Chapter 12, Problem 12.74QP

(a)

Interpretation Introduction

Interpretation: The given question stated that ${\text{C}}_{60}$ reacts with alkali metals to form ${\text{M}}_3{\text{C}}_{60}$ (where $\text{M}$ is $\text{Na}$ or $\text{K}$). The crystal structure of ${\text{M}}_3{\text{C}}_{60}$ contains cubic closest-packed spheres of ${\text{C}}_{60}$ with metal ions in holes. When radius of an ${\text{K}}^{+}$ ion is $138\text{pm}$ then the type of holes ${\text{K}}^{+}$ ion likely to occupy and the density of a crystal of ${\text{K}}_6{\text{C}}_{60}$ is to be explained.

Concept introduction: The cubic closest-packed has eight atoms at the corners of the cube and six atoms at the centers of each face of the cube. The big atoms create the void space in the ccp arrangement and this space is occupied by the small atoms.

To determine: The type of holes ${\text{K}}^{+}$ ion is likely to occupy and the fraction of the holes that will be occupied.

Answer to Problem 12.74QP

Solution

The atomic radii ratio of both the atom is $\underset{\_}{0.27}$.

The fraction of the hole will be $\underset{\_}{27\%}$.

Explanation of Solution

Explanation

Given

The radius of potassium ion is $138\text{pm}$.

The calculated radius of ${\text{C}}_{60}$ is $498.5\text{pm}$.

The smallest unit of a crystal is unit cell which repeat itself in different directions to generate the complete crystal lattice.

The cubic closest-packed has eight atoms at the corners of the cube and six atoms at the center of each face of the cube. The big atoms create the void space in the ccp arrangements and this space is occupied by the small atoms.

The atomic radii ratio is calculated by the formula,

$\text{Atomic}\text{radii}\text{ratio}=\frac{r^{+}}{r^{-}}$

Where,

• $r^{+}$ is the radius of cation.
• $r^{-}$ is the radius of anion

Substitute the value of radius in above formula.

$\begin{array}{c}\text{Atomic}\text{radii}\text{ratio}=\frac{r^{+}}{r^{-}}\\ =\frac{138\text{pm}}{498.5\text{pm}}\\ =0.27\end{array}$

Therefore, the atomic radii ratio of both the atom is $\underset{\_}{0.27}$.

In ccp arrangement if radius ratio is between $0.41-0.73$ then the atoms having smaller atomic radius will occupy the octahedral hole. And if radius ratio is between $0.22-0.41$ then the atoms will occupy the tetrahedral hole.

Therefore, potassium ion will occupy tetrahedral holes.

The fraction of the holes that will be occupied by potassium ion is $\underset{\_}{27\%}$.

(b)

Interpretation Introduction

To determine: The density of a crystal of ${\text{K}}_6{\text{C}}_{60}$

Answer to Problem 12.74QP

Solution

The density of ${\text{K}}_6{\text{C}}_{60}$ is $\underset{\_}{0.0120g/c{m}^3}$.

Explanation of Solution

Explanation

Given

The calculated radius for ${\text{C}}_{60}$ is $11.75\text{pm}$.

The given compound is ${\text{K}}_6{\text{C}}_{60}$.

The smallest unit of a crystal is unit cell which repeat itself in different directions to generate the complete crystal lattice.

The number of atoms per unit cell in bcc lattice is $2$.

The molar mass of given compound is $955.23\text{g}/\text{mol}$.

The density of ${\text{K}}_6{\text{C}}_{60}$ is calculated by formula,

$d=\frac{Z\times M}{a^3\times {\text{N}}_{\text{A}}}$

Where,

• $d$ is the density of unit cell.
• $Z$ is the number of atoms per unit cell.
• $a$ is the edge length.
• ${\text{N}}_{\text{A}}$ is the Avogadro constant.
• $M$ is the molar mass.

The relationship between atomic radius and edge length of bcc lattice is,

$a=\frac{r}{0.4330}$

Where,

• $a$ is the edge length
• $r$ is the radius.

Substitute the value of radius in above formula.

$\begin{array}{c}a=\frac{r}{0.4330}\\ =\frac{11.75}{0.4330}\\ =27.13\text{pm}\end{array}$

Conversion of $1\text{pm}$ into $\text{cm}$ is done as,

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

Therefore, conversion of $27.13\text{pm}$ into $\text{cm}$ is,

$27.13\text{pm}=27.13\times {10}^{-10}\text{cm}$

Substitute the values of $Z$, $a$, $M$, and ${\text{N}}_{\text{A}}$ in the above formula.

$\begin{array}{c}d=\frac{Z\times M}{a^3\times {\text{N}}_{\text{A}}}\\ =\frac{2\times 955.23}{{\left(27.13\times {10}^{-10}\right)}^3\times 6.023\times {10}^{23}}\\ =0.0120\text{g}/{\text{cm}}^3\end{array}$

Therefore, the density of ${\text{K}}_6{\text{C}}_{60}$ is $\underset{\_}{0.0120g/c{m}^3}$.

Conclusion

The density of ${\text{K}}_6{\text{C}}_{60}$ is $\underset{\_}{0.0120g/c{m}^3}$.