The adjacent fig. shows the smaller positive ion of radius `r^(+)` is in contact with three larger negative ions of radii `r^(-)`. It can be seen that `AB = BC = AC = 2r^(-)`, `BD = r^(-) + r^(+)`. Further, the angle ABC is `60�`, and the angle DBE is `30�`.

By trigonometry `Cos 30�` `= (BE // BD).`

`BD =` (`BE //cos 30�`), `r^(+) + r^(-) = r^(-) //cos 30� =` `(r^(-) // 0.866`) `= r^(-) xx 1. 155`, `r^(+) = (1 .155 r^(-)) -r^(-)`

`= 0.155 r^(-).`

Hence `(r^(+) //(r^(-)) = 0. 155.`

Angle ABC is the tetrahedral angle of `109� 28'` and hence the angle ABD is half of this, that is `54�64'`. In the triangle ABD, `sin ABD

`= 0.8164 =AD // AB = (r^(-))/(r^(-) +r^(+))`

Taking reciprocals, `(r^(+) +r^(-))/r^(-) = 1/(0.8164) = 1.225`, rearranging, we get,

` r^(+)/r^(-) =0.225`

A cross section through an octahedral site is shown in the adjacent figure and the smaller positive ion (of radius `r^(+)` touches six larger negative ions (of radius `r^(-)`). (Note that only four negative ions are shown in this section and one is above and one below the plane of the paper).

It is obvious that `AB = r^(+) + r^(-)` and `BD = r^(-)` . The angle ABC is 45� in the triangle ABD. cos ABD = 0.7071 = `(BD // AB)`

`= r^(-)/(r^(+) +r^(-))`

Rearranging, we get,

` r^(-)/r^(+) = 0.414`

`text(Co-ordination number shape Example)`

`x < 0.155`, `2` Linear `BeF_2`

`0.155 < x < 0.225`, `3` Planar Triangle `AlCl_3`

`0.225 < x < 0.414`, `4` Tetrahedron `ZnS`

`0.414 < x < 0.732`, `4` Square planar `PtCl_4^(2-)`

`0.414 < x < 0.732`, `6` Octahedron `NaCl`

`0.732 < x < 0.999`, `8` Body centered cubic `CsCl`