are faster, elimination will not occur. Thus, the optimization of many reactions involving alkyl complexes as intermediates has involved speeding up other processes at the expense of β-hydride elimination—hydrocyanation is a good example.
Synthesis of Alkyl Complexes
The dominant synthetic methods for alkyl complexes are based on nucleophilic attack, electrophilic attack, oxidative addition, and migratory insertion. The first two methods should be intuitive to the organic chemist; the second two are based on more esoteric (but no less important) reactions of organometallic complexes.
Metals bearing good leaving groups are analogous to organic electrophiles, and are susceptible to nucleophilic attack by organolithiums, Grignard reagents, and other polarized organometallics. These reactions can be viewed as a kind of transmetalation, as the alkyl ligand moves from one metal to another. Electron-withdrawing X-type ligands like –Cl and –Br should jump out as good leaving groups. On the other hand, clean substitution of L-type ligands by anionic nucleophiles is much more rare (anionic complexes would result).
Many anionic metal complexes are nucleophilic enough to attack electrophilic sources of carbon such as alkyl and acyl halides in an electrophilic attack mode. An available lone pair on the metal and open coordination site are prerequisites for this chemistry. The charge on the complex increases by one unit (in effect, negative charge is transferred to the electrophile’s leaving group). We can classify these as oxidative ligation reactions—notice that the oxidation state of the metal increases by two units.
Oxidative addition results in the cleavage of a W–Z bond and placement of two new X-type ligands (–W and –Z) on the metal center, with an increase in the oxidation state of the metal and the total electron count by two units. Organic halides are extremely common substrates for this reaction, the first step in the mechanism of cross-coupling reactions. The oxidized metal complex containing new alkyl and halide ligands is the final product. Notice that two open coordination sites are required (not necessarily simultaneously), the metal center must be amenable to two-electron oxidation, and the number of total electrons of the complex increases by two. In essence, the electrons of the W–Z bond join the complex’s party. Take note that there are many known mechanisms for oxidative addition! We’ll explore these different mechanisms in detail in a future post.
Finally, migratory insertion of unsaturated organic compounds is an important method for the synthesis of certain alkyl complexes, and an important step of organometallic reactions that result in addition across π bonds. Migratory insertion is the microscopic reverse of β-hydride elimination. The clever among you may notice that the use of migratory insertion to synthesize alkyl complexes seems inconsistent with our observation that its reverse is ubiquitous for metal alkyls—shouldn’t equilibrium favor the olefin hydride complex? In many cases this is the case; however, there are some notable exceptions. For example, perfluoroalkyl complexes are exceptionally stable (why?), so the insertion of fluoroalkenes is often favored over elimination.
As we noted above, we can still invoke kinetically stable alkyl complexes as intermediates in reactions provided subsequent steps are faster. In the next post, we’ll examine the general classes of reactions in which alkyl complexes find themselves the major players, focusing on the specific mechanistic steps that involve the alkyl complex (reductive elimination, transmetalation, migratory insertion, and [naturally] β-hydride elimination).
Part 3:
In this last post on alkyl ligands, we’ll explore the major modes of reactivity of metal alkyls. We’ve discussed β-hydride elimination in detail, but other fates of metal alkyls include reductive elimination, transmetallation, and migratory insertion into the M–C bond. In a similar manner to our studies of other ligands, we’d like to relate the steric and electronic properties of the metal alkyl complex to its propensity to undergo these reactions. This kind of thinking is particularly important when we’re interested in controlling the relative rates and/or extents of two different, competing reaction pathways.
Reactions of Metal Alkyl Complexes
Recall that β-hydride elimination is an extremely common—and sometimes problematic—transformation of metal alkyls. Then again, there are reactions for which β-hydride elimination is desirable, such as the Heck reaction. Structural modifications that strengthen the M–H bond relative to the M–C bond encourage β-hydride elimination; the step can also be driven by trapping of the metal hydride product with a base (the Heck reaction uses this idea).
On the flip side, stabilization of the M–C bond discourages elimination and encourages its reverse: migratory insertion of olefins into M–H. Previously we saw the example of perfluoroalkyl ligands, which possess exceptionally stable M–C bonds. The fundamental idea here—that electron-withdrawing groups on the alkyl ligand stabilize the M–C bond—is quite general. Hartwig describes an increase in the “ionic character” of the M–C bond upon the addition of electron-withdrawing groups to the alkyl ligand (thereby strengthening the M–C bond, since ionic bonds are stronger than covalent bonds). Bond energies from organic chemistry bear out this idea to an extent; for instance, see the relative BDEs of Me–Me, Me–Ph, and Me–CCH in this reference. I still find this explanation a little “hand-wavy,” but it serves our purpose, I suppose.
Metal alkyls are subject to reductive elimination, the microscopic reverse of oxidative addition. The metal loses two covalent ligands, its formal oxidation state decreases by two units, total electron count decreases by two units, and an R–X bond forms. Reductive elimination is favorable when the R–X bond in the organic product is more stable than the M–R and M–X bonds in the starting complex (a thermodynamic issue). It should be noted, however, that the kinetics of reductive elimination depend substantially on the steric bulk of the eliminating ligands. Concerted reductive elimination of R–H usually possesses a lower activation energy than R–R elimination.
Transmetalation involves the transfer of an alkyl ligand from one metal to the other. An interesting problem concerns the relative reactivity of metal alkyls toward transmetalation. Assuming similar, uncomplicated ligand sets, which of two metal centers is more likely to hold on to an alkyl ligand? Consider the two situations below.
$\ce{MR + M’ <=> M + M’R} \nonumber$
$\ce{MR + M’R’ <=> MR’ + M’R} \nonumber$
The first is a bona fide transmetalation; the second is really a double replacement reaction. The distinction is rarely drawn in practice, but it’s important! The difference is that in the first case, a single-electron transfer of sorts must take place, while in the second case, no redox chemistry is necessary. Favorability in the first case is governed by the relative reduction potentials of M and M’ (the reaction goes forward when M’ is more easily oxidized than M); in the second case, the relative electropositivities of the metals is key, and other factors like lattice energies may be important. The distinction between transmetalation per se and double replacement explains the paradoxical synthetic sequence in the figure below. In practice, both are called “transmetalation.” See these slides (page 6) for a summarizing reference.
This brings our extended look at metal alkyl complexes to a temporary close. Of course, metal alkyls are everywhere in organometallic chemistry…so seeing them again is pretty much inevitable! The next installment in the Epic Ligand Survey series concerns allyl, cyclopentadienyl, and other odd-membered pi systems. These LnX-type ligands can, like arenes, pile as many as six electrons on the metal center at once. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.06%3A_Metal-Carbon_Bonds/13.6.01%3A_Metal_Alkyls.txt |
In a previous post, we were introduced to the N-heterocyclic carbenes, a special class of carbene best envisioned as an L-type ligand. In this post, we’ll investigate other classes of carbenes, which are all characterized by a metal-carbon double bond. Fischer carbenes, Shrock carbenes, and vinylidenes are usually actor ligands, but they may be either nucleophilic or electrophilic, depending on the nature of the R groups and metal. In addition, these ligands present some interesting synthetic problems: because free carbenes are quite unstable, ligand substitution does not cut the mustard for metal carbene synthesis.
General Properties
Metal carbenes all possess a metal-carbon double bond. That’s kind of a given. What’s interesting for us about this double bond is that there are multiple ways to deconstruct it to determine the metal’s oxidation state and number of d electrons. We could give one pair of electrons to the metal center and one to the ligand, as we did for the NHCs. This procedure nicely illustrates why compounds containing M=C bonds are called “metal carbenoids”—the deconstructed ligand is an L-type carbenoid. Alternatively, we could give both pairs of electrons to the ligand and think of it as an X2-type ligand. The appropriate procedure depends on the ligand’s substituents and the electronic nature of the metal. The figure below summarizes the two deconstruction procedures.
The proper method of deconstruction depends on the electronic nature of the ligand and metal.
When the metal possesses π-acidic ligands and the R groups are π-basic, the complex is best described as an L-type Fischer carbene and the oxidation state of the metal is unaffected by the carbene ligand. When the ligands are “neutral” (R = H, alkyl) and the metal is a good backbonder—that is, in the absence of π-acidic ligands and electronegative late metals—the complex is best described as an X2-type Schrock carbene. Notice that the oxidation state of the metal depends on our deconstruction method! Thus, we see that even the oxidation state formalism isn’t perfect.
Deconstruction reveals the typical behavior of the methylene carbon in each class of complex. The methylene carbon of Schrock carbenes, on which electron density is piled through backbonding, is nucleophilic (the 2– charge screams nucleophilic!). On the other hand, the methylene carbon of Fischer carbons is electrophilic, because backbonding is weak and does not compensate for σ-donation from the ligand to the metal. To spot a Fischer carbene, be on the lookout for reasonable zwitterionic resonance structures like the one at right below.
Thanks to the pi-accepting CO ligands, the metal handles the negative charge well. This is a Fischer carbene.
The clever reader may notice that we haven’t mentioned π-acidic R groups, such as carbonyls. Complexes of this type are best described as Fischer carbenes as well, as the ligand is still electrophilic. However, complexes of this type are difficult to handle and crazy reactive (see below) without a π-basic substituent to hold them in check.
Vinylidenes are the organometallic analogues of allenes, and are best described as intermediate in behavior between Fischer and Schrock carbenes. They are electrophilic at the α carbon and nucleophilic at the β carbon—in fact, a nice analogy can be made between vinylidenes and carbon monoxide. Tautomerization to form alkyne π-complexes is common, as the vinylidene and alkyne complexes are often comparable in stability.
Vinylidene tautomerization, and an analogy between vinylidenes and CO.
Take care when diagnosing the behavior of metal carbenes. In these complexes, there is often a subtle interplay between the R groups on the carbene and other ligands on the metal. In practice, many carbenes are intermediate between the Fischer and Schrock ideals.
Synthesis
Metal carbenes present a fascinating synthetic problem. A cursory look at the deconstruction procedures above reveals that these complexes cannot be made using ligand substitution reactions, because the free ligands are far too unstable. Although the synthetic methods introduced here will be new for us, the attuned organic chemist will find them familiar. The conceptual foundations of metal carbene synthesis are similar to methods for the synthesis of alkenes in organic chemistry.
In the post on NHCs, we saw that the free carbene is both nucleophilic (via the lone pair in its σ system) and electrophilic (via its empty 2pz orbital). Organic precursors to metal carbenes and alkenes also possess this property—they can act both as nucleophiles and electrophiles. Fundamentally, this “ambi-electronic” behavior is useful for the creation of double bonds. One bond comes from “forward flow,” and the other from “reverse flow.” Naturally, the other reacting partner also needs to be ambi-electronic for this method to work.
A fundamental paradigm for double bond synthesis: ambi-electronic compounds doing what they do.
What sets carbene precursors apart from free carbenes? What other kinds of molecules may act as both nucleophiles and electrophiles at the same atom? Watch what happens when we tack a third group onto the free carbene…the figure below shows the result in general and a few specific examples in gray.
A "dative ligand" R' is the difference between a carbene and an ylide. Both are ambi-electronic.
An ylide, which contains adjacent positive and negative charges, results from this purely hypothetical process. Ylides (diazo compounds, specifically) are the most common precursors to metal carbene complexes. Like free carbenes, ylides are ambi-electronic. The electrophilic frontier orbital of an ylide is just the σ* orbital of the bond connecting the charged atoms, which makes sense if we consider the positively charged fragment as a good leaving group (it always is). The lone pair is still nucleophilic. The figure above depicts some of the most famous ylides of organic chemistry, including those used for alkene synthesis (Corey-Chaykovsky and Wittig) and cycloaddition reactions (the carbonyl ylide).
Although diazo compounds are most commonly drawn with charges on the two nitrogen atoms, the diazo carbon is a good nucleophile and can attack electrophilic metal centers to initiate metal carbene formation. A slick 15N kinetic isotope effect study showed that C–N bond cleavage is the rate-limiting step of the reaction below. Visualize the carbanionic resonance structure to kick off the mechanism! Don’t think too hard about the structure of rhodium(II) acetate here. Rhodium, copper, ruthenium, and iridium all form carbene complexes with diazo compounds in a similar way.
After association of the nucleophilic carbon to Rh, elimination with loss of nitrogen gas is the slow step of this reaction.
Diazo compounds work well for metal carbene formation when they possess an electron-withdrawing group, which stabilizes the ylide through conjugation. What about Fischer carbenes, which possess electron-donating groups on the carbene carbon? An interesting method that still involves a “push-and-pull” of electron flow (but not ylides) employs metal-CO complexes. Upon addition of a strong nucleophile (“forward flow”) to the carbonyl carbon, a metalloenolate of sorts is produced. Treatment with an electrophile RX that prefers oxygen over the metal (“reverse flow”) results in an OR-substituted Fischer carbene. Reactions reminiscent of transesterification trade out the OR group for an –SR group (using a thiol) or an –NR2 group (using a secondary amine).
Fischer's classical route to L-type carbenes.
As counterintuitive as it may seem, it’s possible to use metal dianions for the synthesis of Fischer carbenes via a method pioneered by Hegedus and Semmelhack. Potassium intercalated in graphite—the mysterious “KC8“—reduces group 6 metal carbonyl complexes to the corresponding dianions, which subsequently unleash a deluge of electrons on a poor, unsuspecting amide to afford NR2-substituted Fischer carbenes after treatment with trimethylsilyl chloride.
One-directional electron flow for Fischer carbene synthesis: the Hegedus-Semmelhack approach.
For other methods for the synthesis of Fischer carbenes, check out this nice review from the Baran group.
Reactions
In many ways, the reactivity of metal carbenes parallels that of ylides. Olefin metathesis catapulted metal carbenes to international stardom, but in many ways, metathesis is conceptually similar to the Wittig reaction, which employs phosphorus ylides. During both mechanisms, an ylide/carbene hooks up with another doubly bound molecule to form a four-membered ring. This step is followed by what we might call “orthogonal breakdown” to yield two new double bonds.
Wittig, Grubbs, and Schrock: Lords of the Chemical Dance. Ambi-electronic molecules are the dancers!
In my opinion, bond insertion reactions are the most interesting processes in which carbenes regularly engage. Bond insertions may be subdivided into cyclopropanation (π-bond insertion) and σ-bond insertion. Evidence suggests that most cyclopropanations take place by a mechanism that overlaps with metathesis—instead of orthogonal breakdown, reductive elimination occurs to release the three carbon atoms as a cyclopropane.
The metallacyclobutane mechanism of cyclopropanation.
σ-Bond insertion involves electron-rich C–H bonds most prominently, suggesting that electrophilic Fischer carbenes should be best for this chemistry. Fischer carbenes incorporating electron-withdrawing groups love to dimerize to form olefins and/or cyclopropanate olefins—how might we put the brakes on this behavior? If we simply tack a π-basic substituent onto the carbene carbon, the reactivity of these complexes is “just right” for C–H insertion. Notably, no covalent organometallic intermediates are involved; the electrophilic carbene carbon snuggles in between C and H in a single step. The transition state of this step resembles the transfer of a hydride from the organic substrate to the carbene, with “rebound” of electron density toward the partial positive charge.
Donor-acceptor carbenoids are the "Goldilocks complexes" of C–H insertion.
Let’s end with a nod to the similarity between Fischer carbenes and carboxylic acid derivatives (esters and amides). Transesterification-type reactions allow the chemist to swap out heteroatomic substituents on the carbene carbon at will (see above). Alkyl substitutents can even be deprotonated at the α carbon, just like esters! When we see the electronic similarities between the M=C bond of a Fischer carbene and the O=C bond of carboxylic acid derivatives, the similar behavior is only natural. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.06%3A_Metal-Carbon_Bonds/13.6.02%3A_Carbenes.txt |
Learning Objectives
In this lecture you will learn the following
• The metal−ligand multiple bonding and their relevance in.
• The Fischer type carbyne complexes.
• The Schrock type carbyne complexes.
The metal−ligand multiply bonded systems even extended beyond the doubly bonded Fischer and the Schrock carbenes to the triply bonded LnM≡CR type Fischer carbyne and the Schrock carbyne complexes. Similar to carbene that exists in a singlet and a triplet spin state, the carbyne also exists in two other spin states i.e in a doublet and a quartet form.
Upon binding to the metal in its doublet spin state as in the Fischer carbene system, the carbyne moiety donates two electrons via its sp hybridized lone pair containing orbital to an empty metal d orbital to yield a LnM←CR type ligand to metal dative bond. It also makes a covalent π−bond through one of its singly occupied pz orbital with one of the metal d orbitals. The carbyne−metal interaction consist of two ligand to metal interactions namely a dative one and a covalent one that together makes the carbyne moiety a LX type of a ligand. In addition to these two types of ligand to metal bonding interactions, there remains an empty py orbital on the carbyne−C atom that can accommodate electron donation from a filled metal d orbital to give a metal to ligand π−back bonding interaction (Figure $1$).
Analogously, in the quartet carbyne spin state in the Schrock carbyne systems three covalent bonds occur between the singly occupied sp, pyand pz orbitals of carbyne−C moiety with the respective singly occupied metal d orbitals (Figure $1$).
Similar to what has been observed earlier in the case of the Fischer carbenes and Schrock carbenes, the Fischer carbyne complexes are formed with metal centers in lower oxidation states for e.g. as in Br(CO)4W≡CMe, while the Schrock carbyne complexes are formed with metals in higher oxidation state, e.g. as in (t−BuO)3W≡Ct−Bu.
Carbyne complexes can be prepared by the following methods.
i. The Fischer carbyne complexes can be prepared by the electrophilic abstraction of a methoxy group from a methoxy methyl substituted Fischer carbene complex.
$\ce{L(CO)4M=C(OMe)Me + 2BX3 -> [L(CO)4M≡CMe]+BX_{4}^{-} + BX2(OMe) -> X(CO)4M≡CMe} \nonumber$
ii. Schrock carbynes can be prepared by the deprotonation of a α−CH bond of a metal−carbene complex.
$\ce{CPCl2Ta=CHR ->[(i)PMe3][(ii)Ph3P=CH2] Cp(PMe3)ClTa≡CR} \nonumber$
iii. by an α−elimination reaction on a metal−carbene complex
$\ce{Cp*Br2Ta=CHt-Bu ->[(i)dmpe][(ii)Na/Hg] Cp*(dmpe)HTa≡Ct-Bu} \nonumber$
iv. by metathesis reaction
$\ce{(t-BuO)3W≡W(Ot-Bu)3 + t-BuC≡Ct-Bu -> 2(t-BuO)3W≡Ct-Bu} \nonumber$
The reactivities of Fischer and the Schrock carbynes mirror that of the Fischer and Schrock carbenes. For example, the Fischer carbyne undergo nucleophilic attack at the carbyne−C atom while the Schrock carbyne undergo electrophilic attack at the carbyne−C atom.
Summary
The theme of metal−ligand multiple bonding extends beyond the doubly bonded Fischer and the Schrock carbene systems to even triply bonded Fischer and the Schrock carbyne systems. The carbyne moieties in these Fischer and the Schrock carbyne systems respectively exist in a doublet and a quartet spin state. The carbyne complexes are generally prepared from the respective carbene analogues by the abstraction of alkoxy (OR), proton (H+), hydride (H) moieties, the α−elimination reactions and the metathesis reactions. The reactivity of the Fischer and the Schrock carbyne complexes parallel the corresponding Fischer and the Schrock carbene counterparts with regard to their reactivities toward electrophiles and nucleophiles.
13.6.04: Carbido and Cumulene Complexes
Carbido ligand: One single carbon
A metal carbido (aka carbide, or simply carbon) complex is a metal complex that contains the ligand, carbon (C). Most molecular carbido complexes are clusters, usually featuring carbide as a six-fold bridging ligand. Examples include \(\ce{[Rh6C(CO)15]^2-}\), and \(\ce{[Ru6C(CO)16]^2-}\). The iron carbonyl carbides exist not only in the encapsulated carbon (\(\ce{[Fe6C(CO)16]^2-}\)) but also with exposed carbon centres as in \(\ce{Fe5C(CO)15}\) and \(\ce{Fe4C(CO)13}\).
In rare cases, carbido ligands are terminal ligands (Figure \(2\), left). Although it is tempting to extend the model of metal-carbon bonds from the alkyl (\(\ce{M-C}\)), carbene (\(\ce{M=C}\), and carbyne (\(\ce{M \bond{#} C}\)) ligands and evoke the idea of a quadruple bond between metal and carbon in the case of a carbido, experimental data is more consistent with a triple bonded to a metal ion ((\(\ce{M \bond{#} C_1}\))). For example, the metal-carbon bond in the \(\ce{RuC(PCy3)2Cl2}\) complex is similar to the bond length of other complexes with \(\ce{C \bond{3} M}\) (Figure \(2\)).
Cumulene ligands: Chains of carbon atoms with consecutive double bonds
Chains of carbon atoms that have consecutive (cumulated) double bonds are intriguing ligands for metal complexes due to their potential utility as molecular wires in nano or optical devices. The first reported complex containing a vinylidene ligand was \(\ce{Ph2C2Fe2(CO)8}\). The first monometallic vinylidene complex was \(\ce{(C5H5)Mo(P(C6H5)3)(CO)2[C=C(CN)2]Cl}\). Just as in the case of conjugated pi systems, the longer the length of a cumulated diene, the closer the HOMO/LUMO gap, and thus the more conductive. To date, the longest cumulenylidene ligand reported is heptahexaenylidene (Figure \(3\)).
13.6.05: Carbon Wires- Polyyne and Polyene Bridges
Organic molecules with conjugated $\pi$ systems are known to have interesting optical and electronic properties. When ligands with conjugated $\pi$ systems are used to bridge two metal centers, the two metals can behave in a cooperative way because the electronic state of one metal ion is conjugated to the other. The most intensely investigated complexes are bimetallic metal complexes (two metal ions) that are bridged by long carbon chains with alternating single and double bonds. Some examples are given below. (This topic is reviewed in Aguirre-Etcheverry D. O’Hare, Chem. Rev., 2010, 110, 4839, doi: 10.1021/cr9003852)
Polyene
Polyenes are organic molecules with alternating double and single bonds. In other words, they contain a series of consecutive $\ce{(-C= C-)_{n)}$ with $n >1$. Aromatic and cyclic polyenes were discussed in a previous section. One of the longest ligands reported as of 2010 was a $\ce{Ru-(CH)14-Ru}$ complex (shown in Figure $1$).
Source: Aguirre-Etcheverry D. O’Hare, Chem. Rev., 2010, 110, 4839, doi: 10.1021/cr9003852
Polyyne ligands
Polyynes are organic molecules with alternating single and double bonds. In other words, they contain a series of consecutive $\ce{(-C \bond{#} C-)_n}$ with $n >1$.
Organometallic polyynes capped with metal complexes are well characterized. As of the mid-2010s, the most intense research had concerned rhenium (ReCnRe, n=6-20), ruthenium (RuRuCnRuRu, n = 8–20), iron (FeC12CFe), platinum (PtCnPt, n = 16–28), palladium (ArCnPd, n = 6–10), and cobalt (Co3CnCo3, n = 14–26) complexes (Figure $1$). | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.06%3A_Metal-Carbon_Bonds/13.6.03%3A_Carbyne_%28Alkylidyne%29_Complexes.txt |
Learning Objectives
In this lecture you will learn the following
• The characterization techniques of organometallic compounds.
• The NMR analysis of these compounds.
• The IR analysis of these compounds.
• The X-ray single crystal diffraction studies of these compounds.
The characterization of an organometallic complex involves obtaining a complete understanding of the same right from its identification to the assessment of its purity content, to even elucidation of its stereochemical features. Detailed structural understanding of the organometallic compounds is critical for obtaining an insight on its properties and which is achieved based on the structure-property paradigm.
Synthesis and isolation
Synthesis and isolation are two very important experimental protocols in the overall scheme of things of organometallic chemistry and thus these needs to be performed carefully. The isolation of the organometallic compounds is essential for their characterization and reactivity studies. Fortunately, many of the methods of organic chemistry can be used in organometallic chemistry as the organometallic compounds are mostly nonvolatile crystalline solids at room temperature and atmospheric pressure though a few examples of these compounds are known to exist in the liquid [(CH3C5H4Mn(CO)3] and even in the vapor [Ni(CO)4] states. The organometallic compounds are comparatively more sensitive to aerial oxygen and moisture, and because of which the manipulation of these compounds requires stringent experimental skills to constantly provide them with anaerobic environment for their protection. All of these necessities led to the development of the so-called special Schlenk techniques, requiring special glasswares and which in conjunction with a high vacuum line and a dry box allow the lab bench-top manipulation of these compounds. Successful isolation of organometallic compounds naturally points to the need for various spectroscopic techniques for their characterizations and some of the important ones are discussed below.
1H NMR spectroscopy
The 1H NMR spectroscopy is among the extensively used techniques for the characterization of organometallic compounds. Of particular interest is the application of 1H NMR spectroscopy in the characterization of the metal hydride complexes, for which the metal hydride moiety appear at a distinct chemical shift range between 0 ppm to −40 ppm to the high field of tetramethyl silane (TMS). This upfield shift of the metal hydride moiety is attributed to a shielding by metal d−electrons and the extent of the upfield shift increases with higher the dn configuration. Chemical shifts, peak intensities as well as coupling constants from the through-bond couplings between adjacent nuclei like that of the observation of JP-H, if a phosphorous nucleus is present within the coupling range of a proton nucleus, are often used for the analysis of these compounds. The 1H NMR spectroscopy is often successfully employed in studying more complex issues like fluxionality and diastereotopy in organometallic molecules (Figure $1$).
The paramagnetic organometallic complexes show a large range of chemical shifts, for example, (η6−C6H6)2V exhibits proton resonances that extend even up to 290 ppm.
13C NMR spectroscopy
Although the natural abundance of NMR active 13C (I = ½) nuclei is only 1 %, it is possible to obtain a proton decoupled 13C{1H} NMR spectra for most of the organometallic complexes. In addition, the off−resonance 1H decoupled 13C experiments yield 1JC-H coupling constants, which contain vital structural information, and hence are very critical to the 13C NMR spectral analysis. For example, the 1JC-H coupling constants directly correlate with the hybridization of the C−H bonds with sp center exhibiting a 1JC-H coupling constant of ~250 Hz, a sp2 center of 160 Hz and a sp3 center of 125 Hz. Similar to what is seen in 1H NMR, a phosphorous−carbon coupling is also observed in a 13C NMR spectrum with the trans coupling (~100 Hz) being larger than the cis coupling (~10 Hz).
31P NMR spectroscopy
The 31P NMR spectroscopy, which in conjunction with 1H and 13C NMR spectroscopies, is a useful technique in studying the phosphine containing organometallic complexes. The 31P NMR experiments are routinely run under 1H decoupled conditions for simplification of the spectral features that allow convenience in spectral analysis. Thus, for this very reason, many mechanistic studies on catalytic cycle are conveniently undertaken by 31P NMR spectroscopy whenever applicable.
IR spectroscopy
Qualitative to semi-quantitative analysis of organometallic compounds using IR spectroscopy are performed whenever possible. In general the signature stretching vibrations for chemical bonds are more conveniently looked at in these studies. The frequency ($\nu$) of a stretching vibration of a covalent bond is directly proportional to the strength of the bond, usually given by the force constant (k) and inversely proportional to the reduced mass of the system, which relates to the masses of the individual atoms.
$\nu = \frac{1}{2\pi C} \:\sqrt {\frac{k}{m_{r}}} \nonumber$
$m_{r} = \frac{m_{1}m_{2}}{m_{1} + m_{2}} \nonumber$
The organometallic compounds containing carbonyl groups are regularly studied using IR spectroscopy, and in which the $\ce{CO}$ peaks appear in the range between 2100−1700 cm-1 as distinctly intense peaks.
Crystallography
The solid state structure elucidation using single crystal diffraction studies are extremely useful techniques for the characterization of the organometallic compounds and for which the X-ray diffraction and neutron diffraction studies are often undertaken. As these methods give a three dimensional structural rendition at a molecular level, they are of significant importance among the various available characterization methods. The X-ray diffraction technique is founded on Bragg’s law that explains the diffraction pattern arising out of a repetitive arrangement of the atoms located at the crystal lattices.
$2d \sin \theta =n \lambda \nonumber$
A major limitation of the X-ray diffraction is that the technique is not sensitive enough to detect the hydrogen atoms, which appear as weak peaks as opposed to intense peaks arising out of the more electron rich metal atoms, and hence are not very useful for metal hydride compounds. Neutron diffraction studies can detect hydrogens more accurately and thus are good for the analysis of the metal hydride complexes.
Summary
Along with the synthesis, the isolation and the characterization protocols are also integral part of the experimental organometallic chemistry. Because of their air and moisture sensitivities, specialized experimental techniques that succeed in performing the synthesis, isolation and storage of these compounds in an air and moisture-free environment are often used. The organometallic compounds are characterized by various spectroscopic techniques including the 1H NMR, 13C NMR and IR spectroscopies and the X-ray and the neutron diffraction studies. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/13%3A_Organometallic_Chemistry/13.07%3A_Characterization_of_Organometallic_Complexes.txt |
Introduction
This chapter is a survey of common organometallic reactions and catalytic cycles. The reactions of organometallic complexes are divided into two main categories: (1) Reactions that involve gain or loss of ligands, and (2) Reactions involving modification of ligands.
14: Organometallic Reactions and Catalysis
Many reactions of organometallic complexes involve change in coordination number. In a dissociation reaction, one or more ligands are eliminated from the complex and the coordination number decreases. In an association, one or more ligands are added to the complex and the coordination number increases. If the coordination number changes while the metal oxidation state remains constant, the reactions are considered association/addition reactions or dissociation reactions. Reactions can also occur where the coordination number changes while the metal oxidation state also changes. These reactions are considered reductive eliminations or oxidative additions (or a combination of these).
Types of reactions involving gain or loss of ligands
• Association or Addition Reaction: a reaction in which the coordination number increases; metal oxidation state is unchanged.
• Dissociation Reaction: a reaction in which the coordination number decreases; metal oxidation state is unchanged.
• Oxidative Addition: a reaction in which coordination number increases while metal oxidation state increases.
• Reductive Elimination: a reaction in which coordination number decreases while metal oxidation state decreases.
14.01: Reactions Involving Gain or Loss of Ligands
As discussed in Chapter 13, ligand substitution is characterized by a continuum of mechanisms bound by associative (A) and dissociative (D) extremes. At the associative extreme, the first step is an Associative reaction; the incoming ligand forms a bond to the metal and creates an intermediate of higher coordination number. Then the departing ligand takes its lone pair and leaves. At the dissociative extreme, the fist step is a Dissociative reaction and the order of events is opposite; a ligand departs and a new complex with lower coordination number forms. Then, a new ligand associates with the complex in the second step. Sometimes, instead of reacting, the "intermediate" is simply a stable complex with a different coordination number than the reactant.
The most common reactions that are used to create new complexes are Dissociations and Substitutions (involving both Dissociation and Association steps). Associative substitution is common for 16-electron complexes (like $d^8$ complexes of Ni, Pd, and Pt), while both Dissociation and Dissociative Substitution are the norm for 18-electron complexes. Then again, reality is often more complicated than these extremes. In some cases, evidence is available for interchange ($I_a$ or $I_d$).
In this subsection, we’ll explore ligand substitution reactions and mechanisms in the context of organometallic chemistry. We’d like to be able to (a) predict whether a mechanism is likely to be associative or dissociative; (b) propose a reasonable mechanism from given experimental data; and (c) describe the results we’d expect given a particular mechanism. Keep these goals in mind as you learn the theoretical and experimental nuts and bolts of substitution reactions.
Dissociation
Carbonyl compounds are known to undergo dissociation reactions upon heating or irradiation with light. For example, upon heating, the the cyclopentadienylmolybdenum tricarbonyl dimer ($\ce{Cp2Mo2(CO)6}$), where Cp is the cyclopentadienyl ligand) loses a CO ligand from each Mo ion, and a triple bond between the Mo ions is formed (Top, Figure $1$). An example of a photoactivated dissociation of CO is the case of neutral iron pentacarbonyl, $\ce{Fe(CO)5}$. Irradiation of $\ce{Fe(CO)5}$ with UV produces $\ce{Fe(CO)4}$ (Bottom, Figure $1$).
However, in each of the cases shown above, the product complex is reactive toward ligand association, usually resulting in an overall substitution. The cyclopentadienylmolybdenum dicarbonyl dimer shown on the top of Figure $1$ binds a variety of substrates across the metal-metal triple bond. And the iron tetracarbonyl complex shown on the bottom in Figure $1$ readily captures a variety of ligands to result in an overall substitution (example given below); when a ligand doe not associate with it, $\ce{Fe2(CO)9}$ is produced.
Sources: Iron Pentacarbonyl and Cyclopentadienylmolybdenum tricarbonyl dimer articles from Wikipedia.
Dissociative Substitution
Associative substitution is unlikely for saturated, 18-electron complexes—coordination of another ligand would produce a 20-electron intermediate. For 18-electron complexes, dissociative substitution mechanisms involving 16-electron intermediates are more likely. In a slow step with positive entropy of activation, the departing ligand leaves, generating a coordinatively unsaturated intermediate. The incoming ligand then enters the coordination sphere of the metal to generate the product. For the remainder of this post, we’ll focus on the kinetics of the reaction and the nature of the unsaturated intermediate (which influences the stereochemistry of the reaction). The reverse of the first step, re-coordination of the departing ligand (rate constant k–1), is often competitive with dissociation.
Reaction Kinetics
Let’s begin with the general situation in which $k_1$ and $k_{–1}$ are similar in magnitude. Since $k_1$ is rate limiting, $k_2$ is assumed to be much larger than $k_1$ and $k_{–1}$. Most importantly, we need to assume that variation in the concentration of the unsaturated intermediate is essentially zero. This is called the steady state approximation, and it allows us to set up an equation that relates reaction rate to observable concentrations Hold onto that for a second; first, we can use step 2 to establish a preliminary rate expression.
$\text{rate} = k_2[L_nM–◊][Li] \tag{1}$
Of course, the unsaturated complex is present in very small concentration and is unmeasurable, so this equation doesn’t help us much. We need to remove the concentration of the unmeasurable intermediate from (1), and the steady state approximation helps us do this. We can express variation in the concentration of the unsaturated intermediate as (processes that make it) minus (processes that destroy it), multiplying by an arbitrary time length to make the units work out. All of that equals zero, according to the SS approximation. The painful math is almost over! Since Δt must not be zero, the other factor, the collection of terms, must equal zero.
$Δ[L_nM–◊] = 0 = (k1[L_nM–L^d] – k–1[L_nM–◊][L^d] – k_2[L_nM–◊][Li])Δt \tag{2}$
$0 = k_1[L_nM–Ld] – k_{-1}[L_nM–◊][Ld] – k_2[L_nM–◊][Li] \tag{3}$
Rearranging to solve for $[L_nM–◊]$, we arrive at the following.
$[L_nM–◊] = k_1 \dfrac{[L_nM–L_d]}{(k_{-1}[L_d] + k_2[Li]} \tag{4}$
Finally, substituting into equation (1) we reach a verifiable rate equation.
$\text{rate} = k_2k_1 \dfrac{[L_nM–Ld][Li]}{(k_{-1}[L_d] + k_2[Li]} \tag{5}$
When $k_{–1}$ is negligibly small, (5) reduces to the familiar equation (6), typical of dissociative reactions like SN1.
$\text{rate} = k_1[L_nM–L_d] \tag{6}$
Unlike the associative rate law, this rate does not depend on the concentration of incoming ligand. For reactions that are better described by (5), we can drown the reaction in incoming ligand to make $k_2[Li]$ far greater than $k_{-1}[Ld]$, essentially forcing the reaction to fit equation (6).
The Unsaturated Intermediate & Stereochemistry
Dissociation of a ligand from an octahedral complex generates an usaturated ML5 intermediate. When all five of the remaining ligands are L-type, as in Cr(CO)5, the metal has 6 d electrons for a total electron count of 16. The trigonal bipyramidal geometry presents electronic problems (unpaired electrons) for 6 d electrons, as the figure below shows. The orbital energy levels come from crystal field theory. Distortion to a square pyramid or a distorted TBP geometry removes the electronic issue, and so five-coordinate d6 complexes typically have square pyramidal or distorted TBP geometries. This is just the geometry prediction process in action!
When the intermediate adopts square pyramidal geometry (favored for good π-acceptors and σ-donors…why?), the incoming ligand can simply approach where the departing ligand left, resulting in retention of stereochemistry. Inversion is more likely when the intermediate is a distorted trigonal bipyramid (favored for good π-donors). As we’ve already seen for associative substitution, fluxionality in the five-coordinate intermediate can complicate the stereochemistry of the reaction.
Encouraging Dissocative Substitution
In general, introducing structural features that either stabilize the unsaturated intermediate or destabilize the starting complex can encourage dissociative substitution. Both of these strategies lower the activation barrier for the reaction. Other, quirky ways to encourage dissociation include photochemical methods, oxidation/reduction, and ligand abstraction.
Let’s begin with features that stabilize the unsaturated intermediate. Electronically, the intermediate loves it when its d electron count is nicely matched to its crystal field orbitals. As you study organometallic chemistry, you’ll learn that there are certain “natural” d electron counts for particular geometries that fit well with the metal-centered orbitals predicted by crystal field theory. Octahedral geometry is great for six d electrons, for example, and square planar geometry loves eight d electrons. Complexes with “natural” d electron counts—but bearing one extra ligand—are ripe for dissociative substitution. The classic examples are d8 TBP complexes, which become d8 square planar complexes (think Pt(II) and Pd(II)) upon dissociation. Similar factors actually stabilize starting 18-electron complexes, making them less reactive in dissociative substitution reactions. d6 octahedral complexes are particularly happy, and react most slowly in dissociative substitutions. The three most common types of 18-electron complexes, from fastest to slowest at dissociative substitution, are:
$d^8$ TBP > $d^10$ tetrahedral > $d^6$ octahedral
Destabilization of the starting complex is commonly accomplished by adding steric bulk to its ligands. Naturally, dissociation relieves steric congestion in the starting complex. Chelation has the opposite effect, and tends to steel the starting complex against dissociation.
I plan to cover the “quirky” methods in a post of their own, but these include strategies like N-oxides for CO removal, photochemical cleavage of the metal–departing ligand bond, and the use of silver cation to abstract halide ligands. Oxidation and reduction can also be used to encourage substitution: 17- and 19-electron complexes are much more reactive toward substitution than their 18-electron analogues.
Associative Substitutions
Despite the sanctity of the 18-electron rule in organometallic chemistry, a wide variety of stable complexes possess fewer than 18 total electrons at the metal center. Perhaps the most famous examples of these complexes are 14- and 16-electron complexes of group 10 metals (Ni, Pt, Pd). Ligand substitution in complexes of this class typically occurs via an associative mechanism, involving approach of the incoming ligand to the complex before departure of the leaving group. If we keep this principle in mind, it seems easy enough to predict when ligand substitution is likely to be associative. But how can we spot an associative mechanism in experimental data, and what are some of the consequences of this mechanism?
The prototypical mechanism of associative ligand substitution. The first step is rate-determining. A typical mechanism for associative ligand substitution is shown above. It should be noted that square pyramidal geometry is also possible for the intermediate, but is less common. Let’s begin with the kinetics of the reaction.
Reaction Kinetics for Associative Reactions
Reaction kinetics are commonly used to elucidate organometallic reaction mechanisms, and ligand substitution is no exception. Different mechanisms of substitution may follow different rate laws, so plotting the dependence of reaction rate on concentration often allows us to distinguish mechanisms. Associative substitution’s rate law is analogous to that of the SN2 reaction—rate depends on the concentrations of both starting materials.
$L_nM–L^d + L_i → L_nM–L_i + L^d$
$\dfrac{d[L_nM–L^i]}{dt} = rate = k_1[L_nM–L^d][L^i]$
The easiest way to determine this rate law is to use pseudo-first-order conditions. Although the rate law is second order overall, if we could somehow render the concentration of the incoming ligand unchanging, the reaction would appear first order. The observed rate constant under these conditions reflects the constancy of the incoming ligand’s concentration ($k_{obs} = k_1[L^i]$, where both $k_1$ and $[Li]$ are constants). How can we make the concentration of the incoming ligand invariant, you ask? We can drown the reaction in ligand to achieve this. The teensy weensy bit actually used up in the reaction has a negligible effect on the concentration of the “sea” of starting ligand we began with. The observed rate is equal to $k_{obs}[L_nM–L^d]$, as shown by the purple trace below. By determining $k_{obs}$ at a variety of $[L^i]$ values, we can finally isolate $k_1$, the rate constant for the slow step. The red trace below at right shows the idea.
In many cases, the red trace ends up with a non-zero y-intercept…curious, if we limit ourselves to the simple mechanism shown in the first figure of this post. A non-zero intercept suggests a more complex mechanism. We need to add a new term (called $k_s$ for reasons to become clear shortly) to our first set of equations:
$rate = (k_1[L_i] + k_s)[L_nM–L^d]$
$k_{obs} = k_1[L_i] + k_s$
The full rate law suggests that some other step (with rate ks[LnM–Ld]) independent of incoming ligand is involved in the mechanism. To explain this observation, we can invoke the solvent as a reactant. Solvent can associate with the complex first in a slow step, then incoming ligand can displace the solvent in a fast step. Solvent concentration doesn’t enter the rate law because, well, it’s drowning the reactants and its concentration undergoes negligible change! An example of this mechanism in the context of Pt(II) chemistry is shown below.
As an aside, it’s worth mentioning that the entropy of activation of associative substitution is typically negative. Entropy decreases as the incoming ligand and complex come together in the rate-determining step. Dissociative substitution shows the opposite behavior: loss of the departing ligand in the RDS increases entropy, resulting in positive entropy of activation.
Stereochemistry of Associative Substitution
As we saw in discussions of the trans effect, the entering and departing ligands both occupy equatorial positions in the trigonal bipyramidal intermediate. Microscopic reversibility is to blame: the mechanism of the forward substitution (displacement of the leaving by the incoming ligand) must be the same as the mechanism of the reverse reaction (displacement of the incoming by the leaving ligand). This can be a confusing point, so let’s examine an alternative mechanism that violates microscopic reversibility.
The figure above shows why a mechanism involving axial approach and equatorial departure (or vice versa) is not possible. The forward and reverse reactions differ, in fact, in both steps. In forward mechanism a, the incoming ligand enters an axial site. But in the reverse reaction, the incoming ligand (viz., the departing ligand in mechanism a) sits on an equatorial site. The second steps of each mechanism differ too—a involves loss of an equatorial ligand, while b involves loss of an axial ligand. Long story short, this mechanism violates microscopic reversibility. And what about a mechanism involving axial approach and axial departure? Such a mechanism is unlikely on electronic grounds. The equatorial sites are more electron rich than the axial sites, and σ bonding to the axial $d_{z^2}$ orbital is expected to be strong. Intuitively, then, loss of ligand from an axial site is less favorable than loss from an equatorial site.
I know what you’re thinking: what the heck does all of this have to do with stereochemistry? Notice that, in the equatorial-equatorial mechanism (first figure of this post), the axial ligands don’t move at all. The configuration of the starting complex is thus retained in the product. Although retention is “normal,” complications often arise because five-coordinate TBP complexes—like other odd-coordinate organometallic complexes—are often fluxional. Axial and equatorial ligands can rapidly exchange through a process called Berry pseudorotation, which resembles the axial ligands “cutting through” a pair of equatorial ligands like scissors (animation!). Fluxionality means that all stereochemical bets are off, since any ligand can feasibly occupy an equatorial site. In the example below, the departing ligand starts out cis to L, but the incoming ligand ends up trans to L.
Associative Substitution in 18-electron Complexes?
Associative substitution can occur in 18-electron complexes if it’s preceded by the dissociation of a ligand. For example, changes in the hapticity of cyclopentadienyl or indenyl ligands may open up a coordination site, which can be occupied by a new ligand to kick off associative substitution. An allyl ligand may convert from its π to σ form, leaving an open coordination site where the π bond left. A particularly interesting case is the nitrosyl ligand—conversion from its linear to bent form opens up a site for coordination of an external ligand.
Summary of Associative Substitution
Associative ligand substitution is common for complexes with 16 total electrons or fewer. The reaction is characterized by a second-order rate law, the possibility of solvent participation, and a trigonal bipyramidal intermediate that is often fluxional. An open coordination site is essential for associative substitution, but such sites are often hidden in the dynamism of 18-electron complexes with labile ligands.
Quirky Substitutions
Over the years, a variety of “quirky” substitution methods have been developed. All of these have the common goal of facilitating substitution in complexes that would otherwise be inert. It’s an age-old challenge: how can we turn a stable complex into something unstable enough to react? Photochemical excitation, oxidation/reduction, and radical chains all do the job, and have all been well studied. We’ll look at a few examples in this post—remember these methods when simple associative or dissociative substitution won’t get the job done.
Photochemical Substitution
Substitution reactions of dative ligands—most famously, CO—may be facilitated by photochemical excitation. We just discussed the photochemically-activated dissociation of CO from ironpentacarbonyl above; that reaction is often used to accomplish substitution of a CO ligand. Other CO complexes also undergo photochemically-activated CO substitution reactions. Two examples are shown below. The first reaction yields only monosubstituted product without ultraviolet light, even in the presence of a strongly donating phosphine.
All signs point to dissociative mechanisms for these reactions (the starting complexes have 18 total electrons each). Excitation, then, must increase the M–L antibonding character of the complex’s electrons; exactly how this increase in antibonding character happens has been a matter of some debate. Originally, the prevailing explanation was that the LUMO bears M–L antibonding character, and excitation kicks an electron up from the HOMO to the LUMO, encouraging cleavage of the M–L bond. A more recent, more subtle explanation backed by calculations supports the involvement of a metal-to-ligand charge-transfer state along with the “classical” ligand-field excited state.
Oxidation/Reduction
Imagine a screaming baby without her pacifier—that’s a nice analogy for an odd-electron organometallic complex. Complexes bearing 17 and 19 total electrons are much more reactive toward substitution than their even-electron counterparts. Single-electron oxidation and reduction (“popping out the pacifier,” if you will) can thus be used to efficiently turn on substitution. As you might expect, oxidation and reduction work best on electron-rich and electron-poor complexes, respectively. The Mn complex in the oxidative example below, for instance, includes a strongly donating MeCp group (not shown).
Reduction works well for electron-poor metal carbonyl complexes, which are happy to accept an additional electron.
There is a two-electron oxidation method that’s also worth knowing: the oxidation of CO with amine oxides. This nifty little method releases carbon dioxide, amine, and an unsaturated complex that may be quenched by a ligand hanging around. The trick is addition to the CO ligand followed by elimination of the unsaturated complex. As the oxidized CO2 and reduced amine float away, the metal complex finds another ligand.
Radical Chain Processes
Atom abstraction from 18-electron complexes produces neutral 17-electron intermediates, which are susceptible to ligand substitution via radical chain mechanisms. The fact that the intermediates are neutral distinguishes these methods from oxidation-based methods. First-row metal hydrides are great for these reactions, owing to their relatively weak M–H bonds. One example is shown below.
After abstraction of the hydrogen atom by initiator, substitution is rapid and may occur multiple times. Propagation begins anew when the substituted radical abstracts hydrogen from the starting material to regenerate the propagating radical and form the product. These quirky methods are nice to have in your back pocket when you’re backed into a synthetic corner—sometimes, conventional associative and dissociative substitution just won’t do the job. In the next post, we’ll press on to oxidative addition.
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An oxidative addition is a reaction in which the coordination number and the oxidation state of the metal ion both increase by 2. In other words, the metal becomes oxidized by two electrons as the coordination number increases by 2 ligands. The microscopic reverse of oxidative addition is reductive elimination (discussed in the next section).
Oxidative addition can occur in a cis- or trans-fashion. A cis-addition means that two ligands are added so that they have cis-orientation to each other after the reaction is complete. A trans-addition implies that two new ligands are in opposite, trans-position after they have been added. We can further distinguish between mononuclear and dinuclear additions. Mononuclear addition involves a complex with a single metal. Dinuclear additions are additions to a metal-metal bond. The addition cleaves the metal-metal bond and adds one ligand to each fragment.
Let's explore a few examples. The first example is a mononuclear cis-addition (Fig. \(1\)). In the reaction we add dihydrogen to a square planar iridium complex. After the reaction there are two additional hydrido ligands, and the coordination number has increased from 4 to 6. We can see that the two hydrido ligands are oriented in cis-fashion relative to each other. We can verify that the addition of the hydrido ligands oxidized the metal by determining the oxidation number of the metal before and after the reaction. There are three neutral ligands and one negatively charged chloro ligand. There is no overall charge at the complex. This means that the oxidation number of Ir is +1. After the reaction, there are the two hydrido ligands that have both a -1 charge. Remember, we get the charge at a ligand by cleaving the metal-ligand bond so that all electrons in the bond get assigned to the ligand. This means the oxidation number must be increased by +2, and is +3 after the reaction.
We can rationalize why the addition occurs in cis-fashion and not in trans-fashion. The two hydrido ligands are the smallest ligands, therefore, steric repulsion is minimized in the complex is minimized when the two hydrido ligands are in cis-position.
The second example is a mononuclear trans-addition (Fig. \(2\)). In this reaction CH3Br is added to the iridium complex. In this case the two new ligands, a CH3 and the Br group are in opposite position. Again, the coordination number increases from four to six. We can again verify that an oxidation has occurred by analyzing the oxidation number of Ir. As previously determined, the oxidation number of iridium in the reactant complex is +1. Let us analyze the product complex. The chloro ligand and the bromo ligand have a 1- charge. In addition to that also the CH3 ligand has a 1- charge assuming that we treat the Ir-C bond as a dative bond. Thus, the oxidation number of Ir must be +3 to give a neutral complex.
The third example is a dinuclear addition (Fig. \(3\)). This addition is neither cis nor trans because it cleaves a metal-metal bond, and the one new ligand is added to each fragment. In the example the Co-Co bond of the dicobalt octacarbonyl gets cleaved upon the reaction with dihydrogen, and two HCo(CO)4 complexes with Co-H bonds form. In this case the coordination number does not increase. Both the reactant and the products have the coordination number 5. We can see that an oxidation of Co has occurred by comparing the oxidation numbers. In the reactant the oxidation number of Co is zero because all CO ligands are neutral and the complex is overall neutral. In the carbonyl hydride product, Co has the oxidation number +1, because the hydrido ligand has a 1- charge.
Oxidative Additions with Intact Ligands
In the previous examples, the ligands that reacted with the complexes lost their integrity. The H-H bond in the H2 molecule cleaved to form the hydrido ligands, and the C-Br bond in CH3Br was cleaved to form separate Br and methyl ligands. Not all ligands lose their integrity upon their addition, in particular when they contain multiple bonds. These ligands are called intact ligands. In these cases the bond order between two atoms in the ligand is decreased, but the bond order is still at least 1. For example, the alkyne ligand shown belo decreases from a carbon-carbon bond order of three to a bond order of two after it binds to Pt.
Examples for intact ligands are are alkenes, alkynes, O2, etc. In an oxidative addition with an alkyne for example, the alkyne binds side-on to the transition metal under the formation of two metal-C single bonds, and the reduction of the bond order within the alkyne from 3 to 2 (Fig. \(4\)). In the shown reaction, a diphenyl ethine molecule adds to tris(triphenyl phosphine) platinum under the formation of two Pt-C single bonds. The C-C triple bond in the ligand becomes a double bond. This reaction occurs under loss of one of the three triphenyl phosphine ligands. The coordination number at the metal increases from 3 to 4. We can again show that the reaction is oxidative by comparing the oxidation numbers of the metal. In the reactant, Pt has the oxidation number 0 because the three triphenyl phosphine ligands have no charge and the complex has overall no charge. After the reaction, the oxidation number of Pt is +2. This is because when we assign the four electrons of the two Pt-C single bonds to the ligand, the ligand becomes a (Ph-C=C-Ph)2- ligand with each C atom carrying a 1- charge. Because the product complex is overall neutral, the oxidation number of Pt is +2.
Nucleophilic Displacement Reactions
Another important reaction is the nucleophilic displacement reaction. This type of reaction can be classified as an oxidative addition although it does not strictly meet the definnition. In these reactions a metal complex (typically an anion) acts as a nucleophile. A ligand is added, but no electrons are added to the complex. The reactions can be extremely useful for the synthesis of organic compounds. The utility results mostly from the fact that an organic electrophile is turned into a nucleophile.
For example, tetracarbonyl ferrate (2-), also known as Collman’s reagent reacts with alkyl halides to form alkyl tetracarbonyl ferrate (-) (Fig. \(5\)). We can check that the addition of the alkyl group did not add any electrons by counting the electrons before and after the reaction. The Fe(CO)42- anion is an 18e complex. Applying the neutral atom method to the electron counting in [R-Fe(CO)4]- means that Fe contributes 8e, the 1- charge adds 1e, the CO ligands 4x2=8 electrons and the alkyl ligand 1e. This gives 8+1+8+1=18 electrons. Also the oxidation number of Fe has not changed upon the addition of the ligand. It is -2 before and after the reaction.
In the alkyltetracarbonyl ferrate (1-) anion, the group R which was formerly an electrophile in R-X, can now act as a nucleophile due to the fact that the Fe-C bond is polarized toward the C-atom. As a nucleophile it can for instance react with protons to form an alkane. It can also react with oxygen to form a carboxylic acid. In the presence of another alkyl halide R’-X it can form a ketone R-C(O)-R’. It can also insert CO into the Fe-C bonding in a so-called migratory insertion reaction, and the reaction product can react with acid to form an aldehyde. With dihalogens X2 it can form acyl chloride. Collman’s reagent cannot only add alkyl halides by also acyl halides to form acyl tetracarbonyl ferrates (-). | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/14%3A_Organometallic_Reactions_and_Catalysis/14.01%3A_Reactions_Involving_Gain_or_Loss_of_Ligands/14.1.02%3A_Oxidative_Addition.txt |
Reductive Elimination Reactions
Reductive elimination is the microscopic reverse of oxidative addition. The reductive elimination can only occur when the two ligands to be eliminated are in cis-position.
For instance the previously discussed oxidative cis-addition of H2 to the iridium complex is reversible, and the reverse is called the reductive elimination (Fig. \(1\)). The cis-orientation of the two hydrido ligands is necessary to form an H-H bond. The reaction can be thought of going along a reaction path in which the Ir-H bonds become gradually larger and and the H-H distance gradually smaller until the H2 molecule is eliminated from the complex.
During reductive elimination, the electrons in the M–X bond head toward ligand Y, and the electrons in M–Y head to the metal. The eliminating ligands are always X-type! On the whole, the oxidation state of the metal decreases by two units, two new open coordination sites become available, and an X–Y bond forms. What does the change in oxidation state suggest about changes in electron density at the metal? As suggested by the name “reductive,” the metal gains electrons. The ligands lose electrons as the new X–Y bond cannot possibly be polarized to both X and Y, as the original M–X and M–Y bonds were. Using these ideas, you may already be thinking about reactivity trends in reductive elimination…hold that thought.
It’s been observed in a number of cases that a ligand dissociates from octahedral complexes before concerted reductive elimination occurs. Presumably, dissociation to form a distorted TBP geometry brings the eliminating groups closer to one another to facilitate elimination.
Square planar complexes may either take on an additional fifth ligand or lose a ligand to form an odd-coordinate complex before reductive elimination. Direct reductive elimination without dissociation or association is possible, too.
Reactivity trends in reductive elimination are opposite those of oxidative addition. More electron-rich ligands bearing electron-donating groups react more rapidly, since the ligands lose electron density as the reaction proceeds. More electron-poor metal centers—bearing π-acidic ligands and/or ligands with electron-withdrawing groups—react more rapidly, since the metal center gains electrons. Sterically bulky ancillary ligands promote reductive elimination since the release of X and Y can “ease” steric strain in the starting complex. Steric hindrance helps explain, for example, why coordination of a fifth ligand to a square planar complex promotes reductive elimination even though coordination increases electron density at the metal center. A second example: trends in rates of reductive eliminations of alkanes parallel the steric demands of the eliminating ligands: C–C > C–H > H–H.
Mechanistic trends for reductive elimination actually parallel trends in mechanisms of oxidative addition, since these two reactions are the microscopic reverse of one another. Non-polar and moderately polar ligands react by concerted or radical mechanisms; highly polarized ligands and/or very electrophilic metal complexes react by ionic (SN2) mechanisms. The thermodynamics of reductive elimination must be favorable in order for it to occur! Most carbon–halogen reductive eliminations, for example, are thermodynamically unfavorable (this has turned out to be a good thing, especially for cross-coupling reactions).
Reductive elimination is an important step in many catalytic cycles—it usually comes near the “end” of catalytic mechanisms, just before product formation. For some catalytic cycles it’s the turnover-limiting step, making it very important to consider! Hydrocyanation is a classic example; in the mechanism of this reaction, reductive elimination of C–CN is the slow step. Electron-poor alkyl ligands, derived from electron-poor olefins like unsaturated ketones, are bad enough at reductive elimination to prevent turnover altogether! Of course, the electronegative CN ligand is not helping things either…how would you design the ancillary ligands L to speed up this step?
