Chapter 12, Problem 12.142MP

Interpretation Introduction

(a)

Interpretation:

The three dimensional structure of one unit cell of the alkali metal fulleride superconductor ${\text{M}}_{\text{3}}{\text{C}}_{\text{60}}$ needs to be drawn.

Concept introduction:

Unit cell is defined as the smaller group of atoms having overall crystal symmetry from which one can build the whole lattice in repetition order in 3D form.

Interpretation Introduction

(b)

Interpretation:

The number of ${\text{C}}_{\text{60}}{}^{3-}$ ions, octahedral holes and tetrahedral holes present in 1 unit cell needs to be determined.

Concept introduction:

The shape of the tetrahedral holes is triangular. When 2 tetrahedral holes combine from different layers, an octahedral hole is formed. Thus, due to alignment of tetrahedral holes of first and second order align together, an octahedral hole is formed.

Interpretation Introduction

(c)

Interpretation:

The fractional coordinates needs to be determined for all the octahedral and tetrahedral holes in the unit cell.

Concept introduction:

Fractional coordinates are defined as fractions of edge length of the unit cell. For example, at the center of the cell, the fractional coordinate of a hole is $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$.

Interpretation Introduction

(d)

Interpretation:

The radii of the octahedral and tetrahedral holes needs to be calculated assuming the $C_{60}^{3-}$ ions are in contact in the unit cell along a face diagonal and radius of the $C_{60}^{3-}$ anion is 500 pm.

Concept introduction:

The face diagonal in the unit cell can be represented as follows:

The length of a face diagonal is $\sqrt{2}a$. Here, a is face length of the cube. The relation between radius of sphere R and face diagonal is as follows:

$2R=\sqrt{2}a$

Now, along the body diagonal there are 2 spheres touching the tetrahedral holes of radius r thus,

$2R+2r_t=\sqrt{3}a$

Here, r_t is radius of the tetrahedral hole.

Or,

$\begin{array}{l}2R+2r_t=\sqrt{3}a\\ 2R+2r_t=\sqrt{3}\left(\frac{2R}{\sqrt{2}}\right)\end{array}$

On rearranging,

$r_t=0.225R$

Here, R is radius of the sphere.

Similarly, an octahedral hole can be represented as follows:

Here, radius of sphere is R and that of an octahedral hole is $r_o$. Pythagoras theorem can be used here as follows:

${\left(R+r_o\right)}^2+{\left(R+r_o\right)}^2={\left(2R\right)}^2$

Or,

$2{\left(R+r_o\right)}^2=2R^2$

Or,

$\begin{array}{l}R+r_o=\sqrt{2}R\\ r_o=\sqrt{2}R-R\\ r_o=0.414R\end{array}$

Here, R is radius of the sphere.

Interpretation Introduction

(e)

Interpretation:

The ion from the given ions that will fit into the octahedral and tetrahedral holes needs to be determined. The ion that will only fit if the framework of ${\text{C}}_{\text{60}}{}^{3-}$ ions expands needs to be determined.

Concept introduction:

The shape of the tetrahedral holes is triangular. When 2 tetrahedral holes combine from different layers, an octahedral hole is formed. Thus, due to alignment of tetrahedral holes of first and second order align together, an octahedral hole is formed.