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If I am given a compound which belongs to, say the cubic $\text{Ia}\bar 3\text{d}$ space group, is there a known way to calculate the lattice parameter of the compound? In general, can lattice parameters be calculate from knowing what space group the compound belongs to?

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  • $\begingroup$ On the other hand, lattice parameters and crystal system can be known independent of space group and atoms by indexing a diffraction pattern. $\endgroup$ – wismuthaft Nov 15 '15 at 10:52
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No - the spacegroup only gives information about symmetry, not the lattice parameters.

The lattice parameters depend on what is in your unit cell, ie your compound - but even if you know your compound, this is not straightforward, as you don't know its arrangement. And if you do, you should already know your lattice parameters ;)

That being said: if you have additional information - e.g. the radius of a protein you crystallise - you might be able to get a rough estimate of what you are expecting as your minimal unit cell parameters. But this does not take into account any stoichiometry in the crystal, which is impossible to know in advance, so multiples thereof are just as likely.

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    $\begingroup$ Indeed, in full support of @Gerhard's answer, I will point out that there are a number of elemental metals that crystallize in the fcc structure, but they all have different lattice parameters. The symmetry of the interatomic potentials is distinct from the length scale of the potential. $\endgroup$ – Jon Custer May 5 '15 at 18:59
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    $\begingroup$ So, the Wyckhoff positions don't determine the coordinates of where the atoms are in a compound that fits in a space group with those Wyckhoff positions? $\endgroup$ – Samuel Reid May 5 '15 at 19:42
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    $\begingroup$ No - the Wyckoff positions are still only part of the symmetry dictated by the spacegroup. $\endgroup$ – Gerhard May 6 '15 at 13:13
  • $\begingroup$ @Sam:I suspect that you are referring to the asymmetric unit: if you know what is in it, you could calculate the lattice parameters. But as I mentioned above, you do not know this in advance - otherwise you would already have determined the structure. $\endgroup$ – Gerhard May 6 '15 at 16:20

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