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10.3.2: Orbital Splitting and Electron Spin

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    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.


    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








    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.


    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

    This page titled 10.3.2: Orbital Splitting and Electron Spin is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Chris Schaller.