Question

Use the relative ionic radii in Fig. 13.8 to predict the structures expected for CsBr and KF. Do these predictions agree with observed structures? The ionic radius of Cs⁺is 169 pm.

Short answer

The ionic radius is used to measure an atom ion present In the crystal lattice. Ionic radius is calculated when we halve the distance between two ions that are barely touching each other.

Step-by-step solution

Given Information

The ionic radius of Cs⁺is 169 pm

Concept Introduction

Ionic radius is defined as the radius of charged ions.

Explanation

For a cubic hole to be filled, the cation to anion radius ratio is between 0.732<r₊/r_-<1.00.

CsBr:

$$\begin{gathered} {\mathrm{Cs}}^{+}\mathrm{radius}=169\mathrm{pm} \\ {\mathrm{Br}}^{-}\mathrm{radius}=195\mathrm{pm} \\ \frac{{\mathrm{r}}_{{\mathrm{Cs}}^{+}}}{{\mathrm{r}}_{{\mathrm{Br}}^{-}}}=\frac{169\mathrm{pm}}{195\mathrm{pm}} \\ \frac{{\mathrm{r}}_{{\mathrm{Cs}}^{+}}}{{\mathrm{r}}_{{\mathrm{Br}}^{-}}}=0.867 \end{gathered}$$

From the radius ratio, Cs⁺should occupy cubic holes. The structure shown should be the CsCl structure. Yes, the prediction agreed with observed structures, but the actual structure is the CsCI structure.

KF:

$$\begin{gathered} {\mathrm{K}}^{+}\mathrm{radius}=133\mathrm{pm} \\ {\mathrm{F}}^{-}\mathrm{radius}=136\mathrm{pm} \\ \frac{{\mathrm{r}}_{{\mathrm{K}}^{+}}}{{\mathrm{r}}_{{\mathrm{F}}^{-}}}=\frac{133\mathrm{pm}}{136\mathrm{pm}} \\ \frac{{\mathrm{r}}_{{\mathrm{K}}^{+}}}{{\mathrm{r}}_{{\mathrm{F}}^{-}}}=0.978 \end{gathered}$$

Again, we would predict that the structure is similar to CsCl, that is, cations in the middle of a simple cubic array of anions. The actual structure is the NaCl structure.

The radius ratio rules fail for KF. Exceptions are common for crystal structures.