What is octahedral and tetrahedral voids?

Answer 1

Octahedral vs. tetrahedral

So far, we've seen the Crystal Field Theory in action in octahedral, tetrahedral and square planar complexes. But 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 ML6will 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 ML4will usually be tetrahedral unless the metal atom has the d8electron 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.

Generally speaking, octahedral complexes will be favoured over tetrahedral ones because:

• It is more favourable to form six bonds rather than four

• The crystal field stabilisation energy is usually greater for octahedral than tetrahedral complexes.

Remember that Δois bigger than Δtet(in fact, Δtetis approximately 4/9 Δo). If we make the assumption that Δtet= 4/9 Δo, we can calculate the difference in stabilisation energy between octahedral and tetrahedral geometries by putting everything in terms of Δo.

For example: for a d3octahedral configuration, the CFSE is -1.2 Δo(refer back to theTableif you like).

For a d3tetrahedral configuration (assuming high spin), the CFSE = -0.8 Δtet

Remember that because Δtetis 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, CFSE = -0.8 x 4/9 Δo= -0.355 Δo.

And the difference in CFSE 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 aTableof Δo, Δtetand the difference between them (we'll ignore their signs since we're looking for the difference between them).

We can then plot these values on a graph.

Notice that:

• The CFSE favours octahedral over tetrahedral in most cases, but the degree of favourability varies with the electronic configuration. In other words, for d1there's only a small gap between the oct and tet lines, whereas at d3and d8there's a big gap

• For d0, d5high spin and d10, there is no CFSE difference between octahedral and tetrahedral.

• The ordering of favourability 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, Δowill 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 Δomight 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, d5or d10configuration 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 the following Table shows, you can find tetrahedral complexes for most configurations, but there are very few for d3and d8.

d0

MnO4-

d5

MnCl42-

d1

TiCl4-

d6

FeCl42-

d2

Cr(OR)4

d7

CoCl42-

That's about it for the crystal field theory.Lecture 4starts by discussing how we actually make transition metal complexes, and this leads on to a section about stability constants.

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This page was written by Dr Mike Morris, March 2001

Last updated March 2012

Answer 2

this is all the magic of arrangement............ of the layers...