Chapter 12, Problem 12.127AP

(a)

Interpretation Introduction

To determine: The distance between the $\text{Si}$ layers.

Interpretation: The distance between the $\text{Si}$ layers is to be calculated.

Concept introduction: The distance between the $\text{Si}$ layers is calculated by the formula,

$n\lambda =2d\times \sin \theta$

Answer to Problem 12.127AP

The distance between $\text{Si}$ layers is $\underset{\_}{598\text{pm}}$

Explanation of Solution

The value of $2\theta$ is $29.86°.$.

The value of wavelength $\left(\lambda \right)$ is $154\text{pm}$.

The value of $n$ is $2$.

The conversion of units of wavelength into $\text{pm}$ is done as,

$\begin{array}{l}1\text{pm}={10}^{-12}\text{m}\\ \text{154}\text{pm}=\text{1}\text{.54}\times {\text{10}}^{-10}\text{m}\end{array}$

The distance between the $\text{Si}$ layers is calculated by the formula,

$n\lambda =2d\times \sin \theta$ (1)

Where,

• $n$ is the number.
• $d$ is the distance between layers.
• $\lambda$ is the wavelength.

Substitute the values of $n,\lambda ,2\text{and}\theta$ in equation (1).

$\begin{array}{c}2\times 1.54\times {10}^{-10}\text{m}=2\times d\times \sin 14.93°\\ 3.08\times {10}^{-10}\text{m}=d\times 0.515\\ d=\frac{3.08\times {10}^{-10}\text{m}}{0.515}\\ =\underset{\_}{5.98\times 10^{-10}m}\end{array}$

Similarly,

The value of $2\theta$ is $45.46°.$.

The value of wavelength $\left(\lambda \right)$ is $154\text{pm}$.

The value of $n$ is $3$.

The conversion of units of wavelength into $\text{pm}$ is done as,

$\begin{array}{l}1\text{pm}={10}^{-12}\text{m}\\ \text{154}\text{pm}=\text{1}\text{.54}\times {\text{10}}^{-10}\text{m}\end{array}$

Substitute the values of $n,\lambda ,2\text{and}\theta$ in equation (1).

$\begin{array}{c}3\times 1.54\times {10}^{-10}\text{m}=2\times d\times \sin 22.73°\\ 4.62\times {10}^{-10}\text{m}=d\times 0.773\\ d=\frac{4.62\times {10}^{-10}\text{m}}{0.773}\\ =\underset{\_}{5.98\times 10^{-10}m}\end{array}$

Similarly,

The value of $2\theta$ is $62.00°.$.

The value of wavelength $\left(\lambda \right)$ is $154\text{pm}$.

The value of $n$ is $4$.

The conversion of units of wavelength into $\text{pm}$ is done as,

$\begin{array}{l}1\text{pm}={10}^{-12}\text{m}\\ \text{154}\text{pm}=\text{1}\text{.54}\times {\text{10}}^{-10}\text{m}\end{array}$

Substitute the values of $n,\lambda ,2\text{and}\theta$ in equation (1).

$\begin{array}{c}4\times 1.54\times {10}^{-10}\text{m}=2\times d\times \sin 31.0°\\ 6.16\times {10}^{-10}\text{m}=d\times 1.03\\ d=\frac{6.16\times {10}^{-10}\text{m}}{1.03}\\ =\underset{\_}{5.98\times 10^{-10}m}\end{array}$

The conversion of units of distance into $\text{pm}$.

$\begin{array}{l}1\text{pm}={10}^{-12}\text{m}\\ \text{5}\text{.98}\times {\text{10}}^{-10}\text{m}=\underset{\_}{598pm}\end{array}$

Therefore, all the values of $d$ are same so distance between $\text{Si}$ layers are $\underset{\_}{598pm}$.

(b)

Interpretation Introduction

To determine: The $\text{Ca}-\text{Si}$ distance, if the ${\text{Ca}}^{2+}$ ion lies exactly the halfway between the layers.

Interpretation: The $\text{Ca}-\text{Si}$ distance, if the ${\text{Ca}}^{2+}$ ion lies exactly the halfway between the layers is to be stated.

Concept introduction: The calcium ion is present at the half distance between the layers, so this is calculated by the formula,

$\text{Ca}-\text{Si}\text{distance}=\frac{\text{Distance}\text{between}\text{Si}\text{layers}}{2}$

Answer to Problem 12.127AP

The distance between $\text{Ca}-\text{Si}$ layer is $\underset{\_}{299pm}$.

Explanation of Solution

The ${\text{Ca}}^{2+}$ ion lies exactly halfway between the layers, thus, the $\text{Ca}-\text{Si}$ distance is calculated by the formula,

$\text{Ca}-\text{Si}\text{distance}=\frac{\text{Distance}\text{between}\text{Si}\text{layers}}{2}$ (2)

Substitute the value of distance between $\text{Si}$ layers in equation (2).

$\begin{array}{c}\text{Ca}-\text{Si}\text{distance}=\frac{598\text{pm}}{2}\\ =\underset{\_}{299pm}\end{array}$

Therefore, the distance between $\text{Ca}-\text{Si}$ layer is $\underset{\_}{299pm}$.

(c)

Interpretation Introduction

To determine: The reason of the given statement, ${\text{CaSi}}_2$ is being used as a “deoxidizer in converting iron ore to steel.

Interpretation: The reason of the given statement that is ${\text{CaSi}}_2$ used as a “deoxidizer in converting iron ore to steel is to be stated.

Concept introduction: The oxidation number of the atom relies upon the atoms attached to it and the charge present on each atom.

Answer to Problem 12.127AP

The most stable oxidation state of silicon is $+4$. The deoxidization of iron results in oxidation of silicon into ${\text{Si}}^{4+}$ state, whereas iron gets reduced to $\text{Fe}$ from ${\text{Fe}}^{3+}$ ions.

Explanation of Solution

The given chemical formula is ${\text{CaSi}}_2$.

The oxidation number of $\text{Ca}$ atom is $+2$.

The oxidation number of silicon atom is assumed to be $x$.

The oxidation number of silicon atom is calculated by the formula,

$\text{Sum}\text{of}\text{oxidation}\text{number}\text{of}\text{constituent}\text{atoms}=\text{overall}\text{charge}\text{of}\text{atom}$ (3)

Substitute the values of oxidation number of constituent atoms in equation (3).

$\begin{array}{c}+2+x+\left(2\right)=0\\ +4+x=0\\ x=-4\end{array}$

Therefore, the oxidation number of silicon atom in ${\text{CaSi}}_2$ is $-4$ but the most stable oxidation state of silicon is $+4$. It has a tendency to gain as well as lose the electrons.

The deoxidization of iron results in oxidation of silicon into ${\text{Si}}^{4+}$ state, whereas iron gets reduced to $\text{Fe}$ from ${\text{Fe}}^{3+}$ ions.

Conclusion

1. a. The distance between $\text{Si}$ layers is $\underset{\_}{598\text{pm}}$
2. b. The distance between $\text{Ca}-\text{Si}$ layer is $\underset{\_}{299pm}$.
3. c. The most stable oxidation state of silicon is $+4$. The deoxidization of iron results in oxidation of silicon into ${\text{Si}}^{4+}$ state, whereas iron gets reduced to $\text{Fe}$ from ${\text{Fe}}^{3+}$ ions.