Chapter 21, Problem 21.63E

Interpretation Introduction

Interpretation:

The volume of the unit cells for the given compounds in Table $21.7$ is to be calculated.

Concept introduction:

A unit cell of the crystal is the three-dimensional arrangement of the atoms present in the crystal. The unit cell is the smallest and simplest unit of the crystal which on repetition forms an entire crystal. Unit cell can be a cubic unit cell or hexagonal unit cell. The classification of a unit cell depends on the lattice site occupied by the atoms.

Answer to Problem 21.63E

The volume of unit cell of colaradoite, ice, hafnia and turquoise are $269.5861{\stackrel{\circ }{\text{A}}}^3$, $130.3938{\stackrel{\circ }{\text{A}}}^3$, $138.2426{\stackrel{\circ }{\text{A}}}^3$, and $460.8275{\stackrel{\circ }{\text{A}}}^3$ respectively.

Explanation of Solution

The volume of a unit cell is given as,

$V=a\times b\times c\times \sqrt{\left(1-{\cos }^2\alpha -{\cos }^2\beta -{\cos }^2\gamma +2\cos \alpha \cos \beta \cos \gamma \right)}$ …(1)

Where,

• $a$ represents the length of the unit cell.

• $b$ represents the breadth of the unit cell.

• $c$ represents the height of the unit cell.

• $\alpha$ represent the unit cell angle between $b$ and $c$.

• $\beta$ represent the unit cell angle between $a$ and $c$.

• $\gamma$ represent the unit cell angle between $b$ and $a$.

In Table $21.7$,

It is given that colaradoite has a cubic unit cell with $a=6.46\stackrel{\circ }{\text{A}}$. The angle $\alpha =90°$, $\beta =90°$ and $\gamma =90°$. The missing data for the colaradoite is the value of $b$ and $c$. The unit cell parameters of the cubic unit cell are represented as,

$\begin{array}{l}a=b=c\\ \alpha =\beta =\gamma =90°\end{array}$

Therefore, the value of $b$ and $c$ can be obtained from the value of $a$.

$a=b=c=6.46\stackrel{\circ }{\text{A}}$

Substitute the value of unit cell parameters of colaradoite in the equation (1).

$\begin{array}{rll}V& =& \left(6.46\stackrel{\circ }{\text{A}}\right)\left(6.46\stackrel{\circ }{\text{A}}\right)\left(6.46\stackrel{\circ }{\text{A}}\right)\times \sqrt{\left(\begin{array}{l}1-{\cos }^{2}90°-{\cos }^{2}90°-{\cos }^{2}90°\\ +2\left(\cos 90°\right)\left(\cos 90°\right)\left(\cos 90°\right)\end{array}\right)}\\ & =& \left(269.5861{\stackrel{\circ }{\text{A}}}^3\right)\times \sqrt{1}\\ & =& 269.5861{\stackrel{\circ }{\text{A}}}^3\end{array}$

Therefore, the volume of unit cell of colaradoite is $269.5861{\stackrel{\circ }{\text{A}}}^3$.

In Table $21.7$,

It is given that ice has a unit cell with $a=& 4.5212\stackrel{\circ }{\text{A}}$ and $c=& 7.366\stackrel{\circ }{\text{A}}$. The missing data for the ice is the value of $b$, $\alpha$, $\beta$ and $\gamma$. Ice has hexagonal lattice. The unit cell parameters of the hexagonal unit cell are represented as,

$\begin{array}{lll}a& =& b\ne c\\ \alpha & =& \beta & =& 90°\\ \gamma & =& 120°\end{array}$

Therefore, the value of $b$ can be obtained from the value of $a$ and the value of angles are constant for all hexagonal lattices.

$a=& b& =& 4.5212\stackrel{\circ }{\text{A}}$

Substitute the value of unit cell parameters of ice in the equation (1).

