VBT, CFT & LFT Theories - 3 | Chemistry Optional Notes for UPSC PDF Download

Tetrahedral complex of a 4th period Transition Metal

• Next, let us consider a tetrahedral complex and see if ligand field theory can explain it well. The point group for a tetrahedron is T_d, and we will need the character table of this point group (Fig. 7.1.32)
  
• We choose the coordinate system like in the crystal field theory. We inscribe the tetrahedron into a cube by occupying every other vertice of the cube with a ligand (Fig. 7.1.34). Then, we let the axes run perpendicular to the faces of the cube. Next, we need to think about the symmetry types of the metal frontier orbitals. For a fourth period transition metal these are the 3_d, the 4s, and the 4p orbitals. Looking into the character table of the point group Td gives us their symmetry types (Fig. 7.1.32 and Fig. 7.1.33). The 4s orbital has the totally symmetric symmetry type A₁, We find the letters x, y, and z in parentheses in the irreducible representation of the type T₂, and this means that the 4p orbitals are triply degenerated and have the symmetry type T₂. The 3d_xy, the 3d_yz, and the 3d_xz orbitals are found in the same irreducible representation, are also triple-degenerated, and have T₂ symmetry. The 3d_x²-_y² and the 3d_z² orbitals have the symmetry type E according to the character table for Td.
  
• Now we need to think about the ligands. Because we need to consider σ-bonding first, we have to find the HOMO of the ligand suitable for σ-bonding. We can choose again a carbonyl ligand as an example ligand, and in this case the HOMOs of the CO ligands would be used for σ-bonding. This is actually not immediately obvious. If we chose the coordinate system of the metal and ligands to be same same, then then the CO bond axis, which we previously defined as the z-axis, would not point toward the metal, and no σ-overlap with a metal orbital would be possible. Therefore, we must give each ligand a different coordinate system with the z-axes pointing toward the metal (Fig. 7.1.34). Only then, the CO molecule would point toward the metal, and would be oriented to make σ-bonding with the metal. Generally, when constructing MOs, the ligands should always be oriented to that bonding is maximized. This is usually the case when orbital overlap for σ-bonding is maximized.
  
• Because we have four ligands we will have four ligand HOMOs that we would group to form four ligand group orbitals. Which symmetry types do they have? In order to obtain the symmetry types we would need to determine the reducible representation first, and then the irreducible representations this reducible representation is composed of. We won’t do this explicitly here and not go through the entire mathematical process, and only consider the results.
• The result is that one ligand group orbital has the symmetry type A₁ and the other three have the symmetry type T₂ (Fig. 7.1.35).
  
• We can now construct the molecular orbital diagram for σ-bonding (Fig. 7.1.36). First, we can plot the metal frontier orbitals according to the energy on the left side of the diagram, and label the orbitals according to their symmetry types. On the right side we plot the four ligand group orbitals, and label them according to their symmetry type. Their energy should be approximately that of the metal d-orbitals. Now we can construct molecular orbitals by combining atomic and ligand group orbitals of the same symmetry type. The 4s orbital and one LGO have A₁ symmetry, and therefore we can combine them to form a bonding 1a₁ and an anti-bonding 2a₁ orbital. Next, let us direct our attention to the t2 symmetry type. The 4p and three metal d orbitals have that symmetry type, and so have three LGOs. That is overall nine orbitals, so we must get nine MOs with t₂ symmetry. Because of triple-degeneracy, there must be three sets of triple-degenerate MOs. We can estimate that one set will be bonding, one will be approximately non-bonding, and one will be anti-bonding. We can label them 1t₂, 2t₂, and 3t₂ respectively. The 2t₂ is actually somewhat anti-bonding in nature. Now only the metal e-orbitals are left. They do not find a partner and remain non-bonding.
• We still need to fill the electrons into the MOs. The four ligand HOMOs are considered full, which gives 4×2=8 electrons. The electrons occupy the bonding 1t₂ and the 1a₁ orbitals explaining the four dative metal-ligand bonds. All metal d electrons, which could be up to ten, would need to go into the e and/or the t₂-orbitals. The e-orbitals are the non-bonding d_z² and d_x²-_y² orbitals, and the 2_t² orbitals are only weakly anti-bonding and have strong d-metal orbital character because they are have been constructed from the d_xy, the d_yz, and the d_xz orbitals, and are fairly similar in energy to those orbitals. We can therefore say that the the e and the 2t₂ orbitals are the metal t₂ and eg orbitals in a tetrahedral ligand field. The energy difference between the e and the 2t₂ orbitals is the tetrahedral ligand field energy Δ_T. You see here again the relationship between ligand and crystal field theory. You can also see that ligand field theory can explain why crystal field theory works as a bonding theory even though it is not an actually bonding theory.

