How do you calculate the coordination number of the ions in a given compound?

Question

How do you calculate it? (Cn being the number ions of the species which surround the other ion
in a lattice, i.e. in CsCl, Cl's Cn is 8, so for every Cs+ ion there are 8 Cl- ions around it)

How do you calculate it? (Cn being the number ions of the species which surround the other ion in a metallic lattice, i.e. in CsCl, Cl's Cn is 8, so for every Cs+ ion there are 8 Cl- ions around it)

right so its dependant on the structure, but how would you find the structure, and therefore coordination no of each species in a given salt? I they are both ions, so VSEPR is no good.

Answer 1

The is no way to calculate it in the way you are talking about. First of all you have to identify the structure of the crystal, there are some rough guildlines as to which one is most likely, but you can consult a table online.

http://cst-www.nrl.navy.mil/lattice/

here they are in 3-d.

Rough guildlines would be:

are the ions a similar size?

If not what is their size ratio?

Can the large ions form a simple structure and the smaller ions fit into the holes? (google: hole size ratio)

Once you know a few off the top of your head you can make a good guess about which structure a crystal has based on its similarity to simple ones.

Then once you know the structure the coordination number is simple to see.

Any questions message me.

Answer 2

oxidation number of what exactly? presumably the metal? Well it is really just like other compounds, but you include the oxidation number of the ligand. eg [CuCl4-] 2- so Cu + (-1 x4) = -2 (charge on ion) so Cu is +2 here or [Cu(en)2]Cl2 so as en (1,2 diaminothane is a neutral molecule) it counts as zero so Cu +(0 x2) + (-1 x2) = 0 so Cu is again +2 another example .. how about Cr in oxalalato-tetrammino- chromium nitrate? Formula Cr(4NH3.C2O4)NO3 Ammonia is neutral molecule, oxalate [ethanedioate ion] is -2 ion, nitrate -1 ion so Cr + (4 x0) + (-2) + -1 = 0 so it is chromium III , 3+

Answer 3

It depends on the crystal habit. Two common crystal habits are body-centered cubic and face-centered cubic. In a body-centered cubic crystal, imagine a cube with a metal ball, M, in the center and Cl- balls at each of the eight corners of the cube. Therefore, the coordination number of M is 8.

Answer 4

not sure