Question

For $B{r^ - }$ ions form a close packed structure. If the radius of $B{r^ - }$ ion is 195 pm, then calculate the radius of cation (A) that just fits into the tetrahedral hole. Can a cation having a radius 82 pm be slipped into the octahedral hole of the crystal ${A^ + }B{r^ - }$?

Answer 1

Hint: The electrostatic interactions between the charged species are responsible for formation of ionic bonding within the crystal of solid. The ionic size of ions can be determined by nuclear separation of the separate contributions of an anion from the cation.

Complete answer: Radius ratio refers to the ratio of smaller ionic radius i.e., radius of cation to the larger ionic radius i.e., radius of anion. Thus, mathematically it can be represented as follows:
Radius ratio $\rho = \dfrac{{{r_ + }}}{{{r_ - }}}\;\;\;\;\;\; - (1)$
According to the radius ratio rule, a stable ionic crystal will be formed in such a way that the larger ions will form the basic unit lattice and smaller ions will tend to fit in the voids of the lattice formed.
Radius ratio rule predicts the location of cations in a crystal lattice as per following table:

Coordination number	Geometry	Radius ratio
2	Linear	$0 - 0.155$
3	Triangular	$0.155 - 0.225$
4	Tetrahedral	$0.225 - 0.414$
6	Octahedral	$0.414 - 0.732$
8	Cubic	$0.732 - 1.0$

Now, as per given in the question, the radius of anion i.e., $B{r^ - }$ ion is 195 pm and we need to fit the cation into its tetrahedral void. From the above table, we know that the minimum radius ratio for a tetrahedral void is 0.225. So, the ratio of cation can be calculated using equation (1) as follows:
${\rho _{{\text{tetrahedral}}}} = \dfrac{{{r_{{\text{cation}}}}}}{{{r_{B{r^ - }}}}}$
$ \Rightarrow 0.225 = \dfrac{{{r_{{\text{cation}}}}}}{{195}}$
$ \Rightarrow {r_{{\text{cation}}}} = 0.225 \times 195$
$ \Rightarrow {r_{{\text{cation}}}} = 43.875\;{\text{pm}}$
Further, if we consider the cation with radius 82 pm then the radius ratio will be as follows:
$\rho = \dfrac{{{r_{{\text{cation}}}}}}{{{r_{B{r^ - }}}}}$
$ \Rightarrow \rho = \dfrac{{82}}{{195}}$
$ \Rightarrow \rho = 0.42$
As the ratio lies in the range of $0.414 - 0.732$ which is given for the octahedral void as per above table. So, we can conclude that the cation with radius 82 pm can be slipped into the octahedral hole of the given crystal.

Note:
It is important to note that the coordination number of a crystal lattice refers to the number of ions immediately surrounding an ion having the opposite charge within a crystal lattice. An increase in the coordination number leads to increase the ionic radius of the ions and thus, the increase in radius ratio is observed.