π-bonding in a tetrahedral complex

• What is the π-bonding in a tetrahedral complex?
  
• First, we need to decide if there are ligand orbitals that are suitably oriented to overlap with metal orbitals in π-fashion. We can see that no ligand orbital is overlapping with a metal d-orbital exactly in π-fashion, however, the e π and 2e π*-orbitals of the ligands still overlap so that at bonding interaction is created, and this orbital overall occurs similarly to π-overlap. Therefore, we can still approximate the bonding interactions as π-bonding. We need to consider though that because of smaller orbital overlap, the π-bonding in tetrahedral complexes is weaker than in octahedral complexes. How many ligand orbitals will we need to consider? Assuming that we will stick with CO as the ligand, there will be four per ligand, and thus there will be 4×4=16 overall. Of these, there will be eight bonding e π-orbitals and eight anti-bonding e π*-orbitals. We group the bonding ones to form a set of eight ligand group orbitals, and group the anti-bonding ones to form another set of eight anti-bonding ligand group orbitals. What will be their symmetry types? We can determine the reducible representations and irreducible representations to find this out. We are not going through the exact process here, but only look at the results. The result is that each set of ligand group orbitals has the two E-type orbitals, three T₁-type, and three T₂-type ligand group orbitals.
  
• Since we now know the symmetry of the ligand group orbitals we can combine them with same-symmetry metal d-orbitals in the tetrahedral ligand field to form π-molecular orbitals. Let us do this for the example of nickel tetracarbonyl, a tetrahedral complex of Ni in the oxidation state 0 and four carbonyl ligand. We can use the MO diagram for σ-bonding which we constructed previously as a starting point, and modify it so that it accounts for π-bonding (Fig. 7.1.38).
• First, we need to add the new ligand group orbitals to the MO diagram. We only need to consider the E and T₂ LGOs and not the T₁ LGOs because no metal orbital has t₁ symmetry. Next, we need to take into account that the CO ligand is a strong π-acceptor ligand. This means that we only need to consider the LGOs formed from the anti-bonding π*-orbitals. We must draw these orbitals above the LGOs for σ-bonding into the MO diagram because the LGOs for σ-bonding have been formed from the CO HOMOs while the anti-bonding LGOs for π-bonding have been formed from the CO LUMOs. We can now combine the π-LGOs and metal d-orbitals of the same symmetry to form π-MOs. We see that we can combine the e-type LGOs with the non-bonding e-type metal d_z2 and d_x²-_y² orbitals to form a pair of bonding MOs of e-symmetry, and an anti-bonding pair of e-symmetry, which we can label 1e and 2e, respectively. Now what is the effect on the T₂-type π-LGOs? We first need to realize that we have already three sets of triple-degenerated t2-MOs generated through σ-interactions. The interaction of the T₂-LGOs with the σ-t₂-MOs must create another set of triply-degenerated orbitals so that the overall number of orbitals with t₂-symmetry remains the same. The interaction will occur mostly between the 2t₂ MO and the T₂-LGOs because the 2t₂ MOs have the most similar energy to the T₂-LGOs. This leads to the lowering of the energy of the 2t₂ MOs, so that these formerly weakly anti-bonding orbitals become weakly bonding. The additionally created t₂-orbitals are weakly anti-bonding. We can label them 3t₂. The former anti-bonding 3t₂ –orbital will be re-labeled 4t₂. We can double-check that we have constructed the t2-MOs correctly by verifying that the number of T₂-LGOs, and that includes σ and π, plus the number of metal T₂ orbitals equals the number of t2 molecular orbitals. We see that the number of metal T₂ orbitals + the number of T₂ LGOs is 6+6=12. The number of t2 MOs is 4×3 which is also 12.
