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Octahedral vs. Tetrahedral Geometries

How do we tell whether a particular complex is octahedral, tetrahedral, or square planar? Obviously if we know the formula, we can make an educated guess: something of the type ML6 will almost always be octahedral (there is an alternative geometry for 6-coordinate complexes, called trigonal prismatic, but it's pretty rare), whereas something of formula ML4 will usually be tetrahedral unless the metal atom has the d8 electron configuration, in which case it will probably be square planar. But what if we take a particular metal ion and a particular ligand? Can we predict whether it will form an octahedral or a tetrahedral complex, for example? To an extent, the answer is yes... we can certainly say what factors will encourage the formation of tetrahedral complexes instead of the more usual octahedral.

The Crystal Field Stabilization Energy (CFSE) is the additional stabilization gained by the splitting of the orbitals according to the crystal field theory, against the energy of the original five degenerate d orbitals. So, for example, in a d1situation such as [Ti(OH2)6]3+, putting the electron into one of the orbitals of the t2g level gains -0.4 Δo of CFSE. Generally speaking, octahedral complexes will be favored over tetrahedral ones because:

  • It is more (energetically) favorable to form six bonds rather than four
  • The CFSE is usually greater for octahedral than tetrahedral complexes. Remember that Δo is bigger than Δtet (in fact, Δtet is approximately 4/9 Δo).

If we make the assumption that Δtet = 4/9 Δo, we can calculate the difference in stabilization energy between octahedral and tetrahedral geometries by referencing everything in terms of Δo.

Example \(\PageIndex{1}\): \(d^3\) Stabilized Structures

Which is the preferred configuration for a d3 metal: tetrahedral or octahedral?

SOLUTION

To answer this, the Crystal Field Stabilization Energy has to be calculated for a \((d^3\) metal in both configurations. The geometry with the greater stabilization will be the preferred geometry.

  • For a d3 octahedral configuration, the Crystal Field Stabilization Energy is

\[3 \times -0.4 \Delta_o = -1.2 \Delta_o\]

  • For a d3 tetrahedral configuration (assuming high spin), the Crystal Field Stabilization Energy is

\[-0.8 \Delta_{tet}\]

Remember that because Δtet is less than half the size of Δo, tetrahedral complexes are often high spin. We can now put this in terms of Δo (we can make this comparison because we're considering the same metal ion and the same ligand: all that's changing is the geometry)

So for tetrahedral d3, the Crystal Field Stabilization Energy is:

CFSE = -0.8 x 4/9 Δo = -0.355 Δo.

And the difference in Crystal Field Stabilization Energy between the two geometries will be:

1.2 - 0.355 = 0.845 Δo.

If we do a similar calculation for the other configurations, we can construct a Table of Δo, Δtet and the difference between them (we'll ignore their signs since we're looking for the difference between them).

Table \(\PageIndex{1}\): Crystal Field Stabilization Energies (not splitting parameters). This table compares the values of the CFSE for octahedral and tetrahedral geometries, assuming high spin configurations. The units are Δo, and we're assuming that Δtet = 4/9 Δo.
  Octahedral Tetrahedral Difference
d0, d5, d10 0 0 0
d1, d6 0.4 0.27 0.13
d2, d7 0.8 0.53 0.27
d3, d8 1.2 0.36 0.84
d4, d9 0.6 0.18 0.42

These values can be plotted:.

Figure \(\PageIndex{1}\): Crystal Field Stabilization Energy as a function of d-electrons for a hypothetical molecule in the octahedral (red dashed curve) and tetrahedral (black solid curve) geometries.

Notice that the Crystal Field Stabilization Energy almost always favors octahedral over tetrahedral in most cases, but the degree of favorability varies with the electronic configuration. In other words, for d1 there's only a small gap between the oct and tet lines, whereas at d3 and d8 there's a big gap. However, for d0, d5 high spin and d10, there is no CFSE difference between octahedral and tetrahedral. The ordering of favorability of octahedral over tetrahedral is:

d3, d8 > d4, d9> d2, d7 > d1, d6 > d0, d5, d10

The units of the graph are Δo. So if we have strong field ligands present, Δo will be bigger anyway (according to the spectrochemical series), and any energy difference between the oct and tet lines will be all the greater for it. A bigger Δo might also push the complexes over to low spin. Similarly, as we saw previously, high oxidation states and metals from the 2nd and 3rd rows of the transition series will also push up Δo.

On the other hand, if large or highly charged ligands are present, they may suffer large interligand repulsions and thus prefer a lower coordination number (4 instead of 6). Consequently if you set out to make something that would have a tetrahedral geometry, you would use large, negatively charged, weak field ligands, and use a metal atom with a d0, d5 or d10 configuration from the first row of the transition series (though of course having weak field ligands doesn't matter in these three configurations because the difference between oct and tet is 0 Δo). As Table 2 shows, you can find tetrahedral complexes for most configurations, but there are very few for d3 and d8.

Table \(\PageIndex{2}\): Tetrahedral complexes of different d electron counts
d0 MnO4- d5 MnCl42-
d1 TiCl4- d6 FeCl42-
d2 Cr(OR)4 d7 CoCl42-

Contributors

  • Dr Mike Morris, March 2001