If we want to compare the stability of a particular electron configuration compared to the imaginary d electron configuration in a spherical electric field, we can calculate the ligand field stabilization energy. Remember, in crystal field theory, we compared the d electrons in a spherical field to the situation in which ligands approached in an octahedral geometry. The e g level is destabilized by 0.6Δ o compared to undifferentiated d orbitals in a spherical field, whereas the t 2g level is 0.4Δ o lower in energy than the d orbitals in a spherical field. Let’s look at the case of a d 4 complex. That’s an interesting case, because as we fill in the d electrons one by one, it is the first example in which there is the possibility of either a high spin or a low spin configuration. If we first consider the high-spin case, then we see that three of the electrons drop into the t 2g level and the last one goes into the e g level. The ligand field stabilization energy is the difference between energy in a spherical field and in an octahedral field. In the high spin d 4 case, that means three electrons are lower in energy and one is higher in an octahedral environment. High spin d 4 : \[ \begin{align*} \text{LFSE} &= [0.6 (1) – 0.4 (3)]Δ_o \\[4pt] &= [0.6 – 1.2]Δ_o \\[4pt] &= -0.6Δ_o \end{align*} \nonumber \] 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 d 4 : \[ \begin{align*} \text{LFSE} &= [0.6 (0) – 0.4 (4)]Δ_o \\[4pt] &= -1.6Δ_o \end{align*} \nonumber \] 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 d 4 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 d 0 , d 5 and d 10 (\(\ce{Ca^{2+}}\), \(\ce{Mn^{2+}}\) 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. Electron count (dn) LFSE (Δo) 0 0.0 1 -0.2 2 -0.4 3 -0.6 4 -0.3 5 0.0 6 -0.2 7 -0.4 8 -0.6 9 -0.3 10 0.0 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 M 3+ ions. Once again, we observe the overall agreement with Coulomb’s Law, and we see that the d 0 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 d 5 , rather than the d 4 \(\ce{Mn^{3+}}\). 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 \(\PageIndex{1}\) Draw a diagram showing the energetic differences between d 7 metal ions in a spherical field, a weak field octahedral environment, and a strong field octahedral environment. Solution Example \(\PageIndex{2}\) Show how you would calculate the ligand field stabilization energy in the following cases: Low-spin d 5 High-spin d 5 Low-spin d 7 High-spin d 7 Solution Low-spin d 5 : LFSE = [0.6 (0) – 0.4 (5)]Δ o = -2.0Δ o High-spin d 5 : LFSE = [0.6 (2) – 0.4 (3)]Δ o = [1.2 - 1.2]Δ o = 0Δ o Low-spin d 7 : LFSE = [0.6 (1) – 0.4 (6)]Δ o = [0.6 - 2.4]Δ o = -1.8Δ o High-spin d 7 : LFSE = [0.6 (2) – 0.4 (5)]Δ o = [1.2 - 2.0]Δ o = -0.8Δ o Example \(\PageIndex{3}\) Calculate the difference between low spin and high spin electronic configurations in the following cases: d 5 d 7 Solution d 5 : ΔE low spin – high spin = (-2.0Δ o + 2Π c + 4Π e ) - (0Δ o + 4Π e ) = -2Δ o d 7 : ΔE low spin – high spin = (-1.8Δ o + 3Π c + 6Π e ) - (-0.8Δ o + 2Π c + 5Π e ) = -Δ o + Π c + Π e