• Finally, we still need to fill the electrons. As previously mentioned, the π-LGOs as empty and therefore the ligand does not contributed any electrons to the π-bonding. The Ni is in the oxidation state 0 and thus contributes 10 electrons because Ni is in the 10th group of the periodic table. Because a neutral Ni atom has the electron configuration 3d84s2 we can draw eight electrons in to the 3d orbitals and two electrons into the 4s orbitals. The 1t₁ and the 1a₁ MOs are already full with ligand electrons due to σ-bonding. Thus, the metal electrons must go into the 2e and the 2t₂ MOs. Both the e-orbitals and the 2t₂-orbitals are bonding, and thus we can say that the metal d-electrons have been stabilized due to the π-acceptor nature of the ligand. We see here an analogy to the octahedral ligand field. Like in the octahedral ligand field π-acceptor ligands tend to lower the energy of the metal d-electrons. We can also ask: What is the effect of a π-acceptor ligand on Δ_T? The answer is: Like Δo gets larger in the octahedral ligand field, Δ_T gets larger in a tetrahedral ligand field. 
• However, the increase is much smaller compared to the octahedral field. Why? this is because in the octahedral ligand field, the energy of the eg orbitals is not affected through the π-acceptor interactions, and only the t_2g orbitals are lowered in energy. In the case of the tetrahedral ligand field both the energy of the e and the t₂ orbitals get lowered, and the energy of the e-orbitals gets only slightly more lowered than the energy of the t₂ –orbitals. Therefore, the Δ_T only slightly increases. This fact can also serve as an additional explanation why tetrahedral complexes never make low-spin complexes, even with strong π-accepting ligands. The effect of the ligand on Δ_T is simply still too small because both the energy of the e and the t2 orbitals have been lowered.
• Finally, let us discuss the results in the context of the 18 electron rule. We can see that the number of electrons involved in σ-bonding are in bonding molecular orbitals. Thus, it is justified to say that the ligand field theory is able to explain the 18 electrons rule.
• Now let us think about how a π-donating ligand influences the magnitude of Δ_T. In this case we only need to consider the bonding E and T₂ ligand orbitals if the ligand has π-bonds. Those orbitals will be energetically below the LGOs for σ-bonding. If the ligand is a simple ion like chloride, then we would consider LGOs from the filled p-orbitals that that have suitable symmetry for π-overlap. Those LGOs would have the essentially the same energy as the σ-LGOs. Let us construct a qualitative MO diagram of TiCl4 which has chloro ligands which are typical π-donating ligands. When constructing the MO diagram for π-bonding we can again start with the MO-diagram for σ-bonding, and then modify this diagram. We can first plot the filled π-T₂ and Eg LGOs with similar energy as the σ-LGOs on the right side of the diagram. The E ligand group orbitals will now interact with the e-metal orbitals. That interaction leads to a pair of bonding MOs of e-symmetry, and a pair of anti-bonding MOs of e-symmetry. This means that effectively, the formerly non-bonding e-metal orbitals become anti-bonding MOs, and move up in energy. 

