In a tetrahedral arrangement of four ligands around a metal ion, none of the ligands lies on any of the three coordinate axes (illustrated in part (a) in Figure \(\PageIndex{2}\)); consequently, none of the five d orbitals points directly at the ligands. Nonetheless, the d_{xy}, d_{xz}, and d_{yz} orbitals interact more strongly with the ligands than do d_{x2−y2} and d_{z2}_{}; this again results in a splitting of the five d orbitals into two groups. The splitting of the energies of the orbitals in a tetrahedral complex (Δ_{t}) is much smaller than that for an octahedral complex (Δ_{o}), however, for two reasons: first, the d orbitals interact less strongly with the ligands in a tetrahedral arrangement; second, there are only four negatively-charged regions rather than six, which decreases the electrostatic interactions by one-third if all other factors are equal. It can be shown that for complexes of the same metal ion with the same charge, the same ligands, and the same M–L distance, \(\Delta_\textrm t=\frac{4}{9}\Delta_\textrm o\). The relationship between the splitting of the five d orbitals in octahedral and tetrahedral crystal fields imposed by the same ligands is shown schematically in part (b) in Figure \(\PageIndex{2}\).

Example \(\PageIndex{1}\): Predicting Structure

For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.

- \([Cu(NH_3)_4]^{2+}\)
- \([Ni(CN)_4]^{2−}\)

**Solution**

Because Δ_{o} is so large for the second- and third-row transition metals, all four-coordinate complexes of these metals are square planar due to the much higher crystal field stabilization energy (CFSE) for square planar versus tetrahedral structures. The only exception is for d^{10} metal ions such as Cd^{2}^{+}, which have zero CFSE and are therefore tetrahedral as predicted by the VSEPR model. Four-coordinate complexes of the first-row transition metals can be either square planar or tetrahedral; the former is favored by strong-field ligands, whereas the latter is favored by weak-field ligands. For example, the [Ni(CN)_{4}]^{2−} ion is square planar, while the [NiCl_{4}]^{2−} ion is tetrahedral.

1.

The copper in this complex is a d^{9} ion and it has a coordination number of 4. So it is probably either square planar or tetrahedral. To estimate which, we need to fill in the CFT splitting diagrams for each with 9 electrons and ask which has a lower energy. Comparing the square planer (Figure \(\PageIndex{1}\)) splitting diagram with tetrahedral (Figure \(\PageIndex{2}\)), suggests that 9 electrons will have a net lower total energy for square planar (since the \(d_{x^2-y^2}\) orbital is high in energy, the others are lower).

For the square planar structure, neither high nor low spin states are possible (only one state) with a single unpaired electron.

2.

The nickle in this complex is a d^{8} ion and it has a coordination number of 4. So it is probably either square planar or tetrahedral. To estimate which, we need to fill in the CFT splitting diagrams for each with 8 electrons and ask which has a lower energy. Comparing the square planer (Figure \(\PageIndex{1}\)) splitting diagram with tetrahedral (Figure \(\PageIndex{2}\)), suggests that 8 electrons will have a net lower total energy for square planar (since the \(d_{x^2-y^2}\) orbital is high in energy, the others are lower).

For the square planar structure, it is a low spin complex since a high spin requires a lot of energy to promote to the \(d_{x^2-y^2}\) orbital. Hence, there are no unpaired electrons