Chapter 15, Problem 15.41P

Interpretation Introduction

Interpretation:

The length of unit cell and nearest neighbor distance in $\text{KF}$ crystal have to be determined.

Concept Introduction:

A unit cell is smallest repeating unit of a crystal lattice. Unit cells are classified as primitive and non-primitive. Unit cells that have atoms at only corners are primitive and unit cells that have atoms at positions other than corners are non-primitive.

Unit cells that are non- primitive are of various types:

Face centered unit cell (fcc), body centered unit cell (bcc), edge centered unit cell, end centered unit cell.

Contribution of an atom at a position is as follows:

  $\begin{array}{cc}\text{position in unit}\text{cell}& \text{contributions}\\ \text{center of body}& \text{1}\\ \text{center of face}& \text{1/2}\\ \text{center of edge}& \text{1/4}\\ \text{corners}& \text{1/8}\end{array}$

Answer to Problem 15.41P

The edge length of unit cell of $\text{KF}$ is $6.7753\times {10}^{-08}\text{cm}$ and nearest neighbor distance in $\text{KF}$ unit cell is $3.3876\times {10}^{-08}\text{cm}$.

Explanation of Solution

In $\text{NaCl}$, the chloride ion is present in a fcc type arrangement that is at corners and face centers and each sodium ion is present at edge center and body center.

There are total of 8 corners with contribution at each corner as $\frac{1}{8}$, 6 face center with contribution at each face center as $\frac{1}{2}$, 1 body center with contribution at each body center as 1 and 12 edge center with contribution at each edge center as $\frac{1}{4}$.

So number of sodium atoms in the unit cell is calculated as follows:

  $\begin{array}{c}\text{Number of sodium}\text{atoms}=\left[\begin{array}{c}\left(\frac{1}{4}\text{atom}\text{per}\text{cell}\right)\left(12\text{atoms}\right)+\\ \left(1\text{atom}\text{per}\text{cell}\right)\left(1\text{atoms}\right)\end{array}\right]\\ =\left(3\text{atoms}\right)+\left(1\text{atoms}\right)\\ =4\text{atoms}\end{array}$

So number of chlorine atoms in the unit cell is calculated as follows:

  $\begin{array}{c}\text{Number of chlorine}\text{atoms}=\left[\begin{array}{c}\left(\frac{1}{8}\text{atom}\text{per}\text{cell}\right)\left(8\text{atoms}\right)+\\ \left(\frac{1}{2}\text{atom}\text{per}\text{cell}\right)\left(6\text{atoms}\right)\end{array}\right]\\ =\left(1\text{atoms}\right)+\left(3\text{atoms}\right)\\ =4\text{atoms}\end{array}$

Since there are 4 sodium and 4 chlorine atoms so 4 $\text{NaCl}$ is formed with $Z$ equals to 4.

So like $\text{NaCl}$ lattice, $\text{KF}$ too have $Z$ equals to 4.

The formula to calculate the density of $\text{KF}$ is as follows:

  $\text{density}=\frac{Z\cdot M}{N_a\cdot a^3}$        (1)

Here,

$Z$ is the number of atoms of $\text{KF}$ per unit cell.

$M$ is the molecular mass of the $\text{KF}$.

$N_a$ is the Avogadro's number.

$a$ is the edge length of cube.

Rearrange equation (1), to calculate the value of $a$.

  $a=\sqrt[3]{\frac{Z\cdot M}{\text{density}\cdot N_{\text{a}}}}$        (2)

Substitute 4 for $Z$, $58.0967\text{g/mol}$ for $M$, $6.023\times {10}^{23}{\text{mol}}^{-1}$ for $N_{\text{a}}$ and $2.481\text{g}\cdot {\text{cm}}^{-3}$ for density in equation (2) to determine $a$.

  $\begin{array}{c}a=\sqrt[3]{\frac{\left(4\right)\left(58.0967\text{g/mol}\right)}{\left(2.481\text{g}\cdot {\text{cm}}^{-3}\right)\left(6.023\times {10}^{23}{\text{mol}}^{-1}\right)}}\\ =\sqrt[3]{1.5551\times {10}^{-22}{\text{cm}}^3}\\ =5.3776\times {10}^{-08}\text{cm}\end{array}$

The edge length of unit cell of $\text{KF}$ is $5.3776\times {10}^{-08}\text{cm}$.

The formula to calculate the nearest neighbor distance in $\text{KF}$ unit cell is as follows:

  $d=\frac{a}{2}$        (3)

Here,

$d$ is nearest neighbor distance.

$a$ is edge length of cube.

Substitute $5.3776\times {10}^{-08}\text{cm}$ for $a$ in equation (3) to determine $d$.

  $\begin{array}{c}d=\frac{5.3776\times {10}^{-08}\text{cm}}{2}\\ =2.6888\times {10}^{-08}\text{cm}\end{array}$

The nearest neighbor distance in $\text{KF}$ unit cell is $2.6888\times {10}^{-08}\text{cm}$.