Chapter 12, Problem 12.132QP

Silicon used in computer chips must have an impurity level below 10⁻⁹ (i.e., fewer than one impurity atom for every 10⁹ Si atoms). Silicon is prepared by the reduction of quartz (SiO₂) with coke (a form of carbon made by the destructive distillation of coal) at about 2000°C.

$SiO2(\text{s})\text{ + }2C(\text{s})\to Si(\text{l})+2CO(\text{g})$

Next, solid silicon is separated from other solid impurities by treatment with hydrogen chloride at 350°C to form gaseous trichlorosilane (SiCl₃H).

$Si(\text{s})\text{ + }3HCl(\text{g})\to SiCl3H(\text{g})+H2(\text{g})$

Finally, ultrapure Si can be obtained by reversing the above reaction at 1000°C.

$SiCl3H(\text{g})\text{ + }H2(\text{g})\to Si(\text{s})+3HCl(\text{g})$

The molar heat of vaporization of trichlorosilane is 28.8 kJ mol and its vapor pressure at 2°C is 0.258 atm. (a) Using this information and the equation

$In\frac{{{\displaystyle \text{P}}}_1}{{{\displaystyle \text{P}}}_2}=\frac{\Delta {{\displaystyle \text{H}}}_{vap}}{\text{R}}\left(\frac{1}{{{\displaystyle \text{T}}}_2}-\frac{1}{{{\displaystyle \text{T}}}_1}\right)$

determine the normal boiling point of trichlorosilane. (b) What kind(s) of intermolecular forces exist between trichlorosilane molecules? (c) Each cubic unit cell (edge length a = 543 pm) contains eight Si atoms. If there are 1.0 × 10¹³ boron atoms per cubic centimeter in a sample of pure silicon, how many Si atoms are there for every B atom in the sample? (d) Calculate the density of pure silicon.

Interpretation Introduction

Interpretation:

(a)

Normal boiling point of trichlorosilane has to be determined.

Concept Introduction:

Boiling point of a liquid substance is defined as the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure.

Molar heat of vaporization is defined as the quantity of heat required to vaporize one mole of substance at its boiling point.

Vapor pressure of the liquid is defined as the pressure of its vapor state when it is in equilibrium with the liquid state.  Vapor pressure of a liquid can be related to molar heat of vaporization of the liquid by “Clausius – Clapeyron” equation as follows –

$\begin{array}{l}\text{ln}\text{P}\text{=}\frac{{\text{-ΔH}}_{\text{vap}}}{\text{R}}\left(\frac{\text{1}}{\text{T}}\right)\text{+}\text{C}\\ \text{Where}\text{T}\text{=}\text{Temperature}\\ \text{ P}\text{=}\text{vapor}\text{pressure}\text{of}\text{the}\text{liquid}\text{}\text{at}\text{temperature}\text{T}\\ {\text{ΔH}}_{\text{vap}}\text{=}\text{Molar}\text{heat}\text{of}\text{vaporization}\\ \text{R}\text{=}\text{Universal}\text{Gas constant}\text{.}\end{array}$

At two different temperature and pressure, the equation is rewritten as –

$\begin{array}{l}\text{ln}\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{=}\frac{{\text{-ΔH}}_{\text{vap}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]\\ \text{Where}{\text{T}}_1{\text{ and T}}_2\text{ are two different Temperature}\\ {\text{ P}}_1{\text{ and P}}_2\text{ are two different Pressure}\\ {\text{ΔH}}_{\text{vap}}\text{=}\text{Molar}\text{heat}\text{of}\text{vaporization}\\ \text{R}\text{=}\text{Universal}\text{Gas constant}\text{.}\end{array}$

Answer to Problem 12.132QP

Boiling point of trichlorosilane is determined as $308K.$

Explanation of Solution

Molar heat of vaporization and vapor pressure of the trichlorosilane are known.  These parameters and other known parameters are substituted in Clausius – Clapeyron equation and boiling point of trichlorosilane is obtained.  Let ${\text{' T}}_{\text{2}}\text{'}$ be the boiling point of trichlorosilane.

