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7.1.1: Tetrahedral and Octahedral Geometries

  • Page ID
    206978

  • Until this point, we have focused on one of the most common cases for metal complexes, the 6-coordinate octahedral ligand field geometry. Now, we will examine other geometries, with a focus on 4-coordinate ligand fields. Almost everything you have already learned about octahedrons can be applied to 4-coordinate metal complexes (effects of metal and ligands on \(\Delta\)); the main difference is the splitting pattern of d-orbitals is different in each 4-coordinate complex than it was in the octahedral cases. Also, the magnitudes of splitting are different in important ways.

    Since CFT is simpler than LFT, let's use CFT to describe and derive these "new-to-you" splitting patterns. Now is a good time to go back and recall how CFT was used to derive the splitting of d-orbitals in an octahedral field so that you can use that as a launching pad for deriving the tetrahedron (harder) and square planar geometries:

    Exercise \(\PageIndex{1}\)

    The d orbital splitting diagram for a tetrahedral coordination environment is shown below. Given this diagram, and the axes in the accompanying picture, identify which d orbitals are found at which level. In the picture, the metal atom is at the center of the cube, and the circle represent the ligands.

    On the left is a cube with an x, y, and z axis. There are four circles are arranged in a tetrahedron on alternating corners of the cube. Each circle represents a ligand. On the right is a diagram of d-orbital splitting for a tetrahedron. There are three higher energy, degenerate orbitals and two lower energy, degenerate orbitals.

    Figure \(\PageIndex{1}\):

    Answer

    The three upper orbitals (dy, dx, dz) interact a little more strongly with the ligands. The two lower orbitals (dz2 and dx2-y2) interact a little more weakly.

    Diagram of d orbitals in a cube with four ligands oriented on alternating corners of the cube. Each ligand is represented by a circle. Two ligands are diagonal from one another in the top corners of the cube. The remaining two are are diagonal from one another in the bottom corners of the cube such that they do not sit directly under the top two. The top three cubes have the dy, dx, and dz orbitals that sit parallel to the front face, side face, and top face, respectively. The bottom two diagrams have the dz-squared and dx-squared-minus-y-squared orbitals inside the cubes. The reason for the difference in the interaction has to do with how close the nearest lobe of a d orbital comes to a ligand. There are really two possible positions: the face of a cube or the edge of a cube. If the ligands are at alternating corners of the cube, then the orbitals pointing at the edges are a little closer than those pointing at the faces of the cube.

    Diagram showing distance from the corner to the edge is smaller than the distance from the corner to the face of a cube. Using Pythagorean's Theorem with two sides of a triangle equalling R, the diagonal is 0.705 times R. Thus the distance from the corner to the center of the face is 0.705 time R. The distance from the corner to the middle of the edge is 0.5 times R.

    Exercise \(\PageIndex{2}\)

    Typically, the d orbital splitting energy in the tetrahedral case is only about 4/9 as large as the splitting energy in the analogous octahedral case. Explain why it is smaller for the tetrahedral case.

    Answer

    The ligands do not overlap with the d orbitals as well in tetrahedral complexes as they do in octahedral complexes. Thus, there is a weaker bonding interaction in the tetrahedral case. That means the antibonding orbital involving the d electrons is not raised as high in energy, so the splitting between the two d levels is smaller.

    Exercise \(\PageIndex{3}\)

    Suppose each of the ions below were in tetrahedral coordination environments. Draw the d orbital diagrams for the high-spin and the low-spin case for each ion. Which one (high-spin or low-spin) is more likely?

    a) Mn2+     b) Co2+     c) Ni2+     d) Cu+     e) Fe3+      f) Cr2+     g) Zn2+

    Answer a)

    Tetrahedral orbital diagrams of low-spin versus high-spin with five electrons. There are three higher energy orbitals and two lower energy orbitals. In the low spin case pairing is more energetically favorable, so there are two electron pairs in the lower orbitals and one unpaired electron in a higher orbital. In the high spin case, there are two unpaired electrons in the lower orbitals and three unpaired electrons in the higher orbitals.  

    High spin is more likely for tetrahedral coordination geometry due to weak interactions between ligand orbitals and metal d-orbitals.

    Answer b)

    Tetrahedral orbital diagrams of low-spin versus high-spin with seven electrons. There are three higher energy orbitals and two lower energy orbitals. The low-spin and high-spin diagrams are identical with two pairs of electrons in the lower energy orbitals and three unpaired electrons in the higher energy orbitals.