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/14%3A_Organometallic_Reactions_and_Catalysis/14.01%3A_Reactions_Involving_Gain_or_Loss_of_Ligands/14.1.03%3A_Reductive_Elimination.txt |
Metal compounds with $d^0$ electron count are able to activate C-H bonds through $\sigma$ bond metathesis reactions. $\sigma$-bond metathesis is a chemical reaction wherein a metal-ligand $\sigma$ bond undergoes metathesis (exchange of parts) with the sigma bond in some other reagent. The reaction is illustrated by the exchange of lutetium(III) methyl complex with a hydrocarbon (R-H) (Figure $1$). The oxidation state of the metal ion does not change during the reaction. This reactivity was first observed by Patricia L. Watson, a researcher at duPont.
The reaction is mainly observed for complexes of metals with $d^0$ electron configuration, e.g. complexes of Sc(III), Zr(IV), Nb(IV), Ta(V), etc. Complexes of the f-block elements also participate, regardless of the number of f-electrons. For metals unsuited for redox, sigma bond metathesis provides a pathway for introducing substituents.
The details of the reaction mechanism are somewhat a matter of debate, however a currently accepted model is that it precedes through a cycloaddition and via a "kite-shaped" transition state (Figure $2$). Indeed, the rate of the reaction is characterized by a highly negative entropy of activation, indicating an ordered transition state.
The reaction attracted much attention because hydrocarbons are normally unreactive substrates, whereas some sigma-bond metatheses are facile. Unfortunately the reaction does not readily allow the introduction of functional groups. It has been suggested that dehydrocoupling reactions proceed via sigma-bond metathesis.
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14.1.05: Application of Pincer Ligands
Pincer ligands are chelating agents that bind tightly to three adjacent coplanar sites of a metal complex. Stoichiometric and catalytic applications of pincer complexes have been studied at an accelerating pace since the mid-1970s.
Pincers often include one anionic, two-electron donor flanked by two neutral, two-electron donor groups. It consists of a rigid, planar backbone, usually consisting of aryl frameworks. The inflexibility of the pincer-metal interaction confers high thermal stability to the resulting complexes. This stability is in part ascribed to the constrained geometry of the pincer, which inhibits ligand exchange and cyclometallation of the pincer. In contrast, cyclometallation is often a significant deactivation process in the case of other types of chelates, in particular limiting their ability to effect C-H bond activation. The pincer ligand also creates a hydrophobic pocket around the reactive coordination site.
There are various types of pincer ligands that are used in transition metal catalysis. Often, they have the same two-electron donor flanking the metal centre, but this is not a requirement. Early examples of pincer ligands were anionic with a carbanion as the central donor site and flanking phosphine donors; these compounds are referred to as PCP pincers. Although the most common class of pincer ligands features PCP donor sets, variations have been developed where the phosphines are replaced by thioethers and tertiary amines. Many pincer ligands also feature nitrogenous donors at the central coordinating group position (see figure), such as pyridines. By altering the properties of the pincer ligands, it is possible to significantly alter the chemistry at the metal centre. Changing the hardness/softness of the donor, using electron-withdrawing groups (EWGs) in the backbone, and the altering the steric constraints of the ligands are all methods used to tune the reactivity at the metal centre.
The Role of Pincer Complexes in Catalysis
Suzuki-Miyaura coupling
Pincer complexes have been shown to catalyse Suzuki-Miyaura coupling reactions, a versatile carbon-carbon bond forming reaction. Suzuki-Miyaura coupling (or Suzuki coupling) is a metal catalyzed reaction, typically with Pd, between an alkenyl (vinyl), aryl, or alkynyl organoborane (boronic acid or boronic ester, or special cases with aryl trifluoroborane) and halide or triflate under basic conditions. This reaction is used to create carbon-carbon bonds to produce conjugated systems of alkenes, styrenes, or biaryl compounds (Scheme 1).
Modified conditions have demonstrated reactivity with less reactive substrates such as alkyl boranes (BR3) or aryl or alkenyl chlorides by amending the base and ligands employed.
Sonogashira Coupling
The Sonogashira reaction (also called the Sonogashira-Hagihara reaction) is the cross coupling of aryl or vinyl halides with terminal alkynes to generate conjugated enynes and arylalkynes (Scheme 1). The reaction typically proceeds in the presence of a palladium(0) catalyst, a copper(I) cocatalyst, and an imine base. Alternative procedures describe Sonogashira coupling reactions performed without the Cu(I) cocatalyst.
Dehydrogenation of alkanes
Alkanes undergo dehydrogenation at high temperatures. Typically this conversion is promoted heterogeneously because typically homogeneous catalysts do not survive the required temperatures (~200 °C) The corresponding conversion can be catalyzed homogeneously by pincer catalysts, which are sufficiently thermally robust. Proof of concept was established in 1996 by Jensen and co-workers. They reported that an iridium and rhodium pincer complex catalyze the dehydrogenation of cyclooctane with a turnover frequency of 12 min−1 at 200 °C. They found that the dehydrogenation was performed at a rate two orders of magnitude greater than those previously reported. The iridium pincer complex was also found to exhibit higher activity than the rhodium complex. This rate difference may be due to the availability of the Ir(V) oxidation state which allows stronger Ir-C and Ir-H bonds.
The homogeneously catalyzed process can be coupled to other reactions such as alkene metathesis. Such tandem reactions have not been demonstrated with heterogeneous catalysts.
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This section will describe organometallic reactions that involve modification of unsaturated ligands.
• 14.2.1: Introduction to Insertion
• 14.2.2: CO Insertions (Alkyl Migration)
• 14.2.3: Migratory Insertion-1,2-Insertions
Insertions of π systems into M-X bonds establish two new σ bonds in one step, in a stereocontrolled manner. As we saw in the last post, however, we should take care to distinguish these fully intramolecular migratory insertions from intermolecular attack of a nucleophile or electrophile on a coordinated π-system ligand. The reverse reaction of migratory insertion, β-elimination, is not the same as the reverse of nucleophilic or electrophilic attack on a coordinated π system.
• 14.2.4: β-Elimination Reactions
In organic chemistry class, one learns that elimination reactions involve the cleavage of a σ bond and formation of a π bond. A nucleophilic pair of electrons (either from another bond or a lone pair) heads into a new π bond as a leaving group departs. This process is called β-elimination because the bond β to the nucleophilic pair of electrons breaks. Transition metal complexes can participate in their own version of β-elimination, and metal alkyl complexes famously do so.
• 14.2.5: Abstraction and Addition
14.02: Reactions Invloving Modification of Unsaturated Ligands
We’ve seen that the metal-ligand bond is generally polarized toward the ligand, making the ligand nucleophilic. The ligand is especially nucleophilic in the case of X-type (anionic) ligands. When a nucleophilic, X-type ligand is positioned cis to a neutral, unsaturated ligand (L) in an organometallic complex, a migratory insertion reaction can occur. In a migratory insertion reaction, an anionic ligand ($\ce{X^-}$) and a neutral unsaturated ligand (L) couple to form a new anionic ligand ($\ce{^- LX}) that is attached to the same metal (Figure \(1$). The overall result is that an M-X bond is broken and M and X are added across a $\pi$ bond of L. There is no change in oxidation state at the metal (unless the ligand is an alkylidene/alkylidyne), but the total electron count of the complex decreases by two during the actual insertion event.
The migratory insertion results in a decrease in coordination number and the creation of an open coordination site on the complex. The open coordination site can be filled by an added ligand (L') (Figure $1$). The reverse of the insertion reaction is called a deinsertion. To drive the chemical equilibrium in the forward (insertion) direction, the reactant neutral ligand (or another neutral ligand) can be added in excess so that the neutral ligand fills the vacancy in the product complex and prevents the reverse (deinsertion) reaction. Typical neutral, unsaturated ligands are CO, olefins, alkynes, carbenes, dioxygen, carbon dioxide, and nitriles. Typical anionic ligands are hydrido, alkyl, aryl, alkoxy, and amido-ligands Figure $1$.
The open site may appear where the unsaturated ligand was or where the X-type ligand was, depending on which group actually moved (see below). It is more common that the anionic X ligand moves (Figure $2$).
We can distinguish between two types of insertions, which differ in the number of atoms in the unsaturated ligand involved in the step. Insertions of CO, carbenes, and other η1 unsaturated ligands are called 1,1-insertions because the X-type ligand moves from its current location on the metal to one spot over, on the atom bound to the metal. η2 ligands like alkenes and alkynes can also participate in migratory insertion; these reactions are called 1,2-insertions because the X-type ligand slides two atoms over, from the metal to the distal atom of the unsaturated ligand (Figure $4$).
This is really starting to look like the addition of M and X across a π bond! However, we should take care to distinguish this completely intramolecular process from the attack of a nucleophile or electrophile on a coordinated π system, which is a different beast altogether. Confusingly, chemists often jumble up all of these processes using words like “hydrometalation,” “carbometalation,” “aminometalation,” etc. Another case of big words being used to obscure ignorance! We’ll look at nucleophilic and electrophilic attack on coordinated ligands in separate posts. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/14%3A_Organometallic_Reactions_and_Catalysis/14.02%3A_Reactions_Invloving_Modification_of_Unsaturated_Ligands/14.2.01%3A_Introduction_to_Insertion.txt |
One of the most common types of migratory insertions is carbonyl insertion (Figure \(3\)). The carbonyl insertion produces an acyl group. Notice in the example below that the complex goes from 18 to 16 total electrons after insertion. A dative ligand comes in to fill the empty coordination site. L can be a neutral or anionic ligand, and it can even be the carbonyl oxygen itself. To drive the reaction, the reaction can be carried out in the presence of a neutral ligand like free CO, which occupies the vacant site.
Reactivity Trends in CO Insertions
Certain conditions must be met for migratory insertion to occur: the two ligands undergoing the process must be cis, and the complex must be stable with two fewer total electrons. Thermodynamically, the formed Y–X and covalent M–Y bonds must be more stable than the broken M–X and dative M–Y bonds for insertion to be favored. When the opposite is true, the microscopic reverse (elimination or deinsertion) will occur spontaneously.
Migratory aptitudes for insertion into CO have been studied extensively, and the general conclusion here is “it’s complicated.” A few ligands characterized by remarkably stable metal-ligand bonds don’t undergo insertion for thermodynamic reasons—the M–X bond is just too darn strong. Perfluoroalkyl complexes and metal hydrides are two notable examples. Electron-withdrawing groups on the X-type ligand, which strengthen the M–X bond, slow down insertion (likely for thermodynamic reasons though…Hammond’s postulate in action).
What factors affect the relative speed (kinetics) of favorable insertions? Sterics is one important variable. Both 1,1- and 1,2-insertions can relieve steric strain at the metal center by spreading out the ligands involved in the step. In 1,2-insertions, the X-type ligand removes itself completely from the metal! Unsurprisingly, then, bulky ligands undergo insertions more rapidly than smaller ligands. Complexes of the first-row metals tend to react more rapidly than analogous second-row metal complexes, and second-row metal complexes react faster than third-row metal complexes. This trend fits in nicely with the typical trend in M–C bond strengths: first row < second row < third row. Lewis acids help accelerate insertions into CO by coordinating to CO and making the carbonyl carbon more electrophilic. For a similar reason, CO ligands bound to electron-poor metal centers undergo insertion more rapidly than CO’s bound to electron-rich metals. Finally, for reasons that are still unclear, one-electron oxidation often increases the rate of CO insertion substantially.
Although the thermodynamics of alkene 1,2-insertion are more favorable for metal-carbon than metal-hydrogen bonds, M–H bonds react much more rapidly than M–C bonds in 1,2-insertions. This fact has been exploited for olefin hydrogenation, which would be much less useful if it had to complete with olefin polymerization (the result of repeated insertion of C=C into M–C) in the same reaction flask! More on that in the next post.
Stereochemistry in CO Insertions
Migratory insertion steps are full of stereochemistry! Configuration at the migrating alkyl group is retainedduring insertion—a nice piece of evidence supporting a concerted, intramolecular mechanism of migration.
What about stereochemistry at the metal center? Migratory insertion may create a stereogenic center at the metal—see the iron example above. Whether the X-type ligand moves onto the unsaturated ligand or vice versa will impact the configuration of the product complex. Calderazzo’s study of this issue is one of my favorite experiments in all of organometallic chemistry! He took the simple labeled substrate in the figure below and treated it with dative ligand, encouraging insertion. Four products of insertion are possible, corresponding to reaction of the four CO ligands cis to the methyl ligand. Try drawing a few curved arrows to wrap your mind around the four possibilities, and consider both CO migration and Me migration as possible at this point.
Note that product D is impossible if we only allow the Me group to migrate—the spot trans to the labeled CO is another CO ligand, so that spot can only pick up L if CO migrates (not if Me migrates). On the other hand, product C must have come from the migration of Me, since the Me group has moved from a cis to a transposition relative to the labeled CO in product C. Calderazzo observed products A, B, and C, but not D, supporting a mechanism involving Me migration. Other experiments since support the idea that most of the time, the alkyl group migrates onto CO. Slick, huh?
I won’t address insertions into alkylidenes, alkylidynes, and other one-atom unsaturated ligands in this post, as insertions into CO are by far the most popular 1,1 insertions in organometallic chemistry. In the next post, we’ll dig more deeply into 1,2-insertions of alkenes and alkynes. Thanks for reading! | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/14%3A_Organometallic_Reactions_and_Catalysis/14.02%3A_Reactions_Invloving_Modification_of_Unsaturated_Ligands/14.2.02%3A_CO_Insertions_%28Alkyl_Migration%29.txt |
Insertions of $\pi$ systems into M-X bonds are appealing in the sense that they establish two new $\sigma$ bonds in one step, in a stereocontrolled manner. As we saw in the last post, however, we should take care to distinguish these fully intramolecular migratory insertions from intermolecular attack of a nucleophile or electrophile on a coordinated $\pi$-system ligand. The reverse reaction of migratory insertion, $\beta$-elimination, is not the same as the reverse of nucleophilic or electrophilic attack on a coordinated $\pi$ system.
Like 1,1-insertions, 1,2-insertions generate a vacant site on the metal, which is usually filled by external ligand. For unsymmetrical alkenes, it’s important to think about site selectivity: which atom of the alkene will end up bound to metal, and which to the other ligand? To make predictions about site selectivity we can appeal to the classic picture of the M–X bond as M + X. Asymmetric, polarized $\pi$ ligands contain one atom with excess partial charge; this atom hooks up with the complementary atom in the M–R bond during insertion. Resonance is our best friend here!
A nice study by Yu and Spencer illustrates these effects in homogeneous palladium- and rhodium-catalyzed hydrogenation reactions. Unactivated alkenes generally exhibit lower site selectivity than activated ones, although steric differences between the two ends of the double bond can promote selectivity.
Reactivity Trends in 1,2-Insertions
The thermodynamics of 1,2-insertions of alkenes depend strongly on the alkene, but we can gain great insight by examining the structure of the product alkyl. Coordinated alkenes that give strong metal-alkyl bonds after migratory insertion tend to undergo the process. Hence, electron-withdrawing groups, such as carbonyls and fluorine atoms, tend to encourage migratory insertion—remember that alkyl complexes bearing these groups tend to have stable M–C bonds.
Insertions of alkenes into both M–H and M–R (R = alkyl) are favored thermodynamically, but the kinetics of M–R insertion are much slower. This observation reflects a pervasive trend in organometallic chemistry: M–H bonds react more rapidly than M–R bonds. The same is true of the reverse, $\beta$-elimination. Even in cases when both hydride and alkyl elimination are thermodynamically favored, $\beta$-hydride elimination is much faster. Although insertion into M–R is relatively slow, this elementary step is critical for olefin polymerizations that form polyalkenes (Ziegler-Natta polymerization). This reaction deserves a post all its own!
As the strength of the M–X bond increases, the likelihood that an L-type $\pi$ ligand will insert into the bond goes down. Hence, while insertions into M–H and M–C are relatively common, insertions into M–N and M–O bonds are more rare. Lanthanides and palladium are known to promote insertion into M–N in some cases, but products with identical connectivity can come from external attack of nitrogen on a coordinated $\pi$ ligand. The diastereoselectivity of these reactions provides mechanistic insight—since migratory insertion is syn (see below), a syn relationship between Pd and N is to be expected in the products of migratory insertion. An anti relationship indicates external attack by nitrogen or oxygen.
Stereochemistry of 1,2-Insertions
1,2-Insertion may establish two stereocenters at once, so the stereochemistry of the process is critical! Furthermore, 1,2-insertions and $\beta$-eliminations are bound by important stereoelectronic requirements. An analogy can be made to the E2 elimination of organic chemistry, which also has strict stereoelectronic demands. For migratory insertion to proceed, the alkene and X-type ligand must be syncoplanar during insertion; as a consequence of this alignment, X and MLn end up on the same face of the alkene after insertion. In other words, insertions into alkenes take place in a syn fashion. Complexes that have difficulty achieving a coplanar arrangement of C=C and M–X undergo insertion very slowly, if at all.
This observation has important implications for $\beta$-elimination, too—the eliminating X and the metal must have the ability to align syn.
Insertions of Other $\pi$ Systems
To close this post, let’s examine insertions into $\pi$ ligands other than alkenes briefly. Insertions of alkynes into metal-hydride bonds are known, and are sometimes involved in reactions that I refer to collectively as “hydrostuffylation”: hydrosilylation, hydroesterification, hydrogenation, and other net H–X additions across the $\pi$ bond. Strangely, some insertions of alkynes yield trans products, even though cis products are to be expected from syn addition of M–X. The mechanisms of these processes involve initial syn addition followed by isomerization to the trans complex via an interesting resonance form. The cis complex is the kinetic product, but it isomerizes over time to the more thermodynamically stable trans complex.
The strongly donating Cp* ligand supports the legitimacy of the zwitterionic resonance form—and suggests that the C=C bond may be weaker than it first appears!
Polyenes can participate in migratory insertion, and insertions of polyenes are usually quite favored because stabilized $\pi$-allyl complexes result. In one mind-bending case, a coordinated arene inserts into an M–Me bond in a syn fashion!
Have you ever stopped to consider that the addition of methyllithium to an aldehyde is a formal insertion of the carbonyl group into the Li–Me bond? It’s true! We can think of these as (very) early-metal “insertion” reactions. Despite this precedent, migratory insertion reactions of carbonyls and imines into late-metal hydride and alkyl bonds are surprisingly hard to come by. Rhodium is the most famous metal that can make this happen—rhodium has been used in complexes for arylation and vinylation, for example. Insertion of X=C into the M–R bond is usually followed by $\beta$-hydride elimination, which has the nifty effect of replacing H in aldehydes and aldimines with an aryl or vinyl group. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/14%3A_Organometallic_Reactions_and_Catalysis/14.02%3A_Reactions_Invloving_Modification_of_Unsaturated_Ligands/14.2.03%3A_Migratory_Insertion-12-Insertions.txt |
In organic chemistry class, one learns that elimination reactions involve the cleavage of a σ bond and formation of a π bond. A nucleophilic pair of electrons (either from another bond or a lone pair) heads into a new π bond as a leaving group departs. This process is called β-elimination because the bond β to the nucleophilic pair of electrons breaks. Transition metal complexes can participate in their own version of β-elimination, and metal alkyl complexes famously do so. Almost by definition, metal alkyls contain a nucleophilic bond—the M–C bond! This bond can be so polarized toward carbon, in fact, that it can promote the elimination of some of the world’s worst leaving groups, like –H and –CH3. Unlike the organic case, however, the leaving group is not lost completely in organometallic β-eliminations. As the metal donates electrons, it receives electrons from the departing leaving group. When the reaction is complete, the metal has picked up a new π-bound ligand and exchanged one X-type ligand for another.
In this post, we’ll flesh out the mechanism of β-elimination reactions by looking at the conditions required for their occurrence and their reactivity trends. Many of the trends associated with β-eliminations are the opposite of analogous trends in 1,2-insertion reactions. A future post will address other types of elimination reactions.
β-Hydride Elimination
The most famous and ubiquitous type of β-elimination is β-hydride elimination, which involves the formation of a π bond and an M–H bond. Metal alkyls that contain β-hydrogens experience rapid elimination of these hydrogens, provided a few other conditions are met.
The complex must have an open coordination site and an accessible, empty orbital on the metal center. The leaving group (–H) needs a place to land. Notice that after β-elimination, the metal has picked up one more ligand—it needs an empty spot for that ligand for elimination to occur. We can envision hydride “attacking” the empty orbital on the metal center as an important orbital interaction in this process.
The M–Cα and Cβ–H bonds must have the ability to align in a syn coplanar arrangement. By “syn coplanar” we mean that all four atoms are in a plane and that the M–Cα and Cβ–H bonds are on the same side of the Cα–Cβ bond (a dihedral angle of 0°). You can see that conformation in the figure above. In the syn coplanar arrangement, the C–H bond departing from the ligand is optimally lined up with the empty orbital on the metal center. Hindered or cyclic complexes that cannot achieve this conformation do not undergo β-hydride elimination. The need for a syn coplanar conformation has important implications for eliminations that may establish diastereomeric olefins: β-elimination is stereospecific. One diastereomer leads to the (E)-olefin, and the other leads to the (Z)-olefin.
The complex must possess 16 or fewer total electrons. Examine the first figure one more time—notice that the total electron count of the complex increases by 2 during β-hydride elimination. Complexes with 18 total electrons don’t undergo β-elimination because the product would end up with 20 total electrons. Of course, dissociation of a loose ligand can produce a 16-electron complex pretty easily, so watch out for ligand dissociation when considering the possibility of β-elimination in a complex. Ligand dissociation may be reversible, but β-Hydride elimination is almost always irreversible.
The metal must bear at least 2 d electrons. Now this seems a bit strange, as the metal has served as nothing but an empty bin for electrons in our discussion so far. Why would the metal center need electrons for β-hydride elimination to occur? The answer lies in an old friend: backbonding. The σ C–H → M orbital interaction mentioned above is not enough to promote elimination on its own; an M → σ* C–H interaction is also required! I’ve said it before, and I’ll say it again: backbonding is everywhere in organometallic chemistry. If you can understand and articulate it, you’ll blow your instructor’s mind.
Other β-Elimination Reactions
The leaving group does not need to be hydrogen, of course, and a number of more electronegative groups come to mind as better candidates for leaving groups. β-Alkoxy and β-amino eliminations are usually thermodynamically favored thanks to the formation of strong M–O and M–N bonds, respectively. These reactions are so favored in β-alkoxyalkyl “complexes” of alkali and alkaline earth metals (R–Li, R–MgBr, etc.) that using these as σ-nucleophiles at carbon is untenable. Such compounds eliminate immediately upon their formation. I had an organic synthesis professor in undergrad who was obsessed with this—using a β-alkoxyalkyl lithium or β-alkoxyalkyl Grignard reagent in a synthesis was a recipe for red ink. β-Haloalkyls were naturally off limits too.
The atom bound to the metal doesn’t have to be carbon. β-Elimination of alkoxy ligands affords ketones or aldehydes bound at oxygen or through the C=O π bond (this step is important in many transfer hydrogenations, and an analogous process occurs in the Oppenauer oxidation). Amido ligands can undergo β-elimination to afford complexes of imines; however, this process tends to be slower than β-alkoxy elimination.
Incidentally, I haven’t seen any examples in which the β atom is not carbon, but would be interested if anyone knows of an example!
Applications of β-Eliminations
As with many concepts in organometallic chemistry, there are two ways to think about applications of β-elimination. One can take either the “inorganic” perspective, which focuses on the metal center, or the “organic” perspective, which focuses on the ligands.
With the metal center in focus, we can recognize that β-hydride elimination has the wonderful side effect of establishing an M–H bond—a feat generally difficult to achieve in a selective manner via oxidative addition of X–H. If the ligand from which the hydrogen came displaced something more electronegative, the whole process represents reduction at the metal center. For example, imagine rhodium(III) chloride is mixed with sodium isopropoxide, NaOCH(CH3)2. The isopropoxide easily displaces chloride, and subsequent β-hydride elimination affords a rhodium hydride, formally reduced with respect to the chloride starting material. See p. 236 of this review for more.
With the ligand in focus, we see that the organic ligand is oxidized in the course of β-hydride elimination. Notice that the metal is reduced and the ligand oxidized! A π bond replaces a σ bond in the ligand, and if the conditions are right, this represents a bona fide oxidation (as opposed to a mere elimination). For example, oxidative addition into a C–H bond followed by β-hydride elimination at a C–H bond next door sets up an alkene where two adjacent C–H bonds existed before, an oxidation process. These dehydrogenation reactions are incredibly appealing in a theoretical sense, but still at an early stage when it comes to scope and practicality.
Summary
We already encountered β-hydride elimination in an earlier series of posts on metal alkyl complexes, where we noted that it’s a very common decomposition pathway for metal alkyls. β-Hydride elimination isn’t all bad, however, as it can be an important step in catalytic reactions that result in the oxidation of organic substrates (dehydrogenations and transfer hydrogenations) and in reactions that reduce metal halides to metal hydrides. The general idea of β-elimination involves the transfer of a leaving group from a ligand to the metal center with simultaneous formation of a π bond in the ligand. β-Elimination requires an open coordination site and at least two d electrons on the metal center, and eliminations of chiral complexes are stereospecific. The leaving group is commonly hydrogen, but need not be—the more electronegative the leaving group, the more favorable the elimination. Stronger π bonds in the product also encourage β-elimination, so eliminations that form carbonyl compounds or imines are common.
In the next post, we’ll explore other types of organometallic elimination reactions, which establish π bonds at different positions in metal alkyl or other complexes. α-Eliminations, for example, establish metal-carbon, -oxygen, or -nitrogen multiple bonds, which are generally difficult to forge through other means
14.2.05: Abstraction and Addition
The nucleophilic and electrophilic addition and abstraction reactions can be viewed as ways of activating a ligand toward reaction with an external reagent. The external reagent reacts directly with the ligand, not with the metal center. The external reagent may be a nucleophile (Lewis Base) or an electrophile (Lewis acid).
Nucleophilic Addition and Abstraction
The reaction of an activated ligand with a nucleophilic reagent is favored if the activated ligand is electron poor. When the metal complex is either cationic and/or has spectator ligands that are are electron withdrawing, the reactive ligand becomes activated by being depleted of electron density. The attack of the nucleophile may result in the formation of a new bond between the nucleophile and the activated unsaturated substrate, in which case it is called nucleophilic addition. Alternatively, the reaction may result in an abstraction of a part or the whole of the activated ligand, in which case it is called the nucleophilic abstraction. Some examples of nucleophilic addition and the abstraction reactions are discussed below.
Nucleophilic Addition
An example of a nucleophilic addition reaction is shown in Figure $1$. The olefin is activated by its interaction with platinum. Pyridine is the nucleophile, and the result is a new $\ce{C-N}$ bond.
Nucleophilic Abstraction
An example of a nucleophilic abstraction reaction is shown in Figure $2$.
Electrophilic Addition and Abstraction
Similar to the nucleophilic addition and abstraction reactions, the electrophilic counterparts of these reactions also exist. An electrophilic attack is favored if the ligand-metal fragment is more electron rich. When the complex is anionic, the metal center is at low oxidation state, and/or when the spectator ligands are electron donating, the ligand becomes activated toward reaction with an electrophile. The reaction of activated ligands with electrophilic substrates may result in the formation of a new bond between the electrophile and the activated unsaturated substrate, in which case it is called electrophilic addition, or may result in an abstraction of a part or the whole of the activated ligand, in which case it is called the electrophilic abstraction.
Electrophilic Addition
An example of an electrophile being added to an activated ligand is shown in Figure $3$. The allylic anion ligand uses its $\pi$ electrons to react with hydrogen ion. This adds a hydrogen to the ligand, converting it from an anionic X-type ligand to a neutral L-type ligand .
Electrophilic Abstraction
Depending on whether the abstraction occurs at the $\alpha$ or $\beta$ position with respect to the metal ion, the reaction is either classified as an $\alpha$-abstraction or a $\beta$-abstraction.
An example of an $\alpha$ electrophilic abstraction reaction is shown in Figure $4$.
Two examples of a $\beta$ electrophilic abstraction are shown in Figure $5$. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/14%3A_Organometallic_Reactions_and_Catalysis/14.02%3A_Reactions_Invloving_Modification_of_Unsaturated_Ligands/14.2.04%3A_-Elimination_Reactions.txt |
One of the most important uses of organometalic complexes is in the catalysis of chemical reactions. In this section, we will apply knowledge of reaction types and the skill of electron counting to understand organometalic catalytic cycles.
14.03: Organometallic Catalysts
Catalytic Deuteration of Benzene
Combinations of oxidative additions and reductive eliminations have many applications in the synthesis of organic molecules using an organometallic reactant.
An example is the deuteration of benzene (Fig. \(1\)). Deuterated benzene is an important solvent in NMR spectroscopy. Industrially, the deuteration is done using a dicyclopentadienyltrihydridotantalum(V) catalyst starting out from benzene and D2. The D2 is provided in excess to drive the chemical equilibrium to the right side.
How does this reaction work mechanistically? This 18e Cp2TaH3 is actually a precatalyst that is in chemical equilibrium with H2 and the 16e dicyclopentadienylhydrido tantalum(III) which is the actual catalyst (Fig. \(2\)). Elimination of H2 from the Ta(V) catalyst is a reductive elimination (RE) and the re-addition of H2 to the Ta(III) species an oxidative addition (OA). In the presence of benzene, the Ta(III) species can oxidatively add benzene to form an 18e Ta(V) complex. This complex can reductively eliminate H2 to form a 16e Cp2Ta(III)-Ph complex. In the presence of deuterium this complex can add D2 oxidatively to form an 18e Cp2D2Ta(V)-Ph complex. This species can then reductively eliminate a monodeuterated benzene molecule under formation of Cp2Ta(III)-D. In the presence of enough D2 this monodeuterated benzene can be further deuterated in subsequent catalytic cycles.
14.3.02: Hydroformylation
Catalytic Olefin Hydroformylation
An important industrial reaction is the catalytic hydroformylation reaction, also known as oxo-process (Figure \(1\)). It was discovered in 1938 by Otto Roelen at BASF. In the hydroformylation reaction an H atom and a formyl group are added to an alkene to form aldehydes. The reaction can produce both branched an linear aldehydes from terminal alkenes, CO, and H2 using a carbonyl hydrides such as HCo(CO)4 as a catalyst. The reaction is performed at about 100°C at a pressure of up to 100 atm.
Mechanism of the Catalytic Olefin Hydroformylation by HCo(CO)4
How does the hydroformylation work mechanistically?
The mechanism is illustrated for the hydroformylation of propene (Fig. \(2\):). The actual catalyst HCo(CO)4 is first formed from its precatalyst Co2(CO)8 in the presence of H2 in a dinuclear oxidative addition reaction. The catalyst can undergo a substitution reaction in which a CO ligand is replaced by the olefin that binds side-on to the cobalt. This species can then undergo a migratory olefin insertion reaction. This leads to a mixture of linear and branched alkyl groups attached to the Co. A new CO ligand can add to the vacant site. The alkyl group can then insert into a carbonyl group in another migratory insert step, and the vacant site can be reoccupied by a new CO molecule. Then, H2 is added in an oxidative addition. This is the slowest and rate-limiting step in the catalytic cycle. From the addition product the aldehyde can then be eliminated in a reductive elimination reaction. Addition of CO regenerates the catalyst, and the catalytic cycle can begin again.
Hydrocarbonylations
After the hydroformylation, a number of other hydrocarbonylations were developed, and industrially deployed.
When hydrogen is replaced by H2O hydrocarboxylations of alkenes lead to carboxylic acids (Fig. \(3\):). With an alcohol instead of H2 hydroalkoxycarbonylcations lead to esters. The employment of amines instead of H2 leads to amides in hydroamidocarbonylation reactions.
14.3.03: Monsanto Acetic Acid Process
The Monsanto Acetic Acid Process
Another carbonylation reaction involving an organometallic catalyst is the Monsanto acetic acid process (Figure \(1\)). It has been introduced by Monsanto in the 1970s for the industrial production of acetic acid from methanol. The reaction involves dual catalysis with HI and [RhI2(CO)2]- as a co-catalysts. How does this reaction work?
In the first step methanol reacts with HI to form methyl iodide. The methyl iodide then reacts with the Rh-catalyst in an oxidative addition reaction in which a methyl and an iodo group are added in trans-fashion to the square-planar Rh-complex to give an octahedral complex. The octahedral complex then undergoes a migratory insertion reaction with CO producing an acyl group and a vacant site. A CO molecule can then add to the vacant site. The acetyl iodide can then be eliminated in a reductive elimination to reform the Rh-catalyst thereby closing the catalytic cycle. The acetyl iodide can then react with methanol to form new methyl iodide and acetic acid. The methyl iodide can start a new catalytic cycle with the Rh-catalyst. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/14%3A_Organometallic_Reactions_and_Catalysis/14.03%3A_Organometallic_Catalysts/14.3.01%3A_Catalytic_Deuteration.txt |
See also Wacker Oxidation (Wikipedia)
The Wacker oxidation refers generally to the transformation of a terminal or 1,2-disubstituted alkene to a ketone through the action of catalytic palladium(II), water, and a co-oxidant. Variants of the reaction yield aldehydes, allylic/vinylic ethers, and allylic/vinylic amines. Because of the ease with which terminal alkenes may be prepared and the versatility of the methyl ketone group installed by the reaction, the Wacker oxidation has been employed extensively in organic synthesis.[1]
Introduction
The stoichiometric conversion of ethylene to acetaldehyde by an acidic, aqueous solution of PdCl2 was discovered over a century ago,[2] but fifty years passed between the discovery of this reaction and the development of a catalytic method. In 1959, researchers at Wacker Chemie reported that a similar transformation takes place in an aqueous, acidic solution of catalytic PdCl2 and a stoichiometric amount of CuCl2 through which oxygen is bubbled (Equ. \(1\)).[3]
Since this initial report, the Wacker process has been widely applied in organic synthesis and has been extended to other classes of substrates and products. To encourage mixing of the organic reactants with the aqueous phase, a co-solvent is generally employed along with water. Dimethylformamide (DMF) is a common choice; when DMF is used as a co-solvent with a stoichiometric amount of CuCl under balloon pressure of oxygen, the reaction is called the "Tsuji-Wacker oxidation."[4] Applications of the Wacker oxidation to organic synthesis typically involve the installation of a methyl ketone moiety, which may subsequently undergo nucleophilic addition or deprotonation to form an enolate.
Mechanism and Stereochemistry
Prevailing Mechanism: Water Nucleophile
The mechanism of the Wacker oxidation has been studied both experimentally and theoretically (Eq. 2). The first step of the Wacker oxidation involves coordination of the the alkene to the palladium center to form π-complex 2. Evidence for this step is provided by the relative sluggishness of electron-poor alkenes, which generally require higher catalyst loadings than unactivated alkenes. Hydroxypalladation then occurs to yield either zwitterionic complex 3 or neutral complex 4 depending on the mode of hydroxypalladation (see below). Studies employing deuterated substrates suggest that β-hydride elimination then occurs to afford enol complex 5, which re-inserts into the Pd-H bond to afford complex 6.[5] Computational studies support the involvement of chloride-assisted deprotonation in the subsequent step,[6] which affords the product and palladium(0). Oxidation of palladium(0) by copper(II) then occurs, regenerating palladium(II) species 1. The role of copper(II) in the mechanism is poorly understood at present.
(2)
The mode of hydroxypalladation is an important issue for the Wacker oxidation. Hydroxypalladation may occur either in a syn fashion through an inner-sphere mechanism or in an anti fashion via nucleophilic attack on the coordinated alkene (Eq. 3). Although the stereocenter in 4 is ultimately destroyed upon elimination, the mode of hydroxypalladation can influence the site selectivity of the reaction. Markovnikov-type addition of water to the more substituted carbon of the alkene forms a methyl ketone, while attack of water at the less substituted position ultimately yields an aldehyde. The mode of hydroxypalladation has been shown to affect the distribution of ketone and aldehyde products, and is also important in stereoselective Wacker cyclization reactions.
(3)
At low concentrations of chloride ion, syn-hydroxypalladation appears to be the norm.[7] Water coordinates to the metal and migratory insertion into the Pd-OH bond occurs to generate 4 directly. At high concentrations of chloride ion, chloride competes with hydroxide for binding at palladium, and thus anti-hydroxypalladation may occur.[8]
Prevailing Mechanism: Alcohol Nucleophiles
Alcohols may also be employed as nucleophiles in the Wacker process. The initial steps of the mechanism are similar to those for the Wacker oxidation with water, but the mechanisms diverge at alkoxy complex 11 (Eq. 4). Elimination of palladium occurs to generate an oxocarbenium ion, which may be captured by solvent to generate an acetal. Small amounts of vinyl ether products support the intermediacy of complex 10.[9]
Enantioselective Variants
When the substrate contains a tethered nucleophile, such as a hydroxyl group or protected amine, the nucleophile may react to form a cyclic, allylic or vinylic ether or amine after β-hydride elimination. This process is known as the "Wacker cyclization," and when the product is allylic, a stereocenter may be established. Chiral ligands have been used to render the Wacker cyclization enantioselective. For example, use of tetra(dihydrooxazoline) ligand 12 and benzoquinone (BQ) as a co-oxidant resulted in formation of allylic ether product in an enantiomeric ratio of 98:2 (Eq. 5). [10]
The Wacker process may also establish a stereocenter in 1,1-disubstituted alkenes via acetal formation. A chiral auxiliary in the alkene has been used successfully to generate chiral acetals with high diastereomeric ratio. The acetal forms at the more electron-deficient site of the alkene (Eq. 6).[11]
Scope and Limitations
Terminal alkenes are the most commonly employed substrates for the Wacker oxidation. Oxygen-containing functionality far from the reactive alkene is well tolerated by the reaction because palladium reacts selectively with the alkene in preference to lone pairs on oxygen. Wacker oxidation followed by intramolecular aldol condensation is a convenient method for the rapid synthesis of carbocycles (Eq. 7).[12]
The original Wacker process and Tsuji-Wacker conditions can present problems for substrates containing acid-sensitive functionality, such as acetals and silyl ethers. Use of copper(II) chloride leads to the generation of strongly acidic hydrochloric acid. To mitigate this issue, a method employing copper(II) acetate has been developed. Some acid-sensitive groups are able to withstand the milder acetic acid generated by this method (Eq. 8).[13] Under similar conditions involving copper(I) chloride, a yield of only 56% was obtained.
Unprotected amines coordinate strongly to palladium and thus cause problems for Wacker oxidations. Amines protected with electron-withdrawing substituents often do not interfere in Wacker oxidations,[14] although they may participate in aza-Wacker cyclizations if appropriately positioned in the substrate (see below).
Tri-substituted alkenes are completely inert to the conditions of the Wacker oxidation; only terminal and 1,1-disubstituted alkenes react. Terminal alkenes are usually oxidized selectively over internal alkenes, except under rather specialized circumstances.[15] Eq. 9 represents a typical case of selectivity for terminal alkenes.[14]
Lewis-basic functionality in the allylic or homoallylic position relative to the reactive alkene may cause unpredictable site selectivity in Wacker oxidations. Under standard Tsuji-Wacker conditions, unprotected allylic alcohols form mixtures of methyl ketone and aldehyde products.[16] The catalyst Pd(Quinox)Cl2 was developed to address this problem; allylic alcohols react selectively in the presence of this catalyst to afford methyl ketones (Eq. 10).[17] tert-Butyl hydroperoxide (TBHP) is used as the oxygen source in this reaction.
Protected, homoallylic alcohols generally form the methyl ketone selectively, but results are more unpredictable for unprotected homoallylic alcohols.[18]
Protected allylic amines generally give higher yields of aldehyde than comparable protected allylic alcohols; these results suggest that coordination of palladium to the Lewis base is responsible for the formation of aldehydes. Using a phthalimide group in the allylic position, the Wacker oxidation may be rendered completely selective for aldehydes (Eq. 11).[19]
Site selectivity in reactions of internal alkenes is low unless an allylic or homoallylic Lewis base is present in the substrate. Electron-withdrawing groups can direct oxidation to the β position of the alkene (Eq. 12), but relatively high catalyst loadings are required.[20]
When an appropriately situated nucleophile is present in the substrate, Wacker cyclization may occur to yield an allylic or vinylic ether (Eq. 13).[21] The aza-Wacker cyclization involves protected amino nucleophiles such as sulfonamides.[22]
When an alcohol is used in stoichiometric or solvent quantities or a diol is present in the substrate, acetal formation may occur. Acetal formation, like ketone formation, generally occurs with Markovnikov selectivity. For example, a Michael acceptor forms an acetal at the β position in the presence of (R,R)-2,4-pentanediol (Eq. 14).[23]
Synthetic Applications
The Wacker oxidation has been used extensively in organic synthesis. A number of methods exist for the installation of terminal alkenes and the methyl ketone products of the reaction can be elaborated easily to more complex compounds. In addition, the reaction is aerobic and operationally simple.
The Wacker oxidation was employed in the course of synthesis of the macrolide elaiolide. Oxidation at two terminal alkenes in a C2-symmetric intermediate afforded a diketone, which was subsequently elaborated to the symmetric natural product (Eq. 15).[24] Notably, the internal alkenes and esters in the intermediate appeared not to affect the oxidation.
A modification of Tsuji-Wacker conditions employing copper(II) acetate was used in the course of a synthesis of hennoxazole A (Eq. 16).[25] After the oxidation, deprotection under acidic conditions led to the intramolecular formation of an acetal present in the target compound.
Comparison to Other Methods
Few true alternatives to the Wacker oxidation (that is, other hydration-oxidation reactions) are known. One rather harsh method involves oxymercuration followed by transmetalation to palladium (Eq. 17).[26]
Experimental Conditions and Procedure
Typical Conditions
Tsuji-Wacker conditions represent a good starting point for experimental conditions for this reaction. Water is mixed with an organic co-solvent (often DMF) and 10 mol% PdCl2, stoichiometric CuCl, and 1 atmosphere of oxygen gas. Copper(I) salts are generally superior to copper(II) salts because the former minimize the concentration of chloride in solution, encouraging syn-hydroxypalladation. Copper(I) is rapidly oxidized to copper(II) by dissolved oxygen; a period of 30 minutes prior to the introduction of the substrate is used to ensure that this oxidation is complete. For situations in which Tsuji-Wacker conditions fail, a wide variety of modifications of the reaction conditions may be employed.
Using high oxygen pressure and a judicious choice of solvent, Wacker oxidations can be carried out without the need for a co-oxidant. Certain ligands also promote direct O2-coupled oxidations.[27]
Peroxides may also be used as the oxygen source in Wacker oxidations; these reactions do not require a co-oxidant. In the Pd(Quinox)/TBHP system, silver hexafluoroantimonate serves as a scavenger of chloride ions.
Experimental Procedure[4]
A 100-mL, 3-necked, round-bottomed flask was fitted with a magnetic stirring bar and a pressure-equalizing dropping funnel containing 1-decene (4.2 g, 30 mmol, 1.0 equiv). The flask was charged with a mixture of PdCl2 (0.53 g, 3 mmol, 0.1 equiv), CuCl (2.97 g, 30 mmol, 1.0 equiv) and aqueous DMF (DMF/H 2O = 7:1, 24 mL). With the other outlets securely stoppered and wired down, an oxygen-filled balloon was placed over one neck, and the mixture was stirred at rt to allow oxygen uptake. After 1 h, the 1-decene (4.2 g, 30 mmol) was added over 10 min, and the solution was stirred vigorously at rt under an oxygen balloon. The color of the solution turned from green to black within 15 min and returned gradually to green. After 24 h, the mixture was poured into cold 3 N HCl (100 mL) and extracted with five, 50 mL portions of ether. The extracts were combined and washed successively with 50 mL of saturated aqueous NaHCO3 solution, 50 mL of brine, and then dried over anhydrous magnesium sulfate. The solvent was removed by evaporation and the residue was distilled using a 15 cm Vigreux column to give 2-decanone as a colorless oil (3.0–3.4 g, 65–73%): bp 43.5 °C (1 mm Hg); IR (neat) 1722 cm–1; 1H NMR (CCl4) δ 2.37 (t, J = 7 Hz, 2H), 2.02 (s, 3H), 0.7–1.8 (m, 15H).
References
1. Michel, B. W.; Steffens, L. D.; Sigman, M. S. Org. React. 2014, 84, 2. (doi: 10.1002/0471264180.or084.02)
2. Phillips, F. C. Am. Chem. J. 1894, 16, 255.
3. Smidt, J.; Hafner, W.; Jira, R.; Sedlmeier, J.; Sieber, R.; Ruttinger, R.; Kojer, H. Angew. Chem. 1959, 71, 176.
4. a b Tsuji, J.; Nagashima, H.; Nemoto, H. Org. Synth. 1984, 62, 9.
5. Smidt, J.; Hafner, W.; Jira, R.; Sieber, R.; Sedlmeier, J.; Sabel, A. Angew. Chem. 1962, 74, 93.
6. Eshtiagh-Hosseini, H.; Beyramabadi, S. A.; Morsali, A.; Housaindokht, M. R. J. Mol. Struct.: THEOCHEM 2010, 941, 138.
7. Henry, P. M. J. Am. Chem. Soc. 1964, 86, 3246.
8. Bäckvall, J. E.; Akermark, B.; Ljunggren, S. O. J. Chem. Soc., Chem. Commun. 1977, 264.
9. Balija, A. M.; Stowers, K. J.; Schultz, M. J.; Sigman, M. S. Org. Lett. 2006, 8, 1121.
10. Zhang, Y. J.; Wang, F.; Zhang, W. J. Org. Chem. 2007, 72, 9208.
11. Hosokawa, T.; Yamanaka, T.; Itotani, M.; Murahashi, S.-I. J. Org. Chem. 1995, 60, 6159.
12. Srikrishna, A.; Kumar, P. P. J. Indian Chem. Soc. 1999, 76, 521.
13. Smith, A. B., III; Cho, Y. S.; Friestad, G. K. Tetrahedron Lett. 1998, 39, 8765.
14. a b Reginato, G.; Mordini, A.; Verrucci, M.; Degl'Innocenti, A.; Capperucci, A. Tetrahedron: Asymmetry 2000, 11, 3759.
15. Ho, T.-L.; Chang, M. H.; Chen, C. Tetrahedron Lett. 2003, 44, 6955.
16. Muzart, J. Tetrahedron 2007, 63, 7505.
17. Michel, B. W.; Camelio, A. M.; Cornell, C. N.; Sigman, M. S. J. Am. Chem. Soc. 2009, 131, 6076.
18. Kang, S.-K.; Jung, K.-Y.; Chung, J.-U.; Namkoong, E.-Y.; Kim, T.-H. J. Org. Chem. 1995, 60, 4678.
19. Weiner, B.; Baeza, A.; Jerphagnon, T.; Feringa, B. L. J. Am. Chem. Soc. 2009, 131, 9473.
20. Yan, B.; Spilling, C. D. J. Org. Chem. 2008, 73, 5385.
21. Mitsudome, T.; Umetani, T.; Nosaka, N.; Mori, K.; Mizugaki, T.; Ebitani, K.; Kaneda, K. Angew. Chem., Int. Ed. 2006, 45, 481.
22. Hegedus, L. S.; McKearin, J. M. J. Am. Chem. Soc. 1982, 104, 2444.
23. Hosokawa, T.; Ataka, Y.; Murahashi, S. Bull. Chem. Soc. Jpn. 1990, 63, 166.
24. Barth, R.; Mulzer, J. Tetrahedron 2008, 64, 4718.
25. Yokokawa, F.; Asano, T.; Shioiri, T. Tetrahedron 2001, 57, 6311.
26. Ellis, J. M.; Crimmins, M. T. Chem. Rev. 2008, 108, 5278.
27. Cornell, C. N.; Sigman, M. S. Org. Lett. 2006, 8, 4117.
Contributors
• This page is based on Organic Reactions. The primary collection of full Organic Reactions chapters is maintained by John Wiley & Sons. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/14%3A_Organometallic_Reactions_and_Catalysis/14.03%3A_Organometallic_Catalysts/14.3.04%3A_Wacker_%28Smidt%29_Process.txt |
Organometallic Compounds as Hydrogenation Catalysts
Migratory insertions play an important role in catalysis.
For example a Rh-catalyst called the Wilkinson catalyst is an effective hydrogenation catalyst for olefins. The mechanism of the hydrogenation involves a combination of oxidative additions, olefin migratory insertions, and reductive eliminations (Fig. \(1\)). Wilkinson’s catalyst is the square planar chloro tris(triphenylphosphine) rhodium(I) complex. This molecule is actually a precatalyst that becomes the actual catalyst when it statistically loses a triphenylphosphine ligand producing chloro bis(triphenylphosphine) rhodium(I). The loss of this ligand is a reversible reaction, and thus the catalyst is in chemical equilibrium with the precatalyst. The actually catalyst is in a second chemical equilibrium with its dimer. The chloro bis(triphenylphosphine) rhodium(I) catalyst can undergo an oxidative addition in the presence of hydrogen to form a trigonal bipyramidal chlorodihydrido bis(triphenylphosphine) (III) rhodium complex. This species is in chemical equilibrium with an octahedral chlorodihydrido tris(triphenyl phosphine) rhodium(III) species that can form due to the presence of free triphenylphosphine ligands in the system. The trigonal bipyramidal species can then add an olefin that binds side-on to the Rh. Because the olefin is in cis-position to the hydride ligand it can undergo an olefin insertion. The Rh-C bond can either form with the first or the second carbon in the carbon chain of the olefin, giving a linear and a branched alkyl complex, respectively. The branched complex can undergo a β-hydride elimination thereby reforming the trigonal bipyramidal Rh-complex, and an olefin. This reaction is a side-reaction because the branched alkyl complex is sterically more crowded than the linear complex. The linear alkyl Rh complex can undergo a reductive elimination to form the linear alkane and the RhCl(PPh3)2 catalyst. This completes the catalytic cycle, and a new cycle can start.
14.3.06: Olefin Metathesis
Olefin Metathesis
Olefin metathesis is a reaction which allows to cut and rearrange C=C double bonds in olefins to make new olefins (Fig. \(1\)). Formally, the carbon-carbon bond of the reactant is cleaved homoleptically and the two carbene fragments are combined in a different way. This reaction is typically an equilibrium reaction, and neither the reactants nor the products are clearly favored. This reaction is catalyzed by molybdenum arylamido carbene complexes or ruthenium carbene complexes.
The former are called Shrock catalysts, and the latter Grubbs catalysts named after their discoverers Richard Shrock and Robert Grubbs who received the Nobel prize for Chemistry in 2005 (Fig. \(2\)). The Schrock catalysts are more active, but also very sensitive to air and water. The Grubbs catalysts, while less active, are less sensitive to air and water.
Olefin metathesis often allows for simpler preparation of olefins compared to other methods. Olefin metathesis is particularly powerful when one olefin product is gaseous because then it can be quite easily removed from the chemical equilibrium by purging. This drives the chemical equilibrium to the right side. An example is the preparation of 5-decene from 1-hexene. Cleavage of the C=C double bond in the hexene leads to C5 and C1 carbene fragments (Fig. \(3\)). The two C1 fragments can combine to form ethylene and the two C5 fragments combine to 5-decene. The ethylene is volatile and can be purged from the reaction system thereby driving the chemical reaction to the right side.
The same principles can also be applied to produce polymers from dienes with two terminal C=C double bonds at the chain ends. This is called acylic diene metathesis (ADMET), Fig. \(4\). For instance the cleavage of the two terminal double bonds in a diene with seven C atoms leads to C1 and C5 fragments. The C1 fragments can combine to form ethylene, and the C5 fragments can combine to make an unsaturated polymer of the type [CH(CH2)3CH]n. Again, the reaction can be driven to the right side by removing the gaseous ethylene from the reaction mixture through purging.
Another variation of olefin metathesis is ring-opening metathesis polymerization (ROMP). It allows to make polymers from strained cycloolefins, for example norbornene. The reaction driving force is the relief of the strain. Because the strain is removed in the polymer, the chemical equilibrium lies far on the right side. The reaction product in norbornene is a polymer with 5-membered rings that are interconnected by ethylene -CH=CH- units (Fig. \(5\)).
The opposite of ROMP is ring-closure metathesis (RCM). RCM allows for the preparation of unstrained rings with C=C double bonds from dienes with C=C double bonds that are five or six carbon atoms apart. This distance is suitable to produce unstrained rings. In the shown example a five-membered ring with a C=C double bond is formed from a diene with terminal C=C double bonds that are five atoms apart.
The Mechanism of Olefin Metathesis
What is the mechanism of olefin metathesis?
In the first step, the alkene adds to the the carbene fragment of the catalyst in a 2+2 cycloaddition reaction to produce an unstable intermediate with a highly strained four-membered ring (Fig. \(7\)). This four membered ring can open to produce the first new alkene product R-CH=CH-R and a metal carbene species. This metal carbene can react with another reactant olefin to form another highly stained 4-ring intermediate via a 2+2 cycloaddition reaction. This ring can then reopen again to produce the second alkene metathesis product, in this case ethylene, and the original catalyst. The regenerated catalyst can then start a new catalytic cycle.
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Heterogeneous Catalysis
In heterogeneous catalysis, the catalyst is in a different phase from the reactants. At least one of the reactants interacts with the solid surface in a physical process called adsorption in such a way that a chemical bond in the reactant becomes weak and then breaks. Poisons are substances that bind irreversibly to catalysts, preventing reactants from adsorbing and thus reducing or destroying the catalyst’s efficiency.
An example of heterogeneous catalysis is the interaction of hydrogen gas with the surface of a metal, such as Ni, Pd, or Pt. As shown in part (a) in Figure $2$, the hydrogen–hydrogen bonds break and produce individual adsorbed hydrogen atoms on the surface of the metal. Because the adsorbed atoms can move around on the surface, two hydrogen atoms can collide and form a molecule of hydrogen gas that can then leave the surface in the reverse process, called desorption. Adsorbed H atoms on a metal surface are substantially more reactive than a hydrogen molecule. Because the relatively strong H–H bond (dissociation energy = 432 kJ/mol) has already been broken, the energy barrier for most reactions of H2 is substantially lower on the catalyst surface.
Figure $1$ shows a process called hydrogenation, in which hydrogen atoms are added to the double bond of an alkene, such as ethylene, to give a product that contains C–C single bonds, in this case ethane. Hydrogenation is used in the food industry to convert vegetable oils, which consist of long chains of alkenes, to more commercially valuable solid derivatives that contain alkyl chains. Hydrogenation of some of the double bonds in polyunsaturated vegetable oils, for example, produces margarine, a product with a melting point, texture, and other physical properties similar to those of butter.
Several important examples of industrial heterogeneous catalytic reactions are in Table $1$. Although the mechanisms of these reactions are considerably more complex than the simple hydrogenation reaction described here, they all involve adsorption of the reactants onto a solid catalytic surface, chemical reaction of the adsorbed species (sometimes via a number of intermediate species), and finally desorption of the products from the surface.
Table $1$: Some Commercially Important Reactions that Employ Heterogeneous Catalysts
Commercial Process Catalyst Initial Reaction Final Commercial Product
contact process V2O5 or Pt 2SO2 + O2 → 2SO3 H2SO4
Haber process Fe, K2O, Al2O3 N2 + 3H2 → 2NH3 NH3
Ostwald process Pt and Rh 4NH3 + 5O2 → 4NO + 6H2O HNO3
water–gas shift reaction Fe, Cr2O3, or Cu CO + H2O → CO2 + H2 H2 for NH3, CH3OH, and other fuels
steam reforming Ni CH4 + H2O → CO + 3H2 H2
methanol synthesis ZnO and Cr2O3 CO + 2H2 → CH3OH CH3OH
Sohio process bismuth phosphomolybdate $\mathrm{CH}_2\textrm{=CHCH}_3+\mathrm{NH_3}+\mathrm{\frac{3}{2}O_2}\rightarrow\mathrm{CH_2}\textrm{=CHCN}+\mathrm{3H_2O}$ $\underset{\textrm{acrylonitrile}}{\mathrm{CH_2}\textrm{=CHCN}}$
catalytic hydrogenation Ni, Pd, or Pt RCH=CHR′ + H2 → RCH2—CH2R′ partially hydrogenated oils for margarine, and so forth
14.04: Heterogeneous Catalysts
Another example of an organometallic catalytic reaction is the Ziegler-Natta olefin polymerization. This reaction is of high industrial importance for the production of olefins like polyethylene. There are both heterogeneous and homogeneous Ziegler-Natta catalysts. The mechanism for the homogeneous catalysts is generally well understood. Homogeneous catalysts are typically metallocene catalysts.
An example of a zirconium-based catalyst is shown (Fig. \(1\)). The catalyst is a coordinatively unsaturated complex cation with two cyclopentadienyl rings and a methyl group. The catalyst is formed from its precatalyst, a neutral molecule with an additional chloro ligand. The catalyst oxidatively adds an olefin like an ethylene molecule to the coordinatively unsaturated site. This step is followed an olefin insertion step that produces a propyl group. The migratory insertion leads to the formation of a vacant site, that can be re-oocupied by another ethylene molecule. This molecule can insert into the propyl chain thereby prolonging the propyl chain to a pentyl chain. The olefin insertion step generates another vacant site that can be reoccupied by a new ethylene molecule. Repeating the catalytic cycle many times eventually leads to polyethylene.
The Ziegler-Natta (ZN) catalyst, named after two chemists: Karl Ziegler and Giulio Natta, is a powerful tool to polymerize α-olefins with high linearity and stereoselectivity (Figure 1). A typical ZN catalyst system usually contains two parts: a transition metal (Group IV metals, like Ti, Zr, Hf) compound and an organoaluminum compound (co-catalyst). The common examples of ZN catalyst systems include TiCl4 + Et3Al and TiCl3 + AlEt2Cl.
