The structure of many ionic solids can be accounted for by considering the relative sizes of the positive and negative ions, and their relative numbers.

Simple geometric calculations allow us to workout, as to how many ions of a given size can be in contact with a smaller ion.

Thus, we can predict the coordination number from the relative sizes of the ions.

### Co-ordination Number 3

The adjacent fig. 9(a) 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 × 1.155,

r+ = (1.155 r) – r

r= 0.155 r

Hence (r+ / r) = 0.155

### Co-ordination Number 4 (Tetrahedron)

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

$\large = \frac{r^-}{r^+ + r^-}$

Taking reciprocals,

$\large \frac{r^+ + r^-}{r^-} = \frac{1}{0.8164} = 1.225$

Rearranging, we get,

$\large \frac{r^+ }{r^-} = 0.225$

##### Co-ordination Number 6 (Octahedron) or 4 ( Square Planar)

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)

$\large = \frac{r^-}{r^+ + r^-}$

Rearranging, we get,

$\large \frac{r^+}{r^-} = 0.414$