    High spin is more likely for tetrahedral coordination geometry due to weak interactions between ligand orbitals and metal d-orbitals.

    Answer c)

    Tetrahedral orbital diagrams of low-spin versus high-spin with eight electrons. There are three higher energy orbitals and two lower energy orbitals. The low-spin and high-spin diagrams are identical with two pairs of electrons in the lower energy orbitals. There is one electron pari and two unpaired electrons in the higher energy orbitals.

    High spin is more likely for tetrahedral coordination geometry due to weak interactions between ligand orbitals and metal d-orbitals.

    Answer d)

    Tetrahedral orbital diagrams of low-spin versus high-spin with ten electrons. There are three higher energy orbitals and two lower energy orbitals. The low-spin and high-spin diagrams are identical with two pairs of electrons in the lower energy orbitals and three pairs of electrons in the higher energy orbitals.

    High spin is more likely for tetrahedral coordination geometry due to weak interactions between ligand orbitals and metal d-orbitals.

    Answer e)

    Tetrahedral orbital diagrams of low-spin versus high-spin with five electrons. There are three higher energy orbitals and two lower energy orbitals. In the low spin case pairing is more energetically favorable, so there are two electron pairs in the lower orbitals and one unpaired electron in a higher orbital. In the high spin case, there are two unpaired electrons in the lower orbitals and three unpaired electrons in the higher orbitals.

    High spin is more likely for tetrahedral coordination geometry due to weak interactions between ligand orbitals and metal d-orbitals.

    Answer f)

    Tetrahedral orbital diagrams of low-spin versus high-spin with four electrons. There are three higher energy orbitals and two lower energy orbitals. In the low spin case pairing is more energetically favorable, so there are two electron pairs in the lower orbitals and zero electrons in the higher orbitals. In the high spin case, there are two unpaired electrons in the lower orbitals and two unpaired electrons in the higher orbitals.

    High spin is more likely for tetrahedral coordination geometry due to weak interactions between ligand orbitals and metal d-orbitals.

    Answer g)

    Tetrahedral orbital diagrams of low-spin versus high-spin with ten electrons. There are three higher energy orbitals and two lower energy orbitals. The low-spin and high-spin diagrams are identical with two pairs of electrons in the lower energy orbitals and three pairs of electrons in the higher energy orbitals.

    High spin is more likely for tetrahedral coordination geometry due to weak interactions between ligand orbitals and metal d-orbitals.

    Exercise \(\PageIndex{4}\)

    Usually, tetrahedral ions are high spin rather than low spin. Explain why.

    Answer

    Because the d orbital splitting is much smaller in the tetrahedral case, it is likely that the energy required to pair two electrons in the same orbital will be greater than the energy required to promote an electron to the next energy level. In most cases, the complex will be high spin.

    Exercise \(\PageIndex{5}\)

    The d orbital splitting diagram for a square planar environment is shown below. Given this diagram, and the axes in the accompanying picture, identify which d orbitals are found at which level.

    Diagram of d orbital splitting for a square planar geometry. There are 4 energy levels. The highest has one orbital, the second has one orbital, the third has one orbital, and the lowest has two orbitals.

    Figure \(\PageIndex{18}\):

    Answer

    The orbitals are shown in order of energy.

    Diagram of the five d orbitals positioned at their respective energies for a square planar d orbitals splitting diagram. Each orbital is placed at the center of an x, y, and z axis with ligands oriented in a square plane. Ligands sit on the x and y axes. The highest energy orbital is the d-x-squared-minus-y-squared where each lobe points at a ligand. The second highest energy orbital is the d-z-squared orbitals where the ring points towards the ligands. The third highest energy orbital is the d-x-y orbital where the lobes point between the ligands. The lowest energy orbitals are the d-y-z and d-x-z orbitals whose lobes point between the y-z and x-z axes, respectively.

