Chapter 12, Problem 12.123SP

(a)

Interpretation Introduction

Interpretation:

Normal boiling point of trichlorosilane has to be determined. Boiling point of trichlorosilane is whether constant with the type of intermolecular forces that exist between its molecules has to be given

Concept Introduction:

Clausius-Clapeyron equation:

$\text{ln}\text{P}\text{=}-\left(\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{RT}}\right)\text{+}\text{C}$

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

$\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}\text{-}\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]$

Boiling point of a liquid: The temperature at which external pressure and vapor pressure of the liquid become same.

Normal boiling point: When the external pressure is $\text{760}\text{mm}\text{Hg}$ we can call it as normal boiling point.

Intermolecular force: The attractive force that withholds two molecules is called as intermolecular force. The influence of intermolecular forces depends on molar mass and the functional group present in the molecule.

Dipole-Dipole Interactions: The attractive force that holds two polar molecules with help of dipole moment present in them is called as Dipole-Dipole Interactions.

The partial positive charge end of one molecule is attracted to the partial negative charge of a neighboring molecule.

Explanation of Solution

Given information follows as:

Molar heat of vaporization,${\text{ΔH}}_{\text{vap}}\text{=}\text{28}\text{.8}\text{kJ/mol}$

Vapor pressure of trichlorosilane at $-2°C$ is $0.258$

Normal boiling point of trichlorosilane can be determined by Clausius-Clapeyron equation given below:

$\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}-\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]$

Rearrange the equation to determine normal boiling point of trichlorosilane as follows:

$\begin{array}{c}\frac{1}{T_2}=\frac{R}{\Delta H_{vap}}\ln \frac{P_1}{P_2}+\frac{1}{T_1}\\ \\ {\text{P}}_{\text{1}}\text{=}\text{0}\text{.258}atm\text{,}{\text{P}}_{\text{2}}\text{=}\text{1}\text{.00}\text{atm}\\ \\ {\text{T}}_{\text{1}}\text{=}-2\text{°C}\text{=}\text{271}\text{K}\text{,}\Delta H_{vap}=28.8\times {10}^3\text{J/mol}\end{array}$

Substitute the values as follows:

$\begin{array}{c}\frac{1}{T_2}=\frac{\left(\text{8}\text{.314 J/K}\text{.mol}\right)}{\left(28.8\times {10}^3\text{J/mol}\right)}\ln \frac{\left(\text{0}\text{.258}atm\right)}{\left(\text{1}\text{.00}\text{atm}\right)}+\frac{1}{\left(\text{271}\text{K}\right)}\\ \\ =3.2989\times {10}^{-3}\\ \\ T_2=\frac{1}{3.2989\times {10}^{-3}}\\ \\ =303K=\left(303-273\right)°C\\ \\ =30°C\end{array}$

Normal boiling point of trichlorosilane is $30°C$

$Si$ atom of trichlorosilane $\left({\text{SiCl}}_{\text{3}}\text{H}\right)$ is $s{p}^3$ hybridized and the geometry of trichlorosilane is tetrahedral. Trichlorosilane has dipole moment and is polar. Hence, the main intermolecular force among trichlorosilane molecules is dipole-dipole interactions

Dipole-dipole interactions are weak and thus trichlorosilane is expected to have a very low consistent boiling point

(b)

Interpretation Introduction

Interpretation:

Types of crystals that $Si{O}_2$ and $Si$ form has to be given

Concept Introduction:

• Crystalline solids have well-defined regular, compact, orderly arrangement of their components of very long range order. They are termed as true solids.
• Ionic solids, molecular solids, covalent solids, atomic solids and metallic solids are the types of crystalline solids
• In ionic solids, ions of opposite charges are bind together by electrostatic force and are neatly stacked to form a regular and well-defined structure. Example: $\text{KC1}$, Rutile.
• Molecular solids are formed by covalent molecules which are either polar or non-polar. Molecular solids are low melting solids and poor conductors of heat and electricity whereas metallic solids are good conductors of heat and electricity.
• In Covalent solids the components are atoms bonded by covalent bond repetitively and thus forms huge network form of solid. Network solids are widely formed by group 14 elements and its compounds. Example: Diamond, graphite. They are high melting solids and poor conductors but exceptions prevail.

Explanation of Solution

Silicon belongs to Group 4A. Similar to carbon atoms, silicon atoms are bonded by covalent bond repetitively and thus forms huge network form of solid.

In $Si{O}_2$, oxygen and silicon atoms are bonded by covalent bond repetitively

Therefore, $Si{O}_2$ and $Si$ form covalent crystals and thus they have are high melting and boiling solids but poor conductors.

(c)

Interpretation Introduction

Interpretation:

Number of silicon atoms that are there for every boron atom in the sample has to be given and to determine whether the sample satisfy the ${10}^{-9}$ purity requirement using the given data.

Concept Introduction:

Crystal structure: It is a systematic or periodic arrangement of atoms or ions in a crystalline material. It is useful to predict the properties of material.

Cubic unit cell: The smallest repeating unit in a crystal is called a unit cell.

Unit conversion:

$1pm=1\times {10}^{-10}cm$

Volume of cube can be determined as follows:

$\begin{array}{c}\text{V}\text{=}{\text{a}}^{\text{3}}\\ \\ \text{where,}\\ \text{a}\text{=}\text{edge}\text{length}\end{array}$

Explanation of Solution

Given information follows as:

Edge length is $543pm=543pm\times \frac{1\times {10}^{-10}cm}{1pm}=5.43\times {10}^{-8}cm$

Volume of unit cell is given as:

$\begin{array}{c}V=a^3\\ \\ ={\left(5.43\times {10}^{-8}cm\right)}^3\\ \\ =1.60\times {10}^{-22}c{m}^3\end{array}$

Volume of silicon unit cell is $1.60\times {10}^{-22}c{m}^3$

Number of unit cell in $1c{m}^3$ is given as:

$\begin{array}{c}Numberofunitcell=1c{m}^3\times \frac{1unitcell}{1.60\times {10}^{-22}c{m}^3}\\ \\ =6.25\times {10}^{21}unitcell\end{array}$

Number of unit cell in $1c{m}^3$ is $6.25\times {10}^{21}unitcell$

In each unit cell, there are $8Si$ atoms. Therefore, the total number of silicon atoms can be given as:

$\begin{array}{c}Numberofsiliconatoms=\frac{8Siatoms}{1unitcell}\times 6.25\times {10}^{21}unitcell\\ \\ =5.0\times {10}^{22}Siatoms\end{array}$

Thus, there are $5.0\times {10}^{22}Si$ atoms

There are $1.0\times {10}^{13}Batoms$ in each $c{m}^3$. Thus, the purity of $Si$ crystal is calculated as follows:

$\begin{array}{c}\frac{No.ofBatoms}{No.ofSiatoms}=\frac{1.0\times {10}^{13}}{5.0\times {10}^{22}}\\ \\ =2.0\times {10}^{-10}\end{array}$

For electronic grade silicon, purity should be ${10}^{-9}$. Since, the purity of given silicon crystal is $2.0\times {10}^{-10}$ which is less than ${10}^{-9}$, requirement for purity is satisfied.