Chapter 11, Problem 11.129QP

Silicon used in computer chips must have an impurity level below 10⁻⁹ (that is, 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:

${\text{SiO}}_2(s)+2\text{C}(s)\stackrel{}{\to }\text{Si}(l)+2\text{CO}(g)$

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

$\text{Si}(s)+3\text{HCl}(g)\stackrel{}{\to }{\text{SiCl}}_3\text{H}(g)+{\text{H}}_{\text{2}}(g)$

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

${\text{SiCl}}_{\text{3}}\text{H}(g)+{\text{H}}_{\text{2}}(g)\stackrel{}{\to }\text{Si}(s)+3\text{HCl}(g)$

(a) Trichlorosilane has a vapor pressure of 0.258 atm at −2°C. What is its normal boiling point? Is trichlorosilane’s boiling point consistent with the type of intermolecular forces that exist among its molecules? (The molar heat of vaporization of trichlorosilane is 28.8 kJ/mol.) (b) What types of crystals do Si and SiO₂ form? (c) Silicon has a diamond crystal structure (see Figure 11.28). 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? Does this sample satisfy the 10⁻⁹ purity requirement for the electronic grade silicon?

Figure 11.28 (a) The structure of diamond. Each carbon is tetrahedrally bonded to four other carbon atoms. (b) The structure of graphite. The distance between successive layers is 335 pm.

Interpretation Introduction

Interpretation:

The normal boiling point of trichlorosilane and its adherence to its intermolecular forces have to be found out and discussed.

Concept Introduction:

Boiling point is the temperature at which the vapor pressure will be in equilibrium with the external pressure which is normally $\text{1}\text{atm}$

Vapour pressure: It is the pressure developed on the vapor when it is in contact with its liquid form or solid form.  So, it is also known as equilibrium vapor pressure which establishes equilibrium between its gaseous vapor state and liquid state or solid state.

The trends of boiling points are only due to the strength of the intermolecular forces that are acting between the molecules of the given compounds.

Intermolecular forces are the forces existing between molecules, atoms, ions or dipoles and hence binds all the molecules of a definite physical state such as solid, liquid or gas state.

Explanation of Solution

The normal boiling point of trichlorosilane can be calculated using the Clausisus-Clapeyron equation:

The Clausius-Clapeyron equation: $\frac{1}{{\text{T}}_2}=\frac{\text{R}}{{\text{ΔH}}_{\text{vap}}}\ln \frac{{\text{P}}_1}{{\text{P}}_2}+\frac{1}{{\text{T}}_1}$

$\begin{array}{l}{\text{ΔH}}_{\text{vap}}=\text{Molar}\text{heat}\text{of}\text{vaporization}\text{.}\\ \text{R}\text{=}\text{Gas}\text{constant}\text{=8}\text{.314}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}.\end{array}$

Given:

${\text{ΔH}}_{\text{vap}}=28.8\text{kJ/mol}$

Converting kilojoules into joules:

$\begin{array}{c}28.8\text{kJ/mol}\text{=}\frac{\text{1000}\text{J}}{\text{1}\text{kJ}}\text{×}\text{28}\text{.8}\text{kJ}\\ =\text{28800}\text{.0}\text{J/mol}\end{array}$

$\begin{array}{c}{\text{T}}_1\text{=}{\text{-2}}^{\text{o}}\text{C}\\ {\text{T}}_2=\text{normal}\text{boiling}\text{point}\text{.}\\ {\text{P}}_1\text{=}\text{0}\text{.258}\text{atm}\\ {\text{P}}_2=1.00\text{atm}\Rightarrow \text{external}\text{pressure}\text{. }\end{array}$

Converting temperature from Celsius to kelvin:

$\begin{array}{c}{\text{T}}_1\text{=}{\text{-2}}^{\text{o}}\text{C}\\ \text{=}{\text{-2}}^{\text{o}}\text{C}\text{+}\text{ 273}\\ \text{=}\text{271}\text{K}\end{array}$

Substituting all these values in the Clausius-Clapeyron equation:

$\begin{array}{c}\frac{1}{{\text{T}}_2}=\frac{\text{R}}{{\text{ΔH}}_{\text{vap}}}\ln \frac{{\text{P}}_1}{{\text{P}}_2}+\frac{1}{{\text{T}}_1}\\ \frac{1}{{\text{T}}_2}=\frac{\text{8}\text{.314}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}}{\text{28800}\text{.0}\text{J/mol}}\ln \frac{\text{0}\text{.258}}{1}+\frac{1}{\text{271}\text{K}}\\ =3.2989\times {10}^{-3}{\text{K}}^{\text{-1}}\end{array}$

$\begin{array}{c}{\text{T}}_2=\frac{1}{3.2989\times {10}^{-3}{\text{K}}^{\text{-1}}}\\ =\text{303}\text{.13}\text{K}\end{array}$

Converting the normal boiling point temperature from Kelvin into Celsius:

$\begin{array}{c}\text{303}\text{.13}\text{K}=\text{303}\text{.13}\text{K}-273\\ =\text{30}{\text{.13}}^{\text{o}}\text{C}\end{array}$

Therefore, the normal boiling point of trichlorosilane is $30.13^{o}C$.

Intermolecular forces existing in trichlorosilane$\left(SiC{l}_{3}H\right)$ :

There are three highly electronegative halogen atoms such as chlorine atoms in the trichlorosilane which are attached to the least electronegative atoms such as silicon and hydrogen atoms.  So there will be permanent dipole moment in the molecule, due to the distinguishable electronegativity in it.  Hence, the interactions between such molecules will be dipole-dipole interactions, which are strong intermolecular forces.  Therefore, the normal boiling point is expected to be in positive integer, since more energy is required to break the strong intermolecular forces.  The calculated boiling point of trichlorosilane is $\text{30}{\text{.13}}^{\text{o}}\text{C}$, which is a positive integer.  Thus, the boiling point of trichlorosilane is consistent with the type of intermolecular forces that exist among its molecules.