    Exercise \(\PageIndex{6}\)

    Predict whether each compound will be square planar or tetrahedral.

    a. [Zn(NH3)4]2+

    b. [NiCl4]2+

    c. [Ni(CN)4]2-

    d. [Ir(CO)(OH)(PCy3)2]2+ ; Cy = cyclohexyl

    e. [Ag(dppb)2]+ ; dppb = 1,4-bis(diphenylphosphino)butane

    f. PtCl2(NH3)2

    g. PdCl2(NH3)2

    h. [CoCl4]2–

    i. Rh(PPh3)3Cl

    Answer a)

    [Zn(NH3)4] 2+ 3d metal, d10, sigma donor ligand → tetrahedral

    Answer b)

    [NiCl42+ 3d metal, d8, pi donor ligand → tetrahedral

    Answer c)

    [Ni(CN)42- 3d metal, d8, pi acceptor ligand → square planar

    Answer d)

    [Ir(CO)(OH)(PCy3)2] 2+ 5d metal, d8 → square planar

    Answer e)

    [Ag(dppb)2]1+ 4d metal, d10, sigma donor ligand → tetrahedral

    Answer f)

    [PtCl2(NH3)2] 5d metal, d8 → square planar

    Answer g)

    [PdCl2(NH3)2] 4d metal, d8, M+2, sigma donor ligand → square planar

    Answer h)

    [CoCl42– 3d metal, d7, sigma donor ligand → tetrahedral

    Answer i)

    [Rh(PPh3)3Cl] 5d metal,  d8 → square planar

     

    LFSE and SE for Other Geometries

    Just like in the case of octahedral metal complexes, we can apply CFT (or LFT) to calculate LFSE and SE of 4-coordinate (or any-coordinate) metal complexes. Once again, in terms of the d orbital splitting diagram, the results are similar to what we see from molecular orbital theory. For tetrahedral geometry, which is the most common geometry when the coordination number is four, we get a set of two high-lying orbitals and three lower ones.

    Diagram of tetrahedral orbital splitting with energy increasing from bottom to top. There are three higher energy orbitals at +0.4 times delta-t above the barycenter and two lower energy orbitals at -0.6 times delta-t below the barycenter. The gap between the energy of these orbitals is defined as delta-t.
    Figure \(\PageIndex{11}\):

    The overall splitting is expressed as Δt; the t stands for tetrahedral. However, sometimes it is useful to compare different geometries. In crystal field theory, it can be shown that the amount of splitting in a tetrahedral field is much smaller than in an octahedral field. In general, Δt = 4/9 Δo.

    Diagram of tetrahedral orbital splitting with energy increasing from bottom to top. Values are written relative to delta-o. There are three higher energy orbitals at +0.18 times delta-o above the barycenter and two lower energy orbitals at -0.27 times delta-o below the barycenter. The gap between the energy of these orbitals is defined as 4/9 times delta-o.
    Figure \(\PageIndex{12}\):

     

    Which geometries are most likely? 

    We wouldn't usually use crystal field theory to decide whether a metal is more likely to adopt a tetrahedral or an octahedral geometry. In most cases, the outcome is more strongly dependent on factors other than the d orbital energies. For example, maybe a complex would be too crowded with six ligands, so it only binds four; it becomes tetrahedral rather than octahedral. Or maybe a metal does not have enough electrons in its valence shell, so it binds a couple more ligands; it becomes octahedral rather than tetrahedral.

    However, this comparison between Δo and Δt does help to explain why tetrahedral complexes are much more likely than octahedral complexes to adopt high-spin configurations. The splitting is smaller in tetrahedral geometry, so pairing energy is more likely to become the deciding factor there than it is in octahedral cases.

    Exercise \(\PageIndex{1}\)

    Tetrahedral complexes are pretty common for high-spin d6 metals, even though the 18-electron rule suggests octahedral complexes should form. In contrast, low-spin d6 complexes do not usually form tetrahedral complexes. Use calculations of stabilisation energies to explain why.

    Answer

    high-spin d6

    Comparison of high spin octahedral versus tetrahedral orbital splitting diagram with six electrons. The octahedral diagram has three lower energy orbitals and two higher energy orbitals. In the octahedral diagram there is one set of paired electrons and two unpaired electrons in the lower orbitals and two unpaired electrons in the higher orbitals. The tetrahedral diagram has three higher energy orbitals and two lower energy orbitals. In this diagram there is one pair of electrons and one unpaired electron in the lower orbitals. And there are three unpaired electrons in the upper orbitals.

    octahedral

        SE = [2(0.6) -4(0.4)]Δo + PE

        SE = -0.4Δo + PE

    tetrahedral

        SE = [3(0.4) - 3(0.6](4/9)Δo + PE

        SE = -0.6Δo + PE

                ΔSE = SEoh - SEtd  =  -0.4Δo + PE -(-0.6Δo + PE) = +0.2Δo

                This is a slight preference for tetrahedral.