$\begin{array}{l}\text{Given data:}\\ {\text{Temperature, T}}_{\text{1}}{\text{ = 2}}^{\text{°}}\text{C = 275}\text{.15 K}\\ {\text{vapor pressure of Trichlorosilane, P}}_{\text{1}}{\text{ at T}}_{\text{1}}\text{ = 0}{\text{.258 atm ; P}}_{\text{2}}\text{ = 1 atm (atmospheric pressure)}\\ {\text{ molar heat of vaporization, ΔH}}_{\text{vap}}\text{ = 28}\text{.8 kJ/mol}\\ \text{ln}\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{ = }\frac{{\text{-ΔH}}_{\text{vap}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]\\ \text{ln}\frac{\text{0}\text{.258 atm}}{\text{1atm}}\text{= }\frac{\text{28}\text{.8 kJ/mol}}{\text{8}\text{.314 J/K}\text{. mol}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{\text{275}\text{.15 K}}\right]\\ \\ \text{Simplifying the above equation,}\\ {\text{T}}_{\text{2}}\text{ = 308 K}\end{array}$

Conclusion

The normal boiling point of trichlorosilane is determined.

Interpretation Introduction

Interpretation:

The types of intermolecular force exist between trichlorosilane molecules have to be described.

Concept Introduction:

• Intermolecular force refers to the attractive forces between the molecules of a substance.  It is the force which holds the molecules together.  Many physical properties of the substance such as – melting point, boiling point, surface tension, viscosity etc., are influenced by the strength of intermolecular force present in the substance.
• The three types of intermolecular forces are – London dispersion force, dipole-dipole force and Hydrogen bonding.  They are collectively known as Van der Waals forces.
• London dispersion forces exist in all types of molecules.  This is the force responsible for the condensation of non-polar compounds into liquids or solids under low temperature.
• Dipole-dipole forces exist in polar covalent compounds.  Hydrogen bonding exists in polar covalent compounds containing Fluorine, Oxygen or Nitrogen directly bonded to Hydrogen.
• The strength of intermolecular forces is

  $\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

Answer to Problem 12.132QP

The intermolecular forces present in trichlorosilane are dipole-dipole force and London dispersion force.

Explanation of Solution

Draw the Lewis dot structure of trichlorosilane.  Lewis dot structure of trichlorosilane illustrates the polarity in a bond.

Owing to the significant electronegativity difference between Silicon and Chlorine atoms and the lone pair electrons present on the Chlorine atoms, the $\text{Si-Cl}$ bond is polar covalent bond.  Polarity in the bond makes the molecules exist as dipoles due to the unsymmetrical distribution of electron cloud between the bonded atoms.  The intermolecular force exist in such polar compound is dipole-dipole force which is the interactive force between the dipoles.  Besides, London dispersion force also present.

Conclusion

The types of intermolecular force present in trichlorosilane are described.

The number of Silicon atoms in a sample of Boron incorporated Silicon.

Interpretation Introduction

Interpretation

The number of Silicon atoms for every Boron atom in a sample containing $\text{1}\text{.0 }\times {\text{ 10}}^{13}$ Boron atoms per cubic centimeter of pure Silicon has to be determined.

Concept Introduction:

The simplest and basic unit of a crystalline solid is known as unit cell.  It is cubic in shape.  It is the building block of crystalline solids.  The unit cells repeat themselves to build a lattice. Crystalline solids consist of many of such lattices.  There are three types of unit cell – simple cubic unit cell, body – centered cubic unit cell and face – centered cubic unit cell.

In crystalline solid, the volume of unit cell is equivalent to the cubic value of the edge length.

Answer to Problem 12.132QP

The number of Silicon atoms for every Boron atom is $\text{5 }\times {\text{ 10}}^9\text{.}$

Explanation of Solution

The volume of unit cell is obtained by cubing the given edge length value of the unit cell.