In 1953, German chemist Karl Ziegler discovered a catalytic system able to polymerize ethylene into linear, high molecular weight polyethylene which conventional polymerization techniques could not make.1 The system contained a transition metal halide with a main group element alkyl compound (Figure \(3\)).
Following the catalytic design, Italian chemist Giulio Natta found that polymerization of α-olefins resulted in stereoregular structures,2 either syndiotactic or isotactic, depending on the catalyst used (Figure \(4\)). Because of these important discoveries, Karl Ziegler and Giulio Natta shared the Nobel Prize in Chemistry in 1963.
Advantages over traditional polymerization method
Traditionally, polymerization of α-olefins was done by radical polymerization (Figure \(5\)). Problem with this technique was that the formation of undesired allylic radicals leaded to branched polymers.3 For example, radical polymerization of propene gived branched polymers with large molecular weight distribution. Also, radical polymerization had no control over stereochemistry. Linear unbranched polyethylene and stereoregulated polypropylene could not be fabricated by free radical polymerization. This technique largely limited the potential applications of these polymeric materials.
The invention of ZN catalyst successfully addressed these two problems. The catalyst can give linear α-olefin polymers with high and controllable molecular weights. Moreover, it makes the fabrication of polymers with specific tacticity possible. By controlling the stereochemistry of products, either syndiotactic or isotactic polymers can be achieved.
Mechanism of Ziegler-Natta catalytic polymerization
Activation of Ziegler-Natta catalyst
It is necessary to understand the catalyst’s structure before understanding how this catalyst system works. Herein, TiCl4+AlEt3 catalyst system is taken as an example. The titanium chloride compound has a crystal structure in which each Ti atom is coordinated to 6 chlorine atoms. On the crystal surface, a Ti atom is surrounded by 5 chlorine atoms with one empty orbital to be filled. When Et3Al comes in, it donates an ethyl group to Ti atom and the Al atom is coordinated to one of the chlorine atoms. Meanwhile, one chlorine atom from titanium is kicked out during this process. Thus, the catalyst system still has an empty orbital (Figure \(6\)). The catalyst is activated by the coordination of AlEt3 to Ti atom.
Initiation step
The polymerization reaction is initiated by forming alkene-metal complex. When a vinyl monomer like propylene comes to the active metal center, it can be coordinated to Ti atom by overlapping their orbitals. As shown in Figure \(7\), there is an empty dxy orbital and a filled dx2-y2 orbital in Ti’s outermost shell (the other four orbitals are not shown here). The carbon-carbon double bond of alkene has a pi bond, which consists of a filled pi-bonding orbital and an empty pi-antibonding orbital. So, the alkene's pi-bonding orbital and the Ti's dxy orbital come together and share a pair of electrons. Once they're together, that Ti’s dx2-y2 orbital comes mighty close to the pi-antibonding orbital, sharing another pair of electrons.
The formed alkene-metal complex (1) then goes through electron shuffling, with several pairs of electrons shifting their positions: The pair of electrons from the carbon-carbon pi-bond shifts to form Ti-carbon bond, while the pair of electrons from the bond between Ti and AlEt3’ ethyl group shifts to form a bond between the ethyl group and the methyl-substituted carbon of propylene (Figure \(8\)). After electron shuffling, Ti is back with an empty orbital again, needing electrons to fill it (2).
Propagation step
When other propylene molecules come in, this process starts over and over, giving linear polypropylene (Figure \(9\)).
Termination step
Termination is the final step of a chain-growth polymerization, forming “dead” polymers (desired products). Figure \(10\) illustrates several termination approaches developed with the aid of co-catalyst AlEt3.4
Mechanistic study: kinetic isotope effect experiments
Unlike the mechanism discussed above, there is also a competing mechanism proposed by Ivin and coworkers.5 They proposed that a 1,2-hydrogen shift occurs prior to monomer association, giving a carbene intermediate (Figure \(11\)).
To determine the actual mechanism, Grubbs6 conducted kinetic isotope effect (KIE) experiments. Due to their different weights, carbon-deuterium bond reacts slower than carbon-hydrogen bond (Figure \(12\)). If such bonds are involved in the rate-determining step, the isotopic species should proceed in lower rate.
In Grubbs’s experiment, deuterated ethylene and normal ethylene were catalyzed by Ziegler-Natta catalyst with 1:1 ratio (Figure \(13\)). The result showed that H/D ratio in the resulting polymers was still 1:1.6 This indicated that no carbon-hydrogen bond cleavage or formation is involved in the rate determining step. Therefore, the mechanism proposed by Ivin was excluded.
Regio- and stero-selectivity
Regio-selectivity
For propene polymerization, most ZN catalysts are highly regioselective, favoring 1,2-primary insertion [M-R + CH2=CH(Me)=M-CH2-CH(Me)(R)] (Figure \(14\)), due to electronic and steric effects.7
Stereo-selectivity
Stereochemistry of polymers made from ZN-catalyst can be well regulated by rational design of ligands. By using different ligand system, either syndiotactic or isotactic polymers can be obtained (Figure \(15\)).
The relative stereochemistry of adjacent chiral centers within a macromolecule is defined as tacticity. Three kinds of stereochemistry are possible: isotactic, syndiotactic and atactic. In isotactic polymers, substituents are located on the same side of the polymer backbone, while substituents on syodiotactic polymers have alternative positions. In atactic polymers, substituents are placed randomly along the chain.
Choice of ZN catalyst regulates the stereochemistry. We use propylene polymerization as an example here. Recall the mechanism section, a monomer approaches the metal center and forms a four-membered ring intermediate. The binding of a monomer to the reactive metal-carbon bond should occur in from the least hindered site.8 As shown in Figure \(16\), the trans-complex in which methyl on the growing chain is trans to the methyl group on the incoming monomer, should be energetically favored than the cis-complex, as the trans-complex experience less steric effect.
Following the trans- complex, the methyl group on the newly added monomer is trans to that on the previous monomer. The step repeats so that a syndiotactic polypropylene is obtained (Figure \(17\)a). However, if the metal center is coordinated with bulky ligands (e.g. –iBu, -Et2 groups), as denoted by Y in Figure \(17\)b, the incoming monomer will adopt a cis-conformation to avoid serious steric effect with the bulky ligand. Thus, at the presence of bulky ligand, the propylene monomer is cis to the growing chain. This results in a isotactic product.
Applications
ZN catalysts have provided a worldwide profitable industry with production of more than 160 billion pounds and creation of numerous positions.9 These products have been extensively applied in different areas, largely improving the quality of people’s life. They can catalyze α-olefins to produce various commercial polymers, like polyethylene, polypropylene and Polybutene-1. Polyethylene and polypropylene is reported to be the top two widely used synthetic plastic in the word.10
Among the polyethylene fabricated by ZN catalysts, there are three major classes: high density polyethylene (HDPE), linear low density polyethylene (LLDPE), and ultra-high molecular weight polyethylene (UHMWPE). HDPE, a linear homopolymer, is widely applied in garbage containers, detergent bottles and water pipes because of its high tensile strength. Compared with HDPE whose branching degree is quite low, LLDPE has many short branches. Its better toughness, flexibility and stress-cracking resistance makes it suitable for materials like cable coverings, bubble wrap and so on. UHMWPE is polyethylene with a molecular weight between 3.5 and 7.5 million. This material is extremely tough and chemically resistant. Therefore, it is often used to make gears and artificial joints.
Compared to polyethylene, polypropylene has enhanced mechanical properties and thermal resistance because of the additional methyl group. Moreover, isotactic polypropylene is stiffer and more resistant to creep than atactic polypropylene. Polypropylene has a wide range of applications in clothing, medical plastics, food packing, and building construction.
Substrate scope and limitations
ZN catalysts are effective for polymerization of α-Olefins (ethylene, propylene) and some dienes (butadiene, isoprene). However, they don’t work for some other monomers, such as 1.2 disubstituted double bonds. Vinyl chloride cannot be polymerized by ZN catalyst either, because free radical vinyl polymerization is initiated during the reaction. Another situation that ZN catalysts don’t work is when the substrate is acrylate. The reason is that ZN catalysts often initiate anionic vinyl polymerization in those monomers.11
Contributors and Attributions
• Fangyuan Dong and Ru Deng | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/14%3A_Organometallic_Reactions_and_Catalysis/14.04%3A_Heterogeneous_Catalysts/14.4.01%3A_Ziegler-Natta_Polymerizations.txt |
The water-gas shift reaction (WGSR) describes the reaction of carbon monoxide and water vapor (steam) at very high temperatures to form carbon dioxide and hydrogen:
\[\ce{CO + H2O <=> CO2 + H2}\]
The WGSR is a highly valuable industrial reaction that is used in the manufacture of ammonia, hydrocarbons, methanol, and hydrogen. Its most important application is in the conversion of carbon monoxide or other hydrocarbons to produce hydrogen gas, an important source of fuel for a modern hydrogen economy. If hydrogen could be produced at low cost and using a renewable resource, it could significantly advance the movement to sustainable energy.
Most of the catalysts that are used for this process are heterogeneous catalysts. Unfortunately, the process using heterogeneous catalysts occurs only at extremely high temperatures and pressures. The high energy cost of performing the reaction limits its industrial utility, and there are efforts to produce homogeneous catalysts that catalyze the reaction at more mild conditions. Another limitation is that the reaction requires sufficient quantities of \(\ce{CO}\), which are primarily obtained from non-renewable coal or petroleum sources. However, there are processes that can convert methane gas, which is an abundant greenhouse gas, to \(\ce{}\).
The WGSR has been extensively studied for over a hundred years. The kinetically relevant mechanism depends on the catalyst composition and the temperature. Two mechanisms have been proposed: an associative Langmuir–Hinshelwood mechanism and a redox mechanism. The redox mechanism is generally regarded as kinetically relevant during the high-temperature WGSR (> 350 °C) over the industrial Fe-Cr catalyst. Historically, there has been much more controversy surrounding the mechanism at low temperatures. Recent experimental studies confirm that the associative carboxyl mechanism is the predominant low temperature pathway on metal-oxide-supported transition metal catalysts.
Associative mechanism
In 1920 Armstrong and Hilditch first proposed the associative mechanism. In this mechanism CO and H2O are adsorbed onto the surface of the catalyst, followed by formation of an intermediate and the desorption of H2 and CO2. In general, H2O dissociates onto the catalyst to yield adsorbed OH and H. The dissociated water reacts with CO to form a carboxyl or formate intermediate. The intermediate subsequently dehydrogenates to yield CO2 and adsorbed H. Two adsorbed H atoms recombine to form H2.
There has been significant controversy surrounding the kinetically relevant intermediate during the associative mechanism. Experimental studies indicate that both intermediates contribute to the reaction rate over metal oxide supported transition metal catalysts. However, the carboxyl pathway accounts for about 90% of the total rate owing to the thermodynamic stability of adsorbed formate on the oxide support. The active site for carboxyl formation consists of a metal atom adjacent to an adsorbed hydroxyl. This ensemble is readily formed at the metal-oxide interface and explains the much higher activity of oxide-supported transition metals relative to extended metal surfaces. The turn-over-frequency for the WGSR is proportional to the equilibrium constant of hydroxyl formation, which rationalizes why reducible oxide supports (e.g. CeO2) are more active than irreducible supports (e.g. SiO2) and extended metal surfaces (e.g. Pt). In contrast to the active site for carboxyl formation, formate formation occurs on extended metal surfaces. The formate intermediate can be eliminated during the WGSR by using oxide-supported atomically dispersed transition metal catalysts, further confirming the kinetic dominance of the carboxyl pathway.
Redox mechanism
The redox mechanism involves a change in the oxidation state of the catalytic material. In this mechanism, CO is oxidized by an O-atom intrinsically belonging to the catalytic material to form CO2. A water molecule undergoes dissociative adsorption at the newly formed O-vacancy to yield two hydroxyls. The hydroxyls disproportionate to yield H2 and return the catalytic surface back to its pre-reaction state.
14.5.01: Concept Review Questions Chapter 12
Concept Review Questions
Section 1
1) What is an oxidative addition?
2) What is a reductive elimination?
3) What is a cis and a trans addition, respectively?
4) Does the coordination number in oxidative additions always increase? Explain.
5) What is the reverse reaction of an oxidative addition?
6) What is a migratory insertion?
7) Are migratory insertions reversible reactions?
Section 2
1) Why are transition metal alkyl complexes that have alkyl groups longer than 1 carbon atom often unstable?
2) What is meant by beta-hydride elimination?
3) What is alpha-abstraction?
4) What is a nucleophilic displacement reaction?
5) What is meant by an oxidative addition of an intact ligand?
6) Write the catalytic cyclic for the deuteration of benzene with Cp2TaH3 as a catalyst. Name each reaction step.
7) What is Collman’s reagent and how is it being used?
8) What is Wilkinson’s catalyst, and what reactions can it be used for. Write the catalytic cycle.
9) Write done the reaction mechanism for the Ziegler-Natta olefin polymerization.
10) What is carbene migratory insertion?
11) Write down the catalytic cycle for the hydroformylation of olefins.
12) Name three hydrocarbonylation reactions. What are the reactants?
13) Write down the mechanism for the Monsanto acetic acid process.
14) What is olefin metathesis?
15) What is ring-opening olefin polymerization (ROMP)?
16) What is acyclic diene metathesis?
17) Why are olefins with terminal C=C double bonds particularly useful in olefin metathesis?
18) What is ring-closing metathesis?
19) What are the characteristics of Schrock and Grubbs catalysts, respectively?
20) Write down the catalytic cycle for olefin metathesis.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
14.5.02: Homework Problems Chapter 12
Homework Problems
Section 1
Exercise 1
The coordination compound CH3Co(CO)4 undegoes a migratory insertion. What is the reaction product?
Answer
Section 2
Exercise 1
Identify all oxidative additions in the catalytic cycle of olefin hydroformylation:
Answer
Step D --> E is an oxidative addition
Formation of HCo(CO)4 from Co2(CO)8 is also an oxidative addition.
Exercise 2
1. What are the reaction products of the following olefin metathesis reactions?
a) 2 CH2=CH-Ph -->
b) n CH2=CH-CH=CH2 -->
Answer
a) CH2=CH2 and Ph-CH=CH-Ph
b) CH2=CH2 and (CH=CH)n
Exercise 3
What reaction product would you expect would you expect from the following reaction:
Fe(CO)42- + CH3CH2I -->
Name the reaction.
Answer
Fe(CO)4CH2CH3- + I- Nucleophilic displacement
Exercise 4
Which of the following phosphines would you expect to be most suitable to stabilize metals in low oxidation states:
a) P(CF3)3
b) P(CH3)3
c) PH3
Answer
a) P(CF3)3
Exercise 5
What reaction product would you expect from the following reaction?
Pt(PPh3)4 + 4 NPh3 -->
Answer
There would be no reaction.
Exercise 6
Which of the following complexes would likely be unstable:
a) [PtCl3(C2H5)]2-
b) [PtCl3(CH3)]2-
c) PtCl42-
Answer
a) [PtCl3(C2H5)]2-
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/14%3A_Organometallic_Reactions_and_Catalysis/14.04%3A_Heterogeneous_Catalysts/14.4.02%3A_Water-Gas_Shift_Reaction.txt |
The isolobal analogy (aka isolobal principle) is a strategy used in organometallic chemistry to relate the structure of organic and inorganic molecular fragments in order to predict bonding properties of organometallic compounds. Roald Hoffmann described molecular fragments as isolobal: "if the number, symmetry properties, approximate energy and shape of the frontier orbitals and the number of electrons in them are similar – not identical, but similar." One can predict the bonding and reactivity of a species from that of a better-understood species if the two molecular fragments have similar frontier orbitals; the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO). Isolobal compounds are analogues to isoelectronic compounds that share the same number of valence electrons and structure.
Isolobal species are indicated with an isolobal arrow symbol; a double-headed arrow with half an orbital (Figure \(1\)).
The concept of isolobality was developed by Roald Hoffmann (Figure \(2\)), who won the Nobel prize in chemistry in 1981. It comes from the greek words “isos” meaning “similar” and “lobos” meaning lobe. Hence, “isolobal” means “similar lobes”. In his Nobel speech he defined isolobality as follows: “Molecular fragments are isolobal if the number, symmetry properties, approximate energy and shape of the frontier orbitals and the number of electrons in them are similar - not identical, but similar. “ When molecular fragments are isolobal then they can likely be combined to form a stable molecule. Thus, isolobal fragments are compatible building blocks for the construction of stable molecules. We can also say that the concept of isolobality helps us to predict when a stable bond forms. Whenever two molecular fragments are isolobal, then they will likely form stable bonds between them.
Isolobal Fragments of main group elements
There is a simple means to determine if fragments are isolobal. In main group chemistry the octet rule is a very helpful guide, especially for the period 2 elements (B,C,N,O,F) for which the octet rule holds fairly strictly. The difference between the number "8" and the number of valence electrons (VE) of the molecular fragment is equal to the number of frontier orbitals and the number of electrons in it.
# of frontier orbitals = # of electrons in frontier orbitals = 8 – VE
Thus, when two fragments (of main group elements) have the same number of valence electrons, they can be considered isolobal. For example, the \(\ce{CH3}\) fragment created by homolytic cleavage of a \(\ce{C-H}\) bond from \(\ce{CH4}\) is shown in Figure \(1\). This \(\ce{CH3}\) fragment has 7 VE on the carbon, and therefore it has one frontier orbital with one electron it (Figure \(1\)). The same is true for the \(\ce{NH2}\) (7 VE) and \(\ce{OH}\) (7 VE) fragments shown in XXX. All of these fragments have 7 VE and one frontier orbital (Figure \(3\)).
According to the isolobal analogy, it should be possible to combine isolobal fragments to form stable molecules. Let's test this on the fragments shown in Figure \(3\). All the posible combinations are shown in Figure \(4\). Two \(\ce{CH3}\) fragments can be combined to form an ethane molecule which is known to be stable. The combination of \(\ce{CH3}\) and\(\ce{NH2}\) gives methylamine which is stable, and the combination of \(\ce{CH3}\) and \(\ce{OH}\) gives methanol, which is also stable. Two\(\ce{NH2}\) fragments give hydrazine, H2N-NH2 which exists, the combination of\(\ce{NH2}\)and \(\ce{OH}\) gives the known hydroxylamine molecule \(\ce{NH2OH}\). The combination of two \(\ce{OH}\) fragments gives hydrogen peroxide \(\ce{H2O2}\).
As another example, a \(\ce{CH2}\) fragment is a 6 VE fragment, thus there are 8-6=2 frontier orbitals with overall two electrons in them (Figure \(5\)). An \(\ce{NH}\)-fragment and an O-atom also have 6 VE, and are therefore isolobal to \(\ce{CH2}\). In \(\ce{NH}\) there are 5+1=6 valence electrons, and an oxygen atom has 6 valence electrons. We can combine two \(\ce{CH2}\) fragments to form ethene \(\ce{H2C=CH2}\). Combining \(\ce{CH2}\) with \(\ce{NH}\) and \(\ce{O}\), respectively gives methanimine \(\ce{H2CNH}\), and formaldehyde \(\ce{H2C=O}\), respectively. Methylene imine is stable in the gas phase, and oligomerizes in higher concentrations to form a hexamer, called urotropine. Other oligomers and polymers are also known. The combination of two \(\ce{NH}\) fragments gives the known molecule diazene \(\ce{HN=NH}\), and the combination with an \(\ce{O}\) atom gives \(\ce{HN=O}\), which is known as nitroxyl or azanone, and is stable in the gas phase. The combination of two oxygen atoms gives the well known O2 molecule.
The analogy could also be extended to 5 VE fragments (not shown). The CH fragment has 4+1=5 VE, and an N-atom has 5 VE as well, thus they can be considered isolobal. Two CH fragments give acetylene C2H2, and two N atoms give the dinitrogen molecule N2. The combination of a CH fragment with an N fragment gives H-CN, well known as hydrogen cyanide.
Isolobal fragments of transition elements
In the way the octet rule can help to predict the number of frontier orbitals and the electrons in them for main group element fragments, the 18 electron rule can be used to predict the number of frontier orbitals and electrons for organometallic fragments, including carbonyl fragments. The number of frontier orbitals and the number of electrons in them is 18 minus the number of valence electrons the organometallic fragment has:
# frontier orbitals = # of electrons in frontier orbitals = 18 – VE
For example, a metal complex with a count of 17 electrons, such as \(\ce{Mn(CO)5}\), there is one frontier orbital with one electron. This implies that a 17 VE carbonyl fragment is isolobal to a 7 VE organic fragment such as \(\ce{CH3}\). For a 16 VE fragment such as \(\ce{Fe(CO)4}\) there are two frontier orbitals with one electron in each of them, and for a 15 VE fragment such as \(\ce{(CO)3Co}\) there are three frontier orbitals with overall three valence electrons (Figure \(6\)). Similarly, a 16 VE metal carbonyl fragment is isolobal to a 6 VE fragment such as \(\ce{CH2}\), and a 15 VE fragment is isolobal to a 5 VE fragment such as CH (Figure \(6\)).
It should be possible to combine the isolobal fragments from Figure \(6\) to form stable molecules. Let us check how well this works (Figure \(7\)). It should be possible to combine the two 17 VE electron fragments such as \(\ce{Mn(CO)5}\) to form \(\ce{(CO)5Mn-Mn(CO)5}\). This is a known molecule. Remember we encountered it previously when we discussed the homoleptic carbonyls of metals with an odd number of electrons. Combining this fragment with a 7 VE \(\ce{CH3}\) fragment leads to \(\ce{(CO)5Mn-CH3}\) which is also stable.
Can we also combine two 16 VE electron fragments to form \(\ce{(CO)4Fe=Fe(CO)4}\) with an Fe=Fe double bond? The answer is no. Metal-metal double bonds, and also metal-metal triple bonds in carbonyl complexes are not formed. Instead clusters with single bonds are realized. The number of single bonds a metal makes is equal to the number of its frontier orbitals. In the case of 16 VE fragments trimeric clusters with single bonds are formed. In this cluster the two frontier orbitals of each fragment make two single bonds to two other fragments. In contrast, a 16 VE carbonyl fragment such as \(\ce{(CO)4Fe}\) can be combined with a 6 VE fragment to form a compound with a Fe=C double bond. A compound with a metal-carbon double bond is called a carbene complex. Similarly, the combination of 15 VE fragments of the type \(\ce{Co(CO)3}\) does not lead to a stable \(\ce{(CO)3Co≡Co(CO)3}\) molecule with a \(\ce{Co≡Co}\) triple bond. Instead, nature realizes a tetrameric tetrahedral cluster in which the three frontier orbitals of each 15 VE fragment make three single bonds to the other three 15 VE fragments. In clusters the CO ligands may not only be terminal, they can also be bridging. In the tetrameric Co-cluster, there are are nine terminal and three bridging CO-ligands. The three bridging CO ligands are connecting the three Co atoms at the base triangular face of the tetrahedron. What about the combination of a 15 VE fragment with an organic 5 VE fragment? The combination of a 15 VE fragment such as Co(CO)3 with a 5 VE organic fragment such as CH to does yield stable complexes such as the \(\ce{(CO)3Co≡CH}\) complex with a \(\ce{Co≡C}\) triple bond. Complexes with metal-carbon triple bonds are called carbine complexes.
Synthesis of Terameric Cluster Carbonyls
How can we make a carbonyl cluster like ((Co(CO)3)4?
It can be prepared by heating the Co2(CO)8 to a temperature above 54°C. Above this temperature the reactant loses four CO ligands and rearranges to form the cluster (Fig. \(8\)). Interestingly, the higher homologues of the Co-cluster, the Rh4(CO)12 and the Ir4(CO)12 form spontaneously from the elements. Remember, that we previously determined that the Rh2(CO)8 and the Ir2(CO)8 are not stable. Rh and Ir favor the tetrameric clusters with 12 COs over the dimer with 8 CO ligands.
Structures of Co4(CO)12, Rh4(CO)12 and Ir4(CO)12
Like in the Co-cluster, there are 9 terminal and three bridging CO ligands in the Rh4(CO)12 cluster. By contrast, there are only terminal CO ligands in the iridium cluster (Fig.\(9\)) . This may be explained by the larger Ir-Ir bond length in comparison to the Co-Co and Rh-Rh bond lengths. The observation reflects the general rule that only 3d and 4d elements have bridging CO ligands, while only terminal ligands are observed in metals with 5d electrons.
Isoelectronic Charged Carbonyl Clusters
The previously discussed clusters were charge-neutral. Can we construct charged isoelectronic ones with different metals, and what is their stability? Let us start with the tetrameric Co-cluster made of four 15 VE fragments. The element left to the Co is the Fe. It has one electron less, therefore Fe(CO)3- is the 15 VE electron fragment which is isoelectronic to the neutral Co(CO)3 fragment. Its tetramer would be a Fe4(CO)124- cluster. This cluster is not known. The cluster already carries a 4- charge which is too high to support stable Fe-Fe bonds. Each Fe carries formally a 1- negative charge and there is too much electrostatic repulsion between the Fe atoms. Another possibility to realize a 15 VE fragment with Fe is a (CO)4Fe+ fragment. Here an additional ligand is added contributing two electrons, and one electron is removed resulting in a fragment with a 1+ charge. The fragment could be combined to form an Fe4(CO)164+ cluster. However, such a cluster is also not stable. This is because the coordination number would be 7, which is too high for a carbonyl. In addition, there is the destabilizing effect of the positive charge on the Fe-C bond. Similarly, tetramers of the 15 VE fragments Mn(CO)4, Cr(CO)4- and (CO)5Cr+ are also not known. Also in these cases, the coordination numbers would be too high. The Mn(CO)4 and the Cr(CO)4- would lead to a coordination number of 7, in the case of the Cr(CO)5+ fragment the coordination number would be even 8. We see here the limitations of the concept of isolobality. The presence of isolobality is not always sufficient to make a stable molecule, also other factors like coordination numbers and charge needs to be considered.What about charged isoelectronic clusters made from 16 VE? We saw previously that the neutral 16 VE Fe(CO)4 fragment gave a Fe3(CO)12 trimer. If we go from the Fe to the Mn, the isoelectronic 16 VE fragment would be a Mn(CO)4- fragment because Mn has one electron less than Fe. This fragment can indeed be trimerized to give a stable Mn3(CO)123- cluster with a 3- charge. This charge is not too high yet to destabilize the Mn-Mn bond and the coordination number of 6 is ideal for the stability of carbonyls. We could add a ligand and remove two electrons to produce a 15 VE Mn(CO)5+ fragment. However, in this case the coordination number would become 7 upon trimerization, which is too high to produce a stable cluster. In addition, the positive charge destabilizes the Mn-C bond. Replacing Mn by Cr in Mn(CO)5+ would give a neutral 16 VE Cr(CO)5 fragment, but its trimer is not stable because also in this case the coordination number becomes too high. Replacing Fe in Fe(CO)4 by Co would require a 1+ charge giving a Co(CO)4+ 16 VE fragment. Its trimer would have the coordination number 6, but it would carry a 4+ charge which is too high. A Co(CO)3- fragment would also have 16 VE, but its trimer is also not known, possibly because the relatively small coordination number of 5, which is less favorable than a coordination number of 6. Overall we can say as a rule for the stability for carbonyl clusters that there should not be positive charges, and no high negative charges. There should not be high coordination numbers.
Clusters with 16 VE and 6 VE Fragments
The cluster chemistry of carbonyls is even more versatile due to the possibility to substitute 16 and 15 VE fragments by 6 and 5 VE fragments, respectively. For instance, we can replace one 16 VE Fe(CO)4 fragment by a 6 VE CH2 fragment in the trimeric triiron dodecacarbonyl Fe3(CO)12 cluster (Fig. \(10\)). In the resulting cluster a methylene group bridges two Fe atoms, therefore the complex can be regarded a μ-methylene complex. The substitution reduces the number of cluster valence electrons (CVE) from 48 to 38. A second substitution of another 16 VE tetracarbonyl iron fragment by another 6 VE methylene fragment gives a cluster with two Fe-C bonds and one carbon-carbon bond having 28 cluster valence electrons. This complex can be regarded as a ethene complex in which an ethene molecule binds side-on to a 15 VE Fe(CO)4 fragment. Upon binding, the two Fe-C bonds are formed at the expense of the C-C π-bond and the bond order in the C-C bond is reduced from 2 to 1. If we substitute the last tetracarbonyl iron fragment by a third CH2 unit, then a completely organic molecule, cyclopropane, results. All the molecules are known to be stable molecules, and their stability can be nicely understood by using the concept of isolobality.
Similarly, we can substitute 15 VE tricarbonyl cobalt units by 5 VE CH fragments in tetracobaltdodecacarbonyl Co4(CO)12. A first substitution gives a 50 VE cluster with one CH fragment that makes three bonds to three Co(CO)4 fragments (Fig. \(11\)). Because the CH group bridges three metal atoms it is regarded a μ3-methine complex. A second substitution produces a 40 VE cluster with an organic CH-CH unit that makes four bonds to two Co atoms of two 15 VE Co(CO)3 fragments. The CH-CH fragment can be considered an ethine molecule binding side-on to a dicobalthexacarbonyl fragment, whereby the two π-bonds in ethine are expended to form four Co-C single bonds. The cluster can be considered a μ22-ethine complex because the ethine bridges two Co atoms and both carbons are involved in the bonding with Co. A third substitution leads to a 30 CVE cluster with three CH units binding to one Co(CO)3 fragment. There are three C-C single bonds and three Co-C bonds in the cluster. The three CH units form a cyclopropenylium complex that binds side-on to a Co(CO)4 fragment. A free cyclopropenylium molecule is a 3-ring with three π-electrons that are delocalized in the ring. Therefore, the bond order in a free cyclopropenylium molecule is 1.5. In the complex the three π-electrons are being used to make the three single bonds with the cobalt atoms. The bond order in cyclopropenylium is thereby reduced from 1.5 to 1. Finally, the fourth substitution of a tricarbonyl cobalt fragment by a methine fragment yields the completely organic tetrahedrane molecule. This molecule and the others discussed before are stable. The isolobality concept lets us rationalize these complex structures easily and understand their stability.
Synthesis of Cluster Complexes From Alkynes
The view that the 40 VE clusters are alkyne clusters is not just a formal view. They can be synthesized from alkynes. For instance alkyne-dicobalt clusters are accessible from dicobalt octacarbonyl and alkynes (Fig. \(12\)).
The cluster forms via the donation of the four π electrons of the alkynes into the metal d-orbitals of the dimeric cobalt carbonyl. This leads to a loss of of two CO ligands, and the formation of four Co-C bonds. Under harsher conditions the C-C triple bond can also cleave and this can give access to methine complexes.
Carbonyl Hydrides
Now let us leave the carbonyl cluster compounds and talk about another interesting class of carbonyls: carbonyl hydrides. They can be rationalized by the concept of isolobality as well. A hydrogen atom can be conceived as a species with one frontier orbital containing one electron. In this case the frontier orbital is simply the 1s orbital of the hydrogen. Thus, it should be possible to combine an H atom with a 17 VE carbonyl fragment like tetracarbonyl cobalt.
Indeed, one can combine such fragments to form stable carbonyl hydrides such as tetracarbonylhydrido cobalt (0). This molecule can be synthesized by reduction of bis(tetracarbonyl cobalt) (0) with dihydrogen (Fig. \(13\)). Counter-intuitively, this molecule is a strong acid, it has an acidity similar to sulfuric acid. One would think that the Co-H bond would be polarized toward H based on electronegativity arguments. However, this is not the case. The carbonyl fragment has a frontier orbital which is energetically higher than that of H, and thus the bond is polarized toward H. In another view we can explain the high acidity be the fact that the loss of the proton leads to the very stable 18 VE tetrahedral Co(CO)4- anion. Thus, loss of the proton occurs easily.
Another 17 VE electron fragment is the Mn(CO)5 fragment. Also this fragment can be combined with an isolobal H fragment to form a stable molecule, which has the composition HMn(CO)5 (Fig. \(14\)). It can be prepared from bis(pentacarbonyl manganese) and dihydrogen at 200 bar and 150°C. This carbonyl hydride is a weak acid, it has a similar acidity as H2S. We could argue that this lower acidity may be because the loss of the proton reduces its coordination number from 6 to 5. The coordination number of 6 is the preferred coordination number for carbonyls and thus the tendency to lose the proton is relatively small. The different metal and the number of carbonyl ligands will likely lead to a different energy of the frontier orbital compared to the previous example, which leads to a different polarity.
Overall, one can tune the properties in carbonyl hydrides from highly acidic to hydridic by choice of the metal the coordination number, and also by the choice of additional ligands L other than carbonyl. The electronic and steric properties of the ligands have an influence on the energy of the HOMO of the fragment, and thus on the polarity of the metal-hydrogen bond, and the acidity.
15.02: The Isolobal Analogy
The isolobal analogy has applications beyond simple octahedral complexes. It can be used with a variety of ligands, charged species and non-octahedral complexes.
The isolobal analogy can also be used with isoelectronic fragments having the same coordination number, which allows charged species to be considered. For example, $\ce{Re(CO)5}$ is isolobal with $\ce{CH3}$ and therefore, $\ce{[Ru(CO)5]^+}$ and $\ce{[Mo(CO)5]^−}$are also isolobal with $\ce{CH3}$. Any 17-electron metal complex would be isolobal in this example.
In a similar sense, the addition or removal of electrons from two isolobal fragments results in two new isolobal fragments. Since $\ce{Re(CO)5}$ is isolobal with $\ce{CH3}$, $\ce{[Re(CO)5]^+}$ is isolobal with $\ce{CH3^+}$.
The analogy applies to other shapes besides tetrahedral and octahedral geometries. The derivations used in octahedral geometry are valid for most other geometries. The exception is square-planar because square-planar complexes typically abide by the 16-electron rule. Assuming ligands act as two-electron donors the metal center in square-planar molecules is $d^8$. To relate an octahedral fragment, MLn, where M has a dx electron configuration to a square planar analogous fragment, the formula MLn−2 where M has a dx+2 electron configuration should be followed.
Octahedral MLn Square Planar MLn
$d^6: \quad \ce{Mo(CO)5}$ $d^8: \quad \ce{[PdCl3]^-}$
$d^8: \quad \ce{Os(CO)4}$ $d^10: \quad \ce{Ni(PR3)2}$
Further examples of the isolobal analogy in various shapes and forms are shown | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/15%3A_Parallels_between_Main_Group_and_Organometallic_Chemistry/15.02%3A_The_Isolobal_Analogy/15.2.01%3A_Extensions_of_the_Analogy.txt |
Molecular orbital considerations in Dinuclear Metal Complexes with Multiple M-M Bonds
Let us begin a new chapter and think about dinuclear transition metal complexes with multiple metal-metal bonds. What is the maximum bond order that we could expect? The transition metals have five d-orbitals available, and in order to determine the maximum possible bond order we need to see how they overlap to form molecular orbitals (Fig. 11.1.1).
If we define the M-M bond axis as the z-axis, then two dz2-orbitals can overlap in σ-fashion to form a bonding and an anti-bonding σ-molecular orbital. A dxz, and a dyz can overlap with another dxz and another dyz in π-fashion to form two degenerated bonding π and two degenerated anti-bonding π*-orbitals. π-overlap is smaller than σ-overlap, therefore the split in energy between the bonding and the anti-bonding π-orbitals is smaller than the split between the bonding and the anti-bonding σ-orbitals. The dxy orbital can overlap with another dxy orbital in $\delta$-fashion, and so can the dx2-y2 with another dx2-y2. This gives two bonding $\delta$-orbitals and two anti-bonding $\delta$-orbitals. The $\delta$-overlap is even smaller than the π-overlap, therefore the energy split between the bonding and the anti-bonding $\delta$-orbitals is even smaller than those for the π and π*-orbitals. So what would be the maximum bond order that could be achieved? The maximum bond order would be achieved when all bonding MOs were full, and all anti-bonding MOs were empty. We have overall five bonding MOs that we could fill with ten electrons from two metal atoms with d5-electron configuration. The maximum bond order BO would therefore be BO=5. However, in practice only bond orders up to 4 are well known. This is because the dx2-y2 orbital is usually too involved in the bonding with the ligands thereby becoming unavailable for metal-metal interactions. The dx2-y2 orbital makes the strongest interactions with the ligands because most ligands approach on or near the x and the y-axes.
Electron Configurations and Multiple M-M Bonds
We can easily predict now which electron configuration leads to which metal-metal bond order. The bond order increases from 1 to 4 with the electron configuration changing from d1 to d4. At d4 all bonding MOs are full with eight electrons.
An example of a complex with a bond order of 4 is the tetraacetatodiaquadichromium complex shown (Fig. 11.1.2). Cr is in the oxidation state +2, which makes the chromium a d4 species. We can quickly show this by counting the valence electrons. A neutral Cr atom has 6 VE, and an electron configuration 3d54s1. There are four acetate ligands having a 1- charge each which gives four negative charges overall. The complex is overall neutral which means that each Cr must have formally a 2+ charge, and the electron configuration is 3d4.
With even more electrons in the metal d-orbitals the bond order begins to decrease again as anti-bonding MOs need to be filled (Fig. 11.1.3). The combination of two metals with d5 electron configuration leads to a triple bond, two d6-metals give a double bond, and two d7 metals give a single bond. A metal-metal bond should not exist for two d8-metals because then the bond order is zero.
Evidence for M-M Multiple Bonds
What experimental evidence can support the existence of a particular bond order (Fig. 11.1.4)? One argument is the bond length which can be obtained through the crystal structure determination of the complex. The shorter the bond, the higher the bond order. For instance, in the four tungsten complexes shown the bond lengths decrease from 272 pm, to 248 pm, to 230 pm to 221 pm corresponding to a single, double, triple, and quadruple bond, respectively.
Another hint can be the conformation of a molecule (Fig. 11.1.5). For instance, the two square-planar units of the Re2Me82- complex anion show eclipsed conformation. Steric repulsion arguments would favor the staggered conformation, so there must be a reason why the two ReMe4 units are eclipsed. The rhenium is in the oxidation state +3, thus it is a d4 species, and we would argue that there may be a Re-Re quadruple bond. This quadruple bond can only form when the dxy orbitals are in eclipsed conformation, and this is only possible when the two ReMe4 fragments are in eclipsed conformation. The very short bond length of 218 pm further supports the existence of the quadruple bond.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate.
15.03: Metal-Metal Bonds
Molecular orbital considerations in Dinuclear Metal Complexes with Multiple M-M Bonds
Let us begin a new chapter and think about dinuclear transition metal complexes with multiple metal-metal bonds. What is the maximum bond order that we could expect? The transition metals have five d-orbitals available, and in order to determine the maximum possible bond order we need to see how they overlap to form molecular orbitals (Fig. 11.1.1).
If we define the M-M bond axis as the z-axis, then two dz2-orbitals can overlap in σ-fashion to form a bonding and an anti-bonding σ-molecular orbital. A dxz, and a dyz can overlap with another dxz and another dyz in π-fashion to form two degenerated bonding π and two degenerated anti-bonding π*-orbitals. π-overlap is smaller than σ-overlap, therefore the split in energy between the bonding and the anti-bonding π-orbitals is smaller than the split between the bonding and the anti-bonding σ-orbitals. The dxy orbital can overlap with another dxy orbital in $\delta$-fashion, and so can the dx2-y2 with another dx2-y2. This gives two bonding $\delta$-orbitals and two anti-bonding $\delta$-orbitals. The $\delta$-overlap is even smaller than the π-overlap, therefore the energy split between the bonding and the anti-bonding $\delta$-orbitals is even smaller than those for the π and π*-orbitals. So what would be the maximum bond order that could be achieved? The maximum bond order would be achieved when all bonding MOs were full, and all anti-bonding MOs were empty. We have overall five bonding MOs that we could fill with ten electrons from two metal atoms with d5-electron configuration. The maximum bond order BO would therefore be BO=5. However, in practice only bond orders up to 4 are well known. This is because the dx2-y2 orbital is usually too involved in the bonding with the ligands thereby becoming unavailable for metal-metal interactions. The dx2-y2 orbital makes the strongest interactions with the ligands because most ligands approach on or near the x and the y-axes.
Electron Configurations and Multiple M-M Bonds
We can easily predict now which electron configuration leads to which metal-metal bond order. The bond order increases from 1 to 4 with the electron configuration changing from d1 to d4. At d4 all bonding MOs are full with eight electrons.
An example of a complex with a bond order of 4 is the tetraacetatodiaquadichromium complex shown (Fig. 11.1.2). Cr is in the oxidation state +2, which makes the chromium a d4 species. We can quickly show this by counting the valence electrons. A neutral Cr atom has 6 VE, and an electron configuration 3d54s1. There are four acetate ligands having a 1- charge each which gives four negative charges overall. The complex is overall neutral which means that each Cr must have formally a 2+ charge, and the electron configuration is 3d4.
With even more electrons in the metal d-orbitals the bond order begins to decrease again as anti-bonding MOs need to be filled (Fig. 11.1.3). The combination of two metals with d5 electron configuration leads to a triple bond, two d6-metals give a double bond, and two d7 metals give a single bond. A metal-metal bond should not exist for two d8-metals because then the bond order is zero.
Evidence for M-M Multiple Bonds
What experimental evidence can support the existence of a particular bond order (Fig. 11.1.4)? One argument is the bond length which can be obtained through the crystal structure determination of the complex. The shorter the bond, the higher the bond order. For instance, in the four tungsten complexes shown the bond lengths decrease from 272 pm, to 248 pm, to 230 pm to 221 pm corresponding to a single, double, triple, and quadruple bond, respectively.
Another hint can be the conformation of a molecule (Fig. 11.1.5). For instance, the two square-planar units of the Re2Me82- complex anion show eclipsed conformation. Steric repulsion arguments would favor the staggered conformation, so there must be a reason why the two ReMe4 units are eclipsed. The rhenium is in the oxidation state +3, thus it is a d4 species, and we would argue that there may be a Re-Re quadruple bond. This quadruple bond can only form when the dxy orbitals are in eclipsed conformation, and this is only possible when the two ReMe4 fragments are in eclipsed conformation. The very short bond length of 218 pm further supports the existence of the quadruple bond.
Dr. Kai Landskron (Lehigh University). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/15%3A_Parallels_between_Main_Group_and_Organometallic_Chemistry/15.03%3A_Metal-Metal_Bonds/15.3.01%3A_Metal-Metal_Multiple_Bonds.txt |
Boranes and the Bonding in boranes
Boranes were introduced in Chapter 8 (Section 8.6.1). Boranes are compounds consisting of boron and hydrogen.The most basic example is diborane (\(\ce{B2H6}\)), all boranes are electron-deficient compounds. For \(\ce{B2H6}\) usually 14 electrons are needed to form 2c,2e-bonds, but only 12 valence electrons are present. Because of this there are two B-H-B bonds, which have three centers, but only two electrons (3c, 2e bond). This can be interpreted as a molecular orbital that is formed by combining the contributed atomic orbitals of the three atoms. In more complicated boranes not only B-H-B bonds but also B-B-B 3c, 2e-bonds occur. In such a bond the three B-atoms lie at the corners of an equilateral triangle with their sp3 hybrid orbitals overlapping at its center. One of the common properties of boranes is, that they are flammable or react spontaneously with air. They burn with a characteristic green flame. And they are colorless, diamagnetic substances.
Nomenclature
In neutral boranes the number of boron atoms is given by a prefix and the number of Hydrogen-atoms is given in parentheses behind the name. example: \(\ce{B5H11}\) -> pentaborane(11), \(\ce{B4H10}\) -> tetraborane(10) For ions primarily the number of hydrogen-atoms and than the number of boron-atoms is given, behind the name the charge is given in parentheses. example: \(\ce{[B6H6]^{2-}}\) -> hexahydrohexaborat(2-)
Wades rule, Structures of boranes
Wades rule helps to predict the general shape of a borane from its formula. Ken Wade developed a method for the prediction of shapes of borane clusters; however, it may be used for a wide range of substituted boranes (such as carboranes) as well as other classes of cluster compounds. Wade’s rules are used to rationalize the shape of borane clusters by calculating the total number of skeletal electron pairs (SEP) available for cluster bonding. In using Wade’s rules it is key to understand structural relationship of various boranes.
Wade’s rules:
The general methodology to be followed when applying Wade’s rules is as follows:
1. Determine the total number of valence electrons from the chemical formula, i.e., 3 electrons per B, and 1 electron per H.
2. Subtract 2 electrons for each B-H unit (or C-H in a carborane).
3. Divide the number of remaining electrons by 2 to get the number of skeletal electron pairs (SEP).
4. A cluster with n vertices (i.e., n boron atoms) and n+1 SEP for bonding has a closo structure.
5. A cluster with n-1 vertices (i.e., n-1 boron atoms) and n+1 SEP for bonding has a nido structure.
6. A cluster with n-2 vertices (i.e., n-2 boron atoms) and n+1 SEP for bonding has an arachno structure.
7. A cluster with n-3 vertices (i.e., n-3 boron atoms) and n+1 SEP for bonding has an hypho structure.
8. If the number of boron atoms (i.e., n) is larger than n+1 SEP then the extra boron occupies a capping position on a triangular phase.
Formula Skeletal Skeletal electron pairs type
\(\ce{[B_{n} H_{n}]^{2-}}\) n+1 closo
\(\ce{B_{n} H_{n + 4}}\) n+2 nido
\(\ce{B_{n} H_{n + 6}}\) n+3 arachno
\(\ce{B_{n} H_{n + 8}}\) n+4 hype
The polyhedra are always made up of triangular faces, so they are called deltahedra. Usually there are three possible structure types:
Structural relationship between closo, nido, and arachno boranes (and hetero-substituted boranes). The diagonal lines connect species that have the same number of skeletal electron pairs (SEP). Hydrogen atoms except those of the B-H framework are omitted. The red atom is omitted first, the green atom removed second. Adapted from R. W. Rudolph, Acc. Chem. Res., 1976, 9, 446.
Closo-boranes
• closed deltahedra without B-H-B 3c,2e-bonds
• thermally stable and moderately reactive.
• example: \(\ce{[B5H5]^{2-}}\): the ion builds up a trigonal, bipyramidal polyhedron
Nido-boranes
• closo borane with one corner less and addition of two hydrogen-atoms instead
• B-H-B-bonds and B-B-bonds are possible.
• thermally stability lies between closo- and arachno-boranes.
• example: \(\ce{B5H9}\) its structure can be assumed as the octahedral deltahedron of \(\ce{[B6H6]^{2-}}\) without one corner tetragonal pyramid
Arachno-boranes
• closo borane deltahedron but with two BH-units removed and two H-atoms added.
• it has to have B-H-B 3c, 2e-bonds.
• thermally unstable at room temperature and highly reactive.
• example: \{\ce{B4H10}\) the structure can be derived from \(\ce{[B6H6]^{2-}}\) -> deltahedron with two corners less.
There exist also other structures like the hypho-boranes, but they are less important.
Wade's Rules
Example \(1\): B5H11
What is the structure of B5H11?
Solution
1. Total number of valence electrons = (5 x B) + (11 x H) = (5 x 3) + (11 x 1) = 26
2. Number of electrons for each B-H unit = (5 x 2) = 10
3. Number of skeletal electrons = 26 – 10 = 16
4. Number SEP = 16/2 = 8
5. If n+1 = 8 and n-2 = 5 boron atoms, then n = 7
6. Structure of n = 7 is pentagonal bipyramid, therefore B5H11 is an arachno based upon a pentagonal bipyramid with two apexes missing.
Ball and stick representation of the structure of B5H11.
Example \(2\): B5H9?
What is the structure of B5H9?
Solution
1. Total number of valence electrons = (5 x B) + (9 x H) = (5 x 3) + (9 x 1) = 24
2. Number of electrons for each B-H unit = (5 x 2) = 10
3. Number of skeletal electrons = 24 – 10 = 14
4. Number SEP = 14/2 = 7
5. If n+1 = 7 and n-1 = 5 boron atoms, then n = 6
6. Structure of n = 6 is octahedral, therefore B5H9 is a nido structure based upon an octahedral structure with one apex missing.
Ball and stick representation of the structure of B5H9.
Example \(3\): B6H62-
What is the structure of B6H62-?
1. Total number of valence electrons = (6 x B) + (3 x H) = (6 x 3) + (6 x 1) + 2 = 26
2. Number of electrons for each B-H unit = (6 x 2) = 12
3. Number of skeletal electrons = 26 – 12 = 14
4. Number SEP = 14/2 = 7
5. If n+1 = 7 and n boron atoms, then n = 6
6. Structure of n = 6 is octahedral, therefore B6H62- is a closo structure based upon an octahedral structure.
Ball and stick representation of the structure of B6H62-.
Table \(1\) provides a summary of borane cluster with the general formula BnHnx- and their structures as defined by Wade’s rules.
Table \(1\): Wade’s rules for boranes.