$\begin{array}{rll}V& =& \left(4.5212\stackrel{\circ }{\text{A}}\right)\left(4.5212\stackrel{\circ }{\text{A}}\right)\left(7.366\stackrel{\circ }{\text{A}}\right)\times \sqrt{\left(\begin{array}{l}1-{\cos }^{2}90°-{\cos }^{2}90°-{\cos }^{2}120°\\ +2\left(\cos 90°\right)\left(\cos 90°\right)\left(\cos 120°\right)\end{array}\right)}\\ & =& \left(150.5702{\stackrel{\circ }{\text{A}}}^3\right)\times \sqrt{\left(1-0-0-0.25+0\right)}\\ & =& \left(150.5702{\stackrel{\circ }{\text{A}}}^3\right)\left(0.8660\right)\\ & =& 130.3938{\stackrel{\circ }{\text{A}}}^3\end{array}$

Therefore, the volume of unit cell of ice is $130.3938{\stackrel{\circ }{\text{A}}}^3$.

In Table $21.7$,

It is given that hafnia has a unit cell with $a=& 5.1156\stackrel{\circ }{\text{A}}$, $b=& 5.17\stackrel{\circ }{\text{A}}$ and $c=& 5.2948\stackrel{\circ }{\text{A}}$. The angle $\beta =& 99.18°$. Hafnia has monoclinic lattice. The unit cell parameters of the monoclinic unit cell are represented as,

$\begin{array}{l}a\ne b\ne c\\ \alpha & =& \gamma & =& 90°\\ \beta \ne 90°\end{array}$

Substitute the value of unit cell parameters of hafnia in the equation (1).

$\begin{array}{rll}V& =& \left(5.1156\stackrel{\circ }{\text{A}}\right)\left(5.17\stackrel{\circ }{\text{A}}\right)\left(5.2948\stackrel{\circ }{\text{A}}\right)\times \sqrt{\left(\begin{array}{l}1-{\cos }^{2}90°-{\cos }^{2}99.18°-{\cos }^{2}90°\\ +2\left(\cos 90°\right)\left(\cos 99.18°\right)\left(\cos 90°\right)\end{array}\right)}\\ & =& \left(140.035{\stackrel{\circ }{\text{A}}}^3\right)\times \sqrt{\left(1-0-0-0.02545+0\right)}\\ & =& \left(140.035{\stackrel{\circ }{\text{A}}}^3\right)\left(0.9872\right)\\ & =& 138.2426{\stackrel{\circ }{\text{A}}}^3\end{array}$

Therefore, the volume of unit cell of hafnia is $138.2426{\stackrel{\circ }{\text{A}}}^3$.

In table $21.7$,

It is given that turquoise has a triclinic unit cell with $a=& 7.424\stackrel{\circ }{\text{A}}$, $b=& 7.62\stackrel{\circ }{\text{A}}$ and $c=& 9.910\stackrel{\circ }{\text{A}}$. The angle $\alpha =& 68.61°$, $\beta =& 69.71°$ and $\gamma =& 65.08°$.

Substitute the value of unit cell parameters of turquoise in the equation (1).

$\begin{array}{rll}V& =& \left(7.424\stackrel{\circ }{\text{A}}\right)\left(7.62\stackrel{\circ }{\text{A}}\right)\left(9.910\stackrel{\circ }{\text{A}}\right)\times \sqrt{\left(\begin{array}{l}1-{\cos }^{2}68.61°-{\cos }^{2}69.71°-{\cos }^{2}65.08°\\ +2\left(\cos 68.61°\right)\left(\cos 69.71°\right)\left(\cos 65.08°\right)\end{array}\right)}\\ & =& \left(560.6174{\stackrel{\circ }{\text{A}}}^3\right)\times \sqrt{\left(1-0.1330-0.1203-0.1775+2\left(0.3647\right)\left(0.3468\right)\left(0.4214\right)\right)}\\ & =& \left(560.6174{\stackrel{\circ }{\text{A}}}^3\right)\left(0.822\right)\\ & =& 460.8275{\stackrel{\circ }{\text{A}}}^3\end{array}$

Therefore, the volume of unit cell of turquoise is $460.8275{\stackrel{\circ }{\text{A}}}^3$.

Conclusion

The volume of unit cell of colaradoite, ice, hafnia and turquoise are $269.5861{\stackrel{\circ }{\text{A}}}^3$, $130.3938{\stackrel{\circ }{\text{A}}}^3$, $138.2426{\stackrel{\circ }{\text{A}}}^3$, and $460.8275{\stackrel{\circ }{\text{A}}}^3$ respectively.