• The bonding e-type MO is created in addition and must have lower energy than the E-type π-LGOs. The interaction of the T₂-LGOs with the metal t₂-orbitals creates an additional set of bonding t₂-orbitals. The 2t₂ orbitals move up in energy, and become more anti-bonding, as a consequence of that. We can just renumber the t₂ and the e MOs. Now we still need to fill the electrons into the MOs. The π-LGOs have overall eight electrons. These electrons will go into the newly formed bonding 1e and 1t₂ molecular orbitals. We see that this leads to a stabilization of these electrons. This is an analogy to the octahedral ligand field. When we have a π-donor ligand, then that the π-ligand electrons. What about the metal electrons. In TiCl₄ the Ti is in the oxidation state +4, therefore it is formally d0, and it does not have any electrons. This explains the stability of the TiCl₄ as complex that does not obey the 18 electron rule. It has only 8 electrons, but the bonding situation is nonetheless ideal. All bonding MOs are full, and all others are empty. If the Ti had d electrons they would need to go into the 2e and the 3t₂ orbitals which are both anti-bonding. This would destabilize the complex. In general π-donating ligands increase the energy of metal d-electrons, and this is another analogy to the octahedral ligand field. Now to the question, what is the effect of a π-donating ligand on Δ_T? Both the metal e and t₂ frontier orbitals have increased in energy but the e-orbitals more so than the t₂ orbitals. That means that the Δ_T has overall decreased. This is a further analogy to the octahedral ligand field. π-donating ligands lead to a smaller Δ_T.

• Overall the ligand field theory can again provide an explanation for the spectrochemical series. π-donors lead to smaller Δ_T ligands, and thus the complex absorbs light of larger wavelengths, π-acceptors lead to larger Δ_T and those complexes absorb light of smaller wavelengths.

Square Planar Complex of a 4th Period Transition Metal

As a last example we will discuss now the molecular orbital diagram of a square planar complex. This will be the most complicated MO diagram we will discuss. The greater complexity stems from the lower symmetry in a square planar complex. It belongs to the point group D_4h whereas tetrahedral and octahedral complexes belong the high symmetry point groups T_d and O_h.
The lower symmetry leads to a reduction in the degeneracy of the molecular orbitals. This creates more energy levels and hence a more complicated molecular orbital diagram. Let us think next about the definition of the axes. I would be sensible to define the xy plane as the plane of the molecule with the x and the y-coordinates going through the bonds. The z-axis would stand perpendicular to the plane of the molecule. What are the symmetry types of the metal frontier orbitals. Assuming a 4th period transition metal the 3d, the 4s, and the 4p orbitals would be the frontier orbitals. Looking into the character table of the point group D4h (Fig. 7.1.10) shows that the symmetry of the 4s orbital is A_1g, the symmetry of the 4p orbitals is A_2u for the 4p_z orbital and eu for the 4p_x and the 4p_y orbitals. The d orbitals have the symmetry types A_1g, B_1g, B_2g, and E_g for the 3d_z², d_x²-_y², 3d_xy, and 3d_xz/3d_yz, respectively.

Next, we to determine the ligand HOMOs suitable for σ-bonding. There are no square planar carbonyl complexes, therefore we will not choose the carbonyl ligand as a example ligand here. Instead we will use the related cyano ligand. The cyano ligand is isoelectronic with the carbonyl ligand. Formally, the O-atom in CO is replaced by N. Because N has one electron less than O, we must add one electron, giving the cyano ligand a 1- negative charge. Like in CO, the cyano ligand has a triple bond between C and N, and there is an electron lone pair at C, and one at N. The MO diagram is similar to that of CO, just the energy difference between C and N orbitals is somewhat smaller than that of the energy difference between C an O orbitals (Fig. 7.1.41). The number, symmetry, and energy order of the MOs of CN- and CO is the same. Therefore, the CN- has the same HOMOs suitable for σ-bonding, and we can select it for the construction of σ-MOs. Each ligand has one HOMO, therefore we have overall four HOMOs.