$\begin{array}{l}\text{given data:}\\ \text{edge length, a = 543 pm = 5}\text{.43}\times {\text{10}}^{-8}\text{ cm}\\ {\text{volume of unit cell, a}}^3\text{ = (5}\text{.43}\times {\text{10}}^{-8}{\text{ cm)}}^3=\text{ 1}\text{.601 }\times {\text{ 10}}^{-22}{\text{ cm}}^{3.}\end{array}$

Determine the number of Boron atoms in the unit cell using the unit cell volume and the

number of Boron atoms per cubic centimetre.

$\begin{array}{l}\text{given data: }\\ \text{no}\text{. of B atoms per cubic centimeter = 1}\text{.0 }\times {\text{ 10}}^{13}\\ \text{no}\text{.of B atoms per cubic centimeter = volume of unit cell }\times \text{ no}{\text{.of Boron atoms/cm}}^3\\ \text{of the volume of a unit cell}\\ \text{ = 1}\text{.601}\times {\text{10}}^{-22}{\text{ cm}}^3\times \text{1}\text{.0 }\times {\text{10}}^{13}{\text{/cm}}^3\text{ = 1}\text{.601}\times {\text{ 10}}^{-9}\text{ B atoms}\text{. }\\ \end{array}$

The number of Boron atoms occupying per centimeter cube of the unit cell is given.  The number of Boron atoms occupying whole unit cell is determined using the given value.

The number of Silicon atoms for every Boron atom.

$\begin{array}{l}\text{No}\text{.of Si atoms for every B atom = }\frac{\text{no}\text{.of Si atoms per unit cell}}{\text{no}\text{.of B atoms per unit cell}}\\ \text{ = }\frac{8}{1.601\times {\text{10}}^{-9}}=5\times {10}^9\text{ Si atoms/B atom}\text{.}\end{array}$

The ratio of Silicon atoms to Boron atoms to gives the number of Silicon atoms for every

Boron atom in the sample.

Conclusion

The number of Silicon atoms for every Boron atom in the sample has been determined.

Interpretation Introduction

Interpretation:

The density of pure Silicon has to be calculated.

Concept Introduction:

Density of a substance refers to the extent of compactness in the substance. It is mass per unit volume and represented as –

$\text{density = }\frac{\text{mass}}{\text{volume}}$

Greater the compactness of a substance less will be the volume it occupies and more will be the density.

Answer to Problem 12.132QP

Density of pure Silicon is calculated as $2.33g/c{m}^3.$

Explanation of Solution

Each unit cell has eight Silicon atoms.  8 times the atomic mass of one Silicon atom gives mass of one unit cell after the conversion of amu unit to grams.

$\begin{array}{l}\text{known data: atomic mass of Si = 28}\text{.09 amu}\\ \text{mass of one unit cell conatining 8 Si atoms = 8}\times \text{28}\text{.09 amu = 224}\text{.72 amu}\\ \text{converting it to grams,}\\ \text{mass of unit cell = }\frac{\text{224}\text{.72 amu}}{6.022\times {\text{10}}^{23}}\text{ = 3}\text{.732 }\times {\text{10}}^{-23}\text{ g}\text{.}\end{array}$

The density of pure Silicon is,

$\begin{array}{l}\text{known data:}\\ \text{volume of unit cell = 1}{\text{.601×10}}^{\text{-22}}{\text{cm}}^{\text{3}}\\ \text{mass of unit cell = 3}{\text{.732 ×10}}^{\text{-23}}\text{g}\text{.}\\ \text{density = }\frac{\text{mass}}{\text{volume}}\\ \text{ = }\frac{\text{3}{\text{.732 ×10}}^{\text{-23}}\text{g}\text{.}}{\text{1}{\text{.601×10}}^{\text{-22}}{\text{cm}}^{\text{3}}}\text{ = 2}{\text{.33 g/cm}}^3.\end{array}$

Volume and mass of the unit cell are calculated in the previous steps.  These parameters are substituted in the formula $\text{density = }\frac{\text{mass}}{\text{volume}}$