    low-spin d6

    Low spin octahedral versus tetrahedral orbitals splitting diagrams for six electrons. In the octahedral, all electrons are paired in the lower energy orbitals. In the tetrahedral there are two pairs of electrons in the lower energy orbitals and two unpaired in the higher energy orbitals.

    octahedral

        SE = [0(0.6) -6(0.4)]Δo + 3PE

        SE = -2.4Δo + 3PE

    tetrahedral

        SE = [2(0.4) - 4(0.6](4/9)Δo + 2PE

        SE = -1.6Δo + PE

                ΔSE = SEoh - SEtd  =  -2.4Δo + 3PE -(-1.6Δo + 2PE) = -0.8Δo + PE

                This is a preference for octahedral, although it would be offset by the pairing energy.

    Crystal field theory has also been used to determine the splitting between the orbitals in a square planar geometry. That is a much more complicated case, because there are four different levels. Once again, the energy levels are often expressed in terms of Δo.

    Orbital splitting diagram from square planar geometry in terms of delta-o. The highest energy orbital is +1.34 times delta-o and the second highest is +0.23 times delta-o. Below the barycenter there is one orbital at -0.43 times delta-o and two orbitals at -0.52 times delta-o.
    Figure \(\PageIndex{13}\):

    We could use a calculation of stabilisation energy to predict whether a particular complex is likely to adopt a tetrahedral or a square planar geometry. Both geometries are possible, so it would be useful to be able to predict which geometry occurs in which case. However, just as in the case of comparing octahedral and tetrahedral gemoetries, there is another factor that is often more important.

    Comparison of square planar angle versus tetrahedral angle. Square planar geometry orients ligands 90-degrees apart. Ligands in a tetrahedral geometry are 109-degree apart.
    Figure \(\PageIndex{14}\):

    That factor is steric crowding. In a tetrahedron, all the ligands are 109o from each other. In a square planar geometry, the ligands are only 90o away from each other. Tetrahedral geometry is always less crowded than square planar, so that factor always provides a bias toward tetrahedral geometry. As a result, we might expect square planar geometry to occur only when sterics is heavily outweighed by ligand field stabilisation energy.

    Exercise \(\PageIndex{2}\)

    The most common examples of square planar complexes have metals that are d8. Use calculations of the stabilisation energy to explain why if the complex is:

    (a) low-spin

    (b) high-spin

    Answer a)

    high-spin d8

    Comparison of square planar geometry versus tetrahedral geometry for high-spin d-8 complex. In the tetrahedral geometry the three lowest orbitals have paired electrons. The highest and second highest energy orbitals each have one unpaired electron. In the tetrahedral geometry the two degenerate lower energy orbitals each have a pair of electrons. The three degenerate higher energy orbitals have one electron pair and two unpaired electrons.

    square planar

        SE = [1.23 + 0.23 - 2(0.43) -4(0.51)]Δo + 3PE

        SE = -1.44Δo + 3PE

    tetrahedral

        SE = [4(0.4) - 4(0.6](4/9)Δo + 3PE

        SE = -0.36Δo + 3PE

                ΔSE = SEsq - SEtd  =  -1.44Δo + 3PE -(-0.36Δo + 3PE) = -1.08Δo

                This is an appreciable preference for square planar.

    Answer b)

    low-spin d8

    Comparison of square planar geometry versus tetrahedral geometry for low-spin d-8 complex. In the tetrahedral geometry the four lowest orbitals have paired electrons. The highest energy orbitals has zero electrons. In the tetrahedral geometry the two degenerate lower energy orbitals each have a pair of electrons. The three degenerate higher energy orbitals have one electron pair and two unpaired electrons.

    square planar

        SE = [0 + 2(0.23) - 2(0.43) -4(0.51)]Δo + 4PE

        SE = -2.44Δo + 4PE

    tetrahedral = same as before (ls = hs for d8 tetrahedral)

          ΔSE = SEsq - SEtd  = -2.44Δo + 3PE -(-0.36Δo + 3PE) = -2.08Δ+ PE

          This is an even more appreciable preference for square planar, although it is offset by pairing energy. Pairing energy would have to be twice as big as Δo in order to completely offset the LFSE.

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