12.92P - Chemistry: Molecular Nature Of Matter And Change

Question

Zinc selenide (ZnSe) crystallizes in the zinc blende structure (see Figure) and has a density of 5.42 g/cm³.

(a) How many Zn and Se ions are in each unit cell?

(b) What is the mass of a unit cell?

(c) What is the volume of a unit cell?

(d) What is the edge length of a unit cell?

Step-by-Step Solution

Answer

1. In each unit cell, Zn and Se ions are 4 and 4 in number respectively.
2. Mass of the unit cell is $\mathbf{9}\mathbf{.}\mathbf{58}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{22}}\mathbf{ }\mathbf{g}$
3. Volume of the unit cell is $\mathbf{1}\mathbf{.}\mathbf{76}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{22}}\mathbf{ }{\mathbf{cm}}^{\mathbf{3}}$
4. Edge length of the unit cell is 3.3 nm

Step 1: Zinc blende structure

Zinc blende is a cubic closed packed structure with the anions occupying the face-centered lattice and the cations are packed in half of the tetrahedral voids.

Step 2: Ions in the unit cell

The face-centered cubic unit cell consists of 8 anions in the corner, thus one anion contributing $\frac{\mathbf{1}}{\mathbf{8}}$ to the unit cell. Along with that, anions are present in the center of the six respective faces of the cell. Hereby, each anion contributing $\frac{\mathbf{1}}{\mathbf{2}}$ to the unit cell.

Thus, each unit cell consists of $\mathbf{8}\mathbf{\times }\left(\frac{1}{8}\right)\mathbf{+}\mathbf{6}\mathbf{\times }\left(\frac{1}{2}\right)\mathbf{=}\mathbf{4}$ , i.e., four anions in the unit cell.

The cations occupy half of the tetrahedral sites. Here in this unit cell, the number of tetrahedral sites is 8. Thus, in this zinc blende structure, the number of sites occupied by cations is four in number.

Step 3: Mass of the unit cell

The mass of the unit cell will be the mass of all the atoms in the unit cell.

Thus, the mass of the unit cell will be $\mathbf{4}\mathbf{\times }\left(65.38 u\right)\mathbf{+}\mathbf{4}\mathbf{\times }\left(78.96 u\right)\mathbf{=}\mathbf{577}\mathbf{.}\mathbf{36}\mathbf{ }\mathbf{u}$ or $\mathbf{9}\mathbf{.}\mathbf{58}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{22}}\mathbf{ }\mathbf{g}$.

Step 4: Volume of the unit cell

$\begin{array}{rcl}\mathbf{Volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{cell}& \mathbf{=}& \frac{\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{cell}}{\mathbf{Density}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{cell}}\\ \mathbf{Volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{cell}& \mathbf{=}& \frac{\mathbf{9}\mathbf{.}\mathbf{58}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{22}}}{\mathbf{5}\mathbf{.}\mathbf{42}}\\ & \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{76}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{22}}\mathbf{ }{\mathbf{cm}}^{\mathbf{3}}\end{array}$

Step 5: Edge of unit cell

$\begin{array}{rcl}\mathbf{Volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{cell}& \mathbf{=}& {\left(\mathrm{Edge} \mathrm{of} \mathrm{unit} \mathrm{cell}\right)}^{\mathbf{3}}\\ \mathbf{Edge}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{cell}& \mathbf{=}& {\left(\mathrm{Volume} \mathrm{of} \mathrm{unit} \mathrm{cell}\right)}^{\frac{\mathbf{1}}{\mathbf{3}}}\\ \mathbf{Edge}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{cell}& \mathbf{=}& {\left(1.76\times {10}^{-22}\right)}^{\frac{\mathbf{1}}{\mathbf{3}}}\\ & \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{ }\mathbf{cm}\end{array}$