Chapter 13, Problem 43E

Interpretation Introduction

Interpretation:

Seven lattice and possible variations

Crystal system	Possible variations	Axial distances or edge lengths	Axial angles	Examples
Cubic	Primitive body centred face centred	$a=b=c$	$\alpha =\beta =\gamma ={90}^0$	NaCl, Zinc blende, Cu
Tetragonal	Primitive Body-centred Face-centred	$a=b\ne c$	$\alpha =\beta =\gamma ={90}^0$	White tin${\text{SnO}}_2$ ${\text{T}}_i{O}_2,\text{Ca}S{O}_4$
Orthorhombic	Primitive Body-centred Face-centred	$a\ne b\ne c$	$\alpha =\beta =\gamma ={90}^0$	Rhombic sulphur ${\text{KNO}}_3$ $BaS{O}_4$
Hexagonal	primitive	$a=b\ne c$	$\begin{array}{l}\alpha =\beta ={90}^{\circ }\hfill \\ \gamma ={120}^{\circ }\hfill \end{array}$	Graphite, $ZnO$ $CdS$
Rhombohedral	primitive	$a=b=c$	$\begin{array}{l}\alpha =\beta ={90}^{\circ }\\ \beta \ne {90}^{\circ }\end{array}$	Calcite$\left(CaC{O}_3\right)$ $HgS\left(cinnabar\right)$
Monoclinic	Primitive	$a\ne b\ne c$	$\begin{array}{l}\alpha =\beta ={90}^{\circ }\\ \beta \ne {90}^{\circ }\end{array}$	Monoclinic sulphur $NaS{O}_4.10H_{2}O$
Triclinic	primitive	$a\ne b\ne c$	$\alpha \ne \beta \ne \gamma \ne {90}^{\circ }$	${\text{K}}_{2}C{r}_2{O}_7$ $CuS{O}_4,5H_{2}O$ $H_{3}B{O}_3$

Concept introduction:

Zinc blende structure has $ccp/fcc$ anion with cation is one set of tetrahedral sites. The $Zn{S}_4$ tetrahedral are linked at their corners and each corner is common to such tetrahedral. The unit cell of $ZnS$ as following:

This is of the form of $\text{MX}$

Coordination number $\text{M}$$=\text{Cordination number of X}$

The general $\text{MX}$ type compound with radius $\left(\frac{r_{+}}{r_{-}}\right)$$=0.225-0.414$ crystalline in zinc blende structure:

$\begin{array}{l}{r}_{+}:\text{ radius of cation }\\ r_{-}:\text{ radius of anion}\end{array}$

To determine: the compound that is most likely to crystallize in the zinc blende structure shown in figure

$\begin{array}{l}\text{CuI}\left(S\right)\left(C{u}^{+}\text{radius}=74{\text{ pm, I}}^{-}\text{radius = 216 pm}\right)\\ \text{KCl}\left(S\right)\left(K^{+}\text{radius}=133{\text{ pm, Cl}}^{-}\text{radius}=181\text{ pm}\right)\\ {\text{MgCl}}_2\left(S\right)\left({\text{Mg}}^{2+}{\text{ radius = 65 pm, Cl}}^{-}\text{ radius = 181 pm}\right)\end{array}$