Type Basic formula Example # of verticies # of vacancies # of e- in B + charge # of bonding MOs
Closo BnHn2- B6H62- n 0 3n + 2 n + 1
Nido BnHn4- B5H9 n + 1 1 3n + 4 n + 2
Arachno BnHn6- B4H10 n + 2 2 3n + 6 n + 3
Hypho BnHn8- B5H112- n + 3 3 3n + 8 n + 4
Contributors and Attributions
Stephen Contakes, Westmont College | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/15%3A_Parallels_between_Main_Group_and_Organometallic_Chemistry/15.04%3A_Cluster_Compounds/15.4.01%3A_Boranes.txt |
UNDER CONSTRUCTION
Low-symmetry groups ($C_1, \; C_s, \; C_i$)
$\begin{array}{|l|c|l|l|} \hline \bf{C_1} & \ E & h=1\ \hline A_1 & 1 \ \hline \end{array} \nonumber$
$\begin{array}{|l|lc|ll|} \hline \bf{C_s} & E & \sigma_h & h=2& \ \hline A & 1 & 1 & x, y , R_z & x^2, y^2, z^2, xy \ A' & 1 & -1 & z, R_x, R_y & yz, xz \ \hline \end{array}\nonumber$
$\begin{array}{|l|cc|ll|} \bf{C_1} & E & i & h=3 & \hline \ \hline A_g & 1 & 1 & R_x, R_y, R_z & x^2, y^2, z^2, xy, xz, yz \ A_u & 1 & -1 & x, y, z & \ \hline \end{array} \nonumber$
The groups $C_n$
$\begin{array}{|l|cc|ll|} \hline \bf{C_2} & E & C_2 & h=2 & \ \hline A & 1 & 1 & z, R_z & x^2, y^2, z^2, xy \ B & 1 & -1 & x, y , R_x, R_y & yz, xz \ \hline \end{array} \nonumber$
$\begin{array}{|l|c|ll|} \hline \bf{C_3} & E \: \: \: \: \: C_3 \: \: \: \: \: C_3^2 & h = 3 & \ \hline A & 1 \: \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: 1 & x, R_z & x^2 + y^2, z^2 \ E & \begin{Bmatrix} 1 & \varepsilon & \varepsilon^* \ 1 & \varepsilon^* & \varepsilon \end{Bmatrix} & (x, y), \; (R_x, R_y) & (x^2-y^2, xy), \; (xz, yz) \ \hline \end{array} \ \nonumber \$
$\varepsilon = e^{(2\pi i)/3}$
$\begin{array}{|l|c|ll|} \hline \bf{C_4} & E \: \: \: \: \: C_4 \: \: \: \: \: C_2 \: \: \: \: \: C_4^3 &h=4& \ \hline A & 1 \: \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: 1 & z, R_z & x^2 + y^2, z^2 \ B & 1 \: \: \: \: -1 \: \: \: \: \: \: \: \: 1 \: \: \: \: -1 & & x^2 - y^2, xy \ E & \begin{Bmatrix} 1 & i & -1 & -i \ 1 & -i & -1 & i \end{Bmatrix} & (x, y), \;(R_x, R_y) & (yz, xz) \ \hline \end{array}$
The groups $C_{nv}$
$\begin{array}{|l|cccc|ll|} \hline \bf{C_{2v}} & E & C_2 & \sigma_v(xz) & \sigma_v'(yz) & h=4 & \ \hline A_1 & 1 & 1 & 1 & 1 & z & x^2, y^2, z^2 \ A_2 & 1 & 1 & -1 & -1 & R_z & xy \ B_1 & 1 & -1 & 1 & -1 & x, R_y & xz \ B_2 & 1 & -1 & -1 & 1 & y, R_x & yz \ \hline \end{array}$
$\begin{array}{|l|ccc|ll|} \hline \bf{C_{3v}} & E & 2C_3 & 3\sigma_v & & \ \hline A_1 & 1 & 1 & 1 & z & x^2 + y^2, z^2 \ A_2 & 1 & 1 & -1 & R_z & \ E & 2 & -1 & 0 & x, y, R_x, R_y & x^2 - y^2, xy, xz, yz \ \hline \end{array}$
The groups $C_{nh}$
$\begin{array}{l|cccc|l|l} C_{2h} & E & C_2 & i & \sigma_h & & \ \hline A_g & 1 & 1 & 1 & 1 & R_z & x^2, y^2, z^2, xy \ B_g & 1 & -1 & 1 & -1 & R_x, R_y & xz, yz \ A_u & 1 & 1 & -1 & -1 & z & \ B_u & 1 & -1 & -1 & 1 & x, y & \end{array} \label{30.9}$
$\begin{array}{l|c|l|l}C_{3h} & E \: \: \: \: \: C_3 \: \: \: \: \: C_3^2 \: \: \: \: \: \sigma_h \: \: \: \: \: S_3 \: \: \: \: \: S_3^5 & & c = e^{2\pi/3} \ \hline A & 1 \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: 1 & R_z & x^2 + y^2, z^2 \ E & \begin{Bmatrix} 1 & \: \: c & \: \: c^* & \: \: 1 & \: \: c & \: \: c^* \ 1 & \: \: c^* & \: \: c & \: \: 1 & \: \: c^* & \: \: c \end{Bmatrix} & x, y & x^2 - y^2, xy \ A' & 1 \: \: \: \: \: \: 1 \: \: \: \: \: \: 1 \: \: \: \: -1 \: \: \: \: -1 \: \: \: \: \: -1 & z & \ E' & \begin{Bmatrix} 1 & c & c^* & -1 & -c & -c^* \ 1 & c^* & c & -1 & -c^* & -c \end{Bmatrix} & R_x, R_y & xz, yz \end{array} \label{30.10}$
The groups $D_n$
$\begin{array}{l|cccc|l|l} D_2 & E & C_2(z) & C_2(y) & C_2(x) & & \ \hline A & 1 & 1 & 1 & 1 & & x^2, y^2, z^2 \ B_1 & 1 & 1 & -1 & -1 & z, R_z & xy \ B_2 & 1 & -1 & 1 & -1 & y, R_y & xz \ B_3 & 1 & -1 & -1 & 1 & x, R_x & yz \end{array} \label{30.11}$
$\begin{array}{l|ccc|l|l} D_3 & E & 2C_3 & 3C_2 & & \ \hline A_1 & 1 & 1 & 1 & & x^2 + y^2, z^2 \ A_2 & 1 & 1 & -1 & z, R_z & \ E & 2 & -1 & 0 & x, y, R_x, R_y & x^2 - y^2, xy, xz, yz \end{array} \label{30.12}$
The groups $D_{nd}$
$\begin{array}{|l|ccccc|ll|} \hline D_{2d} & E & 2S_4 & C_2 & 2C_2' & 2\sigma_d & & \ \hline A_1 & 1 & 1 & 1 & 1 & 1 & & x^2 + y^2, z^2 \ A_2 & 1 & 1 & 1 & -1 & -1 & R_z & \ B_1 & 1 & -1 & 1 & 1 & -1 & & x^2 - y^2 \ B_2 & 1 & -1 & 1 & -1 & 1 & z & xy \ E & 2 & 0 & -2 & 0 & 0 & (x, y), (R_x, R_y) & (xz, yz) \ \hline \end{array} \label{30.14}$
$\begin{array}{l|cccccc|l|l} D_{3d} & E & 2C_3 & 3C_2 & i & 2S_6 & 3\sigma_d & & \ \hline A_{1g} & 1 & 1 & 1 & 1 & 1 & 1 & & x^2 + y^2, z^2 \ A_{2g} & 1 & 1 & -1 & 1 & 1 & -1 & R_z & \ E_g & 2 & -1 & 0 & 2 & -1 & 0 & R_x, R_y & x^2 - y^2, xy, xz, yz \ A_{1u} & 1 & 1 & 1 & -1 & -1 & -1 & & \ A_{2u} & 1 & 1 & -1 & -1 & -1 & 1 & z & \ E_u & 2 & -1 & 0 & -2 & 1 & 0 & x, y & \end{array} \label{30.15}$
The Groups $D_{nh}$
$\begin{array}{|c|rrrrrrrr|cc|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz) & h=8 & \ \hline A_{g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2, \; y^2, \; z^2\ B_{1g} & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & R_z & xy\ B_{2g} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & R_y & zx \ B_{3g} & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & R_x & yz \ A_{u} & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & & \ B_{1u} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \ B_{2u} & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & y & \ B_{3u} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & x & \ \hline \end{array}$
$\begin{array}{|c|rrrrrr|cc|} \hline \bf{D_{3h}} & E & 2C_3 & 3C_2 &\sigma_h & 2S_3 & 3\sigma_v & h=8 & \ \hline A_{1}' & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2, \; z^2\ A_{2}' & 1 & 1 & -1 & 1 & 1 & -1 & R_z & \ E' & 2 & -1 & 0 & 2 & -1 & 0 & (x,\;y) & (x^2-y^2,\; xy) \ A_{1}" & 1 & 1 & 1 & -1 & -1 & -1 & & \ A_{2}" & 1 & 1 & -1 & -1 & -1 & 1 & R_z & \ E" & 2 & -1 & 0 & -2 & 1 & 0 & (R_x,\;R_y) & (xz,\; yz) \ \hline \end{array}$
\begin{array}{|c|rrrrrrrrrr|cc|}
\hline \bf{D_{4h}} & E & 2C_4 & C_2 & 2C_2' & 2C_2" & i & 2S_4 & \sigma_h & 2\sigma_v & 2\sigma_d & h=16 & \
\hline A_{1g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2, \; z^2\
A_{2g} & 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 & R_z & \
B_{1g} & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \
B_{2g} & 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & & xy \
E_{g} & 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 & (R_x,\;R_y) & (xz,\; yz) \
A_{1u} & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & & \
A_{2u} & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \
B_{1u} & 1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 & & \
B_{2u} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & & \
E_{u} & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & & (x, \; y)\
\hline \end{array}
\begin{array}{|c|rrrrrrrr|cc|}
\hline \bf{D_{5h}} & E & 2C_5 & 2C_5^2 & 5C_2 & \sigma _h & 2S_5 & 2S_5^2 & 5 \sigma_h & h=20 & \
\hline A_{1}’ & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2, \; z^2\A_{2}’ & 1 & 1 & 1 & -1 & 1 & 1 & 1 & -1 & R_z & \
E_{1}’ & 2 & 2cos(72 ^{\circ}) & 2cos(144 ^{\circ}) & 0 & 2 & 2cos(72 ^{\circ}) & 2cos(144 ^{\circ}) & 0 & (x, \; y) & \
E_{2}’ & 2 & 2cos(144 ^{\circ}) & 2cos(72 ^{\circ}) & 0 & 2 & 2cos(144 ^{\circ}) & 2cos(72 ^{\circ}) & 0 & & (x^2-y^2,\; xy) \
A_{1}” & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & & \
A_{2}” & 1 & 1 & 1 & -1 & -1 & -1 & -1 & 1 & z & \
E_{1}” & 2 & 2cos(72 ^{\circ}) & 2cos(144 ^{\circ}) & 0 & -2 & -2cos(72 ^{\circ}) & -2cos(144 ^{\circ}) & 0 & (R_x,\;R_y) & (xz,\; yz) \
E_{2}” & 2 & 2cos(144 ^{\circ}) & 2cos(72 ^{\circ}) & 0 & -2 & -2cos(144 ^{\circ}) & -2cos(72 ^{\circ}) & 0 & & \
\hline \end{array}
High-symmetry groups
$\begin{array}{|c|cccccccc|c|c|} \hline \bf{D_{\infty h}} & \mathrm{E} & 2 \mathrm{C}_{\infty}^{\phi} & ... & \infty \sigma_v & i & 2S_{\infty}^{\phi} & ... & \infty C_2 & & \ \hline {A}_{1g} & 1 & 1 & ... & 1 & 1 & 1 & ...& 1 & & x^{2}+y^{2}, \, z^{2} \ {A}_{2g} & 1 & 1 & ... & -1 & 1 & 1 & ...& -1 & R_z & \ {E}_{1g} & 2 & 2\cos \phi & ... & 0 & 2 & -2\cos\phi & ...& 0 & (R_z, \, R_y) & (xz, \, yz) \ {E}_{2g} & 2 & 2\cos 2\phi & ... & 0 & 2 & 2\cos2\phi & ...& 0 & & (x^{2}-y^{2}, \, xy) \ ... & ... & ... & ... & ... & ... & ... & ...& ... & & \ {A}_{1u} & 1 & 1 & ... & 1 & -1 & -1 & ...& -1 & z & \ {A}_{2u} & 1 & 1 & ... & -1 & -1 & -1 & ...& 1 & & \ {E}_{1u} & 2 & 2\cos \phi & ... & 0 & -2 & 2\cos\phi & ...& 0 & (x, \, y) & \ {E}_{2u} & 2 & 2\cos 2\phi & ... & 0 & -2 & -2\cos2\phi & ...& 0 & & \ ... & ... & ... & ... & ... & ... & ... & ...& ... & & \ \hline \end{array}$
$C_{\infty v}$ and $D_{\infty h}$
$\begin{array}{l|cccccccc|l|l} D_{\infty h} & E & 2C_\infty^\Phi & \ldots & \infty \sigma_v & i & 2S_\infty^\Phi & \ldots & \infty C_2 & & \ \hline \Sigma_g^+ & 1 & 1 & \ldots & 1 & 1 & 1 & \ldots & 1 & & x^2 + y^2, z^2 \ \Sigma_g^- & 1 & 1 & \ldots & -1 & 1 & 1 & \ldots & -1 & R_z & \ \Pi_g & 2 & 2cos \Phi & \ldots & 0 & 2 & -2cos \Phi & \ldots & 0 & R_x, R_y & xz, yz \ \Delta_g & 2 & 2cos 2\Phi & \ldots & 0 & 2 & 2cos 2\Phi & \ldots & 0 & & x^2 - y^2, xy \ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & & \ \Sigma_u^+ & 1 & 1 & \ldots & 1 & -1 & -1 & \ldots & -1 & z & \ \Sigma_u^- & 1 & 1 & \ldots & -1 & -1 & -1 & \ldots & 1 & & \ \Pi_u & 2 & 2cos \Phi & \ldots & 0 & -2 & 2cos \Phi & \ldots & 0 & x, y & \ \Delta_u & 2 & 2cos 2\Phi & \ldots & 0 & -2 & -2cos 2\Phi & \ldots & 0 & & \ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & & \end{array} \label{30.16}$
$S_n$ groups
$\begin{array}{l|c|l|l} S_4 & E \: \: \: \: \: S_4 \: \: \: \: \: C_2 \: \: \: \: \: S_4^3 & & \ \hline A & 1 \: \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: 1 & R_z & x^2 + y^2, z^2 \ B & 1 \: \: \: \: -1 \: \: \: \: \: \: \: \: 1 \: \: \: \: -1 & z & x^2 - y^2, xy \ E & \begin{Bmatrix} 1 & i & -1 & -i \ 1 & -i & -1 & i \end{Bmatrix} & x, y, R_x, R_y & xz, yz \end{array} \label{30.17}$
$\begin{array}{l|c|l|l} S_6 & E \: \: \: \: \: C_3 \: \: \: \: \: C_3^2 \: \: \: \: \: i \: \: \: \: \: S_6^5 \: \: \: \: \: S_6 & & c=e^{2\pi/3} \ \hline A_g & 1 \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: 1 & R_z & x^2 + y^2, z^2 \ E_g & \begin{Bmatrix} 1 & \: \: c & \: \: c^* & \: \: 1 & \: \: c & \: \: c^* \ 1 \: \: & \: \: c^* & \: \: c & \: \: 1 & \: \: c^* & \: \: c \end{Bmatrix} & R_x, R_y & x^2 - y^2, xy, xz, yz \ A_u & 1 \: \: \: \: \: \: 1 \: \: \: \: \: \: 1 \: \: \: \: -1 \: \: \: \: -1 \: \: \: \: \: -1 & z & \ E_u & \begin{Bmatrix} 1 & c & c^* & -1 & -c & -c^* \ 1 & c^* & c & -1 & -c^* & -c \end{Bmatrix} & x, y & \end{array} \label{30.18}$
Cubic groups
$\begin{array}{l|c|l|l} T & E \: \: \: 4C_3 \: \: \: 4C_3^2 \: \: \: 3C_2 & & c=e^{2\pi/3} \ \hline A & 1 \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: 1 & & x^2 + y^2, z^2 \ E & \begin{Bmatrix} 1 & c & c^* & 1 \ 1 & c* & c & 1 \end{Bmatrix} & & 2z^2 - x^2 - y^2, x^2 - y^2 \ T & 3 \: \: \: \: \: 0 \: \: \: \: \: \: \: 0 \: \: \: -1 & R_x, R_y, R_z, x, y, z & xy, xz, yz \end{array} \label{30.19}$
$\begin{array}{l|ccccc|l|l} T_d & E & 8C_3 & 3C_2 & 6S_4 & 6\sigma_d & & \ \hline A_1 & 1 & 1 & 1 & 1 & 1 & & x^2 + y^2, z^2 \ A_2 & 1 & 1 & 1 & -1 & -1 & & \ E & 2 & -1 & 2 & 0 & 0 & & 2z^2 - x^-2 - y^2, x^2 - y^2 \ T_1 & 3 & 0 & -1 & 1 & -1 & R_x, R_y, R_z & \ T_2 & 3 & 0 & -1 & -1 & 1 & x, y, z & xy, xz, yz \end{array} \label{30.20}$
Contributors and Attributions
Claire Vallance (University of Oxford)
Curated or created by Kathryn Haas
16.03: Tanabe-Sugano Diagrams
Introduction
Tanabe-Sugano Diagrams are a form of correlation diagrams. These diagrams show relative energies of terms (eg. electronic microstates) by plotting the splitting energy and the field strength in terms of the Racah Parameter, B (plotted as $\frac{E}{B}$ vs $\frac{\Delta}{B}$). The Racah parameter accounts for $d$-electron-electron repulsions that affect the energies of the terms (and thus the transition energy). These diagrams are useful for interpreting electronic spectra. Specifically, the diagrams can be used to qualitatively predict the number of spin-allowed and spin-forbidden transitions, the relative intensities of these transitions, and their relative energies. The diagrams can also be used to estimate the $d$-orbital splitting energy ($\Delta$ or $10 D_q$) and the strength of field necessary to cause transition between high- and low-spin.
The Tanabe-Sugano diagrams shown below can be used to interpret the spectra of octahedral complexes of a given electron configuration. The diagrams for octahedral complexes for $d^2, \; d^3, \; d^4, \; d^5, \; d^6, \; d^7$ and $d^8$ are given below. The Tanabe-Sugano diagrams for $d^0, \; d^1, \; d^9$ and $d^{10}$ are unnecessary. In the case of $d^0$ and $d^{10}$, there are no possible $d-d$ transitions because the $d$-orbitals are either completely empty or completely full. In the case of $d^1$, there is only one free ion term ($^2D$) that is split into a ground state $^2T_{2g}$ and excited state $^2E_g$. There are no electron-electron repulsions to consider, and there is only one possible transition, so the Tanabe-Sugano diagram is unnecessary for $d^1$. The case of $d^9$ is similar and related to $d^1$ by the "positive hole" concept. In the case of $d^9$, the lone $^2D$ free ion term is split into a ground state $^2E_g$ term and an excited state $^2T_{2g}$ term. There is just one transition possible and so the Tanabe-Sugano diagram for $d^9$ is also unnecessary. In some cases, there are two versions given; a complete version and a simplified version. All diagrams are shown in full-page size.
These diagrams can also be used to interpret the electronic spectra of tetrahedral complexes. For a tetrahedral complex with $d^m$, use the diagram given below for $d^{10-m}$. All "$g$" subscripts on the terms are irrelevant for tetrahedrons.
Octahedron with 4 d-electrons
Figure 3 shows a Tanabe-Sugano diagram that includes all terms. A simplified version of the $d^4$ Tanabe-Sugano Diagram is shown below in Figure 4.
Octahedron with 5 d-electrons
Figure 5 shows a Tanabe-Sugano diagram that includes all terms. A simplified version of the $d^5$ Tanabe-Sugano Diagram is shown below in Figure 6. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/16%3A_Appendix/16.02%3A_Character_Tables.txt |
The elements are found in various states of matter and define the independent constituents of atoms, ions, simple substances, and compounds. Isotopes with the same atomic number belong to the same element. When the elements are classified into groups according to the similarity of their properties as atoms or compounds, the periodic table of the elements emerges. Chemistry has accomplished rapid progress in understanding the properties of all of the elements. The periodic table has played a major role in the discovery of new substances, as well as in the classification and arrangement of our accumulated chemical knowledge. The periodic table of the elements is the greatest table in chemistry and holds the key to the development of material science. Inorganic compounds are classi?ed into molecular compounds and solid-state compounds according to the types of atomic arrangements.
01: Elements and Periodicity
All substances in the universe are made of elements. According to the current generally accepted theory, hydrogen and helium were generated first immediately after the Big Bang, some 15 billion years ago. Subsequently, after the elements below iron (Z = 26) were formed by nuclear fusion in the incipient stars, heavier elements were produced by the complicated nuclear reactions that accompanied stellar generation and decay.
In the universe, hydrogen (77 wt%) and helium (21 wt%) are overwhelmingly abundant and the other elements combined amount to only 2%. Elements are arranged below in the order of their abundance,
\[\;_{1}^{1} H > \;_{2}^{4} He >> \;_{8}^{16} O > \;_{6}^{12} C > \;_{10}^{20} Ne > \;_{14}^{28} Si > \;_{13}^{27} Al > \;_{12}^{24} Mg > \;_{26}^{56} Fe\]
The atomic number of a given element is written as a left subscript and its mass number as a left superscript.
1.02: Discovery of elements
The long-held belief that all materials consist of atoms was only proven recently, although elements, such as carbon, sulfur, iron, copper, silver, gold, mercury, lead, and tin, had long been regarded as being atom-like. Precisely what constituted an element was recognized as modern chemistry grew through the time of alchemy, and about 25 elements were known by the end of the 18th century. About 60 elements had been identified by the middle of the 19th century, and the periodicity of their properties had been observed.
The element technetium (Z = 43), which was missing in the periodic table, was synthesized by nuclear reaction of Mo in 1937, and the last undiscovered element promethium (Z = 61) was found in the fission products of uranium in 1947. Neptunium (Z = 93), an element of atomic number larger than uranium (Z = 92), was synthesized for the first time in 1940. There are 103 named elements. Although the existence of elements Z = 104-111 has been confirmed, they are not significant in inorganic chemistry as they are produced in insufficient quantity.
All trans-uranium elements are radioactive, and among the elements with atomic number smaller than Z = 92, technetium, prometium, and the elements after polonium are also radioactive. The half-lives (refer to Section 7.2) of polonium, astatine, radon, actinium, and protoactinium are very short. Considerable amounts of technetium 99Tc are obtained from fission products. Since it is a radioactive element, handling 99Tc is problematic, as it is for other radioactive isotopes, and their general chemistry is much less developed than those of manganese and rhenium in the same group.
Atoms are equivalent to alphabets in languages, and all materials are made of a combination of elements, just as sentences are written using only 26 letters. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/01%3A_Elements_and_Periodicity/1.01%3A_The_origin_of_elements_and_their_distribution.txt |
Wave functions of electrons in an atom are called atomic orbitals. An atomic orbital is expressed using three quantum numbers; the principal quantum number, n; the azimuthal quantum number, l; and the magnetic quantum number, m$ell$. For a principal quantum number n, there are n azimuthal quantum numbers l ranging from 0 to n-1, and each corresponds to the following orbitals.
$\begin{split} \ell & : 0, 1, 2, 3, 4, \ldots \ & \; \; \; s, p, d, f, g, \ldots \end{split}$
An atomic orbital is expressed by the combination of n and l. For example, n is 3 and l is 2 for a 3d orbital. There are 2l+1 m$ell$ values, namely l, l-1, l-2,..., -l. Consequently there are one s orbital, three p orbitals, five d orbitals and seven f orbitals. The three aforementioned quantum numbers are used to express the distribution of the electrons in hydrogen-type atom, and another quantum number ms (1/2, -1/2) which describes the direction of an electron spin is necessary to completely describe an electronic state. Therefore, an electronic state is defined by four quantum numbers (n, l, m$ell$, ms).
The wave function $\psi$ which determines the orbital shape can be expressed as the product of a radial wavefunction R and an angular wave function Y as follows.
$\psi_{n, l, m_{l}} = R_{n, l} (r) Y_{l, m_{l}} (\theta, \phi)$
R is a function of distance from the nucleus, and Y expresses the angular component of the orbital. Orbital shapes are shown in Figure $1$. Since the probability of the electron’s existence is proportional to the square of the wave function, an electron density map resembles that of a wave function. The following conditions must be satisfied when each orbital is filled with electrons.
[The conditions of electron filling]
Pauli principle
The number of electrons that are allowed to occupy an orbital must be limited to one or two, and, for the latter case, their spins must be anti-parallel (different direction).
Hund's rule
When there are equal-energy orbitals, electrons occupy separate orbitals nd their spins are parallel (same direction).
The order of orbital energy of a neutral atom is
$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p \ldots$
and the electron configuration is determined as electrons occupy orbitals in this order according to the Pauli principle and Hund's rule. An s orbital with one m$ell$ can accommodate 2 electrons, a p orbital with three m$ell$ 6 electrons, and a d orbital with five m$ell$ 0 electrons.
Exercise $1$
Describe the electron configuration of a C atom, an Fe atom, and a Au atom.
Answer
Electrons equal to the atomic number are arranged in the order of orbital energies. Since the electrons inside the valence shell take the rare gas configuration, they may be denoted by the symbol of a rare gas element in brackets.
$\begin{split} C & : 1s^{2} 2s^{2} 2p^{2} \quad or \quad [He] 2s^{2} 2p^{2} \ Fe & : 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{6} 4s^{2} \quad or \quad [Ar] 3d^{6} 4s^{2} \ Au & : 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 4f^{14} 5s^{2} 5p^{6} 5d^{10} 6s^{1} \quad or \quad [Xe] 4f^{14} 5d^{10} 6s^{1} \end{split}$
Table $1$ Periodic table of elements. The values are atomic weights
1 2 3 4 5 6 7 8 9
1 1.008
1H
2 6.941
3Li
9.012
4Be
3 22.99
11Na
24.31
12Mg
4 39.10
19K
40.08
20Ca
44.96
21Sc
47.87
22Ti
50.94
23V
52.00
24Cr
54.94
25Mn
55.85
26Fe
58.93
27Co
5 85.47
37Rb
87.62
38Sr
88.91
39Y
91.22
40Z
92.91
41Nb
95.94
42Mo
(99)
43Tc
101.1
44Ru
102.9
45Rh
6 132.9
55Cs
137.3
56Ba
Lanthanoid 178.5
72Hf
180.9
73Ta
183.8
74W
186.2
75Re
190.2
76Os
192.2
77Ir
7 (223)
87Fr
(226)
88Ra
Actinoid
Lanthan-oid 138.9
57La
140.1
58Ce
140.9
59Pr
144.2
60Nd
(145)
61Pm
150.4
62Sm
152.0
63Eu
Actinoid (227)
89Ac
232.0
90Th
231.0
91Pa
238.0
92U
(237)
93Np
(239)
94Pu
(243)
95Am
10 11 12 13 14 15 16 17 18
4.003
2He
10.81
5B
12.01
6C
14.01
7N
16.00
8O
19.00
9F
20.18
10Ne
26.98
13Al
28.09
14Si
30.97
15P
32.07
16S
35.45
17Cl
39.95
18Ar
58.69
28Ni
63.55
29Cu
65.39
30Zn
69.72
31Ga
72.61
32Ge
74.92
33As
78.96
34Se
79.90
35Br
83.80
36Kr
106.4
46Pd
107.9
47Ag
112.4
48Cd
114.8
49In
118.7
50Sn
121.8
51Sb
127.6
52Te
126.9
53I
131.3
54Xe
195.1
78Pt
197.0
79Au
200.3
80Hg
204.4
81Tl
207.2
82Pb
209.0
83Bi
(210)
84Po
(210)
85At
(222)
86Rn
157.3
64Gd
158.9
65Tb
162.5
66Dy
164.9
67Ho
167.3
68Er
168.9
69Tm
173.0
70Yb
175.0
71Lu
(247)
96Cm
(247)
97Bk
(252)
98Cf
(252)
99Es
(257)
100Fm
(258)
101Md
(259)
102No
(262)
103Lr
1.04: Block classification of the periodic table and elements
Starting from hydrogen, over 100 elements are constituted as electrons are successively accommodated into 1s, 2s, 2p, 3s, 3p, 4s, and 3d orbitals one by one from lower to higher energy levels. When elements with similar properties are arranged in columns, the periodic table of the elements is constructed. The modern periodic table of the elements is based on one published by D. I. Mendeleev in 1892, and a variety of tables have since been devised. The long periodic table recommended by IUPAC is the current standard, and it has the group numbers arranged from Group 1 alkali metals through Group 18 rare gas elements (Table \(1\)).
Based on the composition of electron orbitals, hydrogen, helium and Group 1 elements are classified as s-block elements, Group 13 through Group 18 elements p-block elements, Group 3 through Group 12 elements d-block elements, and lanthanoid and actinoid elements f-block elements. (Figure \(2\)). s-Block, p-block, and Group 12 elements are called main group elements and d-block elements other than Group 12 and f-block elements are called transition elements. The characteristic properties of the elements that belong to these four blocks are described in Chapter 4 and thereafter. Incidentally, periodic tables that denote the groups of s-block and p-block elements with Roman numerals (I, II,..., VIII) are still used, but they will be unified into the IUPAC system in the near future. Since inorganic chemistry covers the chemistry of all the elements, it is important to understand the features of each element through reference to the periodic table. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/01%3A_Elements_and_Periodicity/1.03%3A_Electronic_Structure_of_Elements.txt |
Organic compounds are molecular compounds that contain mainly carbon and hydrogen atoms. Since inorganic chemistry deals with all compounds other than organic ones, the scope of inorganic chemistry is vast. Consequently, we have to study the syntheses, structures, bondings, reactions, and physical properties of elements, molecular compounds, and solid-state compounds of 103 elements. In recent years, the structures of crystalline compounds have been determined comparatively easily by use of single crystal X-ray structural analysis, and by through the use of automatic diffractometers. This progress has resulted in rapid development of new areas of inorganic chemistry that were previously inaccessible. Research on higher dimensional compounds, such as multinuclear complexes, cluster compounds, and solid-state inorganic compounds in which many metal atoms and ligands are bonded in a complex manner, is becoming much easier. In this section, research areas in inorganic chemistry will be surveyed on the basis of the classification of the bonding modes of inorganic materials.
Element
Elementary substances exist in various forms. For example, helium and other rare gas elements exist as single-atom molecules; hydrogen, oxygen, and nitrogen as two-atom molecules; carbon, phosphorus, and sulfur as several solid allotropes; and sodium, gold, etc. as bulk metals. A simple substance of a metallic element is usually called bulk metal, and the word “metal” may be used to mean a bulk metal and “metal atom or metal ion” define the state where every particle is discrete. Although elementary substances appear simple because they consist of only one kind of element, they are rarely produced in pure forms in nature. Even after the discovery of new elements, their isolation often presents difficulties. For example, since the manufacture of ultra high purity silicon is becoming very important in science a purification processes have been developed in recent years.
Exercise \(2\)
Give examples of allotropes.
Answer
• Carbon: graphite, diamond.
• Phosphorus: white phosphorus, red phosphorus.
Molecular compounds
Inorganic compounds of nonmetallic elements, such as gaseous carbon dioxide CO2, liquid sulfuric acid H2SO4, or solid phosphorus pentoxide P2O5, satisfy the valence requirements of the component atoms and form discrete molecules which are not bonded together. The compounds of main group metals such as liquid tin tetrachloride SnCl4 and solid aluminum trichloride AlCl3 have definite molecular weights and do not form infinite polymers.
Most of the molecular compounds of transition metals are metal complexes and organometallic compounds in which ligands are coordinated to metals. These molecular compounds include not only mononuclear complexes with a metal center but also multinuclear complexes containing several metals, or cluster complexes having metal-metal bonds. The number of new compounds with a variety of bonding and structure types is increasing very rapidly, and they represent a major field of study in today' inorganic chemistry (see Chapter 6).
Solid-state compounds
Although solid-state inorganic compounds are huge molecules, it is preferable to define them as being composed of an infinite sequence of 1-dimensional (chain), 2-dimensional (layer), or 3-dimensional arrays of elements and as having no definite molecular weight. The component elements of an inorganic solid bond together by means of ionic, covalent, or metallic bonds to form a solid structure. An ionic bond is one between electronically positive (alkali metals etc.) and negative elements (halogen etc.), and a covalent bond forms between elements with close electronegativities. However, in many compounds there is contribution from both ionic and covalent bonds (see Section 2.1 about bondings).
Exercise \(3\)
Give examples of solid-state inorganic compounds.
Answer
• Sodium chloride NaCl
• Silicon dioxide, SiO2
• Molybdenum disulfide, MoS2
The first step in the identification of a compound is to know its elemental composition. Unlike an organic compound, it is sometimes difficult to decide the empirical formula of a solid-state inorganic compound from elemental analyses and to determine its structure by combining information from spectra. Compounds with similar compositions may have different coordination numbers around a central element and different structural dimensions. For example, in the case of binary (consisting of two kinds of elements) metal iodides, gold iodide, AuI, has a chain-like structure, copper iodide, CuI, a zinc blende type structure, sodium iodide, NaI, has a sodium chloride structure, and cesium iodide, CsI, has a cesium chloride structure (refer to Section 2.2 (e)), and the metal atoms are bonded to 2, 4, 6 or 8 iodine atoms, respectively. The minimum repeat unit of a solid structure is called a unit lattice and is the most fundamental information in the structural chemistry of crystals. X-ray and neutron diffraction are the most powerful experimental methods for determining a crystal structure, and the bonds between atoms can only be elucidated by using them. Polymorphism is the phenomenon in which different kinds of crystals of a solid-state compound are obtained in which the atomic arrangements are not the same. Changes between different polymorphous phases with variations in temperature and/or pressure, or phase transitions, are an interesting and important problem in solid-state chemistry or physics.
We should keep in mind that in solid-state inorganic chemistry the elemental composition of a compound are not necessarily integers. There are extensive groups of compounds, called nonstoichiometric compounds, in which the ratios of elements are non-integers, and these non-stoichiometric compounds characteristically display conductivity, magnetism, catalytic nature, color, and other unique solid-state properties. Therefore, even if an inorganic compound exhibits non-integral stoichiometry, unlike an organic compound, the compound may be a thermodynamically stable, orthodox compound. This kind of compound is called a non-stoichiometric compound or Berthollide compound, whereas a stoichiometric compound is referred to as a Daltonide compound. The law of constant composition has enjoyed so much success that there is a tendency to neglect non-stoichiometric compounds. We should point out that groups of compounds in which there are slight and continuous changes of the composition of elements are not rare.
Problem 1.1
Express the isotopes of hydrogen, carbon, and oxygen using the symbols of the elements with atomic and mass numbers and write the number of protons, neutrons, and electrons in parenthesis.
Superheavy elements
The last element in the ordinary periodic table is an actinoid element lawrencium, Lr, (Z = 103). However, elements (Z = 104 – 109) "have already been synthesized" in heavy ion reactions using nuclear accelerators. These are 6d elements which come under the 5d transition elements from hafnium, Hf, to iridium, Ir, and it is likely that their electronic structures and chemical properties are similar. As a matter of fact, only the existence of nuclides with very short lives has been confirmed. The trouble of naming the super heavy elements is that the countries of their discoverers, the United States, Russia and Germany, have proposed different names. The tentative names of these elements are: unnilquadium Une (Z = 104), unnilpentium Unp (Z = 105), unnilhexium Unh (Z = 106), unnilseptium Unq (Z = 107), unniloctium Uno (Z = 108) and unnilennium Une (Z = 108). It has recently been settled that they be named: Rutherfordium 104Rf, Dubnium 105Db, Seaborgium 106Sg, Bohrium 107Bh, Hassium 108Hs, and Meitnerium 109Mt.
"Synthesis" of the element (Z = 110), which should come under platinum, was considered the technical limit, but there is a recent report that even the element (Z = 112) "was synthesized". In any case, the superheavy elements will run out shortly. It is natural that complications are caused by naming of a new element a scientist to have a new element named after him or her. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/01%3A_Elements_and_Periodicity/1.05%3A_Bonding_states_of_elements.txt |
Atomic radii, bond angles, and the valence electrons of the atoms or ions constituting compounds govern the bonding, structure, reactions, and physical properties of the compounds. It is desirable that the chemical properties of known and new compounds can be explained and predicted using universal parameters characteristic of the constituent elements. Inorganic chemistry has developed along with the discovery of new compounds and novel bonding modes. Therefore, it is important to understand the bonding modes, the geometrical and electronic factors governing the bonding, and to learn the basic concepts of molecular orbital theory.
Thumbnail: The central \(\ce{Mn^{3+}}\) metal In \(\ce{Mn(acac)3}\) is ligated by three acac (acetylacetonate) ligands in an approximatt octahedral geometry. (CC BY; Karl Harrison 3DChem.com).
02: Bonding and Structure
The bond in which a pair of electrons bind atoms A and B is called a covalent bond, and it is written as A-B or A:B. Since two pairs of electrons are involved in a double bond and three pairs in a triple bond, they are designated by A=B, A $\equiv$ B or A::B, or A:::B, respectively. The covalent bond is a simple but very useful concept proposed by G. N. Lewis at the beginning of this century and its representation is called the Lewis structure. Unshared pair of valence electrons are called lone pairs, and they are expressed by a pair of dots like A:.
Exercise $1$
Describe the Lewis structures of the nitrogen molecule N2 and the oxygen molecule O2.
Answer
• : N:::N:
• : O::O :
Eight electrons are required to fill an s and three p orbitals, and when the total number of electrons used for the bonds and lone pairs is eight, a stable molecular structure results. This is called the octet rule and is useful when qualitatively discussing the molecular structures of main group compounds. Of course, this rule is not applied to a hydrogen molecule, H2, but is applicable to covalent molecules, such as simple two-atomic molecules O2 or CO and even to complicated organic compounds. For the elements after the 3rd period, the number of covalent bonds is sometimes five (e.g. PCl5) or six (e.g. SF6), and the central atom of these molecules shows hypervalency. In this case, because s and p electrons run short to form more than four 2-electron covalent bonds, it was once believed that d electrons were partly involved. The present view is, however, that these hypervalent bonds use only s and p orbitals but that the bond orders are lower than those of single bonds.
The electrostatic bond between cations (positive ion) and anions (negative ion), such as in sodium chloride, NaCl, is called an ionic bond. Since the total electrical charge in a compound should be zero, the electrical charges of cations and anions are equal. There is a partial contribution from covalent bonds even in an ionic compound, and the ions are not necessarily bonded only by the electrostatic interaction.
Pauling’s electroneutrality principle states that the net electrical charge of each component of a compound is essentially neutral. As will be mentioned later, the structures of many solid compounds are described as an alternate array of cations and anions, and are classified into several representative crystal types.
Metal atoms are bound together by means of the conduction electrons originating from the valence electrons of metal atoms. The bond due to the conduction electrons in a bulk metal is called the metallic bond.
Generally, chemical bonds can be assigned to either of the three kinds mentioned above, but new compounds have been synthesized one after another which cannot always be classified by the simple 2-center electron pair covalent bond. They include electron-deficient bonds in boron hydrides, coordinate bonds in transition metal complexes, the metal-metal bonds in metal cluster compounds, etc., and new concepts have been introduced into bond theory to account for these new kinds of chemical bonds. As has already been described, a weak bonding interaction called the van der Waals interaction has been recognized in neutral atomic or molecular compounds. The potential of this interaction is inversely proportional to the 6th power of the distance between atoms. The adjacent but non-bonded distance between atoms is estimated by the sum of the van der Waals radius assigned to each atom.
The weak interaction X-H-Y that a hydrogen atom forms with the atoms X, Y (nitrogen, oxygen, fluoride, etc..) with larger electronegativity than that of hydrogen is called the hydrogen bond. Hydrogen bonding plays an important role in ice, the structure of the double helix of DNA (deoxyribonucleic acid), etc. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/02%3A_Bonding_and_Structure/2.01%3A_Classification_of_bonding.txt |
Two parameters, radii and the electron attracting power of atoms or ions, determine the bonding, structure, and reaction of elementary substances and compounds. Much effort has been devoted to finding numerical values for these two factors applicable to all materials. It is hoped that the chemical properties of a known compound, and of a still non-existent new material, can be predicted with a combination of suitable numerical values. Firstly, geometrical factors will be described.
Table $1$ Atomic radii (pm)
H
32
Li
123
Be
89
B
82
C
77
N
75
O
73
F
72
Na
154
Mg
136
Al
118
Si
111
P
106
S
102
Cl
99
K
203
Ca
174
Sc
144
Ti
132
V
122
Cr
118
Mn
117
Fe
117
Co
116
Ni
115
Cu
117
Zn
125
Ga
126
Ge
122
As
120
Se
117
Br
114
Rb
216
Sr
191
Y
162
Zr
145
Nb
134
Mo
130
Tc
127
Ru
125
Rh
125
Pd
128
Ag
134
Cd 148 In
144
Sn
140
Sb
140
Te
136
I
133
Cs
235
Ba
198
La
169
Hf
144
Ta
134
W
130
Re
128
Os
126
Ir
127
Pt
130
Au
134
Hg
149
Tl
148
Pb
147
Bi
146
Atomic and ionic radii
The electron density in atoms gradually approaches, but never reaches, zero as the distance from the nucleus increases. Therefore, strictly speaking the radius of an atom or ion is indeterminable. However, it is possible to determine the bond distance between atomic nuclei experimentally. Atomic radii determined experimentally are one of the most important atomic parameters describing the structural chemistry of compounds. It is reasonable to define the metal radius of a bulk metal as half of the distance between metal atoms. Half of the distance between atoms is defined also as the covalent radius of a covalent elementary substance (Table $1$).
Table $2$ Ionic radii (in pm).*
Li+ (4) 59 Be2+(4) 27 B3+(4) 11 N3+(6) 16 O2-(6) 140 F- (6) 133
Na+ (6) 102 Mg2+(6) 72 Al3+(6) 54 P3+(6) 44 S2-(6) 184 Cl- (6) 181
K+ (6) 138 Ca2+(6) 100 Ga3+(6) 62 As3+(6) 58 Se2-(6) 198 Br- (6) 196
Rb+ (6) 152 Sr2+(6) 118 In3+(6) 80 Te2-(6) 221 I- (6) 220
Cs+ (6) 167 Ba2+(6) 135 Tl3+(6) 89
*Numbers in parentheses are the coordination number of the ions.
Since the cations and anions of different elements in an ionic compound are bonded by electrostatic interactions, the bond distance is the sum of ionic radii assigned to the cation and anion. The standard ionic radius of one species is fixed first and is then subtracted from the distance between ions to decide the radius of the partner ion. As the standard, the radius of O2- ion in a number of oxides is set to 140 pm (1 pm = 10-12 m) (R. D. Shannon). Cationic radii in oxides are the difference between the bond distance and 140 pm. After cation radii in oxides are decided, other anion radii can be calculated by subtraction of the cation radii from the distances between the atoms in ionic compounds. By applying such methods to many ionic compounds, ionic radii have been compiled in such a way that experimental and calculated values are generally consistent (Table $2$).
Even ionic compounds have some covalent contribution and it is not expected that calculated and experimental bond distances will agree exactly. Even if the ionic radius assigned to a standard ion is changed, we can still compile a set of ionic radii that are consistent across many compounds. Other examples of the proposed radii of O2- ion are 132 pm (V. M. Goldschmidt) or 60 pm (J. C. Slater). We must also be mindful that the cation-anion distances of the same ion pair become larger as the coordination number of opposite ions increases. Therefore, in any discussion of the structural features of ionic compounds from a viewpoint of ionic radii, a set of the ionic radii calculated using the same standard radius for the compounds with the same coordination number should be used.
Exercise $2$
Which ionic radius is larger, Cs+ or F- ?
Answer
Cs+ (167 pm) > F- (133 pm). The anion radius is not always larger.
The metal and covalent radii, also called the atomic radii, become smaller in the same period of the periodic table as the group of the element goes to the right and then increase again in the next period. The lanthanide contraction is responsible for the 5th period (4d) elements having almost the same atomic radii as those of the 6th period (5d) ones. In the periodic table, the lanthanide elements are inserted before the 5d elements. The atomic radii of lanthanide elements decrease noticeably as the effective nuclear charge increases because the screening effects of the 4f orbitals of lanthanide elements are weak due to their orbital shapes. Consequently, the atomic radii of the elements following lanthanides are very similar to those of the 4d elements.
Lattice enthalpy
Although the stability of a crystal at constant temperature and pressure depends on the Gibbs free energy change of the crystal’s formation from its constituent ions, the stability of a crystal is determined mostly by the enthalpy change alone since the lattice formation is very exothermic, and the entropy term is negligibly small (refer to Section 3.1). Lattice enthalpy, $\Delta H_L$, is defined as the standard enthalpy change of the reaction in which an ionic crystal decomposes into gaseous ions (s is solid, g is gas and L is lattice).
$MX(s) \rightarrow M^{+} (g) + X^{-} (g) \qquad \Delta H_{L}$
Lattice enthalpy is indirectly calculated from the values of the enthalpy change at each stage using a Born-Haber cycle (Figure $1$). Namely, a closed cycle is formed using enthalpy data; standard enthalpy of formation $\Delta H_f$ of an ionic crystal from elements, sublimation enthalpy of an elementary solid, or atomization enthalpy $\Delta\Hatom corresponding to the dissociation enthalpy of a gaseous elementary molecule, the ionization enthalpy \(\Delta$Ηion, which is the sum of the ionization enthalpy of cation formation and electron acquisition enthalpy of anion formation. Lattice enthalpy is calculated using the relation that enthalpy change in a cycle is zero.
$\Delta H_{atom}^{0} + \Delta_{ion}^{0} - \Delta H_{L}^{0} - \Delta H_{f}^{0} = 0$
Madelung constant
The total Coulomb potential energy that exists between the ions in an ionic crystal consisting of ions A and B should be the sum of the individual Coulomb potential energies Vab. Since the locations of the ions in the crystal lattice are decided by the structure type, the total Coulomb potential between all ions is calculated by setting the distance between the ions to d. A is the Madelung constant that is characteristic of each crystal type (Table $3$).
$V = N_{A} \frac{e^{2}}{4 \pi \epsilon_{0}} \left(\dfrac{z_{A} z_{B}}{d} \right) \times A$
NA is Avogadro's constant and zA and zB are the electric charges of the cation and anion. The electrostatic interaction between contiguous ions is the strongest, and the Madelung constant generally becomes larger as the coordination number increases. Because the electrical charges have opposite signs, the potential becomes negative, indicating the stabilization that accompanies the formation of a crystal lattice from well dispersed, gaseous phase ions. Although it is generally true that the lowest electrostatic potential leads to the most stable structure, this is not strictly correct since there are also other interactions to consider.
Table $3$ Madelung constants
Structural type A
Rock-salt 1.748
Cesium chloride 1.763
Sphalerite 1.638
Wurtzite 1.641
Fluorite 2.519
Rutile 2.408
The second largest factor that contributes to the lattice enthalpy is the van der Waals force, and dispersion forces or the London interaction is the main origin of this force. It is an attractive interaction between electric dipoles, which is inversely proportional to the 6th power of the distance d between ions. The van der Waals force is very small.
$V = - \frac{N_{A} C}{d^{6}}$
The value of the constant C is a function of each compound. Since it is at most 1% of the Coulombic force, it may be safely neglected in the calculation of lattice enthalpy.
(d) Structure of metal crystals
If we imagine metal atoms as being hard balls, when densely packed in two dimensions each ball will be in contact with six other balls (A). When another layer of this 2 dimensional arrangement is placed on top of the first, the packing will be densest and the structure most energetically stable when the metal atoms are placed on top of the hollows (B) of the first layer. When a 3rd layer is placed on top of the 2nd layer, there are two possibilities. Namely, the 3rd layer (A) overlaps with the 1st layer (A) or the 3rd layer (C) overlaps with neither (A) nor (B). The ABAB...-type packing is called hexagonally close-packed (hcp) (Figure $2$), and the ABCABC...-type is called cubic close-packed (ccp) (Figure $3$). In both cases, each ball is surrounded by 12 balls, that is, it is 12-coordinated. The coordination polyhedron of hcp is anti-cubooctahedron,
and that of ccp is cubooctahedron. When the lattice is sliced in different planes, the unit lattice of ccp appears to be a face-centered cubic lattice (fcc), containing a ball at each cubical apex and on the center of each face (Figure $4$). The unit lattice of hcp is a rhombohedral prism in which two balls are located in the positions shown in (Figure $5$). There are several different modes of piling up layers other than the normal hcp and ccp, and many examples are known.
The lattice with another ball at the center of a cubic lattice consisting of eight balls is the body centered cubic lattice (bcc), and some metals assume this mode of packing. The ratio of space occupation in a bcc lattice is smaller than that of close-packed ones but the difference is not large. Although the central ball is formally 8-coordinated, it is essentially 14-coordinated since there are a further six balls only 15.5% more distant than the first eight balls. However, because of the smaller ratio of space occupation, bcc appears relatively rarely and pure metals tend to adopt hcp or ccp.
In both hcp and ccp, the cavities among the balls are either the Oh holes enclosed octahedrally by six balls or the Td holes enclosed tetrahedrally by four balls (Figure $6$). (Oh and Td are the symmetry symbols used in group theory.) In ionic solids, if the anions are in hcp or ccp arrangements, cations enter into either of these cavities.
Ionic Crystal
In ionic crystals, such as metal halides, oxides, and sulfides, metal cations and anions are aligned alternately, and the solid is bound together mainly by electrostatic bonding. Many metal halides dissolve in polar solvents, e.g. sodium chloride NaCl dissolves in water; whereas metal oxides and sulfides, in which there is a significant contribution of covalent bonding, are usually insoluble even in the most polar of solvents. The fundamental structure of ionic crystals is that larger ions (usually anions) are close-packed and smaller ions (usually cations) enter into the octahedral or tetrahedral cavities between them. Ionic crystals are classified into several typical structures according to the kinds of cations and anions involved and their ionic radii. Each structure type is called by the name of the typical compound, just as the rock salt structure representing the structures of not only NaCl (rock salt) but also various other compounds. Representative structure types of solid compounds and examples belonging to each type are shown in Table $4$.
Table $4$ Crystal types of solid-state compounds
Crystal type Coordination number Examples of compounds
Rock-salt (6,6) LiCl, NaCl, KBr, RbI, AgCl, MgO, NiO, InP
Cesium chloride (8,8) CsCl, CsBr, CsI, CuZn
Sphalerite (4,4) ZnS, CdS, HgS, CuCl, GaP
Fluorite (8,4) CaF2, SrF2, CdF2, ZrO2, UO2
Rutile (6,3) TiO2, SnO2, RuO2, NiF2
Cadmium iodide (6,3) CdI2, CoI2, Mg(OH)2
Rhenium oxide (6,2) ReO3, WO3, Sc(OH)3
Perovskite (6,2) CaTiO3, BaTiO3, SrTiO3
Rock-salt structure Sodium chloride NaCl is a typical compound in which Cl- anions are arranged in ccp and Na+ cations occupy all the octahedral holes (Oh holes) (Figure $7$). Each Na+ cation is surrounded by six Cl- anions. The same structure results even if the positions of anions and cations are exchanged. In the case of the reversed structure, each Cl- anion is surrounded by six Na+ cations. Namely, each ion is 6-coordinated and it is convenient to describe the structure as the (6,6)-structure. The number of ions in a unit lattice is calculated by summing up the ions shown in Figure $7$. Since there is one ion inside the lattice, the ions on the faces of the lattice are shared by 2, on the edges by 4, and on the corners by 8 lattices, a net of 4 Cl ions belonging to the unit lattice of NaCl is obtained by multiplying the numbers of ions inside the lattice by 1, on the faces by 1/2, on the edges by 1/4 and on the corners by 1/8. The number of Na ions in the unit lattice is also 4 and the ratio of Cl and Na ions agrees with the formula NaCl.
Cesium chloride structure Cesium chloride, CsCl, is a typical example of the structure shown in Figure $8$. There is a Cs+ ion at the center and eight Cl- are located at the eight corners of the cube. Conversely, even if a Cl- comes to the center and eight Cs+ come to the corners, the number of each ion in the unit lattice is the same. Thus, this is referred to as the (8, 8)-structure. Since there is one Cs+ and one Cl- ion belonging to this unit lattice, it coincides with the formula CsCl.
Zinc blende structure Zinc blende has the composition ZnS and its unit lattice is shown in Figure $9$. S2- anions are arranged in ccp and Zn2+ cations occupy half of the tetrahedral holes (Td holes). In this arrangement, each cation is coordinated by four anions, and each anion by four cations. Hence, this is a (4, 4)-structure. There are both four Zn2+ and S2- ions belonging to this unit lattice and the atomic ratio coincides with the formula of ZnS.
Fluorite structure The composition of fluorite is CaF2. Since the number of F- is twice that of Ca2+, all the tetrahedral holes of Ca2+ arranged in ccp are occupied by F-, as shown in Figure $10$. There are four Ca2+ and eight F- ions and the number of ions is 4 times the formula. The anti-fluorite structure exchanges the cations and anions, and is exemplified by potassium oxide K2O, etc.
Exercise $3$
How many cations and anions are there in a unit lattice of zinc blende structure?
Answer
All four cations are included in a unit lattice. The anions occupy the 8 corners and 6 faces and the number is 8 x 1/8 + 6 x 1/2 = 4.
Radius ratio
Generally, the total Coulombic potential energy Ec of the univalent ionic compound MX is expressed by the following formula.
$E_{c} = - \frac{N_{A} e^{2}}{4 \pi \epsilon_{0} R} A$
NA is the Avogadro constant, A the Madelung constant and R the distance between ions. According to this formula, a structure with a larger A / R ratio is more stable. The Madelung constant of an MX compound increases with increasing coordination number. On the other hand, it is advantageous to lower the coordination number and to reduce R in the case of small M, rendering contact between M and X more difficult. In an ionic crystal, the ratio of rM and rX with the anions contacting each other and also with the cations depends on the coordination number.
In a partial structure consisting only of anions, the anions form a coordination polyhedron around a cation. The anionic radius rX is one half of the distance of the edge of the polyhedron and the distance from the cation center to an apex of the polyhedron is the sum of the anion and cation radii rX + rM. The coordination polyhedron of the CsCl structure is a cube, the NaCl structure an octahedron, and the ZnS structure a tetrahedron. The distance from the center of each polyhedron to an apex is $\sqrt{3} r_{X}$, $\sqrt{2} r_{X}$, \frac{\sqrt{6}}{2} r_{X}\). Therefore, the ratios of the cationic and anionic radii rM / rX are $\frac{\sqrt{3} r_{X} − r_{X}}{r_{X}} = \sqrt{3} −1 = 0.732$for CsCl, $\frac{\sqrt{2} r_{X} − r_{X}}{r_{X}} = \sqrt{2} −1 = 0.414$for NaCl, and $\frac{\frac{\sqrt{6}}{2} r_{X} - r_{X}}{r_{x}} = \frac{\sqrt{6}}{2} - 1 = 0.225$for ZnS structures (Figure $11$).
It has been explained that the coordination number decreases when these radius ratios are smaller than the given values since cations and anions do not come into contact with each other, causing instability. On the other hand, the coordination number increases for larger cations, increasing the radius ratio. However, the relation between a coordination number and a radius ratio is not simple. For example, halides of alkali metals adopt the NaCl structures at normal temperatures except cesium chloride CsCl, cesium bromide CsBr and cesium iodide CsI. It is not possible to assume structure types from the radius ratios even in the case of simple ionic compounds like alkali metal halides. However, the qualitative trend that smaller cations have smaller coordination numbers is generally correct.
Variation of the solid structure expression
Many solid-state inorganic compounds have complicated three-dimensional structures. Different structural illustrations for the same compound help our understanding of its structure. In the case of complicated inorganic compounds, drawing bond lines between atoms, as in most organic compounds, causes confusion. The anions in many metal oxides, sulfides, or halides form tetrahedra or octahedra around the central metal cations. Although there is no bond between anions, the structures are greatly simplified if they are illustrated by the anion polyhedra sharing apexes, edges, or faces. In such illustrations, cationic metal atoms are usually omitted. As has been mentioned, ionic solid structures can be thought of as a close packed arrays of anions.
Figures 2.12 and 2.13 illustrate these three representations for molecular phosphorus pentoxide P2O5 (= P4O10) and molybdenum pentachloride MoCl5 (= Mo2Cl10). Polyhedral representations are much easier to understand for the structures of large molecules or solid-state compounds formed by an infinite number of atoms. However, the bond line representation is suitable for molecular compounds such as the above examples. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/02%3A_Bonding_and_Structure/2.02%3A_Geometrical_factors_governing_bonding_and_structure.txt |
The bonding and structure of a compound are determined by electronic properties such as the power of constituent atoms to attract or repel electrons, the molecular orbitals occupied by valence electrons, etc. Geometrical arrangements of atoms are also influenced by the electronic interactions between non-bonding electrons. Here, some fundamental concepts are described.
(a) Effective nuclear charge
Since the positive nuclear charge is generally offset by the negative charge of the internal electrons in the electron shell inside the valence electrons, the nuclear charge that valence electrons feel is smaller than the integer charge, Ze for an atomic number Z. This reduction is expressed by the shielding constant $\sigma$, and the net nuclear charge is called the effective nuclear charge, Zeff e.
$Z_{eff} = Z - \sigma$
The effective nuclear charge varies with different electron orbitals and distances from the nucleus.
(b) Ionization energy
Ionization energy is defined as the minimum energy required to remove an electron from the atom in a gaseous phase (g), as shown in the following equation. Ionization energy is expressed in units of electron volt (eV), where 1 eV = 96.49 kJ mol-1.
$A(g) \rightarrow A^{+} (g) + e^{-} (g)$
The 1st ionization energy, which removes the outermost electron, is the smallest, and the 2nd and 3rd ionization energies, which further ionize cations, increase rapidly. The ionization enthalpy, which is the standard enthalpy change of the ionization process and is used in thermodynamic calculations, is the ionization energy multiplied by RT (R is the universal gas constant 8.31451 J K-1 mol-1 and T is temperature, 2.479 kJ (0.026 eV), at room temperature). The difference between these two parameters is small. The 1st ionization energy varies periodically with atomic number across the periodic table, with the lower left (cesium, Cs) being the smallest and the upper right (helium, He) the largest.
It is understandable that alkali metals generally have the lowest ionization energies because they are stabilized by removal of an s electron to attain the rare gas configuration. Rare gas elements have stable electronic structures, and their ionization energies are the largest. Although the ionization energy increases almost monotonically from alkali metals to rare gases in each period, there are reversals at several places, such as nitrogen N and oxygen O, and phosphorus P and sulfur S. The 1st ionization energies are given in Table $5$.
Table $5$ Electronic parameters of atoms (eV). I: 1st ionization energy, Ae: electron affinity, $\chi_{Μ}$: electronegativity (Mulliken)
Atom I A $\chi_{M}$
H 13.60 0.75 7.18
He 24.59
Li 5.39 0.62 3.01
Be 9.32
B 8.30 0.28 4.29
C 11.26 1.27 6.27
N 14.53
O 13.62 1.46 7.54
F 17.42 3.40 10.41
Ne 21.56
Na 5.14 0.55 2.85
Mg 7.65
Al 55.99 0.44 3.22
Si 8.15 1.39 4.77
P 10.49 0.75 5.62
S 10.36 2.08 6.22
Cl 12.97 3.61 8.29
Ar 15.76
K 4.34 0.50 2.42
Ca 6.11 0.02 3.07
Sc 6.56 0.19 3.38
Ti 6.83 0.08 3.45
V 6.75 0.53 3.64
Cr 6.77 0.67 3.72
Mn 7.44
Fe 7.90 0.15 4.03
Co 7.88 0.66 4.27
Ni 7.64 1.16 4.40
Cu 7.73 1.24 4.48
Zn 9.99
Ga 6.00 0.30 3.20
Ge 7.90 1.23 4.61
As 9.82 0.81 5.31
Se 9.75 2.02 5.89
Br 11.81 3.36 7.59
Kr 14.00
Rb 4.18 0.49 2.34
Sr 5.69 0.11 2.90
Y 622 0.31 3.27
Zr 6.63 0.43 3.53
Nb 6.76 0.89 3.83
Mo 7.09 0.75 3.92
Ru 7.36 1.05 4.26
Rh 7.46 1.14 4.30
Pd 8.34 0.56 4.45
Ag 7.58 1.30 4.44
Cd 8.99
In 5.79 0.30 3.10
Sn 7.34 1.11 4.23
Sb 8.64 1.07 4.86
Te 9.01 1.97 5.49
I 10.45 3.06 6.76
Xe 12.13
Cs 3.89 0.47 2.18
Ba 5.21 0.15 2.68
La 5.58 0.50 3.09
Hf 6.83 0.00 3.42
Ta 7.89 0.32 4.11
W 7.98 0.82 4.40
Re 7.88 0.15 0.40
Os 8.70 1.10 4.90
Ir 9.10 1.60 5.40
Pt 9.00 2.13 5.61
Au 9.23 2.31 5.77
Hg 10.44
Tl 6.11 0.20 3.16
Pb 7.42 0.36 3.89
Bi 7.29 0.95 4.12
(c) Electron affinity
Electron affinity is the negative of the electron-gain enthalpy $\Delta$Heg of an atom in a gas phase, as shown in the following equation and denoted by Ae ( = -$\Delta$Heg) (Table $5$).
$A(g) + e^{-} (g) \rightarrow A^{-} (g)$
It may be regarded as the ionization enthalpy of an anion. Since halogen atoms achieve rare gas electron configurations if an electron is added to them, their electron affinities are large.
(d) Electronegativity
Electronegativity is one of the most fundamental atomic parameters which expresses numerically the tendency to attract electrons to atoms in a molecule. It is very useful in explaining differences in bonding, structure, and reaction from the standpoint of atomic properties. Various schemes have been proposed to explain the theoretical basis of the power of electron attraction, and studies are still actively seeking new numerical values of electronegativity. The Pauling scale, introduced first in 1932, is still the most frequently used , and subsequent new numerical values have been justified if they are close to those of Pauling.
L. Pauling defined electronegativity as the quantitative ionic character of bonds. Originally, the following equation was proposed as a formula to define the ionic character of the bond between atoms A and B.
$\Delta D(AB) - \frac{1}{2} [D(AA) + D(BB)]$
where D is the bond energy of a covalent bond. However, it turned out that ∆ is not necessarily positive, and Pauling modified the definition
$\Delta = D(AB) - \sqrt{D(AA) \times D(BB)}$
and redefined it as the ionic character of the A-B bond. Furthermore, electronegativity $\chi$ was defined in such a way that the difference of the electronegativities of atoms A and B is proportional to the square root of the ionic character. Here, the coefficient
$| \chi_{A} - \chi_{B}| = 0.208 \sqrt{\Delta}$
0.208 is so determined that the electronegativity of hydrogen H becomes 2.1 when bond energies are expressed in kcal mol-1. Since Pauling electronegativities increase the higher the oxidization states of an atom, these values correspond to the highest oxidization number of each element. The electronegativities calculated using recent values of bond energies are shown in Table $6$.
1 2 3 4 5 6 7 8 9
1 H
2.2
2 Li
0.98
Be
1.57
3 Na
0.93
Mg
1.31
4 K
0.82
Ca
1.00
Sc
1.36
Ti
1.54
V
1.63
Cr
1.66
Mn
1.55
Fe
1.83
Co
1.88
5 Rb
0.82
Sr
0.95
Y
1.22
Zr
1.33
Nb
1.6
Mo
2.16
Tc
1.9
Ru
2.2
Rh
2.28
6 Cs
0.79
Ba
0.89
Lanthanoid Hf
1.3
Ta
1.5
W
2.36
Re
1.9
Os
2.2
Ir
2.20
7 Fr
0.7
Ra
0.9
Actinoid
10 11 12 13 14 15 16 17 18
He
B
2.04
C
2.55
N
3.04
O
3.44
F
3.98
Ne
Al
1.61
Si
1.90
P
2.19
S
2.58
Cl
3.16
Ar
Ni
1.91
Cu
2.0
Zn
1.65
Ga
1.81
Ge
2.01
As
2.18
Se
2.55
Br
2.96
Kr
3.0
Pd
2.20
Ag
1.93
Cd
1.69
In
1.78
Sn
1.96
Sb
2.05
Te
2.10
I
2.66
Xe
2.6
Pt
2.28
Au
2.54
Hg
2.00
Tl
2.04
Pb 2.33 Bi
2.02
Po
2.0
At
2.2
Rn
A. L. Allred and E. G.. Rochow defined electronegativity as the electric field Zeff / r2 on the atomic surface. They added a constant in order to make the electronegativity $\chi_{AR}$ as near as possible to the Pauling values by using r for the covalent bond radius of atoms.
$\chi_{AR} = 0.74 + 0.36 \frac{Z_{eff}}{r^{2}}$
It turns out that elements with small covalent radii and large effective nuclear charges have large electronegativities (Table $6$).
R. Mulliken defined electronegativity $\chi_{M}$ as the average of the ionization energy I and electron affinity Ae as follows (Figure $14$).
$\chi_{M} = \frac{1}{2} (I + A_{e})$
As ionization energy is the energy of electronic excitation from the HOMO and electron affinity the energy of electron addition to the LUMO (refer to Section 2.3 (e)), in this definition electronegativity can also be called the average value of the energy levels of the HOMO and LUMO. Those elements which are hard to ionize and easy to attract electrons have large values. Although the electronegativity is defined for the atoms in a valence state in a molecule and has the dimensions of energy, it is treated as a dimensionless number (Table $5$).