We can group them to form LGOs and determine the symmetry types of the LGOs. The result is that there is on ligand group orbital with A_1g symmetry, there are two with E_u symmetry, and there is one with B_1g symmetry (Fig. 7.1.42).

Square Planar Complex: σ-bonding

• We have now the all information necessary to construct a qualitative molecular orbital diagram for σ-bonding (Fig. 7.1.43). As usual we first plot the metal frontier orbitals on the left and the ligand group orbitals on the right side of the MO diagram, and label the symmetry types. Then, we need to combine metal frontier and ligand group orbitals of the same symmetry type to form molecular orbitals. We can for instance begin with the symmetry type A_1g. The 4s orbital and the 3d_z2 orbitals have this symmetry type, and so has one ligand group orbital. We will therefore get three molecular orbitals of that symmetry type, a bonding one, an approximately non-bonding one, and an anti-bonding molecular orbital. 
• We plot them according to their expected energy into the MO diagram, and label the MOs 1a_1g, 2a_1g, and 3a_1g, respectively. Next, we can look at the B_1g orbitals. The 3d_x²-_y² orbital and one LGO of that symmetry type, and thus we can form a bonding and an anti-bonding MO from this combination. We can further see that we can combine the 4px and the 4py orbitals that have Eu symmetry with the remaining two LGOs of the same symmetry. This gives two double-degenerated bonding MOs of eu symmetry, and two anti-bonding bonding ones of the same symmetry. Are there any orbitals with other symmetry types left? Yes, there are the metal dxy, dxz, and dyz orbitals that have B2g, and Eg symmetry, respectively, In addition, there is the pz orbital that has a2u symmetry. We can see that these metal orbitals do not find a partner. There is no LGO of these symmetry types. We therefore have to draw these orbitals as non-bonding orbitals into the MO diagram.
• Now it is the time to fill the electrons into the molecular orbitals. The LGOs are occupied with electrons because they have been formed from the HOMOs of the ligand. Because we have four LGOs we have eight electrons to consider. These electron will therefore fill the 1a₁, the 1b_1g, and the 1eu molecular orbitals. These orbitals are all bonding molecular orbitals. Thus, ligand field theory is able to explain the four bonds and the square planar shape of the molecule. Where would the metal d-electrons go? We could have up to ten d-electrons depending on the metal ion, and these ten electrons would go into the five MOs of the next higher energy. 
• These are the non-bonding b_2g and eg orbitals, the approximately-non-bonding 2a_1g orbital, and the anti-bonding 2b_1g orbital. We can consider these orbitals as the metal-d-orbitals in a square planar ligand field. The b_2g orbitals is the dz₂ orbital in the square planar ligand field, the eg orbitals are dxz and dyz in a square planar ligand field, the 2a_1g orbital is the dz₂ orbital in a square planar ligand field and the 2b1g orbital is the d_x²-_y² orbital in a square planar ligand field. We can see that these results are similar yet not completely analogous to those we obtained in the crystal field theory. Remember, that in the crystal field theory the d_x²-_y² orbital had also had the highest energy. However, the d_z² was lower in energy than the dxy, and the dxy was not degenerate with the dxz and the d_yz orbitals. The difference it due to the fact that in ligand field theory the dxy orbital is considered non-bonding, not interacting with the ligands at all, whereas in crystal field theory a strong electrostatic repulsion between the d_xy and the ligands is assumed because the dxy orbital is in the plane of the molecule.

