Chemistry > Solid State > 5.0 Calculation of Spaces occupied i.e., Packing Fraction
Solid State
1.0 Classification of Solids
2.0 Types of Lattice
3.0 Calculation of number of particles in a Unit Cell
4.0 Close Packing in Crystals
5.0 Calculation of Spaces occupied i.e., Packing Fraction
6.0 Calculation of density of a cubic crystal from its edge
7.0 Classification of Ionic Structures
8.0 Imperfection in Solids
9.0 Properties of Solids
10.0 Silicates
5.1 Radius Ratio Rules
The structure of many ionic solids can be accounted by considering the relative sizes of the cation and anion, and their relative numbers. By simple calculations, we can work out as how many ions of a given size can be in contact with a smaller ion. Thus, we can predict the co-ordination number from the relative size of the ions.
Calculation of some limiting radius ratio values
$1.$ Co-ordination Number $3$:
The figure shows the smaller positive ion of radius $r^+$ in contact with three larger anion of radius $r^–$.
In this figure,$$AB=BC=AC=2r^-$$$$BE=r^-$$$$BD=r^++r^-$$$\begin{equation} \begin{aligned} \angle ABC = {60^ \circ } \\ \Rightarrow \angle BDC = {120^ \circ }or\angle BDE = {60^ \circ } \\ \therefore \angle DBC = {30^ \circ } \\\end{aligned} \end{equation} $
$$\begin{equation} \begin{aligned} BD = \frac{{BE}}{{\cos {{30}^ \circ }}} \\ \Rightarrow {r^ + } + {r^ - } = \frac{{{r^ - }}}{{\cos {{30}^ \circ }}} = \frac{{{r^ - }}}{{0.866}} = 1.155{r^ - } \\ \Rightarrow {r^ + } = 1.155{r^ - } - {r^ - } = 0.155{r^ - } \\ \therefore \frac{{{r^ + }}}{{{r^ - }}} = 0.155 \\\end{aligned} \end{equation} $$
$2.$ Co-ordination Number $4$: (tetrahedron)
Angle ABC is the tetrahedral angle
$$\begin{equation} \begin{aligned} \angle ABC = {109^ \circ }28' \\ \Rightarrow \angle ABD = \frac{1}{2} \times \left( {{{109}^ \circ }28'} \right) = {54^ \circ }44' \\ \sin \left( {\angle ABD} \right) = 0.8164 = \frac{{AD}}{{AB}} = \frac{{{r^ - }}}{{{r^ + } + {r^ - }}} \\ \frac{{{r^ - }}}{{{r^ + } + {r^ - }}} = \frac{1}{{0.8164}} = 1.225 \\ \therefore \frac{{{r^ + }}}{{{r^ - }}} = 0.225 \\\end{aligned} \end{equation} $$
$3.$ Co-ordination Number $6$: (Octahedron)
A cross section through an octahedral site is shown in the figure and the smaller positive ion touches six larger negative ions. (Only four negative ions are shown in the figure but there is one sphere above and the below the plane of paper).
According to the figure, we can write
$$\begin{equation} \begin{aligned} AB = {r^ + } + {r^ - }\;and\;BD = {r^ - } \\ \angle ABC = {45^ \circ } \\ \cos \left( {\angle ABC} \right) = 0.7071 = \frac{{BD}}{{AB}} = \frac{{{r^ - }}}{{{r^ + } + {r^ - }}} \\ \therefore \frac{{{r^ + }}}{{{r^ - }}} = 0.414 \\\end{aligned} \end{equation} $$
In nutshell, we can write the explanation above in table form as
Limiting radius ratio | Co-ordination number | Shape | Examples |
$< 0.155$ | $2$ | Linear | $BeF_2$ |
$0.155 – 0.225$ | $3$ | Trigonal planar | $B_2O_3$ |
$0.225 – 0.414$ | $4$ | Tetrahedral | $ZnS$, $SiO_4^{4–}$ |
$0.414 – 0.732$ | $4$ | Square planar | $PtCl_4^{2–}$ |
$0.414 – 0.732$ | $6$ | Octahedral | $NaCl$ |
$0.732 – 0.999$ | $8$ | Body centred cubic | $CsCl$ |