Although the definition of Mulliken is intelligible since it is directly related to atomic orbitals, generally the values of Pauling or Allred-Rochow are used. As these values are not much different, the Pauling electronegativity is appropriate when choosing only one. Electronegativity values change not only by definition, but are also considerably affected by the bonding state of atoms, and they should be used with considerable caution. The electronegativities of the constituent atoms are fundamental to explaining the differences in bonding, structure, and reactions of compounds. Therefore theoretical chemists continue in their efforts firmly to extend the foundations of this parameter.
Exercise $4$
Describe the numerical tendency of electronegativities of the elements in the periodic table.
Answer
They increase toward the right and decrease down the table. Namely, the electronegativity of alkali metal Cs is the smallest and that of fluorine F is the largest.
(e) Molecular orbitals
The wave functions of electrons in an atom are called atomic orbitals. Since the probability of finding electrons in a molecular orbital is proportional to the square of a wave function, the electron map looks like a wave function. A wave function has domains of positive and negative amplitude called lobes. The overlapping positive lobes or negative lobes of the wave functions of atoms in a molecule amplify each other to form a bond, but the positive and negative lobes cancel each other forming no bond. The extent of this interference effect corresponds to the magnitude of the overlap integral in quantum chemistry.
In the formation of a molecule, atomic orbitals overlap to generate a molecular orbital which is the wave function of the electrons in the molecule. The number of molecular orbitals is the sum of the atomic orbitals and these molecular orbitals are classified into bonding, nonbonding, or antibonding molecular orbitals by the extent of their participation in the bond between atoms. The conditions of the formation of a bonding molecular orbital are as follows.
[Conditions of the formation of bonding molecular orbitals]
1. The lobes of the atomic orbitals of the constituent atoms are suitable for an overlap.
2. The positive or negative sign of the overlapping lobes is the same.
3. The energy levels of atomic orbitals are close.
The simplest case where a molecular orbital is constructed from atomic orbitals A and B is explained here. A bonding molecular orbital is formed between A and B if the above mentioned conditions (1), (2), and (3) are satisfied, but if the sign of one of the atomic orbitals is reversed, condition (2) is not satisfied and an antibonding molecular orbital, in which the signs of the overlapping lobes are different (Figure $15$) results. The energy level of a bonding orbital is lower and the level of an antibonding orbital is higher than those of the constituent atomic orbitals. The larger the energy difference of a bonding and an antibonding orbital, the stronger the bond. When there is no bonding or antibonding interaction between A and B, the resultant molecular orbital is a nonbonding orbital. Electrons occupy the molecular orbitals in order of lowest to highest energy levels. The highest occupied molecular orbital is called the HOMO and the lowest unoccupied one the LUMO. Ken'ichi Fukui (1981 Nobel prize) named these orbitals frontier orbitals.
Two or more molecular orbitals of equal energy are called degenerate orbitals. The symbol of a nondegenerate orbital is a or b, a doubly degenerate orbital e, and triply degenerate orbital t. The symbol g (gerade) is attached as a suffix to the centrosymmetric orbital and u (ungerade) to the orbital which changes sign under inversion around an inversion center. The number before the symmetry symbol is used in order of energy to distinguish orbitals of the same degeneracy. Additionally, they are named sigma ($sigma$) or pi ($\pi$) orbitals according to the orbital character. A sigma orbital has rotation symmetry around the bond axis, and a pi orbital has a nodal plane. Therefore, sigma bonds are formed by the overlap of s-s, p-p, s-d, p-d, and d-d orbitals (Figure $16$) and pi bonds the overlap of p-p, p-d, and d-d orbitals (Figure $17$).
When the wave functions of two atoms are set to $\phi_{A}$ and $\phi_{B}$, a molecular orbital is a linear combination of the atomic orbitals (LCAO) expressed as
$\psi = C_{A} \phi_{A} + C_{B} \phi_{B}$
Only the atomic orbitals of the valence electron shell are used in the simplest molecular orbital method. Construction of a molecular orbital is illustrated below for the simplest case of the two-atom molecules. All the levels below the HOMO are occupied by electrons and the levels above the LUMO are empty.
In a hydrogen molecule, H2, the overlap of the 1s orbital of each hydrogen atom forms a bonding orbital $\sigma_{g}$ if the lobes have equal sign and an antibonding orbital $\sigma_{u}$ if they have opposite signs, and two electrons occupy the bonding orbital $\sigma_{g}$ (Figure $18$).
In the two-atom molecules of the 2nd period, from lithium Li2 to fluorine F2, if the z axis is set as a bond axis, 1$\sigma_{g}$ and 1$\sigma_{u}$ are formed by the overlap of 2s orbital of each atom and 2$\sigma_{g}$ and 2$\sigma_{u}$ from 2pz orbitals and 1$\pi_{u}$ and 1$\pi_{g}$ from 2px, and 2py. The orbital energy levels for the molecules from Li2 to N2 are ordered as 1$\sigma_{g}$ < 1$\sigma_{u}$ < 1$\pi_{u}$ < 2$\sigma_{g}$ < 1$\pi_{g}$ < 2$\sigma_{u}$ and electrons occupy the levels sequentially from the bottom. The example of an N2 molecule with ten valence electrons is shown in Figure $19$. Since the ordering of orbitals is somewhat different in O2 and F2, in which the 2$\sigma_{g}$ orbital comes under that of 1$\pi_{u}$, the molecular orbital of the oxygen molecule, O2, is illustrated in Figure $20$. The 11th and 12th electrons among the 12 valence electrons occupy the doubly degenerate 1$\sigma_{g}$ orbital in the ground state and they have parallel spins under Hund's rule and hence an oxygen molecule has two unpaired electrons.
The molecular orbitals of two different atoms are formed by the overlap of atomic orbitals with different energy levels. The energy level of the orbital of the atom with larger electronegativity generally is lower, and the molecular orbitals are more characteristic of the atomic orbital with the nearer energy level. Therefore, the bonding orbitals have the character of the atom with the larger electronegativity, and the antibonding orbitals that of the atom with the smaller electronegativity.
For example, five molecular orbitals in hydrogen fluoride, HF, are formed from the 1s orbital of hydrogen and the 2s and 2p orbitals of fluorine, as shown in Figure $21$. The bonding 1$\sigma$ orbital has the 2s character of fluorine, and the antibonding 3$\sigma$ orbital the 1s character of hydrogen. Since hydrogen has only one 1s orbital, the overlap with the 2p orbital of fluorine with $\pi$character is not effective, and the fluorine 2p orbital becomes a nonbonding orbital. Since HF has eight valence electrons, this nonbonding orbital is the HOMO.
In carbon monoxide, CO, carbon and oxygen have 2s and 2p orbitals resulting in both sigma and pi bonds, and a triple bond is formed between the atoms. Although 8 molecular orbitals in this case are qualitatively the same as those of the isolectronic nitrogen molecule N2 and 10 electrons occupy the orbital up to 3$\sigma$, the energy level of each orbital differs from that of the nitrogen molecule. The bonding 1$\sigma$ orbital has the 2s character of oxygen because of its larger electronegativity, and the bonding 1$\pi$ orbital also has the 2p character of oxygen. The antibonding 2$\pi$ and 4$\sigma$ orbitals have the 2p character of carbon (Figure $22$).
The bond order between atoms is a half of the number of electrons in the bonding orbitals minus those of the antibonding orbitals. For example, in N2 or CO, it is equal to $\frac{1}{2}$(8 - 2) = 3 and is consistent with the Lewis structure.
Exercise $5$
Why are the atomic orbitals of oxygen atom in the molecular orbital diagram of carbon monoxide, CO, lower than those of carbon?
Answer
It is because the electronegativity of oxygen is larger than that of carbon.
problems
2.1
Using the Pauling equation, calculate the electronegativity of chlorine from the bond energies of the hydrogen molecule H2 (432 kJ mol-1), chlorine molecule Cl2 (239 kJ mol-1), and hydrogen chloride HCl molecule (428 kJ mol-1) and electronegativity of hydrogen ($\chi$ = 2.1).
2.2
Why are the energy levels $\sigma_{g} < \sigma_{u}$ in the orbitals of sigma character and $\pi_{u} < \pi_{g}$ in those of pi character in the molecular orbital diagram of N2 or O2?
Great theory and evaluation
Lewis' valence electron theory proposes that a covalent bond is formed with an electron pair and that there are eight valence electrons around each atom. This is a very important concept with which we understand the bonds between the main group elements. However, the theory was not held in high enough regard for a Nobel prize to be awarded to Lewis. One of the reasons of this disregard seems to be that chemists in the United States, Lewis' homeland, ignored his theory at first, and that a Nobel prize laureate, Langmuir, extended Lewis's theory, which was later known as the Lewis-Langmuir theory. N. Bohr, the eminent physicist who had great influence on the Nobel prize selection, evaluated Langmuir's adsorption theory more highly, which suggests that physicists considered Lewis theory too simplistic.
There is a similar story about the transition state theory of H. Eyring. Physicists and theoretical chemists, who liked mathematical treatment of chemical phenomena, thought Eyrings theory too unsophisticated. For this reason, the most important concept in chemical kinetics was not considered for a Nobel prize. It is an episode in the history of chemistry which reminds us of the comment of R. Hoffmann, who pointed out that simple concepts are more important than deceptively complicated mathematical theories. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/02%3A_Bonding_and_Structure/2.03%3A_Electronic_factors_which_govern_bonding_and_structure.txt |
Inorganic reactions can be described in terms of redox or acid-base concepts. Thermodynamics and electrochemistry are closely related to the analyses of redox and acid-base reactions. Although it appears that the theories of themlodynamics and electrochemistry are described by a number of complicated equations and formulae, only a few equations and parameters are required for a proper understanding. A good grasp of the sign and trend of the parameters in these key equations greatly helps this understanding. A more detailed understanding beyond this general level can be acquired by building on these basic concepts.
03: Reactions
Thermodynamic parameters on changes of state are necessary to describe chemical bonding, structure, and reaction. This is true also in inorganic chemistry, and the most important concepts in thermodynamics are described in this section. Even simple thermodynamic knowledge is considerably useful for judging whether structures of compounds are stable and the likelihood of spontaneous reactions, and for the calculations of reaction heat, determination of reaction mechanism, and understanding of electrochemistry.
Enthalpy
Since enthalpy is the heat content of a system under constant pressure, the change$\Delta$H is negative in an exothermic reaction, and positive in an endothermic reaction. The standard reaction enthalpy, $\Delta$H0, is the enthalpy change between 1 mol of products and reactants in the standard state (105 Pa and 298.15 K). The standard enthalpy of formation, $\Delta$Hf0, of a compound is the standard reaction enthalpy for the formation of the compound from its constituent elements. Since enthalpy is a state function, the standard reaction enthalpy is calculated by defining the standard enthalpy of formation of simple substances to be zero. Namely,
$\Delta H^{0} = \Sigma n \Delta H_{f}^{0}\; (product) - \Sigma n \Delta H_{f}^{0}\; (reactant)$
Entropy
Entropy is a state function, and is a criterion determining whether one state can be reached spontaneously from another state. The 2nd law of thermodynamics states that the entropy, S, of an isolated system increases upon a spontaneous change. Namely,
$\Delta S > 0$
A thermodynamically irreversible process produces entropy. Entropy is related to the disorder of a system in statistical themodynamics as follows:
$S = k \ln W$
k is the Boltzmann constant, and W is the number of the arrangements of atoms or molecules of the system with the same energy, and corresponds to the extent of disorder. As entropy becomes larger, the larger the disorder of a system.
Gibbs energy
This quantity is defined as
$\Delta G = \Delta H - T \Delta S$
A spontaneous reaction occurs when Gibbs energy of a reaction is negative at constant temperature and pressure. The standard Gibbs free energy $\Delta$G0 is related to the equilibrium constant K of the reaction
$A \xrightleftharpoons{K} B \ldotp$
$\Delta G^{0} = - RT \ln K$
K is larger than 1 when $\Delta$G0 becomes negative, and the reaction proceeds to the right.
3.02: Electrochemistry
The standard state is defined as the one corresponding to 25° C (298.15 K), unit activity for all the substances in an electrochemical zero-current cell under 1 bar of pressure (105 Pa). For a reaction in which H+ ions participate, the standard state is pH = 0 (approximately 1 mol acid).
In the hydrogen electrode used as the standard of electrode potential, 1 atm of hydrogen gas (aH+ = 1) is slowly contacted with a platinum-black electrode immersed in a strong acid solution of activity aH2 = 1. The potential is expressed as
$E = E^{0} + \frac{RT}{F} \ln \frac{a_{H^{+}}}{a_{H_{2}}}$
and by definition E0 = 0 in the standard state. The hydrogen electrode in the standard state is called the standard hydrogen electrode, or NHE. Although reduction potential is usually expressed with reference to the NHE standard, the hydrogen electrode is inconvenient to handle. Therefore a saturated calomel (SCE) or an Ag / AgCl electrode is used as a reference electrode for everyday electrochemical measurements and experimental potentials are measured against these electrodes or converted into NHE values. When the NHE value is set to 0, the SCE value is 0.242 V, and the Ag/AgCl value is 0.199 V.
A redox reaction takes place only when redox partners exist and a reactant can be either an oxidant or reductant depending on its reaction partner. The relative redox capability can be expressed numerically by introducing the reduction potentials E0 of imaginary half-reactions (Table $1$). The free energy change $\Delta G^{0}$ of a reaction is related to E0,
$\Delta G^{0} = - nFE^{0}$
where n is the number of transferred electrons and F the Faraday constant 96500 C mol-1.
Table $1$ Standard reduction potentials at 25 °C
Couple E° / V
F2(g) + 2e- ➝ 2F- (aq) +2.87
H2O2(aq) + 2H+ (aq) +2e- ➝ 2H2O(l) +1.77
Ce4+(aq) + e- ➝ Ce3+(aq) +1.72
MnO2- (aq) + 8H+ (aq) + 5e- ➝ Mn2+(aq) + 4H2O(l) +1.51
Cl2(g) + 2e- ➝ 2Cl- (aq) +1.36
O2(g) + 4H+ (aq) + 4e- ➝ 2H2O(l) +1.23
Br2(l) +2e- ➝ 2Br- (aq) +1.09
Fe3+(aq) + e- ➝ Fe2+(aq) +0.77
AgCl(s) + e- ➝ Ag(s) + Cl- (aq) +0.22
Cu2+(aq) + e- ➝ Cu+ (aq) +0.15
2H+ (aq) + 2e- ➝ H2(g) 0
Sn2+(aq) + 2e- ➝ Sn(s) -0.14
Fe2+(aq) + 2e- ➝ Fe(s) -0.45
Zn2+(aq) + 2e- ➝ Zn(s) -0.76
Al3+(aq) + 3e- ➝ Al(s) -1.66
Mg2+(aq) + 2e- ➝ Mg(s) -2.37
Na+ (aq) + e- ➝ Na(s) -2.71
Li+ (aq) + e- ➝ Li(s) -3.04
For example, the two reactions
$\begin{split} 2 H^{+}\; (aq) + 2e^{-} & \rightarrow H_{2}\; (g), \qquad \Delta G_{1}^{0} = -2 FE_{1}^{0} \ Zn^{2+}\; (aq) + 2e^{-} & \rightarrow Zn\; (s), \qquad \Delta G_{2}^{0} = -2FE_{2}^{0} \end{split}$
do not occur independently, but if both H+ (aq) and Zn (s) are present, the redox reaction takes place. The equation for the actual reaction is complete when the latter equation is subtracted from the former.
$2H^{+}\; (aq) + Zn\; (s) \rightarrow H_{2}\; (g) + Zn^{2+}\; (aq)$
The free energy change of the whole redox reaction is the difference between $\Delta G_{1}^{0}, \Delta G_{2}^{0}$ for the respective half-reactions.
$\begin{split} \Delta G^{0} &= \Delta G_{1}^{0} - \Delta G_{2}^{0} \ &= -2F (E_{1}^{0} - E_{2}^{0}) \end{split}$
Because half-reactions are not real and they are used in pairs, the free energy change of $\Delta G_{1}^{0}$ H+ is set to zero for convenience. Since the experimental value of $\Delta G^{0}$ is -147 kJ, equals 147 kJ. Potential $\Delta G_{2}^{0}$ corresponding to $\Delta G^{0}$ of a half-reaction is called the standard reduction potential.
$E^{0} = - \frac{\Delta G^{0}}{nF}$
Therefore,
$\begin{split} E^{0} (H^{+}, H_{2}) &= 0 \quad (\text{by definition}) \ E^{0} (Zn^{2+}, Zn) &= \frac{-147\; kJ/mol}{2 \times 96500\; C/mol} = - 0.76\; V \ (1\; J &= 1\; C \cdot V) \ldotp \end{split}$
The standard potentials of various half-reactions are determined using similar procedures to that mentioned above (Table $1$). The E0 s of redox reactions can be calculated by combining E0 of these half-reactions.
If E0 of a redox reaction is positive, $\Delta G^{0}$ is negative and the reaction occurs spontaneously. Consequently, instead of the free energy change the difference in reduction potentials can be used to judge the thermodynamic spontaneity of a reaction. The higher the reduction potential of a reagent the stronger its oxidation ability. The positive or negative signs are based on the expedient of setting the reduction potential of a proton to 0, and it should be understood that a positive sign does not necessarily mean oxidizing, and a negative sign reducing. The series arranged in the order of redox power is called the electrochemical series. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/03%3A_Reactions/3.01%3A_Thermodynamics.txt |
(a) Oxidation number
The oxidation number is the formal electrical charge of a component atom in a compound or in an ion allocated in such a way that the atom with smaller electronegativity has a positive charge. Since electrical charges do not differ in the case of a molecule composed of the same atoms, the oxidation number of the atoms is the quotient of the net electrical charge divided by the number of atoms. In the case of a compound or an ion consisting of different atoms, the atoms with larger electronegativity can be considered as anions and those with smaller electronegativity as cations. For example, nitrogen is 0 valent in N2; oxygen is -1 in O22-; nitrogen is +4 and oxygen -2 in NO2; but nitrogen is -3 and hydrogen +1 in NH3. That is, the oxidation number can be different for the same atom combined with different partners and the atom is said to be in the formal oxidation state corresponding to that oxidation number. Although this does not express the quantitative deviation of the actual electric charge, it is convenient in counting valence electrons or in dealing with redox reactions.
Exercise $1$
Which halogen has the largest oxidizing power?
Answer
Since the reduction potential of fluorine is the highest, its oxidizing power is the largest.
(b) Redox reactions
Originally, oxidation meant the formation of oxides from elements or the formation of compounds by the action of oxygen, and reduction is the reverse of oxidation. The present definition of reduction is a reaction which gives an electron, and oxidation is the reaction which takes an electron. Therefore, a reagent which gives an electron is a reductant and one which takes an electron is an oxidant. As a result of a redox reaction, a reductant is oxidized and an oxidant is reduced. For example, in the reaction of molybdenum metal and chlorine gas to form molybdenum pentachloride, molybdenum is a reductant and changes its oxidation state from 0 to +5 and chlorine is an oxidant and changes its oxidation state from 0 to -1.
$2\; Mo + 5\; Cl_{2} \rightarrow Mo_{2}Cl_{10}$
(c) Latimer diagram
A Latimer diagram is a diagram in which the chemical species in the highest oxidation state is placed at the left end and a series of the reduced chemical species of the same atom are arranged to the right-hand side in the order of the oxidation states, and the standard reduction potentials (/V) are written above the line which connects each state. This diagram is convenient for discussing a redox reaction. Since electric potential differs between an acidic and a basic solution, different diagrams are required depending on the pH of the solution. Taking the series of the oxides and hydrides of nitrogen in acidic solution as an example,
$\stackrel{+5}{NO_{3}^{-}} \xrightarrow{0.803} \stackrel{+4}{N_{2}O_{4}} \xrightarrow{1.07} \stackrel{+3}{HNO_{2}} \xrightarrow{0.996} \stackrel{+2}{NO} \xrightarrow{1.59} \stackrel{+1}{N_{2}O} \xrightarrow{1.77} \stackrel{0}{N_{2}} \ \xrightarrow{-1.87} \stackrel{-1}{NH_{3}OH^{+}} \xrightarrow{1.41} \stackrel{-2}{N_{2}H_{5}^{+}} \xrightarrow{1.275} \stackrel{-3}{NH_{4}^{+}}$
in a basic solution, the series becomes
$\stackrel{+5}{NO_{3}^{-}} \xrightarrow{-0.86} \stackrel{+4}{N_{2}O_{4}} \xrightarrow{0.867} \stackrel{+3}{NO_{2}^{-}} \xrightarrow{-0.46} \stackrel{+2}{NO} \xrightarrow{0.76} \stackrel{+1}{N_{2}O} \xrightarrow{0.94} \stackrel{0}{N_{2}} \ \xrightarrow{-3.04} \stackrel{-1}{NH_{2}OH} \xrightarrow{0.73} \stackrel{-2}{N_{2}H_{4}} \xrightarrow{0.1} \stackrel{-3}{NH_{3}}$
The additivity of the state function $\Delta$G0 is used in order to calculate the standard reduction potential between remote oxidation states.
$\Delta G^{0} = \Delta G_{1}^{0} + \Delta G_{2}^{0}$
$- (n_{1} + n_{2})FE^{0} = -n_{1} FE_{1}^{0} - n_{2} FE_{2}^{0}$
Where the free energy change and electric potential between adjacent states are $\Delta G_{1}^{0}, E_{1}^{0}, \Delta G_{2}^{0}, E_{2}^{0},$ respectively, and the number of transferred electrons n1, n2. Namely,
$E^{0} = \frac{n_{1} E_{1}^{0} + n_{2} E_{2}^{0}}{n_{1} + n_{2}}$
For example, in the reduction of NO3- to HNO2, two electrons are transferred to form HNO2 via N2O4 and the potential becomes
$E^{0} = \frac{0.803\; V + 1.07\; V}{2} = 0.94\; V$
Exercise $2$
Calculate the reduction potential of the reduction of NO3- to NO2- in a basic solution.
Answer
$E^{0} = \frac{-0.86\; V + 0.867\; V}{2} = 0.004\; V$
In recent years, whenever a new inorganic compound is synthesized, its redox properties are investigated, usually by electrochemical measurements. Cyclic voltammetry is the technique of choice for the study of its redox properties, including the electric potential, the number of transferred electrons, and the reversibility of the reactions, etc. because of the simplicity of the measurements. It is approximately correct to consider that the oxidation potential corresponds to the energy level of the HOMO, because oxidation usually takes an electron from the HOMO and the reduction potential to the level of the LUMO since reduction adds an electron to the LUMO. However, various factors, such as solvent effects, should be taken into consideration during quantitative discussions of redox processes. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/03%3A_Reactions/3.03%3A_Oxidation_and_reduction.txt |
The definition of acid and base has changed over the course of time. This is not a problem of the orthodoxy of one definition but of the convenience of applying the concept to a particular chemical problem. Therefore, ranking the strength of acids and base also depends on the definition of acid and base used.
(a) Arrhenius's acid and base
In 1884, Arrhenius defined that an acid is a substance that gives H+ and a base one that gives OH-. Namely, if an acid is HA and a base BOH, then HA $\rightarrow$ H+ + A- and BOH $\rightarrow$ B+ + OH-. Therefore, when an acid and a base react, water is formed.
(b) Brønsted-Lowry's acid and base
In a new theory submitted in 1923 independently by Brønsted and Lowry, an acid is defined as a molecule or an ion which gives H+ and a molecule or ion that receives H+ from a partner is a base. A base is not only a molecule or an ion which gives OH- but anything which receives H+. Since the acid HA gives H+ to water in an aqueous solution and generates an oxonium ion, H3O+, water is also a kind of base according to this definition.
$HA\; (acid) + H_{2} O\; (base) \rightarrow H_{3} O\; (conjugate\; acid) + A^{-}\; (conjugate\; base)$
Here H3O+ is called a conjugate acid and A- a conjugate base. However, since water gives H+ to ammonia and generates NH4+, it is also an acid, as is shown below.
$H_{2}O\; (acid) + NH_{3}\; (base) \rightarrow NH_{4}^{+}\; (conjugate\; acid) + OH^{-}\; (conjugate\; base)$
That is, water can be an acid or a base dependent on the co-reactant.
Although the definition of Brønsted-Lowry is not much different from that of Arrhenius for aqueous solutions, it is more useful because the theory was extended to non-aqueous acids and bases.
Exercise $3$
Write the molecular formulae of nitric acid, perchloric acid, sulfuric acid, and phosphoric acid as oxo acids together with the formal oxidation number of the central atom.
Answer
• Nitric acid (HO)N5+O2
• Perchloric acid (HO)Cl7+O3
• Sulfuric acid (HO)2S6+O2
• and Phosphoric acid (HO)3P5+O.
Acid strength
A protonic acid gives H+ to water and generates the oxonium ion H3O+. The strength of an acid in a dilute aqueous solution is estimated from the equilibrium constant Ka
$K_{a} = \frac{[H_{3}O^{+}][A^{-}]}{[HA]}$
for the dissociation equilibrium,
$HA + H_{2} O \rightleftharpoons A^{-} + H_{3}O^{+}$
but it is more convenient to use,
$pK_{a} = - \log K_{a} \quad or \quad pH = - \log [H_{3} O^{+}]$
An acid with pKa < 0 is classified as a strong acid and one with pKa > 0 a weak acid. The conjugate base of a strong acid is a weak base. The pKa values of typical acids at 25 °C are shown in Table $2$.
Table $2$ Acidity constants for aqueous solutions of acids at 25 °C.
Acid pKa
HF 3.17
HCl -8*
HBr -9*
HI -10*
H2CO3 6.35
HClO4 < 0
HNO3 < 0
H3PO3 1.5
H3PO4 2.15
H3SO4 < 0
CH3COOH 4.56
C6H5COOH 4.00
NH4+ 9.25
C5H5NH+ 5.25
* Estimated value
Since a solvent also works as an acid or a base, the acidity and its range depend on the solvent dissolving the acid. Full dissociation of an acid stronger than H3O+ gives H+ to water, forming H3O+ in an aqueous solution. For example, both HBr and HI dissociate completely to become H3O+, and their acidities are similar. This kind of phenomenon is called the leveling effect, and all acidities become equal to that of H3O+. In measuring the relative acidity of strong acids, it is necessary to use a solvent whose H+ affinity is smaller than that of water, such as acetic acid or ethanol.
Binary halo acids HX, except for HF, are very strong acids. Although the H3O+ concentration itself is also high in an aqueous solution of HF, the stronger hydrogen bond of F- compared with that of other halide anions decreases the thermodynamic activity of H3O+.
The acidity of oxo acids, such as phosphoric acid, sulfuric acid, nitric acid, and perchloric acid, is related to the formal oxidation number of P, S, N, and Cl. Namely, if the oxo acid HnXOm is denoted by (HO)nXOm-n, the positive charge on X becomes positive (2m- n), and the acidity is higher for larger value of this number. The number parallels the ease of dissociation of OH to give a proton. Acidity is higher in the following order: perchloric acid (HO)ClO3 > sulfuric acid (HO)2SO2 > nitric acid (HO)NO2 > phosphoric acid (HO)3PO. Although phosphoric acid can be written as (HO)3PO, phosphorous acid is not (HO)3P but (HO)2HPO, and has an acid strength comparable to phosphoric acid.
Hammett acidity function
Hydrogen ion concentration and pH are meaningful only in dilute aqueous solutions of acids. The acidity in nonaqueous and concentrated solutions is measured using the Hammett acidity function. This function makes it possible to measure the acidities of various acids in a non-aqueous solvent or of an acid in various non-aqueous solvents.
The Hammett acidity function in the equilibrium,
$B + H^{+} \rightleftharpoons BH^{+}$
is defined by
$H_{0} = pK_{BH^{+}} - \log \frac{[BH^{+}]}{[B]}$
In a very dilute solution
$K_{BH^{+}} = \frac{[B][H^{+}]}{[BH^{+}]}$
and
$H_{0} = - \log \frac{[B][H^{+}]}{[BH^{+}]} - \log \frac{[BH^{+}]}{[B]} = - \log [H^{+}] = pH$
An acid with -H0 over 6 is called a superacid. This is an acid that is 106 times stronger than a 1 molar solution of a strong acid. -H0 for pure sulfuric acid is 12.1, 21.1 for a solution of HF in SbF5, and 26.5 for the combination of HSO3F and SbF5.
Superacids have the ability to remove H- from a hydrocarbon and perform H-D exchange and C-C bond scission, etc.
(c) Lewis acid and base
Whereas the concept of Brønsted acid and base is limited to the transfer of protons, a Lewis acid A is generally defined as an acceptor, and a Lewis base B a donor, of an electron pair. An acid A and a base :B bind together to form an adduct A:B. For example, a Lewis acid BF3 and a Lewis base OEt2 (diethylether) form an adduct F3B:OEt2. The stability increases by the completion of an octet around boron when such an adduct forms. The stability of an adduct is expressed by the equilibrium constant of the reaction
$A + :B \xrightleftharpoons{K_{f}} A - B$
$K_{f} = \frac{[A : B]}{[A][:B]}$
Therefore, the Lewis acidities of a series of acids are measured by comparing Kf against a common base :B. Since a proton is also an electron acceptor, Brønsted acids are the special case of the more general Lewis definition of acids. According to this definition, a co-ordinate bond in a transition metal complex is also an acid-base reaction of a ligand (Lewis base) with a metal center (Lewis acid).
V. Gutmann proposed the negative enthalpy of formation (kcal mol-1 unit) of the adduct (Cl5Sb-Sol) of Sol (solvent) with a standard acid (SbCl5) in dichloroethane as a measure of the Lewis basicity of a solvent. This is called the donor number (D.N.) of a solvent. On the other hand, the 31P NMR chemical shift of Et3P in a solvent is defined as the measure of the Lewis acidity of the solvent and is called the acceptor number (A.N.).
Hard and soft acids and bases
R. G. Pearson classified Lewis acids and bases according to their hardness and softness. This classification is an extension of the original theory of S. Ahrland, J. Chatt, and N. R. Davies, who proposed that metal cations were classified in the order of the stability constants Kf of the formation of the complexes of the metal cations with halide anions. Namely, the order of Kf is I < Br < Cl < F toward metal ions belonging to class a, and F < Cl < Br < I toward those of class b. The class a type metal cations are hard acids, and class b type ones are soft acids. The metal cations which are not much dependent on the kind of halogens have borderline character.
What should be noticed here is that Kf tends to become large for the combination of a hard acid and a hard base, or a soft acid and a soft base. If the concept is extended from simple metal cations and halide anions to general Lewis acids and bases, they can similarly be classified in terms of the hard and soft acid-base affinity. Typical hard acids and bases, and soft acids and bases are shown in Table $3$.
Table $3$ The classification of Lewis acids and bases
Hard Borderline Soft
Acids H+, Li+, Na+, K+
Be2+, Mg2+, Ca2+
Al3+, Ti4+, Cr3+
Fe3+, BF3, Cl7+
Fe2+, Co2+, Ni2+,Cu2+
Zn2+, Sn2+, Pb2+
Sb3+, Bi3+
Cu+, Ag+, Au+
Tl+, Cd2+, Hg+, Hg2+
Pd2+,Pt2+,Pt4+
Bases NH3,H2O,R2O
F-,OH-,O2-
NO3-,SO42-,PO43-
N3-,N2,NO2-
Br-
SO32-
H-,CN-,R-
I-
PR3,SR2,CO
The qualitative expression “softness” is a chemical paraphrasing of the ease of polarization and the larger contribution of covalency than ionicity in bonding. The cations of alkali and alkaline earth metals as well as aluminum are hard and the cations of mercury, copper, silver, and gold, etc. are soft. Whereas oxides are hard, sulfides and phosphorus compounds are soft. In the minerals of the Earth’s crust, aluminum, which is hard and oxophilic is found as an oxide, and cadmium, which is soft and chalcophilic is found as a sulfide.
Exercise $4$
Applying the concept of hard and soft acids for ferric and ferrous ions, what kind of minerals are expected in iron ores?
Answer
Fe3+ is a hard acid and Fe2+ is a borderline acid. Therefore, it is likely that the main iron ores are oxide minerals. Although the main ores are actually the oxides hematite Fe2O3 or magnetite Fe3O4, a Fe2+ sulfide pyrite FeS2 is also widely distributed.
Acid-base and oxidation-reduction
Some people confuse acid-base and redox reactions. This confusion may be caused firstly by the similar terms originating from oxygen and secondly by misunderstanding about electron transfer. Historically, A. L. Lavoisier, who was one of the great founders of modern chemistry in the 18th century, considered that oxygen was the basis of all acids. He also defined oxidation as the formation of oxides from an element and oxygen. It then took a long time before the present definitions of acid-base and redox reactions were proposed and the old definitions were no longer satisfactory. Furthermore, the Lewis acid accepts an electron pair from a base forming a Lewis acid-base complex, and the oxidizing agent accepts electrons from a reducing agent and is reduced. The fact that acids and oxidizing agents are electron acceptors, and that bases and reducing agents are electron donors, is one of the causes of this confusion.
Problems
3.1
The Latimer diagram of oxygen is shown below. Write the oxidation number of oxygen in each compound.
$O_{2} \xrightarrow{+0.70} H_{2}O_{2} \xrightarrow{+1.76} H_{2} O$
Calculate the reduction potential of the reaction which converts an oxygen molecule to water, and judge whether this reaction is a spontaneous reaction.
3.2
Which is the stronger base, ammonia or pyridine?
3.3
The order of the Lewis acidity of boron halides is BF3 < BCl3 < BBr3. Is this order reasonable from the standpoint of the electronegativities of halogens? | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/03%3A_Reactions/3.04%3A_Acid_and_base.txt |
There are about 20 nonmelallic elements which are generally found as either anions in ionic compounds or else as elementary substances. It is possible to learn the names, structures, and main properties of these various compounds following a relatively simple classi?cation. Hydrides, oxides, sul?des, and halides are important, and essential for the study of pure and applied inorganic chemistry of the solid state compounds.
Thumbnail: A sample of sulfur a member of the oxygen group of elements. (Public Domain; Ben Mills).
04: Chemistry of Nonmetallic Elements
(a) Hydrogen
Hydrogen is the simplest element consisting of a proton and an electron, and the most abundant element in the universe. It is next to oxygen and silicon, and about 1 wt% of all the elements on the Earth. Needless to say, most hydrogen exists as water on the Earth. Since its polarity may change freely between hydride (H-), atom (H), and proton (H+), hydrogen also forms various compounds with many elements including oxygen and carbon. Therefore, hydrogen is highly important in chemistry.
Of the three kinds of isotopes of hydrogen, deuterium, D, was discovered by H. C. Urey and others in 1932, and subsequently tritium, T, was prepared from deuterium in 1934. About 0.015% of hydrogen is present as deuterium, and this can be enriched by electrolysis of water. Tritium is a radioactive isotope emitting $\beta$-particles with a half-life of 12.33 years. Since the mass of deuterium and tritium is about twice and three times that of hydrogen, respectively, the physical properties of the isotopes, and compounds containing them, differ considerably. Some properties of the isotopes and water are listed in Table $1$. When the E-H bond in a hydrogen compound is converted into the E-D by deuterium substitution, the E-H stretching frequency in an infrared spectrum is reduced to about $\frac{1}{\sqrt{2}}$, which is useful for determining the position of the hydrogen atom. It is sometimes possible to conclude that scission of the bond with a hydrogen is the rate-determining step when the deuterium substitution shows a marked effect on the rate of reaction of a hydrogen-containing compound.
Since the nuclear spin of hydrogen is 1/2 and given its abundance, it is the most important nuclide for NMR spectroscopy. NMR is widely used not only for identification of organic compounds, but also for medical diagnostic purposes using MRI (magnetic resonance imaging) of water in living bodies. Human organs can now be observed with this non-invasive method.
Table $1$ Properties of isotopic hydrogen and water
Properties H2 D2 T2 H2O D2O T2O
Melting point* 13.957 18.73 20.62 0.00 3.81 4.48
Boiling point 20.39 23.67 25.04 100.00 101.42 101.51
Density (g cm-3, 25°C) 0.9970 1.1044 1.2138
Temp. of maximum density (°C) 3.98 11.23 13.4
* hydrogen (K), water (°C)
There are nuclear-spin isomers in diatomic molecules of the nuclides whose spin is not zero. Especially in the case of a hydrogen molecule, the difference of properties is significant. Spins of para-hydrogen are anti-parallel and the sum is 0 leading to a singlet state. Spins of ortho-hydrogen are parallel and the sum is 1 resulting in a triplet state. Since para-hydrogen is in a lower energy state, it is the stabler form at low temperatures. The theoretical ratio of para-hydrogen is 100% at 0 K, but it decreases to about 25% at room temperature, since the ratio of ortho-hydrogen increases at higher temperatures. Gas chromatography and rotational lines in the electronic band spectrum of H2 can distinguish two hydrogen isomers.
(b) Hydride
Binary hydrides can be classified according to the position of the element in the periodic table, and by the bond characters. The hydrides of alkali and alkaline earth metals among s-block elements are ionic compounds structurally analogous to halides and are called saline hydrides. The Group 13-17 p-block elements form covalent molecular hydrides. No hydride of rare gas elements has been reported. Some of the d-block and f-block transition metals form metal hydrides exhibiting metallic properties. Transition metals which do not give binary hydrides form many molecular hydride complexes coordinated by stabilization ligands, such as carbonyl (CO), tertiaryphosphines (PR3), or cyclopentadienyl (C5H5) (refer to Section 6.1). Typical hydrides of each class are given below.
Saline hydrides
Lithium hydride, LiH, is a colorless crystalline compound (mp (melting point) 680 °C). Li+ and H- form a lattice with a rock salt type structure. Quantitative evolution of hydrogen gas at the anode during the electrolysis of the fused salt suggests the existence of H-. Water reacts vigorously with lithium hydride evolving hydrogen gas. Since it dissolves in ethers slightly, the hydride is used as a reducing agent in organic chemistry.
Calcium hydride, CaH2, is a colorless crystalline compound (mp 816 °C), and reacts mildly with water evolving hydrogen gas. This hydride is used as a hydrogen gas generator, or a dehydrating agent for organic solvents. It is used also as a reducing agent.
Lithium tetrahydridoaluminate, LiAlH4, is a colorless crystalline compound (decomposes above 125 °C) usually called lithium aluminum hydride. The hydride dissolves in ethers, and reacts violently with water. It is used as a reducing and hydrogenating agent and for dehydrating organic solvents.
Sodium tetrahydroborate, NaBH4, is a white crystalline compound (decomposes at 400 °C) usually called sodium borohydride. It is soluble in water and decomposes at high temperatures evolving hydrogen gas. It is used as a reducing agent for inorganic and organic compounds, for preparation of hydride complexes, etc.
Molecular hydrides
All hydrides other than those of carbon (methane) and oxygen (water) are poisonous gases with very high reactivity and should be handled very carefully. Although there are methods of generating the gases in laboratories, recently many are also available in cylinders.
Diborane, B2H6, is a colorless and poisonous gas (mp -164.9 °C and bp -92.6 °C) with a characteristic irritating odor. This hydride is a powerful reducing agent of inorganic and organic compounds. It is also useful in organic synthesis as a hydroboration agent that introduces functional groups to olefins, after addition of an olefin followed by reactions with suitable reagents.
Silane, SiH4, is a colorless and deadly poisonous gas (mp -185 °C and bp -111.9 °C) with a pungent smell, and is called also monosilane.
Ammonia, NH3, is a colorless and poisonous gas (mp -77.7 °C and bp -33.4 °C) with a characteristic irritating odor. Although it is used in many cases as aqueous ammonia since it dissolves well in water, liquid ammonia is also used as a nonaqueous solvent for special reactions. Since the Harber-Bosch process of ammonia synthesis was developed in 1913, it has been one of the most important compounds in chemical industries and is used as a starting chemical for many nitrogenous compounds. It is used also as a refrigerant.
Phosphine, PH3, is a colorless and deadly poisonous gas (mp -133 °C and bp -87.7 °C) with a bad smell, and is called also phosphorus hydride. It burns spontaneously in air. It is used in vapor phase epitaxial growth, in transition metal coordination chemistry, etc.
Hydrogen sulfide, H2S, is a colorless and deadly poisonous gas (mp -85.5 °C and bp -60.7 °C) with a rotten egg odor. Although often used with insufficient care, it is very dangerous and should be handled only in an environment with good ventilation. It is used in chemical analysis for the precipitation of metal ions, preparation of sulfur compounds, etc.
Hydrogen fluoride, HF, is a colorless, fuming, and low boiling point liquid (mp -83 °C and bp 19.5 °C), with an irritating odor. It is used for preparing inorganic and organic fluorine compounds. Because of its high permittivity, it can be used as a special nonaqueous solvent. The aqueous solution is called fluoric acid and is stored in polyethylene containers since the acid corrodes glass.
Metallic hydrides
The hydrides MHx which show metallic properties are nonstoichiometric interstitial-type solids in which hydrogen occupies a part of the cavities of the metal lattice. Usually x is not an integer in these compounds. There are Group 3 (Sc, Y), Group 4 (Ti, Zr, Hf), Group 5 (V, Nb, Ta), Cr, Ni, Pd, and Cu metallic hydrides among the d block elements, but the hydrides of other metals in Group 6 to 11 are not known. Palladium Pd reacts with hydrogen gas at ambient temperatures, and forms hydrides that have the composition PdHx (x < 1). Many metallic hydrides show metallic conductivity. LaNi5 is an intermetallic compound of lanthanum and nickel. It occludes nearly 6 hydrogen atoms per unit lattice and is converted to LaNi5H6. It is one of the candidates for use as a hydrogen storage material with the development of hydrogen-fueled cars.
Exercise $1$
Write the oxidation number of the hydrogen atom in H2, NaH, NH3, and HCl.
Answer
• H2 (0)
• NaH (-1)
• NH3 (+1)
• and HCl (+1).
Hydride complexes
Complexes coordinated by hydride ligands are called hydride complexes. The Group 6 to 10 transition metals that do not form binary hydrides give many hydride complexes with auxiliary ligands such as carbonyl and tertiaryphosphines. Although it was only at the end of the 1950s that hydride was accepted as a ligand, thousands of hydride complexes are known at present. Furthermore, with the synthesis in the 1980’s of molecular hydrogen complexes, the chemistry of transition metal hydrogen compounds took a new turn. Research on the homogeneous catalysis of hydrocarbons in which hydride or dihydrogen complexes participate is also progressing. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/04%3A_Chemistry_of_Nonmetallic_Elements/4.01%3A_Hydrogen_and_hydrides.txt |
(a) Boron
Refined elemental boron is a black solid with a metallic luster. The unit cell of crystalline boron contains 12, 50, or 105 boron atoms, and the B12 icosahedral structural units join together by 2 center 2 electron (2c-2e) bonds and 3 center 2 electron (3c-2e) bonds (electron deficient bonds) between boron atoms (Figure $1$). Boron is very hard and shows semiconductivity.
The chemistry of boranes (boron hydrides) started from the research of A. Stock reported during the period 1912-1936. Although boron is adjacent to carbon in the periodic table, its hydrides have completely different properties from those of hydrocarbons. The structures of boron hydrides in particular were unexpected and could be explained only by a new concept in chemical bonding. For his contribution to the very extensive new inorganic chemistry of boron hydrides, W. N. Lipscomb won the Nobel prize in 1976. Another Nobel prize (1979) was awarded to H. C. Brown for the discovery and development of a very useful reaction in organic synthesis called hydroboration.
Because of the many difficulties associated with the low boiling points of boranes, as well as their activity, toxicity, and air-sensitivity, Stock had to develop new experimental methods for handling the compounds in vacuo. Using these techniques, he prepared six boranes B2H6, B4H10, B5H9, B5H11, B6H10, and B10H14 by the reactions of magnesium boride, MgB2, with inorganic acids, and determined their compositions. However, additional research was necessary to determine their structures. At present, the original synthetic method of Stock using MgB2 as a starting compound is used only for the preparation of B6H10. Since reagents such as lithium tetrahydroborate, LiBH4, and sodium tetrahydroborate, NaBH4, are now readily available, and diborane, B2H6, is prepared according to the following equation, higher boranes are synthesized by the pyrolysis of diborane.
$3 LiBH_{4} + 4 BF_{3} \cdot OEt_{2} \rightarrow 2 B_{2} H_{6} + 3 LiBF_{4} + 4 Et_{2} O$
A new theory of chemical bonding was introduced to account for the bonding structure of diborane, B2H6. Although an almost correct hydrogen-bridged structure for diborane was proposed in 1912, many chemists preferred an ethane-like structure, H3B-BH3, by analogy with hydrocarbons. However, H. C. Longuet-Higgins proposed the concept of the electron-deficient 3-center 2-electron bond (3c-2e bond) and it was proven by electron diffraction in 1951 that the structure was correct (Figure $2$).
It has been elucidated by electron diffraction, single crystal X-ray structure analysis, infrared spectroscopy, etc. that boranes contain 3-center 2-electron bonds (3c-2e bond) B-H-B and
besides the usual 2 center 2 electron covalent bonds (2c-2e bond) B-H and B-B. Such structures can be treated satisfactorily by molecular orbital theory. Boranes are classified into closo, nido, arachno, etc. according to the skeletal structures of boron atoms.
Closo-borane [BnHn]2- has the structure of a closed polyhedron of n boron atoms bonded to n hydrogen atoms, as seen in the examples of a regular octahedron [B6H6]2- and an icosahedron [B12H12]2-. The boranes of this series do not contain B-H-B bonds. Boranes BnHn+4, such as B5H9, form structures with B-B, B-B-B, and B-H-B bonds and lack the apex of the polyhedron of closo boranes, and are referred to as nido type boranes. Borane BnHn+6, such as B4H10, have a structures that lacks two apexes from the closo type and are more open. Skeletons are also built by B-B, B-B-B, and B-H-B bonds, and these are called arachno type boranes. The structures of typical boranes are shown in Figure $3$.
Not only diborane but also higher boranes are electron-deficient compounds that are difficult to explain using Lewis' electronic structure based on simple 2-center 2-electron covalent bonds.
Exercise $2$
Why is diborane called an electron deficient compound?
Answer
It is because there are only 12 valence electrons of boron and hydrogen atoms, although 16 electrons are necessary to assign two electrons each to eight B-H bonds.
K. Wade summarized the relation of the number of valence electrons used for skeletal bonds and the structures of boranes and proposed an empirical rule called the Wade rule. According to this rule, when the number of boron atoms is n, the skeletal valence electrons are 2(n+1) for a closo type, 2(n+2) for a nido type, and 2(n+3) for an arachno type borane. The relationship between the skeletal structure of a cluster compound and the number of valence electrons is also an important problem in the cluster compounds of transition metals, and the Wade rule has played a significant role in furthering our understanding of the structures of these compounds.
(b) Carbon
Graphite, diamond, fullerene, and amorphous carbon are carbon allotropes. Usually a carbon atom forms four bonds using four valence electrons.
Graphite
Graphite is structured as layers of honeycomb-shaped 6 membered rings of carbon atoms that look like condensed benzene rings without any hydrogen atoms (Figure $4$). The carbon-carbon distance between in-layer carbon atoms is 142 pm and the bonds have double bond character analogous to aromatic compounds. Since the distance between layers is 335 pm and the layers are held together by comparatively weak van der Waals forces, they slide when subjected to an applied force. This is the origin of the lubricating properties of graphite. Various molecules, such as alkali metals, halogens, metal halides, and organic compounds intercalate between the layers and form intercalation compounds. Graphite has semi-metallic electrical conductivity (about 10-3 $\Omega$cm parallel to layers and about 100 times more resistant in the perpendicular direction).
Diamond
Its structure is called the diamond-type structure (Figure $5$). A unit cell of diamond contains eight carbon atoms and each carbon atom is 4-coordinate in a regular tetrahedron. Diamond is the hardest substance known, with a Mohs hardness 10. Diamond has very high heat conductivity although it is an electrical insulator. Although previously a precious mineral only formed naturally, industrial diamonds are now commercially prepared in large quantities at high temperatures (1200 °C or higher) and under high pressures (5 GPa or more) from graphite using metal catalysts. In recent years, diamond thin films have been made at low temperatures (about 900 °C) and under low pressures (about 102 Pa) by the pyrolysis of hydrocarbons, and are used for coating purposes, etc.
Fullerene
Fullerene is the general name of the 3rd carbon allotrope, of which the soccer ball-shaped molecule C60 is a typical example (Figure4.6). R. E. Smalley, H. W. Kroto and others detected C60 in the mass spectra of the laser heating product of graphite in 1985, and fullerene’s isolation from this so-called "soot" was reported in 1991. It has the structure of a truncated (corner-cut)-icosahedron and there is double bond character between carbon atoms. It is soluble in organic solvents, with benzene solutions being purple. Usually, it is isolated and purified by chromatography of fullerene mixtures. Wide-ranging research on chemical reactivities and physical properties such as superconductivity, is progressing rapidly. Besides C60, C70 and carbon nanotubes are attracting interest.
(c) Silicon
Silicon is the most abundant element in the earth’s crust after oxygen. Most of this silicon exists as a component of silicate rocks and the element is not found as a simple substance. Therefore, silicon is produced by the reduction of quartz and sand with high-grade carbon using electric arc furnaces. Higher-grade silicon is obtained by hydrogen reduction of SiHCl3, which is produced by the hydrochlorination of low purity silicon followed by rectification. The silicon used for semiconductor devices is further refined by the crystal Czochralski or zone melting methods. The crystal (mp 1410 o C) has a metallic luster and the diamond type structure.
There are three isotopes of silicon, 28Si (92.23%), 29Si (4.67%), and 30Si (3.10%). Because of its nuclear spin of I = 1/2, 29Si is used for NMR studies of organic silicon compounds or silicates (solid-state NMR).
Silicates and organosilicon compounds show a wide range of structures in silicon chemistry. Section 4.3 (c) describes the properties of silicates. Organosilicon chemistry is the most active research area in the inorganic chemistry of main group elements other than carbon. Silicon chemistry has progressed remarkably since the development of an industrial process to produce organosilicon compounds by the direct reaction of silicon with methyl chloride CH3Cl in the presence of a copper catalyst. This historical process was discovered by E. G. Rochow in 1945. Silicone resin, silicone rubber, and silicone oil find wide application. In recent years, silicon compounds have also been widely used in selective organic syntheses.
Although silicon is a congener of carbon, their chemical properties differ considerably. A well-known example is the contrast of silicon dioxide SiO2 with its 3-dimensional structure, and gaseous carbon dioxide, CO2. The first compound (Mes)2Si=Si(Mes)2 (Mes is mesityl C6H2(CH3)3) with a silicon-silicon double bond was reported in 1981, in contrast with the ubiquitous carbon-carbon multiple bonds. Such compounds are used to stabilize unstable bonds with bulky substituents (kinetic stabilization).
Exercise $3$
Why are the properties of CO2 and SiO2 different?
Answer
Their properties are very different because CO2 is a chain-like three-atom molecule and SiO2 is a solid compound with the three dimensional bridges between silicon and oxygen atoms.
(d) Nitrogen
Nitrogen is a colorless and odorless gas that occupies 78.1% of the atmosphere (volume ratio). It is produced in large quantities together with oxygen (bp -183.0 °C) by liquefying air (bp -194.1 °C) and fractionating nitrogen (bp -195.8 °C). Nitrogen is an inert gas at room temperatures but converted into nitrogen compounds by biological nitrogen fixation and industrial ammonia synthesis. The cause of its inertness is the large bond energy of the N$\equiv$N triple bond.
The two isotopes of nitrogen are 14N (99.634%) and 15N (0.366%). Both isotopes are NMR-active nuclides.
(e) Phosphorus
Simple phosphorus is manufactured by the reduction of calcium phosphate, Ca3(PO4)2, with quartz rock and coke. Allotropes include white phosphorus, red phosphorus, and black phosphorus.
White phosphorus is a molecule of composition of P4 (Figure $7$). It has a low melting-point (mp 44.1 °C) and is soluble in benzene or carbon disulfide. Because it is pyrophoric and deadly poisonous, it must be handling carefully.
Red phosphorus is amorphous, and its structure is unclear. The principal component is assumed to be a chain formed by the polymerization of P4 molecules as the result of the opening of one of the P-P bonds. It is neither pyrophoric nor poisonous, and used in large quantities for the manufacturing of matches, etc.
Black phosphorus is the most stable allotrope and is obtained from white phosphorus under high pressure (about 8 GPa). It is a solid with a metallic luster and a lamellar structure. Although it is a semiconductor under normal pressures, it shows metallic conductivity under high pressures (10 GPa).
Phosphorus compounds as ligands
Tertiary phosphines, PR3, and phosphites, P(OR)3, are very important ligands in transition metal complex chemistry. Especially triphenylphosphine, P(C6H5)3, triethyl phosphine, P(C2H5)3, and their derivatives are useful ligands in many complexes, because it is possible to control precisely their electronic and steric properties by modifying substituents (refer to Section 6.3 (c)). Although they are basically sigma donors, they can exhibit some pi accepting character by changing the substituents into electron accepting Ph (phenyl), OR, Cl, F, etc. The order of the electron-accepting character estimated from the C-O stretching vibrations and 13C NMR chemical shifts of the phosphine- or phosphite-substituted metal carbonyl compounds is as follows (Ar is an aryl and R is an alkyl).
$PF_{3} > PCl_{3} > P(OAr)_{3} > P(OR)_{3} > PAr_{3} > PRAr_{2} > PR_{2}Ar > PR_{3}$
On the other hand, C. A. Tolman has proposed that the angle at the vertex of a cone that surrounds the substituents of a phosphorus ligand at the van der Waals contact distance can be a useful parameter to assess the steric bulkiness of phosphines and phosphites. This parameter, called the cone angle, is widely used (Figure $8$). When the cone angle is large, the coordination number decreases by steric hindrance, and the dissociation equilibrium constant and dissociation rate of a phosphorus ligand become large (Table $2$). The numerical expression of the steric effect is very useful, and many studies have been conducted into this effect.
Table $2$ Cone angles ($\theta$°) of tertiary phosphines and phosphites
Ligands Cone angles
P(OEt)3 109
PMe3 118
P(OPh)3 121
PEt3 132
PMe2Ph 136
PPh3 145
PiPr3 160
PtBu3 182 | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/04%3A_Chemistry_of_Nonmetallic_Elements/4.02%3A_Main_group_elements_of_2nd_and_3rd_periods_and_their_compounds.txt |
(a) Oxygen
Dioxygen, O2, is a colorless and odorless gas (bp -183.0 °C) that occupies 21% of air (volume ratio). Since oxygen atoms are also the major components of water and rocks, oxygen is the most abundant element on the Earth’s surface. Despite its abundance, it was established as an element as late as the 18th century. Since an immense quantity of oxygen gas is consumed for steel production now, it is separated in large quantities from liquified air.
The isotopes of oxygen are 16O (99.762% abundance), 17O (0.038%), and 18O (0.200%). 17O has nuclear spin I = 5/2 and is an important nuclide for NMR measurements. 18O is used as a tracer for tracking reagents or for the study of reaction mechanisms. It is also useful for the assignment of absorption lines in infrared or Raman spectra by means of isotope effects.
As already described in section 2.3 (e), dioxygen, O2, in the ground state has two unpaired spins in its molecular orbitals, shows paramagnetism and is called triplet dioxygen. In the excited state, the spins are paired and dioxygen becomes diamagnetic, which is called singlet dioxygen. Singlet dioxygen is important in synthetic chemistry, because it has characteristic oxidation reactivity. Singlet dioxygen is generated in a solution by an energy transfer reaction from a photo-activated complex or by the pyrolysis of ozonides (O3 compounds).
Superoxide ion, O2-, and peroxide ion, O22-, are the anions of dioxygen (Table $3$). They can be isolated as alkali metal salts. There is another state, O2+, called the dioxygen (1+) cation, and it can be isolated as a salt with suitable anions.
Table $3$ Oxidation states of dioxygen
Bond order Compound O-O distance (Å) ν(O-O) (cm-1)
O2+ 2.5 O2[AsF6] 1.123 1858
O2 2.0 1.207 1554
O2- 1.5 K[O2] 1.28 1145
O22- 1.0 Na2[O2] 1.49 842
Ozone, O3, is an allotrope of oxygen that is an unstable gas with an irritating odor. Ozone is a bent three-atom molecule (117°) and has unique reactivities. In recent years it has been discovered that ozone plays an important role in intercepting the detrimental ultraviolet radiation from the sun in the upper atmospheric zone, and in protecting life on the Earth from photochemical damage. It is now clear that chlorofluorocarbons, frequently used as refrigerants or as cleaners of electronic components, destroy the ozone layer, and measures are being taken on a global scale to cope with this serious environmental problem.
(b) Oxides of hydrogen
Oxygen is highly reactive, and direct reactions with many elements form oxides. Water is an oxide of hydrogen and is crucially important for the global environment and life in general.
Water H2O
Ninety-seven percent of water on the Earth is present as sea water, 2% as ice of the polar zone, and fresh water represents only the small remaining fraction. Fundamental chemical and physical properties of water are very significant to chemistry. The main physical properties are shown in Table $1$. Most of the unusual properties of water are caused by its strong hydrogen bonds. Physical properties of water differ considerably with the presence of isotopes of hydrogen. At least nine polymorphs of ice are known and their crystal structures depend on the freezing conditions of the ice.
