Chapter 15, Problem 15A.3E

(a)

Interpretation Introduction

Interpretation:

The distance of separation of $(321)$ planes in an orthorhombic ${\text{SbCl}}_{\text{3}}$ unit cell has to be calculated.

Concept-Introduction:

The separation of planes can be calculated using miller indices as,

    $\frac{\text{1}}{{\text{d}}^{\text{2}}}\text{=}\frac{{\text{h}}^{\text{2}}}{{\text{a}}^{\text{2}}}\text{+}\frac{{\text{k}}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{+}\frac{{\text{l}}^{\text{2}}}{{\text{c}}^{\text{2}}}$

Where,

h, l, and k are miller indices which indicates each plane.

d is distance of separation of plane.

Explanation of Solution

Given,

    $\begin{array}{l}\text{a}\text{=}\text{812pm}\\ \text{b}\text{=}\text{947pm}\\ \text{c}\text{=}\text{637pm}\end{array}$

 The unit cell is orthorhombic and separation of plane equation is given as

    $\begin{array}{l}\frac{\text{1}}{{\text{d}}^{\text{2}}}\text{=}\frac{{\text{h}}^{\text{2}}}{{\text{a}}^{\text{2}}}\text{+}\frac{{\text{k}}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{+}\frac{{\text{l}}^{\text{2}}}{{\text{c}}^{\text{2}}}\text{=}\frac{{\text{3}}^{\text{2}}}{{\text{812}}^{\text{2}}}\text{+}\frac{{\text{2}}^{\text{2}}}{{\text{947}}^{\text{2}}}\text{+}\frac{{\text{1}}^{\text{2}}}{{\text{637}}^{\text{2}}}\\ \\ \frac{\text{1}}{{\text{d}}^{\text{2}}}\text{=}\text{1}\text{.36}\text{×}{\text{10}}^{\text{-5}}\text{+}\text{4}\text{.46×}{\text{10}}^{\text{-6}}\text{+}\text{2}\text{.46}\text{×}{\text{10}}^{\text{-6}}\\ \\ \frac{\text{1}}{{\text{d}}^{\text{2}}}\text{=}\text{2}\text{.05}\text{×}{\text{10}}^{\text{-5}}\text{pm}\\ \\ \text{d}\text{=}\text{220}\text{.86pm}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The distance of separation of plane $(642)$ in an orthorhombic ${\text{SbCl}}_{\text{3}}$ unit cell has to be calculated.

Concept-Introduction:

The separation of planes can be calculated using miller indices as,

    $\frac{\text{1}}{{\text{d}}^{\text{2}}}\text{=}\frac{{\text{h}}^{\text{2}}}{{\text{a}}^{\text{2}}}\text{+}\frac{{\text{k}}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{+}\frac{{\text{l}}^{\text{2}}}{{\text{c}}^{\text{2}}}$

Where,

h, l, and k are miller indices which indicates each plane.

d is distance of separation of plane.

Explanation of Solution

Given,

    $\begin{array}{l}\text{a}\text{=}\text{812pm}\\ \text{b}\text{=}\text{947pm}\\ \text{c}\text{=}\text{637pm}\end{array}$

The unit cell is orthorhombic and separation of plane equation is given as

    $\begin{array}{l}\frac{\text{1}}{{\text{d}}^{\text{2}}}\text{=}\frac{{\text{h}}^{\text{2}}}{{\text{a}}^{\text{2}}}\text{+}\frac{{\text{k}}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{+}\frac{{\text{l}}^{\text{2}}}{{\text{c}}^{\text{2}}}\text{=}\frac{6^{\text{2}}}{{\text{812}}^{\text{2}}}\text{+}\frac{4^{\text{2}}}{{\text{947}}^{\text{2}}}\text{+}\frac{2^{\text{2}}}{{\text{637}}^{\text{2}}}\\ \\ \frac{\text{1}}{{\text{d}}^{\text{2}}}\text{=}5.\text{45}\text{×}{\text{10}}^{\text{-5}}\text{+}1.\text{78×}{\text{10}}^{\text{-5}}\text{+}9.\text{85}\text{×}{\text{10}}^{\text{-6}}\\ \\ \frac{\text{1}}{{\text{d}}^{\text{2}}}\text{=}8.21\text{×}{\text{10}}^{\text{-5}}\text{pm}\\ \\ \text{d}\text{=}110.\text{33pm}\end{array}$