An ionic crystal contains a large number of cations and, anions. Generally, cations are smaller in size than anions. The cations are surrounded by anions and they touch each other. These ions are arranged in space in such a way as to produce maximum stability.

The stability of the ionic crystals may be described in terms of radius ratio, I.e, the ratio of the radius of cation (r) to that of anion (R) (r / R). The range of (r/ R) may be expressed as limiting radius ratio. This value is important to determine the arrangement of the ions in different types of crystals.

Evidently, radius ratio (r/ R) plays a very important role in deciding the stable structure of ionic crystal. Larger cations prefer occupying larger holes (cubic etc.) and smaller cations prefer occupying smaller holes (tetrahedral etc.). The preferred direction of the structure with increase in the radius ratio is as follows:

$\text{Plane triangular} \overset{0.225}{\rightarrow} \text{Tetrahedral} \overset{0.414}{\rightarrow} \text{Octahedral} \overset{0.732}{\rightarrow} \text{Cubic}$

LIMITING RADIUS RATIO FOR VARIOUS TYPES OF SITES

 r/R Coordination number of cation Structural arrangement Example 0.115 – 0.225 3 Plane Trigonal Boron oxide 0.225 – 0.414 4 Tentrahedral ZnS 0.414 – 0.731 4 Square planar = 0.414 – 0.732 6 Ostahedral Nacl 0.732 – 1.000 8 Cubic CsCl

Example 1:

A solid $^+B^-$ has NaCl type close packed structure. If the anion has a radius of 250 pm, what would be the ideal radius for the cation? Can a cation $C^+$ having a radius of 180 pm be slipped into the tetrahedral site of the crystal $A^+ B ^-$ question? Give reason for your answer.

Solution: In crystal each $Na^+ Cl^-$ ion is surrounded by $6Cl^-$ions and vice versa. Thus $Na^+$ ion is placed in octahedral hole.

The limiting radius ratio for octahedral site = 0.414

$Or \dfrac{r}{R} =0.414$

Given that R = 250 pm

$\therefore r = 0.414 R = 0.414 \times 250 pm$

Or r = 103.5 pm

Thus the ideal radius for cation $(A^+)$ is 103.5 pm.

We know that (r / R) for tetrahedral hole is 0.225.

$\therefore \dfrac{r}{R} = 0.225$

or, $r = 0.225R = 0.225 \times 250 = 56.25 pm$

Thus ideal radius for cation is 56.25 pm for tetrahedral hole. But the radius of $C^+$ is 180 pm. It is much larger than ideal radius, i.e., 56.25 pm. Therefore we can not slip cation $C^+$into the tetrahedral site.

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