Water has a bond angle of 104.5° and a bond distance of 95.7 pm as a free molecule. It is described in Section 3.4 (b) that self-dissociation of water generates oxonium ion, H3O+. Further water molecules add to H3O+ to form [H(OH2)n]+ (H5O2+, H7O3+, H9O4+, and H13O6+), and the structures of the various species have been determined.
Hydrogen peroxide H2O2
Hydrogen peroxide is an almost colorless liquid (mp -0.89 °C and bp (extrapolated) 151.4 °C) that is highly explosive and dangerous in high concentrations. Usually it is used as a dilute solution but occasionally 90% aqueous solutions are used. Since it is consumed in large quantities as a bleaching agent for fiber and paper, large-scale industrial synthetic process has been established. This process applies very subtle catalytic reactions to produce a dilute solution of hydrogen peroxide from air and hydrogen using a substituted anthraquinone. This dilute solution is then concentrated.
When deuterium peroxide is prepared in a laboratory, the following reaction is applied.
$K_{2}S_{2}O_{8} + 2 D_{2} O \rightarrow D_{2} O + 2 KDSO_{4}$
Hydrogen peroxide is decomposed into oxygen and water in the presence of catalysts such as manganese dioxide, MnO2. Hydrogen peroxide may be either an oxidant or a reductant depending on its co-reactants. Its reduction potential in an acidic solution expressed in a Latimer diagram (refer to Section 3.3 (c)) is
$O_{2} \xrightarrow{+0.70} H_{2}O_{2} \xrightarrow{+1.76} H_{2} O$
(c) Silicon oxides
Silicon oxides are formed by taking SiO4 tetrahedra as structural units and sharing the corner oxygen atoms. They are classified by the number of corner-sharing oxygen atoms in the SiO4 tetrahedra, as this determines their composition and structure. When the SiO4 tetrahedra connect by corner sharing, the structures of the polymeric compounds become a chain, a ring, a layer, or 3-dimensional depending on the connection modes of adjacent units. Fractional expression is adopted in order to show the bridging modes. Namely, the numerator in the fraction is the number of bridging oxygens and the denominator is 2, meaning that one oxygen atom is shared by two tetrahedra. The empirical formulae are as follows and each structure is illustrated in Figure $9$ in coordination-polyhedron form.
• A bridge is constructed with one oxygen atom. (SiO3O1/2)3- = Si2O76-
• Bridges are constructed with two oxygen atoms. (SiO2O2/2)n2n- = (SiO3)n2n-
• Bridges are constructed with three oxygen atoms. (SiOO3/2)nn- = (Si2O5)n2n-
• Amalgamation of bridging modes with three oxygen and two oxygen atoms. [(Si2O5)(SiO2O2/2)2]n6- = (Si4O11)n6-
• Bridges are constructed with four oxygen atoms. (SiO4/2)n = (SiO2)n
Silicates with various cross linkage structures are contained in natural rocks, sand, clay, soil, etc.
Aluminosilicates
There are many minerals in which some silicon atoms of silicate minerals are replaced by aluminum atoms. They are called aluminosilicates. Aluminum atoms replace the silicon atoms in the tetrahedral sites or occupy the octahedral cavities of oxygen atoms, making the structures more complicated. The substitution of a tetravalent silicon by a trivalent aluminum causes a shortage of charge which is compensated by occlusion of extra cations such as H+, Na+, Ca2+, etc. Feldspars are a typical aluminosilicate mineral, and KAlSi3O8 (orthoclase) and NaAlSi3O8 (albite) are also known well. Feldspars take 3-dimensional structures in which all the corners of the SiO4 and AlO4 tetrahedra are shared.
On the other hand, 2-dimensional layers are formed if [AlSiO5]3- units are lined, and stratified minerals like mica are constructed if 6-coordinate ions are inserted between layers. If the number of oxygen atoms in the layers is not enough to form regular octahedra between layers, hydroxide groups bond to the interstitial Al3+ ions. Muscovite, KAl2(OH)2Si3AlO10, is a type of mica with such a structure and can be eas ily peeled into layers.
Zeolite
One of the important aluminosilicates is zeolite. Zeolites are present as natural minerals and also many kinds of zeolites are prepared synthetically in large quantities. The SiO4 and AlO4 tetrahedra are bonded by oxygen bridges, and form holes and tunnels of various sizes. The structures are composites of the basic structural units of tetrahedral MO4. As shown in Figure $10$, the basic units are cubes with 8 condensed MO4, hexagonal prisms with 12 condensed MO4, and truncated octahedra with 24 condensed MO4.
Silicon or aluminum atoms are located on the corners of the polyhedra and the bridging oxygen atoms on the middle of each edge (it should be noted that this expression is different from the polyhedron model of oxides).
When these polyhdra are bonded, various kinds of zeolite structures are formed. For example, the truncated octahedra called $\beta$ cages are the basic frames of synthetic zeolite A, Na12(Al12Si12O48)] • 27H2O, and the quadrangle portions are connected through cubes. It can be seen that an octagonal tunnel B forms when eight truncated octahedra bind in this way. The structure in which the hexagon portions connect through hexagonal prisms is faujasite, NaCa0.5(Al2Si5O14)] • 10H2O.
Alkali metal or alkaline earth metal cations exist in the holes, and the number of these cations increases with the content of aluminum to compensate for the charge deficiency. The structures of zeolites have many crevices in which cations and water are contained. Utilizing this cation-exchange property, zeolites are used in large quantities as softeners of hard water. As zeolites dehydrated by heating absorb water efficiently, they are also used as desiccants of solvents or gases. Zeolites are sometimes called molecular sieves, since the sizes of holes and tunnels change with the kinds of zeolites and it is possible to segregate organic molecules according to their sizes. Zeolites can fix the directions of more than two molecules in their cavities and can be used as catalysts for selective reactions.
For example, synthetic zeolite ZSM-5 is useful as a catalyst to convert methanol to gasoline. This zeolite is prepared hydrothermally in an autoclave (high-pressure reaction vessel) at ca. 100 °C using meta-sodium luminate, NaAlO2, as the source of aluminum oxide and silica sol as the source of silicon oxide with tetrapropylammonium bromide, Pr4NBr, present in the reaction. The role of this ammonium salt is a kind of mold to form zeolite holes of a fixed size. When the ammonium salt is removed by calcination at 500 °C, the zeolite structure remains.
(d) Nitrogen oxides
A variety of nitrogen oxides will be described sequentially from lower to higher oxidation numbers (Table $4$).
Table $4$ Typical oxides of main group elements
1 2 12 13 14 15 16 17 18
2 Li2O BeO B2O3 CO
CO2
N2O
NO
NO2
3 Na2O
Na2O2
NaO2
MgO Al2O3 SiO2 P4O6
P4O10
SO2
SO3
Cl2O
ClO2
4 K2O
K2O2
KO2
CaO ZnO Ga2O3 GeO2 As4O6
As4O10
SeO2
SeO3
5 Rb2O
Rb2O2
Rb9O2
SrO CdO In2O3 SnO2 Sb4O6
Sb4O10
TeO2
TeO3
I2O5 XeO3
XeO4
6 Cs2O
Cs11O3
BaO HgO Tl2O
Tl2O3
PbO
PbO2
Bi2O3
Dinitrogen monoxide, N2O
Oxide of monovalent nitrogen. Pyrolysis of ammonium nitrate generates this oxide as follows.
$NH_{4}NO_{3} \xrightarrow{250 \;^{o} C} N_{2} O + 2 H_{2} O$
Although the oxidation number is a formality, it is an interesting and symbolic aspect of the versatility of the oxidation number of nitrogen that NH4NO3 forms a monovalent nitrogen oxide (+1 is a half of the average of -3 and +5 for NH4 and NO3, respectively). The N-N-O bond distances of the straight N2O are 112 pm (N-N) and 118 pm (N-O), corresponding to 2.5th and 1.5th bond order, respectively. N2O (16e) is isoelectronic with carbon dioxide CO2 (16e). This compound is also called laughing gas and is widely used for analgesia.
Nitric oxide, NO
An oxide of divalent nitrogen. This is obtained by reduction of nitrite as follows.
$KNO_{2} + KI + H_{2}SO_{4} \rightarrow NO + K_{2}SO_{4} + H_{2} O + \frac{1}{2} I_{2}$
Having an odd number of valence electrons (11 electrons), it is paramagnetic. The N-O distance is 115 pm and the bond has double bond character. The unpaired electron in the highest antibonding $\pi^{*}$ orbital is easily removed, and NO becomes NO+ (nitrosonium), which is isoelectronic with CO. Since an electron is lost from the antibonding orbital, the N-O bond becomes stronger. The compounds NOBF4 and NOHSO4 containing this cation are used as 1 electron oxidants.
Although NO is paramagnetic as a monomer in the gas phase, dimerization in the ondensed phase leads to diamagnetism. It is unique as a ligand of transition metal complexes and forms complexes like [Fe(CO)2(NO)2], in which NO is a neutral 3-electron ligand. Although M-N-O is straight in these kind of complexes, the M-N-O angle bends to 120° ~ 140° in [Co(NH3)5NO] Br2, in which NO- coordinates as a 4-electron ligand. It has become clear recently that nitric oxide has various biological control functions, such as blood-pressure depressing action, and it attracts attention as the second inorganic material after Ca2+ to play a role in signal transduction.
Dinitrogen trioxide, N2O3
The oxidation number of nitrogen is +3, and this is an unstable compound decomposing into NO and NO2 at room temperature. It is generated when equivalent quantities of NO and NO2 are condensed at low temperatures. It is light blue in the solid state and dark blue in the liquid state but the color fades at higher temperatures.
Nitrogen dioxide, NO2
A nitrogen compound with oxidation number +4. It is an odd electron compound with an unpaired electron, and is dark reddish brown in color. It is in equilibrium with the colorless dimer dinitrogen tetroxide, N2O4. The proportion of NO2 is 0.01% at -11 °C, and it increases gradually to 15.9% at its boiling point (21.2 °C), and becomes 100% at 140 °C.
N2O4 can be generated by the pyrolysis of lead nitrate as follows.
$2 Pb(NO_{3})_{2} \xrightarrow{400 \;^{o} C} 4 NO_{2} + 2 PbO + O_{2}$
When NO2 is dissolved in water, nitric acid and nitrous acid are formed.
$2 NO_{2} + H_{2}O \rightarrow HNO_{3} + HNO_{2}$
By one electron oxidation, NO2+ (nitroyl) forms and the O-N-O angle changes from 134° to 180° in the neutral NO2. On the other hand, by one electron reduction, NO2- (nitrito) forms and the angle bends to 115°.
Dinitrogen pentoxide, N2O5, is obtained when concentrated nitric acid is carefully dehydrated with phosphorus pentoxide at low temperatures. It sublimes at 32.4 °C. As it forms nitric acid by dissolving in water, it may also be called a nitric anhydride.
$N_{2} O_{5} + H_{2} O \rightarrow 2 HNO_{3}$
Although it assumes an ion-pair structure NO2NO3 and straight NO2+ and planar NO3- ions are located alternately in the solid phase, it is molecular in the gas phase.
Oxoacids
Oxyacids of nitrogen include nitric acid, HNO3, nitrous acid, HNO2, and hyponitrous acid, H2N2O2. Nitric acid, HNO3, is one of the most important acids in the chemical industry, along with sulfuric acid and hydrochloric acid. Nitric acid is produced industrially by the Ostwald process, which is the oxidation reaction of ammonia in which the oxidation number of nitrogen increases from -3 to +5. Because the Gibbs energy of the direct conversion of dinitrogen to the intermediate NO2 is positive, and therefore the reaction is unfavorable thermodynamically, dinitrogen is firstly reduced to ammonia, and this is then oxidized to NO2.
$\stackrel{0}{N_{2}} \rightarrow \stackrel{-3}{NH_{3}} \rightarrow \stackrel{+4}{NO_{2}} \rightarrow \stackrel{+5}{HNO_{3}}$
Nitric acid, HNO3
Commercial nitric acid is a ca.70% aqueous solution and vacuum distillation of it in the presence of phosphorus pentoxide gives pure nitric acid. As it is a srong oxidizing agent while also being a strong acid, it can dissolve metals (copper, silver, lead, etc.) which do not dissolve in other acids. Gold and platinum can even be dissolved in a mixture of nitric acid and hydrochloric acid (aqua regia). The nitrate ion, NO3-, and nitrite ion, NO2-, take various coordination forms when they coordinate as ligands in transition metal complexes.
Nitrous acid, HNO2
Although not isolated as a pure compound, aqueous solutions are weak acids (pKa = 3.15 at 25 °C) and important reagents. Since NaNO2 is used industrially for hydroxylamine (NH2OH) production and also used for diazotidation of aromatic amines, it is important for the manufacture of azo dyes and drugs. Among the various coordination forms of NO2- isomers now known, monodentate nitro (N-coordination) and nitrito (O-coordination) ligands had already been discovered in the 19th century.
(e) Phosphorus oxides
The structures of the phosphorus oxides P4O10, P4O9, P4O7, and P4O6 have been determined.
Phosphorus pentoxide, P4O10, is a white crystalline and sublimable solid that is formed when phosphorus is oxidized completely. Four phosphorus atoms form a tetrahedron and they are bridged by oxygen atoms (refer to Figure $12$). Since a terminal oxygen atom is bonded to each phosphorus atom, the coordination polyhedron of oxygen is also a tetrahedron. When the molecular P4O10 is heated, a vitrified isomer is formed. This is a polymer composed of similar tetrahedra of phosphorus oxide with the same composition that are connected to one another in sheets. Since it is very reactive with water, phosphorus pentoxide is a powerful dehydrating agent. It is used not only as a desiccant, but also it has remarkable dehydration properties, and N2O5 or SO3 can be formed by dehydration of HNO3 or H2SO4, respectively. Phosphorus pentoxide forms orthophosphoric acid, H3PO4, when reacted with sufficient water, but if insufficient water is used, various kinds of condensed phosphoric acids are produced depending on the quantity of reacting water.
Phosphorus trioxide, P4O6, is a molecular oxide, and its tetrahedral structure results from the removal of only the terminal oxygen atoms from phosphorus pentoxide. Each phosphorus is tri-coordinate. This compound is formed when white phosphorus is oxidized at low temperatures in insufficient oxygen. The oxides with compositions intermediate between phosphorus pentoxide and trioxide have 3 to 1 terminal oxygen atoms and their structures have been analyzed.
Although arsenic and antimony give molecular oxides As4O6 and Sb4O6 that have similar structures to P4O6, bismuth forms a polymeric oxide of composition Bi2O3.
Phosphoric acid
Orthophosphoric acid, H3PO4
It is one of the major acids used in chemical industry, and is produced by the hydration reaction of phosphorus pentoxide, P4O10. Commercial phosphoric acid is usually of 75-85% purity. The pure acid is a crystalline compound (mp 42.35 °C). One terminal oxygen atom and three OH groups are bonded to the phosphorus atom in the center of a tetrahedron. The three OH groups release protons making the acid tribasic (pK1 = 2.15). When two orthophosphoric acid molecules condense by the removal of an H2O molecule, pyrophosphoric acid, H4P2O7, is formed.
Phosphonic acid, H3PO3
This acid is also called phosphorous acid and has H in place of one of the OH groups of orthophosphoric acid. Since there are only two OH groups, it is a dibasic acid.
Phosphinic acid, H3PO2
It is also called hypophosphorous acid, and two of the OH groups in orthophosphoric acid are replaced by H atoms. The remaining one OH group shows monobasic acidity. If the PO4 tetrahedra in the above phosphorus acids bind by O bridges, many condensed phosphoric acids form. Adenosine triphosphate (ATP), deoxyribonucleic acid (DNA), etc., in which the triphosphorus acid moieties are combined with adenosine are phosphorus compounds that are fundamentally important for living organisms.
(f) Sulfur oxides
Sulfur dioxide, SO2
This is formed by the combustion of sulfur or sulfur compounds. This is a colorless and poisonous gas (bp -10.0 °C) and as an industrial emission is one of the greatest causes of environmental problems. However, it is very important industrially as a source material of sulfur. Sulfur dioxide is an angular molecule, and recently it has been demonstrated that it takes various coordination modes as a ligand to transition metals. It is a nonaqueous solvent similar to liquid ammonia, and is used for special reactions or as a solvent for special NMR measurements.
Sulfur trioxide, SO3
It is produced by catalytic oxidation of sulfur dioxide and used for manufacturing sulfuric acid. The usual commercial reagent is a liquid (bp 44.6 °C). The gaseous phase monomer is a planar molecule. It is in equilibrium with a ring trimer ($\gamma$-SO3 = S3O9) in the gaseous or liquid phase. In the presence of a minute amount of water SO3 changes to $\beta$-SO3, which is a crystalline high polymer with a helical structure. $\alpha$-SO3 is also known as a solid of still more complicated lamellar structure. All react violently with water to form sulfuric acid.
Sulfur acids
Although there are many oxy acids of sulfur, most of them are unstable and cannot be isolated. They are composed of a combination of S=O, S-OH, S-O-S, and S-S bonds with a central sulfur atom. As the oxidation number of sulfur atoms varies widely, various redox equilibria are involved.
Sulfuric acid, H2SO4
It is an important basic compound produced in the largest quantity of all inorganic compounds. Pure sulfuric acid is a viscous liquid (mp 10.37 °C), and dissolves in water with the generation of a large amount of heat to give strongly acidic solutions.
Thiosulfuric acid, H2S2O3
Although it is generated if thiosulfate is acidified, the free acid is unstable. The S2O32- ion is derived from the replacement of one of the oxygen atoms of SO42- by sulfur, and is mildly reducing.
Sulfurous acid, H2SO3
The salt is very stable although the free acid has not been isolated. The SO32- ion that has pyramidal C3v symmetry is a reducing agent. In dithionic acid, H2S2O6, and the dithionite ion S2O62-, the oxidation number of sulfer is +5, and one S-S bond is formed. This is a very strong reducing agent.
(g) Metal oxides
Oxides of all the metallic elements are known and they show a wide range of properties in terms of structures, acidity and basicity, and conductivity. Namely, an oxide can exhibit molecular, 1-dimensional chain, 2-dimensional layer, or 3-dimensional structures. There are basic, amphoteric, and acidic oxides depending on the identity of the metallic element. Moreover, the range of physical properties displayed is also broad, from insulators, to semicondutors, metallic conductors, and superconductors. The compositions of metallic oxide can be simply stoichiometric, stoichiometric but not simple, or sometimes non-stoichiometric. Therefore, it is better to classify oxides according to each property. However, since structures give the most useful information to understand physical and chemical properties, typical oxides are classified first according to the dimensionality of their structures (Table $4$, Table $5$).
Table $5$ Typical binary oxides of transition me
Oxidation number 3 4 5 6 7 8 9 10 11
+1 Ti2Ol Cu2O
Ag2O
+2 TiO VO
NbO
MnO FeO CoO NiO CuO
Ag2O2
+3 Sc2O3
Y2O3
Ti2O3 V2O3 Cr2O3 Fe2O3 Rh2O3
+4 TiO2
ZrO2
HfO2
VO2
NbO2
TaO2
CrO2
MoO2
WO2
MnO2
TcO2
ReO2
RuO2
OsO2
RhO2
IrO2
PtO2
+5 V2O5i
Nb2O5
Ta2O5
+6 CrO3c
MoO3l
WO3
ReO3
+7 Re2O7l
+8 RuO4 m
OsO4m
m molecular, c chain, l layer, others 3-demensional. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/04%3A_Chemistry_of_Nonmetallic_Elements/4.03%3A_Oxygen_and_oxides_%28Part_1%29.txt |
Molecular oxides
Ruthenium tetroxide, RuO4,(mp 25 °C and bp 40 °C) and osmium tetroxide, OsO4, (mp 40 °C and bp 130 °C) have low melting and boiling points and their structures are molecular. They are prepared by heating the metal powder in an oxygen atmosphere at about 800 °C. The structures are tetrahedral and they are soluble in organic solvents and also slightly soluble in water. OsO4 is used in organic chemistry especially in the preparation of cis-diols by oxidation of C=C double bonds. For example, cyclohexane diol is prepared from cyclohexene. Since these oxides are very volatile and poisonous, they should be handled very carefully.
1-dimensional chain-like oxide
Mercury oxide, HgO, is a red crystallline compound that is formed when mercury nitrate is heated in air. HgO has an infinite zigzag structure. Chromium trioxide, CrO3, is a red crystalline compound with a low melting point (197 °C) and its structure is composed of CrO4 tetrahedra connected in one dimension. The acidity and oxidizing power of chromium trioxide are very high. It is used as an oxidation reagent in organic chemistry.
Two dimensional stratified oxides
Tetragonal and blue black tin oxide, SnO, and red lead oxide, PbO, are layer compounds composed of square pyramids with the metal atom at the peak and four oxygen atoms at the bottom vertices. The structure contains metal atoms above and below the layer of oxygen atoms alternately and in parallel with the oxygen layers (Figure $11$). Molybdenum trioxide, MoO3, is formed by burning the metal in oxygen and shows weak oxidizing power in aqueous alkaline solutions. It has a 2-dimensional lamellar structure in which the chains of edge-sharing octahedra MoO6 are corner-linked.
3-dimensional oxides
Alkali metal oxides, M2O (M is Li, Na, K, and Rb), have the antifluorite structure (refer to Section 2.2 (e)), and Cs2O is the anti-CdCl2 lamellar structure (refer to Section 4.5 (d)). M2O forms together with peroxide M2O2 when an alkali metal burns in air, but M2O becomes the main product if the amount of oxygen is less than stoichiometric. Alternatively, M2O is obtained by the pyrolysis of M2O2 after complete oxidation of the metal. Peroxide M2O2 (M is Li, Na, K, Rb, and Cs) can be regarded also as the salts of dibasic acid H2O2. Na2O2 is used industrially as a bleaching agent. Superoxide MO2 (M is K, Rb, and Cs) contains paramagnetic ion O2-, and is stabilized by the large alkali metal cation. If there is a deficit of oxygen during the oxidation reactions of alkali metals, suboxides like Rb9O2 or Cs11O3 form. These suboxides exhibit metallic properties and have interesting cluster structures (Figure $12$). Many other oxides in which the ratio of an akali metal and oxygen varies, such as M2O3, have also been synthesized.
MO type metal oxides
Except for BeO (Wurtz type), the basic structure of Group 2 metal oxides MO is the rock salt structure. They are obtained by calcination of the metal carbonates. Their melting points are very high and all are refractory. Especially quicklime, CaO, is produced and used in large quantities. The basic structure of transition metal oxides MO (M is Ti, Zr, V, Mn, Fe, Co, Ni, Eu, Th, and U) is also the rock salt structure, but they have defect structures and the ratios of a metal and oxygen are non-stoichiometric. For example, FeO has the composition FexO (x = 0.89-0.96) at 1000 °C. The charge imbalance of the charge is compensated by the partial oxidation of Fe2+ into Fe3+. NbO has a defective rock salt-type structure where only three NbO units are contained in a unit cell.
MO2 type metal oxides
The dioxides of Sn, Pb, and other transition metals with small ionic radii take rutile-type structures (Figure $13$), and the dioxides of lanthanide and actinide metals with large ionic radii take fluorite-type structures.
Rutile is one of the three structure types of TiO2, and is the most important compound used in the manufacture of the white pigments. Rutile has also been extensively studied as a water photolysis catalyst. As shown in Figure $13$, the rutile-type structure has TiO6 octahedra connected by the edges and sharing corners. It can be regarded as a deformed hcp array of oxygen atoms in which one half of the octahedral cavities are occupied by titanium atoms. In the normal rutile-type structure, the distance between adjacent M atoms in the edge-sharing octahedra is equal, but some rutile-type metal oxides that exhibit semiconductivity have unequal M-M-M distances. CrO2, RuO2, OsO2, and IrO2 show equal M-M distances and exhibit metallic conductivity.
Manganese dioxide, MnO2, tends to have a non-stoichiometric metal-oxygen ratio when prepared by the reaction of manganese nitrate and air, although the reaction of manganese with oxygen gives almost stoichiometric MnO2 with a rutile structure. The following reaction of manganese dioxide with hydrochloric acid is useful for generating chlorine in a laboratory.
$MnO_{2} + 4 HCl \rightarrow MnCl_{2} + Cl_{2} + 2 H_{2}O$
Zirconium dioxide, ZrO2, has a very high melting-point (2700 °C), and is resistant to acids and bases. It is also a hard material and used for crucibles or firebricks. However, since pure zirconium dioxide undergoes phase transitions at 1100 °C and 2300 ºC that result in it breaking up, solid solutions with CaO or MgO are used as fireproof materials. This is called stabilized zirconia.
M2O3-type oxides
The most important structure of the oxides of this composition is the corundum structure (Al, Ga, Ti, V, Cr, Fe, and Rh). In the corundum structure, 2/3 of the octahedral cavities in the hcp array of oxygen atoms are occupied by M3+. Of the two forms of alumina, Al2O3, $\alpha$ alumina and (\gamma\) alumina, $\alpha$ alumina takes the corundum structure and is very hard. It is unreactive to water or acids. Alumina is the principal component of jewelry, such as ruby and sapphire. Moreover, various fine ceramics (functional porcelain materials) utilizing the properties of $\alpha$-alumina have been developed. On the other hand, $\gamma$ alumina has a defective spinel-type structure, and it adsorbs water and dissolves in acids, and is the basic component of activated alumina. It has many chemical uses including as a catalyst, a catalyst support, and in chromatography.
MO3 type oxides
Rhenium and tungsten oxides are important compounds with this composition. Rhenium trioxide, ReO3, is a dark red compound prepared from rhenium and oxygen that has a metallic luster and conductivity. ReO3 has a three-dimensional and very orderly array of ReO6 regular, corner-sharing octahedra (Figure $14$).
Tungsten trioxide, WO3, is the only oxide that shows various phase transitions near room temperature and at least seven polymorphs are known. These polymorphs have the ReO3-type three-dimensional structure with corner-sharing WO octahedra. When these compounds 6 are heated in a vacuum or with powdered tungsten, reduction takes place and many oxides with complicated compositions (W18O49, W20O58, etc.) are formed. Similar molybdenum oxides are known and they had been regarded as non-stoichiometric compounds before A. Magneli found that they were in fact stoichiometric compounds.
Mixed metal oxides
Spinel, MgAl2O4, has a structure in which Mg2+ occupy 1/8 of the tetrahedral cavities and Al3+ 1/2 of the octahedral cavities of a ccp array of oxygen atoms (Figure $15$).
Among the oxides of composition A2+B23+O4 (A2+ are Mg, Cr, Mn, Fe, Co, Ni, Cu, Zn, Cd, Sn, and B3+ are Al, Ga, In, Ti, V, Cr, Mn, Fe, Co, Ni, and Rh) , those in which the tetrahedral holes are occupied by A2+ or B3+ are called normal spinels or inverse spinels, respectively. Spinel itself has a normal spinel-type structure, and MgFe2O4 and Fe3O4 have inverse spinel-type structures. Crystal field stabilization energies (refer to Section 6.2 (a)) differ depending on whether the crystal field of the oxygen atoms is a regular trahedron or octahedron. Therefore, when the metal component is a transition metal, the energy difference is one of the factors to determine which of A2+ or B3+ is favorable to fill the tetrahedral cavities.
Perovskite, CaTiO3, is an ABO3 oxide (the net charge of A and B becomes 6+), and it has a structure with calcium atom at the center of TiO3 in the ReO3 structure (Figure $16$). Among this kind of compounds, BaTiO3, commonly called barium titanate, is especially important. This ferroelectric functional material is used in nonlinear resistance devices (varistor).
(h) Oxides of Group 14 elements
Although GeO2 has a rutile-type structure, there is also a $\beta$ quartz-type polymorphism. There are germanium oxides with various kinds of structures analogous to silicates and aluminosilicates. SnO2 takes a rutile-type structure. SnO2 is used in transparent electrodes, catalysts, and many other applications. Surface treatment with tin oxide enhances heat reflectivity of glasses. PbO2 usually has a rutile-type structure. Lead oxide is strongly oxidizing and used for the manufacture of chemicals, and PbO2 forms in a lead batteries.
(i) Isopolyacids, heteropolyacids, and their salts
There are many polyoxo acids and their salts of Mo(VI) and W (VI). V (V), V (IV), Nb (V), and Ta (V) form similar polyoxo acids although their number is limited. Polyoxoacids are polynuclear anions formed by polymerization of the MO6 coordination polyhedra that share corners or edges. Those consisting only of metal, oxygen, and hydrogen atoms are called isopolyacids and those containing various other elements (P, Si, transition metals, etc.) are called heteropolyacids. The salts of polyacids have counter-cations such as sodium or ammonium instead of protons. The history of polyoxoacids is said to have started with J. Berzelius discovering the first polyoxoacid in 1826, with the formation of yellow precipitates when he acidified an aqueous solution containing Mo (VI) and P (V). The structures of polyoxoacids are now readily analyzed with single crystal X-ray structural analysis, 17O NMR, etc. Because of their usefulness as industrial catalysts or for other purposes, polyoxoacids are again being studied in detail.
Keggin-structure
The heteropolyoxo anions expressed with the general formula [Xn+M12O40](8-n)- (M = Mo, W, and X = B, Al, Si, Ge, P, As, Ti, Mn, Fe, Co, Cu, etc.) have the Keggin structure, elucidated by J. F. Keggin in 1934 using X-ray powder diffraction. For example, the structure of the tungstate ion containing silicon, in which 12 WO6 octahedra enclose the central SiO4 tetrahedron and four groups of three edge-shared octahedra connect to each other by corner sharing, is shown in Figure $17$. The four oxygen atoms that coordinate to the silicon atom of the SiO4 tetrahedra also share three WO6 octahedra. Therefore, the whole structure shows Td symmetry. Although the Keggin structure is somewhat complicated, it is very symmetrical and beautiful and is the most typical structure of heteropolyoxo anions. Many other types of heteropolyoxo anions are known.
Polyoxo anions are generated by the condensation of MO6 units by removal of H2O when MoO42- reacts with a proton H+, as is shown in the following equation.
$12 [MoO_{4}]^{2-} + HPO_{4}^{2-} \xrightarrow{H^{+}} [PMo_{12}O_{40}]^{3-} + 12 H_{2}O$
Therefore, the size and form of heteropolyoxo anions in the crystal precipitation are decided by the choice of acid, concentration, temperature, or the counter cation for crystallization. A number of studies on the solution chemistry of dissolved anions have been performed.
Heteropolyoxo anions display notable oxidizing properties. As heteropolyoxo anions contain metal ions of the highest oxidation number, they are reduced even by very weak reducing agents and show mixed valence. When Keggin-type anions are reduced by one electron, they show a very deep-blue color. It has been proved that the Keggin structure is preserved at this stage and polyoxo anions absorb more electrons and several M(V) sites are generated. Thus, a heteropolyoxo anion can serve as an electron sink for many electrons, and heteropolyoxo anions exhibit photo-redox reactions.
Exercise $4$
What is the major difference in the structures of a polyacid and a solid acid?
Answer
Although polyacids are molecules with definite molecular weights, the usual solid oxides have an infinite number of metal-oxygen bonds. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/04%3A_Chemistry_of_Nonmetallic_Elements/4.04%3A_Oxygen_and_oxides_%28Part_2%29.txt |
(a) Simple substances
Sulfur, selenium, and tellurium are called chalcogens. Simple substances and compounds of oxygen and of the elements of this group in the later periods have considerably different properties. As a result of having much smaller electronegativities than oxygen, they show decreased ionicity and increased bond covalency, resulting in a smaller degree of hydrogen bonding. Because they have available d orbitals, chalcogens have increased flexibility of valence and can easily bond to more than two other atoms. Catenation is the bonding between the same chalcogen atoms, and both simple substances and ions of chalcogens take a variety of structures.
The major isotopes of sulfur are 32S (95.02% abundance), 33S (0.75%), 34S (4.21%), and 36S (0.02%), and there are also six radioactive isotopes. Among these, 33S (I = 3/2) can be used for NMR. Since the isotope ratio of sulfurs from different locations differs, the accuracy of the atomic weight is limited to 32.07+0.01. Because the electronegativity of sulfur ($\chi$ = 2.58) is much smaller than that of oxygen ($\chi$ = 3.44) and sulfur is a soft element, the ionicity in the bonds of sulfur compounds is low and hydrogen bonding is not important. Elemental sulfur has many allotropes, such as S2, S3, S6, S7, S8, S9, S10, S11, S12, S18, S20, and S$\alpha$, reflecting the catenation ability of sulfur atoms.
Elemental sulfur is usually a yellow solid with a melting point of 112.8 °C called orthorhombic sulfur ($\alpha$ sulfur). Phase-transition of this polymorph produces monoclinic sulfur ($\beta$ sulfur) at 95.5 °C. It was established in 1935 that these are crown-like cyclic molecules (Figure $18$). Being molecular, they dissolve well in organic solvents, such as CS2. Not only 8-membered rings but also S6-20 rings are known, and the helix polymer of sulfur is an infinitely annular sulfur. Diatomic molecular S2 and triatomic molecular S3 exist in the gaseous phase. When sulfur is heated, it liquifies and becomes a rubber-like macromolecule on cooling. The diversity of structures of catenated sulfur is also seen in the structures of the polysulfur cations or anions resulting from the redox reactions of the catenated species.
Selenium is believed to have six isotopes. 80Se (49.7%) is the most abundant and 77Se, with nuclear spin I = 1/2 is useful in NMR. The accuracy of atomic weight of selenium, 78.96+0.03, is limited to two decimal places because of composition change of its isotopes. Among many allotropes of selenium, so-called red selenium is an Se8 molecule with a crown-like structure and is soluble in CS2. Gray metallic selenium is a polymer with a helical structure. Black selenium, which is a complicated polymer, is also abundant.
Tellurium also has eight stable isotopes and an atomic weight of 127.60+0.03. 130Te (33.8%) and 128Te (31.7%) are the most abundant isotopes, and 125Te and 123Te with I = 1/2 can be used in NMR. There is only one crystalline form of tellurium, which is a spiral chain polymer that shows electric conductivity.
(b) Polyatomic chalcogen cations and anions
Although it has long been recognized that solutions of chalcogen elements in sulfuric acid showed beautiful blue, red, and yellow colors, the polycationic species that give rise to these colors, S42+, S62+, S64+, S82+, S102+, S192+, or those of other chalcogen atoms, have been isolated by the reaction with AsF5, etc. and their structures determined. For example, unlike neutral S8, S82+ takes a cyclic structure that has a weak coupling interaction between two transannular sulfur atoms (Figure $18$).
On the other hand, alkali metal salts Na2S2, K2S5, and alkaline earth metal salt BaS3, a transition-metal salt [Mo2(S2)6]2-, a complex Cp2W(S4), etc. of polysulfide anions Sx2- (x = 1-6), in which the sulfur atoms are bonded mutually have been synthesized and their structures determined. As is evident from the fact that elemental sulfur itself forms S8 molecules, sulfur, unlike oxygen, tends to catenate. Therefore, formation of polysulfide ions, in which many sulfur atoms are bonded, is feasible, and a series of polysulfanes H2Sx (x = 2-8) has actually been synthesized.
(c) Metal sulfides
Stratified disulfides, MS2, are important in transition metal sulfides. They show two types of structures. One has a metal in a triangular prismatic coordination environment, and the other has a metal in an octahedral coordination environment.
MoS2 is the most stable black compound among the molybdenum sulfides. L. Pauling determined the structure of MoS2 in 1923. The structure is constructed by laminating two sulfur layers between which a molybdenum layer is intercalated (Figure $19$). Alternatively, two sulfur layers are stacked and a molybdenum layer is inserted between them. Therefore, the coordination environment of each molybdenum is a triangular prism of sulfur atoms. Since there is no bonding interaction between sulfur layers, they can easily slide, resulting in graphite-like lubricity. MoS2 is used as a solid lubricant added to gasoline, and also as a catalyst for hydrogenation reactions.
ZrS2, TaS2, etc. take the CdI2-type structure containing metal atoms in an octahedral coordination environment constructed by sulfur atoms.
Chevrel phase compounds
There are superconducting compounds called Chevrel phases which are important examples of the chalcogenide compounds of molybdenum. he general formula is described by MxMo6X8 (M = Pb, Sn, and Cu; X = S, Se, and Te), and six molybdenum atoms form a regular octahedral cluster, and eight chalcogenide atoms cap the eight triangular faces of the cluster. The cluster units are connected 3-dimensionally (Figure $20$). Since the cluster structure of molybdenum atoms is similar to that of molybdenum dichloride, MoCl2, (= (Mo6Cl8)Cl2Cl4/2), the structural chemistry of these compounds has attracted as much attention as their physical properties. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/04%3A_Chemistry_of_Nonmetallic_Elements/4.05%3A_Chalcogens_and_Chalcogenides.txt |
The origin of halogen is the Greek word meaning the production of salt by direct reaction with a metal. Since their reactivity is very high, halogens are found in nature only as compounds. The basic properties of halogens are shown in Table $6$ and Table $7$. The electron configuration of each halogen atom is ns2np5, and they lack one electron from the closed-shell structure of a rare gas. Thus a halogen atom emits energy when it gains an electron. Namely, the enthalpy change of the reaction $X\; (g) + e^{-} \rightarrow X^{-}\; (g)$is negative. Although electron affinity is defined as the energy change of gaining an electron, a positive sign is customarily used. In order to be consistent with the enthalpy change, a negative sign would be appropriate.
Table $6$ Properties of halogens
Ionization energy
(kJ mol-1)
Electronegativity
$\chi_{P}$
Ionic radius
r(X-) (pm)
F 1680.6 3.98 133
Cl 1255.7 3.16 181
Br 1142.7 2.996 196
I 1008.7 2.66 220
Table $7$ Properties of halogen molecules
Interatomic distance
r(X-X) (pm)
mp
°C
bp
°C
Color
F2 143 -218.6 -188.1 Colorless gas
Cl2 199 -101.0 -34.0 Yellow green gas
Br2 228 -7.75 59.5 Dark red liquid
I2 266 113.6 185.2 Dark violet solid
The electron affinity of chlorine (348.5 kJ mol-1) is the largest and fluorine (332.6 kJ mol-1) comes between chlorine and bromine (324.7 kJ mol-1). The electronegativity of fluorine is the highest of all the halogens.
Since halogens are produced as metal salts, simple substances are manufactured by electrolysis. Fluorine only takes the oxidation number -1 in its compounds, although the oxidation number of other halogens can range from -1 to +7. Astatine, At, has no stable nuclide and little is known about its chemical properties.
(a) Manufacture of halogen
Fluorine has the highest reduction potential (E = +2.87 V) and the strongest oxidizing power among the halogen molecules. It is also the most reactive nonmetallic element. Since water is oxidized by F2 at much lower electrode potential (+1.23 V), fluorine gas cannot be manufactured by the electrolysis of aqueous solutions of fluorine compounds. Therefore, it was a long time before elemental fluorine was isolated , and F. F. H. Moisson finally succeeded in isolating it by the electrolysis of KF in liquid HF. Fluorine is still manufactured by this reaction.
Chlorine, which is especially important in inorganic industrial chemistry, is manufactured together with sodium hydroxide. The basic reaction for the production of chlorine is electrolysis of an aqueous solution of NaCl using an ion exchange process. In this process, chlorine gas is generated in an anodic cell containing brine and Na+ moves through an ion exchange membrane to the cathodic cell where it pairs with OH- to become an aqueous solution of NaOH.
Exercise $5$
Why can chlorine be manufactured by electrolysis of an aqueous solution of sodium chloride?
Answer
Despite the higher reduction potential of chlorine (+1.36 V) than that of oxygen (+1.23 V), the reduction potential of oxygen can be raised (overvoltage) depending on the choice of electrode used for the electrolysis process.
Bromine is obtained by the oxidation of Br- with chlorine gas in saline water. Iodine is similarly produced by passing chlorine gas through saline water containing I- ions. Since natural gas is found in Japan together with underground saline water containing I- Japan is one of the main countries producing iodine.
Anomalies of fluorine
Molecular fluorine compounds have very low boiling points. This is due to the difficulty of polarization as a result of the electrons being strongly drawn to the nuclei of fluorine atoms. Since the electronegativity of fluorine is highest ($\chi$ = 3.98) and electrons shift to F, resulting in the high acidity of atoms bonded to F. Because of the small ionic radius of F-, high oxidation states are stabilized, and hence low oxidation compounds like CuF are unknown, in contrast with the compounds such as IF7 and PtF6.
Pseudohalogens
Since the cyanide ion CN-, the azide ion N3-, and the thiocyanate ion SCN-, etc. form compounds similar to those of halide ions, they are called pseudohalide ions. They form psudohalogen molecules such as cyanogene (CN)2, hydrogen cyanide HCN, sodium thiocyanate NaSCN, etc. Fine-tuning electronic and steric effects that are impossible with only halide ions make pseudohalogens useful also in transition metal complex chemistry.
Polyhalogens
Besides the usual halogen molecules, mixed halogen and polyhalogen molecules such as BrCl, IBr, ICl, ClF3, BrF5,IF7 etc also exist. Polyhalogen anions and cations such as I3-, I5-, I3+, and I5+, are also known.
(b) Oxygen compounds
Although many binary oxides of halogens (consisting only of halogen and oxygen) are known, most are unstable. Oxygen difluoride OF2 is the most stable such compound. This is a very powerful fluorinating agent and can generate plutonium hexafluoride PuF6 from plutonium metal. While oxygen chloride, Cl2O, is used for bleaching pulp and water treatment, it is generated in situ from ClO3-, since it is unstable.
Hypochlorous acid, HClO, chlorous acid, HClO2, chloric acid, HClO3, and perchloric acid, HClO4 are oxoacids of chlorine and especially perchloric acid is a strong oxidizing agent as well as being a strong acid. Although analogous acids and ions of other halogens had been known for many years, BrO4- was synthesized as late as 1968. Once it was prepared it turned out to be no less stable than ClO4- or IO4-, causing some to wonder why it had not been synthesized before. Although ClO4- is often used for crystallizing transition metal complexes, it is explosive and should be handled very carefully.
(c) Halides of nonmetals
Halides of almost all nonmetals are known, including fluorides of even the inert gases krypton, Kr, and xenon, Xe. Although fluorides are interesting for their own unique characters, halides are generally very important as starting compounds for various compounds of nonmetals by replacing halogens in inorganic syntheses (Table $8$).
Table $8$ Typical chlorides and fluorides of main group elements
1 2 12 13 14 15 16 17 18
2 LiCl BeCl2 BF3 CCl4 NF3 OF2
3 NaCl MgCl2 AlCl3 SiCl4 PCl3
PCl5
S2Cl2
SF6
ClF3
ClF5
4 KCl CaCl2 ZnCl2 GaCl3 GeF2
GeCl4
AsCl3
AsF5
Se2Cl2
SeF5
BrF3
BrF5
KrF2
5 RbCl SrCl2 CdCl2 InCl
InCl3
SnCl2
SnCl4
SbCl3
SbF5
Te4Cl16
TeF6
IF5
IF7
XeF2
XeF6
6 CsCl BaCl2 Hg2Cl2
HgCl2
TlCl
TlCl3
PbCl2
PbCl4
BiCl3
BiF5
Boron trifluoride, BF3, is a colorless gas (mp -127 °C and bp -100 °C) that has an irritating odor and is poisonous. It is widely used as an industrial catalyst for Friedel-Crafts type reactions. It is also used as a catalyst for cationic polymerization. It exists in the gaseous phase as a triangular monomeric molecule, and forms Lewis base adducts with ammonia, amines, ethers, phosphines, etc. because of its strong Lewis acidity. Diethylether adduct, (C2H5)2O:BF3, is a distillable liquid and is used as a common reagent. It is a starting compound for the preparation of diborane, B2H6. Tetrafluoroborate, BF4-, is a tetrahedral anion formed as an adduct of BF3 with a base F-. Alkali metal salts, a silver salt and NOBF4 as well as the free acid HBF4 contain this anion. Since its coordination ability is very weak, it is used in the crystallization of cationic complexes of transition metals as a counter anion like ClO4-. AgBF4 and NOBF4 are also useful for 1-electron oxidation of complexes.
Tetrachlorosilane, SiCl4, is a colorless liquid (mp -70 °C and bp 57.6 °C). It is a regular tetrahedral molecule, and reacts violently with water forming silicic acid and hydrochloric acid. It is useful as a raw material for the production of pure silicon, organic silicon compounds, and silicones.
Phosphorus trifluoride, PF3, is a colorless, odorless, and deadly poisonous gas (mp -151.5 °C and bp -101.8 °C). This is a triangular pyramidal molecule. Because it is as electron-attracting as CO, it acts as a ligand forming metal complexes analogous to metal carbonyls.
Phosphorus pentafluoride, PF5, is a colorless gas (mp -93.7 °C and bp -84.5 °C). It is a triangular bipyramidal molecule and should have two distinct kinds of fluorine atoms. These fluorines exchange positions so rapidly that they are indistinguishable by 19F NMR. It was the first compound with which the famous Berry's pseudorotation was discovered as an exchange mechanism for axial and equatorial fluorine atoms (refer to Section 6.1). The hexafluorophosphate ion, PF6-, as well as BF4- is often used as a counter anion for cationic transition metal complexes. LiPF6 and R4NPF6 can be used as supporting electrolytes for electrochemical measurements.
Phosphorus trichloride, PCl3, is a colorless fuming liquid (mp -112 °C and bp 75.5 °C). It is a triangular pyramidal molecule and hydrolyzes violently. It is a soluble in organic solvents. It is used in large quantities as a raw material for the production of organic phosphorus compounds.
Phosphorus pentachloride, PCl5, is a colorless crystalline substance (sublimes but decomposes at 160 °C) It is a triangular bipyramidal molecule in the gaseous phase, but it exists as an ionic crystal [PCl4]+ [PCl6]- in the solid phase. Although it reacts violently with water and becomes phosphoric acid and hydrochloric acid, it dissolves in carbon disulfide and carbon tetrachloride. It is useful for cchlorination of organic compounds.
Arsenic pentafluoride, AsF5, is a colorless gas (mp -79.8 °C and bp -52.9 °C). It is a triangular bipyramidal molecule. Although it hydrolyzes, it is soluble in organic solvents. As it is a strong electron acceptor, it can form electron donor-acceptor complexes with electron donors.
Sulfur hexafluoride, SF6, is a colorless and odorless gas (mp -50.8 °C and sublimation point -63.8 °C) It is a hexacoordinate octahedral molecule. It is chemically very stable and hardly soluble in water. Because of its excellent heat-resisting property, incombustibility, and corrosion resistance, it is used as a high voltage insulator.
Sulfur chloride, S2Cl2, is an orange liquid (mp -80 °C and bp 138 °C). It has a similar structure to hydrogen peroxide. It is readily soluble in organic solvents. It is important as an industrial inorganic compound, and is used in large quantities for the vulcanization of rubber etc.
(d) Metal halides
Many metal halides are made by the combination of about 80 metallic elements and four halogens (Table $8$, Table $9$). Since there are more than one oxidation state especially in transition metals, several kinds of halides are known for each transition metal. These halides are most important as starting materials of the preparation of metal compounds, and the inorganic chemistry of metal compounds depends on metal halides. There are molecular, 1-dimensional chain, 2-dimensional layer, and 3-dimensional halides but few of them are molecular in crystalline states. It should be noted that the anhydrous transition metal halides are usually solid compounds and hydrates are coordination compounds with water ligands. As the dimensionality of structures is one of the most interesting facets of structural or synthetic chemistry, typical halides are described in order of their dimensionality.
Table $9$ Typical chlorides and fluorides of transition metals
Oxidation Number 3 4 5 6 7 8 9 10 11
+1 ScCl
YCl
LaCl
ZrCl
HfCl
CuCl
AgCl
AuCl
+2 TiCl2 VCl2 CrCl2
MoCl2
WCl2
MnCl2 FeCl2
RuCl2
CoCl2 NiCl2
PdCl2
PdCl2
CuCl2
+3 ScF3
YCl3
LaF3
TiCl3
ZrCl3
VCl3 CrCl3
MoCl3
WCl3
ReCl3 FeCl3
RuCl3
OsCl3
CoF3
RhCl3
IrCl3
AuCl3
+4 TiCl4
ZrCl4
HfCl4
VCl4
NbCl4
TaCl4
CrF4
MoCl4
WCl4
ReCl4 PtCl4
+5 VF5
NbCl5
TaCl5
CrF5
MoCl5
WCl5
ReCl5 OsF5 IrF5 PtF5
+6 ReF6 OsF6 IrF6 PtF6
+7 ReF7 OsF7
Molecular halides
Mercury(II) chloride, HgCl2
It is a colorless crystal soluble in water and ethanol. It is a straight, three-atomic molecule in the free state. However, in addition to two chlorine atoms bonded to mercury, four additional chlorine atoms of adjacent molecules occupy coordination sites and the mercury is almost hexacoordinate in the crystalline state. The compound is very toxic and used for preserving wood, etc.
Aluminum trichloride, AlCl3.
A colorless crystal (mp 190 °C (2.5 atm) and bp 183 °C) that sublimes when heated. It is soluble in ethanol and ether. It is a Lewis acid and forms adducts with various bases. It is a molecule consisting of the dimer of tetracoordinate aluminium with chlorine bridges in the liquid and gaseous phases (Figure $21$), and takes a lamellar structure when crystalline. It is used as a Lewis acid catalyst of Friedel-Crafts reactions, etc.
Tin (IV) chloride, SnCl4
A colorless liquid (mp -33 °C and bp 114 °C). In the gaseous state, it is a tetrahedral molecule.
Titanium(IV) chloride, TiCl4
A colorless liquid (mp -25 °C and bp 136.4 °C). The gaseous molecule is a tetrahedron similar to tin(IV) chloride. It is used as a component of the Ziegler Natta catalyst (refer to Section 8.1 (a)).
Chain-like halides
Gold (I) iodide, AuI
Yellow white solid. Two iodines coordinate to gold, and the compound has a zigzag 1-dimensional chain structure.
Beryllium chloride, BeCl2
A colorless crystal (mp 405 °C and bp 520 °C). It is deliquescent and soluble in water and ethanol. The tetra-coordinated beryllium forms a 1-dimensional chain via chlorine bridges (Figure $22$). In the gaseous phase, it is a straight two-coordinate molecule. It is a Lewis acid and is used as a catalyst for Friedel-Crafts reactions.
Palladium chloride, PdCl2
A dark red solid. In the $\alpha$ type, the four-coordinate palladium forms a 1-dimensional chain with double bridges of chlorines. The dihydrate is deliquescent and soluble in water, ethanol, acetone, etc. When it is dissolved in hydrochloric acid, it becomes four-coordinate square-planar [PdCl4]2-. It is used as the catalyst for the Wacker process, which is an olefin oxidation process, or in various catalysts for organic syntheses.
Zirconium tetrachloride, (IV) ZrCl4
A colorless crystal (it sublimes above 331 °C). The zirconium is octahedrally coordinated and forms a zigzag chain via chlorine bridges (Figure $23$). It is hygroscopic and soluble in water, ethanol, etc. It is used as a Friedel-Crafts catalyst and as a component of olefin polymerization catalysts.
Stratified halides
Cadmium iodide, CdI2
A colorless crystal (mp 388 °C and bp 787 °C). It has a cadmium iodide structure where the layers of edge-shared CdI6 octahedral units are stratified (Figure $24$). In the gaseous phase, it comprises straight three atomic molecules. It dissolves in water, ethanol, acetone, etc.
Cobalt(II) chloride, CoCl2
Blue crystals (mp 735 °C and bp 1049 °C). It has the cadmium chloride structure. It is hygroscopic and becomes light red when water is absorbed. It is soluble also in ethanol and acetone. The hexahydrate is red and is a coordination compound in which water molecules are ligands.
Iron (II)chloride, FeCl2
Greenish yellow crystals (mp 670-674 °C). It has the cadmium chloride structure, and is soluble in water and ethanol. The hydrates, which are coordinated by various numbers (6, 4, 2) of water molecules, are precipitated from aqueous solutions of hydrochloric acid.
Iron(III) chloride, FeCl3
Dark brown crystals (mp 306 °C and sublimes). It has a lamellar structure in which iron is octahedrally surrounded by six chlorine ligands. In the gaseous phase, it has a dimeric structure bridged by chlorine atoms similar to that of aluminum chloride.
3-dimensional structure halides
Sodium chloride, NaCl
A colorless crystal (mp 801 °C and bp 1413 °C). It is the original rock salt-type structure. In the gaseous phase, this is a two-atom molecule. Although it is soluble in glycerol as well as water, it hardly dissolves in ethanol. Large single crystals are used as prisms for infrared spectrometers.
Cesium chloride, CsCl
A colorless crystal (mp 645 °C, bp 1300 °C). Although it has the cesium chloride type structure, it changes to the rock salt structure at 445 °C. In the gaseous phase, it is a two-atom molecule.
Copper(I) chloride, CuCl
A colorless crystal (mp 430 °C and bp 1490 °C) It has the zinc blende structure and four chlorines tetrahedrally coordinate to copper.
Calcium chloride, CaCl2
A colorless crystal (mp 772 °C and bp above 1600 °C). It has a deformed rutile-type structure and calcium is octahedrally surrounded by six chlorines. It is soluble in water, ethanol, and acetone. It is deliquescent and used as a desiccant. Hydrates in which 1, 2, 4, or 6 water molecules are coordinated are known.
Calcium fluoride, CaF2
A colorless crystal (mp 1418 °C and bp 2500 °C). It has the fluorite type structure. It is the most important raw material for fluorine compounds. Good quality crystals are used also as spectrometer prisms and in photographic lenses.
Chromium(II) chloride, CrCl2
A colorless crystal (mp 820 °C and sublimes). It has a deformed rutile-type structure. It dissolves well in water giving a blue solution.
Chromium(III) chloride, CrCl3
Purplish red crystal (mp 1150 °C and decomposes at 1300 °C). Cr3+ occupies two thirds of the octahedral cavities in every other layer of Cl- ions, which are hexagonally close-packed. It is insoluble in water, ethanol, and acetone.
Exercise $6$
Why do solid metal halides dissolve in water?
Answer
It is because water reacts with halides breaking the halogen bridges in the solid structures and coordinates to the resultant molecular complexes. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/04%3A_Chemistry_of_Nonmetallic_Elements/4.06%3A_Halogens_and_Halides.txt |
In the 18th century, H. Cavendish discovered an inert component in air. In 1868, a line was discovered in the spectrum of sunlight that could not be identified and it was suggested to be due to a new element, helium. Based on these facts, at the end of the 19th century W. Ramsay isolated He, Ne, Ar, Kr, and Xe and by studying their properties demonstrated that they were new elements. In spite of the nearly 1% content of argon Ar in air, the element had not been isolated until then and noble gases were completely lacking in Mendeleev's periodic table. The Nobel prize was awarded to Ramsay in 1904 for his achievement.
Noble gases are located next to the halogen group in the periodic table. Since noble gas elements have closed-shell electronic configurations, they lack reactivity and their compounds were unknown. Consequently, they were also called inert gases. However, after the discovery of noble gas compounds, it was considered more suitable to call these elements "noble gases", as is mentioned in the following chapter.
Although the abundance of helium in the universe is next to that of hydrogen, it is very rare on the Earth because it is lighter than air. Helium originated from solar nuclear reactions and was locked up in the earth’s crust. It is extracted as a by-product of natural gas from specific areas (especially in North America). Since helium has the lowest boiling point (4.2 K) of all the substances, it is important for low-temperature science and superconductivity engineering. Moreover, its lightness is utilized in airships etc. Since argon is separated in large quantities when nitrogen and oxygen are produced from liquid air, it is widely used in metallurgy, and in industries and laboratories that require an oxygen-free environment.
(b) noble gas compounds
Xenon, Xe, reacts with elements with the largest electronegativities, such as fluorine, oxygen, and chlorine and with the compounds containing these elements, like platinum fluoride, PtF6. Although the first xenon compound was reported (1962) as XePtF6, the discoverer, N. Bartlett, later corrected that it was not a pure compound but a mixture of Xe[PtF6]x (x = 1-2). If this is mixed with fluorine gas and excited with heat or light, fluorides XeF2, XeF4, and XeF6 and are generated. XeF2 has chain-like, XeF4 square, XeF6 distorted octahedral structures. Although preparation of these compounds is comparatively simple, it is not easy to isolate pure compounds, especially XeF4.
Hydrolysis of the fluorides forms oxides. XeO3 is a very explosive compound. Although it is stable in aqueous solution, these solutions are very oxidizing. Tetroxide, XeO4, is the most volatile xenon compound. M[XeF8] (M is Rb and Cs) are very stable and do not decompose even when heated at 400 °C. Thus, xenon forms divalent to octavalent compounds. Fluorides can also be used as fluorinating reagents.
Although it is known that krypton and radon also form compounds, the compounds of krypton and radon are rarely studied as both their instability and their radioactivity make their handling problematic.