Interpretation Introduction

Interpretation: The type of crystal formed by $\text{Si}$ and ${\text{SiO}}_2$, have to be found.

Concept Introduction:

The major types of crystals are listed here:

1. 1. Ionic crystals.
2. 2. Covalent crystals.
3. 3. Molecular crystals.
4. 4. Metallic crystals.

Ionic crystals: The crystals that are composed of charged species such as anions and cations.

Covalent crystals: In covalent crystals, all the atoms will be connected in a three-dimensional network by covalent bonds.

Molecular crystals: In molecular crystal, the molecules occupying the lattice points will have attractive forces between them such as van der waals forces or hydrogen bonding.

Metallic crystals: All the lattice points are occupied by the same type of metal.

Explanation of Solution

${\text{SiO}}_2$ is known as silica or silicon dioxide. In this compound, silicon is bonded to oxygen atoms by covalent bonds and thus form a three dimensional network which is known as quartz. So, the type of crystal formed by $\text{Si}$ and ${\text{SiO}}_2$, is covalent crystal.

Interpretation Introduction

Interpretation: The purity requirement for the electronic grade silicon, has to be checked out when boron atom is added to $\text{Si}$ sample

Concept Introduction:

Silicon is being used in various electronic devices today, due its special and important properties such as superconducting property and electrical properties. Silicon can be doped which is a modification in its electrical properties. Silicon used in computer chips must have an impurity level below ${10}^{-9}$(that is fewer than one impurity atom for every ${10}^9$ $\text{Si}$ atoms).

Explanation of Solution

Given:

The edge length of cubic unit cell in the diamond structure of silicon is given as $\text{a}\text{=}\text{543}\text{pm}$.

Converting edge length from picometers into centimeters:

$\begin{array}{c}\text{543}\text{pm}=\text{543}\text{p}\bcancel{\text{m}}\text{×}\left(\frac{\text{1}{\text{×10}}^{\text{-12}}\bcancel{\text{m}}}{\text{1}\text{p}\bcancel{\text{m}}}\text{×}\frac{\text{1}\text{cm}}{\text{1}{\text{×10}}^{\text{-2}}\bcancel{\text{m}}}\right)\\ =\frac{\text{543}{\text{×10}}^{\text{-12}}}{{\text{10}}^{\text{-2}}}\text{cm}\\ =\text{543}{\text{×10}}^{\text{-12}}{\text{×10}}^{\text{2}}\text{cm}\\ =\text{543}{\text{×10}}^{\text{-10}}\text{cm}\end{array}$

Calculating the volume of per unit cell:

The relation between edge length and volume is $V=a^3$

$\begin{array}{c}\text{V=}{\left(\text{543}{\text{×10}}^{\text{-10}}\text{cm}\right)}^3\\ =\text{1}\text{.60}{\text{×10}}^{\text{-22}}{\text{cm}}^{\text{3}}\end{array}$

So, the volume of per unit cell is $\text{1}\text{.60}{\text{×10}}^{\text{-22}}{\text{cm}}^{\text{3}}$.

Calculating the number of unit cells per cubic centimetre:

${\text{1cm}}^{\text{3}}\text{×}\frac{\text{1}\text{unit}\text{cell}}{\text{1}\text{.60}{\text{×10}}^{\text{-22}}{\text{cm}}^{\text{3}}}\text{=}\text{6}{\text{.25×10}}^{\text{21}}\text{unit}\text{cells}$

Therefore, in the volume of ${\text{1cm}}^{\text{3}}$, there will be $\text{6}{\text{.25×10}}^{\text{21}}\text{unit}\text{cells}$.

Calculating the number of $\text{Si}$atoms that are present is $6.25\times 10^{21}unitcells$.

It is given that there are 8 $\text{Si}$ atoms per unit cell.

So,

$\begin{array}{l}\frac{\text{8}\text{Si}\text{atoms}}{\text{1}\text{unit}\text{cell}}\times \text{6}\text{.25×1}{\text{0}}^{\text{21}}\text{unit}\text{cells}\\ \text{=}5.00\times {10}^{22}\text{Si}\text{atoms}\end{array}$

Thus, there will be $5.00\times {10}^{22}\text{Si}\text{atoms}$ in $\text{6}{\text{.25×10}}^{\text{21}}\text{unit}\text{cells}$.

Calculating the purity of $Si$crystal:

Given that $1.0\times {10}^{13}$ Boron atoms are added to per centimetre in a sample of pure silicon.

It is now known that per centimetre sample of pure silicon have $\text{6}{\text{.25×10}}^{\text{21}}\text{unit}\text{cells}$ with

$5.00\times {10}^{22}\text{Si}\text{atoms}$.

So, per centimetre sample of pure silicon have $5.00\times {10}^{22}\text{Si}\text{atoms}$. The purity has to be checked when $1.0\times {10}^{13}$ Boron atoms are added to this $5.00\times {10}^{22}\text{Si}\text{atoms}$.

$\begin{array}{c}\frac{\text{B}\text{atoms}}{\text{Si}\text{atoms}}=\frac{1.0\times {10}^{13}\text{B}\text{atoms}}{5.00\times {10}^{22}\text{Si}\text{atoms}}\\ =2.0\times {10}^{-10}\end{array}$

This implies that there will be fewer than one impurity atom for every ${10}^{10}$$\text{Si}\text{atoms}$.