Square Planar Complex π-bonding

Now let us look at the π-bonding of a square planar transition metal complex (Fig. 7.1.44). If we assume a CN- ligand, then the orbitals that have suitable symmetry to overlap with the metal frontier orbitals in π-fashion are the 1e and 2e π and π* orbitals from the MO diagram of CO. There are two π and two π* orbitals per ligand. Because we have four ligands there are overall eight π and eight π* orbitals. Now we need to consider that four of the π-orbitals are in the xy plane, while the other four are above and below the xy plane. We call the former the parallel orbitals, and the latter the perpendicular orbitals, because they are oriented parallel and perpendicular to the xy plane, respectively. We must distinguish them because they have a different symmetry and overlap differently with the metal d-orbitals. This is illustrated for the bonding π-orbitals in Fig. 7.1.44. The anti-bonding ones behave analogously. Therefore we group the orbitals to form four sets of ligand group orbitals π_⊥ , π_||, π*_⊥ , and π*_||. Our next task is to determine the symmetry types of these orbitals. We again omit the exact process, but only look at the results. Both the π_⊥ and the π*_⊥ orbitals have one A_2u, two E_g and one B_2u ligand group orbitals. The π_|| and π*_|| orbitals have one A_2g, two E_u, and one B_2g ligand group orbitals.

We now need to add the π-molecular orbitals to the molecular orbital diagram that we had constructed previously to account for σ-bonding. First, we need to consider which π-LGOs have the right symmetry to produce π-MOs with metal d-orbitals (Fig. 7.1.45). We can see that among the parallel LGOs the A_1g and the B_1g orbitals find partners, but not the Eu orbitals, For the perpendicular LGOs the Eg orbitals find partners, but not the A_2u and the B_2u orbitals. We therefore only need to consider the A_1g, the B_1g and the Eg LGOs. To make matters more complicated, to get a realistic MO diagram for a square planar complex with cyano ligands it is necessary to consider the interactions of the 4s and the 4p orbitals with the π-LGOs as well. The symmetry of the 4s was A_1g and the symmetry of the 4p orbitals was A_2u and E_u, respectively. The 4s still does not find a partner, but the 4p orbitals do.