Discovery of noble gas Compounds
H. Bartlett studied the properties of platinum fluoride PtF6 in the 1960s, and synthesized O2PtF6. It was an epoch-making discovery in inorganic chemistry when analogous experiments on xenon, which has almost equal ionization energy (1170 kJ mol-1) to that of O2 (1180 kJ mol-1), resulted in the dramatic discovery of XePtF6.
noble gas compounds had not been prepared before this report, but various attempts were made immediately after the discovery of noble gases. W. Ramsay isolated noble gases and added a new group to the periodic table at the end of the 19th century. Already in 1894, F. F. H. Moisson, who is famous for the isolation of F2, reacted a 100 cm3 argon offered by Ramsay with fluorine gas under an electric discharge but failed to prepare an argon fluoride. At the beginning of this century, A. von Antoropoff reported the synthesis of a krypton compound KrCl2, but later he concluded that it was a mistake.
L. Pauling also foresaw the existence of KrF6, XeF6, and H4XeO6, and anticipated their synthesis. In 1932, a post doctoral research fellow, A. L. Kaye, in the laboratory of D. M. L. Yost of Caltech, where Pauling was a member of faculty, attempted to prepare noble gas compounds. Despite elaborate preparations and eager experiments, attempts to prepare xenon compounds by discharging electricity through a mixed gas of xenon, fluorine, or chlorine were unsuccessful. It is said that Pauling no longer showed interest in noble gas compounds after this failure.
Although R. Hoppe of Germany predicted using theoretical considerations that the existence of XeF2 and XeF4 was highly likely in advance of the discovery of Bartlett, he prepared these compounds only after knowing of Bartlett’s discovery. Once it is proved that a compound of a certain kind is stable, analogous compounds are prepared one after another. This has also been common in synthetic chemistry of the later period, showing the importance of the first discovery.
problems
4.1
Write a balanced equation for the preparation of diborane.
4.2
Write a balanced equation for the preparation of triethylphosphine.
4.3
Write a balanced equation for the preparation of osmium tetroxide.
4.4
Describe the basic reaction of the phosphomolybdate method used for the detection of phosphate ions.
4.5
Draw the structure of anhydrous palladium dichloride and describe its reaction when dissolved in hydrochloric acid.
4.6
Describe the reaction of anhydrous cobalt dichloride when it is dissolved in water.
4.7
Draw the structure of phosphorus pentafluoride. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/04%3A_Chemistry_of_Nonmetallic_Elements/4.07%3A_Noble_Gases_and_their_Compounds.txt |
Metals show metallic luster, are good conductors of electricity and heat, and are very malleable and ductile. Such properties are characteristic of bulk metals, although the definition of metal atoms or ions is not simple. Metallic elements form basic oxides or hydroxides in the +1 or +2 oxidation states, and become cations in aqueous acid solutions. All transition elements are metals, but main group elements are classified into metallic and nonmetallic elements. Germanium and polonium may also be included as metals. Boron, silicon, germanium, arsenic, antimony, selenium, and tellurium exhibit some metallic characteristics and they are sometimes called metalloids.
Thumbnail: Pure barium in protective argon gas atmosphere. (Public Domain; Matthias Zepper).
05: Chemistry of Main-Group Metals
Group 1 metals are called alkali metals. Alkali metals are abundant in minerals and sea water. Especially the content of sodium, Na, in the Earth's crust is fourth after Al, Fe, and Ca. Although the existence of sodium or potassium ions was recognized for many years, a number of attempts to isolate the metals from aqueous solutions of their salts failed because of their high reactivity with water. Potassium (1807) and subsequently sodium were isolated by the electrolysis of molten salt of KOH or NaOH by H. Davy in the 19th century. Lithium Li was discovered as a new element (1817), and Davy soon isolated it by molten salt electrolysis of Li2O. Rubidium, Rb and Cesium, Cs, were discovered as new elements by spectroscopy in 1861. Francium, Fr, was discovered using a radiochemical technique in 1939. Its natural abundance is very low.
Table $1$ Properties of group 1 metals
mp
(°C)
bp
(°C)
d(20 °C)
(g cm-3)
E0 (V)
M+ + e-
I
(kJ mol-1)
Li 181 1342 0.534 -3.04 520
Na 98 883 0.968 -2.71 496
K 63 759 0.856 -2.93 419
Rb 39 688 1.532 -2.98 403
Cs 28 671 1.90 -3.03 376
As shown in Table $1$, melting-points, boiling points, and densities of alkali metals are low, and they are soft metals. Since the outer shell contains only one s-electron, the ionization energy is very low, and mono cations of alkali metals form easily. Qualitative analysis of alkali metals is possible by means of flame reactions using characteristic luminescence lines. Especially the orange D-line of sodium is used in the sodium lamp. Alkali metals are oxidized by water evolving hydrogen gas due to their low reduction potentials. Except lithium, the heavier alkali metals react violently with water, and sufficient caution should be exercised in their handling.
Exercise $1$
Describe the reactivity of alkali metals in water.
Answer
The reactivity of lithium is the lowest, sodium reacts violently, and potassium, rubidium, and cesium react explosively.
Alkali metals are also highly reactive to oxygen or halogens. As alkali metals are very reducing, they are used widely as reducing agents. Because of the high affinity of alkali metals to halogens, they are important in organic and inorganic syntheses which produce alkali metal halides as the result of condensation and metathesis reactions. Although it is generally difficult to dissolve metals in solvents to make atomic dipersions, alkali metals can be dispersed in liquid ammonia solutions, amalgams, and as cryptand (Figure $1$), naphthalene, or benzophenone (C6H5)2CO complexes.
Ammonia boils at -33.35 °C but liquid ammonia can be easily handled. Alkali metals dissolve readily in liquid ammonia and dilute solutions are blue but concentrated ones show a bronze color. The metal is recovered when ammonia is evaporated from metal solutions. Alkali metal solutions show the same color irrespective of the kind of alkali metals as the color is due to the solvated electrons. Namely, the dissolution is accompanied by the separation of the alkali metal atoms into metal cations and electrons solvated by ammonia, according to the following equation.
$M + n\; NH_{3} \rightarrow M^{+} [e^{-} (NH_{3})]$
The liquid ammonia solution of an alkali metal is conductive and paramagnetic. The highly reducing solution is used for special reduction reactions or syntheses of alkali metal complexes and polyhalides. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/05%3A_Chemistry_of_Main-Group_Metals/5.01%3A_Group_1_Metals.txt |
Group 2 metals from beryllium Be, to radium, Ra, are also called alkaline earth metals (Table $2$). Beryllium is a component of beryl or emerald. Emerald is a mineral that contains about 2% of chromium, Cr, in beryl, Be3Al2Si6O18. Beryllium metal is silver white and is used in special alloys and for the window of X-ray tubes, or a moderator of nuclear reactors, etc. Compounds of Be2+ resemble the compounds of Mg2+ or Al3+. Since beryllium is a deadly poison, it should be handled with due care.
Table $2$ Properties of group 2 metals
mp
(°C)
bp
(°C)
d(20 °C)
(g cm-3)
E0 (V)
M2++2e-
I
first
(kJ mol-1)
second
Be 1287 2471 1.85 -1.85 899 1757
Mg 650 1090 1.74 -2.37 737 1450
Ca 842 1484 1.55 -2.87 590 1145
Sr 777 1382 2.63 -2.90 549 1064
Ba 727 1897 3.62 -2.91 503 965
Ra 700 5.5 -2.82 509 975
Magnesium, Mg, is mainly produced as carbonates, sulfates, and silicates, and its abundance is between those of sodium and calcium, Ca. It is produced by molten salt electrolysis of magnesium chloride, MgCl2, or the reaction of dolomite, CaMg(CO3)2, with ferrosilicon alloy FeSi. Magnesium metal is silver white and the surface is oxidized in air. At high temperatures, magnesium reacts with nitrogen gas to become nitride, Mg3N2. The metal burns very brightly and is still used for flash lights. The alloy with aluminum is light and strong and used as a structural material in cars and airplanes.
Mg2+ is the central metal ion in the porphyrin ring of chlorophyll, and plays an important role in photosynthesis. The Grignard reagent, RMgX, which F. A. V. Grignard of France synthesized in 1900, is a typical organometallic compound of a main-group metal and is widely used for Grignard reactions. This is an important reagent rewarded by a Nobel prize (1912), and is very useful not only for organic reactions but also for the conversion of metal halides into organometallic compounds.
Calcium is contained in silicates, carbonates, sulfates, phosphates, fluorite, etc. Calcium is a silver white and soft metal that is manufactured by molten salt electrolysis of calcium chloride CaCl2.
Quick lime, CaO, is produced by the calcination of limestone, CaCO3, at 950-1100 °C. Production of quick lime ranks second to sulfuric acid in inorganic chemical industries. Calcium hydroxide, Ca(OH)2, is called also slaked lime. Calcium carbonate is the principal component of limestone and limestone is very important for the production of cement. Gypsum is a dihydrate of calcium sulfate CaSO4 • 2H2O and is obtained in large quantities as a by-product of stack gas desulfurization, and in addition to conventional uses is also used as a building material, etc.
Although calcium is not important in either the chemistry of aqueous solution systems or in organometallic chemistry in organic solvents, the element plays very important roles in living organisms. Not only is calcium the structure material of bones and teeth, calcium ions also have a wide range of functions in biological systems, such as protein stabilization,transfer of hormone action, muscular contraction, nerve communication, and blood coagulation.
Strontium, Sr, is a silver white soft metal. The surface is oxidized by air at room temperature, and it becomes a mixture of oxide, SrO, and nitride, Sr3N2, at high temperatures. In spite of the relatively high content strontium in the Earth's crust, the lement has not been studied widely and its application is limited. There are four natural isotopes and 88Sr (82.58%) is the most abundant. Since the artificial isotope 90Sr is obtained cheaply by nuclear reaction, it is used as a source of $\beta$ particles, and as a radioactive tracer. However, this isotope, as well as 137Cs, has a long half-life (28.8 y) and both are present in the radioactive fallout that accompanies nuclear explosive tests. Both are considered to be very dangerous.
The chemistry of barium, Ba, is unexceptional but BaSO4 is used as a contrast medium for X-ray diagnosis of the stomach because it is insoluble in hydrochloric acid. The Ba2+ ion is highly toxic and water-soluble compounds containing the ion should be handled cautiously.
Although radium, Ra, exists in uranium ores, the content is as low as 10-6 times that of uranium, U. Mr. and Mrs. Curie isolated a trace quantity of uranium chloride from tons of pitchblende in 1898. Elemental uranium was also isolated by Mrs. Curie via an amalgam. Although radium has historical importance in radiochemistry, it is no longer used as a radiation source.
Exercise $2$
Show examples of main group organometallic compounds which are often used in synthetic chemistry.
Answer
• Butyl lithium, LiBu,
• Grignard reagent, RMgBr,
• Triethylaluminum, AlEt3,
• And diethyl zinc ZnEt2.
5.03: Group 12 Metals
Sulfide ores of zinc, Zn, cadmium, and mercury, Hg, of Group 12 metals (Table \(3\)) serve as raw materials in metallurgy. These metals are located immediately after the transition metals in the periodic table but they do not behave like transition metals because their d orbitals are filled, and zinc and cadmium exhibit properties intermediate between hard and soft reactivities of magnesium. Mercury is soft and a liquid and it tends to bond to phosphorus or sulfur ligands. Mercury forms monovalent and divalent compounds but monovalent mercury is actually Hg2+. This is a cationic species which has a Hg-Hg bond, and mercury further catenates to give, for example, Hg4(AsF6)2.
Table \(3\) Properties of group 12 metals
mp
(°C)
bp
(°C)
d(25 °C)
(g cm-3)
E0 (V)
M2++2e-
I
first
(kJ mol -1)
second
third
Zn 420 907 7.14 -0.76 906 1733 3831
Cd 321 767 8.65 -0.40 877 1631 3644
Hg -38.8 357 13.5 0.85 1007 1809 3300
Cadmium and mercury are poisonous, especially organic cadmium and organic ercury compounds are deadly poisons and should be handled carefully. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/05%3A_Chemistry_of_Main-Group_Metals/5.02%3A_Group_2_Metals.txt |
Aluminum, Al, among Group 13 metals (Table $4$) exists as aluminosilicates in the Earth’s crust and is more abundant than iron. The most important mineral for metallurgy is bauxite, AlOx(OH)3-2x (0 < x < 1). Although the Al metal was as valuable as noble metals in the 19th century, the price fell dramatically after it came to be manufactured in large quantities by electrolysis of alumina, Al2O3, melted in cryolite, Na3AlF6. However, because its production requires consumption of a large amount of electrical power, the metallurgy of aluminum is economically feasible only in countries where the price of electrical power is low. Therefore, Japan has withdrawn from aluminum smelting, but Japan’s consumption of the metal is second only to the United States. The properties of aluminum are well known as it is widely used and encountered in every day life, for example in one-yen coins, aluminum foil, cooking pans, aluminum window sashes, etc. Aluminum metal usually exceeds 99% purity, and the metal itself and its alloys, like duralumin, are widely used.
Table $4$ Properties of group 13 metals
mp
(°C)
bp
(°C)
d(20 °C)
(g cm-3)
E0 (V)
M3++3e-
I
first
(kJ mol-1)
second
third
Al 660 2519 2.70 -1.66 577 1816 2744
Ga 29.8 2204 5.90 -0.55 579 1979 2962
In 157 2072 7.31 -0.34 558 1820 2704
Tl 304 1473 11.9 +0.74 589 1971 2877
Aluminum metal dissolves in mineral acids, except concentrated nitric acid, and in aqueous solutions of alkali metal hydroxides evolving hydrogen. Aluminum forms compounds with most nonmetallic elements and shows a rich chemistry, but unlike boron, no cluster hydrides are known. As oxide and halides have already been described (4.3 (c), 4.5 (d)), organo-aluminum compounds will be mentioned here.
Organoaluminum compounds
Organoaluminum compounds are used in large quantities for olefin polymerization, and they are industrially manufactured from aluminum metal, hydrogen, and an olefin as follows.
$\ce{2 Al + 3 H2 + 6 CH2 = CHR \rightarrow Al_{2} (CH2CH2R)6}$
They are dimers except those with bulky hydrocarbyl groups. For example, trimethylaluminum, Al2(CH3)6, is a dimer in which methyl groups bridge aluminum atoms by electron deficient bonds (Figure $2$).
Organnoaluminum compounds are very reactive and burn spontaneously in air. They react violently with water and form saturated hydrocarbons, with aluminium changing to aluminium hydroxide as follows.
$\ce{Al(CH2CH3)3 + 3 H2O \rightarrow 3 C2H6 9 Al(OH)3}$
Therefore, they should be handled in the laboratory under a perfectly inert atmosphere. The Ziegler-Natta catalyst, comprising an organoaluminium compound and a transition metal compound was an epoch-making olefin polymerization catalyst developed in the 1950s, for which the Nobel prize (1963) was awarded.
A transition-metal alkyl compound is formed when an organoaluminum compound reacts with a transition-metal compound. The transition-metal alkyl compound so formed can be isolated when stabilization ligands are coordinated to the metal center.
Gallium, Ga, has the largest temperature difference of the melting point and boiling point among all the metals. Since it melts slightly above a room temperature, the temperature range of the liquid state is very wide and it is used as a high temperature thermometer. In recent years, the metal is used for the manufacture of the compound semiconductors gallium arsenide, GaAs, and gallium phosphide, GaP.
Indium, In, is a soft metal also with a low melting-point. It is the raw material for compound semiconductors InP, InAs, etc. Indium has two stable states, In (I) or In (III), and In (II) compounds are considered to be mixed-valence compounds of monovalent and trivalent In.
Thallium, Tl, also has two stable states, Tl (I) and Tl (III), and Tl (II) is a mixed valence compound of monovalent and trivalent Tl. Since the element is very poisonous the metal and its compounds should be handled carefully. As it is a weak reductant compared to Na(C5H5), thallium cyclopentadienide, Tl(C5H5), is sometimes used for the preparation of cyclopentadienyl compounds, and is a useful reagent in organometallic chemistry.
Exercise $3$
Give the example of the metals for which stable ions differing in oxidation numbers exist.
Answer
In(I), In(III), Tl(I), Tl(III), Sn(II), Sn(IV).
The reaction of organoaluminum compounds
Organoaluminum compounds were synthesized for the first time in 1859, but they were not regarded as important as Grignard reagents or organolithium compounds as synthetic reagents for some time. This is in part due to the low reactivity of the ether adducts, R3Al:OEt2, which were present because of the frequent use of ether as a solvent. The studies of K. Ziegler changed this situation. K. Ziegler also discovered oligomerization of ethylene by organoaluminum compounds and the formation of higher organoaluminum compounds by the insertion of ethylene in aluminum-carbon bonds. Since alcohols were formed by hydrolysis of organoaluminum compounds, these discoveries were important for organic synthesis.
The discovery of the action of trace amount of nickel in the reaction vessel to give only butene from ethylene led to investigation of the effect of transition metals upon this reaction. Many transition metal salts were examined and Ziegler discovered that titanium compounds gave the highest degree of polymerization of ethylene. This was the birth of the so-called Ziegler catalysts. It should be remembered that this great discovery of the 1950s occurred when the petrochemical industry was beginning to develop and revolutionized the chemical industry of higher polymers.
5.05: Group 14 Metals
Of the ten isotopes of tin, Sn, 118Sn (24.22%) and 120Sn (33.59%) are the most abundant. Metallic tin is present as $\alpha$ tin (gray tin), which is stable below 13.2 °C and $\beta$ tin which is stable at higher temperatures. At low temperatures, the phase transition is quick. Divalent and tetravalent compounds are common, and divalent compounds are reducing agents.
208Pb (52.4%) is the most abundant among the four stable isotopes of lead, Pb. Lead is the end product of natural radioactive decay and has 82 protons. The atomic number 82 is important as it is especially stable. Thus, Pb exhibits high abundance for a heavy element. The divalent and tetravalent oxidation states are most common and usually lead is present as Pb2+ except in organometallic compounds. PbO2 is a tetravalent compound and readily becomes divalent, hence it is a very strong oxidizing agent. Although tetraethyl lead was previously used as an anti-knock agent in gasoline, only unleaded gasoline is now permitted for use in Japan.
It has been known since the 1930s that when Ge, Sn, or Pb are reduced by sodium in liquid ammonia, multi nuclear anions such as Ge94-, Sn52-, and Pb94-, are formed. These are called Zintl phases. These multi-atom anions we etc. recently using a cryptand, and cluster structures re crystallized as [Na(crypt)]4 [Sn9] have been elucidated.
problems
5.1
Write a balanced equation for the formation of butyllithium.
5.2
Potassium permanganate is insoluble in benzene but it dissolves in this solvent in the presence of a crown ether which is a cyclic polyether. Why is the solubility of potassium permanganate increased in the presence of a crown ether?
5.3
Why is trimethylaluminum called an electron deficient compound? | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/05%3A_Chemistry_of_Main-Group_Metals/5.04%3A_Group_13_Metals.txt |
Simple substances of transition metals have properties characteristic of metals, i.e. they are hard, good conductors of heat and electricity, and melt and evaporate at high temperatures. Although they are used widely as simple substances and alloys, we typically encounter only iron, nickel, copper, silver, gold, platinum, or titanium in everyday life. However, molecular complexes, organometallic compounds, and solid-state compounds such as oxides, sulfides, and halides of transition metals are used in the most active research areas in modern inorganic chemistry.
• 6.1: Structures of Metal Complexes
Transition elements are metallic elements that have incomplete d or f shells in the neutral or cationic states. They are called also transition metals and make up 56 of the 103 elements. Although Sc and Y belong to the d-block, their properties are similar to those of lanthanoids. The chemistry of d-block and f-block elements differs considerably. This chapter describes the properties and chemistry of mainly d-block transition metals.
• 6.2: Electronic Structure of Complexes (Part 1)
It is necessary to learn a few concepts to understand the structure, spectrum, magnetism, and reactivity of complexes which depend on d electron configurations. In particular, the theory of electronic structure is important.
• 6.3: Electronic Structure of Complexes (Part 2)
The characteristics of transition metal-ligand bonds become clear by an analysis of the molecular orbitals of a 3d metal coordinated by six identical ligands in octahedral complexes [ML6]. As the result of the interaction between the metal d and ligand orbitals, bonding, non-bonding and anti-bonding complex molecular orbitals are formed.
• 6.4: Organometallic Chemistry of d Block Metals (Part 1)
According to the definition of an organometallic compound, at least one direct bond between a metal and a carbon atom should exist, but CN complexes etc. with no organometallic character are usually excluded from organometallic compounds. Metal carbonyl compounds are organometallic in various aspects of their bonding, structure and reactions, and they are a good model system for understanding of the essence of transition metal organometallic chemistry.
• 6.5: Organometallic Chemistry of d Block Metals (Part 2)
• 6.6: Reactions of Complexes
06: Chemistry of Transition Metals
Transition elements are metallic elements that have incomplete d or f shells in the neutral or cationic states. They are called also transition metals and make up 56 of the 103 elements. These transition metals are classified into the d-block metals, which consist of 3d elements from Sc to Cu, 4d elements from Y to Ag, and 5d elements from Hf to Au, and f-block metals, which consist of lanthanoid elements from La to Lu and actinoid elements from Ac to Lr. Although Sc and Y belong to the d-block, their properties are similar to those of lanthanoids. The chemistry of d-block and f-block elements differs considerably. This chapter describes the properties and chemistry of mainly d-block transition metals.
Central Metals
Properties of d-block transition metals differ considerably between the first (3d) and the second series metals (4d), although the differences in properties between the second and the third series (5d) metals is not pronounced. Metallic radii of elements from scandium, Sc, to copper, Cu, (166 to 128 pm) are significantly smaller than those of yttrium, Y, to silver, Ag, (178 to 144 pm) or those of lanthanum, La, to gold, Au, (188 to 146 pm). Further, metal compounds of the first series transition metals are rarely 7 coordinate, whereas transition metals from the second and third series may be 7 to 9 coordinate. Cerium, Ce, (radius 182 pm) ~ lutetium, Lu, (radius 175 pm) fall between La and Hf and, because of the lanthanide contraction, metallic radii of the second and third series transition metals show little variation.
Higher oxidation states in the second and third series transition metals are considerably more stable than those in the first series transition metals. Examples include tungsten hexachloride, WCl6, osmium tetroxide, OsO4, and platinum hexafluoride, PtF6. Compounds of the first series transition metals in higher oxidation states are strong oxidants and thus are readily reduced. On the other hand, whereas M(II) and M(III) compounds are common among the first series transition metals, these oxidation states are generally uncommon in compounds of second and third series metals. For example, there are relatively few Mo(III) or W(III) compounds compared with many Cr(III) ones. Aqua ions (ions with water ligands) are very common among compounds of first series metals but few are known amongst the second and third metal compounds.
Metal carbonyl cluster compounds of first series transition metals with M-M bonds in low oxidation states exist but halide or sulfide cluster compounds are rare. In general, metal-metal bonds are formed much more easily in the 4d and 5d metals than in the 3d ones. Magnetic moments of the first series transition metal compounds can be explained in terms of spin-only values (cf. Chapter 6.2 (d)) but it is difficult to account for the magnetic moments of the second and third series compounds unless complex factors such as spin-orbital interactions are taken into account.
Thus, it is necessary to acknowledge and understand the significant differences in chemical properties that exist between metals of the first and later series metal compounds, even for elements in the same group.
Properties of the d-block transition metals are different not only in the upper and lower positions in the periodic table but also in the left and right groups. The Group 3 to 5 metals are now often referred to as early transition metals and they are generally oxophilic and halophilic. Smaller numbers of d electrons and the hardness of these elements explain their affinity toward hard oxygen and halogens. In the absence of bridging ligands, the formation of metal-metal bonds is difficult for these elements. Organometallic compounds of these metals are known strongly to activate C-H bonds in hydrocarbons. Late transition metals in the groups to the right of the periodic table are soft and have a high affinity toward sulfur or selenium.
The d-block transition metals have s, p, and d orbitals and those with n electrons in the d orbitals are termed ions with a dn configuration. For example, Ti3+ is a d1 ion, and Co3+ a d6 ion. The number of electrons occupying the orbitals split by the ligand field (cf. 6.2(a)) is denoted by a superscript on the orbital symbol. For example, an ion with 3 electrons in t2g and 2 electrons in eg is described as t2g3eg1.
Ligands
Compounds of metal ions coordinated by ligands are referred to as metal complexes. Most ligands are neutral or anionic substances but cationic ones, such as the tropylium cation, are also known. Neutral ligands, such as ammonia, NH3, or carbon monoxide, CO, are independently stable molecules in their free states, whereas anionic ligands, such as Cl- or C5H5-, are stabilized only when they are coordinated to central metals. Representative ligands are listed in Table \(1\) according to the ligating elements. Common ligands or those with complicated chemical formula are expressed in abbreviated forms.
Those ligands with a single ligating atom are called monodentate ligands, and those with more than one ligating atoms referred to as polydentate ligands, which are also called chelate ligands. The number of atoms bonded to a central metal is the coordination number.
Table \(1\) Representative ligands
Name Abbreviation Formula
hydrido H-
carbonyl CO
cyano CN-
methyl Me CH3-
cyclopentadienyl Cp C5H5-
carbonato CO32-
ammine NH3
pyridine py C5H5N
bipyridine bipy C10H8N2
triphenylphosphine PPh3 P(C6H5)3
aqua aq H2O
acetylacetonato acac CH3C(O)CH2C(O)CH3-
thiocyanato SCN-
chloro Cl-
ethylenediaminetetraacetato edta (OOCCH2)2NCH2CH2N(CH2COO)24-
Coordination number and structures
Molecular compounds which consist of d-block transition metals and ligands are referred to as complexes or coordination compounds. The coordination number is determined by the size of the central metal, the number of d electrons, or steric effects arising from the ligands. Complexes with coordination numbers between 2 and 9 are known. In particular 4 to 6 coordination are the most stable electronically and geometrically and complexes with these coordination numbers are the most numerous (Figure \(1\)). Complexes with the respective coordination numbers are described below.
Two co-ordinate complexes
Many electron-rich d10 ions, viz: Cu+, Ag+, and Au+, form linear complexes such as [Cl-Ag-Cl]- or [H3N-Au-NH3]-. A zero-valent complex [Pd(PCy3)2] with very bulky tricyclohexylphosphine ligands is also known. Generally, stable 2-coordinate complexes are known for the late transition metals.
Three co-ordinate complexes
Although [Fe{N(SiMe3)3}3] is one example, very few 3-coordinate complexes are known.
Four co-ordinate complexes
When four ligands coordinate to a metal, tetrahedral (Td) coordination is the least congested geometry, although a number of square planar (D4h) complexes are known. [CoBr4]2-, Ni(CO)4, [Cu(py)4]+, [AuCl4]- are all examples of tetrahedral complexes. There are a few known examples of square planar complexes with identical ligands, such as [Ni(CN)4]2-, or [PdCl4]2-. In the case of mixed ligand complexes, a number of square planar complexes of d8 ions, Rh+, Ir+, Pd2+, Pt2+, and Au3+, have been reported. Examples include [RhCl(PMe3)3], [IrCl(CO)(PMe3)2], [NiCl2(PEt3)2], and [PtCl2(NH3)2] (Et = C2H5).
Cis and trans geometrical isomers are possible for complexes with two different kinds of ligands, and were first noted when A. Werner synthesized 4-coordinate [PtCl2(NH3)2]. As tetrahedral complexes do not give geometrical isomers, Werner was able to conclude that his 4-coordinate complexes were square planar. Recently cis-[PtCl2(NH3)2] (Cisplatin) has been used for the treatment of tumors and it is noteworthy that only the cis isomer is active.
Exercise \(1\)
Write the formal name of cis-[PtCl2(NH3)2].
Answer
cis-diamminedichloroplatinum
Five co-ordinate complexes
Trigonal bipyramidal (D3h) Fe(CO)5 or square pyramid (C4v) VO(OH2)4 are examples of 5-coordinate complexes. Previously, 5-coordinate complexes were rare but the number of new complexes with this coordination is increasing. The energy difference between the two coordination modes is not large and structural transformation readily occurs. For example, the molecular structure and infrared spectrum of Fe(CO)5 are consistent with a trigonal bipyramid structure, but the 13C NMR spectrum shows only one signal at the possible lowest temperature, which indicates that the axial and equatorial carbonyl ligands are fluxional in the NMR time scale (10-1~10-9 s). Structural transformation takes place via a square pyramid structure and the mechanism is well known as Berry’s pseudorotation.
Six coordinate complexes
When six ligands coordinate to a central metal, octahedral (Oh) coordination is the most stable geometry and the majority of such complexes assume this structure. In particular, there are a number of Cr3+ and Co3+ complexes which are inert to ligand exchange reactions, represented by [Cr(NH3)6]3+ or [Co(NH3)6]3+. They have been particularly important in the history of the development of coordination chemistry. [Mo(CO)6], [RhCl6]3-, etc. are also octahedral complexes. In the case of mixed ligands, cis- and trans-[MA4B2] and mer- and fac-[MA3B3] geometrical isomers, and for chelate ligands, \(\Delta\)-[M(A-A)3] and \(\Lambda\)-[M(A-A)3] optical isomers (Figure \(3\)) are possible. The octahedral structure shows tetragonal (D4h), rhombic (D2h), or trigonal (D3h) distortions caused by electronic or steric effects. The tetragonal distortion of [Cu(NH3)6]2+ by an electronic factor is a typical example of the Jahn-Teller effect (refer to 6.2(a)).
Six ligating atoms can assume trigonal prism coordination. Although this coordination is seen in [Zr(CH3)6]2- or [Re{S2C2(CF3)2}3], few metal complexes with this coordination structure are known because octahedral coordination is sterically less strained. This notwithstanding, it has long been known that the bonding mode of sulfur atoms around a metal is trigonal prism in solid-state MoS2 and WS2.
Exercise \(2\)
Write the chemical formula of potassium diamminetetra(isothiocyanato)chromate(III).
Answer
K[Cr(NCS)4(NH3)2]
Higher co-ordinate complexes
Metal ions of the second and third transition metal series can sometimes bond with more than seven ligating atoms and examples are [Mo(CN)8]3- or [ReH9]2-. In these cases, smaller ligands are favorable to reduce steric congestion. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/06%3A_Chemistry_of_Transition_Metals/6.01%3A_Structures_of_Metal_Complexes.txt |
It is necessary to learn a few concepts to understand the structure, spectrum, magnetism, and reactivity of complexes which depend on d electron configurations. In particular, the theory of electronic structure is important.
Ligand field theory
Ligand field theory is one of the most useful theories to account for the electronic structure of complexes. It originated in the application of the crystal field theory of ionic crystals to metal complex systems.
Six co-ordinate octahedral complexes
The five d orbitals of transition metal cations are degenerate and have equal energy.
The spherical negative electric field around a metal cation results in the total energy level being lower than that of a free cation because the electrostatic interactions. The repulsive interaction between the electrons in the metal orbitals and the negative electric field destabilizes the system and compensates for the stabilization to some extent (Figure $4$).
Let us assume that instead of a uniform spherical negative field, the field is generated by six ligands coordinating octahedrally to a central metal. The negative field of the ligands is called the ligand field. Negative charge, in the case of anionic ligands, or a negative end (lone pair), in the case of neutral ligands, exert a repulsive force on the metal d orbitals which is anisotropic depending on the direction of the orbitals. The position of the metal cation is taken as the origin and Cartesian coordinates are constructed (Figure $5$). Then, dx2-y2 and dz2 orbitals are aligned along the directions of the axes and the dxy, dyz, and dxz orbitals are directed between the axes. If ligands are placed on the axes, the repulsive interaction is larger for the eg orbitals (dx2-y2, dz2) than for the t2g orbitals (dxy, dyz, dxz), and the eg orbitals are destabilized and the t2g orbitals are stabilized to an equal extent. In the following discussion, only the energy difference between the t2g and eg orbitals is essential and the average energy of these orbitals is taken as the zero of energy. If the energy difference between the two eg and three t2g orbitals is set to $\Delta_{o}$, the energy level of the eg orbitals is +3/5 $\Delta_{o}$ and that of the t2g orbitals is -2/5 $\Delta_{o}$ (Figure $6$). ($\Delta_{o}$ may also be expressed as 10 Dq. In this case, the energy level of the eg orbitals is +6 Dq and that of the t2g orbitals -4 Dq.)
Transition metal ions have 0 to 10 d electrons and when the split d orbitals are filled from a lower energy level, the electron configuration t2gxegy corresponding to each ion is obtained. With the zero energy level chosen as the average energy level, the energy of the electron configuration relative to zero energy becomes
$LFSE = (-0.4x + 0.6y) \Delta_{o}$
This value is called the ligand field stabilization energy. The electron configuration with smaller value (taking the minus sign into consideration) is more stable. LFSE is an important parameter to explain some properties of d-block transition metal complexes.
A condition other than the orbital energy level is required to explain the filling of electrons being populated into the split t2g and eg orbitals,. Two electrons can occupy an orbital with anti-parallel spins but a strong electrostatic repulsion occurs between two electrons in the same orbital. This repulsive interaction is called pairing energy, P.
When the number of d electrons is less than three, the pairing energy is minimized by loading the electrons in the t2g orbital with parallel spins. Namely, the electron configurations arising are t2g1, t2g2, or t2g3.
Two possibilities arise when the fourth electron occupies either of the t2g or eg orbitals. The lower energy orbital t2g is favorable but occupation of the same orbital gives rise to pairing energy, P. The total energy becomes
$-0.4 \Delta_{o} \times 4 + P = - 1.6 \Delta_{o} + P$
If the fourth electron occupies the energetically unfavorable eg orbital, the total energy becomes
$-0.4 \Delta_{o} \times 3 + 0.6 \Delta-{o} = - 0.6 \Delta_{o}$
The choice of the electron configuration depends on which of the above values is larger. Therefore if $\Delta_{o}$ > P, t2g4 is favoured and this is called the strong field case or the low spin electron configuration. If $\Delta_{o}$ < P, t2g3eg1 is favoured and this is called the weak field case or the high spin electron configuration. A similar choice is required for d5, d6, and d7 octahedral complexes, and in the strong field case, t2g5, t2g6, or t2g6eg1 configurations are favoured, whereas in the weak field case, t2g3eg2, t2g4eg2, or t2geg2 configurations are favoured. The ligand field splitting parameter $\Delta_{o}$ is decided by the nature of the ligands and metal, whereas the pairing energy, P, is almost constant and shows only a slight dependence on the identity of the metal.
Square planar complexes
Complexes with four ligands in a plane containing the central metal are termed square planar complexes. It is easier to understand the electronic energy levels of the d orbitals in square planar complexes by starting from those for hexacoordinate octahedral complexes. Placing the six ligands along the Cartesian axes, the two ligands on the z axis are gradually removed from the central metal and finally only four ligands are left on the x,y plane. The interaction of the two z coordinate ligands with the dz2, dxz, and dyz orbitals becomes smaller and the energy levels of these ligands lower. On the other hand, the remaining four ligands approach the metal and the dx2-y2 and dxy energy levels rise as a result of the removal of the two ligands. This results in the order of the energy levels of five d orbitals being dxz, dyz < dz2 < dxy << dx2-y2 (Figure $7$). Rh+, Ir+, Pd2+, Pt2+, and Au3+ complexes with a d8 configuration tend to form square planar structures because eight electrons occupy the lower orbitals leaving the highest dx2-y2 orbital empty.
Tetrahedral complexes
Tetrahedral complexes have four ligands on the apexes of a tetrahedron around the central metal. [CoX4]2- (X = Cl, Br, I), Ni(CO)4, etc. are all examples of 4-coordination complexes (Figure $5$). When a metal is placed on the origin of the Cartesian axes, as in the octahedral complexes, e orbitals (dx2-y2, dz2) are distant from ligands and t2 orbitals (dxy, dyz, dxz) are nearer ligands. Consequently, the electronic repulsion is larger for the t2 orbitals, which are destabilized relative to the e orbitals. The ligand field exerted by four ligands splits the fivefold degenerate orbitals of the central metal into twofold degenerate e and threefold degenerate t2 sets (Figure $6$). The t2 set has energy of +2/5 $\Delta_{t}$ and the e set -3/5 $\Delta_{t}$ with a ligand field splitting of $\Delta_{t}$. As the number of the ligands is 4/6 = 2/3 of that in hexacoordinate octahedral complexes, and overlap of the ligands with the orbitals is smaller, and the ligand splitting $\Delta_{t}$ is about a half of $\Delta_{o}$. Consequently, only high-spin electron configurations are known in tetrahedral complexes. The ligand field splitting energies calculated by the above method are shown in Table $2$.
Table $2$ Ligand field stabilization energy (LFSE)
Octahedral Tetrahedral
Strong field (LS) Weak field (HS)
dn Example n $\Delta_{o}$ n $\Delta_{o}$ n $\Delta_{t}$
d1 Ti3+ 1 0.4 1 0.4 1 0.6
d2 V3+ 2 0.8 2 0.8 2 1.2
d3 Cr3+,V2+ 3 1.2 3 1.2 3 0.8
d4 Cr2+, Mn3+ 2 1.6 4 0.6 4 0.4
d5 Mn2+, Fe3+ 1 2.0 5 0 5 0
d6 Fe2+, Co3+ 0 2.4 4 0.4 4 0.6
d7 Co2+ 1 1.8 3 0.8 3 1.2
d8 Ni2+ 2 1.2 2 1.2 2 0.8
d9 Cu2+ 1 0.6 1 0.6 1 0.4
d10 Cu1+ 0 0 0 0 0 0
Jahn-Teller Effect
When orbitals of a highly symmetrical nonlinear polyatomic molecule are degenerate, the degeneracy is resolved by distorting the molecular framework to attain lower symmetry and thus lower energy. This is the Jahn-Teller effect and a typical example is seen in the tetragonal distortion of an octahedral coordination structure of hexacoordinate Cu2+ complexes.
They have a d9 configurations and the eg orbitals in the octahedral structure are occupied by three electrons. If the eg orbitals split and two electrons occupy the lower orbital and one electron the upper orbital, the system gains energy of a half of the energy difference, $\delta$, of two split orbitals. Therefore a tetragonal distortion in the z axis becomes favorable. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/06%3A_Chemistry_of_Transition_Metals/6.02%3A_Electronic_Structure_of_Complexes_%28Part_1%29.txt |
Molecular orbital theory of transition metal complexes
The characteristics of transition metal-ligand bonds become clear by an analysis of the molecular orbitals of a 3d metal coordinated by six identical ligands in octahedral complexes [ML6]. As the result of the interaction between the metal d and ligand orbitals, bonding, non-bonding and anti-bonding complex molecular orbitals are formed.
Generally, the energy levels of the ligand orbitals are lower than those of the metal orbitals, bonding orbitals have more ligand character and non-bonding and anti-bonding orbitals have more metal character. The processes of formation of the $\sigma$ and $\pi$ molecular orbitals are described step by step below.
$\sigma$ bond
Firstly, consider the M-L $\sigma$ bond among interactions of the metal s, p, d and ligand orbitals by assuming the position of a metal at the origin of the Cartesian coordinate system and locating ligands on the coordinate axes. As the $\sigma$ bond is a nodeless bond along the bonding axes, the metal s orbital (a1g, non-degenerate), px, py, pz orbitals (t1u, triply-degenerate), and dx2-y2, dz2 orbitals (eg, doubly-degenerate) fit symmetry (+, - signs) and orbital shapes with the ligands’ $\sigma$ orbitals (Figure $9$).
When the ligand orbitals are $\sigma_{1}$ and $\sigma_{2}$ along the x-axis, $\sigma_{3}$ and $\sigma_{4}$ along the y-axis, and $\sigma_{5}$ and $\sigma_{6}$ along the z-axis in Figure $5$, six ligand atomic orbitals are grouped by making linear combinations according to the symmetry of the metal orbitals. Then the orbital to fit with the metal a1g orbital is a1g ($\sigma_{1} + \sigma_{2} + \sigma_{3} + \sigma_{4} + \sigma_{5} + \sigma_{6}$), the one to fit with the metal t1u orbitals is t1u ($\sigma_{1} - \sigma_{2}, \sigma_{3} - \sigma_{4}, \sigma_{5} - \sigma_{6}$) and the one to fit with the metal eg orbitals is eg ($\sigma_{1} + \sigma_{2} - \sigma_{3} - \sigma_{4}, \sigma_{5} + \sigma_{6} − \sigma_{1} - \sigma_{2} - \sigma_{3} - \sigma_{4}$). There is a bonding interaction between the metal eg orbitals and the ligand group orbitals and bonding and anti-bonding molecular orbitals are formed. The relation is shown in Figure $10$.
The levels of the molecular orbitals from the lowest energy are bonding (a1g < t1u < eg) < non-bonding (t2g) < anti-bonding (eg* < a1g* < t1u*). For example, in a complex like [Co(NH3)6]3+, 18 valence electrons, 6 from cobalt and 12 from ammonia, occupy 9 orbitals from the bottom up, and t2g is the HOMO and eg* the LUMO. The energy difference between the two levels corresponds to the ligand field splitting. Namely, the eg set (dx2-y2, dz2) and the ligands on the corner of the octahedron form the bonding σ orbitals but the t2g set (dxy, dyz, dxz) remain non-bonding because the orbitals are not directed to the ligand $\sigma$ orbitals.
$\pi$ bond
When the ligand atomic orbitals have $\pi$ symmetry (i.e. with nodes) through the bond axis, the eg orbitals (dx2-y2) are non-bonding and the t2g orbitals (dxy, dyz, dxz) have bonding interactions with them (Figure $11$). In halide ions, X-, or aqua ligands, H2O, the $\pi$ symmetrical p orbitals have lower energy than the metal t2g orbitals and a bonding molecular orbital, which is lower than the t2g orbital, and an anti-bonding molecular orbital, which is higher than the t2g orbitals, form. Consequently, the energy difference $\Delta_{o}$ between eg and the anti-bonding orbitals becomes smaller. On the other hand, for the ligands having anti-bonding $\pi$ orbitals within the molecule, such as carbon monoxide or ethylene, the $\pi^{*}$ orbitals match the shape and symmetry of the t2g orbitals and the molecular orbitals shown in Fig 6.12 (b) form. As a result, the energy level of the bonding orbitals decreases and $\Delta_{o}$ becomes larger.
Using these simple molecular orbital considerations, the effects of $\sigma$ and $\pi$ orbital interactions between the metal and ligands upon the molecular orbitals are qualitatively understandable.
(c) Spectra
Many transition metal complexes have characteristic colors. This means that there is absorption in the visible part of the spectrum resulting from an electron being excited by visible light from a level occupied by an electron in a molecular orbital of the complex to an empty level. If the energy difference between the orbitals capable of transition is set to $\Delta$Ε, the absorption frequency ν is given by $\Delta Ε = h \nu$. Electronic transitions by optical pumping are broadly classified into two groups. When both of the molecular orbitals between which a transition is possible have mainly metal d character, the transition is called a d-d transition or ligand-field transition, and absorption wavelength depends strongly on the ligand-field splitting. When one of the two orbitals has mainly metal character and the other has a large degree of ligand character, the transition is called a charge-transfer transition. Charge transfer transitions are classified into metal (M) to ligand (L) charge-transfers (MLCT) and ligand to metal charge-transfers (LMCT).
Since the analysis of the spectra of octahedral complexes is comparatively easy, they have been studied in detail for many years. When a complex has only one d electron, the analysis is simple. For example, Ti in [Ti(OH2)6]3+ is a d1 ion, and an electron occupies the t2g orbital produced by the octahedral ligand field splitting. The complex is purple as the result of having an absorption at 492 nm (20300 cm-1) (Figure $13$) corresponding to the optical pumping of a d electron to the eg orbital. However, in a complex with more than one d electrons, there are repellent interactions between the electrons, and the d-d transition spectrum has more than one absorptions. For example, a d3 complex [Cr(NH3)6]3+ shows two d-d absorptions in the 400 nm (25000 cm-1) region, suggesting that the complex has two groups of molecular orbitals between which an electronic transition is possible with a high degree of transition probability. This means that, when three electrons in the t2g orbital are excited to the eg orbital, there are two energy differences due to repellent interactions between the electrons.
Tanabe-Sugano diagrams are constructed from calculations based on ligand field theory and have been widely used in the analysis of absorption spectra of d1 to d9 ions. The analysis becomes increasingly difficult for ions with many electrons. In any case, the existence of a d-d spectrum requires that the energy difference of an occupied orbital and an empty orbital is equivalent to the energy of the UV-visible spectrum, the transition is allowed by the selection rule, and the transition probability is high enough. Generally, a charge-transfer absorption is stronger than a ligand field absorption. An LMCT emerges when ligands have a non-bonding electron pair of comparatively high energy or the metal has empty low energy orbitals. On the other hand, an MLCT tends to appear when the ligands have low energy $\pi^{*}$ orbitals, and bipyridine complexes are good examples of this. Since the lifetime of the excited state of a ruthenium complex [Ru(bipy)3]2+ is extraordinarily long, many studies have been performed on its photoredox reactions.
Spectrochemical series
The magnitude of the ligand field splitting parameter $\Delta_{o}$ is determined mainly by the identity of the ligands. An empirical rule called the spectrochemical series was proposed by a Japanese scientist Ryutaro Tsuchida. The rule was constructed from empirical data collected when spectra of complexes that have the same central metal, oxidation state, coordination number, etc. were measured. It is noteworthy that ligands with $\pi$ acceptor properties are in a higher position in the series.
I- < Br- < S2- < SCN- < Cl- < NO3- < F- < OH- < H2O < NH3 < NO2 < PPh3 < CN- < CO
Although $\Delta_{o}$ does become larger in this order, it is also dependent on the identity of the central metal and its oxidation state. Namely, $\Delta_{o}$ is larger for 4d and 5d metals than for 3d metals and becomes larger as the oxidation number increases. The magnitude of $\Delta_{o}$ is closely related to its absorption position in the electromagnetic spectrum, and is a key factor in determining the position of a ligand in the spectrochemical series. A $\pi$ donor ligand (halogen, aqua, etc.) makes the absorption wavelength longer, and a $\pi$ acceptor ligand (carbonyl, olefin, etc.) shorter by contribution from the $\pi$ bond.
(d) Magnetism
Magnetization, M, (magnetic dipole moment per unit volume) of a sample in a magnetic field, H, is proportional to magnitude of H, and the proportionality constant, $\chi$, depends on the sample.
$M = \chi H$
$\chi$ is the volume susceptibility and the product of $\chi$ and the molar volume Vm of a sample is the molar susceptibility $\chi_{m}$. Namely,
$\chi_{m} = \chi V_{m}$
All substances have diamagnetism, and in addition to this, substances with unpaired electrons exhibit paramagnetism, the magnitude of which is about 100 times larger than that of diamagnetism. Curie's law shows that paramagnetism is inversely proportional to temperature.
$\chi_{m} = A + \frac{C}{T}$
where T is the absolute temperature and A and C are constants. In the Gouy or Faraday methods, magnetic moments are calculated from the change of weight of a sample suspended between magnets when a magnetic field is applied. In addition to these methods, the highly sensitive SQUID (superconducting quantum interference device) has been used recently to carry out such measurements.
Paramagnetism is induced by the permanent magnetic moment of an unpaired electron in a molecule and the molar susceptibility is proportional to the electron spin angular momentum. Paramagnetic complexes of d-block transition metals have unpaired electrons of spin quantum number 1/2, and a half of the number of unpaired electrons is the total spin quantum number S. Therefore, the magnetic moment based only on spins can be derived theoretically.
$\mu = 2 \sqrt{2S(S+1)} \mu_{B} = \sqrt{n(n+2)} \mu_{B}$
Here $\mu_{Β}$ = 9.274 x 10-24 JT-1 is a unit called the Bohr magneton.
Many 3d metal complexes show good agreement between the magnetic moments of paramagnetic complexes measured by a magnetic balance with the values calculated by the above formula. The relationship between the number of unpaired electrons and magnetic susceptibility of a complex is shown in Table $3$. Because of this agreement with theory, it is possible to determine the number of unpaired electrons from experimental values of magnetic measurements. For example, it can be assumed that a Fe3+ d5 complex with a magnetic moment of about 1.7 $\mu_{Β}$ is a low-spin complex with an unpaired spin but a Fe3+ d5 complex with a moment of about 5.9 $\mu_{Β}$ is a high-spin complex with 5 unpaired electrons.
Table $3$ Unpaired electrons and magnetic moments
Metal ion Unpaired electron Spin-only magnetic moment ($\frac{\mu}{\mu_{B}}$)
n Calculated Measured
Ti3+ 1 1.73 1.7~1.8
V3+ 2 2.83 2.7~2.9
Cr3+ 3 3.87 3.8
Mn3+ 4 4.90 4.8~4.9
Fe3+ 5 5.92 5.9
However, the measured magnetic moment no longer agrees with the calculated spin-only value when the orbital angular momentum contribution to the magnetic moment becomes large.. Especially in 5d metal complexes, this discrepancy between the measured and calculated values increases.
Exercise $3$
Calculate the spin-only magnetic moments of high spin and low spin Fe3+ complexes.
Answer
Since they are d6 complexes, a high spin complex has four unpaired electrons with the magnetic moment is 4.90$\mu_{B}$ and a low spin complex has no unpaired electron and is diamagnetic.
Some paramagnetic solid materials become ferromagnetic at low temperatures by forming magnetic domains in which thousands of electron spins are aligned parallel to each other. The temperature at which the paramagnetic-ferromagnetic phase transition occurs is called the Curie temperature. When spins are aligned antiparallel to each other, the material changes to an antiferromagnetic substance, and this transition temperature is called the Néel temperature. The material becomes ferrimagnetic when the spins are incompletely canceled. Recently, attempts have been made to synthesize polynuclear multi-spin complexes with special ligands that make paramagnetic metal ions align to induce ferromagnetic interactions between the spins. This effect is impossible in mononuclear complexes. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/06%3A_Chemistry_of_Transition_Metals/6.03%3A_Electronic_Structure_of_Complexes_%28Part_2%29.txt |
The organometallic chemistry of transition metals is comparatively new. Although an ethylene complex of platinum called Zeise's salt, K[PtCl3(C2H4)], tetracarbonylnickel, Ni(CO)4, and pentacarbonyliron, Fe(CO)5, which today are classified as organometallic compounds, were prepared in the 19th century, their bonding and structures were unknown. The research of W. Hieber and others on metal carbonyl compounds was important in the 1930s, but the results of these studies were limited because of the underdeveloped techniques of structural analyses available at the time.
The discovery of ferrocene, Fe(C5H5)2, in 1951 was epoch-making for the chemistry of this field. The very unique bonding mode of this complex became clear by means of single crystal X-ray structural analysis, NMR spectra, infrared spectra, etc., and served as a starting point for subsequent developments in the field. It was a major discovery that ferrocene exhibited very high thermal stability in spite of the general view that the transition metal-carbon bonds were very unstable. It was also clearly demonstrated that the compound had a sandwich structure in which the five carbon atoms of the cyclopentadienyl groups bonded simultaneously to the central metal iron. While the various coordination modes of hydrocarbon ligands were determined one after another, the industrial importance of organometallic compounds of transition metals increased with the discoveries of olefin polymerization catalysts (Ziegler catalyst), homogeneous hydrogenation catalysts (Wilkinson catalyst), and development of catalysts for asymmetric synthesis, etc. The Nobel prize awarded to K. Ziegler, G. Natta (1963), E. O. Fischer, and G. Wilkinson (1973) was in recognition of this importance.
According to the definition of an organometallic compound, at least one direct bond between a metal and a carbon atom should exist, but CN complexes etc. with no organometallic character are usually excluded from organometallic compounds. Metal carbonyl compounds are organometallic in various aspects of their bonding, structure and reactions, and they are a good model system for understanding of the essence of transition metal organometallic chemistry.
(a) Metal carbonyl compounds
Binary metal carbonyl compounds that consist only of a metal and CO ligands are usually prepared by the direct reaction of the powder of a highly reactive metal and carbon monoxide, or by the reduction of a metal salt to zero valance followed by reaction with high-pressure carbon monoxide. However, tetracarbonylnickel, first discovered at the end of the 19th century, forms by the reaction of nickel metal and carbon monoxide under atmospheric pressure and at room temperature. The preparation of other metal carbonyl compounds, on the other hand, requires high temperatures and high pressures.
Mononuclear metal carbonyl compounds take highly symmetric polyhedral coordination structures. Hexa-coordinate chromium, molybdenum, and tungsten hexacarbonyl, M(CO)6, assume a regular octahedral, penta-coordinate pentacarbonyliron, Fe(CO)5, a triangular bipyramid, and tetracarbonylnickel, Ni(CO)4, a regular tetrahedron coordination structure (Figure $14$). The carbon atoms of carbonyl ligands coordinate to the metal, and the CO moieties are oriented along the direction of the metal-carbon axis. Binuclear metal carbonyl Mn2(CO)10 has an Mn-Mn bond joining two square pyramidal Mn(CO)5 parts. In Fe2(CO)9, two Fe(CO)3 sub-units are bridged by three CO ligands, and in Co2(CO)8, two Co(CO)3 sub-units are connected by both three CO bridges and a Co-Co bond.
There are a number of cluster metal carbonyl compounds with metal-metal bonds joining three or more metals, and terminal CO, $\mu$-CO (a bridge between two metals), and $\mu_{3}$-CO (a bridge capping three metals) are coordinated to the metal frames (refer to Section 6.3 (f)). Many cluster carbonyls are formed by a pyrolysis reaction of mononuclear or binuclear carbonyl compounds. Typical metal carbonyl compounds and their properties are shown in Table $4$.
5 6 7 8 9 10
4 V(CO)6
Black
solid
d.70
Cr(CO)6
White
solid
d.130
Mn2(CO)10
Yellow
solid
mp 154
Fe(CO)5
Yellow
liquid
bp 103
Co2(CO)8
Red
solid
mp 51
Ni(CO)4
Colorless
liquid
bp 42.1
5 Mo(CO)6
White
solid
sublime
Tc2(CO)10
White
solid
mp 160
Ru3(CO)12
Orange
solid
d.150
Rh6(CO)16
Black
solid
d.220
6 W(CO)6
White
solid
sublime
Re2(CO)10
White
solid
mp 177
Os3(CO)12
Orange
solid
mp 224
Ir4(CO)12
Yellow
solid
d.220
Back donation
A metal carbonyl compound consists of carbon monoxide coordinated to a zero valent metal. For a long time, it had been unclear why such bonding was possible, let alone stable at all. The belief that normal coordination bonds were formed by the donation of electrons from highly basic ligands to a metal formed the basis of the coordination theory of A. Werner. Because the basicity of carbon monoxide is very low, and transition metal-carbon bonds are generally not very stable, a suitable explanation for the stability of metal carbonyl compounds was sought. If the shape and symmetry of the metal d orbital and of the CO $\pi$ (antibonding) orbital for the carbon-oxygen bond are suitable for overlap, a bonding interaction between the metal and carbon is expected. The bonding scheme shown in Figure $15$ was proposed from this point of view. The mechanism by which electrons are donated to the vacant carbon monoxide $\pi^{*}$ orbital from the filled metal d orbital is called back donation. Since accumulation of superfluous electrons on a low oxidation state metal atom is prevented, back-donation leads to the stabilization of the M-C bond.
Fig 6.15 - Back donation in metal carbonyls.
A rise in the order of the metal - carbon bond is reflected in the increase of the M-C, and decrease of the C-O, stretching frequencies in vibrational spectra. Infrared spectra are useful because carbonyl frequencies are easily detectable. The lowering of the oxidation state of a metal by the flow of negative charge from its coordinated ligands is reflected in the reduction of the C-O stretching frequencies.
(b) Hydrocarbon complexes
An organometallic compound is one which has metal-carbon bonds, and between one and eight carbon atoms in a hydrocarbon ligand bond to a metal. Hapticity describes the number of atoms in a ligand that have direct coordinative interaction with the metal and the number is added to $\eta$. An example is $\eta^{5}$ (pentahapto)-cyclopentadienyl (Table $5$).
A ligand that donates an odd number of electrons to a metal is formally a radical and it is stabilized by bonding to the metal. A ligand that donates an even number of electrons to a metal is generally a neutral molecule and it is stable even if it is not bonded to the metal. Carbene or carbyne ligands are exceptions to this rule. The chemical formula of an organometallic compound is expressed in many cases without using the square brackets [ ] usual for such a complex, and we shall follow this convention in this book.
Table $5$ Hapticity and number of donating electrons of hydrocarbon ligands
Name Hapticity Number of electrons Example
Alkyl $\eta^{1}$ 1 W(CH3)6
Alkylidene $\eta^{1}$ 2 Cr(CO)5{C(OCH3)C6H5}
Alkene $\eta^{2}$ 2 K[PtCl3(C2H4)]
$\pi$-allyl $\eta^{3}$ 3 Ni($\eta^{3}$-C3H5)2
Diene $\eta^{4}$ 4 Fe(CO)3($\eta^{4}$-C4H6)
Cyclopentadienyl $\eta^{5}$ 5 Fe($\eta^{5}$-C5H5)2
Arene $\eta^{6}$ 6 Cr($\eta^{6}$-C6H6)2
Tropylium $\eta^{7}$ 7 V(CO)3($\eta^{8}$ -C7H7)
Cyclooctatetraene $\eta^{8}$ 8 U($\eta^{8}$ -C8H8)2
Exercise $4$
Describe the difference between cyclopentadiene and cyclopentadienyl ligands.
Answer
The chemical formula of cyclopentadiene is C5H6 and it is bonded to a metal as a $\eta^{2}$ or $\eta^{4}$ ligand. The chemical formula of cyclopentadienyl is C5H5 and it is bonded to a metal as a $\eta^{1}$, $\eta^{3}$, or $\eta^{5}$ ligand.
