Question 2 of 8: CsCl Crystal Structure, Lattice Constant, and Density

Question 2: CsCl Crystal Structure, Lattice Constant, and Density (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $r_{Cs^+}=167$ pm, $r_{Cl^-}=181$ pm; both ions monovalent ($n=1$); atomic masses $M_{Cs}=132.9$, $M_{Cl}=35.5$ g/mol; $N_A=6.02\times10^{23}$ mol$^{-1}$.

Find. (a) Crystal structure type and lattice constant $a$. (b) Density $\rho$. (c) X-ray diffractometer verification procedure.

Approach

Use the cation/anion radius ratio to identify the coordination number and packing geometry (Pauling's radius-ratio rule), fix the lattice constant from the ions touching along the relevant direction, then get density from the mass and volume of one unit cell.

1. Radius ratio ⇒ coordination number. $$\frac{r_{Cs^+}}{r_{Cl^-}}=\frac{167}{181}=\boxed{0.9227}.$$ This falls in the range $0.732\le r_+/r_-<1.0$, which corresponds to coordination number 8: each ion sits at the corners of its own simple-cubic sub-lattice, with the two sub-lattices offset by half the body diagonal (the CsCl-type structure — not true bcc, since the corner and body-centre sites are occupied by different ions).
2. Lattice constant. In the CsCl-type structure the cation and anion touch along the cube body diagonal, so $a\sqrt3 = 2(r_{Cs^+}+r_{Cl^-})$: $$a=\frac{2(167+181)}{\sqrt3}=\boxed{401.8\ \text{pm}}.$$
3. Density. One CsCl unit cell contains $8\times\tfrac18=1$ Cs$^+$ (corners) plus $1$ Cl$^-$ (body centre) $=1$ formula unit: $$\rho=\frac{n(M_{Cs}+M_{Cl})}{N_A\,a^3}=\frac{1\times168.4}{6.02\times10^{23}\times(4.018\times10^{-8}\ \text{cm})^3}=\boxed{4.31\ \text{g/cm}^3}.$$

(c) X-ray diffractometer verification. Mount a powdered CsCl sample in the diffractometer and scan the detector through a range of Bragg angles $2\theta$ while recording diffracted-beam intensity, producing a series of intensity peaks at the angles satisfying Bragg's law $n\lambda=2d_{hkl}\sin\theta$ for a monochromatic source of known wavelength $\lambda$ (e.g. Cu K$\alpha$, $\lambda=154.2$ pm). For the simple-cubic CsCl-type lattice, $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$; indexing the observed peaks (e.g. the (100) reflection) and solving for $a$ from each measured $d_{hkl}$ gives an independently measured lattice constant. Agreement of this measured $a$ with the $401.8$ pm computed from the ionic radii in part (a) confirms both the assumed CsCl-type coordination and the lattice-constant calculation.