MO Diagram of a Square Planar Complex With a Cyano Ligand

• Let us now construct the MO diagram of a square planar complex with a cyano ligand considering both σ and π-bonding for the example of the Ni(CN)₄^2- complex anion. Because of the complexity of the diagram we will not modify the MO diagram for σ-bonding, like we did previously, but construct the complete MO diagram from scratch (Fig. 7.1.46). Like previously we can plot the metal 3d, 4s, and 4p orbitals on the left side of the diagram. On the right side of the diagram are now the LGOs for σ as well as π-bonding. For the Ni(CN)₄^2- complex the σ-LGOs are energetically somewhat below the metal d-orbitals, the bonding π-LGOs have about the same energy as the metal d-orbitals, and the anti-bonding π*-LGOs have an energy significantly higher than the 4p orbitals. We can again construct MOs symmetry type by symmetry type until the MO diagram is complete. We can start with the symmetry type A_1g. We see there is only one σ-LGO that interacts with the 4s and the 3d_z² which also have A_1g symmetry. That leads to a bonding, an approximately non-bonding, and anti-bonding a1g MO which we can label 1a_1g, 2a_1g, and 3a_1g, respectively. Next, we can for instance look at the B2g orbitals. There is one d-orbital that has that symmetry, as well as two LGOs. The interaction is almost only with the bonding B_2g LGO because the energies are similar. This leads to a bonding and an anti-bonding π-MO with B_2g symmetry. 
• The anti-bonding LGO is too high in energy and it does not significantly interact. Therefore we can draw it with the same energy into the MO-diagram. Now let us go to the B1g orbitals. There is a metal d-orbital with B_1g symmetry, and a σ-LGO with B1g symmetry. This produces a bonding and anti-bonding MO with B_1g symmetry. Now let us go to the Eg orbitals. There are two Eg metal d-orbitals, two Eg π-LGOs and two Eg π*-LGOs. Again, the π*-LGOs are too far off in energy, so that the bonding interaction is mostly between metal orbital the π-LGOs. This leads to a pair of bonding MOs and a pair of anti-bonding MOs. The π*-LGOs can be drawn with unchanged energy into the MO-diagram. The 4p orbitals of the metal have symmetry types, we have not considered yet: A2u and Eu. Let us begin with the Eu orbitals. There are two σ-Eu LGOs, two π-Eu-LGOs and two π*-E_u LGOs. Overall we have four pairs of EU-orbitals, and we would therefore expect four pairs of eu-MOs. We can make the approximation that there will be a bonding one, a weakly bonding one, a weakly anti-bonding one, and a strongly anti-bonding one. We can label them 1e_u, 2e_u, 3e_u, and 4eu, respectively. The 4eu has almost only contributions from the anti-bonding π*-EU LGOs due to energy arguments, therefore it is only connected to these via a dotted line, and not to the other E_u orbitals. Now to the A2_u orbitals: There is a π and a π*-orbital of A2_u symmetry that can be combined with the 4p A2_u orbitals. This leads to three A2_u MOs overall, one bonding, one approximately non-bonding, and one anti-bonding one. Now we have used all metal d-orbitals. We notice that there are still some LGOs that we have not used. These are the π and π* A_2g and the B_2u LGOs. They remain non-bonding can be drawn accordingly in the MO diagram.
• We now need to fill the electrons into the MO-diagram. Firstly, there are the σ-electrons to consider. There are overall eight electrons stemming from the four occupied σ-LGOs. We can fill these eight electrons into the four MOs with the lowest energy. These are the 1_a1, the 1e_u and the 1b1. These are all bonding MOs explaining the four σ-bonds within the complex. We can call these four orbitals the σ-bonding orbitals. Next we need to consider the electrons in the π-LGOs. There are overall eight of these orbitals containing 16 electrons. We must fill these electrons into the eight MOs of the next-higher energies. These are the 1b_2g, the 1e_g, the 1a_2u, the 2e_u , the 1b_2u, and the 1a_2g orbitals. We see that these orbitals are either bonding or non-bonding, and thus overall the ligand π-electrons get stabilized due to the π-interactions with the ligand. We can call these orbitals the π-bonding molecular orbitals. Finally we need to look at the metal d-electrons. We assume Ni as the metal with Ni in the oxidation state +2 and d8 electron configuration. We need to fill these eight electrons into the four next higher-energy MOs. This fills the 2b2g, the 2eg, and the 2a1g, orbitals which are either non-bonding or anti-bonding. This means that the metal electrons are overall destabilized. This behavior is consistent with the cyano-ligand acting as a π-donating ligand. 
• We understand from this why square planar complexes prefer 16 over 18 electrons and disobey the 18 electron rule. If we had two more electrons then these would need to go into the energetically much higher 2a_2u orbitals. The energy gap between the 2a_2u and the 2a_1g orbital is significantly higher than that between the 1a_1g, the 2b_2g, and the 2e_g orbitals. Overall we can call the 2au, the 2a1g, the 2eg, and the 2bg orbitals the metal d-orbitals in a square planar ligand field. In a square planar ligand field we have three ∆s. ∆₃ is the energy difference between the 2b_2g and the 2e_g orbitals, ∆₂ is the energy difference between the 2e_g and the 2a_1g orbitals, and ∆₁ is the energy between the 2a_1g and the 2a_2u orbitals. ∆₁ is much larger than the other two. It also almost always larger than the spin-pairing energy, therefore square planar complexes are almost always low-spin complexes.
• Overall, we see that an MO diagram can become very complex even for a relatively simple molecule of relatively high symmetry. It should be noted that for MO diagrams as complex as this one, it is not possible any more determine the exact order of the energy levels just by qualitative inspection, and without exact computations. Nonetheless, you can see that it is possible to readily understand a complex MO diagram by following the symmetry-adapted linear combination of atomic orbitals approach. We just need to go through the symmetry types step by step and construct MOs by combining orbitals of the the same symmetry type until the process is completely. We have now reached the end of the chapter on ligand field theory.