Alkyl ligands
Alkyl or aryl transition metal compounds have M-C single bonds. In spite of many attempts over most of the course of chemical history, their isolation was unsuccessful and it was long considered that all M-C bonds were essentially unstable. Stable alkyl complexes began to be prepared gradually only from the 1950s. Cp2ZrCl(Pr),WMe6, CpFeMe(CO)2, CoMe(py)(dmg)2, (dmg = dimethylglyoximato), IrCl(X)(Et)(CO)(PPh3)2, NiEt2(bipy), PtCl(Et)(PEt3)2 are some representative compounds. Among various synthetic processes so far developed, the reactions of compounds containing M-halogen bonds with main-group metal-alkyl compounds, such as a Grignard reagent or an organolithium compound, are common synthetic routes. Especially vitamin B12, of which D. Hodgkin (1964 Nobel Prize) determined the structure, is known to have a very stable Co-C bond. Metal alkyl compounds which have only alkyl ligand, such as WMe6, are called homoleptic alkyls.
It is gradually accepted that a major cause of the instability of alkyl complexes is the low activation energy of their decomposition rather than a low M-C bond energy. The most general decomposition path is $\beta$ elimination. Namely, the bonding interaction of a hydrocarbon ligand with the central transition metal tends to result in the formation of a metal hydride and an olefin. Such an interaction is called an agostic interaction. Although an alkyl and an aryl ligand are 1-electron ligands, they are regarded as anions when the oxidation number of the metal is counted. The hydride ligand, H, resembles the alkyl ligand in this aspect.
$\pi$ allyl complexes
If an allyl group, CH2=CH-CH2-, is bonded to a metal via a carbon atom, it is a 1-electron ligand like an alkyl group. If the double bond delocalizes, three carbon atoms bond to the metal simultaneously as a 3-electron ligand. This is also an odd electron and formally anionic ligand and is stabilized by being coordinated to the metal.
Pd(C3H5)(Ac)(PPh3), Co(C3H5)3, etc are well-known examples. Since $\eta^{1}$, $\eta^{2}$, and $\eta^{3}$ coordination modes are possible in the catalytic reactions of unsaturated hydrocarbons, various reactions occur.
$\pi$ cyclopentadienl complexes
The cyclopentadie nyl ligand, C5H5, is abbreviated as Cp. C5Me5, in which the hydrogen atoms of Cp are replaced with methyl groups, is a useful ligand called Cp star and is denoted by Cp*. Ferrocene, Cp2Fe, is a very stable orange-colored iron compound in which two cyclopentadienyl groups are bonded to iron. It was discovered independently in two laboratories, but the discoverers proposed incorrect structures. The correct structure was clarified by the group of G. Wilkinson, who won a Nobel Prize (1973). The preparation of ferrocene is usually carried out according to the following reaction path:
$\ce{2 C_{5} H_{6} + 2 Na \rightarrow 2 Na(C_{5}H_{5}) + H_{2}}$
$\ce{FeCl_{2} + 2 Na(C_{5}H_{5}) \rightarrow Fe(C_{5}H_{5})_{2} + 2 NaCl}$
Single crystal X-ray structure analysis showed that the structure of ferrocene is an iron atom sandwiched between two C5H5 rings (Figure $16$). Five carbon atoms bond to the iron simultaneously in ferrocene, and unsaturated C-C bonds are delocalized in the five-membered rings. Since this kind of bond was not known before, it aroused interest, many derivative compounds were prepared, and a wide range of chemistry has since been studied (Table $6$).
4 5 6 7 8 9 10
4 Cp2TiCl2
Red
mp 230
Cp2V
Black
mp 167
Cp2Cr
Scarlet
mp 173
Cp2Mn
Brown
mp 193
Cp2Fe
Orange
mp 174
Cp2Co
Black
mp 173
Cp2Ni
Green
d.173
5 Cp2ZrCl2
White
mp 248
Cp2NbCl2
Brown
Cp2MoCl2
Green
d.270
Cp2TcH
Yellow
mp 150
Cp2Ru
Yellow
mp 200
6 Cp2HfCl2
White
mp 234
Cp2TaCl2
Brown
Cp2WCl2
Green
d.250
Cp2ReH
Yellow
mp 161
Cp2Os
White
mp 229
The cyclopentadienyl ligand is a 5-electron and formally anionic ligand. If only one of the five carbon atoms is bonded to a metal, it is a 1-electron ligand like an alkyl group. It becomes a 3-electron ligand in rare cases and coordinates to a metal as a $\pi$-allyl system that extends over 3 carbon atoms. The Cp group of ferrocene has reactivity analogous to that of aromatic compounds. Since the Cp group has played a significant role as a stabilizing ligand to realize the preparation of new compounds with new metal-ligand bonding modes, it can reasonably be claimed that this ligand has made the greatest contribution to organometallic chemistry of any other ligand. Although two Cp rings are bonded to the metal in parallel in ferrocene, Cp2TiCl2 and Cp2MoH2 have bent Cp ligands and they are called bent-sandwich compounds.
Olefin complexes
Zeise's salt, K[PtCl3(C2H4)], is the oldest known organometallic compound and was synthesized and analyzed in ca. 1825 by Zeise, although its coordination structure was assumed only in 1954 and confirmed by the neutron diffraction in 1975. The mode of coordination of an olefin to a transition metal is described by the Dewar-Chatt-Duncanson model and the bond between the metal and olefin is stabilized by the contribution of d$\pi$-p$\pi^{*}$ back donation. An olefin is a 2-electron ligand and there are many olefin complexes in which the central metal is in a relatively low oxidation state. Dienes or trienes with two or more double bonds coordinate to a metal as 4-electron or 6-electron ligands. Fe(CO)3(C4H6) and Ni(cod)2, in which a butadiene or cyclooctadienes (cod) are coordinated to the metal, are well known examples. Since cyclooctadienes are easily eliminated from Ni(cod)2, it is conveniently used for generating atomic, zero valent nickel. This complex is sometimes called naked nickel.
Arene complexes
Aromatic compounds are 6-electron donors that coordinate to transition metals in the $\eta^{6}$ coordination mode with six carbon atoms. Bisbenzenechromium, Cr(C6H6)2, is a typical example of such a compound. The compound is prepared by reducing chromium chloride in benzene and it has a sandwich structure in which a chromium atom is insertedb etween two benzene rings. When a benzene ligand is replaced by three carbonyls, Cr(CO)3(C6H6) is obtained.
18 electron rule
Counting valence electrons is of utmost importance in chemistry. Changes in the number of valence electrons has a profound influence on the bonding, structure, and reactions of a compound. Since both the metal and organic moieties are involved in organometallic compounds, counting the number of electrons becomes complicated. Hydrocarbyl ligands are classified as either neutral molecules coordinating to the metal or radicals bonding to the metal, and the radicals, such as alkyls and cyclopentadienyl, are generally called anionic ligands. Transfer of one electron from the metal to the radical ligand makes the ligand formally anionic. However, it is less confusing to consider that both the metal and the ligands are neutral when counting the number of valence electrons. The numbers of donor electrons in typical carbon ligands from this viewpoint are listed in Table $5$. It is important to note that even in the same ligand, the number of donor electrons supplied by the ligand differs depending upon the number of ligating atoms that have coordinative interactions with the metal. For example, 1, 3 or 5 electrons can be donated from a cyclopentadienyl ligand, depending on the type of coordinative interactions with the metal.
When the total number of valence electrons of the metal and ligands is 18, a transition metal organometallic compound usually has high thermal stability. For example Cr(CO)6, Fe(CO)5, Ni(CO)4, Fe(C5H5)2, Mo(C6H6)(CO)3 etc. satisfy the 18 electron rule, but the monomeric parts of Mn2(CO)10, Co2(CO)8 or [Fe(C5H5)(CO)2]2 have only 17 electrons and the extra electron comes from the partner metal by forming a metal-metal bond. Unlike the 8 electron rule in main group compounds, applicability of the 18 electron rule is limited. That is to say, it is a sufficient condition but compounds with high thermal stability are not necessarily 18 electron compounds.
Although there are many Group 6 (chromium group) through Group 9 (cobalt group) organometallic compounds with carbonyl or cyclopentadienyl ligands that satisfy the 18 electron rule, many compounds of the early transition metals (Group 3 - 5) and Group 10 (nickel group) fail to conform to this rule. For example, W(CH3)6 (12e), TiCl2(C5H5)2 (16e), and IrCl2(CO)(PPh3)2 (16e), V(CO)6 (17e), Co(C5H5)2 (19e), Ni(C5H5)2 (20e), etc. do not satisfy the 18 electron rule. However, the 18 electron rule provides useful clues as to the bonding modes present in a given complex. For example, Fe(C5H5)2(CO)2 with two pentahapto cyclopentadienyl ligands formally has 22 electrons but if one of the ligands is monohapto, the compound has 18 electrons. Structural analysis has shown that this is the actual coordination of this complex.
Exercise $5$
Calculate the valence electron number of CpMn(CO)3.
Answer
They are a total of 18 electrons from Mn (7), Cp(5) and three CO(6).
(c) Phosphine complexes
Tertiary phosphines, PX3, are very useful as stabilization ligands in transition metal complexes and they coordinate to the metals in relatively high to low oxidation states. Phosphines are frequently used as carbonyl or cyclopentadienyl ligands in the chemistry of organometallic complexes. PX3 are Lewis bases and coordinate to the metal using the lone pair on phosphorus and show $\pi$-acidity when carrying substituents X including Ph, Cl, or F that have strong electron accepting properties. The electronic flexibility of PX3 is the reason it forms so many complexes. Generally, the π-acidity becomes smaller in the order PF3 > PCl3 > PPh3 > PR3. Triphenylphosphine and triethylphosphine are typical substituted phosphines. The tertiaryphosphine complexes mainly of metal halides are listed in Table $7$. Manganese, Mn, and the early transition metals form very few phosphine complexes.
Table $7$ Typical tertiary phosphine complexes (dmpe = 1,2-bisdimethylphosphinoethane; dppe = 1,2-bisdiphenylphosphinoethane)
4 5 6 7 8 9 10 11
4 [TiCl4(PPh3)2] [VCl3(PMe
Ph2)2]
[CrCl2
(dmpe)2]
[Mn(CO)4
(PPh3)]
[FeCl2
(PPh3)2]
[CoCl2
(PPh3)2]
[NiCl2(PEt3)2] [CuBr(PEt3)]4
5 [ZrCl4(dppe)] [NbCl4(PEt
Ph2)2]
[MoCl3(PMe
Ph2)3]
[TcCl3(PMe2
Ph)3]
[RuCl2
(PPh3)3]
[RhCl(PPh3)3] [PdCl2(P
Ph3)2]
[AgCl(PPh3)]
6 [HfCl4(dppe)] [TaCl4(PEt3)2] [WCl4(PPh3)2] [ReCl3(PMe2
Ph)3]
[OsCl3
(PPh3)3]
[IrCl3(PPh3)3] [PtCl2(P
Ph3)2]
[AuCl(PPh3)]
Many derivatives can be prepared by substituting the halogens of the phosphine complexes. A number of the complexes of polydentate phosphines with more than two coordination sites, as well as those of monodentate phosphines, have been prepared, and they are used also as stabilization ligands in hydride, alkyl, dinitrogen, and dihydrogen complexes. The complexes of rhodium or ruthenium, in which optically active phosphines are coordinated, are excellent catalysts for asymmetric synthesis. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/06%3A_Chemistry_of_Transition_Metals/6.04%3A_Organometallic_Chemistry_of_d_Block_Metals_%28Part_1%29.txt |
Small Molecule Complexes
Two or three atomic molecules, such as H2, N2, CO, NO, CO2, NO2, and H2O, SO2, etc., are called small molecules and the chemistry of their complexes is very important not only for basic inorganic chemistry but also for catalyst chemistry, bioinorganic chemistry, industrial chemistry, and environmental chemistry. The complexes of small molecules other than water and carbon monoxide were synthesized comparatively recently. Dihydrogen complexes in particular were reported only in 1984.
Dihydrogen complexes
The oxidative addition reaction of a hydrogen molecule, H2, is one of the methods used to generate the M-H bond of a hydride complex. Schematically, the above reaction is written as
$\ce{M + H2 \rightarrow H-M-H}$
but it was believed that there must be an intermediate complex containing a coordinated dihydrogen. The first example of a stable complex of this sort, [W(CO)3(H2)(PiPr3)2], was reported by G. Kubas in 1984 (Figure $18$). It was proved by the neutron diffraction that the $\ce{H2}$ is coordinated as an $\eta^{2}$ ligand by maintaining the bond between hydrogen atoms with an interatomic distance of 84 pm.
Once this new coordination mode was established, new dihydrogen complexes have been prepared one after another, and dozens of dihydrogen complexes are now known. Dihydrogen complexes are interesting not only from the viewpoint of bond theory but they have also greatly contributed to the study of the activation process of the hydrogen molecule.
Dinitrogen complexes
Since N2 is isoelectronic with CO, the possible stability of dinitrogen complexes analogous in structure to carbonyl complexes was the subject of speculation for many years. These compounds generated great interest because of the parallels with the interaction and activation of nitrogen molecules on the iron catalyst used in ammonia synthesis and the nitrogen fixing enzyme nitrogenase. However, the first dinitrogen complex, [Ru(N2)(NH3)5]X2, was prepared by A. D. Allen (1965) unexpectedly from the reaction of a ruthenium complex and hydrazine. Subsequently, it was discovered by chance that nitrogen gas coordinates to cobalt, and [CoH(N2)(PPh3)3] was prepared in 1967 (Figure $19$). Many dinitrogen complexes have been prepared since these early beginnings.
In most dinitrogen complexes, N2 is coordinated to the metal by one nitrogen atom. That is to say, the M-N$\equiv$N bond is common and there are few complexes in which both nitrogen atoms bond to the metal in the $\eta^{2}$ coordination mode. In 1975, the coordinated dinitrogen in a molybdenum complex was discovered to be protonated by mineral acids to form ammonia, as decribed in the following reaction. The electrons required for the reduction are supplied by the molybdenum in a low oxidation state as this reaction shows.
$[Mo(PMe_{2}Ph)_{4}(N_{2})_{2}] + 6 H^{+} \rightarrow 2 NH_{3} + N_{2} + Mo(V) + \ldots$
In spite of attempts to prepare ammonia and organic nitrogen compounds from various dinitrogen complexes, no nitrogen fixation system which is equal to biological systems has yet been discovered. Ammonia synthesis is a long-established industrial process, and its parameters have been extensively studied and little room for improvement remains. However, elucidating the mechanism of the biological nitrogen fixation reaction at ordinary temperatures and pressures remains one of the major challenges of bio-inorganic chemistry.
Dioxygen complexes
Although it has long been recognized that schiff base complexes of cobalt absorb oxygen, the discovery that Vaska's complex, [IrCl(CO)(PPh3)2], coordinates dioxygen reversibly to form [IrCl(CO)(PPh3)2(O2)] was very significant. In this complex, two oxygen atoms bond to iridium (side-on), and dioxygen has a peroxide character (O22-). However, many superoxide (O2-) complexes in which only one oxygen atom is bonded to the metal are known. There are also binuclear dioxygen complexes in which O2 bridges two metals. The relation between reversible coordination of dioxygen and its reactivity is important in relation to the behavior of dioxygen in living systems (refer to Section 8.2 (a)).
(e) Metal-metal bonds
The concept of the formation of a coordinate bond between ligands and a central metal proposed by A. Werner was the basis for the development of the chemistry of complexes. The bonding mode and structures of known complexes have become the guidepost of the preparation of a much larger number of new complexes. For most of the dinuclear or polynuclear complexes that contain two or more metals in a complex, it was sufficient to take into consideration only the bonds between the metal and ligands.
The concept of direct bonds between metals was born of the necessity of explaining the structural chemistry of the dinuclear metal carbonyls that have a partial structure with an odd number of electrons. Two Mn(CO)5 units in Mn2(CO)10 are connected by a direct Mn-Mn bond (Figure $20$) without the help of bridge ligands. According to X-ray structural analysis (1963), the Mn-Mn distance of 292 pm was significantly longer than twice that of the metal radius of 127 pm but a Mn-Mn direct bond was envisaged in the absence of a bridge carbonyl ligand. This compound’s diamagnetism indicates a structure with an even number of electrons (18 electrons) by sharing electrons between two d7-Mn (17 electrons) moieties, each with five carbonyl ligands.
Similarly, it can be concluded that Co2(CO)8, with two bridging carbonyl ligands, should have a direct Co-Co bond to be compatible with its diamagnetism.
The concept of the single bond between metals introduced for dinuclear metal carbonyl compounds is also very useful in explaining the structure of cluster carbonyl compounds containing two or more metals. The metal-metal bond has been established today as one of the common bonding modes, together with the metal-ligand bond, present in coordination complexes. However, it is not always clear to what extent the interaction between metals exists in the polynuclear complexes which have bridging ligands. As a criterion, the bond order can be evaluated from the bond distance in standard metals (for example, in bulk metals). However, even if the bond distance between metals analyzed by X-ray is sufficiently short, this does not prove the existence of a bond between metals unless the orbital conditions to account for such bonds are also fulfilled.
M-M multiple bonds
There are many dinuclear compounds in which the metal atoms are bound by multiple bonds with bond orders of 2 to 4. The M-M quadrupole bond was proposed first for Re2Cl82-, and this remains the best-known example (Figure $21$). The Re-Re distance in this compound is only 224 pm, which is unusually short compared with the Re-Re distance of 275 pm in rhenium metal. Another unusual feature is that the ReCl4 units assume an eclipsed configuration (chlorine atoms overlap along the direction of the Re-Re bond) even though the staggard configuration (in which chlorine atoms do not overlap along the Re-Re bond direction) should be more stable because the distance between ReCl4 units is very short, resulting in the distances between the chlorine atoms being very short (experimental value of 332 pm). As a result, the repulsive interaction between the chlorine atoms becomes strong.
F. A. Cotton explained this anomaly by introducing the concept of the delta bond between metals in 1964. Namely, if one takes the z-axis in the direction of the Re-Re bond, a $\sigma$ bond is formed between the dz2, the $\pi$ bonds between dyz and dxz orbitals and the $\delta$ bond between dxy orbitals among the five d orbitals. dx2-y2 is mainly used for the Re-Cl bond. The delta bond is formed by a weak sideway overlap of dxy orbitals, when they are located perpendicular to the direction of the metal-metal bond axis and become eclipsed (Figure $22$). Therefore, although the $\delta$ bond is relatively weak among bonding interactions, it is sufficient to maintain the chlorine ligands in their eclipsed positions.
The energy levels of the molecular orbitals of $\sigma$, $\pi$, and $\delta$ bonds decrease in this order, and the energy difference between the bonding and antibonding delta orbitals is small. Therefore, even if one electron is removed (oxidation) from Re2Cl82-, which has a quadruple bond, or one electron is added (reduction) to it, the Re-Re distance should hardly change.
The Mo(II) compound [Mo2(CH3COO)4] which is isoelectronic with Re (III) has a Mo-Mo quadruple bond. [W2Cl9]3- and [W2(NMe2)6] are examples of compounds which have the metal-metal triple bonds. Although the issue of whether such metal-metal multiple bonds really exist has been argued many times, the concept has now been established and hundreds of dinuclear compounds with metal-metal multiple bonds are known at present. Metal-metal distances determined by X-ray analysis are most useful in determining whether a metal-metal bond is a multiple one, but as in the case of metal-metal single bonds, the bond distance alone cannot be the absolute determiner and it is necessary to draw conclusions from molecular orbital calculations.
(f) Metal cluster compounds
Analysis of the structures of newly prepared polynuclear complexes that contain two or more metals was, until recently, very difficult. However, with the progress of single crystal X-ray structural analysis, our understanding of the chemistry of polynuclear complexes is progressing quickly. Metal-cluster complexes are polynuclear complexes built by three or more transition-metal atoms with bonds between the metals coordinated by ligands to form polyhedral frames, such as a triangle, a regular tetrahedron, a regular octahedron, and an icosahedron. Even if there is no strong bond between metals, as long as there is some bonding interaction, they may be included as cluster compounds.
Metal cluster complexes may be broadly classified into groups according to the general character of the associated ligands. They are low oxidation state metal clusters with $\pi$-acceptor ligands like carbonyls (CO), isonitriles (RNC) or phosphines (PR3) and with $\pi$-donor ligands like oxygen (O), sulfur (S), chlorine (Cl) or alkoxides (OR). Many carbonyl cluster and sulfur cluster compounds have been synthesized. Carbonyl cluster compounds are obtained by heating or irradiating mononuclear carbonyl compounds. The chemical properties of cluster compounds such as Fe3(CO)12, Ru3(CO)12, Os3(CO)12, Co4(CO)12, Ir4(CO)12 or Rh6(CO)16 have been studied in detail (Figure $23$).
Since Os3(CO)12 forms many kinds of cluster compounds by pyrolysis, it has been used to study the skeletal structures of osmium cluster compounds and their relationship to skeletal electron numbers. A M-M bond is satisfactorily described by the 2 center 2 electron bond and the 18 electron rule is also applicable to each metal for small clusters such as a triangle and a regular tetrahedron. When clusters become large, the Wade rule that describes the relation between the structures of boranes and skeletal electron numbers, or the Lauher rule that draws the number of the bonding metal-metal orbitals for various metal polyhedral structures from the molecular orbital calculations of bare rhodium clusters without ligands, are more applicable. The relationship between the number of cluster valence electrons and the cluster’s polyhedral shape as shown in Table $8$ has contributed much to the theory of cluster chemistry.
Table $8$ Metal frameworks and cluster valence electrons in metal cluster carbonyl compounds
Metal framework Cluster valence electron Example
Triangle 48 Fe3(CO)12
Tetrahedron 60 Co4(CO)12
Butterfly 62 [Fe4(CO)12C]2-
Trigonal bipyramid 72 Os5(CO)16
Square pyramid 74 Fe5C(CO)15
Octahedron 86 Rh6(CO)16
Trigonal prism 90 [Rh6C(CO)15]2-
Monovalent anions such as halogens, alkoxides, carboxylate ions, and divalent anions such as oxygen and sulfur stabilize the cluster frameworks by helping metals assume oxidation states suitable for cluster formation and connect metal fragments by bridging. Since neutral ligands such as phosphines, carbonyl, or amines can also be coordinated to metals, a variety of cluster complexes have been prepared.
The halide clusters of molybdenum, Mo6X12, tungsten, W6X12, niobium, Nb6X14, and tantalum, Ta6X14, are solid cluster compounds that have been known for many years. The octahedral metal frameworks were shown by X-ray structure analysis more than 50 years ago. The molecular cluster complexes were prepared in the 1960s from solid-state halide clusters by the reaction of ligands such as amines and phosphines, and these cluster compounds generated considerable interest for some time. New halide cluster compounds with octahedral structures have again been prepared recently and they are being studied from new perspectives. The molecular cluster complex [Mo6S8L6] (where L is PEt3, py, etc.), which has similar Mo6 frameworks with those of the superconducting Chevrel phase compounds MxMo6S6 and their tungsten and chromium analogs have been prepared and the relationships between their structures and physical properties attract great interest (Figure $24$).
As will be described in the Chapter on bioinorganic chemistry, clusters such as Fe4S4 are contained in nitrogenase, the nitrogen-fixing enzyme, and also in the active center of ferredoxins, and they play important roles in the activation of dinitrogen or multi-electron transfer reactions. Since R. H. Holm synthesized the Fe4S4(SR)4 cluster (Figure $25$), our understanding of the chemistry of the iron-sulfur cluster has developed considerably.
As the metal species of metal cluster carbonyls are in near-zero valence oxidation states, they had been expected to play a role in specific catalysis Although many organic syntheses using metal cluster compounds as catalysts have been attempted and some interesting reactions were discovered, in most cases the clusters decomposed during the reactions and they turned out to be false cluster catalysts. Despite this, there have been some examples of reactions that pass through several elementary reaction stages on the metal of the cluster. Hence, it is likely that catalytic reactions that employ the multi-center coordination and multi-electron transfer abilities of cluster compounds will be developed in the future.
Metal clusters have been helpful as models of the surfaces of bulk metals, metal oxides, or metal sulfides, and they have been useful in the study of chemisorption and successive reactions on solid surfaces. The fine metal grains which maintain the basic cluster frameworks are deposited by the pyrolysis of metal carbonyl cluster compounds chemically bonded to carriers such as silica and alumina. If used in solid catalysis, it is expected that analysis of the catalytic reaction on a metal cluster framework will be possible. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/06%3A_Chemistry_of_Transition_Metals/6.05%3A_Organometallic_Chemistry_of_d_Block_Metals_%28Part_2%29.txt |
The reactions of complexes are classified into the substitution reaction of ligands, the conversion reaction of ligands, and the redox reaction of the central metal. The substitution and redox reactions in particular have been studied in detail.
(a) Ligand substitution reaction
Ligand substitution reactions of complexes
$L_{n} MX + Y \rightarrow L_{n} MY + X$
are very important for the preparation of various kinds of derivatives. The detailed conditions which complexes and ligands fulfill have been studied in order to understand their stereochemistry and attain practical rates of substitution reactions. As with other types of chemical reactions, we require an understanding of both equilibrium and reaction rates.
Formation constant
The equilibrium constant of a ligand substitution reaction is called a formation or stability constant. The concept and the method of computing successive formation constants were proposed by N. Bjerrum (1941). Equilibrium constants for the replacement of a hydrated ion M by other ligands L in an aqueous solution are
$M + L \rightarrow ML \qquad \qquad K_{1} = \frac{[ML]}{[M][L]}$
$ML + L \rightarrow ML_{2} \qquad \qquad K_{2} = \frac{[ML_{2}]}{[ML][L]}$
$ML_{2} + L \rightarrow ML_{3} \qquad \qquad K_{3} = \frac{[ML_{3}]}{[ML_{2}][L]}$
……………………..
$ML_{n-1} + L \rightarrow ML_{n} \qquad \qquad K_{n} = \frac{[ML_{n}]}{[ML_{n-1}][L]}$
and the overall formation constant $beta_{n}$ is
$\beta_{n} = \frac{[ML_{n}]}{[M][L]^{n}} = K_{1}K_{2}K_{3} \cdots K_{n}$
The thermodynamic stability of a substitution product becomes larger as the formation constant increases.
On the other hand, an understanding of the effect of the leaving ligand, X, and the entering ligand, Y, on the substitution rate and on the intermediate species formed are essential to elucidate the reaction chemistry of metal complexes. It is especially useful to summarize the electronic structure of the central metals, the stereochemistry of complexes, and the correlation between the parameters representing their steric properties and the reaction rate. Generally, reaction mechanisms can be classified into associative, interchange, and dissociative mechanisms according to the differences in the intermediate state (Figure $26$).
Associative mechanism
If the substitution rate of a ligand substitution of a complex is dependant upon the entering ligand, Y, coordinating to the central metal and is insensitive to the leaving ligand, X, it is presumed to take the associative mechanism which increases the coordination number. Such a substitution reaction is often seen in planar tetra-coordinate Pt(II) complexes, and the intermediate species are triangular bipyramidal penta-coordinate complexes. The reaction is first-order with respect to both the tetra-coordinate complex and Y, and is second-order as a whole. Since it is accompanied by a reduction of molecular species in the intermediate stage, thermodynamic measurements of the reaction indicate the activation entropy, $Delta$S, to be negative. The intermediate species in the case of the associative mechanism in hexa-coordinate complexes are hepta-coordinate complexes.
Interchange mechanism
When the life of an intermediate state is very short, the reaction proceeds by the interchange mechanism, as the coordination of Y and elimination of X are considered to occur simultaneously.
Dissociative mechanism
A substitution reaction that is highly sensitive to the identity of the leaving ligand, X, and practically insensitive to the identity of the entering ligand, Y, assumes the dissociative mechanism in which the coordination number decreases in the intermediate state. This is often observed in octahedral hexa-coordinate complexes, and the intermediate states are penta-coordinate complexes that form by the elimination of X. As the elimination is accompanied by an increase of molecular species in the intermediate stage, the entropy of activation, $\Delta$S, becomes positive.
Exercise $6$
The order of the rate of ligand substitution of Pt(II) complexes is H2O < Cl- < I- < PR3 < CN- for entering ligands. Which mechanism, associative or dissociative, do the substitutions take?
Answer
Since they are dependent on the entering ligands, the associative mechanism is more likely.
Trans effect
In square-planar tetra-coordinate complexes typically of Pt(II), the ligand trans to the leaving ligand X governs the substitution rate. This is called the trans effect. The substitution rate increases as the σ donor or π acceptor ability of the trans ligand becomes larger in the order of NH3 < Cl- < Br- < I- < NCS- < PR3 < CN- < CO. An analogous effect may also be seen in octahedral hexa-coordinate complexes, although the effect is usually relatively small.
The H2O exchange rate in aqua ions
Inert, intermediate, and labile are classification of the exchange rate proposed by H. Taube (1952). The exchange rate of aqua ions (ions coordinated by water molecules) of main-group and transition metals differ greatly depending upon the identity of the metal species. Since the rate of water ligand exchange is well correlated withthe exchange rates of other ligands, it is useful for general comparison of the exchange rates in the complexes of different metal ions. For alkali and alkaline earth metals, the exchange rates are very high (105 - 109 s-1), and the complexes of these metals are classified as labile. As the dissociative mechanism is generally found in these cases, ions with smaller ionicity and of larger size attract water ligands less and their exchange rates becomes higher. In Group 12 metal ions Zn2+, Cd2+, Hg2+, Group 13 metal ions Al3+, Ga3+, In3+ and Group 3 metal ions Sc3+, Y3+, rapid water ligand exchange takes place by a dissociative mechanism.
On the other hand, the exchange rates of M (II) ions in d block transition metal ions is medium (10 - 104 s-1), and that of M (III) ions are lower still. The rates of d3 Cr3+ and d6 Co3+ are notably slow (10-1 - 10-9 s-1), and their complexes are termed inert. There has been a great deal of study of ligand-exchange reactions. The exchange rates are smaller the larger the ligand field stabilization energy. Therefore, the ligand-exchange rates of 4d and 5d transition metal complexes are generally slow.
Test tube experiments
Easy chemical or biological reactions performed in test tubes are sometimes called test tube experiments. Solutions in test tubes are mixed at room temperature in air and the mixture is shaken to observe a color change or formation of precipitates and the results of the reactions are speculated on. University professors occasionally attempt these sorts of experiments. Although easy, these simple experiments show only the effects of visible light absorption and solubility. However, since even great discoveries can be born from such experiments, they should not be dismissed.
H. Taube wrote that he found a hint of the inner-sphere electron transfer mechanism from test tube experiments. He mixed Cr2+(aq) and I2 in a test tube in order to clarify the oxidation of Cr2+(aq) and observed the change of color to the one characteristic of [Cr(H2O)6]3+ via green. The green color is due to [(H2O)5CrI]2+ which is unstable and changes to [Cr(H2O)6]3+ + I-. He assumed that this was due to the formation of a Cr-I bond before Cr(II) was oxidized by I2. Subsequently, he performed another test tube experiment using [(NH3)5CoCl]2+ as an oxidant and found that Cr2+(aq) was converted into [Cr(H2O)6]3+ via green [(H2O)5CrCl]2+. This reaction established the inner-sphere electron transfer mechanism in which a Co-Cl-Cr bridge forms between Co3+ and Cr2+ and led to the Nobel Prize in a later year.
(b) Redox reactions
The oxidation number of the central metal in a transition-metal compound can vary in a few steps from low to high. Namely, the oxidation state of a compound is changeable by redox reactions. As a consequence of this, the bond distance and the bond angle between the metal and coordinating elements, or between metals, change, and at times the whole structure of a complex can be distorted remarkably or the compound may even decompose.
The reactions of a metal compound with various reducing or oxidizing agents are also very important from the viewpoint of synthetic chemistry. Especially, reduction reactions are used in the preparation of organometallic compounds, such as metal carbonyls or cluster compounds.
Meanwhile, the study of electron transfer between complexes, especially the redox reactions of transition metal complexes, has developed. Taube won the Nobel Prize (1983) for the study of electron transfer reactions in transition metal complexes, classifying such reactions into two mechanisms. The mechanism of electron transfer in which a bridging ligand is shared between two metals is called the inner-sphere mechanism, and the one involving a direct transfer of electrons between two metals without a bridging ligand is called the outer-sphere mechanism.
Inner-sphere mechanism
When [CoCl(NH3)5]2+ is reduced by [Cr(OH2)6]2+, an intermediate complex, [(NH3)5Co-Cl-Cr(OH2)5]4+, is formed in which the chlorine atom forms a bridge between cobalt and chromium. As a result of an electron transfer from chromium to cobalt through chlorine, [Co(NH3)5Cl]+ , in which cobalt is reduced from a trivalent to a divalent oxidation state and [Cr(OH2)6]3+, in which chromium is oxidized from a divalent to a trivalent oxidation state, are formed. This kind of reaction is a redox reaction via the inner-sphere mechanism. The anions other than halogens suitable for such bridge formation are SCN-, N3-, CN-, etc.
Outer-sphere mechanism
When [Fe(phen)3]3+ (phen is orthophenanthroline) is reduced by [Fe(CN)6]4-, no ligand bridge forms between the metals and an electron moves from the HOMO of Fe(II) to the LUMO of Fe(III) in a very short and direct contact between the two complexes. As the result of the electron transfer, [Fe(phen)3]2+ and [Fe(CN)6]3- form. This kind of reaction is a redox one via the outer-sphere mechanism, and is characteristic of a complex system that has a very slow ligand substitution rate compared with the speed of electron transfer, especially in systems that have the same ligands but different oxidation-numbers, for example, [Fe(CN)6]3- - [Fe(CN)6]4- has a high rate of electron transfer. R. A. Marcus won the Nobel Prize (1992) for his study of this outer-sphere electron transfer mechanism.
problems
6.1
Which cavity, either the octahedral or tetrahedral one, in an array of oxygen atoms do Fe2+ ions tend to occupy in iron oxide Fe3O4 containing both Fe2+ and Fe3+ ions?
6.2
Describe a method of preparing trans-[PtCl(Et)(PEt3)2].
6.3
Propose mononuclear and dinuclear metal complexes containing cyclopentadienyl and carbonyl ligands that satisfy the 18-electron rule.
6.4
Devise a method of selective syntheses of cis-[PtCl2(NH3)2] and trans-[PtCl2(NH3)2] using the trans effect.
6.5
How can it be proven that the reduction reaction of [CoCl(NH3)5]2+ by [Cr(OH2)6]2+ proceeds by the inner-sphere electron transfer mechanism? | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/06%3A_Chemistry_of_Transition_Metals/6.06%3A_Reactions_of_Complexes.txt |
Lanthanoids and actinoids are f-block transition elements, but their general properties differ significantly from those of d-block transition metals. These elements are placed in separate positions in the periodic table showing that the periodicity of their electronic structures differs from the main stream. Although lanthanoids are called rare-earth elements, their abundance in the crust is by 110 means rare and chemistry utilizing their unique properties is likely to develop significantly in the near future. Actinoids are closely related to nuclear chemistry and nuclear energy. Since the amount of superheavy elements "synthesized" in accelerators is very minute, they are very significant from the viewpoint of applied chemistry.
Thumbnail: A billet of highly enriched uranium that was recovered from scrap processed at the Y-12 National Security Complex Plant. (Public Domain).
07: Lanthanoids and Actinoids
The fifteen elements shown in Table \(1\) from lanthanum, La (4f0), to lutetium, Lu (4f14), are lanthanoids. Ln may be used as a general symbol for the lanthanoid elements. Although lanthanoids, scandium, Sc, and yttrium, Y, are sometimes called rare earth elements, they are relatively abundant in the earth’s crust. With the exception of promethium, Pm, which does form a stable isotope, even the least abundant thulium, Tm, and lutetium, Lu, are as abundant as iodine. Because lanthanoids have very similar properties and are difficult to separate from one another, they were not useful for basic research and application, and hence they were regarded as rare elements. Since a liquid-liquid solvent extraction method using tributylphosphine oxide became available in the 1960s, lanthanoid elements have been readily available and widely used not only for chemical research but also as materials in alloys, catalysts, lasers, cathode-ray tubes, etc.
Exercise \(1\)
What is the difference between lanthanoids and lanthanides?
Answer
Fifteen elements La-Lu are lanthanoids and fourteen elements Ce-Lu without lanthanum are lanthanides (meaning the elements similar to lanthanum). Occasionally the names are confused and 15 elements including lanthanum may be called lanthanides.
Table \(1\) Properties of lanthanoids
Atomic number Name Symbol Electron configuration M3+ radius (pm)
57 Lanthanum La 5d16s2 116
58 Cerium Ce 4f15d16s2 114
59 Praseodymium Pr 4f36s2 113
60 Neodymium Nd 4f46s2 111
61 Promethium Pm 4f56s2 109
62 Samarium Sm 4f66s2 108
63 Europium Eu 4f76s2 107
64 Gadolinium Ge 4f75d16s2 105
65 Terbium Tb 4f96s2 104
66 Dysprosium Dy 4f106s2 103
67 Holmium Ho 4f116s2 102
68 Erbium Er 4f126s2 100
69 Thurium Tm 4f136s2 99
70 Ytterbium Yb 4f146s2 99
71 Lutetium Lu 4f145d16s2 98
Because the three stages of ionization enthalpy of lanthanoid elements are comparative low, they are positive elements and readily assume trivalent ionic states. Most compounds of lanthano other than Ce4+ (4f0), Eu2+ (4f7), Yb2+ (4f14), are usually Ln3+ ones. Ln3+ species are hard acids, and since f electrons are buried deeply and not used for bonding, they are hardly influenced by ligands. There is a tendency for atomic and ionic radii to decrease with the increase of the atomic number, and this phenomenon is called the lanthanide contraction. This contraction is due to small shielding effects of 4 f electrons, which causes the atomic nucleus to draw outer shell electrons strongly with an increase of atomic number.
Complexes of lanthanoid metals are 6 to 12 coordinate and especially many 8 and 9 coordinate compounds are known. Organometallic compounds with cyclopentadienyl ligands of the types \(\ce{Cp3Ln}\) or \(\ce{Cp2LnX}\) are also known, all of which are very reactive to oxygen or water. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/07%3A_Lanthanoids_and_Actinoids/7.01%3A_Lanthanoids.txt |
The fifteen elements from actinium, Ac, to lawrencium, Lr, are called actinoids (Table $2$). The general symbol of these elements is An. All the actinoid elements are radioactive and very poisonous. Actinoids that exist in nature in considerable amounts are thorium, Th, protactinium, Pa, and uranium, U, and thorium and uranium are actually isolated from ores and find application. Plutonium metal, Pu, is produced in large quantities in nuclear reactors and its economical efficiency as a fuel for conventional nuclear reactors and fast breeder reactors, as well as its safety, are being examined. As isolable amounts of the elements after americium, Am, are small and their radioactivity is very high, study of their chemical properties is very limited.
Table $2$ Properties of actinoids
Atomic number Name Symbol Electron configuration M3+ radius (pm) Main isotope
89 Actinium Ac 6d17s2 126 227Ac
90 Thorium Th 6d27s2 227Ac
91 Protactinium Pa 5f26d17s2 118 232Th
92 Uranium U 5f36d17s2 117 235U, 238U
93 Neptunium Np 5f57s2 115 237Np
94 Plutonium Pu 5f67s2 114 238Pu, 239Pu
95 Americium Am 577s2 112 241Am,243Am
96 Curium Cm 5f76d17s2 111 242Cm, 244Cm
97 Berkelium Bk 5f97s2 110 249Bk
98 Californium Cf 5f107s2 109 252Cf
99 Einsteinium Es 5f117s2
100 Fermium Fm 5f127s2
101 Mendelevium Md 5f137s2
102 Nobelium No 5f147s2
103 Lawrencium Lr 5f146d17s2
The process of radioactive disintegration of radioactive elements into stable isotopes is of fundamental importance in nuclear chemistry. If the amount of a radionuclide which exists at a certain time is N, the amount of disintegration in unit time is proportional to N. Therefore, radioactivity is
$- \frac{dN}{dt} = \lambda N \qquad (\lambda\; \text{is disintegration constant})$
integration of the equation leads to
$N = n_{0} e^{- \lambda t}$
where N0 is the number of atoms at zero time and the time during which the radioactivity becomes half of N0 is the half life.
$T = \frac{\ln 2}{\lambda} = \frac{0.69315}{\lambda}$
Exercise $2$
How does a nuclide change with α disintegration and $\beta^{−}$ disintegration?
Answer
Because an atomic nucleus of helium atom, 4He, is emitted by $\alpha$ disintegration of a nuclide, its atomic number Z becomes (Z-2) and its mass number A changes to (A-4). In $\beta^{−}$ disintegration, an electron is emitted and Z becomes a nuclide (Z + 1).
Isolation of thulium
Thulium is a rare earth element with the least abundance except promethium, and there were remarkable difficulties in isolating it as a pure metal. P. T. Cleve discovered the element in 1879, but it was only 1911 when the isolation of the metal of almost satisfying purity was reported.
C. James of the United States tried many minerals and found that three ores, ytterspar, euzenite and columbite produced from an island in the northern Norway, were the best source. In order to obtain a purer metal of thulium, chromates of the mixed rare-earth metals obtained by the treatment of a large amount of the ores by sodium hydroxide, hydrochloric acid, oxalic acid, and barium chromate were recrystallized repeatedly from water and water-alcohol. In those days, identification of an element by spectroscopy was already possible, and recrystallizations were repeated 15,000 times over several months, proving that it was not possible to obtain purer metal.
Chemists are requested to repeat monotonous operations even now but it is not likely that patience of this sort still exists. This may hinder the progress of our understanding of the chemistry of rare earth elements.
Although actinoids are similar to lanthanoids in that their electrons fill the 5f orbitals in order, their chemical properties are not uniform and each element has characteristic properties. Promotion of 5f - 6d electrons does not require a large amount of energy and examples of compounds with $\pi$-acid ligands are known in which all the 5f, 6d, 7s, and 7p orbitals participate in bonding. Trivalent compounds are the most common, but other oxidation states are not uncommon. Especially thorium, protactinium, uranium, and neptunium tend to assume the +4 or higher oxidation state. Because their radioactivity level is low, thorium and uranium, which are found as minerals, can be handled legally in a normal laboratory. Compounds such as ThO2, ThCl4, UO2, UCl3, UCl4, UCl6, UF6, etc. find frequent use. Especially uranium hexafluoride, UF6, is sublimable and suitable for gas diffusion and undergoes a gas centrifuge process for the separation of 235U. Thorium is an oxophilic element similar to the lanthanoids.
problems
7.1
What is the reason for the relatively easy separation of cerium and europium among the lanthanoids, which were difficult to isolate?
7.2
Calculate the radioactivity after a period of 10 times as long as the half-life of a given material. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/07%3A_Lanthanoids_and_Actinoids/7.02%3A_Actinoids.txt |
Organic synthesis using complexes and organometallic compounds, homogeneous catalysis, bioinorganic chemistry to elucidate biological reactions in which metals participate, and studying solid state properties such as solid state catalysis, conductivity, magnetism, optical properties are all important fields of applied inorganic chemistry. Basic inorganic chemistry, although previously less developed compared with organic chemistry, is now making fast progress and covers all elements. Construction of general theories of bonding, structure, and reaction covering molecules and solids is a major problem for the near future.
Thumbnail: The discovery and characterization of the structure of ferrocene, \(\ce{Fe(C5H5)2}\) in the early 1950's, led to an explosion of interest in d-block metal carbon bonds and brought about development and the now flourishing study of organometallic chemistry. (CC BY; Karl Harrison 3DChem.com).
08: Reaction and Physical Properties
Catalysts reduce the activation energy of reactions and enhance the rate of specific reactions. Therefore they are crucially important in chemical industry, exhaust gas treatment and other chemical reactions. While the chemical essence of catalysis is obscure, practical catalysts have been developed based on the accumulation of empirical knowledge. However, while gradually we have come to understand the mechanisms of homogeneous catalysis through the development of inorganic chemistry, our understanding of surface reactions in solid catalysts is also deepening.
(a) Homogeneous catalysis
The chemistry of catalysts that are soluble in solvents has developed remarkably since the epoch-making discovery (1965) of the Wilkinson catalyst, [RhCl(PPh3)3]. This complex is a purplish red compound which forms by heating RhCl3 • 3 H2O and PPh3 under reflux in ethanol. When dissolved in an organic solvent, this complex is an excellent catalyst for hydrogenation of unsaturated hydrocarbons by H2 at ambient temperatures and pressures to form saturated hydrocarbons, and hydroformylation reactions of olefins with H2 and CO to form aldehydes.
In the past, the mechanism of catalytic reactions were generally not very clear. Before the Wilkinson catalyst, the Reppe process, which oligomerize aceylenes or the Ziegler-Natta catalysts that polymerize olefins and dienes, had been discovered and detailed studies on homogeneous catalysis had been conducted from the viewpoint of the chemistry of complexes. Consequently, catalytic reactions are now established as a cycle of a combination of a few elementary steps that occur on the metals of catalyst complexes.
Coordination and dissociation
There must be a process in which reactants such as olefins are activated and react with other reactants after being coordinated to the central metal of a complex, and they dissociate from the metal as products.
Oxidative addition
Oxidative addition is one among a few key elementary reactions of metal complexes. This is a reaction of such compounds as alkali metal halides, RX, acids, HX or dihydrogen, H2, to the metal in a complex which then dissociate into R and X, H and X, or H and H, which are bonded to the metal as two fragment anions. If other ligands on the start complex are not removed, the coordination number increases by two. As alkyl, halogen, and hydride ligands are more electronegative than the central metal, they are regarded as formally anionic ligands after coordination. Therefore, the oxidation number of the central metal increases after an addition reaction. As it is an addition reaction accompanied by oxidation of the central metal, it is called oxidative addition.
For example, in the addition reaction of an alkyl halide to a tetra-coordinate iridium(I) complex [IrCl(CO) (PPh3)2],
$[Ir^{I} Cl (CO) (PPh_{3})_{2}] + RI \rightarrow [Ir^{III}(Cl)(I)(R)(CO)(PPh_{3})_{2}]$
iridium becomes hexa-coordinate and undergoes two-electron oxidation from +1 to +3. Since a neutral RI molecule is added, there must be no change in the charge of the whole complex, and if an alkyl and iodine are anions, the oxidation number of the central metal should increase by 2. Similar change occur when two hydride ligands are formed as the result of the addition of dihydrogen.
The reverse reaction is called reductive elimination. Both oxidative and reductive reactions are very important as elementary steps in the mechanism of homogeneous catalysis involving hydrocarbons and dihydrogen.
Exercise $1$
How does the oxidation number of rhodium change with reductive elimination of dihydrogen from [RhCl(H)2(PPh3)2 (Sol)]?
Answer
It changes to Rh(I) from Rh(III).
Insertion reaction
In the reaction of an alkyl or hydride ligand to shift to a carbonyl or olefin ligand coexisting on the central metal, the resultant complex appears as if a carbonyl or an olefin has inserted between the M-R or M-H bond. This is called an insertion reaction.
Reaction of a coordinated ligand
This is the process in which a coordinated reactant reacts to form a product. By coordinating to a metal, the reactants take geometrically and electronically suitable conformations. It is the basis of catalyst design to control these reaction conditions.
Since a reaction is repeated while the complex used as a catalyst remains unchanged by forming a cycle of reactions, the reactants/complex ratio is very small, coinciding with the definition of a catalyst. The catalytic cycle in hydrogenation of ethylene is illustrated in Figure $1$.
If the triphenylphosphine ligand P(C6H5)3 in the Wilkinson catalyst is replaced by an optical active phosphine, asymmetric hydrogenation is realized. Asymmetrical catalysis equivalent to enzyme reactions have been developed by skillful design of asymetrical ligands. In particular, the asymmetric induction of binaphtyldiphosphine (BINAP) has attracted attention.
(b) Solid state catalysis
A solid catalyst is also called a heterogeneous catalyst, and promotes the reaction of reactants in gaseous or liquid phases in contact with a solid material. Since adsorption of reactants on the catalyst surface is the initial step, a large surface area is required for good efficiency of catalysis. Polyphase systems, which carry active catalysts on materials such as zeolites with small pores of molecular sizes, and gamma alumina and silica gel with large surface area, are often used.
Previously, solid state catalysis was explained as arising from a mysterious activation of reactants due to adsorption, but it has become increasingly clear that catalysis is ascribable to surface chemical reactions. Namely, the action of solid state catalysts depends on activation of reactants by surface acids or bases, and by coordination to the metal surface. It is possible to observe these interactions using various spectroscopies (infrared spectroscopy, EXAFS (extended X-ray absorption fine structure), electronic spectra), electron microscopy, or STM (scanning tunnelling microscopy).
Since mechanisms of homogeneous catalysis have been clarified considerably, solid surface reactions can also be analyzed by introducing concepts such as “surface complexes” or “surface organometallic compounds”. However, unlike homogeneous catalysis, in which only one or a few metal centers participate, many active sites are involved in solid state catalysis. Since surface homogeneity and reproducibility are difficult to maintain, major parts of reaction mechanisms are obscure even for such simple reactions as ammonia synthesis.
During the direct production of ammonia from dinitrogen and dihydrogen, reactions occur using iron catalysts containing alkali metal or alkaline earth metal oxides as activators at high temperatures (about 450 °C) and under high pressures (about 270 atm). Prior to the epoch-making discovery of this process by F. Haber (1909), all nitrogen compounds came from natural resources. The realization of this discovery has had an immeasurable influence upon chemical industries, as ammonia is indispensable to the manufacture of fertilizers, gunpowder and other inorganic compounds containing nitrogen. In recognition of this, a Nobel Prize was awarded to F. Haber for this invention (1918). A huge volume of research on the elucidation of the reaction mechanism of ammonia synthesis has been performed up until the present, because the reaction of dinitrogen and dihydrogen on iron catalysts is a good model of solid state catalysis. | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/08%3A_Reaction_and_Physical_Properties/8.01%3A_Catalytic_reactions.txt |
Many biological reactions are known to involve metal ions. There are also metals recognized as essential elements, although their roles in living organisms are not clear. Bioinorganic chemistry, the study of the functions of metals in biological systems using the knowledge and methods of inorganic chemistry, has progressed remarkably in recent years.
The following list shows typical bioactive substances containing metals.
1. Electron carriers. Fe: cytochrome, iron-sulfur protein. Cu : blue copper protein.
2. Metal storage compound. Fe : ferritin, transferrin. Zn : metallothionein.
3. Oxygen transportation agent. Fe: hemoglobin, myoglobin. Cu: hemocyanin.
4. Photosynthesis. Mg: chlorophyll.
5. Hydrolase. Zn: carboxypeptidase. Mg: aminopeptidase.
6. Oxidoreductase. Fe: oxygenase, hydrogenase. Fe, Mo: nitrogenase.
7. Isomerase. Fe: aconitase. Co: vitamin B12 coenzyme.
The basis of chemical reactions of metalloenzymes are
1. Coordinative activation (coordination form, electronic donating, steric effect),
2. Redox (metal oxidation state),
3. Information communication, and, in many cases, reaction environments are regulated by biopolymers such as proteins, and selective reactions are performed.
Examples of actions of metals other than by metalloenzymes include
1. Mg: MgATP energy transfer
2. Na/K ion pumping,
3. Ca: transfer of hormone functions, muscle contraction, nerve transfer, blood coagulation, are some of the important roles of metals.
(a) Oxidation
Oxidation reactions in living systems are fundamental to life, and many studies of these systems have been performed. In particular, the mechanisms of oxygen gas transportation by hemoglobin and mono-oxygen oxidation by the iron porphyrin compounds named P-450 have been studied at length. Oxygen gas transportation, which has been studied for many years, is described below. Iron porphyrins hemoglobin and myoglobin and the copper compound hemocyanin are involved in the transportation of oxygen gas in air to cells in living organisms. The basis of this function is reversible bonding and dissociation of dioxygen to iron or copper ions. In order to perform these functions, metals must be in oxidation states and coordination environments suitable for the reversible coordination of dioxygen. The iron porphyrin compound hemoglobin is found in red bloods of human beings and other animals.
Hemoglobin has the structure of heme iron with four iron porphyrin units combined with a globin protein. Dioxygen is transported in blood by being coordinated to ferrous ions in the hem iron unit. The Fe (II) ion is penta-coordinate with four nitrogen atoms of porphyrin and a nitrogen atom of the polypeptide histidine, and becomes hexa-coordinate when a dioxygen coordinates to it. The spin state of Fe (II) changes from high spin to low spin upon the coordination of dioxygen. The high spin Fe(II) is above the plane of porphyrin because it is too large to fit in the available space. When the Fe(II) ion becomes low spin upon dioxygen coordination, the size of the iron ion decreases and it just fits into the hole of the porphyrin molecule.
This molecular-level movement has attracted interest as an allosteric effect because it affects the whole protein through the histidine coordinate bond and governs the specific bond of a dioxygen molecule. Oxidation of the Fe(II) ion of a hem molecule is prevented by a macromolecular protein, and if the hem iron is taken out of the prote in, Fe(II) ion is oxidized to Fe(III), and two porphyrin rings are bridged by a peroxide $\mu$-O22-, which finally changes to a bridging $\mu$-O2-structure.
When the hem is in this state, it loses the ability to coordinate to the dioxygen molecule. Based on this phenomenon, a synthetic porphyrin that is able reversibly to coordinate to a dioxygen by suppressing dimerization of the iron porphyrin has been developed, and was named the picket fence porphyrin after its three dimensional form.
(b) Nitrogen fixation
The reaction which converts the nitrogen in air into ammonia is basic to all life. Nitrogen fixation, the reaction to fix atmospheric nitrogen to form ammonia, is carried out by Rhizobium in the roots of legumes or by bacteria in algae in an anaerobic atmosphere. All animals and plants, including mankind, were depended on biological nitrogen fixation as a source of nitrogen for protein and other compounds containing nitrogen before the invention of the Harber-Bosch process.
$N_{2} + 8 H^{+} + 8 e^{-} + 16 MgATP \rightarrow 2 NH_{3} + H_{2} + 16 MgADP + 16 Pi \quad \text{(where Pi is an inorganic phosphate)}$
An enzyme named nitrogenase catalyzes this reaction. Nitrogenase contains iron-sulfur and iron-molybdenum sulfur proteins, and reduces dinitrogen by coordination and cooperative proton and electron transfers, while using MgATP as an energy source. Because of the importance of this reaction, attempts to clarify the structure of nitrogenase and to develop artificial catalysts for nitrogen fixation have continued for many years. Recently, the structure of an active center in nitrogenase called iron-molybdenum cofactor was clarified by single crystal X-ray analysis (Figure $2$). According to this analysis, its structure has Fe3MoS4 and Fe4S4 clusters connected through S.
It is believed that dinitrogen is activated by coordination between the two clusters. On the other hand, the portion called P cluster consists of two Fe4S4 clusters. The roles and reaction mechanism of both parts are not yet clear.
(c) Photosynthesis
The formation of glucose and dioxygen by the reaction of carbon dioxide and water is a skillful reaction using photoenergy and in which chlorophyll (Figure $3$), which is a magnesium porphyrin and a manganese cluster complex, plays the central role. A chloroplast contains photosystem I (PSI) and photosystem II (PSII), which use light energy to reduce carbon dioxide and to oxidize water.
Chlorophyll is a fundamental component of PSI. Chlorophyll is a porphyrin complex of magnesium and is responsible for the green colors of leaves. It plays an important role in receiving light energy and transferring it to redox reaction systems. Chlorophyll is excited from the singlet ground state to the singlet excited state by light, the energy of the excited state is transferred to an acceptor within 10 ps, and the resultant energy reduces an iron-sulfur complex and is finally used for reduction of carbon dioxide in subsequent dark reactions. Since charge separation by photochemical excitation is the most important first stage, studies on photoinduced electron transfer are have been actively performed using various kinds of porphyrin compounds as models of chlorophylls. PSI, which obtains oxidizing energy by electron transfer, converts ADP to ATP.
On the other hand, the oxidized form of PSII oxidizes water through a chain of redox reactions of oxo cluster complexes of manganese, and generates oxygen. Since four electrons shift in the reduction of Mn(IV) to Mn (II) in this reaction, at least two manganese species are involved. Probably, a cluster complex which contains two Mn(II) and two Mn(IV) species mediates the electron transfer via four step reactions. However, the details of this reaction are as yet unclear because it is very difficult to isolate this cluster and to analyze its structure. The electron transfer stage is being studied at present by using various manganese complexes as model systems.
Photosynthesis is a very interesting research theme in bioinorganic chemistry as it involves a few metal ions, a porphyrin, sulfide and oxide clusters that constitute a cycle of subtle electron transfer and redox reactions, and generate oxygen gas by photolysis of water and produce carbohydrates from carbon dioxide by reductive dark reactions,. Recently, the reaction center of a photosynthetic bacteria was crystallized and J. Deisenhofer and his colleagues won a Nobel prize for its structural analysis (1988).
Exercise $2$
Give examples of small molecules that are fundamentally important for living things.
Answer
• H2O
• O2
• N2
• CO2 | textbooks/chem/Inorganic_Chemistry/Inorganic_Chemistry_(Saito)/08%3A_Reaction_and_Physical_Properties/8.02%3A_Bioinorganic_chemistry.txt |
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