Question

Rubidium chloride has the sodium chloride structure at normal pressures but assumes the cesium chloride structure at high pressures. What ratio of densities is expected for these two forms? Does this change in structure makesense on the basis of simple models?

Short answer

The density ratio is 1.30.

Step-by-step solution

Determining density on normal pressure      

The ionic radius of ${\mathbf{Rb}}^{+}$and $\mathit{C}{\mathit{l}}^{\mathbf{-}}$is 148 pm.

In normal pressure, ${\mathbf{Rb}}^{+}$and $\mathit{C}{\mathit{l}}^{\mathbf{-}}$touch along the body diagonal (form NaCl structure).

$$\begin{gathered} Cube edge = l = 2(148+181) \\ = 658pm \\ = 6.58 \times {10}^{-8 } cm \end{gathered}$$

Determining density on Higher pressure

In Higher pressure, Rb⁺and Cl^- touch along the body diagonal (form CsCl structure).

$$\begin{gathered} 2r_{- }+ 2r_{+} = 658 pm \\ = body diagonal = l\sqrt{3} \\ l=658\sqrt{3} = 380 pm \end{gathered}$$

$d= \frac{85.47+ 35.45}{6.022 \times {10}^{23}{(3.80\times {10}^{-8})}^3 }=3.66g/c{m}^3$

Determination of the ratio and reason

The high-pressure form has a higher density. The density ratio is 3.66/2.82 = 1.30.

This comes because the effect of pressure is to push things closer together, and thus density increases.